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ADVANCED ALGEBRA 
 
 FOR <^yi-xD^ 
 COLLEGES AND SCHOOLS 
 
 BY 
 
 WILLIAM J. MILNE, Ph.D., LL.D. 
 
 PRESIDENT OF NEW YORK STATE NORMAL COLLEGE, ALBANY, N.Y. 
 
 »>«{o 
 
 NEW YORK-:. CINCINNATI-:. CHICAGO 
 
 AMERICAN BOOK COMPANY 
 
h 
 
 .^* 
 
 \_ 
 
 Copyright, 1901 and 1902, by 
 WILLIAM J. MILNE. 
 
 Entbekd at Stationers' Hall, Londok. 
 advanced algebra. 
 
 B-F 3 
 
PREFACE 
 
 Many colleges and universities offer an optional examination 
 in "Advanced Algebra" in addition to the examination in "Ele- 
 mentary Algebra"' required of all candidates for matriculation. 
 The scientific and technical schools have increased from time to 
 time the amount of algebra that must be mastered before students 
 may pass the entrance examinations, until their requirements at 
 present for admission are substantially equal to what is commonly 
 included under " Advanced Algebra." This work is designed to 
 prepare students in a thorough manner to meet both of these 
 tests. 
 
 The earlier pages of the work are identical with the author's 
 Academic Algebra, but several of the topics discussed in that 
 book have been modified in treatment or enlarged in scope, so 
 that they may the better meet the requirements of an advanced 
 course, and more than one hundred sixty pages of new matter 
 have been added. Consequently this volume contains a complete 
 course in both elementary and advanced algebra. 
 
 The treatment is believed to be sufficiently full and rigorous 
 to meet the demands of college courses in algebra, and to give a 
 scholarly basis for specializing in this science. 
 
 The author desires especially to express his indebtedness to 
 Professor J. H. Tanner of the Department of Mathematics at 
 Cornell University for valuable suggestions regarding the treat- 
 ment of several topics. 
 
 WILLIAM J. MILNE. 
 
 Statk Normal Colleob, 
 Albany, N.Y. 
 
 28177U 
 
p 
 
CONTENTS 
 
 L 
 
 llgebraic Solutions 
 1)kfimtions and Notation 
 Addition . 
 
 nSubtraction 
 
 V^Iui-ti plication 
 
 Division 
 
 v^Review 
 
 >f actor ing 
 
 V Review of Factoring 
 
 V 
 
 Highest CommOx Divisor 
 
 Lowest Common Multiple 
 > Fractions .... 
 
 Review of Fractions . 
 Simple Equations . 
 
 Simultaneous Simple Equations 
 
 Involution 
 
 Evolution 
 y Theory ok J]xponents . 
 '^KRadicals .... 
 
 Review .... 
 
 Quadratic Equations . 
 
 General Review 
 
 Ratio and Proportion . 
 
 variation 
 
6 CONTENTS 
 
 VA6K 
 
 PUOGRESSIONS . « 344 
 
 K Imaginary and Complex Numbers . . . . . . 363 
 
 Inequalities 373 
 
 Variables and Limits 379 
 
 Incommensurable Numbers 388 
 
 Interpretation of Results 393 
 
 Indeterminate Equations 400 
 
 Mathematical Induction 404 
 
 Binomial Theorem 408 
 
 ^ Logarithms 415 
 
 Permutations and Combinations 434 
 
 Probability 445 
 
 Simple Continued Fractions 458 
 
 Theory of Numbers 470 
 
 Determinants • 479 
 
 CONVERGENCY OF SeRIES . .^ 496 
 
 Undetermined Coefficients 509 
 
 )^Exponential and Logarithmic Series 524 
 
 Summation of Series 533 
 
 Functions of a Single Variable 545 
 
 Theory of Equation^ 562 
 
 ^ ^ 
 
ADVANCED ALGEBRA 
 
 oJOic 
 
 ALGEBRAIC SOLUTIONS 
 
 1. Problem 1. A man had 400 acres of corn and oats. If 
 there were 3 times as many acres of corn as of oats, how many 
 acres were there of each ? 
 
 Arithmetical Solution 
 A certain number — the number of acres of oats. 
 Then, 3 times that number = the number of acres of corn, 
 
 and 4 times that number = the number of acres of both ; 
 
 therefore, 4 times that number — 400. 
 
 Hence, the number = 100, the number of acres of oats, 
 
 and 3 times the number = 300, the number of acres of corn. 
 
 Algeuraic Solution 
 
 Let X = the number of acres of oats. 
 
 Then, 3x = the number of acres of corn, 
 
 and 4 X = the number of acres of both ; 
 
 therefore, 4 a; = 400. 
 
 Hence, x = 100, the number of acres of oats, 
 
 and 3 X = 300, the number of acres of corn. 
 
 2. An expression of equality between two numbers or quan- 
 'jities is called an Equation. 
 
 5 a; = 30 is an equation. 
 
 % A question that can be answered only after a course of 
 :?3aso'aing is called a Problem. 
 
 %. The process of finding the result sought is called the Solu- 
 'Oa of' the problem. 
 
 7 
 
8 ['AlqebrMc solutions 
 
 6. The expression in algebraic language of the conditions of a 
 problem is called the Statement of the problem. 
 
 Solve algebraically the following problems : 
 
 2. A horse and saddle cost $ 50. If the horse cost 4 times 
 as much as the saddle, what was the cost of each ? 
 
 3. A bicycle and suit cost $90. How much did each cost, 
 if the bicycle cost 5 times as much as the suit ? 
 
 4. Of 240 stamps that Harry and his sister collected, Harry 
 collected 3 times as many as his sister. How many did each 
 collect ? 
 
 5. If Mr. Brown and his son together had f 220, and Mr. 
 Brown had 10 times as much as his son, how much money had 
 each? 
 
 6. In a room containing 45 students there were twice as many 
 girls as boys. How many were there of each ? 
 
 7. A had 7 times as many sheep as B, and both together had 
 608. How many sheep had each ? 
 
 8. A and B began business with a capital of $ 7500. If A 
 furnished half as much capital as B, how much did each furnish ? 
 
 Suggestion. — Let x = the number of dollars A furnished. 
 
 9. A man bought a cow and a calf for $ 36, paying 8 times as 
 much for the cow as for the calf. What was the cost of each ? 
 
 10. James sold his pony and a saddle for $ 60. If the saddle 
 sold for 1 as much as the pony, what was the selling price of 
 each? 
 
 11. A certain number added to twice itself equals 96. What 
 is the number ? 
 
 12. A farmer raised a certain number of bushels of wheat, 4 
 times as much corn, and 3 times as much barley. If there were 
 in all 4000 bushels of grain, how many bushels of each kind did 
 he raise ? 
 
 13. A boy bought a bat, a ball, and a glove for $ 2.25. If the 
 bat cost twice as much as the ball, and the glove cost 3 times as 
 much as the bat, what was the cost of each ? 
 
ALGEBRAIC SOLUTIONS 9 
 
 14. In a fire B lost twice as much as A, and C lost 3 times 
 as much as A. If their combined loss was $ 300, what was the 
 loss of each ? 
 
 15. A house and lot cost $3000. If the house cost 4 times as 
 much as the lot, what was the cost of each ? 
 
 16. In a business enterprise the joint capital of A, B, and C 
 was $ 2100. If A's capital was twice B's, and B's was twice C's, 
 what was the capital of each ? 
 
 17. John, William, and George together had 120 marbles. If 
 William had twice as many as John, and George had 3 times as 
 many as John, how many had each ? 
 
 18. In an orchard of apple, pear, and cherry trees, containing 
 1690 trees in all, there were 4 times as many cherry trees as 
 pear trees, and twice as many apple trees as cherry trees. How 
 many trees were there of each kind ? 
 
 19. A number plus itself, plus twice itself, plus 4 times itself, 
 is equal to 72. What is the number ? 
 
 20. Charles is twice as old as his younger brother, and half as 
 old as his older brother. If the sum of the ages of the three 
 brothers is 28 years, what is the age of each ? 
 
 21. A farmer had twice as many sheep as horses, and twice as 
 many hogs as sheep and horses together. If there were in all 
 360 animals, how many were there of each kind ? 
 
 22. A tract of land containing 640 acres was divided into three 
 farms, such that the first was 3 times as large as the second, and 
 the third 4 times as large as the first. How many acres did each 
 farm contain ? 
 
 23. Three boys divided 160 marbles among themselves so that 
 one of them received twice as many as each of the others. How 
 many did each receive ? 
 
 24. Divide 30 into two parts, one of which is 14 times the 
 other. 
 
 25. Divide 18 into three parts, such that the first is twice the 
 third, and the second is 3 times the third. 
 
10 ALGEBRAIC SOLUTIONS 
 
 26. Divide 21 into three parts, such that the first is twice the 
 second, and the second is twice the third. 
 
 27. Divide 36 into three parts, such that the first is twice the 
 second, and the third is equal to twice the sum of the first and 
 second. 
 
 28. Three newsboys sold 60 papers. If the first sold twice as 
 many as the second, and the third sold 3 times as many as the 
 second, how many did each sell ? 
 
 29. Henry earned a certain number of dollars per week. With 
 4 weeks' earnings he purchased a rifle, and with 20 weeks' earn- 
 ings, a bicycle. If both together cost $ 72, how much did he earn 
 per week ? How much did the rifle cost ? the bicycle ? 
 
 30. A man sold some ducks for 50 cents each, and the same 
 number of geese for 75 cents each. If for all he received $ 12.50, 
 how many of each did he sell ? 
 
 31. John has 5 times as much money as James. James has 
 24 cents less than John. How much has each ? 
 
 32. A. man had 675 sheep in three fields. If there were twice 
 as many in the first field as in the second, and twice as many in 
 the third field as in both of the others, how many sheep were 
 there in each field ? 
 
 33. A man bequeathed to his daughter twice as much money 
 as to his son, and to his wife 3 times as much as to his daughter. 
 If all together received f 9000, how much did each receive ? 
 
 34. A plumber and two helpers together earned $ 7.50 per 
 day. How much did each earn per day, if the plumber earned 4 
 times as much as each helper ? 
 
 35. What number added to | of itself equals 20 ? 
 
 Solution 
 Let X = the number 
 
 Then, x + |« = 20, 
 
 fx = 20, 
 
 i« = 4. 
 Therefore, x = 12, the number. 
 
ALGEBRAIC SOLUTIONS 11 
 
 36. If ^ of a number is added to the number, the sum is 12. 
 What is the number ? 
 
 37. If ^ of a number is added to twice the number, the sum is 
 36. What is the number ? 
 
 38. The difference between 4 times a certain number and \ of 
 the number is 30. What is the number ? 
 
 39. The difference between | of a certain number and f of it 
 is 16. What is the number ? 
 
 40. After spending i of my money and losing \ of it, I had 
 $ 30. How much had I at first ? 
 
 41. The difference between twice a certain number and | of 
 it is 20. What is the number? 
 
 42. The number 150 can be divided into two parts, one of 
 whicli is f of the other. What are the parts ? 
 
 43. One part of 45 is f of the other. What are the parts ? 
 
 44. Find two parts of 30 such that one is ^ of the other. 
 
 45. To A, B, and C I owe in all $93. If I owe A | as much 
 as C, and B f as much as C, how much do I owe each ? 
 
 46. The length of a field is If times its width, and the distance 
 around the field is 120 rods. If the field is rectangular, what are 
 its dimensions ? 
 
 47. A, B, C, and D buy $ 16,000 worth of railroad stock. How 
 much does A take, if B takes 3 times as much as A, C twice 
 as much as A and B together, and D ^ as much as A, B, and C 
 together ? 
 
 48. In one season an orchard bore 650 bushels of fruit, con- 
 sisting of I as many bushels of pears as of peaches, and twice 
 as many bushels of apples as of pears. How many bushels were 
 there of each ? 
 
 49. A horse, harness, and carriage cost f 340. If the horse 
 cost 3 times as much as the harness, and the carriage cost 1^ 
 times as much as the horse, what was the cost of each ? 
 
DEFINITIONS AND NOTATION 
 
 6. The ideas of number and the knowledge of the processes 
 with abstract numbers that the student has gained in arithmetic 
 are a proper and necessary introduction to his work in algebra; 
 but since number is discussed in a more general way in algebra 
 than in arithmetic, many arithmetical processes, terms, and sym- 
 bols, as 'addition,' 'subtraction,' 'greater,' 'less,' 'exponent,' '+,' 
 '— ,' etc., must now be extended in meaning and application. 
 
 For example, in an arithmetical sense 8 cannot be subtracted 
 from 5, nor does 8» have any meaning ; but in an algebraic sense, 
 as will be shown hereafter, 8 can be subtracted from 5 and 8* is as 
 intelligible as 8^ 
 
 Indeed, the processes and principles of arithmetic are but spe- 
 cial cases of the more fundamental processes and principles of 
 algebra. 
 
 7. A imit or an aggregate of units is called a Whole Number, 
 or an Integer. 
 
 One of the equal parts of a unit or an aggregate of equal parts 
 of a unit is called a Fractional Number. 
 
 Such numbers are called Arithmetical, or Absolute Numbers. 
 
 8. Arithmetical numbers have fixed and known values, and are 
 represented by symbols called numerals; as 1, 2, 3, etc., Arabic 
 figures, and I, V, X, etc., Roman letters. 
 
 9. It is often convenient, in solving a problem, to employ 
 letters, such as x, y, z, to represent the numbers whose values are 
 sought ; and, in stating a rule, to employ letters to represent the 
 numbers that must be given whenever the rule is applied. 
 
 Numbers represented by letters are called Literal Numbers. 
 
 12 
 
DEFINITIONS AND NOTATION 13 
 
 For example, the volume of any rectangular prism is equal to 
 
 the area of the base multiplied by the height. By using v for 
 
 volume, a for area of base, and h for height, this rule is abbreviated 
 
 to 
 
 V = a X h. 
 
 When a = 60 and h = 5, v = 60 x 5 = 300 ; 
 
 when a = 36 and h =10, v = 36 x 10 = 360 ; etc. 
 
 In each problem to which this rule applies a and h represent 
 fixed, known values, but in consequence of being used for all 
 problems of this class, a and h represent numbers to which any 
 arithmetical values whatever may be assigned. Hence, the arith- 
 metical idea of number is extended as follows. 
 
 10. A literal number to which any value can be assigned at 
 pleasure is called a General Number. 
 
 11. A number whose value is known or a number to which 
 any value can be assigned is called a Known Number. 
 
 The numerals, 3 and 4|, and the general numbers a and ft in w = a x h, in 
 § 9, are known numbers. 
 
 Known literal numbers are generally represented by the first 
 letters of the alphabet. 
 
 12. A number whose value is to be found is called an Unknown 
 Number. 
 
 Unknown numbers are usually represented by the last letters of 
 the alphabet. 
 
 ALGEBRAIC SIGNS 
 
 13. The Sign of Addition is +, read ^ plus J 
 
 It indicates that the number following it is to be added to the 
 number preceding it. 
 
 a -I- 6, read ' a plus 6,' indicates that b is to be added to a. 
 
 14. The Sign of Subtraction is — , read ' minus.'' 
 
 It indicates that the number following it is to be subtracted 
 from the number preceding it. 
 
 a — b, read ' a minus 6,' indicates that 6 is to be subtracted from o. 
 
14 DEFINITIONS AND NOTATION 
 
 15. The Sign of Multiplication is x or •, read '■ multiplied hy.^ 
 
 It indicates that the number preceding it is to be multiplied by 
 the number following it. 
 
 o X 6, or a-h, indicates that a is to be multiplied by h. 
 
 The sign of multiplication is usually omitted in algebra, except 
 between figures. 
 
 a X 6, or a-h, may be abbreviated to ab.x x y to xy, i x bioi b, etc. But 
 3x5 cannot be written 35, because 35 means 30 + 5. 
 
 16. The Sign of Division is -r-, read 'divided by.' 
 
 It indicates that the number preceding it is to be divided by 
 the number following it. 
 
 a ~ b indicates that a is to be divided by b. 
 
 Division may be indicated also by writing the dividend above 
 the divisor with a line between them. 
 
 Such indicated divisions are called Fractions. (Cf. § 158.) 
 
 - indicates that a is to be divided by b. 
 b 
 
 17. The Sign of Equality is =, read ' is equal to' ov ' equals.' 
 
 18. The Sign of Inequality is > or <. 
 
 When used between two numbers, it signifies that they are 
 unequal, the greater number being at the opening of the sign, 
 a > 6 is read ' a is greater than 6.' 
 a; < 5 is read ' x is less than 5.' 
 
 - 19. The Signs of Aggregation are : the Parenthesis, ( ) ; the 
 Vincxdum, — ; the Brackets, [ ] ; the Braces, \ \ ; and the Vertical 
 Bar, \. 
 
 They show that the expressions included by them are to be 
 treated as single numbers. 
 
 Thus, each of the forms (a + 6)c, a -ir b-c, [a + 6]c, {a + 6}c, and ale, 
 signifies that the sum of a and b is to be multiplied by c. . + b\ 
 
 When numbers are included by any of the signs of aggregation, they are 
 commonly said to be in parenthesis. 
 
 20. The Sign of Continuation is • • • • or , read ' and so on,' 
 
 or ' and so on to.' 
 
 2, 4, 6, 8, is read ' 2, 4, 6, 8, and so on.' 
 
 21. The Sign of Deduction is .-.. It signifies therefore or hence. 
 
DEFINITIONS AND NOTATION 15 
 
 FACTORS, POWERS. AND ROOTS 
 
 22. Each of two or more numbers which multiplied together 
 produce a given number is called a Factor of the number. 
 
 Since 12 = 2 x 6, or 4 x 3, each of these numbers is a factor of 12. 
 Since 3 a& = 3 x rt x 6, each of these numbers is a factor of 3 ab. 
 
 23. When a factor of a number is considered as the multiplier 
 of the remaining factor, it is called a Coefficient of that factor. 
 
 In 7 X, 5 ax, bxy, and (a — b)x, the coefficients of x are 7, 5 a, by, and 
 (a — 6) ; in bxy, bx is the coefficient of y. 
 
 Coefficients are Numerical, Literal, or Mixed, according as they 
 are composed oificjures, letters, or hoth. figures and letters. 
 
 When no numerical coefficient is expressed, the coefficient may 
 be considered to be 1. 
 
 24. When a number is used a certain number of times as a 
 factor, the product is called a Power of the number. 
 
 Powers are named from the number of times the number is used 
 as a factor. 
 
 When a is used twice as a factor, the product is the second power of a ; 
 when a is used three times as a factor, the product is the third power of a ; 
 four times, the fourth power of a ; n times, that is, any number of times, the 
 nth power of a. 
 
 Tlie second power is also called the square, and the third power the cube. 
 
 The product indicated by a x a x a x a X a may be more briefly 
 indicated by a^. Likewise, if a is to be used n times as a factor, 
 the product may be indicated by a". 
 
 25. A figure or letter placed a little above and to the right of 
 a number is called an Index or an Exponent of the power thus 
 indicated. 
 
 The integers that the student has been using in arithmetic have been 
 positive integers. 
 
 When the exponent is a positive integer, it indicates the num- 
 ber of times that the number is to be used as a factor. 
 
 52 indicates that 5 is to be used twice as a factor ; a* indicates that a is to 
 be used 3 times as a factor. 
 
 When no exponent is written, the exponent is regarded as 1. 
 5 is regarded as the first power of 5, and a^ is usually written a. 
 
16 DEFINITIONS AND NOTATION 
 
 The terms coefficient and exponent should be carefully distin- 
 guished. 
 
 Thus, 5a = a + a + a + a + rt, but a^^axaxaxaxa. 
 
 26. One of the equal factors of a number is called a Root of the 
 number. 
 
 5 is a root of 25 ; a is a root of a* ; 4 a! is a root of 64 y?. 
 
 Koots are named from the number of equal factors into which 
 the number is separated. 
 
 One of the tioo equal factors of a number is its second root ; one of the 
 three equal factors of a number is its third root ; one of the four equal factors, 
 the fourth root; one of the n equal factors, the nth root. 
 
 The second root of a number is also called its square root, and its third 
 root is called its cube root. 
 
 27. The symbol which denotes that a root of a number is sought 
 is y', written before the number. 
 
 It is called the Root Sign, or the Radical Sign. 
 A figure or letter written in the opening of the radical sign 
 indicates what root of the number is sought. 
 It is called the Index of the root. 
 When no index is written, the second, or square root is meant. 
 
 v^ indicates that the third, or cube root of 8 is sought. 
 
 Vox and Va — b indicate the square roots of ax and a — b, respectively. 
 
 The horizontal line used in connection with the radical sign is a 
 vinculum. 
 
 ALGEBRAIC EXPRESSIONS 
 
 28. A number expressed by algebraic symbols is called an 
 Algebraic Expression. 
 
 29. When signs of operation are employed in algebraic ex- 
 pressions, the sequence of operations is determined by the follow- 
 ing conventional law : 
 
 A series of additions and subtractions or of multiplications and 
 divisions are performed in order from left to right. 
 3-1-4-2-1-3= 7-2-1-3 = 6-1-3= 8. 
 3x4H-2x3 = 12-f-2x3 = 6x3 = 18. 
 
 a -\-b — c + d indicates that 6 is to be added to a, then from this result c 
 is to be subtracted, and to the result just obtained d is to be added. 
 
DEFINITIONS AND NOTATION 17 
 
 30. When a particular number takes the place of a letter or 
 general number, the process is called Substitution. 
 
 KuMERicAL Substitutions 
 
 1. When a = 2, ?> = 3, and c = 5, what are the numerical 
 values of 3 c, (?, V8 ab^, or + h-, and (a + 6)^, respectively ? 
 
 Solutions 
 .7c = 3-5=:15. 
 
 c8 = 5 • 5 • 5 = 125. 
 
 VS"^ = V8 • 2 . 3 . 3 = V2 ■ 2 X 2 . 2 X 3 . 3 = 2 x 2 X 3 = 12. 
 
 a2 + 6'^ = 2 • 2 + 3 . 3 = 4 + 9 = 13.' 
 
 (a + 6)2 =(a + 6)(a + 6) = (2 + 3)(2 + 3)= 5 • 5 = 25. 
 
 Find the numerical value of each of the following algebraic 
 expressions, when a = 5, 6 = 3, c = 10, m = 4, ?t = 1 : 
 
 2. 10 a. 11. (ahy. 19. .^/T^i^. 
 
 3. 2 ah. 12. a^ft^. 
 
 20 ^ + ^^ 
 
 4. 3 cm. ,„ /7> ,. 9^ 
 
 13. V2 acn. c — zm 
 
 ^' ^' 14. 3b^cn\ oi . *?-« 
 
 6. ocml 21. c + 
 
 7. 2 a^o. 
 
 f — 2 ?;i 
 22. a'^c. 
 
 8. 3 6ml 16. (a-?.)l 
 
 9. 3a«6. 17. {n + iy. 23. m" ». 
 10. am*. 18. w'^ + 1. 24. (bm)"-^. 
 
 31. An algebraic expression whose parts are not separated by 
 
 + or — is called a Term ; as 2a^, — 5 xyz, and — • 
 
 z 
 
 In the expression 2 a;^ — 5 xyz + ?^ there are three terms. 
 
 z 
 The expression rtt(a + 6) is a term, the parts being in and (a + 6). 
 
 32. Terms that contain the same letters with the same expo- 
 nents are called Similar Terms. 
 
 3 x^ and 12 x^ are similar terms ; also 3(a + 6)2 and 12(a + 6)2 ; also ax 
 and 6x, regarding a and 6 as the coefiBcients of x. 
 
 ALG. — 2 
 
18 DEFINITIONS AND NOTATION 
 
 33. Terms that contain different letters, or tlie same letters 
 with different exponents, are called Dissimilar Terms. 
 
 5 a aiKl 3 by are dissimilar terms ; also 8 d^b and 3 ab'^. 
 
 34. Each literal factor of a term is called a Dimension of the 
 term. 
 
 The number of literal factors or dimensions of a term indicates 
 its Degree. 
 
 abed is a term of the fourth degree, because it is composed of four literal 
 factors, or has four dimensions, x^ is a term of the third degree, since x^ is 
 a convenient way of indicating that x is taken three times as a factor. The 
 expressions 4x:^y2^ and ^xyz* are each of the sixth degree. 
 
 35. The term of highest degree in an expression determines the 
 Degree of the Expression. 
 
 x^ + 3 x- + X + 2 and abc + b-c + ac — b are expressions of the third degree. 
 
 36. When all the terms of an expression are of the same degree, 
 the expression is called a Homogeneous Expression. 
 
 afi + 3 x^y + xy^ -\-2 y^ and abc + b-c + ac- are each homogeneous expressions. 
 
 37. An algebraic expression of one term only is called a 
 Monomial, or a Simple Expression. 
 
 xy and 3 ab are monomials. 
 
 38. An algebraic expression of more than one term is called a 
 Polynomial, or a Compound Expression. 
 
 3 a + 2 6, xy + yz + zx, and a^ + 6^ — c^ + 2 ab are polynomials. 
 
 39. A polynomial of two terms is called a Binomial. 
 Sa + 2b and x^ — y'^ are binomials. 
 
 40. A polynomial of three terms is called a Trinomial. 
 
 a + b + c and 3x — 2y—z are trinomials. 
 
 41. An expression, any term of which is a fraction, is called a 
 Fractional Expression. 
 
 *" — 3 X + - is a fractional expression. 
 a^ X 
 
 42. An expression that contains no fraction is called an Inte- 
 gral Expression. 
 
DEFINITIONS AND NOTATION 19 
 
 5 a2 _ 2 a and 6 x are integral expressions. 
 
 Expressions like x^ -\-\x'^ + \x+\ are sometimes regarded as integral, 
 since the literal numbers are not in fractional form. 
 
 43. An expression that can be written without using a root 
 sign is called a Rational Expression. 
 
 1, 2, 3, •••, a + i. — — , and (a - 6)^ are rational expressions. 
 
 ' x-y 
 V25 is rational, since it can be written 5 without a root sign. 
 
 44. An expression that cannot be written without using a 
 root sign is called an Irrational Expression. 
 
 a + Vh, a + 2 Va + 1, and v^4 are irrational expressions. 
 Va^ is not irrational, however, since it may be written a. 
 
 POSITIVE AND NEGATIVE NUMBERS 
 
 46. For convenience, arithmetical numbers may be arranged 
 in an ascending scale : 
 
 0, 1, 2, 3, 4, 5, ... 
 
 1 1 I I \ I 
 
 The operations of addition and subtraction are thus reduced to 
 counting along a scale of numbers. 2 is added to 3 by beginning 
 at 3 in the scale and counting 2 units in the ascending, or additive 
 direction; and consequently, 2 is subtracted from 3 by beginning 
 at 3 and counting 2 units in the descending, or subtractive 
 direction. In the same way 3 is subtracted from 3. But if we 
 attempt to subtract 4 from 3, we discover that the operation of 
 subtraction is restricted in arithmetic, inasmuch as a greater 
 number cannot be subtracted from a less. If this restriction 
 held in algebra, it would be impossible to subtract one literal 
 number from another without taking into account their arith- 
 metical values. Therefore, this restriction must be removed in 
 order to proceed with the discussion of numbers. 
 
 To subtract 4 from 3 we begin at 3 and count 4 units in the 
 descending direction, arriving at 1 on the opposite, or subtractive 
 side of 0. It now becomes necessary to extend the scale 1 unit 
 in the subtractive direction from 0. 
 
20 DEFINITIONS AND NOTATION 
 
 To subtract 5 from 3 we begin at 3 and count 5 units in the 
 descending direction, arriving at 2 on the opposite, or subtractive 
 side of 0. The scale is again extended, and may be extended 
 indefinitely in the subtractive direction in a similar way. 
 
 For convenience, numbers on opposite sides of are dis- 
 tinguished by means of the small signs "*" and ~, called signs of 
 quality, or direction signs, + being prefixed to those which stand 
 in the additive direction from and ~ to those which stand in 
 the subtractive direction from 0. 
 
 The former are called Positive Numbers, the latter Negative 
 Numbers. 
 
 Hence, the scale of algebraic numbers may be written : 
 
 ..., -5, -4, -3, -2, -1, 0, +1, +2, +3, +4, +5, ... 
 
 I I I \ ! \ I I \ 1 I 
 
 46. By repeating +1 as a unit any positive number may be 
 obtained, and by repeating ~1 as a unit any negative number 
 may be obtained. Hence, positive numbers are measured by 
 the positive unit, "^1, and negative numbers by the negative unit, ~1, 
 or by parts of these units. 
 
 47. If +1 and ~1, or +2 and ~2, or any two numbers numerically 
 equal but opposite in quality are taken together, they cancel 
 each other. For counting any number of units from in either 
 direction and then counting an equal number of units from the 
 result in the opposite direction, we arrive at 0. Hence, 
 
 If a positive and a negative number are united into one mimher, 
 any number of units or parts of units of one cancels an equal number 
 of units or parts of units of the other. 
 
 48. Two concrete quantities of the same kind are sometimes 
 opposed to each other in some sense so that, if united, any 
 number of units of one cancels an equal number of units of the 
 other. For convenience, such quantities are often distinguished 
 as positive and negative. 
 
 If money gained is positive, money lost is negative, for any sum gained 
 is canceled by an equal sum lost. If a rise in temperature is positive, a fall 
 in temperature is negative. If distances )iorth or west or upstream are 
 positive, distances south or east or downstream are negative. 
 
ADDITION 
 
 49. 1. If a man has 10 dollars in one pocket and 15 dollars in 
 another, how much money has he ? 
 
 2. If in algebra money in hand is considered a positive quan- 
 tity, indicate his iinancial condition algebraically. What is the 
 sum of 10 positive units and 15 positive units, that is, of +10 and 
 +15 ? of +4 and +8 ? of +a and +6 ? 
 
 3. If a person owes one man 10 dollars and another 15 dollars, 
 how much does he owe both ? Indicate his financial condition 
 algebraically, regarding a debt as a negative quantity. 
 
 4. What is the sum of 10 negative units and 15 negative units, 
 that is, of -10 and "15 ? of "6 and "14 ? of "a and 'h ? 
 
 5. What sign has the sum of two algebraic numbers that have 
 like signs ? 
 
 6. If a man has 25 dollars and owes 15 dollars, how much of 
 his money will be required to cancel the debt? How many 
 dollars will he have after settlement ? 
 
 7. What is the result when "15 is united with +25, that is, 
 what is the algebraic sum of ~15 and +25 ? of ~20 and ^10 ? of +8 
 and -3? of +6 and "10? 
 
 50. The aggregate value of two or more algebraic numbers is 
 called their Algebraic Sum. 
 
 The process of finding the simplest expression for the algebraic 
 sum of two or more numbers is called Addition. 
 
 51. Principles. — 1. The algebraic sum of two numbers with 
 like signs is equal to the sum of their absolute values with the com- 
 mon sign prefixed. 
 
 21 
 
22 ADDITION 
 
 2. Tlie algebraic sum of two numbers with unlike signs is equal 
 to the difference between their absolute values with the sign of the 
 numerically greater prefixed. 
 
 By successive applications of the above principles any number 
 of numbers may be added. 
 
 Only similar terms can be united into a single term. 
 
 Principle 1 may be established as follows : 
 
 The sum of 5 positive units and 3 positive units is evidently (5 + 3) posi- 
 tive units, or 8 positive units ; tliat is, 
 
 +5 + +3 = +(5 + 3) =+8. 
 
 Similai-ly, whatever absoUite values a and h represent, since a times the 
 unit +1 plus 6 times the unit +1 is equal to (a + 6) times the unit +1, 
 § 46, +a + +6 = +(o + h). 
 
 Again, the sum of 5 negative units and 3 negative units is (5 + 3) negative 
 units, or 8 negative units ; that is, 
 
 -5 + -3 = -(5 + 3) = -8. 
 
 Similarly, whatever absolute values a and h represent, since a times the 
 unit -1 plus h times the unit -1 is equal to (a + b) times the unit ^1, 
 § 46, -a + -6 = -{a + h). 
 
 Principle 2 may be established as follows : 
 
 The sum of 5 positive units and 3 negative units is 2 positive units, since, 
 § 47, the 3 negative units cancel 3 of the positive units and leave 2 positive 
 
 units ; that is, 
 
 +5 + -3 = +(5 -3) = +2. 
 
 The sum of 5 positive units and 7 negative units is 2 negative units, since, 
 
 § 47, the 5 positive units cancel 5 of the negative units and leave 2 negative 
 
 units ; that is, 
 
 +5 + -7 =-(7 -5) =-2. 
 
 Similarly, whatever absolute values a and h represent, 
 if a>6, +a + -6 = +(a -6), 
 
 for, § 47, the h negative units will cancel b of the a positive units and leave 
 (a — b) positive units ; 
 
 but if ft > a, +a + -b = -(b - a), 
 
 for, § 47, the a positive units will cancel a of the b negative units and leave 
 (6 — a) negative units. 
 
 Examples 
 
 -3 + +4 ^_ -2 + +8 + -9. 
 +m + ~n, if m>n. 
 
 Find the value of 
 
 
 
 1. +7 4- +3. 3. 
 
 +7 + -3. 
 
 5. 
 
 2. -7 + -3. 4. 
 
 -7 + +3. 
 
 6. 
 
ADDITION 23 
 
 52. To conform with the ideas already presented, the terms 
 < greater ' and ' less ' must be interpreted as follows : 
 
 An algebraic number is increased, or made greater, when a posi- 
 tive number is added to it, and decreased, or made less, when 
 a negative number is added to it. 
 
 Since, by § 51, -3 + ^1 ="2, -2 + +1 = -f, -f + +1 = 0, 
 +1 -|- +1 = +2, etc., in the scale of algebraic numbers 
 
 ..., -5, -4, -3, -2, -1, 0, +1, +2, +3, +4, +5, ••., 
 
 each number is greater than the number on its left and less than 
 the number on its right ; that is, 
 
 ..., -3<-2, -2<-l, -1<0, 0<+l, +l<+2, +2<+3, .... 
 
 Note. -3 < 2 may be read ' -3 is less than -2 ' or ' -2 is greater than -3.' 
 
 Hence, it follows that : 
 
 1. Any positive number is greater than zero and any negative 
 number is less than zero. 
 
 2. Of two positive numbers that ivhich has the greater absolute 
 value is the greater, and of two negative numbers that which has the 
 less absolute value is the greater. 
 
 63. Abbreviated notation for addition. 
 
 Referring to the scale of algebraic numbers, it is evident that 
 adding positive imits to any number is equivalent to counting 
 them in the positive direction from that number, and adding 
 negative units to any number is equivalent to counting them in 
 the negative direction from that number. Hence, in addition, 
 the signs -|- and — denoting quality have primarily the same 
 meanings as the signs + and — denoting arithmetical addition 
 and subtraction. For example, by the definition of positive and 
 negative numbers, 
 
 +1 means + 1 and "1 means — 1 ; 
 also """5 means 0+5 and ~5 means — 5 ; etc. 
 
 Hence, in addition, but one set of signs, + and — , is necessary, 
 and in finding the sum of any given numbers, the signs + and — 
 may be regarded either as signs of quality or as signs of opera- 
 tion, though it is commonly preferable to regard them as signs 
 of operation. 
 
24 ADDITION 
 
 For brevity, it is customary to omit the sign + before a 
 monomial or before the first term of a polynomial. But the 
 sign — cannot be omitted. 
 
 +5 + +3 + -6 is written 5 + 3 - 6 ; -4 + +8 + "2 is written -4 + 8-2. 
 When there is need of distinguishing between the signs of 
 quality + and — and the signs of operation + and — , the num- 
 bers and their signs of quality may be inclosed in parentheses. 
 
 Thus, if a = + 5, 6 = - 3, and c = - 2, then a + 6 + c=(+5) + (-3) 
 + (-2); a- &-c = (+5)-(-3)-(-2); a6c = (+ 5)(- 3) (- 2); etc. 
 
 54. A term preceded by +, expressed or understood, is called 
 a Positive Term, and a term preceded by — , a Negative Term. 
 
 Thus, in the polynomial 3a + 26 — 5c the first and second terms are 
 positive and the third term is negative. 
 
 Examples 
 Write the following with one set of signs : 
 
 1. +7 + +8. 4. +10 + -2 + --4. 7. +a -f "6. 
 
 2. +6 + -5. 5. -6 + -3 + +16. 8. -a + +& + -c. 
 
 3. -3 + -7. 6. +8 + +4 + -5. 9. -x + -y-[- -%. 
 
 55. 1. How does 5 + 3 — 2 compare in value with 5 — 2 + 3, 
 or with 3-2 + 5, or with - 2 + 3 + 5 ? 
 
 2. How does a-\-h — c compare in value with h — c -{-a, or 
 with h -\- a — c, or with a — c-^-h'i 
 
 3. In what order may numbers be added ? 
 
 Law of Order, or Commutative Law for Addition. — Algebraic num- 
 bers may be added in any order. 
 
 The Law of Order may be established as follows : 
 
 We know from arithmetic that arithmetical numbers may be added in any 
 order. Since, § 51, Prin. 1, algebraic numbers having like signs are added 
 by prefixing their common sign to the sum of their absolute, or arithmetical 
 values, and since in finding this sum the absolute values may be added in 
 any order, it follows that algebraic numbers having like signs may be added 
 in any order. 
 
 If some of the numbers are positive and some are negative, § 47, the same 
 number of positive and negative units will cancel each other, and the same 
 number of one or the other will be left, in whatever order the numbers are 
 added. 
 
ADDITION 25 
 
 Hence, whether the numbers have like or unlike signs, they may be added 
 in any order ; that is, 
 
 a + 6 + c = & + c + a = c + rt + 6 = c+6 + a, etc. , 
 for all values of a, 6, and c. 
 
 56. 1. How are the numbers 4, \, 2, and f grouped in adding? 
 the numbers 25 and 32, or 20 + 5 and 30 + 2 ? 
 
 2. In what manner may the terms of an expression be grouped 
 in addition ? 
 
 Law of Grouping, or Associative Law for Addition. — TJie s%im of 
 three or more algebraic numbers is the same in whatever manner the 
 numbers are grouped. 
 
 The Law of Grouping may be established as follows : 
 
 By the Law of Order, a + 6 + c = 6 + c + a 
 §29, ' =(h + c)+a 
 
 by the Law of Order, — a +{b -\- c). 
 
 Other cases, as a + 6 + c = (a + c) +'6, etc., may be proved similarly. 
 
 Hence, for all values of a, 6, and c, 
 
 rt4-6 + c = rt4-(& + c) = (a4-c)+& = c + (a + 6), etc. 
 
 57. To add similar monomials. 
 
 Examples 
 
 1. Add 4 a and 3 a. 
 
 PROCESS Explanation. — Just as 4 = 1 + 1 + 1 + 1, so 4a=a+a + a+a; 
 4 d just as3 = l + l + l, so 3 a = a + a + a. Therefore, 4 a + 3 a 
 = a + a-\-a-\-a-\-a-\-a + a, the symbol for which is 7 a. 
 Or, the sum may be obtained by adding the numerical coeffi- 
 7 a cients and annexing to their sum the common literal part. 
 
 2. Add Aa, ^a, — 3 a, and i a. 
 
 PROCESS 
 
 4a+-|a — 3a + ia = 4a — 3a + (|a4-|a) = a+-2a = 3(x. 
 
 Explanation. — By the Law of Grouping the sum of f a and ^ a may 
 be added to the sum of 4 a and — 3 a. Just as 4 = 1 + 1 + 1 + 1, so 
 4a = rt + a + a + a; just as — 3=— 1 — 1 — 1, so — 3a=— a — a — a. 
 Therefore, ia — 3a = a + a + a+a — a — a — a= (by the Law of Order) 
 a — a + a — a + a — a + a — + a = a. Just as | = i + i + i, so | a 
 = ^a + ^a + ^a. Therefore, | a + | a = J a + ^ a + | a + ^ a = | a = 2 a. 
 Adding the two groups, rt + 2a = a + a + a = 3a. 
 
 Or, the sum may be obtained by adding in any order or manner the 
 numerical coefficients, and annexing to their sum the common literal part. 
 
 3a 
 
26 ADDITION 
 
 Simplify the following : 
 
 3. 2y -1 y — 5y ~y + 10y — &y + ?>y. 
 
 4. 5a — 3a + 8a — 10a — 5a — lla + 24a. 
 
 5. 3by -5by- 10 by -Uby-^iSby. 
 
 6. 8 a^b + 6 a'b - 11 a^b - 2 aJ'b + 9 a%. 
 
 7. l|ary-ia^2/==-lJi^a^/ + 3iar'/ + r'/. 
 
 8. 5 (xy)2 - 3 (a;^)2 - 15 (xyf + 4 (xyf + 13 (a^)^. 
 
 9. (a — x) + 5 (a — x) + 7 (a — a;) — 3 (a — x) — 2 (a — a;). 
 
 10. 3(a ~\-by-\-6{a+ by-10(a + bf - (a + bf+ 12(a + bf. 
 
 11. 20 Vx - 3 - 8 Va; - 3 - 12 Vx - 3 + Va; - 3 + 7 Va; - 3. 
 
 12. 3 a;(af' - 2 .x- + 3) - a;(a;2 - 2 .t + 3) + 2 a;(ar^ - 2 a; + 3). 
 
 13. 2(a: - 1) - 13(a; - 1) + 5(a; _ 1) + 10(a; - 1) + 6(a; - 1). 
 
 14. i-(a + & - c) - |(a + 6 - c) + i^(a + b - c). 
 
 Since only similar terms can be united into a single term, in 
 algebra dissimilar terms are considered to have been added when 
 they have been w^ritten in succession with their proper signs. 
 
 In algebra many indicated operations are regarded as per- 
 formed. 
 
 Since 5 a, — 3 b, and 2 c cannot be united into a single term, their sum 
 is written 5 a — 3 fe + 2 c. 
 
 15. Add 6a, —5b, —3a, 3b, 2c, and —a. 
 Solution. 6a-5b-3a + Sb + 2c-a = 2a-2b + 2c. 
 
 Add the following: 
 
 16. 2xy, Aab, 3xy, and ab. 
 
 17. ran, —3cd, —6mn, and Acd. 
 
 18. a, — b, 2 c, —2 a, 3 b, and —4 c. 
 
 19. 6x, 3y, —2x,y, —3x,z, and —3y. 
 
 20. 2a, 2 6, 2c, 2c?, —a, —3 6, — c, and —3d 
 
 21. a, —4 a, 2 6, cd, — 2a6, 5 6, and — 3cd. 
 
ADDITION 27 
 
 58. To add polynomials. 
 
 Examples 
 
 1. Add 3a-36 + 5c, — 3a + 2 6, and c - 4 6 + 2 a. 
 
 PROCESS Explanation. — For convenience in adding, simi- 
 
 3a — 36 + 5c lar terms are written in the same column. 
 
 — 3 a + 2 6 '^^^ algebraic sum of the first column is 2 a, of 
 
 2 a — 4 6 4- c *^^ second — 5 6, and of the third + 6 c ; and these 
 
 ^numbers written in succession express in its simplest 
 
 2a — 56 + 6c form the sum sought. 
 
 2. Simplify 1\ a?h - 1 ah' -^ 2 ac" + 10 db - 'i ac^ + b a?h - 4. ab^ 
 + 5 ac=^ + 6^ + 9 a6=^ - 7 a^d - 2 6=^ + 2 a62 - 8 a6 - 6 a?b. 
 
 PROCESS 
 
 Ila2&-7a&2 + 2ac2 + 10a6+ W 
 + 5a26-4a62-4ac2 -263 
 
 -Ta'^ft + Qafe^ + Sac^ 
 -&a?h + 2ab^ - 8a6 
 
 3a''6 +3ac2+ 2a6 
 
 Rule. — Arrange the terms so that similar terms stand in the 
 same column. 
 
 Find the algebraic sum of each column, and write the results in 
 succession with their proper signs. 
 
 3. Add 2 a - 3 6, 2 & - 3 c, 5 c - 4 a, 10 a - 5 6, and 7 & - 3 c. 
 
 4. Add x + y + z, x — y -\-z, y — z — x, z — x — y, and 05 — 2!. 
 
 Simplify the following polynomials : 
 
 5. 7a; — 11 2/4-42 — 72 4-1103 — 4y + 7 y — llz — 4:X-{-y—x—z. 
 
 6. a + 3 6 + 5c-,6a + d + 46-2c-26 + 5a-d-t-a — 6. 
 
 7 . Ax' - 3xy -]- 5y^ + 10 xy -17 y'' -11 x^ - 5xy -{-12x^ - 2xy. 
 
 8. 2xy — 5y^-\- a^y^ — 7xy-}-3y^ — 4: o^y^ + 5 xy + 4 ?/^ + -s^y^. 
 
 9. 2 ai/ — 3 ac — 4 ay + 4 ac — 6 a?/ + 5 ac -H 11 ay — 4 oc — ay. 
 
28 
 
 ADDTTTON 
 
 10. 5 am, — 3 aW + 4 — 4 am + ahn^ — 2 + 5 -f a^m^ —6 + 3 am. 
 
 11. 6V^ — 6^xy + 3Vy — 4Va; + 6y/xy —y/x — -Vy + 3Vy 
 — 2V^. 
 
 Add the following polynomials : 
 
 12. 7a-3b + 5c-10d,2b + d-Sc-4:e, 5c-6a + 2(Z-4e, 
 86 — 7a — 8c— e, a — 5c + 5d + lle, a — 6 + c + 2d + e, and 
 5a — 46 + 2 c. 
 
 13. 5x — 3y — 2z, 4:y--2x-^6z,3a — 2x — 4:y,4b — 2z — 5y, 
 a — 5b, 5y — 6x, Sx + 2y — 5a — 2b, and 6x — y — 2z + 4b. 
 
 14. ?n-|-w — V/n.w, ■\/mn — 2m—3n, 3m-\-2n, 4n — Vmn— 3m, 
 5 Vwin — n, 4 m — ?i — 2 Vmn, 5 w + 2 m — 3 Vmn, and w — 6 m. 
 
 15. 2c — 7d + 6n, 11m — 3c — 5 w, 7/i — 2d — 8c, 8d — 3m 
 + 10c, 4d — 3w — 8m, m — 6n, and 2m — 3d. 
 
 16. 4a:3_2a;2-7a; + l, x-3 + 3a^+ 5ic-6, A.x'-^y? + 2-Qx, 
 2ar»-2ic2 + 8a; + 4, and 2ar' - 3ic2 - 2a; + 1. 
 
 17. a'^ + 5a^6+5a6^ + 6^ a*6 - 2a« + a«62 - 2 6^ a%^-3a^b^ 
 -Aa*b- a% and 2 a^ + a^6 - 2 a^b^ + 2 a26'' - 3 a6* + 6^. 
 
 18. a*-2 a36+3 a^b^ 3 ab^-4: b*-2 aW, 3 a36+4 a*-3 a63+4 b\ 
 5a^b + 7 a'b' -Aab^-S ¥, and a^ - 6 a36 - 8 a^b\ 
 
 19. 5x«-a^ + 7a;-9, 40^* - 3a;'' + 6a^ + 12, x^-5x^-x-l, 
 4-ar=-x«, 4x^-10a;2 + 3a;« + 4, and a;« + ar^ - 3ar^ - 4 a;- 5. 
 
 20. 3(a + 6) + 6(6 + c), 5(a + 6)-10(6 + c), 2(a + 6) + (6 + c), 
 3(6 + c) - (a + 6), 2(6 + c) - 10(a + 6), and 3(a + 6) - 3(6 + c). 
 
 21. a; + 3 (a + 1) — y, — (a + 1) — 2 a; + 4 ?/, and 3 a; — 4 (a + 1). 
 
 22. a^ - 3 a26c - 6 a6% a26 - 6^ - c^ - 3 a6c, aW + b''c + b(?, 
 5 a'bc + 4 a62c + c^, 6^ - a'b - ab\ a' + b'c + be', and 2 a6^c - 2 60^. 
 
 23. .12a;3_4a^.2_^^_^2, .4 a;2 - 4 a; + .4 - a;^^ 3|-a;-.6+3ar2+2a;3^ 
 and l-^x + 1.2x^ + ^0^. 
 
 24. aa; — f ax^ — \ ax^, f ax* — ^ ax^ — |^ 6a;i/, 1 6a;y — f aa;* — ^ ab, 
 1 6a;y — I a6 + ^ ax, and 2 a6 — f aa; + | aa^. 
 
 lab,M 
 
 d 
 
ADDITION 29 
 
 Exercises 
 
 69. 1. If a boy has n marbles and buys 10, how many will he 
 then have ? If he gives away m of these, how many will be left ? 
 
 2. Mary has 25 cents. How many cents will she have after 
 spending 10 cents and earning a cents ? If she has c cents, spends 
 b cents, and earns a cents, how many cents will she have ? 
 
 3. A boy who has p marbles loses q marbles, and then buys 
 r marbles. How many does he then have ? 
 
 4. James is 15 years old. In how many years will he be 21 
 years old ? In how many years will he be x years old ? Harry 
 is y years old. How many years older is he than James? In 
 how many years will Harry be x years old ? 
 
 5. Edith is 14 years old. How old was she 4 years ago? a 
 years ago ? How old will she be 3 years hence ? h years hence ? 
 
 6. William is x years old. How old was he a year ago ? 
 How old will he be in 5 years ? in a years ? After how many 
 years will he be 21 years old ? m years old ? How old will he 
 be when he is twice as old as he is now ? 
 
 7. In a certain family there are five children each of whom 
 is 2 years older than the one next younger. If the youngest is 
 X years old, what are the ages of the others ? 
 
 8. A woman sold some eggs, and with the money bought 8 
 pounds of sugar and 5 pounds of coffee. If the sugar cost a 
 cents a pound, and the coffee cost 6 cents a pound, how much did 
 she receive for the eggs ? 
 
 9. What two whole numbers are nearest to 50? to »? to 
 X -f- 5 ? If 2/ is an even number, what are the nearest even 
 numbers ? 
 
 10. George is a years younger than Henry, and h years 
 younger than John. If John is 16 years old, how old is Henry ? 
 
 11. A man paid two men, whom he owed, in the following 
 manner : To the first he gave an a-dollar bill, and received 
 change amounting to h dollars; and to the second he gave a b- 
 dollar bill, and received change amounting to c dollars. How 
 much did he owe both? 
 
30 ADDITION 
 
 Equations and Problems 
 
 60. 1. Simplify the equation 2x — 3x-\-&x-\-bx — x = 21. 
 
 and find the value of x. 
 
 Solution 
 2a;-3a; + 6x + 5x-x = 27. 
 Uniting terms, 9 x = 27. 
 
 Hence, x = 3. 
 
 Simplify, and find the value of cc : 
 
 2. lOx-1 x + 4.x-Qx + llx-20x-\-12x = 4Q. 
 
 3. 13a; — 6x — 4a; + 7x + lla;-16a;4- 15a: = 20. 
 
 4. 25 .T - 5 X - 7 a; - 2 a; + 14 x - 10 » - 12 X = 36. 
 
 5. 17a; + 2x-6x + 4x-12x-30a; + 40a; = 75. 
 
 6. 10a; + 2x + 3a; + 4a; + lla; + 12a;4-18a; = 60. 
 
 7. 16 a; — 3x — 5 a; — Sx-f 10x + 15x — 15x = 50. 
 
 8. 12a; + 10a;-20a; + 16x-3a!-2a; + 2a; = 75. 
 
 9. 14a; — 11 a; + 26 a; — 35a; — 4 a; + 7 x + 4 a; = 16. 
 
 10. 75 a; - 37 a; - 40 a; + 10 a; - 8 a; - 6 .a; + 9 a; = 21. 
 
 11. 4 x + 10 a; - 60 a; + 48 a; + 12 a; + 5 a; + 2 a; = 63. 
 
 12. 7 a; + 11 a; — 13 a; + 15 a; — 17 a; — 3 a; + 5 a; = 25. 
 
 13. 5 a; -15 a; + 25 a; — 30 a; + 10 a; + 3 a; + 6 a; = 56. 
 
 14. a; + 2 a; + 3 a; + 4 a; + 5 a; + 6 a; + 7 a; + 8 a; =144. 
 
 15. 4 a; + 12 a: — 17 a; — 10 a; + 15 a; + a; + 15 a; = 400. 
 
 16. 3 a; + 10 a; - 20 .T - 4 a; + 12 a; + 3 a; + 11 a; = 300. 
 
 Solve the following problems : 
 
 17. A man bequeathed $10,000 to 3 sons and 4 daughters, so 
 that a son received twice as much as a daughter. What was the 
 share of each daughter, and of each son ? 
 
I 
 
 I 
 
 ADDITION 31 
 
 18. John had twice as many marbles as Henry, and \ as many 
 as Charles. If they had 225 marbles in all, how many had each ? 
 
 19. A had twice as much money as B, who had 3 times as 
 much money as C. If all together had $ 2000, how much money 
 had each? 
 
 20. A merchant owes A a certain sum of money, B i as much, 
 and C twice as much as A. Vaiious persons owe him in all 12 
 times as much as he owes B. If all these debts were paid, he 
 would have $ 10,000. What are the amounts he owes ? 
 
 21. Mr. Jones succeeded in doubling his capital once every 
 5 years. If his capital at the end of 20 years was $ 150,000, with 
 what capital did he begin ? 
 
 22. The distance around a rectangular field 4 times as long as 
 it is wide is 200 rods. What are the dimensions of the field ? 
 
 23. What are the dimensions of a rectangular field whose 
 length is twice its width, if 240 rods of fence are required to 
 inclose it ? 
 
 24. Two boys caught the same number of fish, another caught 
 10 more, and another 10 less. If they caught in all 120 fish, how 
 many did each catch ? 
 
 25. The sum of 3 consecutive whole numbers is 84. What 
 are the numbers? 
 
 Suggestion. — Let x represent the middle number. Then, the other two 
 numbers will be represented by x — 1 and x + 1. 
 
 26. Of what 3 consecutive even numbers is 150 the sum ? 
 
 27. A, B, C, and D together have $ 1500. If A had f 50 more 
 and B $ 50 less, they would each have the same sum as C and i 
 as much as D. How much money has each ? 
 
 28. The ages of 4 brothers differ successively by 2 years. If 
 the sum of their ages is 56 years, what is the age of each ? 
 
 29. Three newsboys sold 270 papers in an evening. If the 
 second sold 5 less than twice as many as the first, and the third 
 5 more than 3 times as many as the first, how many papers did 
 each sell? 
 
SUBTKACTION 
 
 61. 1. What is left when 5xy is taken from 12 xy? What is 
 the sum of ~b xy and +12 xy ? 
 
 2. What is left when +3 mn is subtracted from "•"10 mn ? What 
 is the sum of ~3 mn and """10 mn ? 
 
 3. Instead of subtracting a positive number, what may be 
 done to obtain the same result ? 
 
 4. What is the result when 8 a is subtracted from 10 a ? When 
 (8a— 5a) is subtracted from 10 a ? How does the second result 
 compare with the first ? What effect upon the result has the 
 subtraction of the negative number ~oa? 
 
 5. How does the result of subtracting (5 a; — 2 a;) from 12 a; 
 compare with the result of subtracting 5 x from 12 a; ? What 
 effect upon the result has the subtraction of the negative number 
 -2 a;? 
 
 6. Instead of subtracting a negative number, what may be 
 done to obtain the same result ? 
 
 62. In addition two numbers are given, and their algebraic 
 sum is required; in subtraction the algebraic sum, called the 
 minuend, and one of the numbers, called the subtrahend, are 
 given, and the other number, called the remainder, or difference, 
 is required. 
 
 Subtraction is, therefore, the inverse of addition. 
 The Difference is the algebraic number that added to the sub- 
 trahend gives the minuend. 
 
 63. Principles. — 1. Subtracting a positive number is equiva- 
 lent to adding a numerically equal negative number. 
 
 2. Subtracting a negative number is equivalent to adding a numer- 
 ically equal positive number. 
 
SUBTRACTION 33 
 
 The difference of similar terms, only, can be expressed in one 
 term. 
 
 Prindple 1 may be established as follows : 
 
 Let m represent any minuend and +s any positive subtrahend. 
 It is to be proved that m —+s = m +-s. 
 
 § 62, to find the remainder when +s is subtracted from m is to find the 
 algebraic number that added to +s will give m. 
 
 Since the algebraic sum of +s and -s is 0, by the Associative Law for 
 Addition the algebraic sum of +s and m +-s is m + 0, or m. 
 
 Hence, the algebraic number that added to +s gives m is m +-s. 
 
 .•. m —+s = m +~s. 
 
 Principle 2 may be established as follows : 
 
 Let m represent any minuend and -s any negative subtrahend. 
 
 It is to be proved that m —-s = »i ++s. 
 
 § 62, to find the remainder when -.s is subtracted from m is to find the 
 algebraic number that added to s will give m. 
 
 Since the algebraic sum of -s and +s is 0, by the Associative Law for 
 Addition the algebraic sum of -s and m ++s is m + 0, or m. 
 
 Hence, the algebraic number that added to -s gives m is m + +s. 
 
 .'. m —-S — m ++S. 
 
 64. Since, from the above principles, subtracting algebraic 
 numbers is equivalent to adding them to the minuend with their 
 signs changed, it follows that the Laws of Order and Grouping 
 for Addition hold in the subtraction of algebraic numbers; and 
 that when one or more subtrahends with their signs changed are 
 added to the minuend to form the algebraic sum called the dif- 
 ference, one set of signs, + and — , suffices to denote either quality 
 or operation. 
 
 65. To subtract when the terms are positive. 
 
 Examples 
 
 1. From 10 a; subtract 4 a;. 
 
 PROCESS 
 
 y^ ^ Explanation. — Since subtracting a positive term is equiv- 
 
 . alent to adding a numerically equal negative term (IMn. 1), 
 
 4 X may be subtracted from 10 x by changing the sign of 4 x, 
 and adding 10 x and — 4 %. 
 
 6x 
 
 ALG. — 3 
 
84 
 
 SUBTRACTION 
 
 2. From 10 x subtract 15 a;. 
 
 PROCESS 
 
 10 a; 
 15 a; 
 
 Explanation. — Since subtracting a positive term is 
 equivalent to adding a numerically equal negative term 
 (Prin. 1), 15x may be subtracted from 10 x by changing the 
 sign of 15 X and adding 10 a; and — 15 x. 
 
 3. 4. 5. 6. 7. 8. 
 
 From 12 a 9 am 8ar/ 24 mn^ CVo^ 11 (a + 6) 
 Take 5 a 21am 18 .tV 12 mn^ 15Vaa; 21 (a + &) 
 
 From 9 a + 7 6 
 
 Take 2 a + 3 6 
 
 10. 
 
 5a + 106 
 
 7a 4- Ah 
 
 11. 
 
 10x + 2y 
 Q>x + Ay 
 
 12. 
 
 3m + 3n 
 2 m + 5 n 
 
 13. 14. 15. 16. 
 
 From 15m+n 1 x-\-2y Ax + Ay ^p + Zq 
 
 Take 12m + 2n Ax-\-Ay lx^2y 10p + 2g 
 
 17. From 8jp + 32 subtract lOp + 2. 
 
 18. From 15 m + rt subtract 5 m + 3 n. 
 
 19. From 'dax-\-^hy subtract Aax-\-^hy. 
 
 20. From 8 ahc + 19 mx subtract 20 ahc + 7 rax. 
 
 21. From a + 3& + c subtract a + 6 + 3c. 
 
 22. From 12 a^ + 2 6^ + 14 c^ subtract 3 a- + 13 5^ -|- 3 &. 
 
 23. From & ax-\-hy -\-l cz subtract 2ax-\-'by -\-2 cz. 
 
 24. From 7 ax-\-hy + 2 cz subtract Aax-\-ohy + cz. 
 
 25. From Aah + c subtract or -{■ Ir^ -\- ahc + 2 ah -{■ 2 c. 
 
 26 . From 5 xy subtract o;^ + 2 a;^?/ + 2 a;/ + 3 xi/ + y . 
 
 27. From 1 subtract a;* + 13 ar* + 15 a;^ + 16 a; + 25. 
 
 28. From Ix^ — Ay^ subtract 6 a;^ + 3 a;^/ — 6 ?/l 
 
SUBTRACTION 35 
 
 66. To subtract when some terms are negative. 
 
 Examples 
 1. From 8cc — 3?/ subtract 5x — 7y. 
 
 PROCESS Explanation. — Since subtracting a positive term is equiv- 
 
 8 a; 3 y alent to adding a numerically equal negative term, subtract- 
 
 K J ing 5x from 8 x is equivalent to adding — 5x to 8.r (Prin. 1). 
 
 Since subtracting a negative term is equivalent to adding 
 a numerically equal positive term, subtracting — 7 y from 
 
 Sx-j-Ay —3?/ is equivalent to adding +7y to —3y (Prin. 2). 
 
 Rule. — Change the sign of each teryti of the subtrahend, or con- 
 ceive it to be changed, and add the result to the minuend. 
 
 2. 3. 
 
 From 5 a 6xy 
 
 Take -2a -3xy 
 
 7. 
 From Am — 3n + 2p 
 Take 2m — 5n — j) 
 
 10. 
 From a — b-\- c 
 Take 2a + b — c 
 
 4. 5. 
 
 — Omn -ISVx 
 
 — 4 m7i — -SVa; 
 
 6. 
 
 - 3(a + &) 
 -10(a + 6) 
 
 8. 
 
 Sa-lOb + c 
 6a— 5b — c 
 
 9. 
 
 3x + 2y-z 
 5x — Ay — z 
 
 11. 
 
 8a%-5ac^ + 9a^c 
 3a^b-\-2ac' + 9a^c 
 
 12. 
 
 r — s + t 
 r + s — t 
 
 13. From 5x — 3y-\-z take 2x — y + %z. 
 
 14. From 3 a% + Z^' - a^ take Aa%-8 a'' + 2 /A 
 
 15. From rSa^ + 5b''-Ac^ take Scr + Oft^ + lOc^ 
 
 16. From 15x — 3y + 2z subtract 3x + 8y — dz. 
 
 17. From a^ — ab—b^ subtract cib — 2a'^ — 2b^. 
 
 18. From m^ — mn-\-n^ subtract 2 m^ — 3 mn -\- 2 n^. 
 
 19. From 5x^ — 2xy — y^ subtract 2 x^ -\-2xy — 3y'\ 
 
 20. From 2ax — by — 5xy subtract 2by — 2ax — 3 xy. 
 
36 SUBTRACTION 
 
 21. From 2 a + c subtract a — h+c. 
 
 22. From 2m + n subtract n — 2p. 
 
 23. From x-\-y subtract 3a— 4 + y, 
 
 24. From 2 a^ + 2 xy subtract x^ — xy — y\ 
 
 25. From 2a — 2d subtract a — h + c — d. 
 
 26. From 2 & subtract h — a — c — d. 
 
 'iil. From a'^ + .x'^ subtract a^ — 3 a^ic + 3 ax- — a^. 
 
 28. From a* -\-\ subtract 1 — a + a^ — a'' + a*. 
 
 29. From the sum of 3a^ — 2a& — 6^ and 3a6 — 2a^ subtract 
 a^-ah- b-. 
 
 30. From 3x — y + z subtract the sum of x — Ay + z and 
 2x + 3y-2z. 
 
 31. From a + b + c subtract the sum of a — b — c, b — c — a, 
 and c — a — b. 
 
 32. Subtract the sum of mhi — 2 mn^ and 2 m^n — m^— n^4-2 mn^ 
 from irv^ — n^. 
 
 33. Subtract the sum of 2c — 9a — 36 and 36 — 5a — 5c from 
 h — 3c + a. 
 
 34. From 3 6.1; + 4 oy subtract the sum of Say — Abx and 
 bx -\- ay. 
 
 35. From the sum of 1 -\-x and 1 — a^ subtract 1 — a; + a;- — x*^. 
 
 36. From |a;»— far' + 3x — 7 subtract ^x^ — ^xr + ^x — lO. 
 
 37. From i m^ — \ mhi + 1 ??irr — 27 ^^ subtract n^ — m^ + ^ mn^ 
 
 38. From 5{a + b) —3(x-\-y) +4:(m + n) subtract 4 (a +6) 
 
 + 2(x + y) + {m + n). 
 
 39. From n^ — m^ subtract the sum of 2 mhi^ — 3 mn* and 
 m^ + 4 m^n^ — 2 m^n^ + 5 mrt* — n^. 
 
 40. From the sum of 3a^ — 2a; + l and 2x — 5 subtract the 
 
 sum of a; — a:^ + 1 and 2 a^ — 4 x + 3. 
 
SUBTRACTION 37 
 
 PARENTHESES 
 
 67. The subtrahend is sometimes written within a sign of 
 aggregation preceded by the sign — . 
 
 If a — 6 is to be subtracted from 2 a, it may be written 2 a — (a — ft). 
 
 1. What change must be made in the signs of the terms of the 
 subtrahend, when it is subtracted from the minuend ? 
 
 2. When a number in parenthesis is preceded by the sign — , 
 what change must be made in the signs of the terms, when the 
 subtraction is performed, or when the parenthesis is removed ? 
 
 68. Principles. — 1. A parenthesis preceded by the minus sign 
 may he removed from an expression, if the signs of all the terms in 
 pareyithesis are changed. 
 
 2. A parenthesis jtreceded by the minus sign may be used to inclose 
 an expression, if the signs of all the terms to be inclosed are changed. 
 
 1. When numbers are inclosed in a parenthesis preceded by the plus 
 sign, the parenthesis may be removed without changing the signs of the 
 terms. 
 
 2. Any number of terms may be inclosed in a parenthesis preceded by » 
 plus sign without changing the signs of the terms. 
 
 3. The student should remember that in an expression like — (x — ?/), or 
 — a; — y, the sign of a; is plus, and the expression is the same as if it were 
 written — (+ x — y), or — v x — y. 
 
 Examples 
 Simplify the following : 
 
 1. tt + (6 — c). 8. a — 6 — (c — d). 
 
 2. a — ip—c). 9. a — & — (— c + a). 
 
 3. X — (y — z). 10. a — m — (n — m). 
 
 4. x-{-y-\-z). 11. 5a- 2&- (a -26). 
 
 5. m — n — (—a). 12. a — (6 — c + a) — (c — 6). 
 
 6. m — (n — 2a). 13. 2xy -\- 3y^ — (x^ + xy — y^. 
 
 7. 5x — {2x + y). 14. m + (3 m — n) — (2 ri — m) + n. 
 
38 SUBTRACTION 
 
 Collect in alphabetical order the coefficients of x and y in the 
 following, giving each parenthesis the sign of the first coefficient 
 to be inclosed therein : 
 
 15. ax — by — 3bx-\-2cy—fx — gy. 
 
 PROCESS 
 
 cue — by — 3 bx + 2 cy — fx — gy =(a — 3 b—f)x—(b ~2c-\- g)y 
 
 Suggestion. — The coefficient of y is — b + 2c — g, which is written 
 -(6 -2c + gr) (Prin. 2). 
 
 16. ax — by — bx — cy -{- dx — ey. 21 . bx — cy — 2ay ■+- by. 
 
 17. mx — 2 ny + nx — ry — px + qy. 22. mx — bx — 4y — my. 
 
 18. 5 ax + 3 ay -- 2 dx -{- ny — 5 X — y. 23. rx — ay — sx-\-2 cy. 
 
 19. cx~2bx + 7 ay -\-3ax — Ix — ty. 24. x^ -\- ax — y^ + ay. 
 
 20. bx+ cy — 2 ax + by — ex — dy. 25. a? — ay — ax — y^. 
 
 Group the same powers of x in the following : 
 
 26. an? + ba? — cx-\- ea? — dv? — fx. 
 
 27. x'-^3:)?^3x-ax'-3a'3?^-bx. 
 
 28. a^ — abx — x? — bx^ — ex — mnx^ + dx. 
 
 29. ax'^ — X* — aoi? -\- x^ -\- ax — x — abx? + y?. 
 
 30. Simplify 2 a-\a-\b - i^b -2 a - b)\'\-(b - a). 
 
 When an expression contains parentheses within parentheses, 
 they may be removed in succession, beginning with either the 
 outermost or the innermost, preferably the innermost. 
 
 Solution 
 
 2a-[a-{6-(3 6-2a- 6)}] -(b-a) 
 
 Prin. 1, =2a~la-{b-(3b-2a + b)}]-b + a 
 
 Uniting terms, =3d— [a — {6— (4 6 — 2 a)}] — b 
 
 Prin. 1, = 3 a - [a - {6 - 4 6 + 2 a}] - 6 
 
 Uniting terms, =3a— [a — {— 3& + 2a}]— 6 
 
 Prm. 1, =3a-[a + 36-2a]-6 
 
 Uniting terms, =3a— [— a + 3 6]— 6 
 
 Prin. 1, =Sa + a-Sb — b 
 
 Uniting terms, = 4 a — 4 6. 
 
SUBTRACTION 39 
 
 Simplify the following : 
 
 31. ^a + h — \x + 4.a + h —2y — (x + y)\. 
 
 32. ah — \ab -\- ac — a — (2 a — ac) -{■ {2 a — 2 ac)\. 
 
 33. a-^[t/_|5+4a-(62/ + 3)^-(72/-4a-l)]. 
 
 34. 4m— [2) + 3?i — (m + n) +3 — (6p — 3n — 5m)]. 
 
 35. a + 26 + (14a-56)-^6a + 6&-(5a-4a-46)|. 
 
 36. 12a-{[4-36-(6& + 3c)]+6-8-(5a-2&-6)(. 
 
 37. a+6— {— [a+& — (c+i«)] — [3a— (c— a;+a)-&]+4a|. 
 
 38. x^-lx'-il-x)]-]! +[_:>^ - (1 - x) + ar'J |. 
 
 39. 4 - [ [5 y - (3 - 2 a; - 2)] - [a: + (5 y - a; + 3)] I . 
 
 40. ah -\b + X - {h + c - ah + x)\+[_x - {h - c - 7)]. 
 
 41. a- - 6^ _ ^ad + a^ - (a; + a- - h^) -W\ + 5 ad - (x + 3 ad). 
 
 42. a-(Z>-c)-[a-{6— c-(6+c-a) + (a-&) + (c-a)|]. 
 
 43. — f3ax-[5a;r/-32;] + 2 — (4a;// +[6z + 7aa;]-f 3 2;)*. 
 
 44. 1 - X -\1 — X -\\ — X — {1 — x) — {x — 1)']— X -{-\\. 
 
 45. 1 -a; — {1 -[a;-l + (a;-l) — (1 — ») — a;]+l-a;|. 
 
 46. a;-[— {- (-a;) + a;| — 2a;]. 
 
 47. (a — 6) - { - a - (6 — a) + (a — &)}. 
 
 48. a — 7— [— { — a — (—a — a — 3)1]. 
 
 49. a — « — [— {a + (a; — a) — (a; — 4a)|]. 
 
 50. bxy-l-\{y''-xy)~(xy-y''-2xy)\'\. 
 
 51. 2a-[a- {&-(36~2a-6)n-(6-a)- 
 
 52. a — [— (m — a) — Ja — (m — 2m + 6a)|]. 
 
 53. a — \-h- (c-d)^ + a-[-& + ^-2c- (d-e)f]. 
 
 54. a2 + 5-[2a&-[-(7-3a&) -ab + 2d' - z\- {Za- z)^ 
 
 55. 2 a; + (3 ?/ — 52 X - [?/ + 4 a; - (3 y - a;)] — 2 2/1 - a; — ?/). 
 
 56. l-(-{-[-(-a-a-l)-3]-2|-a)-[a-(a-l)]. 
 
40 SUBTRACTION 
 
 TRANSPOSITION IN EQUATIONS 
 
 69. 1. What number diminished by 2 is equal to 8 ? 
 
 2. If a number increased by 2 is equal to 8, what is the 
 number ? 
 
 3. In the equation a; — 2 = 8, what is done with the 2 in 
 obtaining the value of x ? In the equation ic = 8 + 2, how does 
 the sign of 2 compare with its sign in the previous equation ? 
 
 4. In the equation a; -}- 2 = 8, what is done with the 2 in 
 obtaining the value of x ? In the equation cc = 8 — 2, how does 
 the sign of 2 compare with its sign in the previous equation ? 
 
 5. In changing the 2's from one side of the equation to the 
 other, what change was made in the sign ? 
 
 6. When a term is changed from one side, or member of an 
 equation, to the other, what change must be made in its sign ? 
 
 7. If 3 is added to one member of the equation 2 + 5 = 7, 
 what must be done to preserve the equality? 
 
 8. If 3 is subtracted from one member of the equation 
 2 + 8 = 10, what must be done to preserve the equality ? 
 
 9. If one member of the equation 2 + 5 = 7 is multiplied by 
 5, what must be done to preserve the equality ? 
 
 10. If one member of the equation 10 + 25 = 35 is divided by 
 5, what must be done to preserve the equality ? 
 
 11. If one member of the equation a; = 5 is raised to the second 
 power, what must be done to preserve the equality ? 
 
 12. If the square root of one member of the equation x' = 25 
 is taken, what must be done to preserve the equality? 
 
 13. What, then, may be done to the members of an equation 
 without destroying the equality ? 
 
 70. The parts of an equation on each side of the sign of 
 equality are called its Members. 
 
 The part on the left of the sign of equality is called the First 
 Member, and the part on the right, the Second Member. 
 
SUBTRACTION 41 
 
 71. The process of changing a term from one member of an 
 equation to the other is called. Transposition. 
 
 72. Principle. — A term may he transposed from one member 
 of an equation to the other, provided its sign is changed. 
 
 73. A truth that does not need demonstration is called an 
 Axiom. 
 
 74. Axioms. — 1. Things that are equal to the same thing are 
 equal to each other. 
 
 2. If equals are added to equals, the sums are equal. 
 
 3. If equals are subtracted from equals, the remainders are equal. 
 
 4. If equals are multiplied by equals, the products are equal. 
 
 5. If equals are divided by equals, the quotients are equal. 
 
 6. The same powers of equal numbers are equal. 
 
 7. Tlie same roots of equal numbers are equal. 
 
 Equations and Problems 
 
 75. 1 . li bx — 2 = 'dx + &, find the value of x. 
 
 PROCESS Explanation. — Since the value of x is sought, 
 
 5 ^. _ 2 = 3 aj -J- 6 ^'^ terms containing x must be collected in one 
 
 Sx = 3 X 
 
 member of the equation, and the remaining, or 
 known terms in the other member. 
 ^ ^ ^ ^ '" 3x may be made to disappear from the second 
 
 2 = 2 member by subtracting 3x from both members. 
 
 2 X =8 The result (Axiom .3) is the equation 2 x — 2 — Q. 
 
 . /J. __ 4. — 2 may be made to disappear from the first 
 
 member by adding 2 to both members. The result 
 (Axiom 2) is the equation 2 a; = 8. 
 o X — A =oa; + D Dividing both members of this equation by 2, 
 
 5a; — 3a: = 24-6 the coefiicient of x, the resulting equation (Axiom 
 2 x = 8 5) is X = 4, the value of x sought. 
 
 _ . x =: 4 ^^' since a term may be transposed from one 
 
 member to the other if its sign is changed (Piin.), 
 VERIFICATION 3 x transposed to the first member becomes — 3 x 
 
 20 2 ^ 12 + 6 ^°^ ~ ^ transposed to the second member becomes 
 
 + 2. Therefore, the resulting equation is 
 18 ^ lo 
 
 5x-3x = 2 + 6. 
 
 Uniting terms, 2 x = 8. Dividing both members by 2, the result is x = 4. 
 
42 SUBTRACTION 
 
 The result is verified by substituting the value of x for x in the original 
 equation. If the members are then identical, tlie value found for the un- 
 known number is correct. 
 
 2. If 5 a; - 7 = 30 - 7, find the value of x. 
 
 PROCESS 
 
 5 a; - 7 = 30 - 7. 
 5x = 30. 
 .-, ic = 6. 
 
 Suggestion. — Since, if — 7 were transposed from the first member to the 
 second, it would appear as + 7 and cancel the term — 7 in that member, the 
 two equal terms may be canceled before the transposition. 
 
 RuLK. — Transpose terms so that the unknoivn terms stand in the 
 first member of the equation and the knovm terms in the second. 
 
 Unite similar terms, and divide both members of the equation by 
 the coefficient of the unknown number. 
 
 Verificatiox. — Substitute in the original equation the value of 
 the unknoion number thus found. If the members of the equation 
 are then identical, the value of the unknown number found is correct. 
 
 1. The same term with the same sign in both members of an equation 
 may be canceled (Ax. 2 or 3). 
 
 2. If the signs of all the terms of an equation are changed, the equality 
 will not be destroyed; for (Ax. 3) both members may be subtracted from 
 without destroying the equality. 
 
 Find the value of x and verify : 
 
 3. 5 07 + 3 = 8. 10. 7 0^-10 = 60. 
 
 4. iK + 5 = ll. 11. 7 + 2 a; = 11. 
 
 5. a;- 5 = 11. 12. 3 + 2 a; = 15. 
 
 6. 2 37-3 = 21. 13. 1 + 12 a; = 85. 
 
 7. oa; + 7 = 42. 14. 5 + 3.^ = 11. 
 
 8. 3a;- 2 = 25. 15. 7 + 5.x' = 47. 
 
 9. 2a; + 4 = 10. 16. 2 -|- 9 x = 74, 
 
SUBTRACTION 43 
 
 17. 3 = 5-a;. 26. 7 a; -12 = a; + 13 -|- 5. 
 
 18. 9-5x = -l. 27. 4a;-20 = 5iC-50 + a;. 
 
 19. 7a; + 2 = a; + 14. 28. 3 a; + 16 = 20 -5 a; + 4. 
 
 20. oa; — 5 = 2x + 4. 29. 7 a; - 55 = 18 — 2a; — 1. 
 
 21. 3a; + 2 = a; + 30. 30. -a; -12 = 40 -8 a; + 4. 
 
 22. 5a;- 2 = 2a; + 7. 31. 80 -3 a; = 83 -8 a; + 7. 
 
 23. 2 + 13 a; = 50 -9. 32. 9 a; -90 = 16 -a; + 4. 
 
 24. 10 + a; = 18 -a;. 33. 50 -a; =20 + a;. 
 
 25. 2a; + 2 = 32 -a;. 34. 7 a; + 25 = 30 + 6 a; -3. 
 
 Solve the following problems : 
 
 35. What number increased by 10 is equal to 19 ? 
 
 36. What number diminished by 30 is equal to 20 ? 
 
 37. What number diminished by 111 is equal to — 15? 
 
 38. What number exceeds \ of itself by 10? 
 Suggestion. — Let 3 as = the number. 
 
 39. Five times a number exceeds 3 times the number by 14. 
 What is the number ? 
 
 40. If 5 is subtracted from a certain number, and the differ- 
 ence is subtracted from 3 times the number, the result is 35. 
 What is the number ? 
 
 41. The double of a number is 64 less than 10 times the num- 
 ber. What is the number ? 
 
 42. If 4 is subtracted from a certain number, and the differ- 
 ence is subtracted from 40, the result is 3 times the number. 
 What is the number ? 
 
 43. Three times a certain number is as much less than 72 as 
 4 times the number exceeds 12. What is the number ? 
 
 44. Twice a certain number exceeds \ of the number as much 
 as 6 times the number exceeds 65. What is the number ? 
 
 45. If 16 is added to a certain number, the result is 5Q dimin- 
 ished by 7 times the number. What is the number ? 
 
44 SUBTRACTION 
 
 46. If 6 times a certain number lacks as much of 62 as 3 
 times the number exceeds 19, what is the number ? 
 
 47. Three times a certain number increased by a is equal to 
 the number increased by 9 a. What is the number ? 
 
 48. The sum of 4 numbers in a row is 58. If their common 
 difference is 3, what are the numbers ? 
 
 Suggestion. — Let x = the smallest number. 
 
 Then, at + 3 = the second number, 
 
 X + 6 = the third number, 
 and X + 9 = the fourth number. 
 
 49. A man distributed 1 dollar among 5 boys so that each boy 
 except the youngest received 5 cents more than the boy next 
 younger. If the boys were all of different ages, how much did 
 each receive ? 
 
 50. The common difference of 5 numbers is 2, and their sum 
 is 100. What are the numbers ? 
 
 51. John and James were comparing their earnings. John 
 said, " I have earned 50 cents." James replied, " If I had earned 
 half as much as I have, and 10 cents more, I should have earned 
 the same as you." How much had James earned? 
 
 52. A drover, when asked how many cattle he had, replied, " If 
 I had \ more than I have and 2 more, I should have 200." How 
 many cattle had he ? 
 
 53. The earnings of a mill for 4 years were $46,000. If the 
 books showed an annual increase of $ 1000, what were the earn- 
 ings for each year ? 
 
 54. James had | as much money as John, John 5 cents less 
 than William, and Robert 5 cents more than 3 times as much as 
 James. If they together had $ 1.50, how much had each ? 
 
 Suggestion. — Let 3 x = the number of cents John had. 
 
 55. A speculator who doubled his money by a fortunate invest- 
 ment, afterward lost f 600, but he still had $ 400 more than the 
 original sum. How much had he at first ? 
 
MULTIPLICATION 
 
 76. 1. How many are 3x + 3x + 3a; + 3a; + 3a;? 
 
 2. How many are 5 times 3x? 2 times 3a;? 7 times 2a? 
 
 3. A man saves f 10 a month, indicated by +10. How many 
 dollars will he save in a year ? What sign should be placed be- 
 fore the result to indicate the number of dollars saved ? 
 
 4. How many are 12 times +10? 12 times +10 a? 5 times 
 +3x? 8 times +2 m ? 
 
 5. When a positive number is multiplied by a positive num 
 ber, what is the sign of the product ? 
 
 .6. If a man loses $5 a month, indicated by "5, how many 
 dollars will he lose in a year ? What sign should be placed be- 
 fore the result to indicate the number of dollars lost ? 
 
 7. How many are 12 times "5? 12 times "5 6? 7 a times 
 -3x? 3 times '2 b ? 11 times Sy? 5 times "5 ab ? 
 
 8. When a negative number is multiplied by a positive num- 
 ber, what is the sign of the product ? 
 
 9. If a man's gains in business are $10 a month, indicated 
 by +10, how many dollars less had he 3 months ago, indicated by 
 ~3, than he has now ? Indicate the result algebraically. 
 
 10. How many are +10 multiplied by ~3 ? +10 multiplied by 
 -f) ? +2 multiplied by 3 ? +a multiplied by -& ? 
 
 11. What is the sign of the product when a positive number 
 is multiplied by a negative number ? When a negative number 
 is multiplied by a positive number ? 
 
 12. What, then, is the sign of the product of two numbers 
 having imlike signs ? 
 
 45 
 
46 MUL TIPL ICA TION 
 
 13. If a man who is in debt is getting deeper in debt at the 
 rate of $10 a month, indicated by ~10, how much better off was 
 he 3 months ago, indicated by ~3, than he is now ? Indicate the 
 result algebraically. 
 
 14. How many are ~10 multiplied by ~3? ~10 multiplied by 
 -5 ? -2 multiplied by "3 ? "a multiplied by 'h ? 
 
 15. What is the sign of the product of two negative numbers ? 
 of two positive numbers ? 
 
 16. What, then, is the sign of the product of two numbers 
 having like signs ? 
 
 17. What is the sign of "5 x "2 ? of "5 x "2 x "2 ? of "5 x "2 
 X+2X-2? of +5x-2x-2x 2x-2? of "2 x-3 x+2 x+2 x-2 
 X-2X-2? 
 
 18. What sign has a product, if the number of negative fac- 
 tors is evenf What sign has a product, if the number of negative 
 factors is odd? 
 
 19. What, then, determines the sign of a product? 
 
 20. In the expression a?, what is 3 called ? What does it indi- 
 cate ? In a^ how many times is a used as a factor ? 
 
 21. When a? is multiplied by a*, how many times is a used as 
 a factor in the product ? when a^ is multiplied by a^ ? 
 
 When a" is multiplied by a", what is the product, if m and n 
 are positive integers ? 
 
 22. How, then, is the exponent of a factor in the product de- 
 termined ? 
 
 77. When the multiplier is a positive integer, the process of 
 taking the multiplicand additively as many times as there are 
 units in the multiplier is called Multiplication. 
 
 When the multiplier is any number, multiplication may be 
 defined as the process of finding a number that has the same rela- 
 tion to the multiplicand as the multiplier has to 1. 
 
 The multiplicand and multiplier are called the factors of the 
 product. 
 
 78. Principles. — 1. Law of Signs. — The sign of the prod^ict 
 of tico factors is + when they have like signs, and — when they 
 have unlike signs. 
 
MUL TI PLICA TION 47 
 
 2. Law of Coefficients. — The coefficient of the product is equal to 
 the 2>roduct of the coefficients of the factors. 
 
 3. Law of Exponents. — The exponent of a number in the prod- 
 uct is equal to the sum of its exponents in the factors. 
 
 79. The Law of Signs may be established as follows : 
 
 +3=+l++l++l, (1) 
 
 and -3=-l4.-l +-1 =_+l _+l _+l; (2) 
 
 that is, +3 is obtained from +1 by taking +1 additively three times, and 3 
 by taking +1 subtractively three times. 
 
 Hence, § 77, multiplying any number by +3 is equivalent to taking that 
 number additively three times, and multiplying any number by -3 is equiva- 
 lent to taking that number subtractively three times. 
 
 By (I), +5 X +3 =+5 ++5 ++5 =+15, (3) 
 
 and 5 x+3=-5+-5+-5=-15. (4) 
 
 By (2), +5 x-3 = -+5-+5-+5=-15, (5) 
 
 and -5 X -3 =--5 --5 --5 =+15. (6) 
 
 Similarly, since +(|) = + (i) + + a)++Q) and -(|)=-+(|)-+(|)- + Q), 
 +5x+(|) = + (f) ++(f) + + (|)= + (^); (7) 
 
 and so on, as in (4), (5), and (6). 
 
 In (3) and (6) the product of two algebraic numbers with like signs is 
 positive. 
 
 In (4) and (5) the product of two algebraic numbers with unlike signs is 
 negative. 
 
 (7) shows that like results are obtained when the multiplier is a fractional 
 number (§ 7). 
 
 Passing to general symbols, let a and h be any absolute numbers. 
 First, when h is a whole number. 
 
 Since 
 
 +6=+l ++1 4-+1 + ••• 
 
 to b terms,. 
 
 
 §77, 
 
 +a x+b =+a ++a ++a + •• 
 
 • to b terms 
 
 
 
 =+ab, 
 
 
 (8) 
 
 and 
 
 -a x+b =-a +-a +-a + ••■ 
 
 • to b terms 
 
 
 
 = a6. 
 
 
 (9) 
 
 Since 
 
 -b = -+l -+1 -+1 - 
 
 • •• to b terms, 
 
 
 §77, 
 
 +a x-b = -+rt — +a — +a — 
 
 '•• to b terms 
 
 
 
 =-ab, 
 
 
 (10) 
 
 and 
 
 -a x-b = —-a —~a —~a — 
 
 • •• U)b terms 
 
 
 
 = +ab. 
 
 
 (11) 
 
48 MUL TIPLICA TION 
 
 Second, when b is a fractional number. 
 
 As in (7), the same reasoning applies when 6 is a fractional number. 
 
 Hence, from (8) and (11), the product of any two algebraic numbers with 
 like signs is positive ; and from (9) and (10), the product of any two algebraic 
 numbers with unlike signs is negative. 
 
 When the multiplier is a positive or negative whole or fractional 
 number, it appears from the above proofs that algebraic multi- 
 plication is only abbreviated algebraic addition. Hence, as in 
 addition, but one set of signs + and — is required to denote both 
 quality and operation. 
 
 Hence, the Law of Signs may be expressed as follows : 
 
 + a multiplied by + 6 = + ab, 
 
 — a multiplied by — b = + ab, 
 
 — a multiplied hy -\- b = — ab, 
 and + a multiplied by — b — — ab. 
 
 80. It follows from the Law of Signs, applied repeatedly, that 
 the ^noduct of any number of algebraic numbers is + when the 
 number of negative factors is even, and — when the number of 
 negative factors is odd. 
 
 81. The Law of Exponents or the Index Law for multiplication 
 may be established for positive integral exponents as follows : 
 
 Let m and n be any positive integers. 
 By the definition of a power, § 24, 
 
 a'^ — a X a X a ... to m factors, 
 a^ = a X a X a ... to n factors ; 
 .*. a™ X a" = (a X a X rt ... to Tw factors) (a x a x a ... to n factors) 
 = « X a X a ... to (m + n) factors. 
 Hence, a™ xa"= «"»+". 
 
 In like manner, a'^ x a*^ x a^ = a'"+"+p. 
 
 Thus, a^x a* = a^+* = a^ 
 
 and a3 X a2 X a* = a^+^+* = a^. 
 
 82. 1. How does 2x5 compare with 5 x 2 in value? 3x7 
 with 7x3? 2x5x6 with 2x6x5? 
 
 2. What is the effect upon the value of a product of changing 
 the order of its factors ? 
 
MUL TIPLICA TION 49 
 
 Law of Order, or Commutative Law for Multiplication. — The fac- 
 tors of a product may be taken in any order. 
 
 The Law of Order may be established as follows : 
 
 Since the number of negative factors will not be changed by taking the 
 factors in any order, § 80, the sign of the product is the same in whatever 
 order the factors are taken. 
 
 We know from arithmetic that arithmetical numbers may be multiplied 
 in any order. Hence, the absolute value of the product is the same in what- 
 ever order the factors are taken. 
 
 Since neither the sign nor the absolute value of the product of algebraic 
 numbers is changed by changing the order of the factors, the factors may be 
 taken in any order. 
 
 In general symbols, a x b x c x ••' = b x c x a x ••• = etc. 
 
 83. 1. How does 2 x 3 x 5, or 6 x 5, compare in value with 
 2 X (3 X 5), or 2 x 15 ? with 5x6? a x 6 X c, or (ab) x c, 
 with a X (be) ? 
 
 2. How may the factors of a product be grouped ? 
 
 Law of Grouping, or Associative Law for Multiplication. — The fac- 
 tors of a jyroduct may be grouped in any manner. 
 
 The Law of Grouping may be established as follows : 
 
 By the notation of multiplication, abc denotes that a is to be multiplied 
 by b and then the product ab is to be multiplied by c ; that is, 
 
 abc—(ab)c. (1) 
 
 1. Let it be required to prove that (ab)c = a(bc). 
 By the Law of Order, abc = bca 
 
 by notation, = (bc)a 
 
 by the Law of Order, = a{bc). (2) 
 
 From (1) and (2), (ab)c = a(bc). 
 
 Similarly, it may be proved that (ab)c — b(ac), etc. 
 
 2. Let it be required to prove that abed = (aft) (cd). 
 By notation, abed = ab x c x d 
 
 putting m for ab, = m ■ c • d, or mcd. 
 
 By 1, m • c • d = m(cd). 
 
 Putting ab for m, (ab) • c- d = (ab) (cd) ; 
 
 that is, by notation, abed =:(ab)(cd). 
 
 Similarly, it can be shown that (abc)d = a(bed) = (he) (ad) = (ac) (bd) = 
 (abd)c = (adc)b = c(dba)= etc., the factors being grouped in any manner 
 whatever, 
 
 ALG. — 4 
 
50 M UL T I PLICA TION 
 
 3, In a similar way tlie law may be established for any number of factors, 
 successively for 5, 6, 7, ••• factors. 
 
 Hence, ahc •■■p = a{bc •••p)= b{ac ■••p), etc., for all values of the letters. 
 84. To multiply a monomial by a monomial. 
 
 Examples 
 1. Multiply 5 a^y^ by — 3 xyh. 
 „„^^„„„ Explanation. — Since the multiplier is composed of the 
 
 J, , „ factors — 3, x, y'^, and z, the multiplicand may be multiplied 
 J^ by each successively. — 3 times 5 x-y^ = — 15 x^ij'^ (Prin. 1 
 ~ ^^ and 2); x times - 15 xhf^ = - 15 x^y^ (Prin. 3); y^ times 
 — 15 a^y^z ~ IS x^y^ = - 15 xV (Prin. 3) ; and this multiplied by z is 
 equal to — I5x^y^z (Prin. 3). 
 Or, since the signs of the numbers are unlike, the sign of their product 
 is — ; the coeflBcient of the product is the product of the coefficients 5 and 3 ; 
 and the product of the literal numbers is expressed by affecting each with 
 an exponent equal to the sum of its exponents in the factors. 
 
 Rule. — To the product of the numerical coefficients annex the 
 letters, each with an exponent equal to the s%im of its exponents in 
 both factors. 
 
 Write the sign + before the product when its factors have like 
 signs, and — when they have unlike signs. 
 
 2. 3. 4. 5. 6. 
 
 Multiply -2 6 -7 2 a 2m^ 
 
 By 8 -2 -9 5 6m^ 
 
 
 7. 
 
 8. 
 
 9. 
 
 10. 
 
 11. 
 
 Multiply 
 
 10 a" 
 
 x-y'^ — 
 
 4 abc 
 
 5 a^b(^ 
 
 - 2 xy^ 
 
 By 
 
 5 a' 
 
 xif 
 
 2a^b 
 
 ■ 7 ab'c 
 
 2x'y 
 
 
 12. 
 
 13. 
 
 14. 
 
 15. 
 
 16. 
 
 Multiply 
 
 -Sa'x" 
 
 — 5 im?n^ 
 
 - 6 a^fiVa; 
 
 4 abed 
 
 -3x%f 
 
 By 
 
 -2 ax' 
 17. 
 
 3 mn 
 
 — 4 a'bny^ ■ 
 19. 
 
 -1 
 
 -1 
 
 
 18. 
 
 20. 
 
 21. 
 
 Multiply 
 
 - 2 c^-^ 
 
 - 3 n'y 
 
 4 a^xb'y* 
 
 -1 - 
 
 5 mhi^dhf 
 
 By 
 
 — 4 ax* 
 
 6b'y 
 23. 
 
 3 a^x^b^y 
 24. 
 
 -1 - 
 25. 
 
 2m'«u«c-y 
 
 
 22. 
 
 26. 
 
 Multiply 
 
 5 p(f\)? 
 
 10 mV 
 
 — 2 a^m'n* 
 
 x-yz 
 
 -phX-(i 
 
 By 
 
 -2rq*x 
 
 - Sn^'m* 
 
 8 b'n^ni' 
 
 — x-yz^ 
 
 — ahc 
 
MULTIPLICATION 
 
 61 
 
 
 27. 
 
 28. 
 
 29. 
 
 30. 
 
 31. 
 
 Multiply 
 
 By 
 
 2a'»+' 
 32. 
 
 — 5 a;" 
 
 X 
 
 - x"y" 
 3xy - 
 
 3,n-y-2 
 
 xy 
 
 4 a;"-' 
 
 -2a;"+* 
 
 
 33. 
 
 34. 
 
 35. 
 
 36. 
 
 Multiply 
 By 
 
 by 
 -3.v"-2 
 
 — a" 
 -a" 
 
 a" 
 38. 
 
 -a;'- 
 — a;" 
 
 ym—n+l 
 
 
 37. 
 
 
 39. 
 
 Multiply 
 
 a-b^x^y""^ 
 
 
 a"-i6"-V 
 
 
 m''n'b^y'' 
 
 By 
 
 QnJn-3^2 
 
 
 ^n+1^2g„-l 
 
 
 m}'n%''i/-' 
 
 85. How does 25 x 2 compare in value with 20 x 2 plus 
 5x2? How is 133, or 100 + 30+3, multiplied by 2 ? How is 
 the polynomial a -{-b -'t c multiplied by the monomial m? 
 
 Distributive Law for Multiplication. — The product of a polynomial 
 by a monomial is equal to the algebraic sum of the partial prodxicts 
 formed by multiplying each term of the polynomial by the monomial. 
 
 The Distributive Law may be established as follows : 
 
 Let a + b he the multiplicand and m the multiplier, a, b, and m being 
 positive or negative integral or fractional numbers. 
 
 By the Law of Order the multiplier may change places with the multi- 
 plicand. Hence, (a + 6) x m may be written m(a + 6). 
 
 It is to be proved that m(a + b) = ma + mb. 
 
 First, when m is a positive integer. 
 
 Since m = 1 + 1 + 1 4- ••• to m terms, 
 
 § 77, m(a + b) = (a + b) + (a + b) + (a + b)+ -• to m terms 
 § 56, z=(a + a + a + ••■to m terms) + (6 + 6 + 6 + ••• to m terms) 
 
 §77, 
 
 = ma + mb. 
 
 (1) 
 
 Second, when m is a fractional number. 
 
 P 
 Let m = — , in which p and q are absolute integers. 
 
 |(« + 6) 
 
 by (1), 
 
 p times one qth of (a + 6) 
 = p(a + 6) ^ths 
 = pa qt\\s, +-pb qtha 
 
 = |ofa + |of6 
 = ^a + ^b. 
 
 q q 
 
 (2) 
 
62 MUL TIPLICA TION 
 
 Third, when m is negative. 
 
 Let m = — n, n being any positive whole or fractional number. 
 
 It is to be proved that (^— n)(a + b) = — na — nb. 
 
 By (1) and (2), n(a + b)=na + nb. (3) 
 
 Since, if + n is positive, — (— n) is also positive, substituting — (— n) 
 
 for + n in (3), 
 
 — (— n)(a + 6) = — (— 7i)a — (— n)b 
 
 = — (— va — nb). (4) 
 
 Since both —(— n)(a + b) and —(—na — nb) are now monomial in 
 
 form, both members of (4) may be multiplied by the monomial — 1. 
 
 .'. + ( — n) (a + ft) = + ( — na — nb), 
 
 or ( — n) (a + 6) = — na — nb. (5) 
 
 By (1), (2), and (5), m(a + b)= ma + mb 
 
 for all positive or negative whole or fractional values of m and for all values 
 
 of a and b. 
 
 86. To multiply a polynomial by a monomial. 
 
 Examples 
 1. Multiply Sa^ — y^ by — 4 ?/. 
 
 PROCESS Explanation. — By § 85, each term of the multiplicand 
 
 o , 2 is to be multiplied by the multiplier. 
 
 The product of .3 x'^ and — 4 y is — 12 x^y. But since 
 y the entire multiplicand is 3 x^ — y^, — 4 ?/ times ?/2 must be 
 
 ;[2 x^y -X- 4:ifi subtracted from — 12 x^y. — iy times y^ = — i y^, which 
 
 subtracted from — 12 x'^y gives — 12 x'^y + 4 y^. 
 Or, since a polynomial multiplied by a monomial is equal to the algebraic 
 sum of the partial products formed by multiplying each term of the poly- 
 nomial by the monomial, § 85, 3 a;^ _ j^2 multiplied by — 4 y is equal to 
 - 12 x'^y + 4 2/3. 
 
 Rule. — Multiply each term of the polynomial by the monomial, 
 and find the algebraic sum of the partial products. 
 
 2. 
 
 Multiply 2a2-2a6 + 362 
 By Sab 
 
 3. 4. 
 
 5 m^ — 4 71^ ba? — 2xy — y^ 
 — 2 m^n — a^y 
 
 Multiply : 
 
 5. 3x2- 2 xy ^^ 53^2^ 
 
 6. mhi^ — 3 mn* by 2 mn. 
 
 8. p^q^ — 2p(f by — pq. 
 
 9. 4a2-5&2c by abc^ 
 
 7. 3 a^- 6 a^^ by -2 6. 
 
 10. — 2 ac + 4 aa; by — 5 oca?. 
 
MUL TI PLICA TION 63 
 
 Perform the multiplications indicated : 
 
 1 1 . a-hc (3 a* - 4 a% - 5 a?b' + 2 aft^ - 16 6*). 
 
 12. 2xy{o7?-lQxy-^Qy^-5x + 5y-{-12Qi). 
 
 13. 5 m? (16 m? - 20 m^n + 13 mn^ - 25 «»). 
 
 14. abc {a^h'' - 2 a'c' - 2 6V - a* - 4 5* - c* - 5 ahc). 
 
 15. - 6c(6^ 4- c* - &^ - c^ + 6V _ 4 6^0 + 8 &c2 - 2 he). 
 
 16. — 2 X (ic* — 5 a^y — 16 0^2/^ + 24 .172/^ — 2/* — a:?/ — a; + 4). 
 
 87. To multiply a polynomial by a polynomial. 
 
 To multiply p + q + r by a + 6, 
 
 § 85, {p + q-{-r){a + h)=p{a + h) + q{a + h)+r{a + h) 
 
 §§82,85, =ap + hp + aq + hq-\-ar + hr 
 
 § 55, = ap + aq -\- ar + hp + hq + br. 
 
 Rule. — Multiply every term of the multiplicand by each term of 
 the multiplier, and find the algebraic sum of the partial products. 
 
 Examples 
 1. Multiply x^ — xy by 2 a; + 3 ?/. 
 
 PROCESS 
 
 a? — oey 
 2x +3?/ 
 
 + 2 a; times {a^ — xy) = 2 a^ — 2 a^y 
 
 -\-Sy times (ar^ — xy) = + 3 a^y — 3 xy^ 
 
 .-. (2 X + 3 y) times (a^ — xy) = 2a^4- x^y — 3xy^ 
 
 2. Multiply x^ — 3x^y + 3xy^ — y^ by t? — xy. 
 
 PROCESS 
 
 3?-3x'y + 3xy'' -f 
 ai? — xy 
 
 af-3ai*y + 3a^f- x'f 
 — x*y -\- 3 x^y^ — 3 a^y' + xy* 
 
 0^—4 a;*2/ + 6 x^y^ — 4 a^?/^ + xy* 
 
64 
 
 MUL TIPLICA TION 
 
 3. Multiply Sx + af' — 2aj2 — X* by Sx + l — a^. 
 
 PROCESS 
 
 5x— 2a^ -{- Jt^ — X* 
 1 + 3a; -x" 
 5x- 2 
 15 
 
 C»2+ 1 
 
 3^-1 
 
 x* 
 
 
 - 6 
 
 + 3 
 
 
 -3 
 
 - 5 
 
 + 2 
 
 
 -1 
 
 x* 
 
 + «« 
 
 6 a; + 13 x^ - 10 x^ + 4 a;* - 4 af + x" 
 
 TEST 
 
 = + 3 
 
 = + 3 
 
 = + 9 
 
 Suggestion. — For convenience in writing partial products, both polyno- 
 mials are arranged so that in passing from left to right the several powers of 
 X are either successively higher or lower. In this process, the polynomials 
 are arranged according to the ascending powers of x. 
 
 Test. — Since the correct product of 5 x — 2 x^ + x' — a:* and 1 + 3 x — x^ 
 is the same in form whatever value x represents, it is possible, by assigning 
 an arithmetical value to x, to change the process of multiplying one algebraic 
 expression by the other into a process of multiplying one arithmetical num- 
 ber by another as shown in the test. 
 
 Let 1 be substituted for x. 
 
 Multiplicand = 5x-2x2 + x3-x< = 5-2-fl-l= + 3 
 
 Multiplier = 1 + 3 x - xJ^ = 1 -f - 3 - 1 =4-3 
 
 Product should be equal to 4-9 
 
 5 X 4- 13 x2 - 10 x3 + 4 X* - 4 x5 4- x6 = 5 4- 13 - 10 4- 4 - 4 4- 1 = 4- 9. 
 
 In like manner the multiplication of any two literal expressions may be 
 tested arithmetically by assigning any values we please to the letters. It is 
 usually most convenient to substitute 4- 1 for each letter, since this may be 
 readily done by adding the numerical coefficients. 
 
 Multiply, and test each result : 
 4. 2a;H-3 by x-f-2. 12. Ay — 6b hy 2y-\-b. 
 
 5. 4x-|-l by 3x-|-4. 
 
 6. 5 71 — 1 by 4 ?i -1- 5. 
 
 7. /i + 2fc by 3h-k. 
 
 8. 3r-6sby5r — 2s. 
 
 9. 4r-h2sby2r-f-9s. 
 
 10. 3l + 5t by 2l + 6t. 
 
 11. 4 a -I- 3 X by 4 a — 3 X. 
 
 13. Sx — 2y by 3x + 2y. 
 
 14. 26 + 5c by 5&-2c. 
 
 15. 7x — 2« by 4x-|-2w. 
 
 16. ab — 15 by a& -f 10. 
 
 17. ax -{-by by ax — by. 
 
 18. a^ — ay + y^ by a + y. 
 
 19. 3a^-Qab+3b^ by 2a-3 6. 
 
M UL TI PLICA TION 65 
 
 Multiply : 
 
 20. 2a?-ZW-ab by 3 a='-4ct6-5 &'. 
 
 21. 5a;-5.T2 + 10 by 12-^0x + 2x\ 
 
 22. 3n^ + 3m2 + mw by m^ — 2mn2 + m^n. 
 
 23. 4y2-10 + 2y by 22/^-32/ + 5. 
 
 24. 4 a; - 3 a;2 + 2 a^ by 3 a; - 10 a^ + 10. 
 
 25. a^ + a* + 4 a^ - a^ + a by a + 1- 
 
 26. 31a^-27a^ + 9x-3 by 3a; + l. 
 
 27. 4 0^ - 3 ic^^ + 5 a;^^ - 6 2/3 by 5 tc + 6 y. 
 
 28. a-\-h + c-\-dhja-h — c + d. 
 
 29. a2 + 62_^c2-a& — ac — 6c by a + & + c. 
 
 30. aa^" + a2/^" by a^^' — af''. 
 
 31. aa;"-i + 2/"~' by 3 aaf-^ + 2 .v"-^ 
 
 32. ar^" + 2 a;"^" + 2/'" by ar'" - 2 a;»t/" + 2/'"- 
 
 An indicated product is said to be expanded when the multipli- 
 cation is performed. 
 
 Expand : 
 
 33. (x + y){x + y). 39. (2a2 + 6)(2a2-6). 
 
 34. (a + &)(« + &). 40- ('K" + ?/")(«'" -2/"> 
 
 35. (c3+d3)(c-^ + d3). 41. (3aa: + 2 6y)(3ax + 2 62/). 
 
 36. (a;» + r) (^" + 2/")- *2. (3 aa; + 2 &?/) (3 aa; - 2 &^). 
 
 37. (3 a + 6) (3 a + 6). 43. (4m- 5n)(4m + 5w). 
 
 38. (3a + &)(3a-6). 44. (a + & + c)(a + 6-c). 
 
 45. (a + 6)(a-6)(a + 6)(a-6). 
 
 46. {a' + x^{a'-^)(a' + x')(a'-x^. 
 
 47. (a-&)(a-f6)(a2 + 6^(a* + 6*). 
 
 48. (a" - 6") (a" + 6") («='"' + &'")• 
 
 49. (2a + 36 + 5c)(2a + 36-5c). 
 
 50. (5 m - 2 rt + a;) (5 m - 2 w - a;). 
 
 51 . (a;» - wa;"--Vy + I nx"-y) (x + y). 
 
 52. (a;" + 3 ajp-^z - 6 xP- V) (ar' - ?/^. 
 
66 MULTIPLICATION 
 
 88. When polynomials are arranged according to the ascending 
 or the descending powers of some literal factor, processes may 
 frequently be abridged by using the Detached Coefficients. 
 
 53. Expand (2a;*-3a^ + 3a; + l)(3a; + 2). 
 
 FULL PROCESS DETACHED COEPPICIENTS 
 
 2 a;^-3 0^+3x4-1 2-3+0 +3 +1 
 
 3 a; +2 3+2 
 
 6a^-9.'B* +9i»2+3a; 6 -9 +0 +9 +3 
 
 4a;*-6a^ +6a;+2 4 -6 +0 +6 +2 
 
 6 af-5 x*-6 a^+9 x'+'d x+2 6 a^-5 a;*-6 a^+9 a^+9 x+2 
 
 Since the second power of x is wanting in the first factor, the term, if 
 it were supplied, would be Ox', and the detached coefficient of the term 
 would be 0. 
 
 The detached coefficients of missing terms should be supplied to prevent 
 confusion in placing the coefficients in the partial products and to prevent 
 errors in determining the result. 
 
 54. Multiply a* - 2 a^ft + 3 a^b^ -5ab^ + 5b' by a + 2 &. 
 
 PROCESS TEST 
 
 l_2+3-5+ 5 =2 
 
 1 + 2 =3 
 
 1-2+3-5+ 5 
 
 2-4 + 6-10 + 10 
 
 1-1-0-1+1-5 + 10 =6 
 
 = a' + Oa*b- a?b^ + a?W - 5 a6* + 10 6* 
 = a^ - a?l^ + a?¥ - 5 a6* + 10 b\ 
 
 Observe that the powers of a decrease uniformly from left to right, and 
 that the powers of b increase uniformly from left to right. 
 
 Observe also that the sura of the exponents is the same in every term. 
 
 Expand the following, using detached coefficients, and test the 
 results : 
 
 55. {ix^-\-4:^y -\-Qx^y^ -^-^LXif -\-y*){x->ry). 
 
 56. {5a?-x' -2a^ + x' + ^:i?-l){x + 2). 
 
MULTIPLICATION 67 
 
 57. (a' + 4a='+16a-32)(a3 + a2 + a-Hl). 
 
 58. {p" - 2pq + q") (p' + 2pq + q^. 
 
 59. (a;-l)(a;-2)(a;-3)(a; + 4)(.'K + 2). 
 
 60. (15 r*s -10rs* + 7^ + ^ + 3?V + Si^sP)(r -2s). 
 
 61. (x^^ - a^ + cc^ - x' -{- x^ - x^ -^ X* - a^ + a^ - X + l)(x + 1). 
 
 62. (x^o + 3 x» - 2 a* + 5 a;* + 2 ar^ + 2) (cB« + «* + ar' + 1). 
 
 63. (a^ + a:^ _,_ ^,2 ^ a; ^ j^>) (^^4 _ 3^ _l_ a^ _ g, ^ 1^ (^a; ^ 1^ (a; - 1). 
 
 64. (x^ — 4:2i^y + 5 afy^ + 3 a^y* — x'f + xf - /)(2 a; — S^/). 
 
 Expand : 
 
 65. (2a^ + 4.'B2 + 8a; + 16)(3a;-6). 
 
 66. (a^ + 4:0^ + 5x-24:)(a^- 4.x + 11). 
 
 67. (a^-4a:2^11a;-24)(ar' + 4a; + 5). 
 
 68. (a^ - a; 4- 1) (ar' + a; + 1) (a;* — a^ + 1). 
 
 69. (16a*-8a'' + 4a2-2a + l)(2a + l). 
 
 70. (a^ - 2 a^ + 3 a - 4) (4 a^ + 3 a^ + 2 a + 1). 
 
 71 . (m* + 2m^-\-m''-4m- 11) (m=^ - 2 m + 3). 
 
 72. (m* + 2 m^n + 2 mn- + w^) (^n,^ — 2 m'^w + 2 mn' — n*). 
 
 73. (a:3-La^ + ^a;-T'^)(a; + i). 
 
 74. (^x3 + ^^ + |^4.2)(a;-|). 
 
 76. fa^-^Vsa' + 2aV3a-3\ 
 
 77. (a;"-^ — a;"-2 + a;»-3 — a;"-*) (a; + 1). 
 
 78. (1 - a" + or" - a^") (1 - a"). 
 
 79. (a^" + 2 a"6" + ft^™^ (^^r, _ j„^_ 
 
 80. (ar^" — x^j/™ + 2/^") (a;" + 2/"). 
 
58 MULTIPLICATION 
 
 Equations and Problems 
 89. 1. Given 5 (2 a; -3) =7(3 a; + 5) -72, to find the value of x. 
 
 Solution 
 5(2x-3)=7(3x + 6)-72. 
 Expanding, 10 « - 15 = 21 x + 35 - 72. 
 
 Transposing, 10 x - 21 x = 15 + 35 - 72. 
 
 Uniting terms, — 11 x = — 22. 
 
 Multiplying by - 1, 11 x = 22. 
 
 .-. x = 2. 
 Verification. — Substituting 2 for x in the given equation, 
 5(4 -3)= 7(6 + 5) -72. 
 5 = 77 - 72 = 5. 
 
 Find the value of x, and verify the result, in the following : 
 
 2. 4 = 5-(x-2)-(x-3). 5. 10iK-2(a;-3) = -10. 
 
 3. 2 = 2a; -5 -(x -3). 6. 6a;-3(a;-6)=4(2a:-l) + 2. 
 
 4. 1 =5(2a;-4) + 5.T-f-6. 7. 7(5 - 3a;) = 3(3 - 4a;) - t 
 
 8. 4(2-4a;) = 4-2(a; + 5). 
 
 9. 5 + 3a;-4 = 13 + 4(a;-4). 
 
 10. 49-2(2a; + 3) = 9 + 2(2a;-3). 
 
 11. 3a;-2(4a;-5) = -2(6 + 2a;). 
 
 12. 3(a; + l)-2(2a; + 5) = 6(3-a;). 
 
 13. 2(a;-5) + 7 = a; + 30-9(a;-3). 
 
 14. 5 + 7(a;-5) = 15(a; + 2-36). 
 
 15. (a; -2) (a; -2) = (a; -3) (a; -3) + 7. 
 
 16. (a;-4)(a; + 4) = (a;-6)(a; + 5) + 25. 
 
 17. 4ar'-4(a.-3-a;2_^3,_2) = 4a^. 
 
 18. 7(2a;-3 6) = 26-3(2a; + &). 
 
 19. lla=3(.T-2a)-5(2a;-2a). 
 
 20. 3(2&-4a;)-(a;-6)=-6 6. 24. 3 (a;-a-2 6)=3 ft. 
 
 21. 4a;-x-2=a;(2— a;)+2a. 25. 5 ft=3 (2 a;— 6)— 4 6. 
 
 22. 2(a; + d)=10c. 26. 13 (a:-a)=5 (2 a;+a). 
 
 23. 5c=b{x+a-b). 27. 5 (4 a;-3 a)-6 (3 a;-2 a)=3 a. 
 
I 
 
 MULTIPLICATION 59 
 
 Solve tlie following problems and verify the solutions : 
 
 28. George and Henry together had 46 cents. If George had 
 
 4 cents more than half as many as Henry, how many cents had 
 
 each? 
 
 First Solution 
 
 Let X = the number of cents George had. 
 
 Then, « — 4 = the number of cents George had less 4, 
 
 and 2 (x — 4) = the number of cents Henry had ; 
 
 /. X + 2 (a: - 4) = 46. 
 
 Expanding, x + 2x — 8 = 46; 
 
 .-. X = 18, the number of cents George had, 
 
 and 2 (x — 4) = 28, the number of cents Heniy had. 
 
 Second Solution 
 Since George had 4 cents more than half as many as Henry, 
 let 2 X = the number of cents Henry had ; 
 
 then, X 4- 4 = the number of cents George had, 
 
 and 2 X + X + 4 = 46 ; 
 
 .-. X = 14, 
 
 2 X = 28, the number of cents Henry had, 
 and X + 4 = 18, the number of cents George had. 
 
 Verification 
 
 The answers obtained should be tested by the conditions of the problem. 
 If they satisfy the conditions of the problem, the solution is presumably 
 correct. 
 
 1st condition : — They had together 46 cents. 
 
 18 + 28 = 46. 
 2d condition : — George had 4 cents more than half as many as Henry. 
 18 = I of 28 + 4. 
 
 29. In a certain election at which 8000 votes were polled, B 
 received 500 votes more than half as many as A. How many 
 votes did each receive ? 
 
 30. A had $40 more than B; B had f 10 more than one third 
 as much as A, How much money had each ? 
 
60 MULTIPLICATION 
 
 31. Mary is 25 years younger than her mother. If she were 
 one year older, she would be \ as old as her mother. What is 
 the age of each ? 
 
 32. If John had 3 more marbles, he would have 3 times as 
 many as Clarence. Both have 41 marbles. How many has 
 each? 
 
 33. Two boys together had $8.20, and one had 50 cents less 
 than half as much as the other. What amount had each ? 
 
 34. If 5 is subtracted from twice a certain number, and the 
 difference is multiplied by 3, the result is 9 less than 5 times the 
 number. What is the number ? 
 
 35. A is f as old as B; 8 years ago he was ^ as old as B. 
 What is the age of each ? 
 
 Suggestion. — Let 5 x = the number of years in B's present age. 
 
 36. In 2 years A will be twice as old as he was 2 years ago. 
 How old is he ? 
 
 37. Two wheelmen start at the same time from A to ride to B. 
 One rides at the rate of 10 miles an hour, and rests 3 hours ; the 
 other rides at the rate of 8 miles an hour, and by resting only 
 1 hour arrives at B as soon as the faster rider. How far is it 
 from A to B, and how many hours are occupied in making the 
 trip ? 
 
 38. A man had two flocks of sheep with the same number «f 
 sheep in each. After selling 100 sheep from one flock, and 20 
 from the other, the numbers remaining were as 2 to 3. How 
 many sheep had he in each flock at first ? 
 
 39. Mary bought 17 apples for 61 cents. For a certain num- 
 ber of them she paid 5 cents each, and for the rest she paid 3 
 cents each. How many of each kind did she buy ? 
 
 40. George is \ as old as his father ; a years ago he was \ as 
 old as his father. What is the age of each ? 
 
 41. A rug is 3 feet longer than it is wide. When it is placed 
 on the floor of a certain room, it leaves a margin of 2 feet on 
 every side. If the area of the floor is 172 square feet greater 
 than that of the rug, what are the dimensions of the floor ? 
 
MULTTPLICA TION 
 
 61 
 
 SPECIAL CASES IN MULTIPLICATION 
 
 90. The square of the sum of two numbers. 
 
 (a + h){a + h)=a? + 2ah + h\ 
 {x + y){x -^y)=x' + 2xy + f. 
 
 1. When a number is multiplied by itself, what power is 
 obtained ? What is the second power, or square of (a + 6) ? 
 
 of (.x- + t/)? 
 
 2. How are the terms of the square of the sum of two numbers 
 obtained from the numbers ? 
 
 3. What signs have the terms ? 
 
 91. Principle. — The square of the sum of two numbers is equal 
 to the square of the first number, plus twice the product of the first 
 and second, plus the square of the second. 
 
 Since 5 a^ x 5 a^ = 25 a«, 3 a*b^ x 3 a*b' = 9 a^ft^", etc., it is evident 
 that in squaring a number the exponents of literal factors are 
 doubled. 
 
 Examples 
 
 Expand by inspection : 
 
 1. 
 
 (m + n){m + n). 
 
 13. 
 
 (3b + cy. 
 
 2. 
 
 (P + <i)(P + Q)' 
 
 14. 
 
 (2a + 3 6)2. 
 
 3. 
 
 (r + s){r + s). 
 
 15. 
 
 {Bx + 2yy. 
 
 4. 
 
 (a + x){a + x). 
 
 16. 
 
 (72 + 3 cf. 
 
 5. 
 
 {x -\- 4:){X + A). 
 
 17. 
 
 (3 6 + 10 xf. 
 
 6. 
 
 (m 4- 5)(m + 5). 
 
 18. 
 
 (a^ + by. 
 
 7. 
 
 (a + 6)(a + 6). 
 
 19. 
 
 (a' + by. 
 
 8. 
 
 (2/ + 7)(2/ + 7). 
 
 20. 
 
 (a» + 6")'. 
 
 9. 
 
 (z+l){z + l). 
 
 21. 
 
 (x^+y")'. 
 
 10. 
 
 (2 a + x)(2 a + x). 
 
 22. 
 
 (3 a* + 5 6^1 
 
 11. 
 
 (3 m -1- n)(3 m + n). 
 
 23. 
 
 (1 + 2 a'bY. 
 
 12. 
 
 {5 x + z){o X + z). 
 
 24. 
 
 (^-i + y-iy. 
 
62 MUL TIP Lie A TION 
 
 25. Find the square of 41. 
 
 Solution 
 412 = (40 + 1)^ = 402 + 2 X 40 X 1 + 12 = 1681. 
 
 Square : 
 
 
 
 
 
 
 
 26. 21. 
 
 29. 
 
 45. 
 
 32. 
 
 22. 
 
 35. 
 
 81. 
 
 27. 24. 
 
 30. 
 
 83. 
 
 33. 
 
 72. 
 
 36. 
 
 91, 
 
 28. 25. 
 
 31. 
 
 65. 
 
 34. 
 
 43. 
 
 37. 
 
 101, 
 
 38. Find the square of 1\. 
 
 • Solution 
 
 (7J)2=(7 + 1)2 = 72 + 2 X 7x i+(l)2 = 49+7 + J = 56|. 
 
 Observe that the middle term of the square of any number expressed by an 
 integer and the fraction \ is equal to the integer. Hence, the square of such 
 a number is equal to the square of the integer, + tlie integer, + tiie square 
 of the fraction. Observe also that the sum of the first two terms of the square 
 may be found by multiplying the integer by tlie integer increased by 1. 
 
 Thus, 
 
 (3i)2 = 9 + 3 + i = 12i, 
 
 
 or 
 
 (3^)2 = 3 X 4 + i = 12i. 
 
 
 Find the square of 
 
 
 
 39. 5^. 41. 
 
 2f 43. 1.5. 
 
 45. 6.5. 
 
 40. 41 42. 
 
 12^. 44. 5.5. 
 
 46. 8.6. 
 
 92. The square of the difference of two numbers. 
 
 (a - h){a -b)=a' -2ab-\- h\ 
 {x - y){x -y)=;t?-2xy -\- y\ 
 
 1. What is the square of (a — h) ? of {x — y) ? 
 
 2. How is the square of the difference of two numbers obtained 
 from the numbers ? 
 
 3. How does the square of (a — h) differ from the square of 
 (a+6)? 
 
 93. Principle. — The square of the difference of two numbers is 
 equal to the square of the first mimber, minus twice the product of 
 the first and second, lilus the square of the second. 
 
\ 
 
 
 MULTIPLICA TION 
 
 
 68 
 
 
 
 Examples 
 
 
 
 i)xpa 
 
 ind by inspection: 
 
 
 
 
 
 1. 
 
 [x — m) (a- — m). 
 
 10. 
 
 (2 a- x)\ 
 
 19. 
 
 (Sx-2y. 
 
 2. 
 
 (m — n) (m — n). 
 
 11. 
 
 (3 m- n)\ 
 
 20. 
 
 (2x-5yy. 
 
 3. 
 
 (x _ 6) (a; - 6). 
 
 12. 
 
 {4x-yy. 
 
 21. 
 
 (2x-4.yy. 
 
 4. 
 
 (i>-8)(p-8). 
 
 13. 
 
 (5 m — ny. 
 
 22. 
 
 (5 m — 3 ny. 
 
 5. 
 
 (5 -7) (9 -7). 
 
 14. 
 
 (m — 4 n)^ 
 
 23. 
 
 i3p-oqy. 
 
 6. 
 
 (a — c) (a — c). 
 
 15. 
 
 (p-3qy. 
 
 24. 
 
 (a" _ 6")2. 
 
 7. 
 
 (^•-0(»--0- 
 
 16. 
 
 (a - 7 6)2. 
 
 25. 
 
 {x^-yy. 
 
 8. 
 
 'a — x) (a — cc). 
 
 17. 
 
 (4 a -3)2. 
 
 26. 
 
 (a--x_2/»-i)«. 
 
 9. 
 
 {:x-l){x-l). 
 
 18. 
 
 (5ic-4)2. 
 
 27. 
 
 (ma;™ — wy")*. 
 
 28. Find the square of 19. 
 
 Solution 
 192 = (20 - 1)2 = 202 - 2 X 20 X 1 + 12 = 361. 
 
 Find the square of 
 
 29. 49. 32. 29. 35. 67. 
 
 30. 69. 33. 38. 36. 89. 
 
 31. 79. 34. 48. 37. 99. 
 
 38. 998. 
 
 39. 999. 
 
 40. 595. 
 
 94. The square of any polynomial. 
 
 By actual multiplication, 
 
 (a+b + cy=a^+b^+c'-\-2ab-\-2ac+2bc. 
 
 (a+b-c+dy'=a^+b--\-cr+cl^+2ab-2ac+2ad-2bc-]-2bd-2cd. 
 
 (a-\ \-k-\ \-my=a^-^ \-J(^-\ [-m^ + 2ak-\ 
 
 + 2amA \-2km-i . 
 
 1. In the square of each polynomial what terms are squares? 
 
 2. How are the other terms formed from the terms of the 
 polynomial ? 
 
 3. What signs have the squares ? How are the signs of the 
 other terms determined ? 
 
64 
 
 MUL TIPLICA TION 
 
 95. Principle. — The square of a polynomial is equal to the 
 smn of the squares of the terms and tioice the product of each term 
 by ea^h term that follows it. 
 
 Examples 
 
 Expand by inspection : 
 
 4. (x-y + zf. 7. (a-2b + cy. 
 
 5. (^x + y-3zy. 8. (2a-b-cY. 
 
 6. (x-y + 3zy. 9. (b-2a + cy. 
 
 15. (2a-5b + 3cy. 
 
 16. (2 m — 4: n — ry. 
 
 17. (12-2x + Syy. 
 
 18. (a-\-m + b + ny. 
 
 19. (a — m + b — ny. 
 
 96. The product of the sum and difference of two numbers. 
 
 (a + 6) (a - 6) = a2 _ ^2^ 
 (x" + y") (x" — ?/") = a^" — /". 
 
 1. Since a + 6 represents the sum and a — b the difference of 
 two numbers, to what is the product of the sum and the difference 
 of two numbers equal ? 
 
 2. How are the terms of the product obtained from the 
 numbers ? 
 
 3. What sign connects the squares ? 
 
 97. Principle. — TJie product of the sum and difference of two 
 numbers is equal to the difference of their squares. 
 
 1. 
 
 [x + y + zy. 
 
 2. 
 
 [x + y-zy. 
 
 3. 
 
 [x-y- zy. 
 
 10. 
 
 [ax — by + czy. 
 
 11. 
 
 [ma + n6 — rzy. 
 
 12. 
 
 [qb — pc — rdy. 
 
 13. 
 
 [ac — bd — dey. 
 
 14. 
 
 [3x-2y + 4z) 
 
 Examples 
 
 Expand by inspection ; 
 
 1. (x + y)(x-y). 
 
 2. (a + c)(a— c). 
 
 3. {m + n) (m — n). 
 
 4. (p + 9)(i>-^). 
 
 5. (2> + 5)(p-5). 
 
 6. (r + s) (r — s). 
 
 7. (a; + 1) (a; - 1). 
 
 8. {:)^ + l){;t^-l). 
 
 9. {^+l){o? -1). 
 10. (a;* - 1) (»* + 1). 
 
MUL TIPL ICA TION 65 
 
 11. (ar> - 1) (ar* -f- 1). 20. {2a^ + 5y')(2a^ - 5f). 
 
 12. (x^ + f)(x^^-f). 21. {3a^ + 2y')(Sx'-2f). 
 
 13. {ab + 5){ab-5). 22. {20^ -h2b-'){2a' -2 b'). 
 
 14. (cd + d^) (cd - d^). 23. (-5n-26)(-5n + 26). 
 
 15. (2a;+32/)(2a;-3y). 24. (- cc- 2y)(- a; + 2?/). 
 
 16. (3m + 4n)(3OT-4w). 25. (- 5 - 3m)(- 5 + 3m). 
 
 17. (12-\-xy){12-xy). 26. (ic'»-^ + ^»+i)(a;'»-i - ?/»+^). 
 
 18. (3 m^w - 6) (3 m^'n + 6). 27. (3 x" + 7 y") (3 x™ - 7 y"). 
 
 19. (a& + cd) (a6 - cd). 28. (5 a^b^ + 2 x*) {5 a^b^ - 2 x*). 
 One or both of the numbers may consist of more than one term. 
 
 29. Expand (a + m — w) (a — m + n). 
 
 Solution 
 
 a + m — n = a + (m — n). 
 
 a — m + n = a — (m — n). 
 
 .'. [_a-{-m — n}la — »» + n] = [a + (m — n)'\la — (m — n)]. 
 
 Prin., = a^ — (m — n)^ 
 
 § 93, = a2 - (m2 - 2 mn + n«) 
 
 = a2 - m2 4- 2 mn - n^. 
 Expand : 
 
 30. (a-\-x-y){a-x-]-y). 36. (y + c -{-d){y + c -d). 
 
 31. (x + c — d)(iB — c + d). 37. (a-\-x + y){a + x—y). 
 
 32. (r+i3-9')(r-i3 + ^). 38. (o^ + 2x + l)(a^ + 2a; - 1). 
 
 33. (r + p + q)(r-p-q). 39. (ar* + 2 x - 1) (x^ _ ^ a, ^ ^-^ 
 
 34. (x + b + n)(x-b-n). 40. (x^^Sa;- 2)(r' - 3a; + 2). 
 
 35. (2/ + c + d)(2/-c-d). 41. (a;2 + 3 a; + 2) (ar' - 3 a; + 2). 
 
 42. (m*-2m' + l){m* + 2m' + l). 
 
 43. (2x + Sy-4:z){2x-\-3y + 4:Z). 
 
 44. (2x*-a;2/ + 32/2)(2ar' + xi/-3 2r^. 
 
 45. (ar* + ajy + 2/^) (ar* - a^ + 2/^. 
 
 ALG. — 5 
 
66 
 
 MULTIPLICATION 
 
 46. [(a + &)+(c-f d)][(a + 6)-(c + d)]. 
 
 47. {a-{-h + x + y){a-\-h — x — y). 
 
 48. {a + h + m — n){a -^ b — m -\- n). 
 
 49. (x — m -{- y — n) (x — m — y -\- n). 
 
 50. {p — q -\-r-\- s){p — q — r — s). 
 
 51. (a — m — 6 — w) (a + m — 6 + w). 
 
 52. {a-\-x + h ~y){a~x + b -{-y). 
 
 53. Find the product of 32 x 28. 
 
 Solution 
 32 X 28 =(30 + 2) (30 -2) 
 
 = 302 - 22 = 900 - 4 = 896. 
 
 Find the product of 
 
 54. 31 X 29. 57. 48 X 52. 60. 98 x 102. 
 
 55. 42 X 38. 58. 57 X 63. 61. 99 x 101. 
 
 56. 69 X 71. 59. 95 X 85. 62. 95 x 105. 
 
 63. What is the square of 95 ? 
 
 Solution 
 (a + 6)(a-6)=a2_52. ^Ij 
 
 Transposing, etc., a'^ = (a + b) (a — 6) + 6^. f2) 
 
 Let a = 95 and 6 = 5. 
 
 Equation (2) becomes 952 _ (95 + 5) (95 _ 5) + 52 
 
 = 100 X 90 + 25 = 9025 
 
 64. What is the square of 48 ? 
 
 Solution 
 Let a = 48 and 6 = 2. 
 
 Equation (2) becomes 482 _ (43 + 2) (48 - 2) + 2* 
 
 = 50 X 46 + 4 = 2304. 
 
 Square by a similar process : 
 
 65. 98. 67. 93. 69. 58. 71. 87. 73. 68. 
 
 66. 96. 68. 97. 70. 49. 72. 79. 74. 129. 
 
MUL TIPLICA TION 67 
 
 98. The product of two binomials that have a common term. 
 
 {x 4- 2){x + 5)= af + 2 X -1- 5 a; 4- 10 
 
 ^x' + l x + 10. 
 {x + 2){x - 5)= x- + 2 X - 5 a; - 10 
 
 = af' - 3 a; - 10. 
 {x - 2){x -b)=x'-2x-hx + 10 
 
 = a.-^ - 7 a: + 10. 
 (x + a)(x — 6) = .x*^ -f ax — bx — ab 
 
 = XT -\-{a — b)x — ab. 
 
 1. How many terms are alike in each pair of factors ? 
 
 2. How is the first term of each product obtained from the 
 binomial factors ? 
 
 3. How is the third term of each product obtained from the 
 factors ? 
 
 4. How is the second term of the product in the first example 
 obtained from the factors ? in the second example ? in the third 
 example ? in the fourth example ? 
 
 5. How can the formation of the second, or middle term be 
 described so as to apply to all of the examples ? 
 
 99. Principle. — Tlie product of two binomials that have a com- 
 mon term is the algebraic sum of the square of the common term, the 
 common term multiplied by the algebraic sum of the unlike terms, 
 and the algebraic product of the unlike terms. 
 
 Examples 
 Expand by inspection : 
 
 1. (x + 5)(x-^6). 7. (a; - 5)(a; - 1). 
 
 2. (a; + 7) (a; 4- 8) 
 
 3. (a;-7)(a; + 8) 
 
 4. (a; + 7)(a;-8) 
 
 5. (x - 5)(x — 4) 
 
 6. (a;-3)(a;-2) 
 
 8. (x + 5)(x + 8). 
 
 9. (p-4)(p + l). 
 
 10. (r-3)(r-l). 
 
 11. (m-6)(m + 5). 
 
 12. (m - 2)(m 4- 10). 
 
68 
 
 M UL TIP Lie A TION 
 
 13. 
 
 (n-8)(ri -12). 
 
 
 
 24. 
 
 (jj-2a){y + ^h). 
 
 14. 
 
 (ii-6)(?6 + 15). 
 
 
 
 25. 
 
 (2-4 a) (2 + 3 a). 
 
 15. 
 
 (ar' + 5)(ar-3). 
 
 
 
 26. 
 
 (2 a; + 5) (2 a; + 3). 
 
 16. 
 
 (a^-7)(ar' + 6). 
 
 
 
 27. 
 
 (2 a; -7) (2 a; + 5). 
 
 17. 
 
 (x-*-3)(ir* + 9). 
 
 
 
 28. 
 
 (3.v-l)(32/ + 2). 
 
 18. 
 
 (x + c){x + d). 
 
 
 
 29. 
 
 (4a^+l)(4ar^-7). 
 
 19. 
 
 (m + a) (m + 6). 
 
 
 
 30. 
 
 (a&-6)(a6+4). 
 
 20. 
 
 (r + a)(r-6). 
 
 
 
 31. 
 
 (ar^/-a)(a^y^ + 2a). 
 
 21. 
 
 (s — a) (s + w). 
 
 « 
 
 
 32. 
 
 (2 m - ab) (2 m + 3 a6). 
 
 22. 
 
 (a;" - 5) (x" + 4). 
 
 
 
 33. 
 
 (5i?-ac2)(2ao2 + 5i>). 
 
 23. 
 
 (a;" - a) (x^ - b). 
 
 
 
 34. 
 
 (3xy-\-9/){y^-xy). 
 
 
 35. (a 
 
 + b + 5){a 
 
 + 6 + 2). 
 
 
 36. (a 
 
 -h- 
 
 -4)(o 
 
 i-h 
 
 + 10). 
 
 
 37. (x 
 
 + y- 
 
 -1)(^ 
 
 + 2/ + 2). 
 
 
 38. (x 
 
 39. (^ 
 
 -y- 
 
 -2){x 
 
 -i)C 
 
 -y 
 
 -8). 
 
 
 + x 
 
 ^ + 
 
 x + 3). 
 
 40. (2 m + w - 3) (2 m + » + 4). 
 
 By an extension of the method given above, the product of 
 any two binomials may be written. 
 
 41. Expand (3x + 2y)(5x — 4:y). 
 
 Solution 
 (3x + 2y)(5x-^y)- 15x2+ \Oxy - \2zy-8y' 
 = 15 x2 - 2 aiy - 8 y^. 
 
 Expand by inspection : 
 
 42. (2x + 5y)(3x-\-4.y). 
 
 43. (3x-4:y)(2x + 5y). 
 
 44. (3a-66)(2a + 3&). 
 
 45. (3 a2 _ 1) (2 a^ + 3). 
 
 46. (m^ + n) {2 m^ - n). 
 
 47. (a + 26)(c-2d). 
 
MULTIPLICATION 69 
 
 Exercises 
 
 100. 1. In a certain family there are three children each of 
 whom is 2 years older than the one next younger. When the 
 youngest is x years old, what are the ages of the others ? When 
 the oldest is y years old, what are the ages of the others ? 
 
 2. What two whole numbers are nearest to the whole num- 
 ber X? 
 
 3. Mary read 10 pages in a book, stopping at the top of page a. 
 On what page did she begin to read ? 
 
 4. A man made three purchases of a, b, and 2 dollars, respec- 
 tively, and tendered a 10-dollar bill. Express the number of dol- 
 lars change due him. 
 
 5. A sold B grain, hay, and potatoes for a, h, and c dollars, 
 respectively ; but some of the grain becoming damaged, and 
 some of the potatoes having been frozen, he deducted x + y 
 dollars from B's indebtedness. If B offered a 100-dollar bill in 
 payment, what was the amount due him in return ? 
 
 6. What is the cost of 5 apples at b cents each? What will 
 a apples cost at b cents each ? 
 
 7. How many cents are there in a dollars ? How many 
 dimes are there in b dollars ? in ax dollars ? 
 
 8. If a man earns $2 per day, how much will he earn in 
 a days ? in c days ? in a — c days ? 
 
 9. How much will a man whose wages are a dollars per day 
 earn in b days ? in c days ? in a days ? in a days ? 
 
 10. If" a man earns a dollars per month and his expenses are 
 b dollars per month, how much will he save in a year ? 
 
 11. How far can a wheelman ride in a hours at the rate of 
 b miles an hour ? How far will he have ridden after a hours, if 
 he stops c hours of the time to rest ? 
 
 12. A has X dollars and B y dollars. If A gives B m dollars, 
 how much will each then have ? 
 
 13. The number 25 may be written 20 + 5. Write the number 
 whose tirst digit is x and second y. 
 
70 MULTIPLICATION 
 
 14. A book contained x pages. If they averaged y lines to a 
 page and z letters to a line, how many letters were there in the 
 book? 
 
 15. Hov/ many square rods are there in a square field one of 
 whose sides is 2 & rods long ? x rods long ? 
 
 16. What is the number of square rods in a rectangular field 
 whose length is 30 rods and width 20 rods ? What will be the 
 area in square rods, if the length is a rods and the width 6 rods ? 
 
 17. A fence is built across a rectangular field so as to make 
 the part on one side of the fence a square. If the 'field is a rods 
 long and b rods wide, what is the area of each part ? 
 
 18. A man who had s dollars bought h bales of hay at n cents 
 a bale and a bushels of oats at m cents a bushel. How many 
 cents had he left ? 
 
 19. A speculator bought a car loads of wheat at m dollars a 
 car, and sold h car loads of it at n dollars a car. How much did 
 he gain by the transaction, if he sold the rest of the wheat for 
 c dollars a car ? 
 
 20. A sold a farm which had cost him n dollars an acre to two 
 men, a acres to one and h acres to the other, at the uniform price 
 of m dollars an acre. How much did he gain ? 
 
 21. In a library there were p + q volumes that averaged p + q 
 pages per volume, p -\- q words per page, and 7 letters per word.' 
 How many letters were there in all these books ? 
 
 22. A wheelman who had a journey of x miles to make rode 
 the first a hours at the rate of h miles an hour, when he was 
 obliged to stop c hours for repairs. After that he rode 2 miles 
 an hour faster, so that he made the whole journey in 10 hours. 
 What was the length of the journey ? 
 
 23. A wheelman rode a hours at the rate of m miles an hour, 
 then decreased his speed 5 miles an hour for 3 hours, and finished 
 his journey in 6 hours more, increasing his first rate 2 miles an 
 hour. How far did he ride ? 
 
 If a = 4, 6 = 2, and m = 10, how many miles did he ride, and 
 in what time did he accomplish the journey ? 
 
DIVISION 
 
 101. 1. Since + 2 times + 10 is + 20, if + 20 is divided by 
 + 10, what is the sign of the quotient ? 
 
 2. What is the sign of the quotient when a positive number 
 is divided by a positive number ? 
 
 3. Since + 2 times - 10 is - 20, if - 20 is divided by - 10, 
 what is the sign of the quotient ? if — 40 is divided by — 5 ? 
 
 4. What is the sign of the quotient when a negative number is 
 divided by a negative number ? 
 
 5. What is the sign of the quotient when the dividend and 
 divisor have like signs ? 
 
 6. Since + 2 times — 10 is — 20, if — 20 is divided by + 2, 
 what is the sign of the quotient ? if — 20 is divided by + 5 ? 
 
 7. Since - 2 times - 10 is + 20, if + 20 is divided by - 2, 
 what is the sign of the quotient ? if + 20 is divided by — 4 ? 
 
 8. What is the sign of the quotient when the dividend and 
 divisor have unlike signs ? 
 
 9. How many times is 2 a contained in 6 a ? in 10 a? 
 How is the coefficient of the quotient obtained ? 
 
 10. Since a^ x a^ = a^, if a^ is divided by a", what is the quo- 
 tient ? What is the quotient, if a* is divided by a^ ? 
 
 What is the quotient, if b'' is divided by &^ ? by 6* ? 
 
 How is the exponent of a number in the quotient obtained ? 
 
 11. What is the exponent of a in the quotient of a* -i- a*? of 
 a' -j- a^ ? How many times is a* contained in a* ? a^ in a^ ? 
 
 12. What is the value of any expression whose exponent is ? 
 
 71 
 
72 DIVISION 
 
 102. In multiplication two numbers are given and their product 
 is to be found. The inverse process, finding one of two numbers 
 when their product and the other number are given, is called 
 
 Division. 
 
 10 -^ 2 = 5 and D -^ d = q 
 
 are inverses of 5 x 2 = 10 and q x d = D. 
 
 The dividend corresponds to the product, the divisor to the mul- 
 tiplier, and the quotient to the multiplicand. Hence, the quotient 
 may be defined as that number ivhich multiplied by the divisor 
 produces the dividend. 
 
 The quotient of a divided by b, indicated by (a -^ b), or -, 
 
 is defined for all values of a and b by the relation 
 
 - X b = a. 
 b 
 
 103. Principles. — 1. Law of Signs. — The sign of the quotient 
 is + when the dividend and divisor have like signs, and — when 
 they have unlike signs. 
 
 2. Law of Coefficients. — The coefficient of the quotient is equal to 
 the coefficient of the dividend divided by that of the divisor. 
 
 3. Law of Exponents. — TJie exponent of a letter in the quotient 
 is equal to its exponent in the dividend diminished by its exponent in 
 the divisor. 
 
 An expression whose exponent is is equal to 1. 
 
 The Law of Signs may be established as follows : 
 
 Since + a x + b = + ab, + ab -^ + b = + a. 
 
 Since + a X — b = — ab, — ab -. — b = -{■ a. 
 
 Since — a x + ft = — aft, — aft -f- + 6 = — a. 
 
 Since — ax — 6=+ aft, +a6^ — ft=— a. 
 
 The Law of Exponents or the Index Law for Division may 
 be established as follows, m and n being positive integers and m 
 being greater than n : 
 
 By § 24, a" = a X a X a ••• to m factors, 
 a" = a X a X a ••• to ?i factors ; 
 .-. a"» -4- a" = (a X a X a ••• to m factors) -h- (a x a x a ••• to n factors) 
 = a X a X a • • • to (m — n) factors. 
 Hence, a"" -i- a^ = o*"-". 
 
DIVISION 73 
 
 104. Commutative, Associative, and Distributive Laws for Division. 
 
 1. The Commutative Law may be established as follows: 
 Since, by the definition of division, § 102, a = a -^ c x c, 
 
 axb-7-c = a-^cxcxb-i-c 
 §82, = a -i- c X b X c-i-c 
 
 = a ^ c X b. (1) 
 
 Also, § 102, o-7-6-T-c = a-f-cxc-^6^c 
 
 by notation, § 29, =(^a-^cxc-^b)-i-c 
 
 by (1), =(a -T-c-i- ft X c)-f-c 
 
 by notation, =a-i-c-^bxc-i-c 
 
 = a^c^b. (2) 
 
 The Commutative Law for division is expressed by (2). 
 
 (2) may be written - -;- c = - -h 6. (1) may be written — = - x 6. 
 be c c 
 
 It follows from (1) and (2) that in a succession of multiplications and 
 
 divisions the multipliers and divisors may be arranged in any order. 
 
 2. The Associative Law may be established as follows: 
 By the Commutative Law just proved, 
 
 axb -^ c = b ^ ex a 
 
 by notation, § 29, ={b ^ c)x a 
 
 by the Commutative Law, = a x (6 -=- c). (3) 
 
 Also a-4-6-^c = 6xc-4-(6xc)xa-i-6-f-c 
 
 by the Commutative Law, —b^bxc^cxa-^ipxc) 
 
 = a^{b xc). (4) 
 
 The Associative Law for division is expressed by (4). 
 
 (4) may be written - -r- c = — • (3) may be written — = a x — 
 b be c c 
 
 It follows from (3) and (4) that in a succession of multiplications and 
 
 divisions the multipliers and divisors may be grouped in any manner, each 
 
 element keeping its own sign, x or -i-., if the sign x precedes the sign of 
 
 grouping, but changing it, if the sign -f- precedes the sign of grouping. 
 
 3. The Distributive Law may be established as follows : 
 
 §85, {a^m + b-i-m)xm = a-^mxm + b-. 
 
 - m xm 
 
 = n + 6. 
 Dividing both members by m, 
 
 
 a-i-m + b-^m=:(a + b)-i-m; 
 
 
 atis, « + &^.«^6. 
 
 m mm 
 
 
74 DIVISION 
 
 105. The Reciprocal of a number is 1 divided by the number. 
 The reciprocal of 5 is -; of 6, - ; of (a + 6), 
 
 6 h a + b 
 
 106. a-i- b = a xl -i-b 
 by the Associative Law, = a x (1 -^ 6). 
 
 Hence, dividing by a number is equivalent to multiplying by the 
 reciprocal of the number. 
 
 107. To divide a monomial by a monomial. 
 
 Examples 
 
 1. Divide -ISa'b^ by 6ab^ 
 
 Explanation. — Since the dividend and divisor have 
 PKOCESS unlike signs, the sign of the quotient is — (Prin. 1). 
 
 6 ab^ ) - 18 a^b^ Then, - 18 divided by 6 is - 3 (Prin. 2); a^ divided 
 
 3a<5 by a is a* (Prin. 3); and b^ divided by b^ is b (Prin. 3). 
 
 Therefore, the quotient is — 3 a*b. 
 
 Rule. — Divide the numerical coefficient of the dividend by the 
 numerical coefficient of the divisor, and to the result annex the letters, 
 each with an exponent eqital to its exponent in the dividend minus 
 its exponent in the divisor. 
 
 Write the sign + before the quotient when the dividend and divisor 
 have like signs, and the sign — when they have unlike signs. 
 
 
 2. 
 
 3. 
 
 4. 
 
 5. 
 
 6. 
 
 Divide 
 
 12 a; 
 
 -120^^/ 
 
 35 xy^ 
 
 -14x3^2 
 
 - 26 a'b'c' 
 
 By 
 
 4x 
 
 4aV 
 
 -7xy 
 
 -2x^y 
 
 13 ab-c 
 
 Find the quotient of 
 
 7. 2S a*b-c^ -7 abc. 10. - U a^f:i* -i- 7 a^fz\ 
 
 8. — 16 x^y^^ -=- 4 xyz. 11. — 27 m'n^p^ -?- 9 m^n^p^. 
 
 9. - 36 a^mV --- 9 mV. 12. - 3^ q^r'p ^ - 13 qp. 
 
 108. To divide a polynomial by a monomial. 
 
 Examples 
 1. Divide 5 x*y* — IQ s^if + h x^rf by bx^j^. 
 
 PKOCESS. § 104, 3, 5 a? f)5 xY - 10 s^y" + 5 x^f 
 
 x^y — 2xy^ -f t/* 
 
DIVISION 75 
 
 Rule. — Divide each term of the dividend by the divisor, and Jind 
 the algebraic sum of the jjartial quotients. 
 
 Find the quotient of 
 
 4,a^W-\2a%- + \%a*b ^ Amhi-Smhi^ -^ 4:mn^ 
 
 4 a^b 4 mn 
 
 24 a%- + 32 a'W - 40 a*b* 5 x*y - 10 xY + 20 x'f 
 
 8a'b^ ' ' 5xy 
 
 - 35 a^y^z^ + 45 xyz- ^ -a-&-c-(?-e 
 
 5 a^y'z — 1 
 
 — 39 oc^yV' + 65 x^y^z' —a — a-b — a^c — a * d — a^e 
 
 — I'Sx^yh^ ' —a 
 
 25 7-«/ _ 125 ?V - 75 r^s'" 
 
 10. 
 
 11. 
 
 5 ?-V 
 
 3 ckl- 
 
 12. (34 a^xV - 51 a*xy - 68 a^'x'y^) -i- 17 a'x^. 
 
 13. (8 a'b^ - 28 a«6^ - 16 a'b' + 4 a^6«) -=- 4 a*ly^. 
 
 14. [a (6 - c)3 - & (& - c)- + c(& - c)] -- (6 - c). 
 
 15. [(a; - ^j) - 3 (x - yf + i x(x - yf] ^(x- y). 
 
 16. {X' - 2 X''+'^ - 5 a;''+2 _ a.a+3 _,_ 3 ^+4^, ^ ^a 
 
 17. (^y«+l _ 2 ?/»+2 - 7/»+3 _ 3 7f+* + ?/"+') -- ?/" + '. 
 
 18. (a;" — x"-^ + x"-^ — ic"-^ + x""* — a;""^) -=- a^. 
 
 109. To divide a polynomial by a polynomial. 
 
 Examples 
 1. Divide 3 a^ + 35 + 22 a; by x + 5. 
 
 
 PROCESS 
 
 3ar*+22x + 35 
 . 3a^ + 15a; 
 
 7a; + 35 
 7 a; + 35 
 
 a; + 5 
 
 TEST 
 
 + 60 -1- + 6 
 
 a;(x + 5) . 
 
 3« + 7 
 
 = + 10 
 
 7(a; + 5) 
 
 
 
76 DIVISION 
 
 Explanation. — For convenience, the divisor is written at the right of 
 the dividend, and both are arranged according to the descending powers of x. 
 
 Since the dividend is the product of the quotient and divisor, it is the 
 algebraic sum of all the products formed by multiplying each term of the 
 quotient by each term of the divisor. Therefore, the term of highest degree 
 in the dividend is the product of the terms of highest degree in the quotient 
 and divisor. Hence, if Sx'^, the first term of the dividend as arranged, 
 is divided by x, the first term of the divisor, the result, 3 x, is the term of 
 highest degree, or the first term, of the quotient. 
 
 Subtracting 3x multiplied by (x + 5), or 3x times (x + 5) from the divi- 
 dend, the remainder is 7 x + 35. 
 
 Since the dividend is the algebraic sum of the products of each term of the 
 quotient multiplied by the divisor and since the product of the first term of the 
 quotient multiplied by the divisor has been canceled from the dividend, 
 the remainder, or new dividend, is the product of the other part of the quotient, 
 multiplied by the divisor. 
 
 Proceeding, then, as before, 7 x -^ x = 7, the next term of the quotient. 
 7 multiplied by (x + 5), or (x + 5) multiplied by 7 equals 7x + 35. Sub- 
 tracting, there is no remainder. Hence, all of the terms of the quotient have 
 been obtained, and the quotient is 3 x + 7. 
 
 Test. — Let x = 1. 
 
 Dividend = 3 x2 + 22 x + 35 = 3 + 22 -f 35 = + 60. 
 
 Divisor =x-|-5 =1 + 5 =+6. 
 
 Quotient should be equal to + 10. 
 
 Quotient =3x + 7 =3 + 7 =+10. 
 
 Similarly, the result may be tested by substituting any other value for x. 
 When the value substituted for x gives the result -^ or for a divisor, 
 some other value should be tried. 
 
 Rule. — Arrange both dividend and divisor according to the 
 ascending or the descending poioers of a common letter. 
 
 Divide the first term of the dividend by the first term of the divisor, 
 and write the resxdt for the first term of the quotient. 
 
 Multiply the whole divisor by this term of the quotient, and sub- 
 tract the product from the divideird. The remainder will be a new 
 dividend. 
 
 Divide the new dividend as before, and continue to divide in this 
 way until the first term of the divisor is not contained in the first 
 term of the new dividend. 
 
 If there is a remainder after the last division, write it over the 
 divisor in the form of a fraction, and add the fraction to the part 
 of the quotient previously obtained. 
 
DIVISION 
 
 11 
 
 2. Divide 2a* -ba?h + %o?h'' -^aW + h*' by a?-ah + h^ 
 
 PROCESS 
 
 4a63+5* 
 
 2 a^ - 3 a6 + 6^ 
 
 2 a^ - 5 a^ft + 6 a?h^ 
 2 a*-2a?h + 2 a?b^ 
 
 - 3 a^d + 4 d'b'' - 4 aft^ 
 
 - 3 a^6 + 3 a?b'' - 3 a6^ 
 
 a26^ - aW + 6* 
 ^252 - db^ + 6* 
 
 3. Divide a* + 9a2 4-81 by a2_3a-f9. 
 
 a6 + 62 
 
 TEST 
 
 = 
 
 a* + 9 a^ + 81 
 a4_3a3_^9a2 
 
 PROCESS 
 
 g^ - 3 g + 9 
 
 3 a' + 81 
 
 3 g'^ - 9 g' 4- 27 g 
 
 g' + 3 g + 9 
 
 TEST 
 
 91-5-7 
 = 13 
 
 9 g2 - 27 g -+- 81 
 9a' -27 a + 81 
 
 Divide, and test the results : 
 
 4. x^ — x — 20 by ic — 5. 
 
 5. a^ -I- 7 a; -h 12 by .^-^3. 
 
 6. m'^ — 3 m — 18 by m — 6. 
 
 7. ;4_6Z2_16 by V-\-2. 
 
 8. cd.-0? + 2(? hy c-\-d. 
 
 9. cc2 _ 11 a; + 10 by a; - 10. 
 
 10. x' + lbx + 5A: \>Y x + 6. 
 
 11. r^" -f- 11 r" + 30 by r" + 6. 
 
 12. gW — 4 gm^ + 3 m^ by gm — 1. 
 
 13. 6g2 + 13a& + 662 by 3g + 26. 
 
 14. g^ + 16 + 4 g2 by 2 g + g" + 4. 
 
 15. g* + g^ + g* + g2 + 3 g — 1 by g + 1. 
 
 16. 20aj22/-25a^-182/3 + 27ay^ by 62/-5». 
 
78 DIVISION 
 
 17. ay? — aV — hx^+h^ by ax ~ b. 
 
 18. a* -41 a -120 by a^ + Aa + S. 
 
 19. x^-61x-60 by ar' - 2a; - 3. 
 
 20. 25ar'-a;3_g^_2a..2 by 5ar'-4a;. 
 
 21. 4^*-9?/2 + 6?/-l by 22/2 + 3y-l. 
 
 22. a* -Aa^x + 6 a-x- — 4 aa;^ + a;* by a^ — 2 aa; + a^. 
 
 23. a'' - 1 
 
 a^ + a* 
 
 — a' 
 
 a+ 1 
 
 -a + l+ ~^, 
 a+1 
 
 a*-a^ + a^- 
 
 a^-l 
 
 a^ + a^ 
 
 a-1 
 a + 1 
 
 24. a:^ + x - 25 
 a:3-3a;2 
 
 3x2+ aj 
 3ar'-9a; 
 
 
 10 a;- 
 lOx- 
 
 -25 
 -30 
 
 -2 
 a;-3 
 
 ar' + 3a; + 10 + 
 
 25. a;<-3«3+ a^ + 2a;-l 
 
 aj* — a^ — 2 a^ 
 
 -2a^ + 3ar' + 2a; 
 -2a;3 + 2a^ + 4a; 
 
 ar'-2a;-l 
 a?- x-2 
 
 x"- 
 
 - x-2 
 
 -x + 1 
 -x~2 
 
 ^- 
 
 -2a; + l+-f 
 
 XT 
 
 x + l 
 
DIVISION 
 
 79 
 
 Divide : 
 
 26. trv' -\-n^ by m-\-n. 
 
 27. a^ + 32 by x + 2. 
 
 28. ^ + y^ by x^ + y^. 
 
 29. a*' + 5a*-a=' + 2a + 3 by a-1. 
 
 30. 2 n' - 4 w* - 3 n=^ + 7 n'' - 3 n -I- 2 by n - 2. 
 
 31. a? + 1^ -{- ^ — S xyz hy X -\- y -\- z. 
 
 32. m^ + n^ + ar* + 3 m^n + 3 mn^ by m + n + aj. 
 
 33. a^ - 6 a^ + 12 a - 8 - 6" by a - 2 - 6. 
 
 34. 2/^ + 32/^ + 5y3 + 3y2 + 32/ + 5 by y + 1. 
 
 35. 2a^,-x^ + 2a;*-a^ + ar* + 5 by ic + 1. 
 
 36. x' + 2^-2x* + 2ii?-l by a; 4-1. 
 
 37. a'' — 2 a^c + 4 ac? — aa^ — 4 c^ic + 2 ca^ by a — a;. 
 
 38. a^ - 6^ + c^ + 3 a6c by a^ -|- 6^ _|. (.2 _^ ^5 _ ^^ + he. 
 
 39. a;" 4- ?/" by x-{-y to five terms of the quotient. 
 
 40. a^ — 6 a; + 5 by a^ — 2 a; + 1, using detached coefficients. 
 
 1+0+0+0+0-6+5 
 1-2 + 1 
 
 PROCESS 
 
 1-2 + 1 
 
 2- 
 
 -1+0 
 
 
 
 2- 
 
 -4 + 2 
 
 
 
 
 3-2 + 
 
 
 
 3-6-r.5 
 
 
 
 4- 
 
 -3- 
 
 -6 
 
 
 4- 
 
 -8 + 4 
 
 
 
 5- 
 
 -10 + 6 
 
 
 
 5- 
 
 -10 + 5 
 
 1+2+3+4+5 
 
 a:(* + 2a!* + 3ar^ + 4x + 5 
 
80 DIVISION 
 
 Divide, using detached eoefl&cients when convenient : 
 
 41. a;^ + 8a; + 7 by a^ + 2ic + l. 
 
 42. a*' -f 38 a +12 by a + 2. 
 
 43. m'' — 19 m — 6 by m-\-2. 
 
 44. m^ - 32 m^ - 4 m + 8 by m - 2. 
 
 45. a*' + 27a2-9a-10 by a^-3a + 6. 
 
 46. 21a;* -29a;3-8a,'2+6a; + 4 by 3ic-2. 
 
 47. 2a;*-lla^ + 16ic2-12a; + 9 by 2a;-3. 
 
 48. 30.T*-62a^ + 60a^-36a; + 8 by 5a;-2. 
 
 49. 21x*-SSo(? + ^Qx'-lVdx + b5\iy 9x-5. 
 
 50. x^ -2x'-a?~10x-3Q hy x-2. 
 
 51. «* — 4a;3+ 5a^ — 4a; + l by a;^ — ic + l. 
 
 52. x^-^-lOx' + lx + l^hj x^-2x-3. 
 
 53. 2a^ + 7a^- 27 a^-8a; + 16 by a^ + 5a;- 4. 
 
 54. 28a;* + 6a^ + 6a;2-6a;-2 by 4ic2 4-2a; + 2. 
 
 55. 7a^-6iK* + 2a;2_a,_2 by 6ar= + 5a; + 2. 
 
 56. 25x'-20a? + Zx* + Ux-&hj Sx'-%x + 2. 
 
 57. 3a;* + 7a^ + 6ar^ + 3a;-l by ar' + x + l. 
 
 58. 6a^-23a:3 + 30a.-2-18a; + 4 by 2i»2_5^_^2. 
 
 59. 24a;* + 32a^-16a^-25a;-4 by 6ar^-a;-4. 
 
 60. ar*-2a;* + T^a^ + |a^ + yV^'+f by a;-f. 
 
 61. ar^ - f a* + If a^ - l^or' + f a; - ^ by a; - |. 
 
 62. a*-iar> + |aJ«-^a^ + e^-H^ + A by ^-^x+h 
 
 63. |a^ + ^^ + z3-^a;2/2 by |a; + ^2/-|-a. 
 
 64. to five terms. 
 
 1 + a; 
 
 65. to five terms. 
 
 1 ~ X 
 
 66. a;'"-^ + 2/^"+^ by x"~^ + y"'*'\ 
 
DIVISION 
 
 81 
 
 SPECIAL CASES IN DIVISION 
 110. By actual division, 
 
 x-y 
 a? — 1^ 
 
 x-y 
 
 X* — y* 
 x-y 
 
 I x-y 
 
 = x-\-y, 
 
 = x^ + xy+y'. 
 
 = Qc^ + x'y + xy^ + y*. 
 
 = X* -\- x^y + x^y^ + xy^ + y*. 
 
 From the above we infer that the difference of the same powers 
 of two numbers is divisible by the difference of the numbers. 
 
 x-\-y 
 ^ — 'i^ 
 
 x + y 
 
 x^-y 
 x + y 
 
 = x-y. 
 
 = x^ — xy -{-y^, Rem., — 2 2/*. 
 
 = 3? — x'y + xy^ — ?/*. 
 
 = x* — !3?y + ci?y^ — xy^ ^y^, Rem., —2?/*. 
 
 From the above we infer that the difference of the same powers 
 of two numbers is divisible by the sum of the numbers only when 
 the powers are even. 
 
 {x' + y'' 
 
 ^x + y, Rem., 22/^ 
 
 = 2i?-\-xy + y^, Rem., 2 f. 
 
 = 21? + x^y + xy^ + 'f, Rem., 2 y*. 
 
 ^_ =x* + x'y + a?y'- + xy^ + y*, Rem., 2?/*. 
 
 From the above we infer that the sum of the same powers of 
 two numbers is not divisible by the difference of the numbers. 
 
 ALG. 6 
 
 3. 
 
 x — y 
 ^ + f 
 
 x-y 
 
 x^ + y' 
 
 x-y 
 
82 
 
 DIVISION 
 
 01? + y- 
 ^ -|— ,-r = ^ — y> Rem., 2 y\ 
 
 a^ + y^ 
 a^ + ?/^ 
 
 = a^ - .Ti/ + y''. 
 
 = u? — x^y -\- xif — y^, Rem., 2 ?/^ 
 
 = a;'' — ar'y 4- a;y — xy^ + r/*. 
 
 Observe that the sum of the same powers of two numbers is 
 divisible by the sum of the numbers only when the powers are odd. 
 
 111. Hence, when n is a positive integer. 
 Principles. — 1. x^ — y^ is always divisible by x — y. 
 
 2. x" — y" is divisible by x -\- y only when n is even. 
 
 3. xf -f 2/" is never divisible by x — y. 
 
 4. ic" + y" is divisible by x + y only when n is odd. 
 
 112. From § 110, the following law of signs may be readily 
 inferred : 
 
 When x — y is the divisor, the signs in the quotient are x>lus. 
 When x + y is the divisor, the signs in the quotient are alternately 
 plus and minus. 
 
 113. The following law of exponents may also be inferred : 
 When a;" ± ?/" is divided by x ±y, the quotient is homogeneous, 
 
 the exponent of x decreasing and that of y increasing by 1 in each 
 successive term. 
 
 114. Proofs of preceding principles. 
 
 Fbinciple 1 
 
 1st Rem., 
 2(1 Kern., 
 
 a;"- 
 
 -r 
 
 
 
 
 X"- 
 
 -oc"- 
 
 'y 
 
 
 
 
 x»- 
 
 'y- 
 
 yn 
 
 
 
 a!»- 
 
 'y- 
 
 Xn- 
 
 -2y2 
 
 x-y 
 
 + x»--y + 
 
 ji;n-2^2 _ yn 
 
 nth Rem., 
 
 Q^n-nyn _ yn 
 
 xOyn - yn -yn ^ yn = Q 
 
DIVISION 83 
 
 By dividing until several remainders are obtained, it is found that the 
 first term of the first remainder is a;»-'y ; of the second, x''-^ J of the third, 
 a;n-3y8 . of the fourth, x"-*j/* ; and consequently of the nth, x"-"?/". But 
 x«-" = x", which, § 103, equals 1. Therefore, the first term of the «th re- 
 mainder reduces to y". 
 
 Since the second term of the nth remainder is — j/", the entire nth re- 
 mainder is ?/» — 2/», or ; that is, there is no remainder, and the division is 
 exact. 
 
 Therefore, x" — y" is divisible hy x — y when x and y represent any two 
 numbers and n is any positive integer. 
 
 Principle 2 
 
 x" — y" 
 X" + x"-ij/ 
 
 1st Rem., — x^-iy — j/» 
 
 — x"-'j/ — x^-h/"^ 
 
 
 2d Rem., x"-2y2 _ ^n 
 
 x"-2j/2 + x^-Sy* 
 
 3d Rem., - x^-^y^ - y« 
 
 — x"~^^ — x"~*y* 
 
 4th Rem., x"-V - y" 
 
 Suggestion. — Since the second term of each remainder is negative, no 
 remainder can reduce to unless its first term is positive. Show for what 
 values of n such remainders reduce to when n is a positive integer. 
 
 Prove Principle 3. 
 
 Prove Principle 4. 
 
 Examples 
 
 Write by inspection the quotient of 
 1. 1.^^. 6. ^'-< 11. 
 
 12. 
 
 a? 
 
 -b^ 
 
 a 
 
 -b 
 
 mP 
 
 -f-n" 
 
 m 
 
 + n 
 
 ^ 
 
 -./ 
 
 X 
 
 + y 
 
 r' 
 
 -s' 
 
 r 
 
 — s 
 
 a> 
 
 + b' 
 
 m - 
 
 -n 
 
 n'- 
 
 1 
 
 n — 
 
 1 
 
 &- 
 
 d? 
 
 c — 
 
 d 
 
 1 + 
 
 a» 
 
 1 -\-a 
 
 0*- 
 
 1 
 
 13. 
 
 14. 
 
 a;2-9 
 
 a; + 3 
 
 a^-S 
 
 a-2 
 
 ar*-32 
 
 a;-2 
 
 c3 + 27 
 
 c + 3 
 
 a' 4- 128 
 
 10. ^^^ ^- 15. 
 
 a + 6 a;.-j-l a + 2 
 
84 DIVISION 
 
 16. Find five exact binomial divisors of a^ — «*. 
 
 Solution 
 
 «« — a* is divisible by a — a; (Prin. 1). 
 
 0(6 _ a« is divisible by a + x (Prin, 2). 
 
 Since a^ — x^ = (^a^y _ (x^)^, a^ — ofi may be regarded as the difference of 
 tv70 odd powers, and is, therefore, divisible by a? — x^ (Prin. 1). 
 
 Since a^ — ifi — {a^Y — (x3)2, a^ — x® may be regarded as the difference of 
 two squares, and is, therefore, divisible by a? — x^ (Prin. 1). 
 
 Since a^ — x^ =(a^y — (x^)^, a^ — x^ may be regarded as the difference 
 of two squares, and is, therefore, divisible by a^ + x^ (Prin. 2). 
 
 Therefore, the exact binomial divisors of a^ — x^ are a — x, a + x, a^ — x^, 
 a^ — x^, and a^ + x^. 
 
 17. Find an exact binomial divisor of a® + a^. 
 
 Solution 
 
 Since a^ + x^ =(0^)^ +(x^y, a^ + ofi may be regarded as the sum of the 
 cubes of a^ and x^, and is, therefore, divisible by o^ + x^ (Prin. 4). 
 
 Find exact binomial divisors : 
 
 18. a^ — ml 24. ic^ + a\ 30. a* - 6*, four. 
 
 19. a^-m^ 25. a}^ -\- b^'^. 31. a« - 1, five. 
 
 20. b^ + x^. 26. a}" + 6*. 32. a^ — 6^ six. 
 
 21. «^- a'. 27. ai2_,_5i2^ 33. a^o _ ?,io^ five. 
 
 22. c^-hn\ 28. a^ - 27. 34. a^® - 6^^ eight. 
 
 23. a«4-6^ 29. a*'-27. 35. a^^ _ 512^ nine. 
 
 Equations and Problems 
 115. 1. Find the value of a; in the equation 603 — 6^ = ca — c^. 
 
 Solution 
 
 bx — b^ = cx — c'. 
 Transposing, bx-cx = b^ - c^. 
 
 Collecting coefficients of x, (b — c)x = ft'-^ — c^. 
 
 52 _ g2 
 
 • Dividing by 6 - c, x = — = + c. 
 
DIVISION 85 
 
 2. Find the value of x in the equation x — o? = 2 —ax. 
 
 Solution 
 X — a^ = 2 — ax. 
 Transposing, ax + a; = a^ + 2, 
 
 Collecting coefficients of x, (a + l)x = o^ + 2. 
 
 Dividing by a + 1, x = ^!-±-? = a2 _ ^ + j ^ . 
 
 a+ 1 a+ 1 
 
 Solve the following equations : 
 
 3. 1 a — 10 = a? — ax-\-5x. 6. co; — c^ — d^ -f- da; = 0. 
 
 4. a; — 1 — c = ca; — c^ — c*. 7. a^ -- ax — 2 ab -\- hx + h^ = 0. 
 
 5. 2m^ — ma;-f wa; — 2w^ = 0. 8. 2%"^ -\-bn-{-x = n^ — nx — 2. 
 
 9. ri^a; — 3 7n?n^ -\- nx -\- Z w? + x = 0. 
 
 10. a^a; - a'' + ^a^ + Sa;- 5a + 10 = 0. 
 
 11. Zah- a^ -2hx = 2b'^ - ax. 
 
 12. 9 a^ 4- 4 ma; = - (3 aa; - 16 m^). 
 
 13. c?/-c^-2c3-2c2 = 2c-y + l. 
 
 14. ay — 2by + 3cy = a — 2b + 3c. 
 
 15. z + Gw* — 4n^ = l — 3n2!-)-2w — nl 
 
 16. X - 3b^ -192 b^c^- 4. ex + 16 c'x = 0. 
 
 17. 8&3-1862_576-2&a; + 7a; + 77 = 0. 
 
 Solve the following problems : 
 
 18. A drover, who had 5 times as many sheep as oxen and ^ as 
 many oxen as horses, sold all for $ 2300, — the horses at f 35 a 
 head, the oxen at f 25 a head, and the sheep at $ 4 a head. What 
 was the number of each ? 
 
 19. A man paid yearly a certain amount of money for taxes 
 and twice that amount for improvements, and received for rent 
 3 times as much as he paid out for improvements. If his net 
 gain per year was $ 300, what were his taxes per year ? 
 
 20. A owed B a certain sum of money and C twice as much. 
 D owed A 3 times as much as A owed B, and E owed A 5 times 
 the sum A owed B. A found that if he could settle with them all 
 he would have $ 5000. How much did he owe B and C ? 
 
86 DIVISION 
 
 21. After taking 3 times a number from 11 times the number 
 and adding to the remainder 7 times the number, the result was 
 12 less than 117. What was the number ? 
 
 22. A merchant failed in business, owing A 3 times as 
 much as B, C twice as much as A, and D as much as A and B. 
 If the entire debt to A, B, C, and D was f 28,000, how much did 
 he owe each ? 
 
 23. At a certain election there were three candidates for the 
 office of mayor. A received half as many votes as B and 4 times 
 as many as C. If the total vote lacked 25 votes of being 2300, 
 how many votes did each receive ? 
 
 24. Three boys together had 140 marbles. If the second boy 
 had twice as many as the first and half as many as the third, how 
 many had each ? 
 
 25. In a certain school of 600 students there were twice as 
 many Sophomores and 3 times as many Freshmen as Juniors, 
 and 40 more Seniors than Juniors. How many students were 
 there in each class? 
 
 26. Divide 25 into three parts such that the first is one third of 
 the second and 5 greater than the third. 
 
 27. A, B, and C divided $40 so that for every $2 A received, 
 B and C each received $ 3. What was the share of each ? 
 
 28. Divide $2200 among A, B, and C, so that B shall have 
 twice as much as A and $200 less than C. 
 
 29. Divide $ 351 among three persons so that for every dime 
 the first receives the second shall receive 25 cents and the third a 
 dollar. 
 
 30. A man gave equal amounts of money to a school and to a 
 library, and ^ the same amount to a hospital. If to all he gave 
 $ 28,000, what sum did he give each ? 
 
 31. When wheat was worth 85 cents a bushel, oats 35 cents a 
 bushel, and corn 60 cents a bushel, a man bought a quantity of 
 wheat, oats, and corn for $67. If he bought twice as many 
 bushels of oats as of wheat, and also three times as many bushels 
 of corn as of wheat, how many bushels of each did he buy ? 
 
REVIEW 
 
 87 
 
 REVIEW 
 116. Simplify; 
 
 1. a^ + 2 aVxy — 3 mn + 4 mn — 4 a^ — 5 a^/xy + 3 «^ + 4 a^/xy 
 
 ■ 2mn. 
 
 2. h 01? + 3 3?y -\- ^ xy^ — '^ — V^ + 6 ?/^ + a^— Vy — 5a?y — 7 a? 
 
 ■ 5xy^ + x^ + 2^x + a^y — 6 2/^ 4- V?/ — Va; — 2 a.*^?/ + 3 ajy'^ + y?. 
 
 3. (f a - 3 6c + i c - 7 6) - (f a + ^ 6c + ^ c + 3 6). 
 
 4. (a-ar^ — 4 ay + 4 6c + ax) — (6V — 4 6y — aa; + 6c). 
 
 5. V? -ix" -hx'^y ^X^ix'f -Vd^f ^hxy" -f). 
 
 6. |a-|a;-(fa-ia:)-(36--V-a;-|a) + ^a. 
 
 7. (a^ + 3 a^^/ + 3 ay^) (a' - 2 ay + y^- 
 
 8. (a^" + 2 a;»y» + y*") (a^" — 2 a;"?/" + /»). 
 
 9. {y^^^lxy^\xf){\7?-\xy^\y% 
 10. (.2a2-.8a + 1.6)(.la2+.4a+.8). 
 
 Expand : 
 
 11. (y-3)(2/ + 4). 21. 
 
 12. (y + 7)(y-8). 22. 
 
 13. (.v-l)(2/ + 2). 23. 
 
 14. (y-5)(y-9). 24. 
 
 15. (y+8)(?/-4). 25. 
 
 16. (m — a;) (m + x). 26. 
 
 17. (m4-a)(m — 6). 27. 
 
 18. (a; — m) (a; + w). 28. 
 
 19. (x2 _,_ j.^ (^ ^ 2). 29. 
 
 20. (x2 + 4) (ar^ - 3). 30. 
 
 31. (a - 6) (a + 6) (a== + 62). 
 
 32. (l-a;)(l+a;)(l + a^)(l + X*). 
 
 33. (1 — a;)(l +a;)(l— a;)(l + x). 
 
 34. (m + ^0 (w, -f w) i^n — w) (m — w) 
 
 (a; + m) (x + m). 
 
 (af' + ic^) (a; + 1). 
 
 (a;-l)(l+x). 
 
 (a; + 3i!/)(x-2y). 
 
 (a» + 6"')(a"-6'»). 
 
 {or + 6") (a" - 6"). 
 
 (a + 6 + c) (a + 6 — c). 
 
 {X -\- y ^ z){x - y ^ z). 
 
 (r + s ~ t) (s — t — r). 
 
 {m + n —p)(m — n -^ p). 
 
88 REVIEW 
 
 35. (a*-\-a^ + a^ + a + 1) (a - 1). 
 
 36. {a^ + a^y + xy- + f)(x-y). 
 
 37. (ic*-a;* + a^-a^ + a;-l)(a; + l). 
 
 38. (a^ + 2 a* + 4.a^ + 80^ + 16a + 32)(a - 2). 
 
 Square : 
 
 39. 2x-oy. 42. 100-5. 45. a + b + c + d. 
 
 40. x* — aa^. 43. w^" — m". 46. 2 a — 3 6 — 4 c. 
 
 41. 50 — 1. 44. 7x — b^y 47. a;"-i - ?/ - al 
 
 Expand : 
 
 48. (5a-4.v)(5a-3y). 51. (2 a^a; - 5 ft^y) (4 a^x - 3 6^)- 
 
 49. (6x~4y)(3x + 5y). 52. (6 amn + 5 j?) (6 amw - 3p). 
 
 50. (3 a; + a^) (3 a; 4- 62/). 53. (3a"+i-26»-i)(2a"+i- 3 6"-»). 
 
 54. (a; + 2/)(a;-2/)(a^ + y^(a^+2/')(aJ« + 2/«). 
 
 55. (m8 + 1) (m* + 1) (m^ + 1) (m + 1) (m - 1). 
 
 56. (16a;* + l)(4a:2 + l)(2a; + l)(2a;-l). 
 Divide : 
 
 57. ;r* - 2 a^ + 2 a^ + 12 a;2 - a; - 8 by a; + 1. 
 
 58. a;* - 4 a^ + 5 a.-2 - 4 a; + 1 by a;- - 3 a; + 1. 
 
 59.. a^-45a^ + 45a;*-18a;2_,_9^_l ^^ a^- 4ar' + 3a;- 1. 
 
 60. a'-12a^-a + 12 hy a^-2a' + 4:a-3. 
 
 61. 6«-10&2_56 + 4 by 63_262 + 36-l. 
 
 62. m^" - 6 w^ + 5 m - 2 by m* + 2 m^ - 3 m - 2. 
 
 63. a^-160aH127a3_i00a=^-20a+16 by a3-6a2+5a-4. 
 
 64. 610+29 6*-170 63-61 624.210 6-22 by 6^+2 6^-5 6-11. 
 
 65. 6a^-HffaV-ffa/ + |i/ by a^ + i a'^y - i a^/^ + i y^ 
 
 66. a^c—ab^+acd~ad^ -abc-\-b^-bcd+bd'-a(^-\-cb^-(^d-{-cd^ 
 by ac — 6^ + cd — d^ 
 
REVIEW 89 
 
 67. Divide 1 by 1 — a; to six terms. 
 
 Simplify : 
 
 68 . 4 a^ + 3 6^ - (2 a^ - 3 6^) - (7 6^ - 2 a^ - ( - a' - 6«). 
 
 69. a- - (b' _ c2) - (62 + c^ - a') ^{c'-b'-a^- (a" - 6^ - c*). 
 
 70. a? - {2xy - y") - {a? + xy - y"^ - x" -2xy - y^ + ^y\ 
 
 71. m + |2m-[7i + 3|?-(4p-3n)-57i + 2m]— 7p|. 
 
 72. X — d y + \2 z — {5 y — ^x — 1 z) — {x - y) — z\ — X. 
 
 73. 1 -{l-[a^-3-(2a;2-4)4-3r'^l-]_(a;2_4)|-l. 
 
 74 . a - (2 6 + 5 a) - 2 6 - [3 a - (6 6 - 5 a) - a] + 12 a. 
 
 75. x — \m—{x + {3m — 2x)-^5x—(2x + m)']—2x + m\. 
 
 76. x-55a;-[6a;-(7a;-8a;-9a;)-10a5]+lla;|4-9a;. 
 
 77. 3 -fc-[5- (2c- 7 -3c- 11) + 5c]- 6c -20|. 
 
 78. .T + 13y+[4a;-(2y-7a;)-32/]-(102/ + 4a;) + ^y|- 
 
 79. 1 _ I _ [_ (1 _ a;) - 1]- 1| - )x - (5 - 3 a;) - 7 ^ aJ^ 
 
 Collect the coefficients of x, y, and z : 
 
 80. ax+ ay + az — bx — by — bz. 
 
 81. cue — 2?/ + C2: + 6?/ — 12a;4- 42. 
 
 82. Smx — nx-{-by — y + Scz — 4:Z. 
 
 83. p?/ — y — 42 + 62 — a; + ?na; — wic — 2. 
 
 84. ex — 6?/ — 3 a2 4- a; — y — 4 2 + 2 — y. 
 
 85. 16ny — 16mx + ax + by + cx — 2y. 
 
 86. ma; + ny + az + 2ax — 2 my + 2 n2. 
 
 87. X — y — az -{- S mx -\-aby — x^ -\- y^ -\- z. 
 
 88. a^a; + 6^.v — 2 ax — 2 C2 + c22 + a; + y + «• 
 
 89. m^x — nhj + m^y — n^x — 2 mnx — 2 mwy — n''? 4- z- 
 
 90. 4 (ax — 6i/ + C2) — 2 (6x — a?/ — C2) — 2 (x — y + z). 
 
FACTORING 
 
 117. The numbers that multiplied together produce a given 
 number are called its Factors. 
 
 The factors of 12 a are 2, 2, 3, and a; or 4, 3, and a ; or 2, 6, and a ; 
 or 12 and a ; or 2 and 6a; or 4 a and 3, etc. 
 
 118. A number that has no integral factors except itself and 1 
 is called a Prime Number. 
 
 119. A number that has integral factors besides itself and 1 is 
 called a Composite Number. 
 
 120. A factor that is a prime number is called a Prime Factor, 
 
 121. The process of separating a number into its factors is 
 called Factoring. 
 
 122. To factor a monomial. 
 
 The factors of the numerical coefficient are found as in arith- 
 metic, but the factors of the literal part are evident. 
 
 Thus, in a^, the three factors are as evident as if they were written axay-a. 
 
 It is seldom necessary to resolve a monomial into its simplest 
 factors, but the following problem often occurs : 
 
 Given one factor of a monomial, to find the other. 
 
 Rule. — Divide the monomial by the given factor. 
 
 1. In each of the following, if xy is one factor, find the other: 
 6 or*?/, 15 x*y^, 2 a^?/^, d^y^h'^y^, — mnxy, — xy. 
 
 2. In each of the following, if abc is one factor, find the other: 
 a^hc, ah\ abc?, — a%^<?, — a^bc, — \ abc. 
 
 3. Find two equal positive factors of o^ ; of 9 aV ; of 64 m*. 
 
 4. Find two equal negative factors of 25 a;*; of 16 a*; of 9 a*. 
 
 90 
 
FA CTORING 91 
 
 123. To factor a polynomial whose terms have a common factor. 
 
 Examples 
 
 1. What are the factors of 3 a?xy — 6 aa?y + 9 axy^ ? 
 
 PROCESS Explanation. — By examining the terms 
 
 3 a^xv 6 ax^v 4- 9 axi/^ °^ ^^^ polynomial, it is seen that 3 axy is a 
 
 = 3 axy (a — 2x + 3v) ^^^*°^ '^^ ^"^^^^ *^^™* ^i^i'^^'S ^y *^i^ ^o™' 
 '^ mon factor, the other factor is found. 
 
 Hence, the factors are 3 axy, the monomial factor, and (a — 2x + 3y^, the 
 
 polynomial factor, since, by the Distributive Law for multiplication, 
 
 3 axy (a — 2x + 3y) = 3 a^xy — 6 az^y + 9 axt/^. 
 
 Find the factors of each of the following polynomials : 
 
 2. 5 x^ — 5 x^. 14. ac — be — cy — abc. 
 
 3. 80-^ + 20;*. 15. 3 a^//3 - 3 ar'/ + 12 a;?/. 
 
 4. 3x3-6a^2/- 16. 16 a^/Zc* - 24 a'ftV + 32 a^&V. 
 
 5 . 4 a^ — 6 a6. 1 7 . 60 m^n^i'^ — 45 mVr' + 90 m*n^i^. 
 
 6. 5m2-3m7i. 18. 12 a% - 18 a6y + 24 a^^y. 
 
 7. 3 if^?/^ — 3 xry^. 19. 14 a^mn^— 21 a^m^n^— 49 a*mn^. 
 
 8. 5m%-10mV. 20. 12 afy^z^ - 16 afy^z^ - 20 a^yh^ 
 
 9. la^^-Ga^ftl 21. 25 c^dar' + 35 c-^dV - 55 c^d^a^. 
 
 10. 5x^-10a^-5a;l 22. 51 xy^z' ~68 a^fz^ + 85 xYz*. 
 
 11. 3a*-2a'b + aVA 23. 52 a^ftV - 65 a362g2 _ 9^ ^2^2^ 
 
 12. x^2 + X-" + a;'" - x». 24. 44 aV/ 4- 66 aVy' + 88 aVy*. 
 
 13. 3 m^ - 12 mV+ 6 m?i*. 25. 84ary~36a^?/'-f 60a^?/*-48a;/. 
 
 124. To factor a polynomial whose terms may be grouped to show 
 a common polynomial factor. 
 
 Examples 
 1. Factor ax -\- ay + bx -\- by. 
 
 Solution 
 
 ax + ay + bx + by = a(x + y) + b(x + y) 
 = ia + b){x + y). 
 
92 
 
 FACTORING 
 
 2. Factor ax — ay — hx + by. 
 
 Solution 
 ax — ay — bx -{■ by 
 = a(x -y)- b{x - y) 
 = ia-h){x-y). 
 
 Observe that, when the first two terms are factored, (x — y) is found to 
 be the binomial factor. Since {x — y) is to be a factor of the other two 
 terms, the monomial factor is — 6, not -f h. 
 
 3. Factor ex + y — dy -\- cy — dx + x. 
 
 Solution 
 
 GK + y — dyA-cy — cbc-\-x 
 = ex — dx + x -\- cy — dy + y 
 = (c-d+l)x + (c-d+ l)y 
 = (c-d+l)(a; + 2/). 
 
 Arranging terms, 
 
 Factor the following ; 
 
 4. am — an + mx — nx. 
 
 5. bc — bd + cx — dx. 
 
 6. pq —px — rq+ rx. 
 
 7. ay — by — ab -f- b\ 
 
 8. a? — xy — bx + 5y. 
 
 9. b- — bc + ab — ac. 
 
 10. ic^ + x'y — aa; — ay. 
 
 11. c^ — 4c + ac — 4a. 
 
 12. 2 a; — ?/ + 4ar' — 2 a;?/, 
 
 13. 1 — m + n — mn. 
 
 14. 22> + ^ + 6p^ + 3pg 
 
 15. ar — rs — ab + bs. 
 
 28 
 29 
 
 16. a.-^ 4- a;^ + a; + 1. 
 
 17. _j^ + 2/'-3t/-3. 
 
 18. a;^ + af' 4- a;^?/ + y. 
 
 19. 2-2n-n'' + n\ 
 
 20. .^'^ — a; — a + aa;. 
 
 21. S2(?-15x + 10y-2a?y. 
 
 22. 12a''-8a&-3a* + 2a''6 
 
 23. 3 m^n — 9 mn^ + am — 3 an. 
 
 24. loab--^bh-3oab +21 be. 
 
 25. 16 ax- + 12 ay - 8 6x - 6 by. 
 
 26. aa."^— aaj— aa^y + ay+a;— 1. 
 
 27. a^y+a;— 3 t/^— 3 y— 41/— 4. 
 aa; — a — 6a; + 6 — ca; + c. 
 mx—nx—x—my+ny-\-y. 
 
 30. a^— a— a&+6— 2ac+2c. 
 
FACTORING 93 
 
 31 . mp^ — np^ + TTiq — nq + m — n. 
 
 32. aa^ — bixr — ax + bx+a — b. 
 
 33. 2 ma^ — Tix^ -{- n -\- 2 nx — 4^ mx — 2 m. 
 
 34. bx^ — b — xy — y-\- ya? — bx. 
 
 35. a^x — a^y — ay — y + x + ax. 
 
 36. 2-3& + 3a6-2a-|-4a2-6a26. 
 
 37. m^ -\- mn + mn + n^ -{■ m -\- n. 
 
 125. To factor a trinomial that is a perfect square. 
 
 1. Since (a + b)(a + b) = a^ + 2ab + b^, what are the factors of 
 a^ + 2ab + b^? How are they obtained from a^ + 2ab + b^? 
 
 2. Since (a — 6)(a — b)=a^ — 2ab-\- b^, what are the factors of 
 a^ — 2ab + b^? How are they obtained from a^ — 2 a6 + 6^ ? 
 
 3. What determines the sign that connects the terms of each 
 factor ? 
 
 126. One of the two equal factors of a number is called its 
 Square Root. 
 
 127. In a trinomial that is a perfect square one term is equal 
 to twice the product of the square roots of the other two terms. 
 
 25 a;2 — 20 xy + 4 y" is a perfect square, for twice the product of the square 
 root of 25 x^ and the square root of 4 y^ is 20 xy. 
 
 Rule. — Connect the square roots of the terms that are squares 
 with the sign of the other term, and indicate that the result is to 
 be taken twice as a factor. 
 
 From any expression that is to be factored, the monomial 
 factors should usually first be removed. 
 
 Thus, 2a3-4a2 + 2a = 2 a{a^ -2a + l) = 2a(a- 1)2. 
 
 Examples 
 Factor the following : 
 
 1. x^ + 2xy + y\ 5. x' + Gx-^Q. 
 
 2. p^-2pq^q\ 6. m''-8m.+ 16. 
 2. c' + 2cd+d?. 1. x^-2x + l. 
 
 4. m^-2mn-\-n\ ■ 8. a^- 16 a + 64. 
 
94 FA CTORING 
 
 9. ar' + 4a; + 4. 20. 9 + 42 6^ _|. 49 fts 
 
 10. 4-4a + a2. 21. 9m^-6m^ + l. 
 
 11. 4a-4a'' + a'. 22. 4 a.^ - 20 rci/ + 25. 
 
 12. ^x' + Qxy + ^y^ 23. 4 ar' + 12 a;?/z + 9 /2«. 
 
 13. 2m2 — 4mw + 2nl 24. 9aW — 6a'm + l. 
 
 14. 5ar + 30a; + 45. 25. 2 x + 20 a^a; + 50 a^x 
 
 15. 10 a^- 20 a; + 10. 26. 18 a^6 4- 60 aft^ + 50 6'. 
 
 16. 16p2_24i) + 9. 27. a''af-2ax'bf + bY. 
 
 17. 9a.-2-42a; + 49. 28. 25 a'™ - 60 a'-ft" + 36 &*•. 
 
 1 8. 1 + 4 & + 4 6^. 29. a^" - 2 x"y"z" + ^^"z^". 
 
 19. l-6a3-f9a«. 30. 81 a'b^c? + IS ab'c'd -\- b'c?d\ 
 
 When either or both of the squares are polynomials, the expres- 
 sion may be factored in a similar manner. 
 
 31. Factor a^ + 6x(x-y) + 9{x-7jy. 
 
 Solution 
 
 352 + 6 a; (x - ?/) + 9 (x - y)2 
 = [x + 3(x-y)][x + 3(x-y)] 
 = (x + 3x-3j/)(x + 3x-3y) 
 = (4x-3y)(4x-32/). 
 
 32. Factor (a - 6)^ + 2 (a - b)(b -c) + {b- cf. 
 
 Solution 
 
 (a - 6)2 + 2 (a - 6)(6 - c) + (& - c)2 
 = [(«-6) + (b-c)][(a-6) + (6-c)] 
 = (a-6 + 6-c)(a-6 + 6-c) 
 — (a — c)(a — c). 
 
 Factor : 
 
 33. a^ + 2x(x-y) + (x — yy. 
 
 34. a2-4a(a-l) + 4(a-l)^ 
 
 35. c2-6c(a-c) + 9(a-c/. 
 
 36. m^ + 2 m (m — n) + (wi — n)^- 
 
 37. 16-24(a-6)+9(a-&)2. 
 
 38. a^ + 25 (y3 - xf -\-10x(f- x). 
 39 14a(a;-t/) + (a;-2/)2 + 49a2. 
 40. 10m(m — 4) + 25m*+(m — 4)* 
 
FACTORING 96 
 
 41. {a-^hy-2{a + h){h-\-c) + {h + cf. 
 
 42. {a -2 xf + A{a -2x){2x -h) + ^(2x -b)\ 
 
 43. 16(a - xf + 32(a - x) {x + h) + \Q{x + hf. 
 
 44 . (a + 3 6)2 - 4 (a + 3 6) (3 6 - 2 c) + 4 (3 & - 2 c)". 
 
 45. {x" + X + iy +2{x + l){ci? + X + 1) + {x + If. 
 
 46. (a + & + c)2 + 2(a + 6 — c) (a + 6 + c) + (a + 6 - c)». 
 
 47. (ar' - x'")'' + 2 (a;^ _ ^^^ (a; + 1) + (a^ + 2 a; + 1). 
 
 128. To factor the difference of two squares. 
 
 1. Since {a-{-h){a — h) = a?~-h^, what are the factors of a? — V? 
 How do these two factors differ ? 
 
 2. Since (a^ + 6^ (a^ — If) = a* — 6^ what are the factors of 
 a* —b*? How do these two factors differ ? 
 
 Rule. — Find the square roots of the ttvo terms, mid make their 
 sum one factor and their difference the other. 
 
 Sometimes the factors of a number may themselves be factored. 
 
 Examples 
 
 1. Factor 6^ - y'^. 
 
 Solution. V^ — y'^ ={b -\- y){h — y). 
 
 2. Factor x^ — 1. 
 
 Solution. x^ — \ ={x + \){x — \). 
 
 3. Factor x* — 1. 
 
 Solution. x* - 1 = (a;2 + 1) (a;2 _ i) 
 
 =:(x'! + l)(a;+l)(a;-l). 
 
 Resolve into their simplest factors : 
 
 4. ar'-ml 9. 25 -c^. 14. a^^-b\ 
 
 5. a^-y\ 10. a* -49. 15. 4a^-25t/l 
 
 6. a^-ie. 11. x^-81. 16. 9 a^- 49 61 
 
 7. y^-a?. 12. a* -16. 17. a^x^-Ac?. 
 
 8. a2-9. 13. a^-6*. 18. m* - 16 n*. 
 
96 FACTORING 
 
 19. 25a:2_i 25 400a^-100t/l 31. 5a^-5. 
 
 20. 81m*-l. 26. 2a^^2h\ 32. 3a^-3a. 
 
 21. 36 a* -25. 27. 4m*-4&*. 33. x^-xf. 
 
 22. 12162_c2. 28. 3 a;* -3/. 34. 5a^y-5ay. 
 
 23. 400 a''- 81 yl 29. 5a;*-5y^". 35. a^" - y^\ 
 
 24. lOOa^" — 1. 30. 8x^^-Sf. 36. cc2"+i - a;?/^". 
 
 When either or both of the squares are polynomials, the 
 expression may be factored in a similar manner. 
 
 37. Factor 25 a^ - (3 a + 2 bf. 
 
 Solution 
 
 One factor is 5 o + (3 a + 2 6), and the other is 5 a - (3 a + 2 6). 
 
 5 a + (3 a + 2 6) = 5a + 3a + 2b = Sa + 2b = 2(4: a + b). 
 
 5 a - (S a + 2 b)= 6 a - S a - 2 b = 2 a - 2 b = 2(a - b). 
 
 .-. 26 a2 - (3 a + 2 6)2 = (5 a + 3 a + 2 6) (5 a - 3 a - 2 6) 
 
 = (8a + 2b)(2a-2b) 
 
 = 2(4a + 6).2(a-6) 
 
 = 4(4a + 6)(a-6). 
 Factor : 
 
 38. a^'-ia + bf. 42. 9 6^ _ (a - a;)* 
 
 39. b^-(2a+by. 43. 9 a^- (2 a -5)2. 
 
 40. a2_(5_^c)2 44^ a;* - (3 a^ - 2 2/)'^. 
 
 41. 4c2-(6 + c)2. 45. 49 a^- (5a -4 6)2. 
 
 46. Factor (3 a - 2 6)^- (2a -5 6)2. 
 
 Solution 
 (3a -2 6)2 -(2a -56)2 
 = [(3 a - 2 6) + (2 a - 5 6)][(3 a - 2 6) - (2 a - 5 6)] 
 = (3a-26 + 2a-56)(3a-26-2a + 56) 
 = (5a-7 6)(a + 36). 
 Factor : 
 
 47. (2a + 36)2-(a + 6)2. 49. (2 a; + 5)^ - (5 - 3 a;)^. 
 
 48. (5a -36)2 -(a -6)2. 50. (a- 2 6)2 - (a - 5)2. 
 
FACTORING 97 
 
 51. {2x-Zyf-{^y + zy. 54. {^x + &yf -{4.x-'6yf. 
 
 52. (5&-4c)2-(3a-2c)2. 55. {a? + x^f - (2 x + 2)\ 
 
 53. {A.x-^yy-{2x-^af. 56. (a + 6 + c)'-(a-6 -c)^ 
 
 57. Factor a^ + 4 — c^ — 4 a. 
 
 Solution 
 a2 + 4 - c2 - 4 a 
 Arranging terms, =a2-4a + 4-c2 
 
 = (a-2)2-c2 
 = (a-2 + c)(a-2-c). 
 
 58. Factor a2 + 62_c2_4_ 2a6 + 4c. 
 
 Solution 
 a2 ^. 52 _ c2 _ 4 - 2 a6 + 4 c 
 Arranging terms, = a^ - 2 a& + &^ - c^ + 4 c - 4 
 
 = (a2 - 2 a6 + 62) _ (c2 _ 4 c + 4) 
 = (a-6)2-(c-2)2 
 = (a - & + c - 2) (a - 6 - c + 2). 
 Factor : 
 
 59. a^ - 2 ax + ar* - n^. 67. (? - a" -h^ -2ab. 
 
 60. &2 4.2 6?/-j-2/2-w2. 68. h''-x^-y'^ + 2xy. 
 
 61. l—4:q + Aq^ — a^. 69. ic^ — x^ — ^f — 2xy. 
 
 62. ?-2 - 2 ra; + a:^ - 16 «^ 70. 9 c^ - ar' - 1/^ + 2 a;y. 
 
 63. 9a26-6a62+63_45c2 ^^^ a^ - a^aj - 4 ft^a; - 4 a&a;. 
 
 64. 4 a^c + 12 a&c + 9 6^0- 4 c». 72. bc^ - 9 a^b - b^ - 6 ab^. 
 
 65. 3a!2|/-12ay + 12y3_3^^ 73^ ab' - 4: a^ - 12 a'c - 9 ac\ 
 
 66. 4aw*-16aV+16a3-4an«. 74. 27 0^- 12a2 + 36 a& - 27 6'' 
 
 75. a'-2ab + b^-c? + 2cd-cP. 
 
 76. ar' - 2 a;?/ + ^='-m2 + 10 m — 25. 
 
 77. 4a^ + 9 — 12 a; + 10 mn — m^ — 25 w^. 
 
 78. ar' - a2 + ^'^ - 6^ + 2 xn/ - 2 a6. 
 
 ALO. — 7 
 
98 FACTORING 
 
 129. To factor a quadratic trinomial. 
 
 (a; + 3)(a;+ 5) = o^ + 8 a; + 15. 
 {x-S){x- 5) = x'- 8a; + 15. 
 (« + 1) (a; + 15) = a^ + 16 a; + 15. 
 (a; - 1) (x + 15) = a^ + 14 a; - 15. 
 (a;-3)(a;+ 5) = a^+ 2a; -15. 
 (a; + 3) (a;- 5) = ay' - 2 a; -15. 
 
 1. How may the first term of each factor be found from the 
 product ? 
 
 2. When the last term of the product has the sign +, how do 
 the signs of the last terms of the factors compare ? How, when 
 the last term of the product has the sign — ? 
 
 3. How does the coefficient of the middle term of the product 
 compare with the algebraic sum of the last terms of the factors ? 
 
 130. A trinomial of the form cn^ -{-bx + c, in which x^ is the 
 square of any number, c the product of two numbers, and b their 
 algebraic sum, b and c being either positive or negative, is called 
 a Quadratic Trinomial. 
 
 Rule. — Arrange the trinomial according to the descending jwivers 
 of one of the letters. 
 
 For the first term of each factor take the square root of the first 
 term of the trinomial; and for the second terms, such numbers that 
 their product is the third term of the trinomial, and their algebraic 
 sum multiplied by the first term of the factor loill be equal to the 
 second term. 
 
 Examples 
 
 1. Resolve a^ — 13 a; — 48 into two binomial factors. 
 
 Solution. — The first term of each factor is evidently x. 
 Since the product of the second terms of the two binomial factors is — 48, 
 the second terms must have opposite signs ; and since their algebraic sum, 
 
 — 13', is negative, the negative term must be numerically larger than the 
 positive term. 
 
 The two factors of — 48 whose sum is negative may be 1 and — 48, 2 and 
 
 — 24, .3 and — 16, 4 and — 12, or 6 and — 8. Since the algebraic sum of 3 
 and — 16 is — 13, 3 and — 16 are the factors of — 48 sought. 
 
 .-. x2 _ i3x _ 48 =(a; + 3)(a: - 16). 
 
FACTORING 99 
 
 2. Factor rn? + m — 72. 
 
 Solution. — Since + 9 and — 8 are the only two factors of — 72 whose 
 algebraic sum is + 1, the coefficient of the middle term, the required factors 
 are (m + 9) and (m — 8). 
 
 .-. r/i2 + m - 72 =(w + 9)(m - 8). 
 
 Separate the following into their simplest factors : 
 
 3. x^ + 7x + 12. 18. a^ + 5aa; + 6al 
 
 4. ;y2_7 2/ + 12. 19. o?-&ax-\-ba?. 
 
 5. p2_8^_^12. 20. y^ - i by- 12 b\ 
 
 6. ,-2 4.8r + 12. 21. /-3wy-28ji2. 
 
 7 . m^ + 5 m — 14. 22.2;^ — anz — 2 a-wl 
 
 8. a^-2a-15. 23. «* + 19 cx^-f 90 c^. 
 
 9. 6- + &-12. 24. »« + 12 aa^ -I- 20 a". 
 
 10. »^ + r-30. 25. a;^«-lliV + 24 6*. 
 
 11. c^ — c — 72. 26. 5 wa;''' — 55 wa: + 150 n. 
 
 12. c2_5c-14. 27. 3tt26aj2_3a2to-6a26. 
 
 13. x2_a;-110. 28. 12 m^a;^ - 60 m^fea; + 72 m^^^. 
 
 14. a2 + 9a-52. 29. 4 aa; + 2 aa;- - 48 a. 
 
 15. a2 + 8a-128. 30. 11 a-;«- 55 aa; + 66a;. 
 
 16. ar^- 25a; + 100. 31. 20 6a; + 10 6^ - 630 a;l 
 
 17. ar' + 12 a; -85. 32. x^ -{-(b — a)x ~ ab. 
 131. To factor trinomials of the form ax^ + bx -\- c. 
 
 Examples 
 1. Separate 3 a;^ + 11 a; + 6 into two binomial factors. 
 
 Solution. — Since 3 x^ is the product of the first terms of the factors, and 
 6 is the product of their last terms, and since the only factors, each contain- 
 ing X, that 3 x2 can have are 3 x and x, one factor is 3 x plus a factor of 6, and 
 the other is x plus the other factor of 6. By trial, it is found that 2 and 3 
 are the factors of 6, and that if 2 is added to 3 x and 3 to x, the middle term 
 of the product (3 x + 2) (x + 3) will be 11 x. 
 
 .-. 3x2+ llx + 6=:(3x + 2)(x + 3). 
 
100 FACTORING 
 
 2. Factor 9«2 + 30a; + 16. 
 
 Solution 
 
 9 a;2 + 30 a; + 16 
 = (3a;)2+10(3x)+ 16 
 Put m for 3 r, = m^ + 10 m + 16 
 
 § 130, = (m + 2) (m + 8) 
 
 Put 3 a; form, =(3x + 2)(3x + 8). 
 
 Suggestion. — When the coefficient of x^ is a square, and when the square 
 root of the coefficient of yfi is exactly contained in the coefficient of x, the 
 trinomial may be factored by the method illustrated above. 
 
 3. Factor 4a^ — 5x — 6. 
 
 Solution 
 
 4x2 - 5x -6 =(4x2 - 5x - 6) X * = 1^^^^^20x^^ 
 ^ ^4 4 
 
 _ (4x)2 - 5(4 x) - 24 ^ (4 X - 8)(4 X + 3) 
 4 ■ 4 
 
 = ^(^-Y^ + ^) =(x-2)(4x + 3). 
 
 Explanation. — Although the first term is a square, its square root is not 
 exactly contained in the second term. 
 
 But if such a trinomial is multiplied by the coefficient of x^, the resulting 
 trinomial will be one whose second term exactly contains the square root of 
 its first term. 
 
 Multiplying the given trinomial by 4, factoring as in example 2, and divid- 
 ing the result by 4, the factors of the given trinomial are (x— 2) and (4 x+3). 
 
 4. Factor 24 x^ + 14. x - 5. 
 
 Solution 
 
 24x2-|-14x-5=(24x2+ 14 x - 5) x - = ^M^Jl^I^^iJO 
 
 6 6 
 
 _ (12 x)2 + 7(12 X) - 30 _ (12 X + 10) (12 x - 3) 
 6 6 
 
 = ^(^^ + t^-^i^^-l) =(6x + 5)(4x-l). 
 
 A X o 
 
 Suggestion. — When the first term is not a square, it may always be made 
 & square whose square root will be exactly contained in the second term by 
 multiplying the trinomial by the coefficient of x^. 
 
FACfORmG 101 
 
 After factoring, the result should be divided by the coefficient of x^. 
 Frequently the multiplier can be taken smaller than the coefficient of Z'^, 
 as in the above example. 
 
 Separate into their simplest factors : 
 
 5. 2x' + x-15. 17. 9 a?* - 10 or* - 16. 
 
 6. 9 ^^-42 a; + 40. 18. 27&*-3 62_l4. 
 
 7. 5a^ + 1.3a; + 6. 19. 10 a;« - 2 a.^ - 44. 
 
 8. ^x^-llx + lQ. 20. 2a? + 5xy + 2j^. 
 
 9. 25ar' + 15a; + 2. 21. 20^= + 3a;^ - 2/. 
 
 10. 16a^ + 20a;-66. 22. 3 a^ - 10 a;?/ + 3 j/^. 
 
 11. 36 x-2 - 48 a; - 20. 23. 6a^-lla;-35. 
 
 12. 9a;- + 43a;- 10. 24. 6 a;^ _ ^3 3. _,_ g 
 
 13. 25a;2^25a;-24. 25. 15 a;^ - 14 a; - 8. 
 
 14. 49a;2-42a;-55. 26. Wx^ + llx-i:. 
 
 15. 16x-2 + 50a;-21. 27. 21 a^- a- 10. 
 
 16. 4a.-2-4a;-35. 28. 18 ar^ - 3 a; - 36. 
 
 132. To factor the sum of the same odd powers of two numbers. 
 
 Examples 
 
 1. Factor dJ + h^. 
 
 Solution 
 
 §§ 111-113, a^ + 6^ =(a + 6) (06 - a^b + a^h"^ - a^h^ + a^ft* - ah^ + b^). 
 
 2. Factor m'' + 32ar*. 
 
 SOLDTION 
 
 m^ + 32 a;5 = ms + (2 a;)^ 
 §§ 111-113, = (m + 2 a;)(m* - 2 mH + 4 rrfir? - 8 mx^ + 16 x*). 
 
 3. Factor »* + /. 
 
 Solution 
 
 X« + 2/«=(x2)8+(y2)8 
 §§ 111-113, = (X2 + 2/2) (X* - X2y2 + 2/4). 
 
102 FACTORING 
 
 Factor the following : 
 
 4. m^ + n^. 7. ar* + l. 10. a" + 32. 
 
 5. a? + x\ 8. r^ + s^ 11. p^ + 21. 
 
 6. a^ + ft'*. 9. x>-{-y^. 12. a;" + 128. 
 
 133. To factor the difference of the same odd powers of two 
 numbers. 
 
 Examples 
 
 1. Factor a^ — &^ 
 
 Solution 
 
 §§ 111-113, a' -W ={a- b)(a^ + a^b + a*b^ + a^b^ + a'^b* + a¥ + b^). 
 
 2. Factor m^ - 32 rc^. 
 
 Solution 
 
 §§ 111-113, wi5 - 32 x8 = m5 - (2 «;)6 
 
 = (m - 2 X) (m* + 2 ??i3a; + 4 ,^23:2 + 8 nix^ + 16 x*). 
 
 3. Factor a'" — 6^ 
 
 Solution 
 
 §§ 111-113, alO - 65 =(a2)5_ 65=,(a2 _ ft)(«8 + ^65 + ^452 + ^263 + 64), 
 
 Factor the following : 
 
 4. a'-6^ 7. 1/-8. 10. ar^ -?/'". 
 
 5. a^-6''. 8. a^-32. 11. a.-« - ?/3 
 
 6. ?/^-al 9. a^-l. 12. ar'-243. 
 
 134. To factor the difference of the same even powers of two 
 numbers. 
 
 Examples 
 1. Factor a^ — b^ 
 
 First Solution 
 
 §§ 111-113, a6 - &6 =(a - b)(a^ + a^b + a^b^ + a'^b^ + ab* + b^) 
 = (a-b) {cfi + aH)^ + a*6 + a6< + a%'^ + b^) 
 = {a- b)\_d\a^ 4- fts) + ab{a^ + &3)+ 62(^3 + js)] 
 = (a - 6) (a2 + aft + 62) (a8 + 63) 
 
 § 132, = (a - 6) (a2 + a6 + 62) (« + 6) {cfi - a6 + 62). 
 
FACTORING 103 
 
 Second Solution 
 § 128, a6 - 66 = (a3)2 _ (63)2 ^ (^ta ^ ^3) (^8 _ ^S) 
 
 §§ 132, 133, = (a + 6) (a^ - a6 + &2) (a _ ft) (a2 + a& + ft^)- 
 
 In each of the above solutions a^ — h^ is first separated into two factors. 
 Thus, a^ — b^ is divisible by a — 6 ; also by a^— 6^, since a** — 66=(a')2— (ft*)2. 
 
 When the even powers are regarded as squares, as in the second solution, 
 the process may be regarded && factoring the difference of two squares. 
 
 Separate the following into their simplest factors : 
 
 2. 
 
 x" — /. 
 
 5. x*-16. 
 
 8. 
 
 l-6«. 
 
 3. 
 
 x' - 1. 
 
 6. a;* -81. 
 
 9. 
 
 64-/. 
 
 4. 
 
 a« - &«. 
 
 7. a' -625. 
 
 10. 
 
 1-^8. 
 
 Fac 
 
 itor the following by previous principles : 
 
 
 
 11. 
 
 y-q\ 
 
 17. ?'«-s«. 
 
 23. 
 
 1+a^. 
 
 12. 
 
 p^-h<f. 
 
 18. / + s«. 
 
 24. 
 
 X + 2C^. 
 
 13. 
 
 p" - ^. 
 
 19. m^ — n^ 
 
 25. 
 
 32 n - n« 
 
 14. 
 
 r* - s*. 
 
 20. m' + w^ 
 
 26. 
 
 m^ — a^. 
 
 16. 
 
 r'-^. 
 
 21. m^ — n^ 
 
 27. 
 
 6^ - a^ft*. 
 
 16. 
 
 1^ + ^. 
 
 22. a«-l. 
 
 28. 
 
 125 + a\ 
 
 135. To factor by the Factor Theorem. 
 
 1. If 5 aj = 0, what must be the value of 05 ? 
 
 2. If 5 (a; — 3) = 0, which factor reduces the first member to 
 zero ; that is, what vabie of x reduces it to zero ? 
 
 3. Since the expression 5 (a;— 3) reduces to zero when a; — 3 
 reduces to zero, and a; — 3 is reduced to zero by substituting 3 
 for X, what value substituted for x will reduce to zero every expres- 
 sion of which a; — 3 is a factor ? 
 
 4. How, then, may it be discovered whether a; — 3 is a factor 
 of a given expression containing x ? 
 
 5. What value of x will reduce the expression a.*' — 8 to zero ? 
 What factor, then, has a^ — 8 ? 
 
 6. What value of x will reduce the expression a^ — 8 a; + 7 to 
 zero ? What factor, then, has a^-^ — 8 a; 4- 7 ? 
 
104 FA CTORING 
 
 7. When does a rational integral expression containing x have 
 the factor x-1? x-2? x-S? x-a? 
 
 136. Factor Theorem. — If a rational integral expression containing 
 X redvbces to zero when a is substituted for x, it is exactly divisible 
 by x — a. 
 
 Demonstration. — Let D represent any rational integral expression con- 
 taining X, and let D reduce to zero when a is substituted for x. 
 
 It is to be proved tliat D is exactly divisible by x — a. 
 
 Suppose that the dividend D is divided by x — a until the remainder does 
 not contain x. Denote the remainder by B and the quotient by Q. 
 
 Then, D = Q(x - a)+ B. (1) 
 
 But, since, by supposition, D reduces to zero when x = a; that is, when 
 X — a = 0, equation (1) becomes 
 
 = 0+ R; 
 whence, ^ = 0. 
 
 That is, the remainder is zero, and the division is exact. 
 
 Note, a is the known number substituted for x, and it may be either 
 positive or negative. 
 
 Examples 
 
 1. Factor a;' — 8 x^ 4- 17 ic — 10 by the factor theorem. 
 
 Solution 
 
 jc3 _ 8 a;2 + 17 X - 10 
 
 = (a;-l)(x2_7a;+10) 
 
 = (a;-l)(a;-2)(x-5). 
 
 Since, when 1 is substituted for x, x^ — Sx^ + 17x — 10 reduces to 0, 
 X — 1 is a factor of the expression. Dividing by x — 1, the other factor is 
 x^ — 7 X + 10, whose factors, § 130, are (x — 2) and (x — 5). 
 
 2. Factor a^ + a^ — 16a^ — 4a;-|-48 by the factor theorem. 
 
 Solution 
 
 X* + x3 - 16 x2 - 4 X + 48 
 = (X - 2)(x8 + 3 x2 - 10 X - 24) 
 = (x - 2) (x + 2)(x2 + X - 12) 
 = (x-2)(x4-2)(x-3)(x + 4). 
 
 Since, when 2 is substituted for x, the expression reduces to 0, x — 2 is 
 a factor of the expression. Dividing by x — 2, the other factor is x^ + 3 x^ 
 — lOx - 24. Since, when — 2 is substituted for x, x^ + 3x* — lOx — 24 
 reduces to 0, x + 2 is a factor of x' + 3 x^ — 10 x — 24. 
 
FACTORING 105 
 
 Dividing by x + 2, the other factor is oc^ + x - 12, a quadratic trinomial 
 whose factors, § 130, are (a; - 3) and (x + 4). Hence, the factors of the 
 given expression are (« - 2), (x + 2), (x - 3), and (x + 4). 
 
 Factor the following polynomials by the factor theorem : 
 
 3. x'-dlx + dO. 21. y?-lx + Q. 
 
 4. 4a;2-7a; + 3. 22. a^- 19 a; + 30. 
 
 5. 26a^-10a;-16. 23. a.-^ - 67 a; - 126. 
 
 6. 48 a^ - 31 a; - 17. 24. a^- 39 a; -70. 
 
 7. 36 a^ - 61 a; + 25. 25. a'' + 4a2- 11a -30. 
 
 8. a^- 9x2 + 23a; -15. 26. a'^ + 9a2 + 26a + 24. 
 
 9. 3^-13x2 + 47 a; -35. 27. m'^ - 6m2 - m + 30. 
 
 10. a^-14a:2 + 35a;-22. 28. 6^ - 5 6^ -29& + 105. 
 
 11. 0^-4^2 -7a; + 10. 29. a'^ + lOa^ - 17a- 66. 
 
 12. a;3-6a;2-9a; + 14. 30. m=' + 7m2 + 2m - 40. 
 
 13. a;^ - 12 a;2 ^ 41 ^ _ 30. 31. &3_^ 16 6^ + 736 + 90, 
 
 14. .r' - 11 a;' + 31 a; - 21. 32. «» + 12 11^ + 41 n + 42. 
 
 15. ar^-10a;2^29a;-20. 33. a;* - 15 ar' + 10 a; + 24. 
 
 16. a;3- 16 3^^ + 71 a; -56. 34. a;* - 25 a;^ ^ 60 ^. ._ 36 
 
 17. a;' -57 a; + 56. 35. a;* + 13 ar^ - 54 a; + 40. 
 
 18. a;^- 21 a; + 20. 36. a;* + 22 a;^ + 27 a; - 50. 
 
 19. ar'- 31 a; -30. 37. x^ - ^ x^y"" - 4: xf + 12 1/. 
 
 20. ar'- 13 a; + 12. 38. x* -'da?y'' + 12xif - 4.y^. 
 
 39. a;*-a;»-7ar' + a; + 6. 
 
 40. a;^ - 9 a;3 + 213^^ + 0; -30. 
 
 41. a;* + 8 ar^ + 14 3^^-8 a; -15. 
 
 42. a;^-4a;^ + 19a;--28a; + 12. 
 
 43. a;«-18a^ + 30.'»2-19a; + 30. 
 
 44. a;«-10a;* + 40a;3_30ar'+80a;-32. 
 
106 FACTORING 
 
 SPECIAL APPLICATIONS AND DEVICES 
 
 Examples 
 
 137. 1. Factor 
 
 a^ + W + c^ + d^ + 2ah -2 ac+2ad -2bc + 2bd -2cd. 
 
 Solution. — Since the polynomial consists of the squares of four inirabers 
 together with twice the product of each of them by each succeeding number, 
 the polynomial is the square of the sum of four numbers, § 95, and may be 
 separated into two equal factors containing a, 6, c, and d with proper signs. 
 
 Since — 2 ac, — 2 be, and —2 cd, the products that contain c, are negative, 
 while 2 ab, 2 ad, and 2 bd, the products that do not contain c, are positive, 
 it is evident that the sign of c is the opposite of that of a, b, and d. 
 
 Therefore, the factors are either 
 
 {a + b — c + d) (a + b — c + d), 
 of (— a — b + c — d)(— a — b + c — d). 
 
 Factor the following : 
 
 2. 9x^ + 4:y- + 25z^-12xy + 30xz-20yz. 
 
 3. 25 m^ -\- 36 n^ •{- p^ — 60 mn — 10 7np + 12 np. 
 
 4. a" -^16 x' + S6 if -Sax" + 12 ay -4.8 x^y. 
 
 5. x' + ia^ + b^ -\- y^ -\- Aax — 2 bx + 2 xy ~ 4:ab + 4:ay - 2by. 
 
 6. m^ + 4 ?i^ + a^ + 9 — 4 mn — 2 am + 6 m + 4 an — 12 )i — 6 a. 
 
 138. The principle by which the difference of two squares is 
 factored has many special applications. 
 
 1. Factor a* + a^b^ + b\ 
 
 Solution. — Since «■* + a^b"^ + b* lacks + a^^a of being a perfect square, 
 and since the value of the polynomial will not be changed by adding a'^6* 
 and also subtracting a^b'^, the polynomial may be written 
 
 a* + 2 a262 + ^4 _ ^52^2^ 
 which is the difference of two squares. 
 
 a* + a'^b^ + 6* = a* + 2 a%'^ + b* - a%^ 
 = {a^ + b^)2-(aby 
 = (a2 + ab + b-^) (a^ - ab + ¥). 
 
 2. Factor 4 a* -13 a- + 9. 
 
 Solution. 4 a* - 13 a'^ + 9 - 4 a* - 12 a"^ + 9 - cfi 
 
 = (2 a2 - 3)2 - «2 
 = (2 a- +. a - 3) (2 a- - a - 3). 
 
FACTORING 107 - 
 
 3. Factor a* + 4. 
 
 Solution. a* + 4 = a* + 4 a^ + 4 - 4 a^ 
 
 = (a^ + 2)2 - (2 ay- 
 = (a2 + 2a + 2)(a2-2a + 2). 
 
 Factor the following : 
 
 4. X* -\- x'y'' + y\ 11. a;* + a^ + l. 
 
 5. a« + a*6* + &^. 12. 7t« + ?i* + l. 
 
 6. p*+pY + <1*- 13. 16a;* + 4a.V + ?/^. 
 
 7. 9 a^ + 20 ar'/ + 16 2/*. 14. a^6* - 21 a^d^ + 36. 
 
 8. 4 a* + 11 a^fe^ + 9 6*. 15. c* + c^d V + c? V. 
 
 9. 16a^-17aV + a;^ 16. 25 a^ - 14 a^ft* + 61 
 
 10. 25 a;* -29 0^2/2 + 4?/*. 17. 9 a* + 26 a^ft^ + 25 5*. ' 
 
 18. &^ + 64. 21. a* + 324. 24. x^ + Uy\ 
 
 19. a* + 4 6*. 22. a«-16. 25. 4 a* + 81. 
 
 20. m^ + 4. 23. nv" + 4 mn*. 26. a^/ + 4 xy^. 
 
 139. Many polynomials may be written as quadratic trinomials 
 in which o? and x are replaced by polynomials. 
 
 1. Factor 9^^ ^ 4^2 ^ 1222 + 21 icz + 14?/z + i2xy. 
 
 Solution. 9 x^ + 4 2/2 + 12 2;2 + 2I a;g + 14 yz + 12 xy 
 
 = (9a;2 + 12x?/ + 4j/2) + (21a;2 + 14t/0)+1222 
 = (3 a; + 2 j/)2 + 7 2(3 X + 2 y) + 4 2 . 3 « 
 
 §130, =(.Tx + 2j/ + 4«)(3x + 22/ + 3«). 
 
 Factor the following : 
 
 2. a2 + 2 a& + &2 + 8 ac + 86c + 15 c2. 
 
 3. 0^2 _ 6 a;?/ + 9 / + 6 a;z — 18 2/2 + 5 z^. 
 
 4. m? + n^ — 2mn -\-l mp —1 np — 30 p*. 
 
 5. 16 n^ + 55 - 64 w - 16 m + m2 + 8 wn. 
 
 6. 9m* + fc2 _ 30 + 39m'' + 13 A; + 6 m?k. 
 
 7. 25a2 + 2/2 + 10x2 + 10a?/-35ax-7a^. 
 
 8. 4 a;2 _|_ ^2 _ g 2,2 _ 4 ^.^^ 4. 2 a;2! — 2/2. 
 
 9. a2 + 62 + c2 + 2a6 + 2ac + 26c + 5a + 56 + 5c f 6. 
 
108 FACTORING 
 
 REVIEW OP FACTORING 
 140. Factor the following : 
 
 1. ire-n\ 11. p* + 4:. 21. 4a;*-4aj. 
 
 2. x^-l. 12. 1 + 0^2. 22. 7?/* -175. 
 
 3. y*-l. 13. y-a*y. 23. 8-27aV. 
 
 4. 1-0^. 14. x'y-f. 24. 32a; -2a^. 
 
 5. ajio-l. 15. a^3_^5i2 25. 6 &* + 24. 
 
 6. x^-1. 16. a* -256. 26. a* + 27 a-. 
 
 7. l-a^. 17. 3a3-3a. 27. ft^ - 196. 
 
 8. 1-6*. 18. 64-2?/^. 28. 450 -2a2. 
 
 9. a-dJ. 19. 7?i^-7n. 29. 4m3 + .004. 
 10. H' + h. 20. a^-9a. 30. 125-8iB«. 
 
 31. ic2_25a; + 100. 45. y'^-xy-42x'. 
 
 32. x'-xy-lS^yK 46. ar* - aa; - 72 a^. 
 
 33. aa^-3aa; — 4a. 47. n^—an — ^Oo?. 
 
 34. ar'' + 5a^-6a;. 48. a%'' -\- ah - 5%. 
 
 35. 3ar' + 30.^ + 27. 49. 10 a2c + 33 ac - 7c. 
 
 36. 128 a^- 250 a^ 50. 60w/-61 wy- 56n. 
 
 37. 5a;i« + 10af-15. 51. 25 a^ + 60 a^ + 36 1/^ 
 
 38. 6 .^2 - 19 a; + 15. 52. Q ay? + 5 axy - (^ ay\ 
 
 39. a-2" + 2af3/^ + /^. 53. 169 a^ - 26 aaf + aV. 
 
 40. 7:i?-nxy-Mf. 54. aV + a^&V+ftl 
 
 41. /-25?/a; + 136a«. 55. 16 a;* + 4 a^i/^ + y*. 
 
 42. 9ar^-24a;?/ + 16?/2. 56. 6*0-13 6^0 + 420. 
 
 43. 289 a^- 34 an/ + 2/2 57 2 a'' - 6 a6 - 140 6=*. 
 
 44. 3 6a^ + 6a;2/ - 10 6?/^. 58. m«» - 21 mn^ + 80 n». 
 
FACTORING 109 
 
 59. 17 0^ + 25.^-18. 74. x' + x'y-^lxif-lObf. 
 
 60. 5x^-2Qxy + by\ 75. a? -ex -\-2dx-2cd. 
 
 61. y2 _j_ 16 ay - 36 a*. 76. o^y + ^x^y -^Ixy — Idy. 
 
 62. 8a2-21a6-962. 77. a^- 3aa; + 4te-12a&. 
 
 63. 60a^ + 8aa;-3a*. 78. oar' - 9 ar= + 26 aa; - 24 a. 
 
 64. 30 ar' - 37 a; - 77. 79. 12aa;-8&a;-9ai/ + 6 6y. 
 
 65. 2a^ + 28a,-2 + 66a;. 80. 25 a^ - 9 y^ _ 24 ?/2 - 16 2^. 
 
 66. a2 + 62_c2_2a&. 81. x'-z'' + y''-a''-2xy + 2az. 
 
 67. aar' + 10ax-39a. 82. 2h'^m-3aW-\-2hmx-^abx. 
 
 68. n* + wW6* + a*6» 83. 0^+1^+0^-2 ah -2 ac-\-2 he. 
 
 69. aV + aV + al 84. a^t/ + 14 a^y + 43 a^ + 30 y. 
 
 70. a2-16a-17. 85. o^y - 15 ar'?/ + 38 a^ - 24 y. 
 
 71 . a^x^ — 4 aaj + 3. 86. aha? + 3 aha? — ahx — 3 ah. 
 
 72. &8 + 6y + 2/^ 87. 3&ma; + 2&m — 3ana;— 2aM. 
 
 73. a;^-2a;« + a;. 88. 20 aa^ - 28 aa^+ 5 a^x- 7 al 
 
 89. x' + ^f + 26z^-Qxy-Wxz + ^0yz. 
 
 90. 9ar' + 2/^ + 16z2_6a;2/-8.V2 + 242a;. 
 
 91. a?yh^ + a-h^ + l + 2ahmz + 2xyz + 2ah. 
 
 92. a^"^ + h^(? + <?d? - 2 ahh + 2 ahcd -- 2 he'd. 
 
 93. a^ + nV + w» + 27iV + 2nV + 2wV. 
 
 94. a^Wa? - a^ft* - 6V + h"" - a^a? + a" + a? - 1. 
 
 95. (aH-6)«-l. 100. 3a^ + 96a;. 
 
 96. a^- 2 0^ + 1. 101. (a -2)'^+ (a -1)1 
 
 97. &3_452_,_8. 102. 12a^ + 3a^-8a;-2. 
 
 98. a^-10ar' + 125. 103. 2 a:^ _,_ ^q a; + aa; + 5 a. 
 
 99. 8a;*-6ar'-35. 104. a.-^ + 5 ar^ - 29 a; - 105. 
 
 105. 7n?n^ + a^h^ + lM + 2hmv? + 2ah''n + 2abmn. 
 
110 FACTORING 
 
 106. a* + a% + a^h"" + arb^ + ah* + &^ 
 
 107. a262_4a6a;-4cc + 2a& + 4a^. 
 
 108. (a+6)='(a;-y)-(a + &)(a^-y2). 
 
 109. 1 — a^ + abx' + ^a;'' — 6a; — a&. 
 
 110. 31? — 2(? -{• x^y — xy -\- 2(?y — xy^. 
 
 111. a^"-2 + 6y + 2 a;"-i&v. 
 
 112. :i? + Ws^ + 15x + 12b. 
 
 113. 4 (a6 -I- cdf - (a^ -|- 52 _ ^2 _ ^2^2^ 
 
 114. x^ — a^. 125. a;^ + 4a:. 
 
 115. {a^ + ¥-cy-4.o?b\ 126. x^ - ar^ - a;* + a;3_ 
 
 116. a*62 + 0,26-12. 127. (a + hy-(h-c)\ 
 
 117. x^ — xy — x^y + y^. 128. 3 «&(« + &) + a'' + 6^ 
 
 118. .T* - 4 x-/ f 2 x^ - 16 ia''. 129. (x + ^)3 + (a; - 2/)^ 
 
 119. a*-h*-{a + h){a-h). 130. a^' - (a + Syi 
 
 120. x^- 6^2 + 12 a; -8. 131. x'-l\^xY + y*- 
 
 121. 1000 x^ - 27 ?/3. 132. vv" + m- - mn - myi\ 
 
 122. (a + a;)* — a;^ 133. (a? — y^ — (a? — xyf. 
 
 123. l+(a;+l)3. 134. x^ - f - 3 a^y^x- - y^). 
 
 124. ab - bx" + x^y'" — ay"'. 135. (x^+6x+9y-(x^+ox+6y. 
 
 136. ar^ + (a + 6 — c)a^ + (0,6 — ac — 6c) .t — a6c. 
 
 137. Factor 32 — a;^ by the factor theorem. 
 
 138. Factor 16 + 5 x — 11 ar' by the factor theorem. 
 
 139. If n is odd, factor a" — «" by the factor theorem. 
 
 140. If n. is odd, factor x" + ?•" by the factor theorem. 
 
 141. Factor a? — 6 bx^ + 12 b'\v — 8 6^ by the factor theorem. 
 
 142. Discover by the factor theorem for what values of n, 
 between 1 and 20, x" + «" has no binomial factors. 
 
FACTORING 111 
 
 EQUATIONS SOLVED BY FACTORING 
 141. 1. Find the value of a; in o^ + 1 = 10. 
 
 a^ + 1 = 10 
 
 FIRST PROCESS EXPLANATION. — Transposing the 
 
 known term 1 to the second member, 
 the first member contains the second 
 x* = 9 power, only, of the unknown number. 
 
 X'X =iZ '3 .*. a; = 3 Separating each member into two equal 
 
 factors, 
 
 a.a: = 3.3 ora!;-x = — 3- — 3. 
 
 or a; • a; = —3 • —3 .-. x = —3 
 
 .-. a; = ± 3 
 
 Since, if a; = 3, x • a; = 3 • 3, and if 
 
 a; = — 3, a;-x = — 3-— 3, the value of x that makes x'^ = 9, or that makes 
 
 x2 4- 1 = 10, is either + 3 or - 3 ; that is, x = ± 3. 
 
 Find the values of x in the following equations : 
 
 2. a^ + 3 = 28. 7. a^- 24 = 120. 
 
 3. ar^ + l = 50. 8. ar^ + ll = 180. 
 
 4. x2-5 = 59. 9. ar^-ll = 110. 
 
 5. a^-7 = 29. 10. a^-62=a2-2a6. 
 
 6. a:^ + 3 = 84. 11. ar* — 4 w^ = m'^ — 4mw. 
 
 12. Find the value of a; in a^ + 1 = 10. 
 
 Explanation. — The first process 
 SECOND PROCESS is given in example 1. 
 
 o , H H rv In the second process, all terms 
 
 are brought to the first member, 
 ^ — 9 = U which is factored as the difference of 
 
 /jp _ 3"\ (a; -f 3) = the squares of two numbers. 
 
 n rx ^ o Since the product of the two fac- 
 
 ••. a; — 3 = 0, whence x = 6 . . ^ * *, • i +^ n 
 
 ' tors IS 0, one of them is equal to 0. 
 
 or a; + 3 = 0, whence a; = —3 This gives x-3 = or x + 3 = 0; 
 • a; = + 3 whence x=:3 or x = — 3j that is, 
 
 X = ± 3. 
 
 Solve the following equations ; 
 
 13. x2 + 35 = 39. 16. a^-31*=0. 
 
 14. a^-50 = 50. 17. a^- 4 6^ = 0. 
 
 15. a^ + 90 = 91. 18. ar^-9w2 = 0. 
 
112 FACTORING 
 
 19. or' -21 =4. 24. .32-0^ = 28. 
 
 20. ar^-56 = 8. 25. 65 - a:^ ^ jg 
 
 21. a^-3a2 = 6a*. 26. 40^! - 862= Sd^. 
 
 22. x'-\-bh* = Qh\ 27. ar' + 25 = 25 + m". 
 
 23. x2_4o = 24. 28. ar' - 30 = 2 (2 6'^ - 15). 
 
 29. Solve a^ + 2 atn = a^ -\- m^. 
 
 Solution 
 
 x^ + 2 am = a"^ + in^. 
 
 x2 = a2 _ 2 am + m*. 
 X • 05 = (a — jw) (a — m) , 
 or X • X = — (« — m) • — (a — m). 
 
 .•. X = ±(a — m). 
 
 Solve the following equations : 
 
 30. a^-c2 = d2-2cd. 36. ar^ - c^ = 36 - 12 c. 
 
 31. a;2-62 = 46c + 4c2. 37. a^ - 4 6^ = 36 - 24&. 
 
 32. a^-w2 = 6w + 9. 38. aj2-a2 = 9 -6a. 
 
 33. a^ + 10a = a2 + 25. 39. a,'^ - &^ = 4 -4ft2. 
 
 34. a?-a? = 2a + l. 40. a^ - a^ft" = 2a& + 1. 
 
 35. ar'-m2 = 8m + 16. 41. a? - r* =^ h* - 2 7^h\ 
 
 42. Find the value of a; in a^ + 4 a; = 45. 
 
 FIRST PROCESS SECOND PROCESS 
 
 a^4-4x = 45 ar* + 4aj = 45 
 
 /c2 + 4a;-45=0 a^ + 4x + 4 = 49 
 
 (a;-5)(a; + 9)= (a; + 2)(a; + 2)=7 • 7 or -7- -7 
 
 .-. a;-5= .-. a; + 2 = 7 or -7 
 
 or a; + 9=0 a; = 7-2or-7-2 
 
 .*. a; = 5 or — 9 /. a; = 5 or — 9 
 
FACTORING 113 
 
 Explanation. —The explanation given for example 12 will serve for the 
 first process. 
 
 In the second process, it is seen that, by adding 4 to each member of the 
 equation, the first member will become the square of the binomial (x + 2). 
 Solving for (a: + 2) as for x in previous examples, x + 2—±l; whence 
 a; = db7-2 = 6or -9. 
 
 Sdggestion. — In the following examples, when the coefficient of the first 
 power of the unknown number is even, either of the above processes may be 
 used ; but when it is odd, the first process is simpler. 
 
 Solve the following equations : 
 
 43. a.-2-6a; = 40. 62. ar^ + 4a; + 3 = 0. 
 
 44. a^- 8 a; = 48. 63. ar^ + 6ar + 8 = 0. 
 
 45. .x-2-5a; = -4. 64. x2-9x + 20-0. 
 
 46. x''-lx = l^. 65. ic2 + lla; + 30 = 0. 
 
 47. x^ -}- 10 a; = 56. 66. a;^ + a; - 132 = 0. 
 
 48. a;'-' + 12 a; = 28. 67. Z2 = 4.x + x'. 
 
 49. a^~3ai = 40. 68. 3a: = 88 -ar^. 
 
 50. ar'-9a; = 36. 69. 160 = .x-^ - 6 a:. 
 
 51. a:^ _|_ 11 a; = 26. 70. Ay = y--1^2. 
 
 52. x^ - 12 x = 4.5. 71. 600 = ^2 _ 10,^^ 
 
 53. y^-Wy = 5i. 72. ^+16x-36 = 0. 
 
 54. y^-21y = A6. 73. a;^ + 15a; - 34 = 0. 
 
 55. a:2_i0a; = 96. 74. ?/2 - 8 ?/ - 84 = 0. 
 
 56. f-20y = 96. 75. y^ -2ay + a^ = 0. 
 
 57. 2/2 + 12y = 85. 76. a;2 + 2^ + 6^=0. 
 
 58. 2/^ + 42 = 13 y. 77. x^ + Aax + Aa^= 0. 
 
 59. 2/- + 63 = 16t/. 78. 2=' + 22z + 121=0. 
 
 60. v^-60 = 11 V. 79. X- -(a + h)x + a& = 0. 
 
 61. y^ + 140 = 72 y. 80. x^ +{c-^ d)x + ccZ = 0. 
 
114 FACTORING 
 
 81. V+(a + 2)a; + 2a = 0. 83. x" -{a - d)x - ad = Q. 
 
 82. /-(c-n)?/ — nc = 0. 84. a? —{h -\-'!)x + 1 h = 0. 
 
 85. (2a; + 3)(2a;-5)-(3a;-l)(a;-2)=l. ' 
 
 86. (2a;-6)(3a;-2)-(5a;-9)(x-2)=4. 
 
 87. Solve 6x2 + 5a; -21=0. 
 
 Solution 
 6 x2 + 6 a; - 21 = 0. 
 Factoring, § 131, (2 a; - 3) (3 a; + 7) = 0. 
 
 .-. 2 a; - 3 = 0, 
 or 3 X + 7 = 0. 
 
 .•. X = I or — |. 
 
 Solve the following equations: 
 
 88. 3ic2 + 2x-l = 0. 93. Ix' + Qx-l^a. 
 
 89. 5x2_,_4a;_i =0. 94. 2 1)2- 9v- 35 = 0. 
 
 90. 3!/'' + 2/ -10 = 0. 95. 6/ -222/ + 20 = 0. 
 
 91. 3?/2_4y-4 = 0. 96. 30^* + 13a; -30 = 0. 
 
 92. 41/2 + 92/ -9 = 0. 97. 4a;2^-L3a,_i2 = o. 
 
 98. Solve the equation a? — 2oi? — 6x-^Q = 0. 
 
 Solution 
 
 a;3_ 2x2-5x4-6 = 0. 
 Factoring by the factor theorem, (x — l)(x — 3)(x + 2) = 0. 
 X - 1 = 0, or X - 3 = 0, or X + 2 = 0. 
 .•. X = 1, or 3, or — 2. 
 
 Solve the following equations : 
 99. ar^ - 15 a.-^ + 71 a; - 105 = 0. 101. a;' - 12 a; + 16 = 0. 
 
 100. a;3 _^ 10 a;2 + 11 a; - 70 = 0. 102. a;^ - 19a; - 30 = 0. 
 
 lOS. a;* + a.-^ - 21 a;2 _ a. _,_ 20 = 0. 
 
 104. a;*-7a;3 + a;* + 63a;-90 = 0. 
 
 105. a;*+8a;3-a;2_(58a; + 60 = 0. 
 
 106. ar>-lla;* + 45a;3-85a,-2 + 74a;-24 = 0. 
 
HIGHEST COMMON DIVISOR 
 
 142. 1. Name all the numbers that will exactly divide both 
 a* and a?. Which of these is of the highest degree ? 
 
 2. What is the highest divisor common to 6^ and h^? to a? 
 and X? 
 
 3. Since the highest divisor common to a* and a? is o?, to h^ 
 and W is h^, and to x^ and x is x, what is the highest divisor com- 
 mon to a*&V and a?h^x ? 
 
 4. What is the highest common divisor of 36a^& and 90 a6^? 
 What prime factors, or divisors, are common to 36 a^h and 90 ab^ ? 
 How may the highest common divisor of 36a^6 and 90 a6^ be 
 found from their factors ? 
 
 143. A number that exactly divides each of two or more alge- 
 braic expressions is called a Common Divisor of them. 
 
 The common divisors of 12 a^ and 4 a^ are 2, a, 4, a^, 2 a, 4 a, 2 a^, 
 and ia^. 
 
 An exact divisor of an expression is a factor of it. 
 
 Two expressions whose only common divisor, or factor, is 1 are 
 said to be prime to each other. 
 
 3 X and 2 a are prime to each'other ; also x + y and x — y. 
 
 144. That common divisor, or factor, of two or more algebraic 
 expressions which is of the highest degree is called their Highest 
 Common Divisor, or Highest Common Factor. 
 
 The highest common divisor, or factor, of 12 a^ and 4 a^ jg 4 ^2, 
 The abbreviation H. C. D. is used for Highest Common Divisor. 
 The highest common divisor in algebra corresponds to the greatest com- 
 mon divisor in arithmetic. . But there would be an inaccuracy in applying 
 the term greatest common divisor to literal numbers, since letters may repre- 
 sent any numbers, as, for instance, fractions. 
 
 116 
 
116 HIGHEST COMMON DIVISOR 
 
 Thus, if ffl = I, then a? = \, a? = \, and the higher the degree of the literal 
 number, the less will be its arithmetical value. Consequently, it is inaccurate 
 to speak of a^, the highest common divisor of a^ 2 a*b, and 3 a^, as their 
 greatest common divisor, because a may represent a proper fraction. 
 
 145. Pkinciple. — The highest common divisor, or factor, of two 
 or more algebraic expressions is the product of all their common 
 prime factors. 
 
 146. To find the highest common divisor of expressions that may 
 be factored readily by inspection. 
 
 Examples 
 1. What is the highest common divisor of 12 a^h^c and 8 a^W(? ? 
 
 FIRST PROCESS SECOND PROCESS 
 
 12 a*b''c = 3 X 2 X 2 X aaaa x bb x c 12 a*b'c = 4 a^b'^c x 3 a" 
 8a'6V = 2 X 2 X 2 X aa X 666 X ccc 8a^b^(f = ^a'Wc x2b<? 
 
 H. C. D. = 2 X 2 X aa X 66 X c ^^a'bh H. C. D. = 4 d'b^ 
 
 Explanation. — Since the highest common divisor of the expressions is 
 the product of all their common prime factors (Prin.), and since the only 
 prime factors common to the given expressions are 2, 2, a, a, b, 6, and c, 
 their product, 'ia^b^c, is the highest common divisor. 
 
 Suggestion. — Frequently the work may be abridged by grouping com- 
 mon factors, as in the second process. Since 3 a^ and 2 bc^ are prime to 
 each other, 4 a^b'^c must contain all the common factors, and be the highest 
 common divisor. 
 
 2. What is the H. C. D. of 3x^-3xy^ and a^ - 2 ar'y + a?j^ ? ' 
 
 PROCESS 
 
 3a:^ — 3xy^ =3x(x + y)(x — y) 
 
 a? —2 3?y 4- xy^ = x(x — y)(x — y) 
 
 .-. H.C.D.= x(x-y) 
 
 Explanation. — For convenience in selecting the common divisors, the 
 expressions are resolved into their simplest factors. 
 
 Since the only common prime factors are x and (x — y), the highest 
 common divisor sought is their product, x{x — y) (Prin.). 
 
HIGHEST COMMON DIVISOR 117 
 
 Rule. — Separate the expressions into their p>rime factors. 
 
 The product of all the common prime factors, each factor being 
 taken the least number of times it occurs in any of the given expres- 
 sions, is the highest common factor. 
 
 The factors that enter into the H. C. D. can often be selected without 
 actually separating the expressions into their prime factors. 
 
 3. What is the H. C. D. of 5 aV - 5 6V and a^a? - Wv? + aV 
 
 -6y? 
 
 Solution 
 
 5 a2c2 _ 5 h2c2 = 5 c2(a2 _ 62) 
 
 d^x^ - &%s + aV - foy = (x« + f) {d^ - b') 
 .-. H. C. D. = a2 - 62 
 
 Find the highest common divisor of 
 
 4. 10 2(?y^, 10 a-y, and 15 xy*z. 
 
 5. l()a%\ 21 a*b*, and 35 a*b\ 
 
 6. Sm^n^, 28 m%*, and 56m*w^, 
 
 7. 4:b^cd, 6 6V, and 24 aftc^. 
 
 8. 10 (x — 2/) V and 15 (z — y)(x — y)^. 
 
 9. 4(a + bf{a - h) and b{a+ h)\a - 6)2. 
 
 10. 3 (a^ - 62)2 and a (a - 6) (a^ - 6^). 
 
 11. x-2 - 2 a; - 15 and a^ - a; - 20. 
 
 12. x* — y*, a? — }f, and x-\-y. 
 
 13. a^ + 7 a + 12 and a'^ + 5 a + 6. 
 
 14. a^ + 2/^ and x^ + 2xy -\- y^. 
 
 15 . a^ — a? and a^ — 2 aa; + a:^. 
 
 16. a2-62 and a2 + 2a6 + 62. 
 
 17. x* + a:2y2 ^ yi a^jj(j a^ ^_ a-y _^ ^2. 
 
 18. a^ + 2/^, a^ + i/', and a^?/ + xy^- 
 
 19. a^ + a26* + 6» and 3 a^ - 3 a62 + 3 6*. 
 
 20. a^ — a?, a^-\-2ax + x', and a^ + a^. 
 
 21 . ax — y + xy — a and aa;'^ + a^y — a — y. 
 
 22. a^6 — 6 — a'^c + c and a6 — ac — 6 + c. 
 
118 HIGHEST COMMON DIVISOR 
 
 23. 1-4:X^, l + 2x, and A a -16 ax'. 
 
 24. (a - b) (h - c) and (c - a) {a? - 6-). 
 
 25 . 24 af*/ + 8 x'f and 8 a^i/^ _ g r^^^ 
 
 26. 6 if2 + a; - 2 and 2 a^ - 11 a; + 5. 
 
 27. 16a:2_25 and 20ar'-9a;-20. 
 
 28. a^ - 6 .T + 5 and a^-5a^ + 7a;-3. 
 
 29. a? -A and a^ - lOa^ + 31a;- 30. 
 
 30. x'-^ and a;« - 12 a;^ ^ 41 3. _ 42, 
 
 31. a;3 _ 4 a; + 3 and ar' + ar — 37 a; + 35. 
 
 147. To find the highest common divisor of expressions that can- 
 not be factored readily by inspection. 
 
 1. What are the exact divisors of a&? Will they be factors 
 of 2 times a& ? of a times cib ? of m times ah ? 
 
 2. If a number is an exact divisor of an expression, what will 
 be its relation to any number of times the expression ? 
 
 3. What common divisor have ctiX and ay ? any number of times 
 ax and ay, as tn • ax and n • ay ? 
 
 4. If two numbers have a common divisor, what divisor has 
 their sum ? their difference ? the sum or difference of any n\im- 
 ber of times the numbers ? 
 
 5. What is the highest common divisor of 2am{x+y') and 
 Shm{x + y)? How will it be affected, if the second number is 
 multiplied by 7 or 2: ? by 2 or a ? How will it be affected, if the 
 first number is multiplied by 5 ? hy b? 
 
 6. By what numbers may one of two expressions be multi- 
 plied without affecting their highest common divisor ? 
 
 7. How will the highest common divisor of 2am(a; + ^) and 
 3bm{x + y) be affected, if the first number is divided by 2 ? 
 by a ? by m ? by (a; + y)? How, if the second number is divided 
 by 6 ? by m ? 
 
 8. By what numbers may one of two expressions be divided 
 without affecting their highest common divisor ? 
 
HIGHEST COMMON DIVISOR 119 
 
 148. Principles. — 1. A divisor of an expression is a divisor 
 of any number of times the expression. Hence, by § 85, 
 
 2. A common divisor of two expressions is a divisor of their sum, 
 of their difference, and of the sum or the difference of any number 
 of times the expressions ; also, 
 
 3. The highest common divisor of two expressions is not affected 
 by mxdtiplying or dividing either of them by numbers that are not 
 factors of the other. 
 
 Examples 
 1. Find the H. C. D. of x' + bx + Q and 4a^ + 21 x^ + SOa; -f 8. 
 
 PROCESS 
 
 ar^ + 5 a; + 6)4 ar^ + 21 x2 4- 30 a; + 8(4 » + 1 
 4 a^ + 20 ar* + 24 a; 
 
 x^+ 6x + 8 
 a?+ 5a; + 6 
 
 X + 2)v? + 5 a; + 6(a; + 3 
 x^ + 2x 
 
 3x + 6 
 .-. H.C.D. = a; + 2. 3 x- + 6 
 
 Explanation. — Since the highest common divisor cannot be higher than 
 05^ + 5 X + 6, it will be a;2 ^ 5 a; ^ 6, if x^ + 5 x + 6 is exactly contained in 
 4 x^ + 21 x2 + 30 X + 8. By trial, it is found that it is not exactly contained 
 in 4 X* + 21 x^ -I- .30 x + 8, since there is a remainder of x + 2. Therefore, 
 x^ + 5 X + 6 is not the highest common divisor. 
 
 Since x'^ + 5 x + 6 contains the highest common divisor, (4 x + 1) times 
 x2 + 5x + 6 will also contain the highest common divisor (Prin. 1); and 
 since both 4 x" + 21 x^ + 30 x + 8 and (4 x + 1) (x^ + 5 x + 6) contain the 
 highest common divisor, their difference, x + 2, must contain the highest 
 common divisor (Prin. 2). Hence, the highest common divisor cannot be 
 higher than x + 2. 
 
 X + 2 will be the highest common divisor, if it is exactly contained in 
 x2 + 6x + 6, since, if it is contained in x2 + 5x + 6, it will be contained 
 in any number of times x^ + 5 x + 6, as (4 x + 1) (x^ -f 5 x + 6) (Prin. 1); 
 and in the sum of (4 x + l)(x2 + 5 x + 6) and x + 2, or 4 x^ + 21 x^ + 30 x + 8 
 (Prin. 2). By trial, x + 2 is found to be exactly contained in x^ + 5 x + 6. 
 Therefore, x + 2 is the highest common divisor of the given expressions. 
 
120 HIGHEST COMMON DIVISOR 
 
 2. Find the H. C. D. of Oa^ + 33a;-63 and 2a;3_,_llar'-a;-30. 
 
 PKOCESS 
 
 3)6 a? + 33 X - 63 
 2 a^ + 11 a; - 21 
 
 2cc3 + llic2_ a._ 3Q(a. 
 2 x^ -f 11 x^ _ 21 a; 
 
 10 )20 a; - 30 
 
 2 a;- 3)2 a;2 ^ 11 a; - 21(a; + 7 
 2ar'- 3a; 
 
 H. C. D. = 2 a; - 3. 
 
 14 a; - 21 
 14 a; - 21 
 
 Suggestion. — Since only common factors are sought, factors that ar« 
 not common to the given expressions, as 3 p.nd 10, may be rejected from any 
 expression before it is used as a divisor (Prin. 3). 
 
 3. Find the H. C. D. of 
 
 2 a;3 _^ 5 ar^ _ 22 a; + 15 and ba? + l^v? -dZx + lQ. 
 
 PROCESS 
 
 2a;3_^5aJ!_22a; + 15)5 a.-^ + 18 a;'' - 
 
 9 
 
 33 a; + 10 
 
 10a;3_^36a^_ 66 a; + 20(5 
 10 a:^ + 25 a;^ - 110 g; + 75 
 11 )11 a;^+ 44 a; -55 
 01? ^r 4a;— 5 
 
 ar^ + 4 a; - 5)2 a;3 ^ 5 3,2 _ 22 a; + 15(2 a; - 3 
 2 a;3 + 8 a;2 - 10 a; 
 
 H. C. D. = a;2 _^ 4 a- _ 5. 
 
 - 3 a;2 - 12 a; + 15 
 
 - 3 a;^- 12 a; + 15 
 
 Suggestion. — When the first term of the divisor is not contained in the 
 first term of the dividend an integral number of times, fractional quotients 
 may be avoided by nmltiplying the polynomial taken for the dividend by 
 some number not a factor of the divisor (Prin. 3). In the above example 
 the simplest factor that may thus be introduced is 2, if 5 x^+lS a:'- — 33x + 10 
 is taken for the dividend ; or 5, if 2 x^ + 5 a;^ _ 22 x + 15 is taken for the 
 dividend. 
 
HIGHEST COMMON DIVISOR 121 
 
 4. Find the H. C. D. of 
 
 30 amx^ — 21 amx — 99 am and 42 ahoi? + 33 ab:i? — 45 ahx, 
 
 PROCESS 
 
 30 am2(? — 21 amo; - 99 am ) 42 aha? + 33 a&a;^ — 45 ahx 
 
 Eeject m (Prin. 3). Eeject hx (Prin. 3). 
 
 30 aar^ - 21 ox - 99 a)42 ax" + 33 aa; - 45 a 
 
 Reserve the common factor 3 a as a factor of the H. G. D. 
 
 lOar'- Ix- ^Z)Ux'-\-llx- 15 
 _7 5 
 
 70ar'-49a;-231)70a,-2 + 55a;- 75(1 
 70 ar^- 49 a; -231 
 
 52 ) 104 a; + 156 
 
 2a;+ 3 
 2a; + 3)10x2 -7a;- 33(5a;_ll 
 .-. H.C.D. = 3a(2a; + 3). 10a;2-7a;- 33 
 
 ScGGESTioN. — Since each of the polynomials contains a factor not found 
 in the other, these two factors may be rejected (Prin. 3). Consequently, m 
 is rejected from the first polynomial, and bx from the second. 
 
 To simplify the process the common factor 3 a is removed and reserved as 
 a factor of the H. C. D. 
 
 5. Find the H.C.D. of 9a;2-35a; + 24 and 29»-8a;«-15. 
 
 PROCESS 
 
 - 8 a;2 ^ 29 a; - 15 
 9a;2-35a; + 24 
 
 a?- 6a;+ 9)9a;2 - 35« + 24(9 
 
 9a;2_54a; + 81 
 
 19 ) 19 a; -57 
 
 X- 3)ar'-6a; + 9(a;-3 
 .-. H. C. D. = a; - 3. a;'-6a; + 9 
 
 ScGGBSTioN. — Since the sum or the difference of two expressions con- 
 tains their highest common divisor (Prin. 2), it is evident that at the outset 
 a simpler expression that will contain the highest common divisor may be 
 obtained by adding the given expressions, giving oj'^ — 6 x + 9. 
 
122 HIGHEST COMMON DIVISOR 
 
 Find the H. C. D. of 
 
 6. a^ + 2 a; - 24 and 2 a^ + 7 X - 30. 
 
 7. 2 x^ _ a. _ 21 and 4x2 + 4 a; -63. 
 
 8. 3x2+.10x-8 and 6x2-7x + 2. 
 
 9. 2x3-6x2 + 7x-6 and 2x3 + 4ar'-3x + 9. 
 10. x3^9aJJ4.26x + 24 and 2x3_^14a^_^20x. 
 
 11. Find the H. C. D. of 
 3 aa? — 4 ax^ — 13 ax + 14 a and 3 ah:x? + 5 ab^ — 10 abx — 42 ah. 
 
 FIRST PROCESS 
 
 3 ox^ — 4 aa^ - 13 ax + 14 a)3 aha? + 5 ahx^ — 10 a6x — 42 ah 
 
 Reserve the common factor a as a factor of the H. C. D. 
 
 3a^_4ar=_13a;+14)36x3+56x2-10 6x-42 6(6 
 36x3-46x2- 13 & X4-146 
 6 )96x2+ 'Sbx-bQb 
 9x2 _j. 3^ _56)3a^_ 4x2-13x+14 
 
 3 
 
 9.r'-12x2-39x+42(x 
 9x^+ 3x2-56x 
 
 -15x2+17x+42 
 -15x2+17x+42)9x2+ 3x- 56 
 
 _5 
 
 46x2+15x-280(-3 
 45x2-51 x-126 
 22 )66 X- 154 
 
 3x- 7)-15x2+17x+42(-5x-6 
 
 -15x2+3 5x 
 
 -18X+42 
 .-. H.C.D. =a(3x-7). -18x+42 
 
 Since the arrangement of the dividend, divisor, and quotient 
 may be either: Divisor) Dividend (Quotient; or Quotient) Divi- 
 dend ( Divisor ; by using these two arrangements alternately, the 
 above process may be more compactly written as follows : 
 
HIGHEST COMMON DIVISOR 
 
 123 
 
 SECOND PROCESS 
 
 3 aoi? — 4 (vx- — 13 ax + 14 a)3 ahx^ + 5 ahy? — 10 a&ic — 42 ah. 
 Reserve the common factor a as a factor of the H.C.D. 
 
 
 3ar5_ 4 a,-^ - 13 a; + 14 
 3 
 
 9 a^ - 12 ar^ - 39 a; + 42 
 9ar'+ 3a^-oGa;' 
 
 3bs(^ + 5ba^- 10 &x - 42 & 
 3 6ar* - 4 ftar^ - 13 6a; + 14 & 
 
 
 h)9ba^+ 3te-56& 
 
 X 
 
 9a^-i- 3x -56 
 5 
 
 
 - 15 a;-' + 17 a; + 42 
 
 - 15 a^ + 35 X 
 
 45ar' + 15a; -280 
 45a;2-51a; -126 
 
 
 22)66 a; -154 
 
 — 5x 
 
 3a; -7 
 .-. H.C.D. = a (3 a; -7). 
 
 -6 
 
 - 18 a; + 42 
 
 - 18 a; + 42 
 
 -3 
 
 Suggestion. — When the quotient consists of more than one term, for 
 convenience each term is placed opposite the corresponding part of the divi- 
 dend or product. 
 
 Rule. — Divide one expression by the other, and if there is a 
 remainder, divide the divisor by it; then divide the preceding divisor 
 by the last remainder, and so on, until there is no remainder. The 
 last divisor will be the highest common divisor. 
 
 If any remainder does not contain the letter of arrangement, the 
 expressions have no common divisor in that letter. 
 
 If more than two expressions are given, find the highest common 
 divisor of any two, then of this divisor and another, and so on. Tlie 
 last divisor will be the highest common divisor. 
 
 1. If either expression contains a monomial factor not found in the other, 
 it should be rejected before beginning the process. 
 
 2. A common factor of the expressions should be removed before begin- 
 ning the division, but it must appear as a factor of the highest common divisor. 
 
 3. When necessary, to avoid fractional quotients, any dividend or divisor 
 may be multiplied or divided by any number not a factor of the other. 
 
 4. The highest common divisor has an ambiguous sign. For, if a positive 
 divLsor is contained in a dividend, the same negative divisor also will be con- 
 tained in that dividend, but the signs of the quotient will be changed. It is 
 not customary to write both divisors. 
 
124 HIGHEST COMMON DIVISOR 
 
 149. The principle, that the exact divisor reached by the process 
 given in the rule is the highest common divisor, may be proved as 
 follows : 
 
 Let A and B represent any two polynomials freed of monomial factors, 
 the degree of B being not higher than that of A. 
 
 Divide A by B, and let the quotient be m and the remainder D ; divide B 
 by D, and let the quotient be n and the remainder E ; divide D by E, and 
 let the quotient be r and the remainder zero ; that is, let E be an exact 
 divisor of D. 
 
 It is to be proved that E is the highest common divisor of A and B. 
 
 PROCESS 
 
 B)A{m 
 mB 
 D)B(n 
 nP 
 E)D{r 
 rE 
 
 
 Since the minuend is equal to the subtrahend plus the remainder, 
 A = mB + D, and A - mB = D ; 
 B= nD + E, and B- nD = E; and D = rE. 
 
 Since the division has terminated, E is a, common divisor of D and nD 
 (Prin. 1) ; also of D and nD+E, or B (Prin. 2) ; also of B and mB (Prin. 1) ; 
 also of B and mB + D, or A (Prin. 2)! That is, JE" is a common divisor of B 
 and A. 
 
 Every common divisor of A and B is a divisor of mB (Prin. 1) ; and of 
 A — mB, or D (Prin. 2), Therefore, every common divisor of A and B is a 
 divisor of nD (Prin. 1) ; and of B — nD, or E (Prin. 2). 
 
 But, since no divisor of E can be of higher degree than E itself, E is the 
 highest common divisor of A and B. ' 
 
 150. The principle, that the highest common divisor of several 
 expressions may be obtained by finding the highest common 
 divisor of two of them, then of this result and a third expression, 
 and so on, may be proved as follows : 
 
 Let P be the highest common divisor of A and B, and Q the highest com- 
 mon divisor of P and a third expression C. 
 
 Then, since P contains all the common factors of A and B, and Q con- 
 tains of these particular factors only such as are factors of C also, Q is the 
 highest common divisor of A, B, and C. 
 
 This method may be extended to embrace any number of expressions. 
 
HIGHEST COMMON DIVISOR 
 Find the H. C. D. of 
 
 125 
 
 12. 2a:3-7aj2 + 2a; + 3 and 2.'K3 + 7a^-5a;-4. 
 
 13. 9a^ + 18a:2-a;-10 and 3a^ + 13a;2 + 2a;-8. 
 
 14. l-2a;-5a^ + 6a^ and l + 5a; + 2if2-8af'. 
 
 15. l-4a; + a^ + 6a^ and l + 3a;-6a^-8a^. 
 
 16. \-x-Ux'-{-24.v? and d&a?-2^a? + x + l. 
 
 17. m^ — 4 nv? — 20 m + 48 and m^ — m^ — 14 m + 24. 
 
 18. 3a'' + 20a2-a-2 and 3a»+17a2 + 21a-9. 
 
 19. 8ax2 + 22aa; + 15a and 66a^ + ll 6a; + 3&. 
 
 20. 20 ft^c - 2 &c - 4 c and 8 a^ft^c - 4 a?bc + a^c. 
 
 21. 21 a.r — 17 aa!^ — 5 aa^ + aa;* and 7 aa; + 34 aa^ — 5 aa^. 
 
 22. a^-7a; + 6, a-* - 2aj3 - 9a^ + 18a;, a;3 + a;^ -4a;-4. 
 
 23. x^-5x + 4:, x*-2x'^ + l, a;« + 4a^-3a;-2. 
 
 24. l+4a^ + 5a^, 2 + 5a; + 3x*, a;« - 4 a;* + 5 ar^ - 2. 
 
 25. 3 + ar-8a^ + 4a^, 3 -8a;-8a^4-8a^, 16 a;* - 48 a^ + 81. 
 
 26. c^-6a^-5x-U, x3-10a:2^20a: + 7, a;* - 310 a; - 231. 
 
 27. FindtheH.C.D. of a;«+a^+a^-a;-2 and 2a^+a;^-a^-a:*-l. 
 
 PBOCESS BY DETACHED COEFFICIENTS 
 
 1+0+1+1+0-1-2 
 
 1-2-3-1+2+3 
 
 2+4+2-2-4-2 
 
 2-4-6-2+4+6 
 
 8 )8 + 8 + 0-8-8 
 
 1+1+0-1-1 
 
 .-. H.C.D. = a^+a^-a;-l. 
 
 2+1+0-1-1+0 
 
 2+0+2+2+0-2 
 
 1-2-3-1+2+3 
 
 1+1+0-1-1 
 -3-3+0+3+3 
 -3-3+0+3+3 
 
 Find the H. C. D. of 
 
 28. ar*-a^-2a:3_a^ + a, + 2 and a^ + 3a^ + 3a^ + a^-a;-l. 
 
 29. af + a;*-a^-7a;-4 and 2ar^ + 3a;* + 3ar' + 3ar' - 7a;- 4. 
 
126 LOWEST COMMON MULTIPLE 
 
 30. a^-2x*-2x''-llx'-x-15sind2aT'-7x*+A2^-15x^+x-10. 
 
 31. a'-3a*-3a^-3a'-19a-15sLnd a'+3a*-3a^+9a^-a-15, 
 
 32. 5 a^-\-a*- 11 a^ +9 a^-Sa+4: and 2 a^-a^-iT a^-f 8 a--4 a. 
 
 33. x^ — 5x + 4: and ic* — a:' — 3 a^ — 5 a; — 12. 
 
 34. a' + 3 a^ - 2 a - 6 and a^ + 4 a* + 4 a'' + 4 a^ - a - 12. 
 
 35. 1 - 4 a^ + 3a* and 1 + a - a'^ - 5a^ + Aa*. 
 
 36. 2 — a + 3a^ + 5a^-a* and 4 — 4 a + a^ — 9 a*. 
 
 37. f-\-13y^ + 20y-U and 7 - 3?/ - 20?/=' + 2?/- y*. 
 
 38. 6x3-lla^-35a;, 30 ar^ - 115 x + 35, 23^ - dx" - ox-7. 
 
 oV»ic 
 
 LOWEST COMMON MULTIPLE 
 
 151. 1. What number exactly contains 2, 5, a, and b, or is a 
 multiple of 2, 5, a, and b ? 
 
 2. What different prime factors must enter into every number 
 that will contain 4 a^b, a^b^, and 10 ab^, or must be found in 
 every common multiple of Aa^b, a^b'^, and 10 a6^? 
 
 3. What is the lowest power of a that common multiples 
 of 4 a^b, d^b^, and 10 a¥ can contain ? What is the lowest power 
 of fe ? of 2 ? of 5 ? 
 
 What, then, is the lowest common multiple of 4 a?b, a^b', and 
 10 aW? 
 
 To what is the lowest common multiple of two or more expres- 
 sions equal ? 
 
 152. An expression that exactly contains each of two or more 
 given expressions is called a Common Multiple of them. 
 
 6 abx is a common multiple of a, 3 6, 2 x, and 6 abx. These numbers 
 may have other common multiples, as 12 abx., 6 a^6%, 18 a*6x'^, etc. 
 
LOWEST COMMON MULTIPLE 127 
 
 153. The expression of knvest degree that will exactly contain 
 each of two or more given expressions is called their Lowest 
 Common Multiple. 
 
 6 abx is the lowest common multiple of a, 3 ft, 2 a;, and 6 abx. 
 
 The abbreviation L. C. M. is used for Lowest Common Multiple. 
 
 The lowest common multiple in algebra corresponds to the least common 
 multiple in arithmetic. But, since letters may represent any numbers, as, 
 for instance, numbers not prime to each other or fractions, the term least 
 is not applicable to algebraic common multiples. 
 
 Thus, the algebraic lowest common multiple of a^b'^, ab^, and bx is a^b'x. 
 If a = 4, 6 = 3, and x = 2, a'^b^x, the lowest common multiple of the given 
 expressions, is equal to 864. If, however, the values of a, ft, and x are sub- 
 stituted for those letters, the given expressions become 144, 108, and 6 ; and 
 their least common multiple is 432. 
 
 It is thus seen that tlie lowest common multiple of two or more expressions 
 is not necessarily their least common multiple. 
 
 154. Principle. — The lowest common multiple of two or more 
 algebraic expressions is the product of all their different prime fac- 
 tors, using each factor the greatest number of times it occurs in any 
 of the expressions. 
 
 155. To find the lowest common multiple of expressions that may 
 be factored readily by inspection. 
 
 Examples 
 1. What is the L. C. M. of 12 x'yz*, 6 a^xf, and 8 axyz^ ? 
 
 PROCESS 
 
 12x^yz' =2'2'3-x^-y-^ 
 6 a^xy'^ =2 • 3 • a^ • X • y'^ 
 8 axyz^ =2 -2 '2 ^ a • x - y • '^ 
 
 L. C. M. = 2 . 2 . 2 . 3 . a=^ • ar' • / • «* = 24 aV^z* 
 
 Explanation. — The lowest common multiple of the numerical coefficients 
 is found a.s in arithmetic. It is 24. 
 
 The literal factors of the lowest common multiple are each letter with the 
 highest exponent it has in any of the given expressions (Prin.). They are, 
 therefore, a'^, x^, j/2^ and 2*. 
 
 The product of the numerical and literal factors, 24 a'^x^y'^z*, is the lowest 
 common multiple of the given expressions. 
 
128 LOWEST COMMON MULTIPLE 
 
 2. What is the L. C. M. of or' - 2 »?/ + /, x? - y^, and a^ -f 2/»? 
 
 PROCESS 
 
 x^ - 2xy + y^ = (x - y)(x - y) 
 a^-y^ ={x-y) {x + y) 
 a^ + f — {x + y){x^ — xy + f) 
 
 L. C. M. =(x-y)\x+y){x'-xy,+ f) 
 
 = (a; - y)\:»? + f) 
 
 Rule. — Factor the expressions as far as may he necessaTy to 
 discover their different prime factors. 
 
 Find the product of all their different prime factors, using each 
 factor the greatest number of times it occurs in any of the given 
 expressions. 
 
 The factors of the L. C. M. may often be selected without separating the 
 expressions into their prime factors. 
 
 Find the L. C. M. of 
 
 3. a^x^y, a^xtf, and aa^y. 
 
 4. 10 a^ft-V, 5 aft^c, and 25 6Vd3. 
 
 5. 16a'b\ 24 d^de, and 36 a^b^d^^. 
 
 6. IS a'^br-, 12p^g^r, and b^ab^p^q. 
 
 7. a/*"*!/^, ic^-y, aj^'V, and x'^+^y. 
 
 8. a? — y^ and a? -\- 2 xy -\- y^. 
 
 9. a? — y^ and a? — 2xy + y^. 
 
 10. X? — y^, X? -\-2 xy + y^, and .r^ — 2 a;?/ + y^. 
 
 11 . a^ — n^ and 3 a^ + 6 a^n + 3 an^. 
 
 12. a;"-! and a2a^ + a'-6^a;2-62 
 
 13. a^ + 1, ab — b,a^ + a, and a"^ — 1. 
 
 14. 2x -\- y,2xy — y"^, and Ax^ — y^. 
 
 15. 1 + a;, a; — a^, 1 + a^, and a^(l — x). 
 
 16. 3 + a, 9 — a^, 3 — a, and 5 a + 15. 
 
 17. a — b,h — c,b + a, and a^ — b^. 
 
LOWEST COMMON MULTIPLE 129 
 
 18. 2x + 2, 5x — 5, Sx — 3, and ccF — l. 
 
 19. 3x-9y, 3a^ + 27r, and 2x-\-6y. 
 
 20. 16 62 _ 1, 12 62 _|_ 3 5, 20 6 - 5, and 2 6. 
 
 21. 1-23^ + 3;', (l-ic)2, and l + 2a; + a^. 
 
 22. 1 - a, 1 + a, 1 + a^, 1 + a*, and 1 + al 
 
 23. xy — 'tf, a? + xy, ocy-^y^, and a^ + 2/^. 
 
 24. a? — y^, x'-\-xy-\-'if, and a? — xy. 
 
 25. 6- - 5 6 + 6, 6^ _ 7 5 + 10, and 6^ _ lo 6 + 16. 
 
 26. a^+ 7x — 8, ic^ _ 1^ a; ^aj2^ and 3cw^— 6ax + 3a. 
 
 27. a^ — x^, a — 2x, a^ -\-2 ax, and a^ — 3 a^o; + 2 aa^. 
 
 28. m^ — a^, y/i^ + 7n«, m^ + mx + aj^, and (m + 3;)a^. 
 
 29. a^-3a; + 2, ar' + 4a; + 4, a^ + 3a; + 2, and ar* — 1. 
 
 30. a;2 — y^, X* + a?y^ -\- y*, a^ + i/^, and x^ + xy + y\ 
 
 31. a;^ + a;2y + x'?/2 + ?/3 and oi? — x^y + xy^ — y^. 
 
 32. a2^4a + 4, 0^ — 4, and a* — 16. 
 
 33. a^-{p-\- cf, 62 - (c + a)2, and (?-{a^ Vf. 
 
 34. 1 — a + a^, 1 + a + a^, and 1 + a^ -f a*. 
 
 35. a^ 4- 4 and a* — 2 a^ + 4 a — 4. 
 
 36. a« - 6^ and a* + a^62 + 6*. 
 
 37. a;^ + y^ and c^^ — b'^y'^ + a^?/^ — 62a;2. 
 
 38. a^-2 a?b + a262 - 9 6^ and a* + 5 a'b'' + 9 6*. 
 
 39. a* — a? + 1, a^ + 1, a* + a^ + 1, and a* — 1. 
 
 40. Find the lowest common multiple of aj^ + 6a^ + 5a; — 12 
 
 and a:3_8a^ + 19a;-12. 
 
 Ik 
 
 Solution 
 x8 + 6a;2+ 5a;-12=(«-l)(x2 + 7a;+ 12) 
 
 = (x-l)(x + 3)(« + 4). 
 a;8 _ 8 x2 + 19 X - 12 = (x - l)(x2 - 7 X + 12) 
 
 = (x-l)(x-3)(x-4). 
 .-. L. C. M. =(x-l)(x + 3)(x-3)(x + 4)(x-4) 
 
 = (x-l)(x2-9)(x2-16). 
 
 ALG. — 9 
 
180 LOWEST COMMON MULTIPLE 
 
 Suggestion. — In solving the following the Factor Theorem will be found 
 useful. 
 
 41. a^-6x^ + llas-6 2ind x^-9x' + 26x-24:. 
 
 42. ar* - 5 x2 _ 4 a; + 20 aud a^ -h 2ar^ - 25 a; - 60. 
 
 43. cc^ -f-3a^ — 4 and a^ + a^ — ic — 1. 
 
 44. a^-4:a^ + 5x-2 Siud a^-8x^ +21X-18. 
 
 45. a^ + 5a;2 + 7a; + 3 and a^ — 7 x^ — 5x + 75. 
 
 46. a^ + 2a^-4a;-8, a^-a^-8a;+ 12, a^ + 4a^- 3a;- 18. 
 
 47. ar'-9a;2_^23a;-15, a^ + x'-lT x + 15, x^+yar^ + jaj-m 
 
 48. a^ + Tx' + Ux + S, a.-3 + 3a^-6a;-8, x^ + x^ -lOx -^S. 
 
 156. To find the lowest common multiple of expressions that can- 
 not be factored readily by inspection. 
 
 x'-Sx^ 2 = (x-l){x-2). (1) 
 
 a^_5a;+ 6 = (x-2)(x-3). (2) 
 
 a;2_7a; + 12 = (a;-3)(a;-4). (3) 
 
 L. C. M.. = (x- l){x -2)(x- 3)(x - 4). 
 
 1. Find the lowest common multiple of expressions (1) and (2) 
 from their factors ; from the product of their factors. By what 
 factor of the two expressions must the product be divided to 
 obtain the lowest common multiple? 
 
 2. How, then, may the lowest common multiple of two expres- 
 sions be found ? 
 
 3. Since (x — l)(x — 2) (x — 3) is the lowest common multiple 
 of the first two expressions, what factor of the third expression 
 must the lowest common multiple of all the expressions contain ? 
 
 157. Principles. — 1. The lowest common multiple of two ex- 
 pressions is equal to their product divided by their highest common 
 divisor; or, it is equal to either of them multiplied by the quotient 
 of the other divided by the highest common divisor. 
 
 2. The lowest common multiple of several expressions may be 
 obtained by finding the lowest common multiple of two of them ; then 
 of this result and a third expression; and so on. 
 
LOWEST COMMON MULTIPLE 131 
 
 Proof of Principle 1. 
 
 Let F be the highest common divisor, or factor, of A and B, and L their 
 lowest common multiple. Let F be contained a times in A and 6 times in B, 
 or let A = aF and let B = bF. 
 
 It is to be proved that L - ^ ^ ^ , or ^ x — , or jB x — • 
 ^ F F F 
 
 Since F contains all the common factors of A and B, a and 6 have no 
 
 common factors ; consequently, since A — aF and B = bF, 
 
 L = abF. 
 
 Multiplying by F, FL = abFF ; 
 
 but Ax B = aFxbF= abFF. 
 
 Therefore, Ax. 1, FL = Ax B, 
 
 and L = Aii^ or ^ x — , or ^ x — • 
 
 F F F 
 
 Proof of Principle 2. 
 
 Let L be the lowest common multiple of A and B, and ilf the lowest com- 
 mon multiple of L and a third expression C 
 
 It is to be proved that M is the lowest common nmltiple of A, B, and C. 
 
 Since L is the expression of lowest degree that is exactly divisible by both 
 A and B, and M is the expression of lowest degree that is exactly divisible 
 by both L and C, M is the expression of lowest degree that is exactly divis- 
 ible by ^, B, and C 
 
 Examples 
 
 1. Find the L. CM. of a? + 6x?-\-llx + 6 aii6.a^-4:x'-\-x-[-6. 
 
 PROCESS 
 
 Prin. 1, L. C. M. = (^ + 6.^ + 11. + 6)g-4a^ + . + 6) 
 
 ^ (x + l)(x' + 5x + 6)(x + l)(x' - 5x + 6) 
 
 x-\-l 
 = {x + l)(a^ + 5x + 6){x'-5x + 6) 
 
 or (x + l)(a: 4- 2)(x + S)(x- 2)(a;- 3) 
 
 or (a; + l)(a^-4)(a;='-9) 
 
 Find the L. C. M. of 
 
 2. 4a3 + 7a2 + 10a-3 and 4a8 + Ga^ + Ua + 3. 
 
 3. 2a3-lla* + 18a-14 and 2a' + Sa"- 10a + 14. 
 
 4. 5a^-lla;* + 3a;-l-12 and 5a;3-19ar' + 27a;-12. 
 
132 LOWEST COMMON MULTIPLE 
 
 5. 4a^- 14x2 + 22 a; -8 ^nd 2ci^ -3a^ - x^ + 12x. 
 
 6. 6a3 + 3a2-15a-75 and 2a» + lla2 + 25a + 25. 
 
 7. 4a^-27a^-2a + 15 and 2a* -9a^ -2Sa^ -15a. 
 
 8. 3(f-llc'-32c-16 and Sc^ - 19c2 + 8c + 16. 
 
 9. 4x* - 7 a^ + 7 x" ~llx + 6 and 2a!* + a^- a^- x- 6. 
 
 10. a;*-ar^ — 3a; + 9 and Sax* — 3 ax^ — 18 ax^ + 45 ax -27 a. 
 
 11. 20a^ + 40^2^ 25a; + 125 and ea:^.^ 7^^ ^ 10x + 25. 
 
 12. 12m3-18m2 + 26m-10 and 15 Trv" - 9 iri" + 19 m -{- 10. 
 
 13. 6 a^a; — 5 a^x — 18ax — Sx and 6 a^b — 13 a^b — 6 ab + S b. 
 
 14. 4a;3 + 4a!2y-5a;2/=' + 252/3 and 4 a;3_ 16^^.2^ ^25 a;/ -25 2^. 
 
 15. 10a^ + 29a^-36a + 9 and Sa^ -\-34:a^ + 9a -9. 
 
 16. 4 a;^ — 17 x^/ + 42/* and 2x* — a^y — 3 a^y^ — 5xy^ — 2y*. 
 
 17. 5x*+8a^-27 x'+Ux-lO and 3a;*+4a;3_i7a^_,_14a,_10. 
 
 18. 2 a;*-9 ar' + 18 .^•2- 18 .T + 9 and 3.T*-lla;3^17a^_12a;+6. 
 
 19. 3a*+13a='-19a2+12a-4 and 4a* + 22a3-2a2 + 2a + 4. 
 
 20. 63^ + 5x-6, 8a;2^i0a;-3, 10a;2_^9^_9 
 
 21. x*-2c(^ + a^-l, x*-x'-\-2x-l, x*-3a^ + l. 
 
 22. a;*-7a;2 + 9, a;* + 2 a;^ _^ ar^ - 9, a;*- a;^ - 6a;- 9. 
 
 23. X*- 42^ + 40^-16, a;*-12ar^+16, a;* - 4 ar^ + 16 a; - 16. 
 
 24. a;*-4ar'' + 4a;2_25^ a;* - 4 a;^ _^ 20 a; - 25, a;* - 14 a;^ _,_ 25. 
 
 25. 4a;* + 5a;2-a;-l, 6a;* + a;3_|.8ic2_l, 36^*-13a;2^1 
 
 26. 10a;*+7a;3-33a;2^26a;-10 and 2a;*+7a.'3+5ar'-4a;-10. 
 
 27. 16a;* + 16ar'-48a;2-36a; + 27 and 24 a;* + 20 a;^ - 74 a;* 
 
 — 45 a; + 45. 
 
 28. 10a;* + 7a;« + 2.T2-a;-2 and 6a^ + 5a^ + 4x + 1. 
 
 29. 5a;* + 3af* + r)ar' + a; + 3 and 15 a;^ ^ ^4 ^.2 _j_ ^. ^ j^2. 
 
 30. 2a.'3-a;2_3a._,_2^ 4a;3_^ 6a^_ 2a;_ 4, 4a^-5x + 2. 
 
 31. a;3-l, 2af + 2ar'-5a; + l, a;3-3a; + 2. 
 
FRACTIONS 
 
 158. A fraction is expressed by two numbers, one called the 
 numerator, written above a line, and the other the denominator, 
 
 written below the line. Thus, - is a fraction. 
 
 6 
 
 If a and b represent positive integers, as 3 and 4, the fraction 
 - is equal to - ; that is, it represents 3 of the 4 equal parts of 
 
 anything. This is the arithmetical notion of a fraction. 
 
 But, since a and b may represent any numbers, positive or 
 
 negative, integral or fractional, rational or irrational, - may repre- 
 
 4 ^ 
 
 sent an expression like — . Since a thing cannot be divided into 
 
 of 
 5§ equal parts, algebraic fractions are not accurately described by 
 
 the definition commonly given in arithmetic. But, since an ex- 
 pression like ^^', regarded as 20 fourths, is equivalent to 5, or 
 20 -j- 4, it is evident that the numerator of a fraction may be re- 
 garded as a dividend, and the denominator as its divisor ; and this 
 interpretation of a fraction is broad enough to include the fraction 
 
 Y when a and 6 represent any numbers whatever. Hence, 
 
 The expression of an unexecuted division, in ivhich the dividend 
 
 is the numerator and the divisor the denominajtor, is an Algebraic 
 
 Fraction. 
 
 The fraction - is read, ' a divided by 6.' 
 b 
 
 159. The numerator and denominator of a fraction are called 
 its Terms. 
 
 160. An expression, some of whose terms are integral and 
 
 some fractional, is called a Mixed Number, or a Mixed Expression. 
 
 a — b r^ /j2 1 
 
 c — , — — 2 -t- — , and a — b -\ — are mixed expressions. 
 
 c a^ x^ ab 
 
 133 
 
134 FRACTIONS 
 
 REDUCTION OP FRACTIONS 
 
 161. The process of changing the form of an expression with- 
 out changing its value is called Reduction. 
 
 162. To reduce fractions to higher or lower terms. 
 
 163. A fraction is in its Lowest Terms when its terms have no 
 common divisor. 
 
 1 X 3 a; 
 
 164. 1. How many eighths are there in -? in -? in — ? 
 
 . 5a6o • 4iCn . 9ar> • 8a;Vo 
 in ? m — ? in—-? in — ^? 
 
 4 16 24 32 
 
 2. How many tenths are there in -? in — ? in — ^? 
 
 2 5 20 
 
 3. If a dividend is multiplied by any number, as 2, and the 
 divisor is multiplied by the same number, how is the quotient 
 affected ? 
 
 4. If a dividend is divided by any number, as 2, and the divisor 
 is divided by the same number, how is the quotient affected ? 
 
 5. Since a fraction may be regarded as an indicated division, 
 what may be done to the terms of a fraction without changing 
 the value of the fraction ? 
 
 165. Principle. — Multiplying or dividing both terms of afrao' 
 tion by the same number does not change the value of the fraction. 
 
 The proof of the principle is as follows : 
 
 Let a and b be any two numbers, a the dividend, b the divisor, and - 
 the quotient. Also, let m be any number. 
 
 It is to be proved that «:=!??:«. 
 mb 
 
 Since the quotient multiplied by the divisor equals the dividend, 
 
 ? X 6 = a. (1) 
 
 b 
 
 Multiplying (1) by m, Ax. 4, - x mb = ma. (2) 
 
 Dividing (2) by mb, Ax. 5, q^m^, ^8) 
 
 mo 
 
FRACTIONS 135 
 
 Hence, the terms of any fraction, as -, may be multiplied by any num- 
 
 6 
 
 ber, or the terms of any fraction, as — , may be divided by any number, 
 
 mb 
 
 without changing the value of the fraction. 
 
 Examples 
 
 1. Reduce — - — to a fraction whose denominator is o? — bK 
 
 a + b 
 
 PROCESS 
 
 (a2 _ 62) -4- (a + &) = a - 6 
 
 a _ a(a — b) _ a^ — ab 
 •*• a + b ~ {a + b){a-b)~ a?-b^ 
 
 Explanation. — Since the required denominator is {a — b) times the 
 given denominator, both terms of the fraction must be multiplied by (a — 6) 
 (Prin.). 
 
 2. Reduce ^^^-^ — ^ to its lowest terms. 
 
 30 a^osz 
 
 Explanation. — Since a fraction is in its lowest 
 PROCESS terms when its terms have no common divisor, the given 
 
 2 „ fraction may be reduced to its lowest terms by remov- 
 
 — ' J = --L ing in succession all common divisors of its numerator 
 
 oO a xz 10 az ^t^^ denominator (Prin.), as, 3, a, a, and x ; or by divid- 
 ing the terms by their highest common divisor, 3 d^x, 
 
 3. Change — to a fraction whose denominator is 4 61 
 
 u 
 
 5 a 
 
 4. Change — to a fraction whose denominator is 42. 
 
 5. Change to a fraction whose denominator is 656. 
 
 3a 
 
 6. Change to a fraction whose denominator is 84 xy. 
 
 14 a; 
 
 7. Change to a fraction whose denominator is 20 m'. 
 
 62/ 
 
136 FRACTIONS 
 
 „ q 
 
 8. Change to a fraction whose denominator is {x — 1)'. 
 
 ar — 1 
 
 9. Change ''^~ — to a fraction whose denominator is (2 x-\-by. 
 
 2x-\-5 
 
 10. Change — - — to a fraction whose numerator is 3 a + a}. 
 
 o — a 
 
 11. Reduce ^ ~ to a fraction whose denominator is a? — b\ 
 
 a + b 
 
 12. Eeduce — — ^ to a fraction whose numerator is ic^ — y\ 
 
 2x + y 
 
 13. Reduce to a fraction whose denominator is a — 6. 
 
 b — a 
 
 14. Reduce -^^^ — to a fraction whose denominator is 4 — oj^. 
 
 x-2 
 
 Reduce the following to their lowest terms : 
 
 15. ^^. 22. -^^^y"^' - 
 
 a^xy — 100 x*y^ 
 
 am^n* x'^y* 
 
 o?b^3(? _. xT-^a? 
 
 Wxy"^ »"• 
 
 16 mhiy?z^ x"+^y 
 
 40 am^ya^ xy"*'^^ 
 
 jg 210&c^d 26 ^"""^^ 
 
 750 a&^c' ■ ax 
 
 42 ab^cd* 3 a^^ft 
 
 121 a^b^c' ' 2ar]f 
 
FRACTIONS 137 
 
 „„ ax"-»+i _ a" -11a + 24. 
 29. — 45. — — • 
 
 ftaj"-" a^ — a—6 
 
 n(n-2)ab ' a^ + 2x'~35x 
 
 31 a'-6' 4- 7a;-2g^-3 
 
 a2 + 2a6 + &2' ' 2ar' + 7a;-4' 
 
 g'^ - 2 a6 + ft'' a(a + 2 6)* 
 
 a2_52 * • 6(a2_4j2^2 
 
 8 a' + 27 ar*' ' a* - 2 g'^ft^ + ^^4* 
 
 ,^ 3a* + 3g6 ^- a^-2x* + a^ 
 
 o4. — • 50. • 
 
 g* + ab^ x^ — or 
 
 3 x'y — 6xy - af-{-5ot^ — 6x 
 
 o^y -8xy' ' 2x^-2 
 
 oo. — — -• 5<o. 
 
 2g3&-2&* ' x*-10ar' + 9 
 
 4 g^ - ab^ ar' - 21 a; + 20 
 
 8a* + g6«' * a^*-26a^ + 25' 
 
 3g 2a^y^-8y* ^^ a^ + 3a^ + 3a; + l 
 
 ' 4:0!^y — 32y* ' a^ + x^ — 4a; — 4 
 
 a*bc - b'c a? - 3 a?b + 3 g6^ - W 
 
 ' '6a%-\-3V' ' 3a'b-3ab^ 
 
 10 7ia; + 10j«v _„ 3 g- 4-4 ga; — 4a;^ 
 
 40. '-^' 5d. • 
 
 25 iix" - 25 ny^ 9 g" - 12 ga; + 4 x^ 
 
 2 ga; — gy — 4 6a; 4- 2 6y 
 4 ga; — 2 gy — 2 6a; + 6y 
 
 ^„ g"+^-gY >o 9.r3-13g2a;-4g« 
 
 42. — • 58. 
 
 a»+3 + a"+'/ 3 6a; + 3 a;y — 4 g6 — 4 gy 
 
 a;*y + x^y^ + y° gg m — m^ — n-\- mn 
 
 a;6 _ y6 ^ _ ^^ ^ ^2 _ j^ 
 
 . . x*y — a;^y^ + y ^ gm — g/i — m + ?^ 
 
 a;® 4- J/* am — an + m — n 
 
 ™n+2 ^n 
 
 41. ±_ ±.. 57. 
 
 /j»w4-3 ^_ /j»H 
 
138 FRACTIONS 
 
 61. Keduce -— — — ^ ^ ' -— to its lowest terms. 
 
 Solution 
 
 The process of finding the H. C. D. of the terms of the fraction can be 
 shortened in some instances by finding the sum or the difference of the 
 terms, since the result will either be the H.C.D. or some multiple of it, § 148. 
 
 3 a:3 - 16 x2 + 25 a: - 12 
 3 a:3 - 8 a;2 - 7 a; + 12 * 
 
 Subtracting the numerator from the denominator, 
 
 8 a;2 - 32 X + 24 
 
 = 8(a;2-4a; + 3). 
 
 By trial, a;2 - 4 x + 3 is found to be the H. C. D. 
 
 Dividing the terms of the fraction by x"^ —'ix-\- Z, the fraction in its 
 
 lowest terms is -• 
 
 3x + 4 
 
 Reduce the following to their lowest terms : 
 a,-3 + 5a^-9a;-45 
 
 62. 
 
 63. 
 
 64. 
 
 65. 
 
 66. 
 
 67. 
 
 68. 
 
 69. 
 
 0^ + 3x2-25 a;-75 
 
 a^ + 2a^-23a;-60 
 :i?-llx'-10x + 200* 
 
 4,a?j^qx' + l{)x-^ 
 
 x' + ^x' + ^x+Q 
 a,-3 + 3a^ + 4a; + 2' 
 
 5af-17x2 + 16a;-4* 
 
 5x^- 14a^ + 22a; + 5 
 5a^-18ar' + 34.'B-15* 
 
 a^ + 6 3i?y —■ 2 xy^ — 5 y* 
 
 g" + 6^ 4- 2 c^ + 2 a6 + 3 ac + 3 6c 
 a' + b' + c' + 2ab + 2ac-\-2bc' 
 
70. 
 
 71. 
 
 72. 
 
 73. 
 
 FRACTIONS 139 
 
 ^2 + 62 -I- c2 -f 2 a6 - 2 ac - 2 6c 
 
 a2 + 62. 
 
 - c^ + 2 a6 
 
 
 a2 -1- 62 + c^ - 
 
 ■ 2 a6 — 2 ac + 2 6c 
 
 
 a2 + 62 + 5 c^ - 
 
 - 2 a6 - 6 ac + 6 6c 
 
 
 4a2 + 9 6^ + 16 c2 + 12a6 + 16ac 
 
 + 24 6c 
 
 4a2-962 + 16c2 + 16ac 
 
 ab (x^ — y^ + xy{d^ — 6^ 
 
 166. Signs in fractions. 
 
 167. The sign written before the dividing line of a fraction is 
 called the Sign of the Fraction. 
 
 It belongs to the fraction as a whole, and not to either the 
 numerator or the denominator. 
 
 In — — the sign of the fraction is — , while the signs of x and 3 z are + . 
 
 168. An expression like ^^ indicates a process in division, in 
 
 — 6 
 which the quotient is to be found by dividing a by 6 and prefixing 
 the sign according to the law of signs in division ; that is, 
 
 — a,a -\-a , a 
 
 — 6 6 +6 6 
 
 — a _ a + a _ a 
 + 6~ 6' -b~ b 
 
 By comparing the above fractions and their values the following 
 principles may be deduced : 
 
 169. Principles. — 1. TJie signs of both the numerator and the 
 denominator of a fraction may be changed without changing the sign 
 of the fraction. 
 
 2. Tlie sign of either the numerator or the denominator of a frac- 
 tion mmj be changed, provided the sign of the fraction is changed. 
 
 When either term of a fraction is a polynomial, its sign is changed by 
 changing the sign of each of its terms. Thus, the sign of a — 6 is changed 
 by writing it — a + 6, or 6 — a. 
 
140 FRACTIONS 
 
 Examples 
 Reduce to fractions having positive numbers in both terms : 
 —3 „ —a—x _ —a—h „ —2—n 
 
 o. -— • /. 
 
 -4 2x c-\-d 2 + n 
 
 2. -^. 4. -4^ . 6. ^2_. ^ -4(a + &). 
 
 -5 -b — y —d — y 5{—x — y) 
 
 170. Since, from the laws of signs, changing the signs of an 
 even number of factors does not change the sign of the product, it 
 follows that : 
 
 Principle 1. — The signs of an even number of factors of the 
 numerator or of the denominator may be cJianged without changing 
 the sign of the fraction. 
 
 Since, from the laws of signs, changing the signs of an odd 
 number of factors changes the sign of the product, it follows from 
 Prin. 2, § 169, that : 
 
 Pbinciple 2. — The signs of an odd number of factors of the 
 numerator or of the denominator may be changed, provided tht> sign 
 of the fraction is changed. 
 
 
 
 Examples 
 
 
 1. 
 
 Show that 
 
 -b b 
 b—a a—h 
 
 
 2. 
 
 Show that 
 
 — a a 
 
 
 6 — a + c a — h—c 
 
 
 3. 
 
 Show that 
 
 2 2 
 
 
 a{b — a) a(a — b) 
 
 
 4. 
 
 Show that 
 
 1 1 
 
 
 {a-b){c-h) (a-b)(b-c) 
 
 
 5. 
 
 Show that 
 
 m — n m — n 
 
 
 (a — c)(b — a) (c — a)(a — b) 
 
 
 6. 
 
 Show that 
 
 1 1 
 
 
 (6-a)(c-6)(a-c) (a -&)(&- c)(c- 
 
 -a) 
 
 7. 
 
 Show that 
 
 n—m m—n 
 
 . 
 
 (y -x){z- y) {x -z) {x- y) (y -z){z- x) 
 
FRACTIONS 141 
 
 171. To reduce a fraction to an integral or a mixed expression. 
 
 1. How many units are there in ^-^-^ in ^\^? in ^ ? 
 
 2. How many units are there in ^ c m r 
 
 Examples 
 1. Reduce ^^'^ to a mixed number. 
 
 X 
 
 PROCESS Explanation. — Since, § 158, a fraction may be re- 
 
 , , garded as an expression of unexecuted division, by per- 
 
 "* =z a -\ — forming the division indicated the fraction is changed 
 
 ^ ^ into the form of a mixed number. 
 
 "When the degree of the numerator is lower than the degree of the denomi- 
 nator, the fraction cannot be reduced to an integral or mixed expression. 
 
 Reduce the following to integral or mixed expressions : 
 
 2. 
 
 3. 
 4. 
 
 5. 
 6. 
 
 7. 
 
 a?- 
 
 3a^4-a-l 
 
 
 
 a' 
 
 
 4a^ 
 
 -Sx^ + 2x- 
 
 1 
 
 
 2x 
 
 
 ah - 
 
 -bc-cd + cP 
 
 
 
 b 
 
 
 aV 
 
 — ax^ — x — 1 
 
 
 
 ax 
 
 
 x"- 
 
 x-15 
 
 
 X 
 
 -4 
 
 
 c^- 
 
 2xy-f 
 
 
 10. 
 
 11. 
 
 1'> 
 
 a;3 4.5ar! + 3a;-6 
 
 
 a; + 2 
 
 13. 
 
 a' + 9 a^ + 24 a + 22 
 
 a + 3 
 
 14. 
 
 ar5_6a:2 + 14a;-9 
 
 x-1 
 
 15. 
 
 cB*-3a^ + 5a;-l 
 
 x-Z 
 
 16. 
 
 a* -f 3 a^b"" + &* 
 
 a2 + &' 
 
 17. 
 
 4 a;2 + 22 a; + 21 
 
 2a; + 4 
 
 18. 
 
 a^-3a^ + 4a;-3 
 
 a; — 4 
 
 19. 
 
 a^ + 2 a& + 6^ + c^ 
 
 a + 6 + c 
 
 f>(\ 
 
 a3-6a26 + l2a&2-10 6' 
 
 a^ — 6 xy + 4 y^ ^ 
 2xy 
 
 a2_2a-26 
 a + 5 
 
 a« + 2 a& + h\ 
 
 a-\-h a — 26 
 
 x^ + f 21. a;^ + 4a:^y + 6a^/ + 4a;y^ ^ 
 
 x'-xy + f ' x-\-y 
 
142 FRACTIONS 
 
 172. To reduce an integral or a mixed expression to a fraction. 
 
 1 . How many fifths are there in 6 ? in 10 ? in a ? in 3 6 ? 
 
 2. How many fifths are there in 64 ? in a + — - ? in 2 a + - ? 
 
 5 5 
 
 Examples 
 
 1 . Reduce a + - to a fractional form, 
 c 
 
 PROCESS Explanation. — Since 1 = c -h c, a = ac-^c. 
 
 Since, § 158, 
 
 _ac I, 5 
 
 <* — "T - means 6-^c, a + - = ac-^c + 6-HC. 
 
 c c c 
 
 b _ac b _ac-\-b § 104, 3, = (ac + b)^c. 
 
 Rule, — Multiply the integral part by the denominator of the 
 fraction; to this product add the numerator when the sign of the 
 fraction is plus, sid)tract it when the sign of the fraction is minus, 
 and write the result over tJie denominator. 
 
 If the sigji of the fraction is -, the signs of all the terms in the niunera- 
 tor must be changed when it is subtracted. 
 
 Reduce the following to fractional forms : 
 
 a^ — db 
 
 2. 
 
 -1- 
 
 
 3. 
 
 ._|. 
 
 
 4. 
 
 5c + — 
 
 c 
 
 5. 
 
 36-^ 
 
 + 26^ 
 b 
 
 g 
 
 4a; -1-''^ 
 
 — X 
 
 7. 
 
 40:+ ^ 
 , , , 2 6c - 1 
 
 8. a — 
 
 9. a — 
 
 10. fe- 
 
 ll, a 
 
 b 
 
 a — b—c 
 2 
 
 b — a — c 
 2 
 
 c — a — b 
 
 12. ^L±A±i_c. 
 
 13. 6 + ^'+^^ 
 
 ab — 6* 
 
a — X 
 
 
 a^ + 4: 
 
 
 X-4: 
 
 
 A.h2 _ 
 
 W 
 
 FRACTIONS 143 
 
 14. a + x 17. ^•\-xy + if-\ — 2l — 
 
 " ~ x-y 
 
 15. x + 5--^^^r^- 18. 3a-6&-^^^'-^^. 
 
 a+2h 
 
 16. a^-ah + b^ " — 19. x-a?-^ ~. 
 
 a+b 1+x ■ 
 
 173. To reduce dissimilar fractions to similar fractions. 
 
 1. Into what fractions having the same denominator may |, 
 |, and f be changed ? 
 
 12 4 
 
 2. Express -— , — -, and -—- by fractions whose common de- 
 
 sx 5x 15a; 
 
 nominator is the lowest common multiple of the given denomi- 
 nators. 
 
 174. Fractions that have the same denominator are called 
 Similar Fractions. 
 
 175. Fractions that have different denominators are called 
 Dissimilar Fractions. 
 
 176. Principle, — The lowest common denominator oj two or 
 more fractions is the lowest common multiple of their denominators. 
 
 The abbreviation L.C.D. is used instead of Lowest Common Denominator. 
 
 Examples 
 
 1. Eeduce - — and - — - to similar fractions having their 
 3bc ooJ) 
 
 lowest common denominator. 
 
 PROCESS Explanation. — Since the L. C. D. of the 
 
 f> n 2 given fractions is the lowest common multi- 
 
 —7- = —-- ~ = pie of their denominators (Prin.), the lowest 
 
 60c 3bc X ^a b abc common multiple of their denominators must 
 
 c __ c X c _ (^ be found. This is 6 abc. 
 
 6ab ~ 6ab X c ~ 6 abc '^^ reduce the fractions to equivalent frac- 
 
 tions having the common denominator 6 abc, 
 § 165, the terms of each fraction must be multiplied by the quotient of 6 abc 
 divided by the denominator of the fraction. 
 
144 FRACTIONS 
 
 Rule. — Find the lowest common multiple of the denominators 
 of the fractions for the lowest common denominator. 
 
 Divide this denominator by the denominator of the first fraction, 
 and multiply the terms of the fraction by the quotient. 
 
 Proceed in a similar manner with the other fractions. 
 
 All mixed expressions should first be reduced to the fractional form, and 
 all fractions to their lowest terms. 
 
 Reduce to similar fractions having their L. C. D. : 
 
 2. ^ and ^. 
 
 2 6 
 
 „ ab 1 a? 
 
 3. — and — 
 
 4 y 
 
 4. « and ^. 
 
 3 6y 
 
 2a ■, 3x 
 
 5 6 4 a a^y*' a^y^' a^ 
 
 6. ^ and ^- 12. ^±i^, ^ ^ " y, '^.^. 
 c'd cd^ . 2 4 6 
 
 7. -1- and ^. 13. ''-^=^, ^^^, a-^ -^ 
 2xy 4ay x y 3x 
 
 14. 
 
 15. 
 
 8. 
 
 m — n 
 a 
 
 2 * . 
 m-\- n 
 
 9. 
 
 4 6c 
 
 3ac 
 
 7a6 
 
 3aV 
 
 4 6V 
 
 6ac2 
 
 
 
 3a6 
 
 8aV 
 
 la" 
 ^l)'c 
 
 5 
 
 
 a36c2 
 
 .1. 
 
 3 
 
 -6 
 
 3 
 
 16. 
 
 17. 
 
 18. 
 
 19. 
 
 13. - 
 
 4 ay 
 
 X 
 
 -5 ■) 
 
 y 
 
 01? X X 
 
 
 
 or^ — l' a; + l' x — \ 
 
 
 a^ a 2 a 
 
 
 
 a* -16' a2 + 4' a=^-4 
 
 
 4 a 3 6 1 "^^ 
 
 
 
 a - 6' 6 + a' a^ - ? 
 
 
 
 a X —ax 
 
 
 
 1 — ax' 1 + ax ax + 1 
 
 
 1 1 
 
 ^ 
 
 1 
 
 x' + lx-^lQ' x' + x-2' 
 
 h4a; 
 
 a+5 a-2 
 
 
 a -, X 
 
 a2_4a + 3 a'^-Sa+lS a^-ea+S 
 
FRACTIONS 145 
 
 ADDITION AND SUBTRACTION OP FRACTIONS 
 
 T.- J XI- 1 „ 4 3 7 2 a & 1 2 
 
 177. 1. Find the value of - + -; : -+ : - + — 
 
 X X a ay y x y 
 
 2. What kind of fractions can be added or subtracted without 
 changing their form ? 
 
 3. What must be done to dissimilar fractions before they can 
 be added or subtracted? How are dissimilar fractions made 
 similar ? 
 
 178. Principle. — Only similar fractions may he united by 
 addition or subtraction into one term. 
 
 Examples 
 
 1. Find the algebraic sum of - + 
 
 b b b 
 
 I 
 
 PROCESS 
 
 J. ^A^ o a,cda + c — d 
 
 §104,3, - + ___ = ___ 
 
 2. What is the sum of —, — , and ^? 
 4 ' 10 12 
 
 PROCESS 
 
 Zx , 7x , 5v 
 
 — - 
 
 4 10 12 
 
 ^\.,x 42a; 25y 
 
 ^ 60 60 
 
 ^ 87a; + 25y 
 
 - 60 \ 
 
 Explanation, — Since the fractions are dissimilar, they must be made 
 "'IT befotP-they can be united into one term (Prin.). 
 
 •^^mmon denominator is 60. 3x ^ 45* 7* ^ 42^ 5j/ ^ 25]/ 
 , ^ 4 60 10 60 12 60 
 
 Therefore; the sum is ^^J + Ji^ + gil = ^5x + 42 x + 25y ^87 x + 25y. 
 60 60 60 60 60 
 
 ALG. 10 
 
146 FRACTIONS 
 
 3. Imd the algebraic sum of 
 
 ^ 8 7 4 
 
 PROCESS 
 
 bx-1 3a;-2 a;-5 ^ 35a;-7 (24a;-16) 14a;-70 
 8 7 4 56 56 56 
 
 ^ 35 a; - 7 - (24 a; - 16) + 14 a; - 70 
 
 56 
 ^ 35a;_7_24a;4-16 + 14a;-70 
 
 56 
 ^ 25a;-61 
 
 m 
 
 Suggestion. — When a fraction is preceded by the sign — , it is expedient 
 for the beginner to inclose the numerator in parenthesis, if it is a polynomial, 
 as is shown above. 
 
 Rule. — Reduce the fractions to similar fractions having their 
 lowest common denominator. 
 
 Change the signs of all the terms of the numerators of fractions 
 preceded by the sign — , then find the algebraic sum of the numera- 
 tors, and write it over the common denominator. 
 
 1. Reduce the resulting fraction to its lowest terms, if necessary. 
 
 2. The integral and fractional parts of mixed expressions may be united 
 separately. 
 
 3. An integer may be expressed as a fraction whose denominator is 1. 
 
 Find the algebraic sum of 
 . a—b,b—c 
 
 4. 1 • 
 
 ab be 
 
 _ a + b , a — b 
 
 o. 1 • 
 
 a — & a + b 
 
 6. -J^ + a. 
 a — x 
 
 7. a + 6 + 
 
 a — b 
 
 9. 
 
 b-c 
 
 a — c 
 
 be 
 
 ac 
 
 10. 
 
 a-\-b 
 
 a — b 
 
 a — b 
 
 a-hb 
 
 11. 
 
 5a' + b 
 a'-b' 
 
 2 
 
 --2. 
 
 12. 
 
 x + y- 
 
 o^-hy'. 
 
 x-y 
 
 19 
 
 X 
 
 x-2 
 
 8 a; + l 1 , 
 
 * x' + x + l x-1 ' x-2 x + 2 
 
FRACTIONS 147 
 
 Simplify : 
 
 2a; + la;-2 a; — 3 5 — a; 
 3 4 6 2* 
 
 x-2 x-4. 2-Zx 2x + l 
 6 9 4 12 ' . 
 
 x — 1 X — 2 4a; — 3 1 — X 
 " "3 18 2f~ 6 
 
 ,„ 2-6a; , 4a;-l 5x-3 1-x 
 5 2 6 3 
 
 a; + 3 a; — 2 a; — 4 a; + 3 
 4 5 10 6 * 
 
 a; — 4 x — 6 , ^ x-{-S 
 
 „^ 2-3a; 3-2a; , l-4a; 
 
 20. — + x 
 
 3 4 o 
 
 ^, l-2a,2a-l 2a-a2 + l 
 
 21. r 
 
 22. 
 
 5 4 8 
 
 S + x-a^ 1-x + x^ l-2a;-2a;' 
 4 6 3 ' 
 
 23. §^ + ^-3+i. 30. ^-& +^+^' ^ 
 
 5 2 6 2(a+6) a^-fi^ a-b 
 
 „. a — ba + b,6ab _, a-f-1 6 ,10 
 24. — — 7+-^^ — n- 31. — — — P + 
 
 a + 6 a — 6 a^ — b^ o.^ — 9 a + 5 a + 3 
 
 25. a±x^a-x_^_2a^^ 32_ 2 a - 3 6 - i^^^±^'. 
 a — a;a4-a;a^ — a;^ 2a + 3b 
 
 , i , ^ — ^ ooQ o 8a^ — 4a; 
 
 26, a;4-lH -• 33. 3a — 2x — — -— • 
 
 a; — 1 6a-\- 2x 
 
 ^^ ^_ax-bx + ab 34_ m-^' + ^^' + n. 
 
 a;^ m — w 
 
 28 g + l ■ Ct-l 35 _J: 1___l1. 
 
 • a^ + a + l a'-a + l ' 2(a;-l) 2(a; + l)^ar' 
 
 29. 3a;+ — -f2a; + AY 36. ^4-l4--l^-2. 
 
 ax V <*^/ ^ 1 + a-' 
 
148 FRACTIONS 
 
 07 O' ft — 2 , 3 
 
 a — 2 a + 2 4 — a- 
 SuGGESTiON.— By Prill. 1, §109, ^ -3 
 
 4 - a2 a-' - 4 
 
 38. 5L±1 + ^^JZ1+ 2a'' 
 
 a — 1 a + 1 1— a' 
 89. «i±2+ 2 3 
 
 £c2-4a;-2 2-a; 
 
 ^^ x(a + x) 3ax-af ^ ^^ 
 a — X X — a 
 
 41 cf + & g^ + &" . & — g 
 g — 6 b^ — a^ a + b 
 
 42. ^L^ + ^^ + _i 
 
 a; (a; — g) g (g — x) x — 
 
 43. 1-1^+. 1 
 
 g3 + 8 8-g3 4-g-^ 
 a; — 1 , a^ 3 
 
 44. ^; — r + 
 
 x^ — 5x -\-6 X — 2 x + 1 
 _ 2 , 3 2a;-3 , 1 
 
 X l-2a; 43^*-! l + 2a; 
 ..a^ + g^, a; + g 2 
 
 4o. — --J-- 
 
 •3? — c^ a;^ + ga;4-g^ x- 
 .„ 3 m 771 + 2 a; 
 
 48. 
 
 (m — 2 x)^ (m + .t) (m — 2 a;) m-\-x 
 
 3 2 
 
 if — m?/ — 12 tii^ y^ — 5 m?/ -\-^'ni^ 
 
 ._ g6 g?) ^ 
 
 49. 1. 
 
 g2 - g6 + 62 ^^ + g6 + 6^ 
 
 50. „ 1 „- . 1 „+ 1 
 
 51. 
 
 52. 
 
 x'2-3a; + 2 a^ + 2a;-3 a;2 + a;-6 
 2 11 
 
 ar' + SaJ + e a;' + 6 a; + 8 a;2 + 7a;4-12 
 5(a;-3) 2(a; + 2) x-\ 
 
 a^-x-2 x-2 + 4a; + 3 6-a;-a^ 
 
FRACTIONS 149 
 
 _„ a — h — c a — h + c. 4rtc 
 
 5o. — ; ; h 
 
 a + h -{- c a + 6 — c (a-\-hy — c^ 
 54. Simplify „ — —z:—2 + —» — 7. — —7' 
 
 + 1 a' + 2a + 
 
 Solution 
 
 ff -^ - 2 a + 1 a^ + 2 a + 1 \ a^ - 2 a + 1 / \ a^ + 2 a + 1 / 
 
 4a 4a 16a^ 
 
 a2-2a+l a2+2a+l (a-l)2(a+l)2 
 
 Suggestion. — Frequently, by reducing one or more of the given fractions 
 to mixed numbers, the integers cancel each other and the numerators are 
 thus simplified. 
 
 Simplify : 
 
 55. 
 
 56. 
 
 57. 
 
 58. 
 
 59. 
 
 60. 
 
 61. 
 
 a — b a + b a^ + b^ a* -\- b* 
 
 Solution 
 Combining first two fractions, -^ ^ = ^ "^ . (1) 
 
 Combining (1) with the third fraction, -^^ ^^^ = ^"^^ • (2) 
 
 Combining (2) with the fourth fraction, -^^ i^ =A^^. (3) 
 
 Hence, « « ^ «* 4 a68 _ 8a67 
 
 a2+62 
 
 1+ ^<'b . 
 
 a^ + 3ab + 2b^ 
 
 a"- 13 b^ 
 
 a'-^Sab-Ab' 
 
 a'-16lr' 
 
 x^-\-x + l x" 
 
 — x-{-l 
 
 af-x + 1 a^- 
 
 -2a;+l 
 
 a^ + a^ + x + 1 
 
 1 ^ 
 
 a^ — a^ + x — l 
 
 ^ x-1 
 
 X + 1 x—1 
 X — 1 x + 1 
 
 x+2 x-2 
 
 x-2 x+2 
 
 x-{-3 x~3 
 x—3 x+3 
 
 X-\-4: X — 4: 
 X — 4: X+ 4: 
 
 a a 
 
 2ab 4.ab^ 
 
 a-b a + b a^ + 62 a* + b* a^ - 6" 
 
150 FRACTIONS 
 
 -„ a + h a — b 4 a6 , Sab^ 
 D<6. ;; ~r 
 
 a-b a + b a' + b^ a*+b* 
 
 a-b a + b a^ -{- b^ a* -\- b* 
 
 _ . 1 , _1 x__ __ 7? + ^X 
 
 ' x — 1 x + l x^ + 1 0^" + !' 
 
 a-\-x g^ + a^ _a — a3_a^ — a^_4 a^x -}- 4 aa? 
 a — a; a^ — x^ a + x a^ + x^ a* — x* 
 
 66. Simplify -^ + . 
 
 (6 — c){c— a) {a — c) (a — 6) (b — a)(c — b) 
 
 Solution 
 a b 
 
 + ■ 
 
 (6 — c)(c — a) (a — c)(ffl — 6) (6 — a)(c — 6) 
 § 170, = ^^ + ^^ + *" 
 
 (6 — c)(c — a) (c — a)(a — b) (a — 6) (6 — c) 
 
 _ a(a— b)+ b(b — c)4- c(c — a) _ 
 (6 — c)(c — a) {a — b) 
 
 Simplify : 
 
 67. = \ 1 . 
 
 (& — c)(a — c) (c — a) (a — b) (b — a) (6 — c) 
 
 68. « + l I b + 1 c + 1 
 
 69. 
 
 70. 
 
 (a — 6) (a — c) (6 — c) (6 — a) (a — c) (6 — c) 
 
 c^a& &^ca a^bc 
 
 (c — a) (6 — c) (6 — a) (6 — c) (a — b)(a — c) 
 
 b — c c — a a + b 
 
 (b — a) (a — c) {b — c){a — b) (a — c) (b — c) 
 
 c + a b + c , a + b 
 
 71- -. rrr. r — 7 tt; r + 
 
 72. 
 
 (a — b)(b — c) (c — a)(b — a) (c — b)(a — c) 
 
 c + a c — b a — b 
 
 {a + b){b — c) (c — a) (a + b) (b — c) (a — c) 
 
 _ c + a — b . b + c—a a + b + c 
 
 (a — 6) (6 — c) (c — a) (a — 6) (6 — c) (a — c) 
 
FRA CTIONS 151 
 
 MULTIPLICATION OP FRACTIONS 
 
 179. 1. How much is 5 times f? 2 x ^? 6x^? 3 x — ? 
 
 4x^? 3x^? ex?? ax^? 
 5 4 4 d 
 
 2. Express 5 x ^^ in its lowest terms ; 3 x | ; 4 x ^r^; 2 x -— ; 
 
 Q K 7 12 6 
 
 Uy—- 7 X — • 10 X — • 8 X 
 ^"^22' ^14' ^"^20' "^16 
 
 3. In what two ways, then, may a fraction be multiplied by an 
 integer ? 
 
 4. How much is ^ of \% or ^ ^ 5 ? i of ^, or — -- 4 ? 
 
 5. How much is ^ of |, or | -=- 5 ? ^^ of -— , or =- 4 ? 
 
 6. In what two ways, then, may a fraction be divided by an 
 integer ? 
 
 180. Principles. — 1. Multiplying the numerator or dividing 
 the denominator of a fraction by any number multiplies the fraction 
 by that number. 
 
 2. Dividing the numerator or multiplying the denominator of a 
 fraction by any number divides the fraction by that number. 
 
 Examples 
 
 1. Multiply I by 5. 
 d 
 
 PROCESS Explanation. — To multiply - by - is to find c times 
 
 b d 
 a c__ac 
 
 b d bd *^® P*'^ o* -• - P*"^ of - = — (Prin. 2), and c times 
 d b d b bd 
 
 -^ = ^ (Prin. 1). Therefore, 7 x ^ = ^• 
 bd bd^ ^ b d bd 
 
 OB 
 
 To find the product of ^ x |- 
 u 
 
 Let »-ix|- (1) 
 
 a 
 
152 FRACTIONS 
 
 Multiplying each member of the equation by 6 x d, Ax. 4, 
 
 xxb X d = -x-xb X d 
 b d 
 
 § 104, 1, =^xbx-xd = ac. 
 
 b d 
 
 xbd = ac. 
 
 Dividing by bd, Ax. 5, « = — • (2) 
 
 bd 
 
 .-. Eq. (1) and (2), Ax. 1, ^x| = f§. 
 
 b d bd 
 
 Similarly, «x:^x? = «^; 
 
 b d f bdf 
 
 and so on for any number of fractions. 
 
 Rule. — Multiply the numerators , together for the numerator of 
 the product, and the denominators for the denominator of the 
 product. 
 
 1. Cancel equal factors from the numerator and denominator. 
 
 2. Reduce integral and mixed expressions to the fractional form before 
 multiplying. 
 
 2. Simplify ^^±^ X ^^^ X —^ 
 
 ^ ^ x + 3 x-\-5 'Jaf-lOx 
 
 Solution 3x^ + 16x ^ gj-g ^ 1 
 
 X4-3 x + 5 2x2_i0x 
 
 ^ 3x(x + 5) ^^ (x + S)(x-3) ^^ 
 
 x + S x + 6 2x(x-5) 
 
 Canceling, ^ 3 (x - 3) 
 
 ^ 2(x - 5) 
 
 Multiply : 
 
 3. ^ bv -^^ 6 i^ bv -^^^ 
 
 4:xy -^ 3a^ Sxy ^ IGm^ 
 
 4 ^ bv ^-^- 7 ^^ bv -1^ 
 
 ' 2 ac -^ 10 ?/2 ' 12 by ^ x' ' 
 
 _ 4a6 , 3 6c , _ a'"6" , 6.t« 
 
 lOc^ -^ a» 4 a; "^ a'"-^^*" 
 
FRACTIONS 153 
 
 9. <^'by^'. n. ^^ by 25-10 »:. 
 
 b"'+^ 'or 20 -8 a; 
 
 x-y 
 
 -_ a , h ,„1 — or<; + 6a^,2 
 
 a-\-h a — h 2—^x-\-x^ 1—x 
 
 Simplify the following : 
 
 13. (a-6)% . b ^ Aa-hbY 
 a + b o? — ah o? — b'^ 
 
 14 ^* ~ '^^ X ^ + ^ X g^ — aa; + ar^ 
 a'' + a:^ a^ — o? (a + a;)^ 
 
 ,_4a — & 2a 4a:^ — ?/2 
 
 15. 7i X— -X —^i- 
 
 2x + y Aa^ — ab 4 
 
 j6. £±4 X y - ^i X — ^ 
 
 a; -3 2^2-8 j9x + 3i> 
 17. i^'-g" X ^~^ X -^— . 
 
 18 a^ + 8 a'^ + 2a + 4 
 • a3_8 a2-2a + 4' 
 
 jg g* + g V + a;^ ^ 
 
 20. 
 
 g* — ga;* g^ — ga; + a^ 
 
 g* + 4 ^ g^ + g + l 
 
 g* + g^ + 1 g2 + 2 g + 2 
 
 21 g^ + gx + e ^^ a^' + Ta^ + lO 
 ar' + Gaj + S a^4-7a; + 12 
 
 g^ + 3 a; - 10 g^ - 10 a; + 21 
 
 a^-4a;-21 ar' + Taj + lo' 
 
 g^4-g^+gc4-?>c a^—ax-{-ay—xy x^—x{y—d)—ay 
 ax~ay — x^-\-xy a^-\-ac+ax+cx a^—a(j/—b)—by 
 
 a^-5x^ + 8x-4: a;« - 10 ar^ + 33 a; - 36 
 • a:3_8ar' + 19x-12 a^-Gx' + llx-G ' 
 
 a^-3ar^-23a^ + 75a;-50 a^- lOar' + 29a;- 20 
 ' aj*-oa;«-21a;2 + 125a;-100 a;^ _ ^2 a;^ + 45 a; - 50' 
 
164 FRACTIONS 
 
 DIVISION OF FRACTIONS 
 
 181. 1. How many times is \ contained in 1 ? ^ in 1 ? ^ in 1 ? 
 
 2. If the numerator of a fraction is 1, how does the number of 
 times the fraction is contained in 1 compare with the denominator ? 
 
 3. How many times is - contained in 1 ? in 1 ? 
 
 d x + y 
 
 o _ 
 
 4. How many times is ^ contained inl? -inl? -inl? 
 
 d d 
 
 182. The reciprocal of a fraction is the fraction inverted. 
 
 For, since - multiplied by - produces 1, § 102, if 1 is the dividend and 
 a b 
 
 - the divisor, - is the quotient. 
 h a 
 
 Examples 
 
 1. Divide - by — 
 b ^ d 
 
 Explanation. — The fraction - is contained in 1, 
 
 PROCESS ^ 
 
 c 1 
 
 c a d ad ^ times ; and - is, therefore, contained in 1, - part 
 
 b d b c be t J ^- d .. 
 
 of d times, or - times, 
 c 
 
 Since - is contained - times in 1, it will be contained - times -, or — 
 d c ' h c be 
 
 times in -• 
 
 ^ OR 
 
 To find the quotient of ^-r--. 
 b d 
 
 § 102, X X - = ^• 
 
 d b 
 
 Multiplying each member of this equation by -, 
 
 c 
 c d a d 
 
 XX-X- — -X-- 
 
 d c b c 
 
 That is, x = ^x^' (2) 
 
 be 
 
 Hence, eq. (1) and (2), Ax. 1, ^ ^ - = - x -• 
 b d b e 
 
FRACTIONS 155 
 
 EuLE. — Multiply the dividend by the reciprocal of the divisor. 
 
 1. Change integral and mixed expressions to the fractional form. 
 
 2. An integer may be expressed as a fraction by writing 1 for its de- 
 nominator. 
 
 3. When possible, use cancellation. 
 
 Divide : 
 
 2. 1 by — 4. 1 by 
 
 3. Ibyi '-'^y^^s 
 
 Write the reciprocal of 
 
 10. 
 
 b n b —y 
 
 7. ^. 9. ^. U. 4- 
 
 p om ab 
 
 Simplify : 
 
 ,„ 5mn , 10 m^ „, a? + x — 2 
 
 Qbx 
 
 ' 3aa^ 
 
 12 a'b 
 
 4ax 
 
 25 ac 
 
 ■ 24 c2 
 
 Sabm 
 
 7 
 
 -i-a^x. 
 
 5ab 
 
 25 b^ 
 
 3aV 
 
 ' 15 a' 
 
 7af . 
 
 21 x'f • 
 14 a 
 
 my — 
 
 f . f 
 
 ■ 25ac ■ 24c2 
 
 14. ^-^^^abx. 23. 
 
 7 
 
 ,^ 5a& 25 &2 
 
 15. —-TT^ -s- ^ ^ „ • 24. 
 
 ar^ — 5a; + 4 
 a* -6* 
 
 a^- 
 
 - X — 
 
 6 
 
 0^ + 
 
 X — 
 
 20 
 
 . a' 
 
 + b' 
 
 
 a'-2ab + b' a'- ab 
 
 x^ + f . a^-\-ocy + y' 
 x^ — y' x — y 
 
 a?+h^ gg _ aft 4- 1/ 
 a2_4&2 • a -26 
 
 7_^_^21^V. ' 25 m^ — if m^ + m'^y + my^ 
 
 my — y* my^ + y^ 
 
 17. . "'■^-y ^ 2^ 26. 
 
 m*a; + m* m^a^ — ma;* 
 
 (m + yY m' — if m^x — ma^ im?v? + a^ 
 
 20 
 
156 FRACTIONS 
 
 x^-6x^+llx~6 . 0^-13 a; + 12 
 x^ + 2a^~19x-20 ' a^ + lOa^ + 29a; + 20* 
 
 ar^-15a^ + 74a;-120 . a^ -9a^ + 26x-24: 
 3^-5x2-07 + 5 ■ a^-ex' + llx-e' 
 
 a2 + 62_c2 + 2a6 . g' -b'' + (f -2ac 
 a^^h^-(? + 2bc ' a?-h^ + c' + 2ac 
 
 a^ -\- x^ — y^ -\- 2 ax _^ a? — a? + y^ -\- 2 ay 
 a^ — x^ + y^ — 2ay a^ — x^ — y^ — 2ocy 
 
 30. 
 
 31. 
 
 32. 
 
 33. 
 
 34. Simplify (3,-0. + ^^^1 + 1 
 
 Solution 
 
 Simplify : 
 
 (--7)^(1-^) 
 
 _y^ — xy + x"^ . x^ + y^ 
 
 y 
 
 X2y2 
 
 
 
 _x^-xy + y^^^ 
 
 
 a;V 
 
 
 y (• 
 
 x-^y)( 
 
 x^ — xy + J/*) 
 
 _ x^y 
 
 
 
 
 x + y 
 
 
 
 
 
 H^ 
 
 4:X - 
 
 0^- 
 
 -^} 
 
 f.,^'' + ''l'. 
 
 '^^f. 
 
 :-\-S-4 
 
 1 
 
 37. (af-^^fx-'^ 
 
 38. 
 
 40. 
 
 i + i + iUri+l+i 
 
 y^ y*. 
 
 y y 
 
 39. fl-n-(l-^ + -^ 
 x^J \ y y\ 
 
 2x 
 
 2f 
 
 ^■\-f. 
 
 ^ 1- 
 
 1]L 
 
 ^ + y. 
 
FRACTIONS 157 
 
 COMPLEX FRACTIONS 
 
 183. A fraction one or both of whose terms contains a fraction 
 is called a Complex Fraction. 
 
 It is simply an expression of unexecuted division. 
 
 Examples 
 
 - 
 
 1. Simplify the expression -• 
 
 y 
 
 PROCESS 
 
 £_a_^x_a y _ay 
 X b y b X bx 
 
 y 
 
 Simplify the following expressions : 
 
 x + y 2 + — 
 
 ab 4 b 
 
 — « + - 
 
 ab^ 3 
 3. * 6. j. 9. 
 
 1 
 
 X 
 
 X 
 
 7. Y" 1®' 
 
 11. Simplify the expression ^ — 
 
 -2 + - + 1 
 
 y^ y 
 
 Solution. — Multiplying the numerator and denominator of the fraction 
 by p% the L. C. D. of the fractional parts of the numerator and denominator, 
 
 the depression becomes ^ — ^V +y . 
 
 x'^ + xy +y' 
 
 h + 
 
 c 
 
 
 a 
 
 
 Sm 
 
 m - 
 
 
 
 
 X 
 
 
 X 
 
 X - 
 
 
 
 
 m 
 
 6- 
 
 c 
 2 
 
 
 
 b 
 
 
 
 
 2 
 
 c 
 
 
 
 ax 
 
 2 
 
 
 
 a' 
 2 
 
 -ax 
 
 
 x + y 
 
 aj 
 
 + y 
 
 y 
 
 
 
 X 
 
 
 1 
 
 1 
 
 
 
 2^ 
 
 X 
 
 
158 FRACTIONS 
 
 Simplify the following : 
 
 12. -^^. 14. , "^^ „ • 16. 
 
 1 1 
 
 + 
 
 
 x^-irf 
 
 
 xy 
 
 ar' 
 
 ~xy-\-f 
 
 1 
 
 a + 1 
 
 1 
 
 1 
 
 x' + y 
 
 2 
 
 X 
 
 22/ 
 
 a; 
 
 y 
 
 a; 
 
 
 13. ^: yf- 15. ^+! . 17. 1:1^. 
 
 a; 2/ + 2! a + 1 1 + a 
 
 a;-2+— ^ 6a-l-- 
 
 18. ^1. 20. -^ ^• 
 
 x + 2+-^ ^^-Zl 
 
 x-2 3a 
 
 14 4 a; -5 ^24 
 
 ,^ X x^ a? 2 X 
 
 19. ^. 21. 
 
 1 + ^V^„ 
 
 3a;-9 
 
 .'c a;- X 
 
 ^ + 1 a; + l 1 — a; 
 22. —. h H 
 
 \-\-x \ — x \-\-x 
 
 X—\ V — \ z — 1 
 
 3 xyz X y 
 
 23 ^ -^ 
 
 ?/z + zx + an/ 1 1 1 
 
 X 2/ ^ 
 
 12 9 a^ + (a + &)a; + a6 
 
 x + y X — y ^x — y a^ — (a -\-b)x + ab 
 
 24. — ^-^ f ^. 25. ^ J^ ;„ 
 
 — 8y a.-^ — ft'' 
 
 2/^ — 9 a^ * ^ — (^ 
 
 i+ 1 
 
 a 6 + c 
 26 
 
 1 1_ • ^ y' + d'-a? 
 
 a h +c 2 he 
 
FRA CTIONS 159 
 
 x + y a^ + y^ 
 
 x — y _ a? — -if oj* + xh^ + 2^ 
 x + y a^ + f 
 
 a — hh — c ^ c 
 
 ^^ 1 + a& 1 + 6c 
 
 29. 
 
 ^ (a-6)(6-c) -1 
 (1 + db) (1 + 6c) a "^ 
 
 (g + yf -z" {x-y + zy 
 f 4. 
 
 184. An expression of the form is called a 
 
 Continued Fraction. b-\ 
 
 d + 
 
 /+ 
 
 30. Simplify 
 
 1+^ 
 
 ^-l 
 
 Solution 
 1 
 
 1 + -^ 1 + 
 
 14-^ ^ + i 
 
 X X 
 
 1 
 
 1 + 
 
 x+1 
 _ x + 1 
 ~ x + 1 +z 
 
 _ a; + 1 , 
 ~2x+i 
 
 SuGGESTTON. — In the above example, the part first simplified is the last 
 
 complex part r, which is reduced to a simple fraction. 
 
 1 + - 
 
 X 
 
 Every continued fraction may be simplified by successively reducing its 
 last complex part to a simple fraction. 
 
160 FRACTIONS 
 
 Simplify the following : 
 1 
 
 31. 
 
 32. 
 
 SS. 
 
 .+ 1 
 
 1+^ + 1 
 
 3 — a; 
 
 a 
 
 a + lH 
 
 a + 1-- 
 a 
 
 2- 
 
 34. 
 
 x-2 
 
 X ^ a; 
 
 35. 
 36. 
 
 x~2 
 1 
 
 .a + - 
 a 
 
 1 1 " 
 
 ' 1+0+ 2c 
 
 2-^ 1 + 1 
 
 2 — a; c 
 
 REVIEW OF FRACTIONS 
 185. Reduce to their lowest terms : 
 
 ^ a^-f a^ + a;-3 a;^ + a:^ - 22 a; - 40 
 
 a^ + Sx' + Sx + S ' a^-7a^ + 2a; + 40' 
 
 2 ^-i^-x-2 a!« + 10a^ + 7a;-18 . 
 '«» + 3ar' + 3a; + 2* ' a^-Sx'-llx + lS 
 
 3 a^ + 4a^ + 8a; + 5 2a^ + 7a;''-9a!-9 
 
 ■ a^ + 3a;2 + 7a; + 5' ' 2a^ + 9aj2 + x-3 ' 
 
 ^ a^ + 3ar' + 4a; + 2 8g^ - 22a^ + 17a; -3 
 
 •a;3_3aJ!_8a;-10* ' 6a^ -17 a^ -i-Ux-3 
 
 Simplify : 
 
 9 ^ + y - y-^ 
 
 ■ 2y-1^2y + l l-Ay^ 
 
 10. — ^! ( g +-^ l^l^Y 
 
 4(l-a)2 V8(l-a) 8(a + l) 4(a + l)y 
 
 11 2a;-l 3a;4-l 3a;-l 2a; + l 
 x — 1 x + 1 x — 2 x + 2' 
 
 12. ^^^ + ^— + "" 
 
 {a — 6) (a — c) (6 — c) (6 — a) (c — a) (c — 6) 
 
22. 
 
 13. 
 14. 
 15. 
 16. 
 
 FRA CTIONS 
 
 m- 4- m — 2"* 
 
 161 
 
 fl + 2 \ / m^ + m - 2 Y 
 \^ m — ly\ m^ + m J 
 
 1 
 
 a + a' 
 
 
 af X 
 
 a 
 
 y 
 
 + 1 
 
 17. 1 + 
 
 a — b a + 6 a^ + 6'^ 
 
 a;" 
 
 4 a6 \ a^ — 6' 
 
 y. 
 
 
 18. 
 
 19. 
 
 20. 
 
 x+l+^ + h^ix + 1 
 X ctr^ 
 
 a + & a^ + 6^ /a + 6 
 
 X x^J 
 
 a — b a^ — b^J\a — b 
 
 63 
 
 x^ + y' 
 
 \x-y J 
 
 21. (a^-/ 
 
 z^ 
 
 2yz)^ 
 
 x-\-y-\-z 
 
 1 -f a 4ct 8 a 
 
 1-al + a'l-a^ 1 + a 
 
 
 1- 
 
 l + g' 4 a'' 
 
 a" 1 + a^ 1 + 
 
 23. 
 
 a^a^ . abx' _ aca^ _ &^ . a^ _ a 
 6d (^d cP ccP be d 
 
 ax b 
 c d 
 
 24. (x'-3xy-2y'+ ^^^ \^(3x-6y- ^^ Y 
 
 25. 
 
 26. 1 
 
 m — 3 ri\ A 4 n \ ^ /^^ , o _ ^^ ^'^A 
 
 2a; + 5ar (a^+x 
 
 2(x + iy 
 
 2x- 
 
 27. 1 + 
 
 a — a;y va; + a 
 
 ALG. — 11 
 
 / 3-3a; \ 
 
 X' 
 
 2 ay' -\- 2 ax — a 
 3c^ + 3ax + 2a^ 
 
162 FRACTIONS 
 
 Expand : 
 
 Simplify : 
 
 1 4_i ±.— 6a 4 
 
 34. ^. 38. -^ -^. 
 
 1 1\ a2 + 3a + 2 
 
 — 7>l 1^1 
 
 
 35. -^^ , .^ / ^— 39. 1- 
 
 m^ — n^ g^ + 5 g + 4 
 
 d m^ — w^ A 2 w 
 
 .+ 
 
 3g_ (g + 1)'' (g + iy ^^ m« + n' 
 
 
 u, 1 ' -J 2m?i 
 
 (g + 1)* (g + 1)* m^ — mn + w^ 
 
 37. z 41. 
 
 1- 
 
 3 
 
 4- ' 
 
 1 — X 6 — X 
 
 ' \x — y x + y) ' \2(x — y) 2x+2yj 
 
 (i+-+-+-Mi--+-> 
 
 \ar x a ay \oir x aj 
 
 ' Qt?f-a?y'^ 2 4 a;?/ + 1 
 
 xy x^y^ 
 
SIMPLE EQUATIONS 
 
 ONE UNKNOWN NUMBER 
 
 186. An Equation has been defined, § 2, as an expression of 
 equality between two numbers or quantities. 
 
 187. An equation all of whose known numbers are expressed 
 by figures Us called a Numerical Equation. 
 
 188. An equation one or more of whose known numbers is 
 expressed by letters is called a Literal Equation. 
 
 189. - An equation that does not involve an unknown number 
 
 in any denominator is called an Integral Equation. 
 
 2 X 
 sc + 5 = 8 and h 5 = 8 are integral equations. The second equation 
 
 is integral ; for though it contains a fraction, the unknown number x does 
 not appear in the denominator. 
 
 190. An equation that involves an unknown number in any 
 denominator is called a Fractional Equation. 
 
 8 Ix 
 
 a; + 5 = - and = 7 are fractional equations. 
 
 x x — \ 
 
 191. An equation whose members are identical, or such that 
 they may be reduced to the same form, is called an Identical 
 Equation, or an Identity. 
 
 a + b=a + b ; a^ — b^ = (a -i- b)(a — b) are identical equations. 
 
 An equation whose members are numerical is evidently an 
 identical equation. 
 
 10 = 6 + 4; 8x2 = 6 + 12 — 2 are identical equations. 
 
 163 
 
164 SIMPLE EQUATIONS 
 
 A literal equation that is true for all values of the letters in- 
 volved is an identical equation, or an identity. 
 
 (x + yY = x^ + 2xy + y^ is au identity, because it is true for all values of 
 X and y. 
 
 192. An equation that is true for only certain values of its 
 letters is called an Equation of Condition. 
 
 An equation of condition is usually termed simply an Equation. 
 
 X 4- 4 = 10 is an equation of condition, because it is true only when the 
 value of a; is 6. a;^ _ 9 jg ^q equation of condition, because it is true only 
 when the value of a; is + 3 or — 3. 
 
 193. When an equation is reduced to an identity by the substi- 
 tution of certain numbers for the unknown numbers, the equation 
 is said to be satisfied. 
 
 When « = 2, the equation 3 x + 4 = 10 becomes 6 + 4 = 10, an identity ; 
 consequently, the equation is satisfied. 
 
 Any number that satisfies an equation is called a Root of the 
 equation. 
 
 2 is a root of the equation 3 x + 4 = 10. 
 
 Finding the roots of an equation is called solving the equation. 
 
 194. An integral equation that involves only the first power of 
 one unknown number in any term when the similar terms have 
 been united is called a Simple Equation, or an Equation of the First 
 Degree. 
 
 3 X + 4 = 10 and x + 22/ = z + 8 are simple equations. 
 
 195. Two equations that have the same roots, each equation 
 having all the roots of the other, are called Equivalent Equations. 
 
 By the axioms in § 74, if the members of an equation are 
 equally increased or diminished or are multiplied or divided by 
 the same or equal numbers, the resulting numbers are equal and, 
 § 186, form an equation. But it does not necessarily follow that 
 the equation so formed is equivalent to the given equation. 
 
 For example, if both members of the equation x + 2 = 5, whose only root 
 is X = 3, are multiplied by x — 1, the resulting numbers (x + 2)(x — 1) and 
 5(x — 1) are equal and form an equation (x + 2) (x — 1) = 5(x — 1) ; but this 
 equation is not equivalent to the given equation, since it is satisfied by x = 1 
 as well as by a; = 3. 
 
SIMPLE EQUATIONS 165 
 
 196. Principles. — ^- If the same expression is added to or 
 subtracted from both members of an equation, the resulting equation 
 is equivalent to the given equation. 
 
 2. If both members of an equation are multiplied or divided by 
 the same Tcnown number, except zero, the resulting equation is equiva- 
 lent to the given equation. 
 
 3. If both members of an integral equation are midtiplied by the 
 same unknown integral expression, the resulting equation has all the 
 roots of the given equation and also the roots of the equation formed 
 by placing the multiplier equal to zero. 
 
 It follows from Prin. 3 that if the same unknown factor is removed 
 from both members of an equation, the resulting equation has all the roots 
 of the given equation except those obtained from the equation formed by 
 placing the factor removed equal to zero. 
 
 Principle 1 may be established as follows : 
 
 Let A = B (1) 
 
 be any equation and C any expression to be added or subtracted. 
 
 It is to be proved that A± C= B ±C . (2) 
 
 is equivalent to (1), the given equation. 
 
 All the values of the unknown number or numbers that satisfy (1), that 
 is, make A identical with B, make A + C identical with B + C and A — 
 identical with B — C, that is, satisfy (2). Hence, (2) has all the roots of (1). 
 
 For the same reason A±CT C= B± CT C (3) 
 
 has all the roots of (2). But by the Associative Law for addition (3) may be 
 written A + = B + 0, or A = B. Hence, (1) has all the roots of (2). 
 
 Since (2) has all the roots of (1) and (1) has all the roots of (2), (2) is 
 equivalent to (1), the given equation. 
 
 Principles 2 and 3 may be established as follows : 
 
 Let A = B (1) 
 
 and MA = MB. (2) 
 
 From (1), by Prin. 1, A-B = 0. (8) 
 
 From (2), by Prin. 1, M(A - S)= 0. (4) 
 
 Since the first member of (4) can reduce to zero only when one or both of 
 its factors become 0, (4) is satisfied by those values of the unknown number 
 that make A — B = 0, that is, by the roots of (3), or, Prin. 1, of (1); and 
 also by those values of the unknown number that make M=0, that is, by 
 the roots of M = 0, but by no other values. 
 
 If M is any known number, not zero, M cannot be placed equal to zero 
 and then (4), or (2), is equivalent to (3), or to (1). 
 
166 SIMPLE EQUATIONS 
 
 If Jf is an unkuown expression, (4), or (2), lias the roots of itf = in addi- 
 tion to the roots of ^ — ^ = 0, or of (1). 
 
 197. By § 196, Prin. 1 and 2, every simple equation involving 
 one unknown number may be reduced to an equivalent equation 
 having the form x = a, a being a fixed known number. Hence, 
 
 Every simple equation involving one unknown number has one 
 root and only one ; also, by § 196, Prin. 3, 
 
 The equation (x — a) {x — h) {x — c) ••• (x — r) = is equivalent 
 to the simple equations x — a = 0, x — b=:0, x — c=0, '■• x — r = 0, 
 and has as many roots as factors of the first degree involving x. 
 
 CLEARING EQUATIONS OF FRACTIONS 
 
 198. 1. If one third of a number is 10, what is the number ? 
 
 2. li \x = l, what is the value of aj? li \x = 5, what is the 
 value of a; ? If i a; = 6, what is the value of a; ? 
 
 3. If I a; = 6, what is the value of 2 a;? If ^a; = 10, what is 
 the value of 5 a; ? 
 
 4. If - = 5, what is the form of the resulting equation when 
 
 o 
 
 both members are multiplied by 3 ? by 6 ? by 9 ? by any multiple 
 of the denominator ? 
 
 199. The process of changing an equation containing fractions 
 to an equation without fractions is called Clearing the Equation of 
 Fractions. 
 
 200. Principles. — 1. An equation may he cleared of fractions 
 by multiplying both members by some common multiple of the denomi- 
 nators of the fractions. (Ax. 4.) 
 
 2. If both members of a fractional equation are midtiplied by the 
 eocpression of loioest degree required to clear the equation of fractions, 
 the resulting equation is equivalent to the given equation. 
 
 Principle 2 may be established as follows : 
 
 By § 196, Prin. 1, all the terms of the second member may be transposed 
 to the first member. Hence, uniting the terms of the first member into one 
 and reducing this to its lowest terms, any fractional equation may be reduced 
 to an equivalent equation of the form 
 
 4 = 0) 
 
SIMPLE EQUATIONS 167 
 
 in which A is prime to J5, and B is the expression of lowest degree required 
 to clear the equation of fractions. 
 
 Since A and B have no common factors, A and B cannot reduce to zero 
 at the same time for any value of the unknown number. Hence, eq. (1) 
 is satisfied by every value of the unknown number that makes ^ = 0, and by 
 no other values ; that is, the equation ^ = 0, obtained by multiplying both 
 members of the given equation by the expression of lowest degree required 
 to clear it of fractions, is equivalent to the given equation. 
 
 Examples 
 
 1. Given '- =6 , to find the value of x. 
 
 4 3 
 
 PROCESS 
 
 a; — 4 _^ X 
 
 Clearing of fractions, 3a; — 12 = 72 — 4a; 
 
 .-. a; = 12 
 
 Explanation. — Since the first fraction will become an integer if the 
 members of the equation are multiplied by 4 or any multiple of 4, and since 
 the second fraction will become an integer if the members of the equation are 
 multiplied by 3 or any multiple of 3, the equation may be cleared of fractions 
 in a single operation by multiplying its members by some common multiple 
 of 4 and 3 (§ 196, Prin. 2, § 200, Prin. 1). 
 
 Since the terms derived from the numerators will be least or lowest when 
 the multiplier is the least or lowest common multiple of the denominators, 
 the members of the equation should be multiplied by the least common mul- 
 tiple of 4 and 3, which is 12. 
 
 The resulting equation is 3 a; — 12 = 72 — 4 a; ; .-. a; = 12. 
 
 X — 1 
 2. Find the value of x in '■ 
 
 x-2 2 
 3 3 
 
 X-3 
 4 
 
 2 x-S 
 
 
 3 4 
 
 
 Solution. 
 
 2 3 
 
 Clearing of fractions, 6(x - 1) - 4(x - 2) = 8 - 3(a; - 3). 
 
 6a;-G-4a; + 8 = 8-3« + 9. 
 
 .-. X = 3. 
 
168 SIMPLE EQUATIONS 
 
 Rule. — Multiply both members of the equation by the least or 
 lowest common multiple of the denominators. 
 
 1. The multiplier of lowest degree required to clear an equation of frac- 
 tions so that the resulting equation is equivalent to the given equation is 
 usually the L. C. D. But if fractions having like denominators have not 
 been united and every fraction reduced to its lowest terms, the multiplier 
 required may be of lower degree than the L. C. D. 
 
 Thus, given ^^ = ^J^±^ + ^^±^. 
 
 % — 'I X — 2 3 
 
 Uniting terms, 2x2-.Sa;-2 ^ or 2 a; + 1 = 'i*-±-§. 
 X — 2 3 
 
 Multiplying by 3, 6 a; + 3 = 5 x + 8. 
 
 .*. X = 5. 
 
 Had the given equation been cleared of fractions by multiplying by 
 3(x — 2) instead of by 3, the resulting equation simplified, which is 
 aj'^ — 7 X + 10 = 0, or (x — 5) (x — 2) = 0, would have been satisfied both by 
 X = 5 and by x = 2. The given equation is satisfied by x == 5 but not by 
 X = 2. Hence, the latter value is not a root of the given equation, but has 
 been introduced by using (x — 2) times the necessary multiplier, 3. Boots 
 so introduced may be discovered by verification and rejected. 
 
 2. If a fraction has the minus sign before it, the signs of all the terms 
 of the numerator must be changed when the denominator is removed. 
 
 Find the value of x, and verify the result : 
 
 3. 2x + -^~. 7. ^ + ^ = 1^. 
 
 3 3 2 6 3 
 
 4. - + 10 = 13. 8. 7|- — =-. 
 
 5. - + 2x = 26. 9. ^ + ^ = 24. 
 6 5 7 
 
 6. 3a;-- = 14. 10. 
 
 
 3 
 
 11. 
 
 XX X 3x o X _j 
 2 3 4 "^10 12 ~ ■ 
 
 
 25 a; 5a;2« 5a;_o 
 18 9 3 6 ~ ■ 
 
 12. 
 
 2 X 5 _ a; 
 6~4* 
 
 2 X ^' _x , 4^ _ ^ 
 
 ' Y~4:~2^T~U 
 
SIMPLE EQUATIONS 1(39 
 
 2x 7a; 5a; a; __ 4 
 T 8 18 24 ~ 9* 
 
 T^16 2" 16 8' 
 
 ,„ 15 a; , 5 a; 11a; , 19 a; « 
 
 16. = ^« 
 
 7 6 3 14 
 
 ,„ 2a; , 5a; 4x , a; 1 
 
 17. — = — 
 
 15 25 9 6 9 
 
 3a; 7x^11 a; 8a;3 
 4 12 36 9 2* 
 
 19 3a; JL^_3a; X 4^^ 
 ^^' T^20"10^5+15 
 
 20. I^±i-2a; + ^^-^±^ = l. 
 5 7 
 
 21. 
 
 a; — la; — 2x — 3 5 a; — 1 
 2 3 4 6 
 
 
 22. 
 
 x + 1 a; + 4a; + 3_-|^g 
 3 5*4 
 
 
 23. 
 
 7a; + 2 12-a;a; + 2_^ 
 6 4 2 * 
 
 
 24. 
 
 a;-3a; + r> a; + 2 ^ 
 7 3 6 
 
 
 25. 
 
 3 a; - 5 7 a; - 13 _ 3 a; + 3 
 4 6 2 
 
 
 26. 
 
 2a;-5 3a;-2_-,^ a; + 2_ 
 5 7 6 
 
 
 27. 
 
 l-2a; 7-2a; ll-2a; 
 3 4*6 
 
 7 
 12' 
 
 28. 
 
 a; + 4,2-2a; a;+l qi 
 3 ' 6 2 ^* 
 
 
170 SIMPLE EQUATIONS 
 
 29 9a; + 5 8a;-7 ^ 36 a; + 15 10^ 
 14 "^ 6a; + 2 56 14* 
 
 Suggestion. — The equation may be written 
 
 §104,3, 9» j_ 8a;-7 ^36^ 15 41^ 
 
 14 14 6 a; + 2 66 56 56 
 
 Simplify as much as possible before clearing of fractions. 
 „_ 3x-2 , 8a7-21 6a;-22 
 
 31. 
 
 32. 
 
 33. 
 
 34. 
 
 35. 
 
 2 a; -5 5 10 
 
 4a; + 3 ^ 8a; + 10 7 a; -29 
 9 18 5 a; - 12* 
 
 6a; + l 2a;-4 ^ 2a;-l 
 15 7a;-13~ 5 
 
 10 a; + 17 5 a; - 2 ^ 12 a; - 1 
 18 9 ~lla;-8' 
 
 6a; + 3 3a;-l ^ 2a;-9 
 15 5x-25~ 5 
 
 2a; + l 8 ^ 2a;-l 
 
 2a;-l 4a;2-l~2a; + l' 
 
 36. Solve the equation ^Jzl + ?J:z:§ = £lZ^ + ^JH^. 
 ^ a;-2a;-7 a;-6a;-3 
 
 Solution. — It will be observed that if the fractions in each member were 
 connected by the sign — , and the members were simplified, the numerators 
 of the resulting fractions would be simple. The fractions can be made to 
 meet this condition by transposing one fraction in each member. 
 
 Consequently, it is sometimes expedient to defer clearing of fractions. 
 
 Transposing, *~^ x-2 x-b x- 
 
 Uniting terms, 
 
 X - 
 
 -1 
 
 -2 
 
 X — 
 
 2 
 
 X - 
 
 X — 
 
 3 
 
 
 — 
 
 1 
 
 
 x-6 x-1 
 
 -1 
 a;2 _ 5 X + 6 «^ - 13 a; + 42* 
 
 Since the fractions are equal and their numerators are equal, their denomi- 
 nators must be equal. 
 
 x2 _ 5 X + 6 = a;2 _ 13 X + 42. 
 
 .-. X = 4i. 
 
39. 
 
 40. 
 
 SIMPLE EQUATIONS 171 
 
 X—Z X — Q 03—6 x — 4 
 
 „Q x — ^x — 1 x — Q X — ^ 
 
 on. -\ - — — — H -• 
 
 iK — 4 x — Q X — 7 X — 5 
 
 x + 2 _ x + 3 _ x + 5 _ x-\-6 
 x + 1 x-\-2~ x + 4: x + 5' 
 
 x-\-l x + 6 _ x + 2 x + 5 
 x + 2 x-\-7~x + 3 x + G 
 
 .^ a; — 5_a; — 10_ir — 4_a; — 9 
 
 'x + 5 a; + 10~ic + 4 x + d 
 
 I^_2 5a;-- ^-2 
 
 - A^ r^=V-i- 
 
 43. ^nil + '^^ + ^.zie.s^o. 
 
 oj — 2 x — 3 X — 4: 
 
 44. ■ = \- 2x. 
 
 x-1 x + 1 x^-l 
 
 45. —- T = —o — T — 2 a; 
 
 a; + l a; — 1 ar — 1 
 
 46. i-^ = Ii^ + ?('?- A 
 
 2 3 2l^a; y 
 
 A-16 A + 6 
 47 bx ox _4^ 
 
 24 60~^y 
 
 48. |(2-a;)-|(3-2a;)=^±12. 
 
 2x(\-^^ 3x{\-^ 
 
 V x) \ xj a; -4 
 
 *'• 3 + 4 -J- 
 
 50. l.-2(^-3)=4-|(^^l). 
 
 61 (2a; + l)' (4a;-iy ^l5 3(4 a; + 1) 
 6 20 8 40 ' 
 
172 SIMPLE EQUATIONS 
 
 52. l.^±l + 2 = l + '^-l ^ 
 
 2a;-l 2a; + l l-a:^ 
 
 174_? i+l§ 21_^ 100 5 
 
 2x 
 
 54. K^-4) 4a;-16 ^3 
 
 2 
 
 10 
 
 + 5 
 
 G 5 f 
 
 2 + 
 
 55. 1+ T = jr' 56. = — T^- 
 
 1 4-1 1+^ x_^ x + 5 
 
 X X x 3 
 
 3 3a;-l 
 
 1. Solve the equation 
 
 Literal Equations 
 
 X — W X — a^ 
 
 a b 
 
 Solution 
 a; - 62 X - a2 
 
 a b 
 
 Clearing of fractions, bx — b^ = ax — a". 
 
 Transposing, etc., ax — bx = a^ — b^. 
 
 (a — b)x = a^ — b^. 
 Dividing by (a — 6) , x — a^ + ab + b^. 
 
 Solve the following equations : 
 
 2 ^ — ^ I ^^^ _ 1 , g a; — 2a& l_ a; — 3c 
 
 na; ca; c cx x abx 
 
 ^ _a6__7 __49^ _ a? — a |2a;_q 6 b 
 
 X ab abx ha a 
 
 4 .^_2a!=i_?_^. 8 «' [ b'' ^ a + h 3(a + 6) 
 
 a6^ ft^a; a^a; bx ax ah x 
 
 K ^ a;4-26 _a ^^ _ a^ + 6^ a — h _b 
 
 b a ""6 ' ' 2bx ~ 2bx'~ x 
 
SIMPLE EQUATIONS 173 
 
 iA I h{x + g) _ 2 a5 a; — 2a , x _ a- -\-h^ 
 
 10. ft H — " 11. f- — — — • 
 
 X — a X— a a o ao 
 
 12. Q X + l^(l -^= a{x - a). 
 
 13. 6(2a;-9c-14&)=c(c-a;). 
 
 14. a(x-a-26)+6(x- &)+c(x + c)=0. 
 
 15. (ft — x){x — 5) + (ft + a;)(a; — 6) = (ft — 6)-. 
 a — ft + c h — a -\- c 
 
 16. 
 
 a; + ft x — a 
 
 17. ^^ + -^ = 0. 
 
 ft(6 — x) b{c — x) a{c — x) 
 
 1-A 
 
 ft — 1 X- — a- 
 
 a - 1 x-1 (ft - l)(.r - 1) 
 
 a + x 2x x^(x — ft) _ 1^ 
 
 ft ft + a; ft(ft^ — a:^) 3 
 
 20 ^ + « I -^ + c a; + ?) _ ft 6 c j^^ 
 
 6 ft c b c a 
 
 21. 
 
 22. 
 
 a:^ — fta? — 6a; -f- ft5 a^ — 2 6.t + 2 />^ c^ 
 
 a; — ft X — b X — c 
 
 1 2 m?j 7n X — lb 
 
 m -f w (m + 7(-)''* (m + ny (m + ^0^ 
 
 23. + =a? + b'' + c' + 2db. 
 
 a -{■ b + c a -\-b — c 
 
 a; 4-26 a; + 3 6 ^ a; 4- 6 , a; + 26 
 X — b X + b X — Sb a;4-56 
 
 -_ 2a; + 3ft,3a;4-7ft 2a ,m 
 
 25. ' 1 —: = ' 1-5- 
 
 a;4-a a; + 2a a; + 4a 
 
 „- a; + 7a. X — a x + 7 a x — a 
 ^D. 1 — = 1- 
 
 a;-f6ft x — 3ft a; + ft ic + 2ft 
 27. (ft — 6)(a! — c) — (6 — c)(x — a) = (c — a)(x — 6), 
 
174 SIMPLE EQUATIONS 
 
 Problems 
 
 Directions for Solving. — Represent one of the unknown 
 numbers by x, and from the conditions of the problem find an 
 expression for each of the other unknown numbers. 
 
 Find from the problem two expressions that are equal, and write 
 them as an equation. 
 Solve the equation. 
 
 1. A man bought a farm, a house, and a barn for $12,600. 
 If the house cost twice as much as the barn, and the farm twice 
 as much as the house and barn together, how much did each 
 cost? 
 
 Suggestion. — There are three unknown quantities — the cost of the 
 house, the cost of the barn, and the cost of the farm. It is evident that, if 
 the cost of the barn were known, the cost of the house could be found from 
 it, and from both the cost of the farm. 
 
 Accordingly, represent the cost of the barn as x dollars, and express the 
 cost of the house and the cost of the farm in terms of x. Discover two 
 expressions for the total cost, and equate them. 
 
 2. If A is twice as old as B, and B twice as old as C, how 
 old is each, if the sum of their ages is 140 years ? 
 
 3. Mr. Henry bought three building lots for $36,000. If the 
 third cost twice as much as the second, and the second 3 times 
 as much as the first, what was the cost of each ? 
 
 4. A man left $63,000 to his wife, two sons, and a daughter. 
 If each son received twice as much as the daughter, and half as 
 much as the wife, what was the share of each ? 
 
 5. A man left ^ of his property to his wife, \ to his son, \ to 
 his daughter, and the rest, which was $2000, to a hospital. What 
 was the value of his property ? 
 
 6. A owed B, C, and D $27,000 in all. If he owed B 4 times 
 as much as C, and D f as much as B, what sum did he owe 
 each ? 
 
 7. A person spends \ of his annual income for board, \ for 
 clothes, and $260 for other expenses. If he saves ^ of his 
 income, what is his income ? 
 
SIMPLE EQUATIONS 175 
 
 8. A table and a chair cost $ 11. The table and a picture 
 cost $ 14. If the chair and the picture together cost 3 times as 
 much as the table, what was the cost of each article ? 
 
 9. A and B together had $ 50, and A and C had f 60. After 
 each had spent $5, A had \ as much as B and C together. 
 How much had each at first? 
 
 10. A gave his age as follows : " f of my age less ^ of what it 
 will be a year hence is equal to \ of my age 5 years ago." What 
 was his age ? 
 
 11. James is 5 years older than his sister, and 5 years hence 
 he will be 3^ times as old as his sister was 5 years ago. What is 
 the age of each ? 
 
 12. A's daily wages are f of B's, and C's are f of A's. If A 
 and C together earn 25 cents more a day than B, what are the 
 daily wages of each ? 
 
 13. A man paid a debt of $8.00 with an equal number of 
 6, 10, and 25-cent pieces. How many of each were there ? 
 
 14. A man bought equal quantities of white and brown sugar, 
 paying 6^ cents a pound for the former and 5 cents a pound for 
 the latter. How many pounds of each did he buy, if the whole 
 quantity cost him $ 1.80 ? 
 
 15. A field is twice as long as it is wide. By increasing its 
 length 20 rods and its width 30 rods, the area will be increased 
 2200 square rods. What are its dimensions ? 
 
 16. Three fifths of a certain number exceeds ^ of it by 7. 
 What is the number? 
 
 17. One third of a number added to 3 times the number is 
 equal to 50. What is the number ? 
 
 18. After spending f of my income and $300 more, I had 
 \ of it left. What was my income? 
 
 19. The sum of ^ of a number, \ of it, and | of it is 4 more 
 than f of the number. What is the number ? 
 
 20. The sum of a number, its half, its third, and its fourth, 
 and 16 is 66. What is the number ? 
 
176 SIMPLE EQUATIONS 
 
 21. A man deposited f of his month's wages in a bank, and 
 paid out ^ of the remainder for groceries and $ 9 for dry goods. 
 If he had $ 3 left, how much money had he at first ? 
 
 22. Divide 500 into two parts, such that the greater decreased 
 by \ of the smaller is 5 times as much as the smaller decreased 
 by ^ of the larger. 
 
 23. Divide 61 into two parts, such that, if the greater is 
 divided by the less, the quotient will be 3 and the remainder 1. 
 
 24. Divide 111 into two parts, such that the first diminished 
 by 3 is equal to the second divided by 3. 
 
 25. Divide 40 into two parts, such that the first divided by 3 
 is equal to the second divided by 5. 
 
 26. Divide 40 into two parts, such that the first is 10 greater 
 than twice the second. 
 
 27. Divide 44 into two parts, such that, if the greater is divided 
 by 5 and the less by 7, the difference of the quotients will be 4. 
 
 28. Divide 54 into two parts, such that ^ of the first part is 
 equal to \ of the second. 
 
 Solution 
 Let a; = J of the first part or ^ of the second part. 
 
 Then, 4 x = the first part, 
 
 and 6 X = the second part ; 
 
 4a; + 5x = 54; 
 whence, x = 6, 
 
 4 X = 24, the first part, 
 and 5 X = 30, the second part. 
 
 29. Divide 40 into three parts, such that ^ of the first, ^ of the 
 second, and \ of the third are equal. 
 
 30. Find three parts of 60, such that the first divided by 5, 
 the second multiplied by 2, and the third increased by 5 are 
 equal. 
 
 31. Divide 72 into four parts, such that the first divided by 2, 
 the second diminished by 2, the third multiplied by 2, and the 
 fourth increased by 2 are equal. 
 
SIMPLE EQUATIONS 177 
 
 32. A can do a piece of work in 8 days. If B can do it in 10 
 days, in how many days can both working together do it ? 
 
 Solution 
 Let X = the required number of days. 
 
 Then, - = the part of the work both can do in 1 day, 
 
 X 
 
 I = the part of the work A can do in 1 day, 
 ^ = the part of the work B can do in 1 day ; 
 
 X 8 10 40' 
 Solving, X = ^, or 4|, the required number of days. 
 
 33. A can do a piece of work in 10 days, and B can do it 
 in 15 days. How long will it take both to do it ? 
 
 34. Three pipes empty into a cistern. One can fill the cistern 
 in 5 hours, another in 6 hours, and the third in 10 hours. How 
 long will it take the three pipes together to fill it ? 
 
 35. A can do a piece of work in 10 days, B can do it in 12 
 days, and C can do it in 8 days. In how many days can all 
 together do it? 
 
 36. A can pave a walk in 6 days, and B can do it in 8 days. 
 How long will it take A to finish the job after both have worked 
 3 days ? 
 
 37. A can build a wall in 15 days, but with the aid of B and 
 C, the wall can be built in 6 days. If B does as much work 
 in 1 day as C does in 2 days, in how many days can B and C 
 separately build the wall ? 
 
 38. A and B can dig a ditch in 10 days, B and C can dig it in 
 6 days, and A and C in 7^ days. In what time can each man do 
 the work ? 
 
 Suggestion. — Since A and B can dig ^ of the ditcli in 1 day, B and C 
 I of it in 1 day, and A and C ^ of it in 1 day, i^ + i + ^ is twice the part 
 they can all dig in 1 day. 
 
 39. A and B can load a car in 2y\ hours, B and C in 3f hours, 
 and A and C in 3^ hours. How long will it take each alone to 
 load it ? 
 
 ALG. — 12 
 
178 SIMPLE EQUATIONS 
 
 40. A boy bought some oranges at the rate of 30 cents a dozen. 
 He sold f of them for 4 cents each, and the rest for 3 cents each. 
 If he gained 90 cents, how many oranges did he buy ? 
 
 41. Find a fraction whose value is ^ and whose denominator 
 is 15 greater than its numerator. 
 
 42. Find a fraction whose value is f and whose numerator is 
 3 greater than half of its denominator. 
 
 43. The numerator of a certain fraction is 8 less than the de- 
 nominator; and if each term of the fraction is decreased by 5, 
 the value of the fraction becomes \. What is the fraction ? 
 
 44. The units' digit of a number expressed by two digits ex- 
 ceeds the tens' digit by 5. If the number increased by 63 is 
 divided by the sum of its digits, the quotient is 10. What is the 
 
 number ? * 
 
 Solution 
 
 Let z = the digit in tens' place. 
 
 Then a; 4- 6 = the digit in units' place, 
 
 and 10 re + (a; + 5), or 11 a; -I- 5 = the number ; 
 
 2a; + 5 
 whence, x = 2, 
 
 and a; + 6 = 7. 
 
 Therefore, the number is 27. 
 
 45. The tens' digit of a number expressed by two digits is 
 3 times the units' digit. If the number diminished by 33 is 
 divided by the difference of the digits, the quotient is 10. What 
 is the number ? 
 
 46. The tens' digit of a number expressed by two digits is ^ 
 of the units' digit. If the number increased by 27 is divided by 
 the sum of its digits, the quotient is 6^. What is the number ? 
 
 47. In a purse containing $1.45 there are ^ as many quarters 
 as 5-cent pieces and f as many dimes as 5-cent pieces. How 
 many pieces are there of each kind ? 
 
 48. A woman spent $10 more than f of her money; then 
 $10 more than | of the remainder. If she had $2 left, how 
 much money had she at first ? 
 
SIMPLM EQUATIONS 179 
 
 49. A man spent $ 1 less than ^ of his money and had left 
 $ 1 less than f of it. How much money had he at first ? 
 
 50. A girl found that she could buy 12 apples with her money 
 and have 5 cents left, or 10 oranges and have 6 cents left, or 6 
 apples and 6 oranges and have 2 cents left. How much money 
 had she ? 
 
 51. A boy spent \ of his money and |- a cent more, then \ of 
 the remainder and ^ a cent more, then i of what he had left and 
 ^ a cent more, when he found that he had 2 cents remaining. 
 How much had he at first ? 
 
 52. Five boys bought a boat, agreeing to share the expense 
 equally. But one of them having left $ 1 of his share unpaid, 
 each of the others had to pay ^^ ^lore than one fifth of the ex- 
 pense. What was the cost of the boat ? 
 
 53. A sum of money was divided among A, B, C, and D so 
 that A received \ as much as all the others, B received \ as much 
 as all the others, C received \ as much as all the others, and D 
 received $ 2800 less than A. What sum did each receive ? 
 
 54. In an alloy of 90 ounces of silver and copper there are 6 
 ounces of silver. How much copper must be added that 10 
 ounces of the new alloy may contain ^ of an ounce of silver ? 
 
 55. If 80 pounds of sea water contain 4 pounds of salt, how 
 much fresh water must be added that 45 pounds of the new solu- 
 tion may contain If pounds of salt ? 
 
 56. An officer, attempting to arrange his men in a solid square, 
 found that with a certain number of men on a side he had 34 men 
 over, but with 1 man more on a side he needed 35 men to com- 
 plete the square. How many men had he ? 
 
 Suggestion. — With x men on a side, the square contained x^ men ; with 
 X -I- 1 men on a side, there were places for (a; + 1)2 men. Since the number 
 of men was the same under both arrangements, x^ + 34 = (x + 1)^ — 35. 
 
 57. A regiment drawn up in the form of a solid square lost 60 
 men in battle. Afterward, when the men were arranged in a 
 solid square with 1 man less on a side, it was found that there 
 was 1 man over. How many men were there in the regiment at 
 first? 
 
180 . SIMPLE EQUATIONS 
 
 58. A regiment drawn up in the form of a solid square was 
 reenforced by 240 men. When the regiment was formed again in 
 a solid square, there were 4 more men on a side. How many 
 men were there in the regiment at first ? 
 
 59. A man was hired for 40 days under the following condi- 
 tions : for every day he worked he was to receive $ 3 besides his 
 board, while for every day he was idle he was to receive nothing, 
 but was to be charged $ 1.20 for his board. If at the end of the 
 period he received $ 57, how many days did he work ? 
 
 60. A man invested $800, a part at 6% and the rest at 5%. 
 If the total annual interest was $ 45, how much did he invest at 
 each rate ? 
 
 Sdggkstion. — Let x = the number of dollars invested at 6 %. 
 Then, 800 — aj = the number of dollars invested at 5 % ; 
 
 61. A man has f of his property invested at 4%, \ at 3%, and 
 the remainder at 2%. How much property has he, if his annual 
 income is $ 860 ? 
 
 62. A man put out $4330 in two investments. On one of 
 them he gained 12%, and on the other he lost 5%. If his net 
 gain was $ 251, what was the amount of each investment ? 
 
 63. There were distributed among 20 men and 25 women $ 160 
 in such a way that the sum of what a man and a woman received 
 was $ 7. How much did the men receive, and how much did the 
 women receive ? 
 
 64. At what time between 5 and 6 o'clock will the hands of a 
 clock be together ? 
 
 Solution 
 Let X = the number of minute spaces that the minute hand travels 
 
 after 5 o'clock before they come together. 
 Then, — = the number of minute spaces that the hour hand travels in 
 
 the same time. 
 Since they are 25 minute spaces apart at 6 o'clock, 
 
 «- — = 25; . 
 
 12 ' ^ 
 
 .'. X = 27^, the number of minutes after 5 o'clock. 
 
 65. At what time between 1 and 2 o'clock will the hands of a 
 clock be together ? 
 
SIMPLE EQUATIONS 181 
 
 66. At what time between 6 and 7 o'clock will the hands of a 
 clock be together ? 
 
 67. At what time between 10 and 11 o'clock will the hands of 
 a clock point in opposite directions ? 
 
 68. At what times between 4 and 5 o'clock will the hands of a 
 clock be 15 minute spaces apart ? 
 
 69. When after 9 o'clock and before 10 o'clock will the hands 
 of a clock be at right angles to each other ? 
 
 70. A man rows downstream at the rate of 6 miles an hour 
 and returns at the rate of 3 miles an hour. How far downstream 
 can he go and return within 9 hours ? 
 
 71. At the rate of 3 miles an hour uphill and 4 miles an hour 
 downhill a woman can walk 60 miles in 17 hours. How much 
 of the distance is uphill, and how much is downhill ? 
 
 72. A hare pursued by a hound takes 4 leaps while the hound 
 takes 3 ; but 2 of the hound's leaps are equal to 3 of the hare's. 
 If the hare has a start equal to GO of her own leaps, how many 
 leaps must the hound take to catch the hare ? 
 
 Solution 
 Let 3 a; = the number of leaps taken by the hound. 
 
 Then, 4 x = the number of leaps taken by the hare. 
 
 Suppose a = the number of feet in one leap of the hare. 
 
 q « 
 
 Then, — = the number of feet in one leap of the hound, 
 
 — X 3 a; = _ax ^^^ jmmber of feet the hound runs, 
 2 2 
 
 and a X 4 a; = 4 aa;, the rmmber of feet the hare runs. 
 
 Since the hare has a start equal to 60 times a feet, or 60 a feet, the hare 
 runs 60 a feet less than the hound. 
 
 Therefore, 
 
 Aax = ^ ** 60 a. 
 
 2 
 
 Dividing by a. 
 
 4« = — -60. 
 
 2 
 
 Therefore, 
 
 X = 120, 
 
 and 
 
 3 X = 360, the number of leaps taken by the hound. 
 
182 SIMPLE EQUATIONS 
 
 73. A fox is 70 leaps ahead of a hound and takes 5 leaps while 
 the hound takes 8 ; but 3 of the hound's leaps equal 7 of the 
 fox's. How many leaps must the hound take to catch the fox ? 
 
 74. A rabbit makes 5 leaps -while a dog makes 4 ; but 3 of the 
 dog's leaps are equal to 4 of the rabbit's. If the rabbit has a 
 start of 20 leaps, how many leaps will each take before the rabbit 
 is caught ? 
 
 75. A hound is 39 of his leaps behind a rabbit that takes 7 
 leaps while the hound takes 8. If 6 leaps of the rabbit are equal 
 to 5 leaps of the hound, how many leaps must the hound take to 
 catch the rabbit ? 
 
 76. A wheelman and a pedestrian leave the same place at the 
 same time to go to a point 54 miles distant, the former traveling 
 3 times as fast as the latter. The wheelman makes the trip and 
 returning meets the pedestrian in 6f hours from the time they 
 started. What is the rate of each ? 
 
 77. If 1 pound of lead loses -^ of a pound, and 1 pound o£ 
 iron loses -^-^ of a pound when weighed in water, how many pounds 
 of lead and of iron are there in a mass of lead and iron that 
 weighs 159 pounds in air and 143 pounds in water ? 
 
 78. If 97 ounces of gold weighs 92 ounces when it is weighed 
 in water, and 21 ounces of silver weighs 19 ounces when it is 
 weighed in water, how many ounces of gold and of silver are 
 there in a mass of gold and silver that weighs 320 ounces in air 
 and 298 ounces in water ? 
 
 79. A merchant increases his capital annually by \ of it, and 
 at the end of each year takes out $ 800 for expenses. At the end 
 of three years, after taking out his expenses, he finds that his 
 capital is $ 6325. What was his original capital ? 
 
 80. A merchant added annually to his capital \ of it, and at 
 the end of each year took out f 1000 for expenses. If at the end 
 of the third year, after taking out the last $ 1000, he had f of 
 his original capital, what was his original capital ? 
 
 81. A cistern can be filled by one pipe in 20 minutes, by 
 another in 15 minutes, and it can be emptied by a third in 10 
 
SIMPLE EQUATIONS 183 
 
 minutes. If the three pipes are running at the same time, how 
 long will it take to fill the cistern ? 
 
 82. A man walked from A to B at the rate of 2 miles an hour, 
 and rode back at the rate of 3^ miles an hour, being gone 13 
 hours. How far is it from A to B ? 
 
 83. An express train whose rate is 40 miles an hour starts 
 1 hour and 4 minutes after a freight train and overtakes it in 
 1 hour and 36 minutes. How many miles does the freight train 
 run per hour ? 
 
 84. The distance from Albany to Syracuse is 148 miles. A 
 canal boat leaves Albany for Syracuse, moving at the rate of 
 
 3 miles in 2 hours ; at the same time another leaves Syracuse for 
 Albany, moving at the rate of 5 miles in 4 hours. How far from 
 Albany do they meet ? 
 
 85. A steamer goes 5 miles downstream in the same time that 
 it goes 3 miles upstream ; but if its rate each way is diminished 
 
 4 miles an hour, its rate downstream will be twice its rate up- 
 stream. How fast does it go in each direction ? 
 
 86. A, B, and C together can do a piece of work in 3^ days ; 
 B can do \ as much as A, and C can do | as much as B in a day. 
 In how many days can each do the work alone ? 
 
 87. A can do a piece of work in 6 days that B can do in 14 
 days. A began the work, and after a certain number of days B 
 took his place and finished the work in 10 days from the time 
 it was begun by A. How many days did B work ? 
 
 88. In a certain weight of gunpowder the niter composed 10 
 pounds more than | of the weight, the sulphur 3 pounds more 
 than y^2) 3,nd the charcoal 3 pounds less than ^V ^^ ^^® niter. 
 What was the weight of the gunpowder ? 
 
 89. A library containing 16,000 volumes was divided into five 
 departments. In the department of history there were twice as 
 many volumes as in the department of science, and 500 more 
 than \ as many volumes as in the juvenile department. Of 
 fiction there were 1\ times as many volumes as of science, and 
 500 less than 8 times as many as in the reference department. 
 How many volumes were there in each department ? 
 
184 SIMPLE EQUATIONS 
 
 90. An estate was divided among four heirs, A, B, C, and D. 
 If the value of the estate had been $ 1000 less, what A received 
 would have been \ of it, and what B received ^ of it ; if the value 
 of the estate had been \ greater, what C received would have 
 been \ of it, and what D received \ of it. What sum did each 
 receive ? 
 
 91. A father takes 3 steps while his son takes 5 ; but 2 of the 
 father's steps are equal to 3 of the son's. How many steps will 
 the son require to overtake his father, who is 36 steps ahead ? 
 
 92. A purse contained some money and a ring worth $10 
 more than the money. If the purse was worth ^ as much as the 
 money it contained, g,nd the purse and the money together were 
 worth \ as much as the ring, what was the value of each ? 
 
 93. Brass is 8| times as heavy as water, and iron 1\ times as 
 heavy as water. A mixed mass weighs 57 pounds, and when 
 immersed displaces 7 pounds of water. How many pounds of 
 each metal does the mass contain ? 
 
 94. A man began business with $4725, and annually added to 
 his capital 4 of it. At the end of each year he put aside a 
 certain sum for expenses. If at the end of the third year, after 
 taking out the sura for expenses, his capital was $3800, what 
 were his annual expenses ? 
 
 95. The sum of two numbers is s, and their difference d. 
 What are the numbers ? 
 
 Solution 
 Let X = the greater number. 
 
 Then, X — (Z = the less number, 
 
 and X + X - d = s ; 
 
 .-. X = ^"*" , the greater number, (1) 
 
 and — d = - — -, the less number. (2) 
 
 If the sum of two numbers is 30, and their difference is 6, 
 what are the numbers ? 
 
 By (1), the greater number is — ^, or 18 ; 
 
 by (2), the less number is '- — ^^—i or 12. 
 
SIMPLE EQUATIONS 185 
 
 A problem in which the numbers assumed to be known are 
 represented by letters to which any values may be assigned is 
 called a General Problem. 
 
 Problem 95 is a general problem. 
 
 The results obtained in solving a general problem may be con- 
 sidered formvlce for solving similar problems. 
 
 96. Divide c cents between two boys so that one shall have d 
 cents more than the other. If c — 50 and d = 10, how much will 
 each receive ? 
 
 97. A horse and a saddle are together worth a dollars, and the 
 horse is worth m times as much as the saddle. What is the value 
 of each ? What, if a = 160 and m = 3 ? 
 
 98. Divide 6 into two parts one of which represents m times 
 the other. What will they be, if b represents 100, and m, 4 ? 
 
 99. An estate of a dollars is divided between two heirs in the 
 proportion of m to n. What is the share of each ? What is the 
 share of each, if a = 40,000, m = 5, and ?t = 3 ? 
 
 100. If A can do a piece of work in a days, and B in 6 days, 
 in what time can both do it working together ? Give the result, 
 if a = 10 and b = 15. 
 
 101. An alloy of two metals is composed of m parts of one to 
 n parts of the other. How many pounds of each are required in 
 the composition of a pounds of the alloy ? 
 
 Bell metal is an alloy of 5 parts of tin and 16 parts of copper. 
 How many pounds of tin and of copper are there in a bell which 
 weighs 4200 pounds ? 
 
 102. A wheelman set out from B at the rate of r miles an hour. 
 a hours later another started in pursuit at the rate of p miles an 
 hour. How far from B will the second wheelman overtake the 
 first ? What will be the distance, if r = 10, p = 12, and a = 8 ? 
 
 103. A man traveled from home at the rate of a miles an hour 
 and returned at the rate of b miles an hour. If he made the 
 entire journey in h hours, how far from home did he go ? How 
 far, if a = 4, & = 3^, and h = 15? 
 
SIMULTANEOUS SIMPLE EQUATIONS 
 
 TWO UNKNOWN NUMBERS 
 
 201. 1. It x + y=12, what is the value of x? oty? How 
 many values may x have ? How many may y have ? 
 
 2. In the expression x -\- y = 12, x and y each may have an 
 indefinite number of values, but if, at the same time, x — y = 4, 
 what is the value of a; ? of y? 
 
 3. Although one equation containing two unknown numbers 
 has an indefinite number of values for each unknown number, or 
 is indeterminate, what can be said about the values of the un- 
 known numbers, when two equations are given involving the same 
 values of the unknoAvn numbers, but in difEerent relations, that is, 
 when two independent equations are given ? 
 
 202. Two or more equations in which the unknown numbers 
 have the same values are called Simultaneous Equations. 
 
 If X and y represent the same numbers in2a;-t-3?/ = 19as they represent 
 inSx — j/=22, 2a: + 32/ = 19 and 5 a; — y = 22 are simultaneous equations. 
 
 203. Equations that represent different relations between the 
 unknown numbers, and so cannot be reduced to the same form, are 
 called Independent Equations. 
 
 3x + Sy = IS and 2 x + 2 1/ = 12 really express but one relation between 
 X and y ; viz., that their sum is 6. Hence, both equations may be reduced to 
 the same form, as x + y = 6. But 3 x + 3 v = 18 and x + Sy = 14 express 
 different relations between x and y, and cannot b« reduced to the same form. 
 Hence, they are independent equations. 
 
 204. An equation whose unknown numbers may have an in- 
 finite number of values is called an Indeterminate Equation. 
 
 X + 1/ = 6 is an indeterminate equation, because, if x = 2, y — 4 ; if « = 3, 
 y = 3 ; if X = 2|, J/ ^ 3^ ; etc. 
 
 186 
 
SIMULTANEOUS SIMPLE EQUATIONS 187 
 
 205. Principle. — Every single equation involving two or more 
 unknown numbers is indeterminate. 
 
 206. The process of deriving from a set of simultaneous equa- 
 tions, equations involving a less number of unknown numbers 
 than is found in the given equations is called Elimination. 
 
 207. Two sets of simultaneous equations each having all the 
 roots of the other set are called Equivalent Systems of equations. 
 
 Thus I x + y = % \ ^ r2x + 2y = 16| 
 
 Itius, \ 3 a; + 2 2/ = 21/ ^^"^ \3x + 2?/ = 2l| 
 
 are equivalent systems, for each system is satisfied by the same set of values, 
 X = 5 and ?/ = 3, and neither is satisfied by any other set of values. 
 
 208. Any equation in a set of simultaneous equations may be 
 transformed into an equivalent equation by employing the prin- 
 ciples of equivalency stated in § 196, since these principles apply 
 to all equations. Hence, it remains to seek the principle by which 
 simultaneous equations are combined in the process of elimination 
 without introducing or losing roots. 
 
 This Principle of Elimination may be stated as follows, a and h 
 being known multipliers, not zero, and either positive or negative : 
 
 If any equation' of a system is replaced by the stim or difference 
 of a times that equation and b times another equation of the system, 
 the resulting system is equivalent to the given system. 
 
 The proof is as follows : 
 
 By § 196, Prin. 1, all the terms of the second member of an equation may 
 be transposed to the first member. 
 
 Then, let ^ = 
 
 B = 
 
 (1) 
 
 be the given system, and let a and 6 be any knovfn multipliers except zero 
 and either positive or negative. 
 It is to be proved that the system 
 
 aA + bB = 0^ 
 
 5 = (2) 
 
 is equivalent to the given system (1). 
 
 Since a and b are knovsrn multipliers, not zero, by § 196, Prin. 2, every set 
 of values of the unknown numbers that makes ^ = makes aA = 0, and 
 every set of values that makes .6 = makes bB = 0. Hence, every set of 
 
188 SIMULTANEOUS SIMPLE EQUATIONS 
 
 values that satisfies (1) makes rt^ + hB equal to zero and thus satisfies (2), 
 since all of the equations of (1) and (2) except the first are the same. 
 
 Again, since every set of values that satisfies (2) makes B, or bB equal to 
 zero and also makes aA + hB = 0, every such set of values makes aA = 
 and therefore, by § 196, Prin. 2, makes ^ = and satisfies (1). 
 
 Since all the roots of (1) are roots of (2) and all the roots of (2) are roots 
 of (1), (1) and (2) are equivalent systems. 
 
 209. Elimination by Addition and Subtraction. 
 
 1. It 2 X + 2y = 10 and 3x — 2 y = 5 are added, what is the 
 resulting equation ? When may an unknown number be elimi- 
 nated by addition ? 
 
 2. If a; + 2 ?/ = 5 is subtracted from 3 x -\- 2 y = 11, what is 
 the resulting equation ? When may a number be eliminated by 
 subtraction ? 
 
 Principle. — A literal number having the same coefficient in 
 two equations may be eliminated by adding the equations, if the 
 coefficients have unlike signs, or by subtracting one equation from the 
 other, if the coefficients have like signs. 
 
 Examples 
 
 1. Find the value of x and of ?/ in 2 x+3 y = 7 and 3 x+4: y=10. 
 
 Explanation. — Since x has not the same co- 
 eflBcients in both equations, the equations are 
 
 2 X 4- 3y = 7 (1) multiplied by such numbers as will make the co- 
 
 3 a; -4- 4 V = 10 ^2^ efficients of x alike. Multiplying eq. (1) by 3 and 
 
 eq. (2) by 2 gives equations (3) and (4). From 
 
 6 X -{- 2 y = 21 (3) these x is eliminated by subtraction, and the value 
 6 a; + 8 ?/ = 20 (4) of y is found. 
 
 The value of x may be found by multiplying 
 
 2/ ~ ■*■ \^) eq. (1) and (2) by such numbers as will make 
 2 a; + 3 = 7 (6) the coefficients of y equal and then subtracting 
 X = 2 (7) the resulting equations. 
 
 Or, the value of x may be found by substituting 
 the value of y for y in one of the given equations, as eq. (1). 
 
 Proof of the Equivalence. — By the principle of elimination-, the system 
 (1, 6), obtained by subtracting (4) from (3), is equivalent to the given system 
 (1, 2). Hence, the only value of y in the given system is 1. 
 
 Since y represents 1 in the given system, (1) is equivalent to (6), which 
 by § 196 is equivalent to (7). Hence, the given system, being equivalent to 
 (1, 5), is equivalent to (7, 5), whicj^ is satisfied by one and only one set of 
 values, X = 2 and y = 1. 
 
SIMULTANEOUS SIMPLE EQUATIONS 
 
 189 
 
 Rule. — If necessary, multiply the equations by such numbers as 
 will cause the coefficients of one letter to be numerically equal in the 
 resulting equations. 
 
 When the signs of these coefficients are unlike, add the equations; 
 when the signs are alike, subtract one equation from the other. 
 
 10. 
 
 11. 
 
 12. 
 
 Solve by addition or subtraction ; 
 
 7 a; — 5 y = 52, 
 
 2 a; + 5 y = 47. 
 
 3 a? + 2 2/ = 23, 
 
 x + y = ^. 
 
 3 .r - 4 ?/ = 7, 
 
 x + 10?/ = 25. 
 
 r 2 a; - 10 2/ = 15, 
 2a;-4y = 18. 
 
 (- 3 a; - y = 4, 
 U+3i/ = -2. 
 
 4 a; — y = 19, 
 a; + 3 y = 21. 
 
 a; + 2 2/ = 5, 
 2x + y = l. 
 
 2x + 3y = 17, 
 
 3 a; + 2 ?/ = 18. 
 
 3 a: -f 4 7/ = 25, 
 4.x-\-Sy = 31. 
 
 5x + 6y = 32, 
 7x-Sy = 22. 
 
 3x + 6y = S9, 
 9 a; — 4 2/ = 51. 
 
 13, 
 
 14. 
 
 15. 
 
 16. 
 
 17. 
 
 18. 
 
 19. 
 
 20. 
 
 (7x-9y = 6, 
 
 [x + 2y = U. 
 
 13x-y = 20, 
 4:x + 2y = 20. 
 
 (Sx-3y^U, 
 \7x-5y = 29. 
 
 6x-5y = 33, 
 4 a; + 4 y = 44. 
 
 a; + 14 y = 38, 
 14 a; + y = 142 
 
 5x + y = 12, 
 X -\- 5 y = 36. 
 
 3 a; + 11?/ = 67, 
 5x — 3y = b. 
 
 - + ^ = 12, 
 
 4 2 ' 
 
 | = _2. 
 
 21. < 
 
 3 3 
 
 7, 
 
 + ^ = 6i 
 
 X 
 
 6^2 
 
190 SIMULTANEOUS SIMPLE EQUATIONS 
 
 210. Elimination by Comparison. 
 
 1. If, in the simultaneous equations x — y = ^ and a; + 4 ?/ =13, 
 the terms containing y are transposed, what will be the resulting 
 equations ? 
 
 2. Since the second members of these derived equations are 
 each equal to x, how do they compare with each other ? 
 
 3. If these second members, then, are placed equal to each 
 other, how many itnknown numbers Avill this equation contain ? 
 
 4. How may an unknown number be eliminated from two 
 simple simultaneous equations by comparison? 
 
 Examples 
 1. Find the value of x and of 2/ in 2 x—Z ;y=10 and 5 a;+2 ?/=6. 
 
 Explanation. — Solving eq. (1) as if 
 X were the only unknown number, the 
 value of X is given as in eq. (3). In like 
 manner, solving eq. (2) for x, another 
 expression is found for the value of x, as 
 in eq. (4). 
 
 Since the equations are simultaneous, 
 
 6 — 2y ^ . X the unknown numbers have the same 
 
 /•^ yy? g ^ / values in each, and the two values of x 
 
 ^(\ "X k O form an equation, as eq. (5), from which 
 
 .'. — — ^ = ^ (5) X has been eliminated. Solving eq. (5), 
 
 ^ '^ the value of y is found to be — 2, eq. (6). 
 
 .*. 2/ = — 2 (6i) By a similar process the value of X may 
 
 10 — 6 /Q\ ^6 found ; or, substituting the value of y 
 
 2 foj" y ^^ 0116 of the preceding equations, 
 
 _ 2 /j\ as (.3), the value of x is found to be 2, 
 
 ~ eq. (7). 
 
 Proof of the Equivalence. — By § 196, (3) is equivalent to (1) and (4) 
 is equivalent to (2). Hence, the system (3, 4) is equivalent to the given 
 system. 
 
 Since (5), which by § 196 is equivalent to (6), may be obtained by sub- 
 tracting (3) from (4) and transposing, by the principle of elimination, § 208, 
 (6, 1). or (6, 3), is equivalent to (4, 3), or to (2, 1), the given system. 
 
 But since — 2 is the only value of y in the given system, by § 196, (7) is 
 equivalent to (3), or to (1). Hence. (7, 6), which is satisfied by one and 
 only one set of values, x = 2 and y = — 2, is equivalent to the given system. 
 Therefore, the given system is satisfied by x = 2, y = — 2 and by no other 
 values of x and y. 
 
 
 PROCESS 
 
 
 2x 
 
 -3y = 
 
 :10 
 
 (1) 
 
 5x 
 
 + 2y = 
 
 : 6 
 
 (2) 
 
 
 X = 
 
 10+32/ 
 2 
 
 (3) 
 
SIMULTANEOUS SIMPLE EQUATIONS 
 
 191 
 
 KuLE. — Find an expression for the value of the same unknown 
 number in each equation, equate the two expressions, and solve the 
 equation thus formed. 
 
 Solve by comparison : 
 [x + y = 70. 
 
 (5x + y = 22, 
 
 [X + 5y =14:. 
 
 4. \ 
 
 (2x + 3y = 24:, 
 
 [5x-3y=lS. 
 
 (3x + 5y=14:, 
 ' {2x-3y=3. 
 
 g (3x + 2y==36, 
 \5x-9y = 23. 
 
 7. \ 
 
 l2x + 7y = S, 
 
 [3x + 9y = 9. 
 
 g Ux + 6y = 19, 
 \3x-2y=^^. 
 
 10. \ 
 
 4a; + 3?/ = 34, 
 
 11 a; + 5?/ = 87. 
 
 (4:x-13y = 5, 
 
 11. 
 
 13. 
 
 14. 
 
 15. 
 
 [3x-\-lly=-17. 
 
 lSx-3y = Ay, 
 1- 4a; + 3?/ = 27. 
 
 12. i 
 
 (7y-X = 0, 
 
 \x + 2y = lS. 
 
 (3y + 9 = 5x, 
 |l6-2a; = 5y. 
 
 5 a; — 40 = y, 
 
 5y — 60 = X. 
 
 (2y-llx = 67, 
 
 I 2a; + 5?/ =20. 
 
 211. Elimination by Substitution. 
 
 1. In an equation containing two unknown numbers, if the 
 value of one is found, how is the value of the other obtained ? 
 
 2. Express the value of x in the first of the simultaneous 
 equations x + y = 5 and x + 2y = 7 by transposing y to the 
 second member. 
 
 3. When this expression for the value of x is substituted for 
 X in the second equation, how many unknown numbers does the 
 resulting equation contain ? 
 
 4. How may an unknown number be eliminated from simul- 
 taneous equations by substitution ? 
 

 PKOCESS 
 
 
 
 X 
 
 + 5y. 
 
 = 9 
 
 
 
 (1) 
 
 3a; 
 
 -2y. 
 
 = 10 
 
 
 
 (2) 
 
 
 X 
 
 = 9- 
 
 6y 
 
 (3) 
 
 3(9- 
 
 -52/) 
 
 -2y = 
 
 = 10 
 
 
 (4) 
 
 
 
 .'. 2/ = 
 
 = 1 
 
 
 (5) 
 
 
 
 x = 
 
 = 9- 
 
 ■5 
 
 (3) 
 
 192 SIMULTANEOUS SIMPLE EQUATIONS 
 
 Examples 
 1. Find the value of x and of y in ic+r>?/=9 and 3x—2y=10. 
 
 Explanation. — Solving eq. (1) 
 for x, X = 9 — 5 y. 
 
 Since the given equations are 
 simultaneous, x has the same value 
 in eq. (2) as in eq. (1). 
 
 If 9 — 5 ?/ is substituted for x in 
 eq. (2), the resulting equation will 
 be true. Substituting and solving, 
 
 y = l. 
 
 Substituting the value of y for y 
 x = 4 (6) in eq. (3), x = 4, eq. (6). 
 
 Proof of the Equivalence. — By § 196, (3) is equivalent to (1). In the 
 process, (4) is obtained by substituting the expression equal to x in (3) for x 
 in (2), or by substituting 3(9 — 5 ?/) for 3x in (2). Since the substitution 
 of 3(9 — 5 y) for 3 x may be performed by subtracting 3 a; = 3(9 — 5 «/) from 
 (2), by the principle of elimination (4), or the equivalent equation (5), may 
 take the place of (2) in the system (3, 2) equivalent to the given system. 
 Hence, (3, 6) is equivalent to (1, 2). 
 
 Since the only value of y in the system (3, 5) is 1, (6) is equivalent to (3); 
 and (6, 5), which is satisfied by only one set of values, a; = 4, y = 1, is equiva- 
 lent to (3, 5) and therefore to (1, 2). 
 
 Rule. — Find an expression for the value of one of the unknown 
 mimbers in one of the equations. 
 
 Substitute this value for that unknown number in the other equof 
 tion, and solve the resulting equation. 
 
 Solve by substitution : 
 
 2 \x-y = 4:, ri7 = 3a;-f2, 
 
 \4:y-x^U. ' \l = 3z-2x. 
 
 ^ |x + 2/ = 10, g_ {4.y = \0-x, 
 
 [6a;— 72/ = 34. {y — x = b. 
 
 4. 
 
 6. 
 
 3a;-4?/ = 26, ^ r7z-3x = 18, 
 x-^y = 22. ' \2z-bx = l. 
 
 62/-10a; = 14, ^^ r3-152/=-a;, 
 y — x = 3. { 3 + 15 ?/ = 4 a;. 
 
 (y + l = 3x, ^^ (l-x = 3y, 
 
 [5x + 9 = 3y. ' 1 8(1- a;) = 40-2/. 
 
SIMULTANEOUS SIMPLE EQUATIONS 
 
 193 
 
 Solve by any method, eliminating without clearing of fractions, 
 when possible : 
 
 12. 
 
 x + | = ll, 
 
 + 3y = 21. 
 
 19. 
 
 3 4 
 
 13. 
 
 ^ + ^ = 21, 
 4 5 
 
 3 5 
 
 20. 
 
 2 3 ' 
 
 2a;-l . 3y-1 ^5 
 2 3 6* 
 
 fa; 
 
 14. •! 
 
 3 2 
 
 3^ 7 
 
 21. 
 
 ' 1 + x 2a; — « o E? 
 5y^ + i^3^18-5ar. 
 
 15. 
 
 ^f + 42/ = 15, 
 
 6 3 
 
 22. 
 
 5 _ 12 = ^4- 8, 
 2 4 ' 
 
 ^ + y , a; _ 2y-a; ^k 
 
 16. 
 
 f a;-! 
 4 
 
 a;-l 
 
 I 4 
 
 + 2/ = 3, 
 + 4y = 9. 
 
 23. 
 
 f 1 
 
 x — 1 a; + y 
 3 
 
 = 0, 
 
 x-y 
 
 + 3 = 0. 
 
 17. 
 
 3 a; 2 .V _ 9^ 
 ^ + -3--20, 
 
 24. 
 
 ^ 1 9 ^ y + 32 
 2 " 4 ' 
 
 y 3a;-2y _o- 
 8"^^5~~-'^^- 
 
 18. 
 
 ^ _y 
 
 3~2' 
 
 ?_^ = 1 
 3 3 
 
 ALG. — 13 
 
 25. 
 
 ■2 y + .5 ^ .49 a; - .7 
 
 1.5 4.2 ' 
 
 .5 a; -.2 ^41 1.5y-ll 
 
 1.6 16 8 ' 
 
 fA 
 
194 
 
 SIMULTANEOUS SIMPLE EQUATIONS 
 
 26. 
 
 27. 
 
 28. 
 
 29. 
 
 30. 
 
 31. 
 
 a; + i(3a; - 2/ - 1) = ^ + |(y _ 1), 
 |(4a; + 32/) = J^(72/ + 24). 
 
 ' 6^ + 9 , 3^ + 5 V oi , 3a;-h4 
 
 Ly + 7 6a; -3y ^ 4.^-9 
 10 "^2(y-4) "^ 5 
 
 3^-5.v 2a;-8y-9 31 
 3 12 ~ 12' 
 
 f+!+^^ 
 
 0-( 
 
 4 a; - 2^ - 25 
 
 ^-.9n_ 2y-a; _ 2a;-59 
 23 - a; 2 
 
 y-lziy_3o= gy-73 . 
 
 a; - 18 3 
 
 3a; + 6 x-\-5 Qx — 2 
 
 3y-5 
 
 14 
 
 32. ^ 
 
 2y-3 3y + 4 ^3y + 5 
 6 "^5a;-7 9 
 
 'a; 16 — a; on , 5w + 2a; 
 
 2y-3 , 83-8y .^ 
 yf8- + "8 =^'-^- 
 
 23/- 
 
 4a;4- 
 
 33. i 
 
 17 -3 a; 
 
 2 ^ 16a;-l 
 
 50 y-^ -9., I 147-24y 
 
SIMULTANEOUS SIMPLE EQUATIONS 
 
 195 
 
 34. Solve 
 
 U 
 
 X 
 
 14 
 
 2 5^25^ 
 !.« y 3 
 
 SOLCTIOK 
 
 4_ 3 _14 
 
 X y b 
 
 2 5 ^25 
 X 2/ 3 
 
 (2) X 2, 
 (3)-(l), 
 
 4 10, 
 « J/ 
 
 13 
 
 60 
 3 
 
 208 
 16 
 
 16 
 16 
 
 ^ 16 
 Substituting the value of - in equation (1), 
 
 48. 
 15" 
 
 1^ 
 
 X 
 
 14 
 6 
 
 3 
 2 
 
 x=?. 
 3 
 
 (1) 
 
 (2) 
 (3) 
 (4) 
 (5) 
 (6) 
 
 (7) 
 (8) 
 (9) 
 
 In fractional equations in which the denominators are simple expressions 
 like the above, an unknown number should be eliminated before clearing of 
 fractions. 
 
 35. 
 
 36. 
 
 2 3 ' 
 
 ^4-^ = 12. 
 2 3 
 
 ^ + ^ = 13, 
 
 3 3 ' 
 
 f "3 = -^- 
 
 37. 
 
 38. 
 
 - + - = 64, 
 
 X y 
 
 ^ + 5 = 73^. 
 X y 
 
 X y 
 X y 
 
196 
 
 SIMULTANEOUS SIMPLE EQUATIONS 
 
 39. 
 
 40. 
 
 41. 
 
 r5 3 
 
 X 
 
 y 
 
 = -% 
 
 2^ + 1 = 6. 
 X y 
 
 2 3. 
 
 = 5, 
 
 X y 
 
 X y 
 
 4,3 9 
 
 X y o 
 
 3 4^11 
 X y 12' 
 
 42. -! 
 
 43. 
 
 44. 
 
 i + 5 = 30, 
 
 X y 
 
 1 + ^ = 30. 
 
 2x y 
 
 — + - = 23. 
 2a; y 
 
 8j; 3y ' 
 
 
 
 6cc 
 
 lly 
 
 = 17. 
 
 LITERA.L Simultaneous Equations 
 
 1. Solve \<^ + 'by = ^, 
 Kcx + dy = n. 
 
 (1) X d, 
 
 (2) X b, 
 (3) -(4), 
 
 (l)xc, 
 (2) X a, 
 (7) -(6), 
 
 Solution 
 ax + 6y = m 
 ex + dy = n 
 adx + 6^2/ = dm 
 
 hex + bdy = bn 
 {ad — 6c)a: = dm — bn 
 
 . _ dm — &» 
 ad — be 
 
 acx + fecy 
 acx + ady 
 
 cm 
 an 
 
 (ad — bc)y ■■ 
 
 ■'■ y 
 
 an 
 
 cm 
 
 ad — be 
 
 (1> 
 (2) 
 (3) 
 
 (4) 
 
 (5) 
 
 (6) 
 (7) 
 
 (8) 
 
 In literal simultaneous equations, elimination is usually performed by the 
 method of addition and subtraction. 
 
SIMULTANEOUS SIMPLE EQUATIONS 
 
 197 
 
 3. 
 
 5. 
 
 6. 
 
 9. 
 
 10. 
 
 11. 
 
 ■ ax +hy = m, 
 bx — ay=:c. 
 
 f ax — by = m, 
 [ex — dy = r. 
 
 f ax = by, 
 \x + y = ab. 
 
 ( X —ay = n, 
 [ bx + y = p. 
 
 (a(x-y) = 5, 
 [ bx — cy = n. 
 
 ( a(a — x) = b(y^b), 
 \ax= by. 
 
 \x + y = b—a, 
 \bx — ay + 2ab = 0. 
 
 x y a 
 11^1 
 
 ( X y b 
 
 X y 
 
 b _«__i 
 
 X y 
 
 a b 
 
 l ab ab 
 
 12. 
 
 i a b 
 
 [ bx — ay = 0. 
 
 13. 
 
 14. •( 
 
 a b 
 6 a~2 
 
 1 . 1 
 
 — + r- = c> 
 ax by 
 
 15. 
 
 6a; ay 
 
 a; y 
 
 r a;4-l _ a + 6 + l 
 16. •(2/ + 1 a-b + 1 
 [x-y = 2b. 
 
 x-{-y_x — y 
 
 17. 
 
 18. 
 
 
 a 
 
 
 & 
 
 
 x 
 
 -y - 
 
 a 
 1 
 
 L 
 
 1_ 
 
 a 
 
 X 
 
 — a 
 
 a 
 
 — 
 
 y 
 
 X 
 
 + 2/_ 
 
 a 
 
 
 
 19. \ 
 
 20. 
 
 a;-?/ 
 
 a b 
 be 
 
 + 
 
 _a 
 
 a + a; 6 — 2/ & 
 6 , a 
 
 .a + X b — y a 
 
198 SIMULTANEOUS SIMPLE EQUATIONS 
 
 Problems 
 
 1. There are two numbers such that if twice the first is added 
 to 3 times the second, the sum will be 130 ; but if 5 times the 
 first is diminished by the second, the remainder will be 70. 
 What are the numbers ? 
 
 
 Solution 
 
 Let 
 
 X — the first number, 
 
 and 
 
 y = the second number. 
 
 Then, 
 
 2 a; + 3 y = 130, 
 
 and 
 
 5 X - 2/ = 70. 
 
 Eliminating y, 
 
 17 x = 340, 
 
 
 a; = 20. 
 
 Whence, by substitution 
 
 y = 30. 
 
 2. A drover sold 3 cows and 7 horses to one person for $ 600, 
 and to another person, at the same prices, 3 cows and 3 horses for 
 $ 300. How much per head did he get for each ? 
 
 3. With $ 30 a man can buy 20 yards of one kind of cloth and 
 50 yards of another ; with $ 23 he can buy 30 yards of the first 
 kind and 20 yards of the second kind. What is the price of each 
 per yard ? 
 
 4. If 45 bushels of wheat and 37 bushels of rye together cost 
 $ 62.70, and 37 bushels of wheat and 25 bushels of rye, at the 
 same prices, cost $ 48.30, what is the price of each per bushel ? 
 
 5. Henry expended 95 cents for apples and oranges, paying 
 5 cents for each orange and 4 cents for each apple. If he had 22 
 of both, how many of each did he buy ? 
 
 6. Five years ago A was ^ as old as B, and 10 years hence he 
 will be ^ as old as B.. What are their ages ? 
 
 7. A said to B, " If you were twice as old, and I were ^ as old, 
 or if you were ^ as old, and I were 3 times as old, the sum of 
 our ages would be 70." How old was each ? 
 
 8. A boy is given 28 cents to buy a dozen cakes. He finds 
 that some cost 2 cents each and some 3 cents each. How many 
 of each kind can he purchase ? 
 
SIMULTANEOUS SIMPLE EQUATIONS 199 
 
 9. A said to B, " Give me $ 20, and I shall have 3 times as 
 much money as you." B replied, " Give me $ 5, and I shall have 
 twice as much money as you." How much money had each ? 
 
 Solution 
 
 Let -x = the number of dollars A had, 
 
 and y = the number of dollars B had. 
 
 Then, a; + 20 = 3(y - 20), 
 
 and y+ 5 = 2(x— 5). 
 
 Solving, X = 25, the number of dollars A had, 
 
 and y = 35, the number of dollars B had. 
 
 10. If A gives B $ 100, B will have 4 times as much money as 
 A ; but if B gives A $ 200, A will have 4 times as much money 
 as B. What sum of money has each ? 
 
 11. A said to B, " Give me 20 cents of your money, and I shall 
 have, half as much as you." B replied, "Give me 25 cents of 
 your money, and I shall have 5 times as much as you." How 
 much had each ? 
 
 12. If A had $300 more, he would have twice as much as B; 
 if B had $ 300 less, he would have \ as much as A. How much 
 money has each ? • 
 
 13. If 1 is added to each term of a fraction, its value will 
 be f ; if 1 is subtracted from each term of the fraction, its value 
 will be ^. What is the fraction ? 
 
 Solution 
 Let - represent the fraction. 
 
 y 
 
 Then, =«=+l-2 
 
 and 
 
 y+1 3' 
 
 x-l 1 
 
 y-1 2 
 Solving, a; = 3, 
 
 and y = 5, 
 
 3 
 
 That is, - is the fraction. 
 
 6 
 
200 SIMULTANEOUS SIMPLE EQUATIONS 
 
 14. If 1 is added to the numerator of a certain fraction, its 
 value becomes f ; if 2 is added to the denominator, its value 
 becomes ^. What is the fraction ? 
 
 15. Find a fraction that is equal to \ when its terms are 
 diminished by 2, and is equal to f when its terms are increased 
 by 2. 
 
 16. A certain number expressed by two digits is equal to 7 
 times the sum of its digits ; if 27 is subtracted from the number, 
 the difference will be expressed by reversing the order of the 
 digits. What is the number? 
 
 
 SOLCTION 
 
 Let 
 
 05 = the digit in tens' place. 
 
 and 
 
 y = the digit in units' place. 
 
 Then, 
 
 10 a; + y = the number, 
 
 and 
 
 10 y + a; = the number with its digits reversed ; 
 
 
 .-. 10« + j^ = 7(x + y), 
 
 and 
 
 10 X + y - 27 = 10 1/ + X. 
 
 Solving, 
 
 x = 6, 
 
 and 
 
 2/ = 3. 
 
 Hence, 
 
 10 a; + !/ = 60 + 3, or 63, the number. 
 
 17. The sum of the two digits of a certain number is 12, and 
 the number is 3 greater than 6 times the sum of its digits. 
 What is the number ? 
 
 18. When a certain number expressed by two digits is divided 
 by the sum of its digits, the quotient is 8 ; and when the first 
 digit is diminished by 3 times the second, the remainder is 1. 
 What is the number? 
 
 19. The sum of the two digits of a number is 12. If the 
 order of the digits is reversed, the number will lack 12 of being 
 doubled. What is the number ? 
 
 20. A farmer bought 100 acres of land for $ 3250. If part of 
 it cost him $ 40 an acre and the rest of it ^ 15 an acre, how many 
 acres were there of each kind ? 
 
SIMULTANEOUS SIMPLE EQUATIONS 201 
 
 21. The admission to an entertainment was 50 cents for adults 
 and 35 cents for children. If the proceeds from 100 tickets 
 amounted to $ 39.50, how many tickets of each kind were sold ? 
 
 22. A man paid a bill of $16 in 25-cent pieces and 5-cent 
 pieces. If the number of coins was 80, how many of each kind 
 were there ? 
 
 23. A man paid $ 1 for some apples at 3 cents each and some 
 oranges at 5 cents each. He sold ^ of the apples and \ of the 
 oranges at cost for 34 cents. How many of each did he buy ? 
 
 24. A and B together can do a piece of work in 12 days. 
 After A has worked alone for 5 days, B finishes the work in 26 
 days. In what time can each alone do the work ? 
 
 25. A blacksmith and his son had a contract to make a certain 
 number of horseshoes. If both had worked together, they could 
 have done the work in 6 days. But the father worked 8 days, 
 and the son finished the work in 3 days. In how many days 
 could each have done the work ? 
 
 26. A man and his two sons can dig a ditch in 6 days ; if the 
 man and either son work 7 days, the other son can complete the 
 ditch by working 2 days. In what time can each alone dig 
 the ditch? 
 
 27. A certain number of persons agree to share equally the 
 expense of hiring a coach. If each paid 75 cents, there would be 
 $ 1.25 over; but if each paid 50 cents, there would be $ 2.50 lack- 
 ing. What is the number of persons and the expense of hiring 
 the coach ? 
 
 28. A train ran a certain distance at a uniform rate. Had the 
 rate been increased 5 miles an hour, the journey would have been 
 2 hours shorter; but had the rate been diminished 5 miles an 
 hour, the journey would have been 2\ hours longer. What was 
 the distance and the rate of the train ? 
 
 Suggestion. — Let x miles per hour be the actual rate of the train and y 
 the number of hours required to complete the journey. 
 
 29. A sum of money was divided equally among a certain 
 number of persons. If there had been 4 persons more, the 
 share of each would have been $ 3 less ; but if there had been 
 
202 SIMULTANEOUS SIMPLE EQUATIONS 
 
 2 persons less, the share of each would have been $2 more. 
 Among how many persons was the money divided and what was 
 the share of each ? 
 
 30. A dealer had eggs to sell and wished to buy potatoes. 
 He found that 6 dozen eggs were worth 10 cents moie than 2 
 bushels of potatoes ; and that 10 dozen eggs were worth 10 cents 
 less than 4 bushels of potatoes. How much were eggs and 
 potatoes worth ? 
 
 31. If a rectangular floor were 2 feet wider and 5 feet longer, 
 its area would be 140 square feet greater ; if it were 7 feet wider 
 and 10 feet longer, its area would be 390 square feet greater. 
 What are its dimensions ? 
 
 32. If 54 is added to a certain number, expressed by two 
 digits whose sum is 8, the order of the digits will be reversed. 
 What is the niimber ? 
 
 33. If 13 is added to a certain number, the sum will be equal 
 to 5 times the sum of the two digits of the number ; and if 36 is 
 added to the number, the order of its digits will be reversed. 
 What is the number ? 
 
 34. A and B can do a piece of work in a days, or if A works 
 m days alone, B can finish the work by working n days. In how 
 many days can each do the work ? 
 
 35. A can build a wall in c days, and B can build it in d days. 
 How many days must each work so that, after A has done a part 
 of the work, B can take his place and finish the wall in a days 
 from the time A began ? 
 
 36. One cask contains a mixture of 20 gallons of wine and 
 30 gallons of water, and another contains a mixture of 12 gallons 
 of wine and 15 gallons of water. How many gallons must be 
 drawn from each cask to form a mixture that will contain 8 gallons 
 of wine and 11 gallons of water ? 
 
 Suggestion. — If x gallons of the mixture are drawn from the first cask, 
 I X gallons of it will be wine. 
 
 If y gallons of the mixture are drawn from the second cask, | y gallons of 
 it will be wine. 
 
SIMULTANEOUS SIMPLE EQUATIONS 203 
 
 37. "If I had received 3 oranges more for my money," said 
 A, " they would have cost me 1 cent less each ; but if I had 
 received 2 less, they would have cost me 1 cent more each." 
 How many oranges had he bought, and at what price each ? 
 
 38. A merchant mixes two kinds of tea. If he mixes it in 
 parts proportional to 7 and 5, the value of the mixture is 46 cents 
 a pound. If he mixes it in parts proportional to 5 and 7, the 
 value of the mixture is 50 cents a pound. What is each kind of 
 tea worth per pound ? 
 
 39. A man invested $4000, a part at 5% and the rest at 4%. 
 If the annual income from both investments was $ 175, what was 
 the amount of each investment ? 
 
 40. A man invested a dollars, a part at ?•% and the rest at s% 
 yearly. If the annual income from both investments was b 
 dollars, what was the amount of each investment ? 
 
 41. A sum of money, at simple interest, amounted to h dollars 
 in t years, and to a dollars in s years. What was the principal, 
 and what was the rate of interest ? 
 
 42. A had a certain sum invested at a certain rate per cent, 
 and B had f 100 less invested at a rate 2% higher. B's annual 
 income was $ 56 greater than A's ; but if B's rate upon his invest- 
 ment had been only 1% higher than A's his annual income would 
 have been only $ 25 greater than A's. How much was invested 
 by each man, and at what rate ? 
 
 43. A crew can row 8 miles downstream and back^ Oi- 12 miles 
 downstream and half the way back in 1| hours, What is the 
 rate of rowing in still water and the velocity of the stream ? 
 
 44. A man rows 15 miles downstream and oack in 11 hours. 
 The current is such that he can row 8 miles downstream in the 
 same time as 3 miles upstream. What is his rate of rowing in 
 still water, and what is the velocity of the stream ? 
 
 45. A box will hold 18 quires of paper and 18 bunches of 
 envelopes, or 20 quires of paper and 15 bunches of envelopes. 
 How many quires of paper will the box hold ? How many 
 bunches of envelopes will it hold ? 
 
204 SIMULTANEOUS SIMPLE EQUATIONS 
 
 46. A shelf will hold 20 arithmetics and 24 algebras oi 14 
 arithmetics and 36 algebras. How many arithmetics will the 
 shelf hold ? How many algebras will it hold ? 
 
 47. Two men had a certain distance to row and took turns in 
 rowing. Whenever the first rowed, the boat moved at a rate suf- 
 ficient to cover the entire distance in 10 hours, and whenever the 
 second rowed, in 14 hours. If the journey was completed in 12 
 hours, how many hours did each row ? 
 
 48. A train ran 1 hour and 36 minutes, and was then detained 
 40 minutes. It then proceeded at f of its former rate and reached 
 its destination 16 minutes late. If the detention had occurred 10 
 miles farther on, the train would have arrived 20 minutes late. 
 At what rate did the train set out, and what was the whole dis- 
 tance traveled ? 
 
 49. A certain number of people charter an excursion boat, 
 agreeing to share the expense equally. If each pays a cents, 
 there will be h cents lacking from the necessary amount ; and if 
 each pays c cents, d cents too much will be collected. How many 
 persons are there, and how much should each pay ? 
 
 50. A sum of money was to be divided equally among a certain 
 number of persons, but a persons more than were expected ap- 
 peared to claim a share, and in consequence each received h dol- 
 lars less. If there had been c persons less than were expected, 
 each would have received d dollars more. How many persons 
 were there, and how much did each receive ? 
 
 Give the results when a = 5, h — 100, c = 4, and d = 125. 
 
 51. A and B working together can do a piece of work in a 
 days. But finding it impossible to work at the same time, A 
 works h days, and later B finishes the work in c days. In how 
 many days can each do the work alone ? 
 
 If a = 5y^y, 6 = 5, and c = 6, in how many days can each do 
 the work alone ? 
 
 52. A purse holds c crowns and d guineas; a crowns and 6 
 
 guineas will fill — th of it. How many will it hold of each ? 
 n 
 
 How many, if c = 12, d = 6, a = 4, 6 = 6, wi = 1, and n = 2'i 
 
SIMULTANEOUS SIMPLE EQUATIONS 205 
 
 53. A mine is emptied of water by two pumps which together 
 discharge m gallons per hour. Both pumps can do the work in h 
 hours, or the larger can do it in a hours. How many gallons per 
 hour does each pump discharge ? What is the discharge of each 
 per hour when a = 5, & = 4, and m = 1250 ? 
 
 54. Two trains are scheduled to leave A and B, m miles apart, 
 at the same time, and to meet in h hours. If the train that leaves 
 B is a hours late and runs at its customary rate, it will meet the 
 first train in c hours. What is the rate of each train ? 
 
 What is the rate of each, if m = 800, c = 9, a = If, and 6 = 10 ? 
 
 55. If a quarts of good wine is mixed with h quarts of poorer 
 wine, the mixture will be worth c cents a quart ; if h quarts of 
 the better wine is mixed with a quarts of the poorer, the mixture 
 will be worth d cents a quart. What is each kind of wine worth 
 per quart ? What is each kind of wine worth per quart, if a = 40, 
 6 = 20, c = 100, and rf = 80 ? 
 
 212. Discussion of the general solution of a system of two simul- 
 taneous simple equations involving two unknown numbers. 
 
 Let ax + by — c (1) 
 
 and * a'x + b'y = c' (2) 
 
 be any two simultaneous simple equations. 
 
 (1) X 6', ab'x + bb'y = b'c (3) 
 
 (2) X b, a'bx + bb'y = be' (4) 
 
 (3) - (4), iab'- a'b}x = b'c - be (5) 
 
 (1) X a', aa'x + a'by = ca' (6) 
 
 (2) X a, aa'x -t- ab'y = c'a (7) 
 
 (7)-(6), (ab'-a'b)y = c'a-ca' (8) 
 
 By the principles of equivalence the given system may be replaced by the 
 
 equivalent system (5, 8), in which (5) involves x alone and (8) involves y 
 
 alone. By § 197, each of these simple equations involving one unknown 
 
 number has one and only one root, which can be found except when the 
 
 common coefficient of x and y is equal to zero. Hence, when ab'— a'b is not 
 
 equal to zero, the given system is satisfied by one and only one set of 
 
 values 
 
 ^^b^c-bc' ^j^^ y ^ C'a - ca' 
 
 ab'— a'b ab' — a'b 
 
 * In algebraic notation a', a", a'", etc., are read 'a prime,' 'a second,' 
 ' a third,' etc. 
 
206 SIMULTANEOUS SIMPLE EQUATIONS 
 
 If ah' — a'b = 0, that is, if ab'= a'b, the first members of (3) and (4) are 
 identical, and therefore the second members must be equal. The same is 
 true of equations (6) and (7). Hence, (4) is only a different form of (3), 
 and (7) is only a different form of (6); that is, (1) and (2) are not inde- 
 pendent equations. 
 
 But if ab'— a'b is not equal to zero, neither (3) and (4) nor (6) and (7) 
 are reducible to the same form ; that is, (1) and (2) are independent 
 equations. 
 
 It is evident from the Distributive Lav? for multiplication that the equations 
 (3) and (4), and also (6) and (7), cannot be combined by addition or sub- 
 traction unless X and y have the same values in (2) as in (1); that is, that 
 (1) and (2) cannot be solved unless they are simultaneous equations. 
 
 It is evident that equations (5) and (8), and therefore (1) and (2), cannot 
 be solved if ab' — a'b = 0, since § 196, Prin. 2, the members cannot be divided 
 by a knovrn expression equal to ; and it has been shown that ( 1 ) and (2) 
 are dependent or independent equations according as ab' — a'b is or is not 
 equal to zero. 
 
 Hence, it follows that : 
 
 Two simple equations involving two unknoicn numbers cannot he solved 
 unless the equations are simultaneous and independent. 
 
 Evei'y system of two independent simultaneous simple equations involving 
 two unknown numbers can be solved, and is satisfied by one, and only one, 
 set of values of its unknown numbers. 
 
 THREE OR MORE UNKNOWN NUMBERS 
 
 213. 1. In the equations x + 2 y -\- z = S and 2x + y — z = 1, 
 how may z be eliminated ? 
 
 2. If one of the unknown numbers in the above equations is 
 eliminated, how many unknown numbers will be left ? 
 
 3. How many independent equations are necessary before the 
 values of two unknown numbers can be found ? 
 
 4. How many independent equations containing the same two 
 unknown numbers can be formed by combining the equations 
 in (1) ? 
 
 5. Since only one derived equation containing two unknown 
 numbers was obtained from the given equations by eliminating z, 
 and since we must have two such equations before we can find 
 the values of x and y, if another independent equation involving 
 aj, y, and z is combined with either of the equations in (1), how 
 
SIMULTANEOUS SIMPLE EQUATIONS 207 
 
 many independent equations containing x and y only will be 
 available for finding the values of x and y? 
 
 6. When the values of x and y are found, how may the value 
 of z be found ? 
 
 7. Then, how many independent equations containing three 
 unknown numbers must be given, so that the values of the un- 
 known numbers may be found ? How many to find the values 
 of four unknown numbers ? 
 
 214. Principle. — Every system of independent simultaneous 
 simple equations involving the same number of unknown numbers as 
 there are equations can be solved, and is satisfied by one and only 
 one set of values of its unknown numbers. 
 
 The above principle may be established as follows : 
 
 From the given system of n equations involving n unknown numbers, a 
 second system of w — 1 equations involving n — \ unknown numbers may be 
 derived by eliminating one of the unknown numbers ; from the second system 
 a third system of m — 2 equations involving n — 2 unknown numbers may be 
 derived ; and this process may be continued until the wth system, a single 
 simple equation involving only one unknown number, is obtained. 
 
 Since, § 197, this equation has one and only one root, by substituting this 
 value in either of the two equations of the next preceding system and solving, 
 one and only one value of the other number in that equation is obtained ; by 
 substituting these two values in any one of the three equations of the next 
 preceding system, one and only one value of the remaining unknown number 
 in that equation is obtained ; arid by continuing this process, the value of 
 each of the other unknown numbers is obtained. 
 
 By the principles of equivalent equations, the following system of n equa- 
 tions may be substituted for the given system : the single equation finally 
 derived by elimination and composing the nth system ; either of the two 
 equations of the preceding, or (n — l)th system ; any one of the three equa- 
 tions of the system preceding that, or of the (n — 2)th system ; and so on to 
 any one of the n equations in the 1st or given system. 
 
 But each of the n equations just described has one and only one value 
 of an imknown number. Hence, the given system can be solved, and is 
 satisfied by one and only one set of values of its unknown numbers. 
 
 If the number of unknown numbers is greater than the number of inde- 
 pendent simultaneous equations., the last equation obtained by repeated elimi- 
 nations is indeterminate, and hence the system is indeterminate. 
 
 If the number of unknown numbers is less than the number of independent 
 simultaneous equations, say n — p, any n — p of the equations involving the 
 n — p unknoion numbers form a determinate system. 
 
208 SIMULTANEOUS SIMPLE EQUATIONS 
 
 Examples 
 
 1. Find the values of x, y, and 2; in | 2 a; + y -(- 2 2 = 10, 
 
 I 3 a; + 4?/ -32! = 2. 
 
 Solution 
 a; + 2y + 3z = 14 
 2x + y + 2z = \0 
 33: + 4y-32 = 2 
 Eliminating z by combining (1) and (3), 
 (l) + (3), 4a: + 6y = 16 
 
 Eliminating z by combining (2) and (3), 
 (2)x3, 6x + 3y + 62 = 30 
 
 (3)x2, 6a; + 82/-60=4 
 
 Adding, 12 a; +11?/ = 34 
 
 Eliminating x by combining (7) and (4), 
 (4)x3, 12a; + 18y = 48 
 
 (8) -(7), ly = U 
 
 .-. 2/ = 2 
 Substituting in (4), 4 a; + 12 = 16 
 
 .-. X = 1 
 Substituting the values of x and y in (1), 
 
 1 + 4 + 32 = 14 (13) 
 
 .-.2 = 3 (14) 
 
 Explanation. — Eliminating z from (1) and (3) and from (2) and (3), 
 two simultaneous equations, (4) and (7), are obtained involving x and y. 
 By the principle of elimination, § 208, the new systeiii (1, 4, 7), or (2, 4, 7), 
 or (3, 4, 7), is equivalent to the given system. 
 
 Eliminating x from (4) and (7), a simple equation involving but one 
 unknown number y is obtained, and from this equation the value of y is 
 found, equation (10). Hence, the system (1, 4, 7) has been replaced by the 
 equivalent system (1, 4, 10), which is, therefore, equivalent to the given 
 system. 
 
 Substituting 2 for y in (4), the value of x is found, giving a new system 
 (1, 12, 10) equivalent to the given system. Substituting the values of both 
 X and y in (1), the value of ^ is found, giving the desired system (14, 12, 10) 
 equivalent to the given system. 
 
SIMULTANEOUS SIMPLE EQUATIONS 
 
 209 
 
 Solve the following : 
 
 r2ic- y + 2z = 12, 
 2. |iB+3?/+ 2 = 41, 
 
 3. 
 
 9. 
 
 10. 
 
 11. 
 
 f3a; + 5y — 2 = 8, 
 
 I 4a; + 32/ + 22 = 47, 
 
 ' 03 + 3 ?/ — z = 10, 
 2a; + 5y + 42! = 57, 
 3a;-2/ + 2z! = 15. 
 
 a; + 2/ + z = 53, 
 a; + 22/ + 3z = 105, 
 a; + 32/ + 4z; = 134. 
 
 a; - 2/ + z = 30, 
 3 y - a; - 2 = 12, 
 qz — y + 2x= 141. 
 
 |'8a;-52/ + 2z = 53, 
 J a; + y - z = 9, 
 [l3a;-9y + 3z = 71. 
 
 ra; + 32/ + 4z = 83, 
 I a; + 2/ + z = 29, 
 l6a; + 82/ + 3z = 156. 
 
 2x + 3?/ + 4z = 29, 
 
 3a; + 22/+5z = 32, 
 
 .4a; + 32/ + 2z = 25. 
 
 f 2a; -3?/ + 42-^ = 4, 
 4a; + 2?/ — z + 2'u = 13, 
 a;-2/ + 2z + 3v = 17, 
 3aj + 22/-2 + 4y = 20. 
 
 ALG. — 14 
 
 12. 
 
 13. 
 
 14. 
 
 15. 
 
 16. 
 
 17. 
 
 18. 
 
 3a;-22/ + z = 2, 
 2x^5y + 2z = 21, 
 a; + 3?/ + 3z = 25. 
 
 4a;-52/ + 3z = 14, 
 X + 7 2/ — z = 13, 
 2a; + 52/ + 5z = 36. 
 
 f2a; + y-3z + 4w = 44, 
 3 a; — 2 2/ + z — «; = — !, 
 4a; — 2/ + 2z + w = 55, 
 5a; — 32/ + 4z — w = 39. 
 
 7a; -1 = 32/, 
 11 2 _ 1 = 7 ?;, 
 
 ]4z-l = 7?/, 
 19a;_l=3u 
 
 \x + \y + \z=^12. 
 
 ria;-i2/ + iz = 3, 
 ia;-i2/ + iz = l, 
 ia;-i2/ + iz = 5. 
 
 ■^Jhl + 3z = 29, 
 3 
 
 -^^-?/ + 2z = 22, 
 2 
 
 3a;-2/ = 3(z-l). 
 
 3a;+?/-z + 2v = 0, 
 3?/ — 2a; + z — 47; = 21, 
 a; — 2/ + 2z — 3-^ = 6, 
 4 a; + 2 2/ - 3 z + V = 12. 
 
210 
 
 SIMULTANEOUS SIMPLE EQUATIONS 
 
 { u-\-v-\-x-{-y—z=5, 
 u+v+x—y \-z=l, 
 19. Solve \ u-\-v—x-{-y^z=9, 
 
 u—v+x+y-\-z=H, 
 , v—u-\-x-\-y-\-z=13. 
 Solution. — Adding the equations, 
 
 Dividing by 3, u + v ->r x -\- y + z = \b. 
 
 Subtracting each of the given equations from this equation, 
 22- = 10, 2!/ = 8, 2x = 6, 2» = 4, 2w = 2; 
 . •. « = 5, S/ = 4, X = 3, V — 2, M = 1. 
 
 Solve the following : 
 
 f a; + ?/ = 9, 
 
 2^+2 = 7, 
 
 z -\- x = 5. 
 
 ■v + x + y=15, 
 x-\-y + z = 18, 
 
 y + Z + V = 17, 
 
 2 + -y + a; = 16. 
 
 20 
 
 21. ^ 
 
 24. 
 
 25. 
 
 22. 
 
 6, 
 10, 
 
 23. 
 
 X y 
 y z 
 
 Z X 
 
 _xy_^l 
 x-\-y 5' 
 
 yz ^1^ 
 y + z 6' 
 
 26. 
 
 27. 
 
 «» 
 
 z + x 7 
 Suggestion. — If 
 
 ( X + 3 y-\-z = 14:, 
 x + y + 3z=16, 
 3x + y-\-z = 20. 
 
 ' y + z -\-v — x = 22, 
 z -\- V + X — y = 18, 
 V -^x + y — z = 14, 
 x-\-y + z — v = 10. 
 
 1+^-1 = 0, 
 
 X y 
 
 1 + 1 + 3 = 0, 
 
 y z 
 
 - + --2 = 0. 
 
 z X 
 
 _xy_^l^ 
 
 x + y a 
 
 yz ^1^ 
 
 2/ + 2 b 
 
 zx _1 
 
 z + X c 
 
 xy 
 x + y' 
 
 1 x+y 5 , lie 
 
 1 775 = zr ; whence, - + - = 6. 
 
 5 xy 1 y X 
 
SIMULTANEOUS SIMPLE EQUATIONS 
 
 211 
 
 
 X 
 
 b 
 
 y 
 
 z 
 
 a, 
 
 28. Solve 
 
 hzx 
 
 — 
 
 cxy 4- ayz — hxyz, 
 
 
 a 
 
 h 
 
 c _ 
 
 c. 
 
 
 X 
 
 y 
 
 z 
 
 Solution 
 
 (l) + (3), 
 
 
 
 
 X 
 
 .■.x= 2«. 
 
 a + c 
 
 (2) ^ xyz. 
 
 
 
 b_ 
 
 y 
 
 -5 + « = 6. 
 
 Z X 
 
 (5) -(3), 
 
 
 
 
 2fe ». ^ 
 =0 — c. 
 
 26 
 
 b-c 
 Substituting the values of x and y in (1), and solving, 
 
 2c 
 
 (1) 
 (2) 
 (3) 
 
 (4) 
 (6) 
 
 29. 
 
 30. 
 
 31. 
 
 32. 
 
 axy — x — y = 0, 
 
 bzx — z — x = 0, 
 
 . cyz — y — z = 0. 
 
 x + y — z = 0, 
 X — y = 2 b, 
 ,x + z = 3a + b. 
 
 ' V -\- X = 2 a, 
 x-{-y = 2a — z, 
 y + z = a + b, 
 V — z = a -\- c. 
 
 'y + z — 3x = 2a, 
 z-\-x — Sy = 2b, 
 x + y — 3z = 2c, 
 
 [2x + 2y-\-v = 0. 
 
 f abxyz -\- cxy — ayz — bzx = 0, 
 33. ^bcxyz+ayz—bzx—cxy=0, 
 
 [ caxyz + bzx — cxy — ayz = 0. 
 
 ix + y + z = a + b + c, 
 b-\-2c, 
 & + 3c. 
 
 'v-{-x + y = a + 2b-{-c, 
 x + y + z = 3b, 
 y + z+v = a + b, 
 
 ,z-\-v-\-x = a + 3b — c. 
 
 ' ax + by -{- cz = 3, 
 a + b 
 
 34. i x-{-2y + 3z 
 
 36. 
 
 36. i 
 
 x + y = 
 y + z = 
 
 ab 
 
 6 + c 
 
 be 
 
212 SIMULTANEOUS SIMPLE EQUATIONS 
 
 Peoblems 
 
 215. 1 . Three men bought grain at the same prices. A paid 
 $ 4.80 for 2 bushels of rye, 3 bushels of wheat, and 4 bushels of 
 oats ; B paid f 6.40 for 3 bushels of rye, 5 bushels of wheat, and 
 
 2 bushels of oats; and C paid $5.30 for 2 bushels of rye, 4 
 bushels of wheat, and 3 bushels of oats. What was the price of 
 each? 
 
 2. A dealer was asked his price for 10 bushels of wheat, corn, 
 and rye. He replied, " For 5 of wheat, 2 of corn, and 3 of rye, 
 $ 6.60 ; for 2 of wheat, 3 of corn, and 5 of rye, $ 5.80 ; and for 
 
 3 of wheat, 5 of corn, and 2 of rye, f 5.60." What prices had 
 he in mind ? 
 
 3. Divide 90 into three parts such that the sum of i of the 
 first, \ of the second, and \ of the third shall be 30; and the 
 first shall be twice the third diminished by twice the second. 
 
 4. There are three numbers such that the sum of \ of the 
 first, \ of the second, and \ of the third is 12 ; of ^ of the first, 
 \ of the second, and \ of the third is 9, and the sum of the num- 
 bers is 38. What are -the numbers ? 
 
 5. There are three numbers whose sum is 72. If the sum of 
 the first two is divided by the third, the quotient is If ; and if 
 the third is subtracted from twice the first, the remainder will be 
 \ of the second. Find the numbers. 
 
 6. A and B can do a piece of work in 10 days ; A and C can 
 do it in 8 days ; and B and C can do it in 12 days. How long 
 will it take each to do it alone ? 
 
 7. A certain number is expressed by three digits whose sum is 
 14. If 693 is added to the number, the digits will appear in 
 reverse order. If the units' digit is equal to the tens' digit 
 increased by 6, what is the number ? 
 
 8. The third digit of a number of three digits is as much larger 
 than the second digit as the second is larger than the first. If 
 the number is divided by the sum of its digits, the quotient is 15. 
 What is the number, if the order of its digits may be reversed by 
 adding 396 ? 
 
SIMULTANEOUS SIMPLE EQUATIONS 213 
 
 9. Find three numbers such that the first increased by \ of 
 the sum of the other two shall be 36 ; the second increased by ^ 
 of the sum of the other two shall be 40 ; and the third increased 
 by \ of the sum of the other two shall be 44. 
 
 10. Divide 800 into three parts such that the sum of the first, 
 \ of the second, and f of the third shall be 400 ; and the sum of 
 the second, | of the first, and \ of the third shall be 400. 
 
 11. Three cities, A, B, and C, connected by straight roads, are 
 situated at the vertices of a triangle. From A to B by Avay of C 
 is 130 miles ; from B to C by way of A is 110 miles ; and from C 
 to A by way of B is 140 miles. How far apart are the cities ? 
 
 12. Find three numbers such that the first with \ of the sum of 
 the second and third is 340 ; the second with \ of the sum of the 
 first and third is 600; and the third with \ of the remainder 
 when the first is subtracted from the second is 450. 
 
 13. A merchant has three kinds of tea. He can sell 2 pounds of 
 the first kind, 3 of the second, and 4 of the third for $ 4.70 ; or 4 
 of the first, 3 of the second, and 2 of the third for f 4.30. If a 
 pound of the third kind is worth 5 cents more than f of a pound 
 of the first kind and |^ of a pound of the second kind, what is the 
 value of 1 pound of each kind ? 
 
 14. A, B, and C have certain sums of money. If A gives B 
 $ 100, they will have the same amount ; if A gives C $ 100, C 
 will have twice as much as A; and if B gives C $100, C will 
 have 4 times as much as B. What sum has each ? 
 
 15. A quantity of water sufiicient to fill three jars of different 
 sizes will fill the smallest jar 4 times ; the largest jar twice with 
 4 gallons to spare ; or the second jar 3 times with 2 gallons to 
 spare. What is the capacity of each jar ? 
 
 16. A gave to B and C as much as each of them had ; B then 
 gave to A and C as much as each of them had ; and C then gave 
 to A and B as much as each of them had, after which each had 
 $ 8. How much had each at first ? 
 
 17. Three boys. A, B, and C, each had a bag of nuts. After 
 each boy had given each of the others \ of the nuts in his bag, 
 they counted and found that A had 740, B 580, and C 380. How 
 many had each at first ? 
 
INVOLUTION 
 
 216. 1. How many times is a number used as a factor in pro- 
 ducing its second power ? its third power ? its fourth power ? its 
 fifth power ? its /ith power, when n is a positive integer ? 
 
 2. What is the meaning of 2^? of (-2)3? of a'? oi (axf? 
 of cc", when n is a positive integer ? 
 
 3. What sign has (+ ay? (+ of ? (+ a)", or a7iy power of a ? 
 What sign has any power of a positive number ? 
 
 4. What sign has (-a)2? (-af? (-a)*? (-«)«? 
 
 What sign have the even powers of a negative number ? What 
 sign have the odd powers ? 
 
 5. What is the fourth power of a^ ? of a? ? of a^" ? of a", when 
 w is a positive integer ? What are the fifth powers of these num- 
 bers ? the sixth powers ? the mth powers, when m is a positive 
 integer ? 
 
 6. How does 8^ compare in value with 2^x4^? with 22x2^x22? 
 32 with 6- -=- 2^ ? 5=^ with 10=^ -=-22? 
 
 217. The process of finding any required power of an ex- 
 pression is called Involution. 
 
 218. Principles. — 1. Law of Signs. — All p owers of a positive 
 number are positive; even powers of a negativ e number are positive, 
 and odd powers are negative. 
 
 2. Law of Exponents. — The exponent of a power of a number is 
 equal to the exponent of the number multiplied by the exponent of the 
 power to which the number is to be raised. 
 
 3. Any p)oiver of a product is equal to the product of its factors 
 each raised to that poiver. 
 
 4. Any power of the quotient of two numbers is equal to the quo- 
 tient of the numbers each raised to that power. 
 
 214 
 
INVOLUTION 215 
 
 The above principles may be established as follows : 
 Principle 1 follows directly from the law of signs for multiplication. 
 Principle 2. When m and n are positive integers, 
 § 24, (a*")" = fl" X a" X a™ ••• to n factors 
 
 ;-; 0m+in+m+— to n terms 
 
 = a™". 
 Principle 3. When n is a positive integer, 
 § 24, (o6)"= ab X ab X ab ••• to n factors 
 
 § 83, =(aaa ••• to n factors) {bbb ••• to n factors) 
 
 = a»6». 
 Principle 4. When n is a positive integer, 
 
 « ., « 
 
 § 24, (^)" = - X r X " - to n factors 
 
 §180, 
 
 6 b 
 aaa ••• to n factors 
 bbb ••• to n factors 
 
 6" - 
 219. Involution of monomials. 
 
 Examples 
 
 1. What is the third power of 4 a^h ? 
 Solution. (4 a^b)^ = 4 a^b x 4 a^b x 4 a^b = 64 a'6'. 
 
 2. What is the fifth power of - 2 aft^ ? 
 
 Solution. (- 2o62)5 = - 2a?;2 x -2ab^ x - 2o&2 x - 2«62 x - 206' 
 = - 32 asfti". 
 
 To raise an integral term to any power : 
 
 Rule. — Raise the numerical coefficient to the required power and 
 annex to it each letter with an exponent equal to the product of its 
 exponent by the exponent of the required power. 
 
 Prefix the sign + to any power of a positive number or to an even 
 power of a negative number; the sign — to an odd power of a nega- 
 tive number. 
 
216 
 
 INVOLUTION 
 
 To raise a fraction to any power : 
 
 Rule. — Raise both numerator and denominator to the required, 
 power and prefix the proper sign to the result. 
 
 Raise to the power indicated : 
 
 3. 
 
 (a6V)2. 
 
 4. 
 
 [a^hy. 
 
 5. 
 
 {2a'cf. 
 
 6. 
 
 [1 ahny. 
 
 7. 
 
 {-ly. 
 
 8. 
 
 {-aby. 
 
 9. 
 
 {-3cy. 
 
 10. 
 
 (-10ar^» 
 
 11. 
 
 {-Qa'a^y. 
 
 12. 
 
 (-4cy)^ 
 
 13. 
 
 {-2l*m'dy. 
 
 14. 
 
 {-a^afy"-y. 
 
 27. 
 
 What is the square of — 
 
 15. 
 
 [abcxy. 
 
 16. ( 
 
 ;2eV)«. 
 
 17. ( 
 
 ;3 bey. 
 
 18. ( 
 
 '2 aV)». 
 
 19. ( 
 
 -^T- 
 
 20. ( 
 
 - If. 
 
 21. ( 
 
 -l)^ 
 
 22. ( 
 
 _ 5)2"+l. 
 
 23. ( 
 
 ^_52^)2„+l_ 
 
 24. ( 
 
 [— a^'b^'c" Hy. 
 
 25. ( 
 
 '— a2ny3p^4ry^ 
 
 26. ( 
 
 [- a"^'b"-'cf. 
 
 5a^a^ , 
 7b^c ' 
 Solution 
 
 / 6a%2\2^ 
 V 7 b-^c I 
 
 "b'^c 
 
 7 62c 
 
 = + 
 
 Raise to the power indicated ; 
 
 28. 
 
 {M 
 
 33. 
 
 ab 
 
 29. —- 
 
 30. 
 
 31. 
 
 32. 
 
 2x 
 
 '2xy 
 
 Syj' 
 
 2 6" 
 
 34. -^ 
 
 35. -^ 
 
 36. 
 
 37. 
 
 2 
 3x^ 
 
 3^Y 
 
 2yj 
 
 _2ay 
 x^y) 
 
 25a6x* 
 49 6*c2' 
 
 38. 
 
 39. 
 
 40. 
 
 41. 
 
 42. 
 
 _2_a 
 b 
 
 bW 
 a^xj 
 
 xy*^ 
 
 r,n— 1 /^n\ n 
 
INVOLUTION 217 
 
 220. Involution of polynomials. 
 
 § 91, (a + by = (i^ + 2ah + b^ 
 
 § 93, (a - 6)2 = a2 - 2 a6 + b^. 
 
 § 95, (a + 6 + c)2 = a2 + 62 + c2 + 2 a6 + 2 «c + 2 6c. 
 
 Raise the following to the second power : 
 
 1. 2a + 6. 5. 3x — 4.f. 9. a—h + x — y. 
 
 2. 2a-b. 6. 5m* — 11. 10. a" + a;" — 2/"+^ 
 
 3. a^-Sft". 7. l — 3abc. 11. 2a + 3& — 4c. 
 
 4. a^ — 2ar^. 8. 4a^ + 5. 12. 5a2— l-f-4n'. 
 
 Raise to the required power by multiplication : 
 
 13. (x + yf. 15. (x + yy. 17. (x + yy. 
 
 14. (x — yy. 16. (x — yy. 18. {x — yy. 
 
 221. Involution of binomials by the Binomial Theorem. 
 By multiplication, 
 
 (a + xy=a? + Sa^x + Saa^ + 0^. 
 (a - xy= a?-3o?x + 3a2i? + af». 
 (a + a;)* = a* + 4 o?x + 6 a^'ar' + 4 aa^ + a^. 
 (a — cb)* = a"* — 4 a^a; + 6 a^x^ — 4 a^ + a:*, 
 (a + a;)^ = a'' + 5 a*x + 10 a'^a^ + 10 a?a? + 5 aa;< + a^. 
 (a - a;/ = a^ - 5 a^o; + 10 a^'a;- - 10 a?a? + 5ax^- a^. 
 Examine carefully the above powers of (a + a;) and (a — x). 
 
 1. How does the number of terms in a power of a binomial 
 compare with the exponent of the binomial ? 
 
 2. What terms of the power contain the first term of the bino- 
 mial ? the second term of the binomial ? both terms ? 
 
 3. What is the exponent of the first term of the binomial in 
 the first term of the power ? in the second ? in the third, etc. ? 
 
218 INVOLUTION 
 
 4. What is the exponent of the second term of the binomial in 
 the second term of the power ? in the third ? in the fourth, etc. ? 
 
 5. What is the coefficient of the first term of the power? 
 How does the coefficient of the second term compare with the 
 exponent of the binomial ? 
 
 6. If the coefficient of any term is multiplied by the exponent 
 of the first term of the binomial found in that term, and the 
 product is divided by the number oft the term, how does the 
 quotient compare with the coefficient of the succeeding term ? 
 
 7. What are the signs of the terms in any power of (a + &) ? 
 What terms are negative in any power of (a — b)? 
 
 222. Principles. — 1. The number of terms in a positive in- 
 tegral power of a binomial is one greater than the index of the 
 required power. 
 
 2. Tlie first term of the power contains only the first term of the 
 binomial; the last term of the power, only the second term of the 
 binomial; all other terms of the power contain as factors both terms 
 of the binomial. 
 
 3. The exponent of the first term of the binomial in the first term 
 of the poiver is the same as the index of the required power, and it 
 decreases 1 in each succeeding term. The exponent of the second 
 term of the binomial in the second term of the power is 1, and it 
 increases 1 in each succeeding term. 
 
 4. Tlie coefficient of the first term of the jwwer is 1. The co- 
 efficient of the second term is the same as the index of the required 
 poiver. 
 
 5. Tlie coefficient of any term may be found by multiplying the 
 coefficient of the preceding term by the exponent of the first term of 
 the binomial found in that term, and then dividing the result by the 
 number of the term. 
 
 6. If both terms of the binomial are positive, all the terms of any 
 power of the binomial will be positive. 
 
 7. If the second term of the binomial is negative and the first term 
 positive, the terms of any power of the binomial will be alternately 
 positive and negative. 
 
INVOLUTION 
 
 219 
 
 Examples 
 1. Find the fifth power of (& — y) by the binomial theorem. 
 
 Solution 
 
 Letters and exponents, b^ b*y b^y^ b^y^ by* j/* 
 
 Coefficients, 15 10 10 5 1 
 
 Signs, + _ + _ + _ 
 
 Combined, 
 
 65-5 b*y + 10 6%2 _ jq b'Y + 5by*-y^ 
 
 In every term of a power of a binomial the sum of the exponents of the 
 terms of the binomial is equal to the index of the required power. 
 
 Expand : 
 
 2. 
 
 (x + yy. 
 
 13. 
 
 (c - ny. 
 
 3. 
 
 (m + n)*. 
 
 14. 
 
 {x - ay. 
 
 4. 
 
 (m — ny. 
 
 15. 
 
 {d-yy. 
 
 5. 
 
 (a -0)3. 
 
 16. 
 
 {b + yy. 
 
 6. 
 
 (a + by. 
 
 17. 
 
 (m + ny. 
 
 7. 
 
 (b + ay. 
 
 18. 
 
 (p-qy- 
 
 8. 
 
 (q-ry. 
 
 19. 
 
 {s + ty. 
 
 9. 
 
 (c + ay. 
 
 20. 
 
 ix + 2y. 
 
 10. 
 
 (x-^yy. 
 
 21. 
 
 {a + sy. 
 
 11. 
 
 (x - yy. 
 
 22. 
 
 (a; + 4)3. 
 
 12. 
 
 {^-yy- 
 
 23. 
 
 (x + 5y. 
 
 35. 
 
 Expand {2h'^ - 
 
 S 
 
 OLUTION 
 
 Let 
 
 
 2 62 = a 
 
 , and Sy 
 
 Then, 2 b^ 
 
 -^y = a 
 
 -X, 
 
 and 
 
 (262- 
 
 SyY^O 
 
 2-xY 
 
 24. 
 
 [x-2y. 
 
 25. 
 
 {x + ry. 
 
 26. 
 
 [b-cy. 
 
 27. 
 
 [p + qy. 
 
 28. 
 
 [a - by. 
 
 29. 
 
 {a + 6c)*. 
 
 30. 
 
 [ab - cy. 
 
 31. 
 
 (m — pny. 
 
 32. 
 
 [m — any. 
 
 33. 
 
 [ax — byy. 
 
 34. 
 
 [ax - byy. 
 
 z=a* -4a^x + 6 a^x'^ - 4 aaj' + a^ 
 Restoring values, = (2 6-^)* - 4 (2 62)3(3 y) + 6 (2 62)2(.3 y)^ 
 
 -4(2 62)(3j/)8 + (3y)* 
 = 16 68 _ 96 b^y + 216 6*1/2 - 216 62y8 4. gl yi. 
 
220 INVOLUTION 
 
 36. Expand (1 + x^f- 
 
 Solution 
 
 (1 + a;2)3 = 13 + 3(l)2(x2)+ 3(l)(a;2)2 +(a;2)8 
 
 41. (1-Zx^\ 45. {1-xy. 
 
 42. (5x^-ab)\ 46. {l-2xf. 
 
 43. {l + a?hy. 47. (a;-i)«. 
 
 44. {2ax — hf. 48. (^a; — iy)*. 
 
 52. f3a2 + -Y. 55. f— -2 a; 
 
 Expand : 
 
 
 37. (a; + 
 
 22/)*. 
 
 38. (2 a; 
 
 -2/)^- 
 
 39. (2a; 
 
 -5)3. 
 
 40. (a;^- 
 
 - 10)«. 
 
 Expand : 
 
 
 49. (2 a 
 
 -ij 
 
 50. i^- 
 
 \y 
 
 
 "• c^- 
 
 1)- 
 
 6y V2a; 
 
 53. M+3^Y. 66. (l-a^' 
 
 -• (i4y- - ("+r 
 
 58. Expand (a-h-cf. 
 
 Solution 
 
 (a — 6— c)3=(a — 6 — c)3, a binomial form. 
 
 (a^-c)3=(a-6)3-3(a-6)2c + 3(a-6)c2-c3 
 
 = a3-3a26+3a62-63_3c(a2-2a6 + ft2) + 3ac2-3 6c2-c8 
 = a3-3 ffl25 + 3 ah^-h^-Z d^c+Q a6c-3 62c+3 ac2-3 6c2-c3. 
 
 59. Expand {a + h — c— df. 
 
 Suggestion. (o + 6 — c — d)3 = (a + 6 — c + d)3, a binomial form. 
 
 Expand : 
 
 60. {a + x-y)\ 66. (a + 2b-^cf. 
 
 61. (a — m — «)'. '67. {a + h + x + y) 
 
 62. (a — a; + 2/)^. , 68. {a + h — x — y) 
 
 63. (a — a; — 3/)^. 69. {a — h + x — y) 
 
 64. (a + a; + 2y. 70. (a - & - a; + 1/)' 
 
 65. (a -a; -2)'. 71. (a-b-x-y) 
 
EVOLUTION 
 
 223. 1. Of what two equal numbers is 16 the product ? What 
 is the square root of 16 ? Since 16 is equal also to — 4 x — 4, 
 what other square root may 16 have ? What is the square root 
 of 25 ? of 64 ? What is the fourth root of 16 ? of 81 ? 
 
 2. What is the sign of an even root of a, positive number? 
 
 3. Can the square root of — 16 be found ? of — 25 ? of — 64? 
 the fourth root of — 16 ? of — 81 ? Can an even root of any 
 negative number be found ? 
 
 4. What is the cube root of 8? of 27? of 64? of -8? 
 of - 27 ? of - 64 ? What is the fifth root of 32 ? of - 32 ? 
 
 5. How does the sign of an odd root of a number compare 
 with the sign of the number ? 
 
 6. Since a^=a^ x a^ X a^, what power of a^ is a®? What is 
 the cube root of a^ ? of a^ ? of a^^ ? How is the exponent of a in 
 the cube, or third, root of any power of a found ? What is the 
 fourth root of a8 ? of a'^? 
 
 7. How is the exponent of a root of a power obtained from 
 the index of the power and the index of the root ? 
 
 8. How does V4 x 25 compare in value with V4 x V25 ? 
 V4^ with V4 X V9? \/8 x 1000 with -^8 x ^1000? In each 
 case how does the root of the product compare in value with the 
 product of the roots of the factors ? 
 
 9. How does Vl00-=-4 compare in value with Vl00-=-V4? 
 VW^9 with V36 -^ V9 ? -^64 --8 with ^/6i -- ^/8 ? In each 
 case how does the root of the quotient compare in value with the 
 quotient of the roots of the dividend and the divisor ? 
 
 221 
 
222 EVOLUTION 
 
 224. The process of finding any required root of an expression 
 is called Evolution. 
 
 225. Since the product of two numbers having like signs is 
 positive, every positive number has two square roots, numerically 
 equal, hut with opposite signs. It will be seen later that every 
 number has two square roots, three cube roots, four fourth roots, 
 five fifth roots, and, in general, q gth roots. Of these roots the 
 positive roots of positive numbers and the negative odd roots of 
 negative numbers are called Principal Roots. 
 
 \/25 = + 5 or — 5, and + 5 is the principal square root. \/l6 = + 2 or 
 —.2 or, as will be seen later, + V— 4 or — V— 4, but + 2 is the principal 
 root. The principal cube root of 8 is + 2 and of — 8 is — 2. 
 
 226. A number that is or can be expressed as an integer or as 
 a fraction with integral terms is called a Rational Number. 
 
 a, 3, 5|, a^ _|_ ^^2^ V25, and ..333 are rational numbers. 
 
 A number that cannot be expressed as an integer or as a frac- 
 tion with integral terms is called an Irrational Number. 
 
 When the indicated root of a number cannot be exactly obtained, 
 the root is irrational. 
 
 The indicated roots V2, -y/i, VoM^, v^, \/x^, and, in general, the 
 gth root of a number that is not the gth power of some rational number, are 
 irrational numbers. 
 
 227. A rational arithmetical number is called a Commensurable 
 Number, and an irrational arithmetical number is called an Incom- 
 mensurable Number. 
 
 2, f, .54, and .666 are commensurable, but V2 is incommensurable. 
 
 In algebra, commensurable and incommensurable numbers may 
 be either positive or negative. 
 
 The terms rational and irrational applied to algebraic numbers 
 relate to their forms, while the terms commensurable and incom- 
 mensurable relate to their arithmetical values. 
 
 3 a, a + b, x — 3, x^, are rational but not necessarily commensurable. 
 For a may represent \/2, b may represent \/5, etc. Again, Vx is irrational, 
 but if X = 16, Vx is commensurable. 
 
 Incommensurable numbers obey the Commutative, Associative, and Dis- 
 tributive Laws, but the proof is too complicated to be given here. 
 
EVOLUTION 223 
 
 228. Since the wth power of a number is the product of n equal 
 factors, and since one of these factors is a root of the power, it 
 follows that ( >/«)"= a, and if the principal root is meant, -\/a"=a, 
 when 11 is a positive integer. 
 
 In the statement of the following principles and iri Ax. 7 the 
 term root means principal root. 
 
 229. Principles. —1. Law of Signs. — An odd root of a number 
 has the same sign as the number ; an even root of a jJositive number 
 is positive ; an even root of a negative number is impossible, or 
 imaginary. 
 
 2. Law of Exponents. — Tlie exponent of any root of a number is 
 equal to the exponent of the given number divided by the index of the 
 root. 
 
 3. Any root of a product is equal to the product of that root of 
 each of the factors. 
 
 4. Any root of the quotient of two numbers is equal to the qiiotient 
 of that root of each of the numbers. 
 
 Even roots of negative numbers will be discussed later. 
 
 The above principles may be established as follows : 
 
 Pbinciple 1 follows from the Law of Signs for multiplication. 
 
 Principle 2. When m and n are positive integers, 
 § 218, Prin. 2, a"*" = (a")". 
 
 Taking the Jith root, § 228, Va^ = a^ ; 
 
 .-. y/a^ = a"""^ = a™. 
 
 Thus, \^aP = a2. 
 
 Principle 3. When n is a positive integer, 
 § 228, § 218, Prin. 3, ab = ( y/a)« x ( v^)» = ( 7a x V6)» 
 
 Taking the nth root, Vab = Va x Vb. 
 
 Thus, v/27a9 = ^/2^ x v^ = 3 a«. 
 
 Principle 4. When n is a positive integer, 
 
 §228, §218, Prin. 4, ? = i^"=f-fiy. 
 
 Taking the nth root, r'k = "^. 
 
^4 EVOLUTION 
 
 230. Evolution of monomials. 
 
 Examples 
 
 1. What is the square root of 36 a^6^ ? 
 
 Solution. — Since, in squaring a monomial, § 218, the coefficient is 
 squared and the exponents of the letters are multiplied by 2, to extract the 
 square root, the square root of the coefficient must be found, and to it must 
 be annexed the letters each with its exponent divided by 2. 
 
 The square root of 36 is 6, and the square root of the literal factors is a^h. 
 Therefore, the principal square root of 36 a^h'^ is 6 a^b. 
 
 The square root may also be — 6 a^fe, since — 6 a% x — 6 a^b = 36 a^V^. 
 
 .: V36a6p = ± 6 aSft. 
 
 2. What is the cube root of - 125 a^y^ ? 
 Solution. y/-l25a^y^>- = - 5 x^y''. 
 
 To find the root of an integral term : 
 
 Rule. — Extract the required root of the numerical coefficient, 
 annex to it the letters each with its exponent divided by the index of 
 the root sought, and prefix the proper sign to the residt. 
 
 To find the root of a fractional term : 
 
 Rule. — Find the required root of both numerator and denomi- 
 nator and prefix the proper sign to the resulting fraction. 
 
 Find the indicated root : 
 
 3. 
 
 -^/a^^c"^. 
 
 4. 
 
 Va«6^V*. 
 
 5. 
 
 S/a^V^. 
 
 6. 
 
 A/a^"6«c^. 
 
 7. 
 
 Vic^Yz^". 
 
 8. 
 
 A/-8a«6«. 
 
 9. 
 
 ^-32aJY« 
 
 10. 
 
 V16 xy. 
 
 11. 
 
 ■</ - a'^b^x'*. 
 
 12. 
 
 A/-243y*». 
 
 13. 
 
 a/IG mV. 
 
 14. 
 
 15. 
 
 16. 
 
 17. 
 
 18. 
 
 r l(x-yy* 
 \ 128 a;" 
 
 5/ -32aV» 
 \ 243 yi« 
 
 8 256 ar8 
 
 6561 
 
 3 / 125 a; V 
 \ 1728 c' 
 
 \ ^4r 
 
 2»6* 
 
 2.4ny2n 
 
 ' \2"ar^"^' 
 
EVOLUTION 225 
 
 231. To extract the square root of a polynomial. 
 
 1. Since o? + 2ab + b'^ is the square of (a + 6), what is the 
 ■ square root of a^ -{-2iab +¥? 
 
 2. How may the first term of the square root be found from 
 a^ + 2 a6 + 6' ? 
 
 3. How may the second term of the square root be found from 
 2 ah, the second term of the power ? 
 
 4. What are the factors of o?+2db + h'^? 
 
 5. Since 2ab + b^ is equal to b(2a + b), what are the factors 
 of the last two terms of the square of a binomial ? 
 
 6. By what divisor, then, must the last two terms of 
 a^ -\- 2 ab -\- b^ be divided so that the quotient may be the second 
 term of the square root ? 
 
 Examples 
 
 1. Find the process for extracting the square root of 
 a'' + 2ab + b\ 
 
 PROCESS 
 
 a^ + 2 a6 + 6" I a + & 
 
 a 
 
 Trial divisor, 2 a 
 Complete divisor, 2a-\-b 
 
 2ab-{-b^ 
 2ab-\-b^ 
 
 Explanation. — Since a^ + 2 a6 + b'^ is the square of (a + 6), we know 
 that the square root of a^ -{- 2 ab + b^ is a + b. 
 
 Since the first term of the root is a, it may be found by taking the square 
 root of a^, the first term of the power. Subtracting a^, there is a remainder 
 of 2 ah + 62. 
 
 The second term of the root is known to be b, and that may be found by 
 dividing the first term of the remainder by twice the part of tlie root already 
 found. This divisor is called a trial divisor. 
 
 Since 2ab + b^ is equal to 6(2 a + b), the complete divisor which multi- 
 plied by 6 produces the remainder 2 a6 + 6^ is 2 a + 6 ; that is, the complete 
 divisor is found by adding the second term of the root to twice the root 
 already found. 
 
 Multiplying the complete divisor by the second term of the root and sub- 
 tracting, there is no remainder, consequently, a + 6 is the required root. 
 
 ALG. — 16 
 
226 EVOLUTION 
 
 2. Find the square root of 9oc? -SQxy -^-25^. 
 
 PKOCESS 
 
 9 x2 - 30 xt/ + 25 y2 [ 3a;-5y 
 
 9 or 
 
 6a; 
 
 6x — 5y 
 
 - 30 x?/ + 25 / 
 
 - 30 ic^ + 25 / 
 
 Since, in squaring a + h -\-c,a + h may be represented by x, and 
 the square of the number by ar^ + 2 a;c + c^, it is obvious that the 
 square root of a number whose root consists of more than two 
 terms may be extracted in the same way as in Example 1, by con- 
 sidering the terms already found as one term. 
 
 3. Find the square root of 4 a;* + 12 a;^ — 3 a.-^ - 18 a; + 9. 
 
 PROCESS 
 
 4a;< + 12ar»- 3a;2_i8a; + 9 | 2a;^ + 3a;-3 
 
 4 a!* 
 
 12x3- 3a^ 
 
 4x2 
 
 4x2 + 3^ 
 
 12x3+ g^ 
 
 4x2 + 6x |-12x2-18x + 9 
 
 4x'4-6x-3 I -12x2-18x + 9 
 
 Explanation. — Proceeding as in the previous example, the first two 
 terms of the root are found to be 2 x^ + 3 x. 
 
 Considering (2 x^ + 3 x) as the first term of the root, the next term of the 
 root is found as the second term of a root is found, by dividing the remainder 
 by twice the part of the root already found. Hence, the trial divisor is 
 4 x^ + 6 X, and the next term of the root is — 3. Annexing this, as before, 
 to the trial divisor already found, the entire divisor is 2 x'^ + 3 x — 3. This 
 multiplied by — 3 and the product subtracted from — 12 x^ — 18 x + 9, leaves 
 no remainder. Hence, the square root of the number is 2 x'-^ + 3 x — 3. 
 
 Rule. — Arrange the terms of the polynomial with reference to 
 the consecutive powers of some letter. 
 
 Extract the square root of the first term, write the result as the 
 first term of the root, and subtract its square from the given 
 polynomial. 
 
EVOLUTION 227 
 
 Divide the first term of the remainder by twice the root already 
 found, as a trial divisor, and the quotient will he the next term of 
 the root. Write this result in the root, and annex it to the trial 
 divisor to form the complete divisor. 
 
 Multiply the complete divisor by this term of the root, and sub- 
 tract the product from the first remainder. 
 
 Find the next term of the root by dividing the first term of the re- 
 mainder by the first term of the trial divisor. 
 
 Form the complete divisor as before and continue in this manner 
 until all the terms of the root are found. 
 
 Find the square root of the following : 
 
 4. 4x2 + 12a; + 9. 7. 3^ + xy-^\if. 
 
 5. 4 ^2 ^ 20 a; + 25. 8. 4. x? - 52 x + 169. 
 
 6. 25a^ + 40a; + 16. 9. (a + 6)2- 4(a + 6) +4. 
 
 10. a;^ + 4a^ + 2a;^ + 9a;2 — 4a; + 4. 
 
 11. 9 x* - 12 or' + 10 ar*- 4 a; -f-1. 
 
 12. x*-6ci^y-\-13x'y^-12xf-\-4:y*. 
 
 13. a^ + 2 a^x^ - a'x* - 2 a^af + a\ 
 
 14. 25a;* + 4 -12a;- soar' + 29x2. 
 
 15. l + 6x + 15x'-\-20a^ + 15x* + 6x^-{-a^. 
 
 16. l-2x + 3x^-4:X^-\-3x!' -2x^ + 3^. 
 
 17. a*-2a^b + 2a^c^-2b(^ + b^ + c*. 
 
 18. 4 a" - 12 a& +16 ac + 9 6^ + 16 c^ - 24 be. 
 
 19. 9a;2_^25/ + 9z2_30a;?/ + 18a;;?-30^2. 
 
 20. a;2_^2a;-l-- + -. 
 
 X XT 
 
 21. x* + a?-\-'^^ + ^-\-—- 
 
 20 5 25 
 
 22. - + aa; + -^+ ^ f- 
 
 23. a;8 + 4a;'-2a;«-20ar*-3a;* + 32a;' + 4a;2_iga._,_4 
 
228 EVOLUTION 
 
 24. Find four terms of the square root of 1 + x. 
 
 Solution 
 
 14 
 
 X\ l + |x-jtx2 + A** 
 
 1 
 
 2 + ix 
 
 X 
 
 X+ ix2 
 
 2 + a; - ^ x2 
 
 - ix2 
 
 2 + x- 1x2+^^x3 I 1x3 -^V a;* 
 Find the square root of the following to four terms : 
 
 25. 1-a. 27. ar^-1. 29. ?/ + 3. 
 
 26. a? + l. 28. 4 - a. 30. a^ + 2b. 
 
 SQUARE ROOT OP ARITHMETICAL NUMBERS 
 
 1- = 1 10'' = 100 100^ = 10000 
 
 9" = 81 992 = 9801 9992 = 998001 
 
 232. 1. How many figures are required to express the square 
 of a number expressed by 1 figure ? 2 figures ? 3 figures ? 4 
 figure's ? 
 
 2. How does the number of figures in the square of a number 
 compare with the number of figures in the number ? 
 
 3. How many figures are there in the square root of a number 
 that is expressed by 4 figures ? by 3 figures ? by 5 figures ? 
 by 6 figures ? by 7 figures ? 
 
 4. How, then, may the number of figures in the square root of 
 a given number be found ? 
 
 Principles. — 1. The square of a number is expressed by twice 
 as many figures as there are in the number itself, or by one less than 
 twice as many. 
 
 2. TJie orders of units in the square root of a number correspond 
 to the number of ji^riods of two figures each into ivhich the number 
 can be separated, beginning at units. 
 
EVOLUTION 229 
 
 233. If the number of units expressed by the tens' digit is 
 represented by t and the number of units expressed by the units' 
 digit by w, the square of a number consisting of tens and units 
 will be represented by (t + uY, or f ■\-2tu-{- v?. 
 
 Thus, 25 = 2 tens + 5 units, or 20 + 5 units, 
 
 and 252 ^ 202 + 2(20 + 5) + 52 = 625. 
 
 Examples 
 1. What is the square root of 2809 ? 
 
 Explanation. — Accordinff to Prin. 
 
 FIRST PROCESS ^ „ „„, , , . . . , 
 
 2, § 231, the orders of unus m the 
 
 28-09 I 50 + 3 square root of a number may be deter- 
 
 t^ = 25 00 mined by separating the number into 
 
 ^Q periods of two figures each, beginning 
 
 2« = 100 
 
 u = 3 
 
 2t + ^l = 10^ 
 
 at units. Separating 2809 thus, there 
 are found to be two orders of units in 
 the root ; that is, it is composed of 
 o /-kq tens and units. 
 
 Since the square of tens is hundreds, 
 
 and the hundreds of the power are less than 36, or 6^, and more than 25, 
 or 52, the tens' figure of the root must be 5. 5 tens, or 50, squared is 2500, 
 and 2500 subtracted from 2809 leaves 309, which is equal to 2 times the 
 tens X the units + the units2. 
 
 Since two times the tens multiplied by the units is much greater than the 
 square of the units, 309 is a little more than 2 times the tens multiplied by 
 the units. Therefore, if 309 is divided by 2 times the tens, or 100, the trial 
 divisor, the units are found to be 3. And since the complete divisor is found 
 by adding the units to twice the tens, the complete divisor is 100 + 3, or 103. 
 This multiplied by 3 gives as a product 309, which subtracted from 309 leaves 
 no remainder. Therefore, the square root of 2809 is 53. 
 
 SECOND PROCESS 
 
 ^o«uy I OtJ Explanation. — In practice it is usual to 
 
 place the figures of the same order in a col- 
 309 umn, and to disregard the ciphers on the 
 
 right of the products. 
 
 «== 25 
 
 2«=100 
 u = 3 
 2t + u = 103 
 
 309 
 
 Since any number may be regarded as composed of tens and 
 units, the processes given above have a general application. 
 Thus, 346 = 34 tens + 6 units ; 2377 = 237 tens + 7 units. 
 
230 
 
 EVOLUTION 
 
 2. Find the square root of 104976. 
 
 Solution 
 
 Trial divisor = 2 x 30 =60 
 Complete divisor = 60 + 2 =62 
 Trial divisor = 2 x 320 = 640 
 Complete divisor = 640 + 4 = 644 
 
 10.49.76 
 ^ 
 
 1 49 
 
 124 
 25 76 
 25 76 
 
 324 
 
 Rule. — Separate the number into periods of two figures each, 
 beginning at units. 
 
 Find the greatest square in the left-hand period and ivrite its root 
 for the first figure of the required root. 
 
 Square this root, subtract the result from the left-hand period, and 
 annex to the remainder the next period for a new dividend. 
 
 Double the root already found, with a cipher annexed, for a trial 
 divisor, and by it divide the dividend. The quotient or the quotient 
 diminished will be the second figure of the root. Add to the trial 
 divisor the figure last found, midtiply this complete divisor by the 
 figure of the root found, subtract the product from the dividend, and 
 to the remainder annex the next period for the next dividend. 
 
 Proceed in this manner until all the periods have been used. The 
 result will be the square root sought. 
 
 1. When the number is not a perfect square, annex periods of decimal 
 ciphers and continue the process. 
 
 2. Decimals are pointed off from the decimal point toward the right. 
 
 3. The square root of a fraction may be found by extracting the square 
 root of both numerator and denominator separately or by reducing it to a 
 decimal and then extracting its root. 
 
 Extract the square root of the following : 
 
 3. 529. 9. 57121. 15. 125316. 
 
 4. 2209. 10. 42025. 16. 455625. 
 
 5. 4761. 11. 95481. 17. 992016. 
 
 6. 7921. 12. 186624. 18. 2480.04. 
 
 7. 17424. 13. 165649. 19. 10.9561. 
 
 8. 19321. 14. 134689. 20. .001225. 
 
EVOLUTION 
 
 231 
 
 21. 10201. 
 
 22. 95481. 
 
 23. 363609. 
 30 
 31 
 
 625 
 
 24. 1332.25. 
 
 25. 101.0025. 
 
 26. 111.0916. 
 32. HI- 34 
 33- TT5^* 35 
 
 Extract the square root to four decimal places : 
 
 38. f. 40. |. 42 
 
 39. f. 41. .6. 43 
 
 576 
 
 27. 
 
 540.5625. 
 
 28. 
 
 1.018081. 
 
 29. 
 
 13003236. 
 
 Ml- 
 
 36. m- 
 
 361 
 400- 
 
 37. m- 
 
 places : 
 
 
 5 
 
 44. f 
 
 1- 
 
 45. ^. 
 
 234. To extract the cube root of a polynomial. 
 
 1. Since a^ + 3 a^6 + 3 ab^ + &' is the cube of (a + h), what is 
 the cube root of o? + 3 a^S + 3 ab^ + 6^ ? 
 
 2. How may the first term of the root be found from a? + 3 a^h 
 
 3. How may the second term of the root be found from the 
 second term of the power, 3 a?b ? 
 
 4. What are the factors of 3 aj^h + SaW + W? 
 
 5. Since 3 a^^fi + 3 ah'^ + W is equal to 6 (3 a^ + 3 a& + h% by 
 what number must the last three terms of a^ + 3 a?h + 3 a6^ + If 
 be divided so that the quotient may be the second term of the 
 cube root? 
 
 Examples 
 
 1. Find the process for extracting the cube root of a'-'t^a^h 
 
 PROCESS 
 
 a^ + 3 a?h + 3 aft^ + h^ \a + b 
 a? 
 
 Trial divisor, 3 a? 
 
 Complete divisor, 3 a^+ 3 a6 + 6^ 
 
 Za?b + SaJf+W 
 
 Explanation. — Since a^ + 3 a'^6 + 3 aV^ + h^ is the cube of (a + 6), we 
 know that the cube root of a' + 3 a% -{■ Z ah^ + b^ is a + h. 
 
 Since the first term of the root is a, it may be found by taking the cube 
 root of a^, the first tern? of the power. Subtracting, there is a remainder of 
 3 a26 + 3 ab^ + ¥. 
 
232 EVOLUTION 
 
 The second term of the root is known to be 6, and that may be found by 
 dividing the first term of the remainder by 3 times the square of tlie part of 
 the root already found. This divisor is called a trial divisor. 
 
 Since 3 a% + 3 ah^ + b^ is equal to h (.3 a"^ + ^ah + b^), the complete divi- 
 sor, which multiplied by b produces the remainder 3 a^b + 3 ab^ + b^, is 
 S a^ + Sab + b'^ ; that is, the complete divisor is found by adding to the 
 trial divisor 3 times the product of the first and second terms of the root and 
 the square of the second term of the root. 
 
 Multiplying the complete divisor by the second terra of the root, and sub- 
 tracting, there is no remainder ; consequently, a + 6 is the required root. 
 
 Since, in cubing a-f-6 + c, a + b may be expressed by x, its 
 cube will he a^ + 3 x'c + 3 xc' + c^. Hence, it is obvious that the 
 cube root of an expression whose root consists of more than two 
 terms may be extracted in the same way as in example 1, by con- 
 sidering the terms already found as om term. 
 
 2. Find the cube root ot b^ - 3 b' + 5b^ -3b -1. 
 
 PROCESS 
 
 b^-3b'>+5b^-3b-l[b^~b-l 
 ^ 
 
 Trial divisor, 3 b* 
 
 Complete divisor, 3b*-3b^+b^ 
 
 -3b'+5b^ 
 -3b'+3b'-W 
 
 Trial divisor, 3 &*-6 W-\-3 b^ 
 
 Complete divisor, 3 b*-6 5^+3 & + 1 -3 6^-1-6 6^-3 6-1 
 
 -3 6^-^-6 63-3 6-1 
 
 Explanation. — The first two terms are found in the same manner as in 
 the previous example. In finding the next term, b^ — b is considered as one 
 term, which we square and multiply by 3 for a trial divisor. Dividing the 
 remainder by this trial divisor, the next term of the root is found to be — 1. 
 Adding to the trial divisor 3 times (b^ — b) multiplied by — 1, and the square 
 of — 1, we obtain the complete divisor. This multiplied by — 1, and the 
 product subtracted from —Sb* + 6b^ — 36 — 1, leaves no remainder. Hence, 
 the cube root of the polynomial is 6^ _ ^ _ i. 
 
 Rule. — Arrange the polynomial with reference to the consecutive 
 powers of some letter. 
 
 Extract the cube root of the first term, write the result as the first 
 term of the root, and subtract its cube from the given polynomial. 
 
 Divide the first term of the remainder by 3 times the square of 
 the root already found, as a trial divisor, and the qxiotient will be 
 the next term of the root. 
 
EVOLUTION 233 
 
 Add to this trial divisor 3 times the product of the first and second 
 terms of the root, and the square of the second term. TJie result 
 will be the complete divisor. 
 
 Multiply the complete divisor by the last term of the root found, 
 and sxibtract this product from the dividend. 
 
 Find the next term of the root by dividing the first term of the 
 remainder by the first term of the trial divisor. 
 
 Form the complete divisor as before, and continue in this manner 
 until all the terms of the root are found. 
 
 Find the cube root of 
 
 3. or' — 3 x^y -\- 'S xy^ — i^. 
 
 4. m^-9m24-27m-27. 
 
 5. 8 m^ - 60 mhi + 150 mn^ - 125 n\ 
 
 6. 27 a^ - 189 a;"^/ + 441 0^2/2 - 343 y'. 
 
 7 . 125 a^ + 675 a^x + 1215 aa? + 729 ^. 
 
 8. 1000 j9« - 300 p*q + 30 p'^q' - q\ 
 
 9. m^ + 6 m^ + 15 m'^ + 20 m^ + 15 m^ 4- 6 m + 1. 
 
 10. a:^ - 6 ar* + 15 a;^ - 20 a^ + 15 a^- 6 a; + 1. 
 
 11. a^ + 3 ar* + 9 a* 4- 13 a^ + 18 ar^+ 12 a; + 8. - 
 
 12. a^ + 12 ar^ + 63 a;* + 184 \i? + 315 ar' + 300 x + 125. 
 
 13. a;« 4- 6 a;^ - 18 a;* - 1000 + 180 a^ - 112 x? + 600 x. 
 
 14. a^-12ar^ + 54a;-112 + ^-^ + -^. 
 
 a; ar ar 
 
 15. 1 - 6 a + 21 a" - 44 a^ + 63 a^ - 54 a« + 27 a\ 
 
 16. 64 - 144p + 156 p'' - 99i)3 + 39 p* - Qp" +p\ 
 
 ^^ a^Wx> _<M Zacx^ 3 a^bx^ 
 
 c? b^ b 
 
 c 
 
 18. a:«+15x2 + ^ + 20 + -^ + ^ + 6a^, 
 ar a;* ar 
 
234 EVOLUTION 
 
 CUBE ROOT OF ARITHMETICAL NUMBERS 
 
 1^ = 1. 10^ = 1000. 100^ = 1000000. 
 
 38 = 27. 303 = 27000. 3003 = 27000000. 
 
 93 = 729. 993 = 970299. 999^ = 997002999. 
 
 235. 1. How many figures are there in the cubes of numbers 
 that have 1 figure ? 2 figures ? 3 figures ? 4 figures ? 
 
 2. What places, then, belong to the cube of units ? of tens ? 
 of hundreds ? of thousands ? 
 
 3. How many figures are there in the cube root of a number 
 expressed by 4 figures ? 5 figures ? 6 figures ? 7 figures ? 
 
 4. How may the number of figures in the cube root of a 
 number be found? 
 
 Principles. — 1. Tlie cube of a number is expressed by three 
 times as many figures as the number itself, or by one or two less 
 than three times as many. 
 
 2. The orders of units in the cube root of a number correspond to 
 the number of periods of three figures each, into which the number 
 can be separated, beginning at units. 
 
 236. If the number of units expressed by the tens' digit is 
 represented by t, and the number of units expressed by the units' 
 digit by m, the cube of a number consisting of tens and units will 
 be represented by (t + uf, or ^ + 3 t^u +3tu^ + u^. 
 
 Thus, 25 = 2 tens + 5 units, or 20 + 5 units, 
 
 and 253 = 203 4. 3 (202 x 5) + 3 (20 x 52) + 58 = 15625. 
 
 Examples 
 1. What is the cube root of 12167 ? 
 
 FIRST PROCESS 
 
 12-167 [20 + 3 
 «3 = 8 000 
 
 Trial divisor, Sf = 1200 
 
 Stu^ 180 
 m2= 9 
 
 Complete divisor, = 1389 
 
 4 167 
 
 4 167 
 
EVOLUTION 
 
 235 
 
 Explanation. — By Prin. 2, § 235, the orders of units may be determined 
 by separating the number into periods of three figures each, beginning at 
 units. Separating 12167 thus, there are found to be two orders of units in 
 the root ; that is, the root is composed of tens and units. 
 
 Since the cube of tens is thousands, and the thousands of the power are 
 less than 27, or 3^, and more than 8, or 2^, the tens' figure of the root is 2. 
 2 tens, or 20, cubed is 8000, and 8000 subtracted from 12167 leaves 4167, 
 which is equal to 3 times the tens2 x the units + 3 times the tens x the units2 
 + the units^. 
 
 Since 3 times the tens2 x the units is much greater than 3 times the tens 
 X the units2, 4167 is a little more than 3 times the tens2 x the units. If, 
 then, 4167 is divided by 3 times the tens2, or by 1200, the trial divisor, the 
 quotient 3 will be the units of the root, provided proper allowance has been 
 made for the additions necessary to obtain the complete divisor. 
 
 Since the complete divisor is found by adding to 3 times the tens2 3 times 
 the tens x the units and the units2, the complete divisor is 1200 + 180 + 9, or 
 1389. This multiplied by 3 gives 4167, which subtracted from 4167 leaves no 
 remainder. Therefore, the cube root of 12167 is 20 + 3, or 23. 
 
 SECOND PROCESS 
 12-167 
 
 e = 8 
 
 St"" =1200 
 
 3tu= 180 
 
 w«= 9 
 
 4 167 
 
 1389 
 
 4 167 
 
 23 
 
 Explanation. — In practice it is usual to 
 place figures of the same order in a column, 
 and to disregard the ciphers on the right of the 
 products. 
 
 2. What is the cube root of 1740992427 ? 
 
 
 «8 = 
 
 3«2 =3(10)2 
 3«u = 3(10 X 2) 
 m2 = 22 
 
 = 
 
 
 1.740.992427 
 1 
 
 1203 
 
 0.9 
 
 300 
 
 60 
 
 4 
 
 740 
 
 
 
 3 f2 = 3(120)2 
 
 
 364 
 4320 
 
 
 
 728 
 12992 
 
 12992427 
 12992427 
 
 
 III 
 
 *2. 
 
 3 «2 = 3(1200)2 
 3 «M = 3(1200 X 
 m2 = 32 
 
 3) = 
 
 4320000 
 
 10800 
 
 9 
 
 
 
 
 
 433080 
 
 9 
 
 
 Since a root expressed by any number of figures may be regarded as com- 
 posed of tens and units, the processes of example 1 have a general application. 
 Thus, 120 = 12 tens + units ; and 1203 = 120 tens + 3 units. 
 
236 
 
 EVOLUTION 
 
 Since the third figure of the root is 0, it is not necessary to form the com- 
 plete divisor, inasmuch as tlie product to be subtracted will be 0. 
 
 Rule. — Separate the numbers into periods of three figures each, 
 beginning at units. 
 
 Find the greatest cube in the left-hand period, and write its root 
 for the first term of the required root. Cube the root, subtract the 
 result from the left-hand period, and annex to the remainder the next 
 period for a new dividend. 
 
 Take 3 times the square of the root already found, with two ciphers 
 annexed, for a trial divisor, and by it divide the dividend. The 
 quotient, or quotient diminished, will be the second figure of the root. 
 
 To this trial divisor add 3 times the product of the first part of 
 the root with a cipher annexed, multiplied by the second part, and 
 also the square of the second part. Tlieir sum will be the complete 
 divisor. 
 
 Multiply the complete divisor by the second part of the root, and 
 subtract the product from the dividend. 
 
 Continue thus until all the figures of the root have been found. 
 
 1. When there is a remainder after subtracting the last product, annex 
 decimal ciphers, and continue the process. 
 
 2. Decimals are pointed off from the decimal point toward the right. 
 
 ?>. The cube root of a common fraction may be found by extracting the 
 cube root of both numerator and denominator separately, or by reducing it 
 to a decimal and then extracting its root. 
 
 Extract the cube root of 
 
 3. 
 
 29791. 
 
 9. 2406104. 
 
 
 
 15. 
 
 .000024389. 
 
 4. 
 
 54872. 
 
 10. 69426531. 
 
 
 
 16. 
 
 .001906624. 
 
 5. 
 
 110592. 
 
 11. 28372625. 
 
 
 
 17. 
 
 .000912673. 
 
 6. 
 
 300763. 
 
 12. 48.228544. 
 
 
 
 18. 
 
 .259694072. 
 
 7. 
 
 681472. 
 
 13. 17173.512. 
 
 
 
 19. 
 
 926.859375. 
 
 8. 
 
 941192. 
 
 14. 95.443993. 
 
 
 
 20. 
 
 514500.058197. 
 
 Ex 
 
 :tract the cube root to three decimal 
 
 places : 
 
 
 21 
 
 . 2. 
 
 23. .8. 
 
 25. 
 
 A- 
 
 
 27. f 
 
 22 
 
 . 5. 
 
 24. .16. 
 
 26. 
 
 2 
 
 
 28. ^. 
 
EVOLUTION 237 
 
 237. To extract any root of a polynomial. 
 
 To find a formula for obtaining the complete divisor in extract- 
 ing the fourth, fijih, sixth, or any required root of a polynomial, 
 raise (a + b) to the required power, and separate all the terms 
 after the first into two factors one of which shall be the second 
 term of the root. The other factor will be the formula for the 
 complete divisor. 
 
 (a + 6)5 = ffl5 4. 5 a*b + 10 a^b^ + 10 a^b^ + bab* + b^. 
 
 Trial divisor, 5 a*. 
 
 Complete divisor, 5 a* + 10 a% + 10 a^b'^ + 5 a¥ + b*. 
 
 (a + 6)T ^ a7 + 7 a<ib + 21 a^b'^ + 35 a*b^ + 35 a^b* + 21 a^b^ + 7 ab^ + b\ 
 
 Trial divisor, 7 a^. 
 
 Complete divisor, 7 a« + 21 a^b + 35 a^fe^ + 35 a^b^ + 21 a^fc* + lab^ + 6«. 
 
 Since the fourth power is the square of the second power, the 
 sixth power the cube of the second power, etc., any root whose 
 index is 4, 6, 8, 9, etc., may be found by extracting successively 
 the roots corresponding to the factors of the index. 
 
 Tlie fourth root may be obtained by extracting the square root of the 
 square root ; the sixth root, by extracting the cube root of the square root, 
 or the square root of the cube root ; the eighth root, by extracting the square 
 root of the square root of the square root. 
 
 Examples 
 
 1. Find the fourth root of 16 - 32 a; + 24 a^ - 8 a^ -f a^. 
 
 2 . Find the fourth root of a^ + 12 a^y + 54 a^/ + 108 xf + 81 2/*. 
 
 3. Find the fourth root of 16 m* - 32 m^ + 24 m^ - 8 m + 1. 
 
 4. Find the fifth root of 32 a;* + 80 a^ + 80 a^ + 40 x-^ + 10 a; + 1. 
 
 5. Find the fifth root of a»''+15a8+90a«+270a*+405a2+243. 
 
 6. Find the sixth root of a^ - 12 ar> + 60 a;^ - 160 a^ + 240 x" 
 
 - 192 a; + 64. 
 
 7. Find the sixth root of 64 a^ - 576 3? + 2160 x^ - 4320 y? 
 
 + 4860a^2_29i6a; + 729. 
 
 8. Find the sixth root of a^ + 6 acx' + 15 a'^c-x'' -|- 20 a^(?x' 
 
 + 15 a''c''Qi? \ 6 aVa; + aV. 
 
238 
 
 EVOLUTION 
 
 238. To extract any root of an arithmetical number. 
 
 Examples 
 1. Find the cube root of 42875. 
 
 Solution 
 By factoring, 42875 = 5x5x5x7x7x7. 
 
 ;. §§ 26, 229, Prin. 3, Vi2875 = 5 x 7 = 35. 
 Find, by factoring, the roots indicated ; 
 
 2. ^/3375. 
 
 3. a/1296. 
 
 4. •v/531441. 
 
 5. ^759375. 
 
 6. V4084101. 
 
 7. ^262144. 
 
 8. Find the sixth root of 1771561. 
 Solution. — The square root of 1771561 is 1331. 
 The cube root of 1331 is 11. 
 
 § 237, V1771561 = 11. 
 
 Find the roots indicated : 
 9. -^50625. 11. a/531441. 
 
 10. V46656. 
 
 12. V5764801. 
 
 13. V24137569. 
 
 14. -v/10604499373. 
 
 239. Factoring by evolution. , 
 
 Examples 
 
 1. Factor a;* + 4a^ + 8a^ + 8a;-5. 
 
 Solution. — Extracting the square root as far as possible, the root ob- 
 tained is a;2 + 2 a: 4- 2 with a remainder of — 9. 
 
 Therefore, a^ + 4a;3 + 8x2 + 8a;-6 
 
 = (a;2 + 2x4-2)2-9 
 
 §128, =(x2 + 2a; + 2 + 3)(x2 + 2x + 2-3) 
 
 = (x2 + 2x + 5)(x2 + 2x- 1). 
 
 Factor the following polynomials : 
 
 2. iK* + 6a;« + llx2^g3,_8, 
 
 3. a;« + 2a;' +6a;^ + 8a;» + 8ar^ + 8a; + 3. 
 
 4. x«-4ar* f 6a;* + 6a^-19ic2 + 10a; + 9. 
 
 5. 4a^ + 12.x' + 25cc* + 40a^ + 40ar' + 32a; + 15. 
 
THEORY OF EXPONENTS 
 
 240. Thus far the exponents used have been positive integers 
 only, and consequently the laws relating to exponents have been 
 obtained in the following restricted forms : 
 
 1 . a™ X a" = a""*"" when m and n are positive integers. 
 
 2. a"* -r- a" = a™"" when m and n are positive integers and 
 m > n. 
 
 3. (a"*)" = a"" when m and n are positive integers. 
 
 4. -v/o^ = a"*^" when m and n are positive integers, and m is a 
 multiple of n. 
 
 5. {ahy — a"b" when n is a positive integer. 
 
 If all restrictions are removed from m and n, we may then have 
 expressions like a~^ and a^. But such expressions are, as yet, 
 unintelligible, because — 2 and f cannot show how many times a 
 number is used as a factor. 
 
 Since, however, these forms may occur in algebraic processes, 
 it is important to discover meanings for them that will allow 
 their use in accordance with the laws already established, for 
 otherwise great complexity and confusion would arise in the pro- 
 cesses involving them. 
 
 Assuming that the law of exponents for multiplication, 
 
 a^ X a''^ «"'+", (I) 
 
 is true for all values of m and n, the meanings of zero, negative, 
 and fractional exponents may be readily discovered. 
 
 Then, having verified the remaining laws of exponents for these 
 exponents, all the laws will have been shown to be of general 
 application for all commensurable exponents. 
 
 239 
 
240 THEORY OF EXPONENTS 
 
 241. The meaning of a zero exponent. 
 
 1. What is the exponent of a in the product of a^ x a^? of 
 a^ X a^ ? of a' x a**, if the law of exponents for multiplication is 
 true for all values of m and n ? 
 
 2. Since a^xa'* = a^, what is the value of a**? What is the 
 value of a^? of 6«? of /? of m"? 
 
 3. What, then, is the value of the zero power of any number ? 
 
 242. Principle. — Any number with a zero exponent is equal 
 to 1. 
 
 The above principle may be established as follows : 
 
 It is to be proved that a" = 1. 
 
 Since Law I is to be true for all values of m and n, § 240, when » = 0, 
 
 ar^ X a'^ = a^+o = a™. 
 
 Dividing by a", a" = — = 1. 
 
 a" 
 
 243. The meaning of a negative exponent. 
 
 1. What is the exponent of a in the product of a* x a^? of 
 a'^ xa? of a* X a" ? of a* x a"\ if the law of exponents for multi- 
 plication is true for all values of m and n? of a* x a'''' ? 
 
 2. Since a* x a~^ — a^ and since a* x - = a% to what fraction is 
 
 a 
 
 a~^ equal ? Since a* x a~^ = a^, to what fraction is a~^ equal ? 
 
 3. What is another expression for a~^ ? for x~^ ? for y~*? 
 
 4. What is the equivalent of any number with a negative 
 exponent ? 
 
 5. Since a~^ = —, to what is a~^6 equal ? ? 
 
 a^ c 
 
 6. Since — - = 1 -^ — , or a^, to what is - — - x - , or — — , equal ? 
 
 a~'^ a^ a - c a~x 
 
 7. Without changing the value of the fraction, transfer a from 
 the numerator to the denominator m ; in — ; from the 
 
 denominator to the numerator in ^ 
 
 h b 
 
 a~^c a'c 
 
THEORY OF EXPONENTS 241 
 
 8. How may a factor be transferred from one term of a frac- 
 tion to the other without changing the value of the fraction ? 
 
 244. Principles. — 1. Any immher with a negative exponent is 
 equal to the reciprocal of the same number with a numerically equal 
 positive exponent. 
 
 2. Any factor may be transferred from one term of a fraction to 
 the other, without changing the value of the fraction, if the sign of 
 the exponent is changed. 
 
 The above principles may be established as follows : 
 
 Principle 1. It is to be proved that a~" = — 
 
 a» 
 
 Since § 240, Law I, a'" x a^ = 0™+", is to hold for all values of m and n, 
 when m = — n, a-n x «» = a-"+" = a° ; 
 
 but, § 242, ao = 1 ; 
 
 .". Ax. 1, / a" X a" = 1. 
 
 Dividing by a", a " = — • 
 
 Principle 2. It is to be proved that ~ — = — 
 
 6-» a™ 
 
 By Prin. 1, a-"» = — and 6-» = — • 
 
 J_ 
 
 Therefore, «I" = ^ = ±x^ = ^. 
 
 b-" J^ a" 1 a" 
 
 b» 
 
 Examples 
 Write the following with negative exponents : 
 
 1. In-a. 3. 1 -h a". 5. c~(j^x^. 
 
 2. 1 -T- a^ 4. a-':-'^?. Q. am? -=- to". 
 
 7. Write ^x~^y^ with positive exponents. 
 Solution. — Prin. 1, 5 x-^j/^ ::3 5 yi — £_M_. 
 
 Write the following with positive exponents : 
 
 8. 2x-^. 11. a-'6-\ 14. 4aV. 
 
 9. 5a~^ 12. x-^y~^. 15. 3ax~^. 
 10. 3 6-". 13. a-^b'^c-\ 16. a»6-«". 
 
 ALG. — iU 
 
 I 
 
242 THEORY OF EXPONENTS 
 
 q ~2 
 
 17. Write — — without a denominator. 
 
 or 
 
 Solution. — Prin. 2, ^ = 3 a'^x-\ 
 
 Write the following without denominators : 
 
 18. — . 23. -X-.- 28. ^^ 
 
 by a~^6^ \y 
 
 19. ^. 24. A- 29. r^Y- 
 
 20. ^. 25. ^^^'. 30. (^ 
 . a?h^ x^ \bj 
 
 21. '^. 26. A. 31. ^'. 
 6^/1^ a;"'* 6~' 
 
 22. — . , 27. — . 32 ^ 
 
 aj"^ ?/ (aby 
 
 245. The meaning of a fractional exponent. 
 
 1. What is the cube root of a^? How is its exponent ob- 
 tained ? Express the root with this division indicated. 
 
 2. In a^ what does 6 express with reference to a? What 
 does 3 express with reference to a^ ? 
 
 3. What is the fifth root of b^^? Express the root with a 
 fractional exponent. 
 
 4. In b ^ what does 15 express with reference to 6 ? What 
 does 5 express with reference to 6'^ ? 
 
 5 . Express the cube root of a with a fractional exponent ; the 
 fourth root of a^; the third root of the seventh power of a. 
 
 6. What does the numerator of a fractional exponent indi- 
 cate ? What, the denominator ? 
 
 7. Since a^ x a^ X a^ = a", if the law of exponents for multi- 
 plication holds true for all values of m and n, of what is a^ the 
 cube root ? 
 
 Since a^ x a^ x a^ x a^ x a^ x a^ = a^, of what is a^ the sixth 
 power ? What two meanings may a* have ? a^? a^? 
 
 8. What does a fractional exponent indicate ? 
 
THEORY OF EXPONENTS 243 
 
 246. Principles. — 1. The numerator of a positive fractional 
 exponent indicates a poiver and the denominator a root. 
 
 2. A positive fractional exponent indicates a root of a power or a 
 power of a root. 
 
 Prin. 2 refers to principal roots only (§ 225). 
 
 The above principles may be established as follows : 
 
 Let p and q be any positive integers, and a any positive number. 
 
 — p p p 
 Since 3— = -» - may represent any positive fraction. 
 
 p 
 1. It is to be proved that in a», p indicates a power and q a root. 
 Since Law I, or a"^ x a" = a"»+", is to hold for all values of m and n, 
 
 when 
 
 also 
 
 ? ? - _ 
 
 Hence, a* x a* x a* ••• to g factors = a « = qp. 
 
 Taking the gth root, § 26 and Ax. 7, 
 
 p 
 
 a« = i/af. 
 ? p 
 
 Hence, in a', p indicates a power and q a root, and a' indicates the qth 
 
 root of the pth. power of a. 
 
 E 
 
 2. It is to be proved that a' = v^a", or ( \/a)i». 
 
 p 
 
 By the previous proof, a' = VaK (1) 
 
 I 
 It p = l, a» = Va. 
 
 Raising to the pth power, 
 1 1 1 
 a' X a' X a* ••• to p factors = ( Vay, 
 
 
 
 m 
 
 
 : n 
 
 = 
 
 3' 
 
 
 
 
 
 
 
 
 p 
 
 X 
 
 p 
 
 — 
 
 p + p, 
 a" " 
 
 _ , 
 
 2p 
 
 
 
 p 
 
 
 p 
 
 
 p 
 
 
 ?p 
 
 
 p 
 
 
 ^ 
 
 
 «» 
 
 X 
 
 a' 
 
 X 
 
 a» 
 
 = 
 
 a« 
 
 X 
 
 a« 
 
 = 
 
 ffl«, 
 
 etc. 
 
 or, Law I, o«=(Va)J'. (2) 
 
 p 
 
 Hence, (1) and (2), a' indicates the gth rooi 0/ «/ie pth power of a, or the 
 
 pth power of the qth root of a. 
 
 -? 1 
 It follows from § 244 that a » = — • 
 
 p 
 
 a' 
 
244 THEORY OF EXPONENTS 
 
 Examples 
 
 1. Express s/a'bc'* with positive fractional exponents. 
 Solution. \/a^bc^* = ahh~^ = ^^^. 
 
 Express with positive fractional exponents : 
 
 2. Va6^. 5. (V^)^ 8. (^/a^)-2. 
 
 3. -Vxy. 6. {-^yy. 9. 5Vx-hj-\ 
 
 4. V^. 7. (^a&)3. 10. 2-v/(a + 6)=^. 
 
 In the following, large numbers may be avoided by extracting the root first. 
 Find the value of 
 
 11. 8^. 15. 64^ 19. 64"l 
 
 12. si 16. 32l 20. (-8)"^. 
 
 13. 8"*. 17. 25\ 21. (-32)-i 
 
 14. (-8)1 18. 8ll 22. 16"*. 
 
 Simplify : 
 
 23. ^/c^' + x^ + 8^+3x^-5^x-'^/2r. 
 
 24. 4.^x + 5af-Sx-^ + 2\/^'-8^-2xi 
 
 25. -v/^--v/a^+-^a6^_5_^^_^4^^_4^i5§^^_ 
 
 Express with radical signs : 
 
 26. ak 29. a^fti 32. a^ -^ .t^ 
 
 27. x^. 30. iK^i 33. m^-T-nk 
 
 28. .^'i 31. aV^. 34. .^•^ -=- ?/! 
 
 247. It now remains to complete the proof that the other laws 
 of exponents are of general application for commensurable ex- 
 ponents by showing that they apply when negative^nd fractional 
 exponents are employed with the meanings just obtained. 
 
THEORY OF EXPONENTS 245 
 
 248. To prove that the law of exponents for division is general. 
 
 It is to be proved that a"^ ^ a'^ = a™-» for all values of m and n. 
 
 Since dividing by a" is equivalent to multiplying by its reciprocal — , 
 
 a" 
 
 § 106, o™ -r- a" = «" X — 
 
 § 244, Prin. 2, = a" x a"". 
 
 Hence, for all values of m and ?t, a'" ^ a^ = a"*"". (II) 
 
 249. To prove that the law of exponents for involution is general. 
 
 It is to be proved that (a™)" = a™" for all values of m and n. 
 
 Case 1 . — Let m represent any value and n a positive integer. 
 
 Then, (a"*)" = «"• x a"* x a"* ••• to w factors- 
 
 Law I, =: Qm+m+m ... to n terms 
 
 P 
 
 Case 2. — Let m represent any value, and let ?j = — , p and q being 
 positive integers. 
 
 Then, § 246, Prin. 2, (a")? = </{a'^)p 
 
 Case 1, = Va^ 
 
 mp 
 
 § 246, Prin. 1, = aJ. 
 
 Case 3. — Let m represent any value, and let n = — r, r being a positive 
 integer or a positive fraction. 
 
 1 
 
 Then, (a™)" 
 
 («")' 
 
 Case 1, =— _ 
 
 a""- 
 
 § 244, Prin. 2, = a-""". 
 
 Hence, for all values of m and n, («"")" = a"*". (HI) 
 
 250. To prove that the law of exponents for evolution is general. 
 
 It is to be proved that y/a"" = a"*^" for all values of m and n. 
 
 Since (a™)" = a""' for all values of m and n, it is true when ~ is sub- 
 
 71 
 
 stituted for n. 
 
 1 - mx- 
 
 Substituting - for n, (a*")" = a" ", 
 
 or, § 246, Va"" = a™^". (IV) 
 
246 THEORY OF EXPONENTS 
 
 251. To prove (ab)" = a"b" for all values of n. 
 
 P 
 Case 1. — Let n = -, p and q being positive integers. 
 
 Then, § 249, Case 1, since g is a positive integer, 
 
 [(a6)»]«=(a&)«''*=(a6)p 
 § 240, 5, = aPhP. (1) 
 
 Also, § 249, Case 1, 
 
 p p p p p p 
 
 {a^h^Y = a»6» x a«6« ••• to q factors 
 
 p p p p 
 
 § 83, = (a« X a« ••• to g- factors) (6* x 6« ••• to 5- factors) 
 
 = aPhP. (2) 
 
 (1) and (2), Ax. 1, [(a&)']» = (a»6»)«. 
 
 p p p 
 Taking the qih root, (a6)« = a^6«. 
 
 Case 2. — Let « = — r, »• being a positive integer or a positive fraction. 
 Then, § 244, Prin. 1, (aft)"'' = — ^ 
 
 Case 1, = -i- 
 
 § 244, Prin. 2, = a-'-fi-'. 
 
 Hence, for all values of n, (a&)" = a"6". (V) 
 
 Examples 
 252. Multiply: 
 
 1. a^ by a-2. 3. a* by a"*. 5. a^ by a". 
 
 2. a^ by a-^ 4. a by a~*. 6. a;^ by x^. 
 
 7. a^6^ by a^si • 10. w"^ by ani 
 
 8. m^n by m^w~\ 11. a""" by a""''. 
 
 OT+n w»-« 
 
 9. a*&' by a~*6*. 12. a ^ by a ^ . 
 
 13. Multiply x^y"^ + cc* + x^y^ -f a;^?/ + 2/ by a;*?/*. 
 
 14. Multiply 2/" + x-y +' + a;- V+2 + a;-32/"+3 by ary-\ 
 
THEORY OF EXPONENTS 247 
 
 15. Expand (ah~^ -^1 + arh^){ah~^ -1 + a'h^. 
 
 First Solution 
 
 a^b~^ + 1 + a~h^ 
 ah~^ - 1 + a~h^ 
 
 ah-} + ah~^ + oOfeo 
 
 -ah~^- 1 -a~h^ 
 
 + a^b'^ + a~^b^ + a^b 
 
 ah-^ +1 + a~^b 
 
 Second Solution 
 
 (ah~^ + 1 + a~h^)(ah~^ - 1 + a'h^) 
 
 = ah-^ + 2 a^fto + a'h - 1 
 = ah-^ + 2 + a'h - 1 
 = ah-^ + 1 + a~h. 
 Expand : 
 
 16. {a^+b^)(a^ — b^). 21. {x^—xhj~^+y~^){x^ + y~^). 
 
 17. {x^+y^)(x^—y^). 22. {a^ — b~^+ah~^ + l)(a^-b^). 
 
 18. (x-^ + 10){x-^-l). 23. (l-x+o^)(x-^+x-^+x-'). 
 
 19. (a;^-4)(a:^+5). 24. (a-i4-&~^+c^)(a-^ + &~^+2c*). 
 
 20. (x*-?/*)(x*+2/^). 25. {a'b'-ab^+b^a-'b-^+a-^b-'+a-*). 
 
 Divide : 
 
 26. a' by a*. 28. a'^ by a-\ 30. a;^ by a;i 
 
 27. a^ by a". 29. x^ by ic~i 31. a;""^ by a;"-". 
 
 32. Divide a:* + a;^?/^ + i/* by a^^t/^. 
 
 33. Divide a "» + a-^'d + fe'' by a-^b. 
 
 34. Divide a^ + 2 a»^ + 3 a^ar^ + a^a; — a* by a*x*. 
 
248 THEORY OF EXPONENTS 
 
 y/ 35. Divide b'' + S a"'^ - 10 a'^b by ah-^-2. 
 
 Solution 
 
 aVi - 2)6-1 + 3 a"^ - 10 a-^b(a~^ + 5 a-^b 
 6-1 - 2 a~^ 
 
 5 a 2-10 «-i6 
 5 a~2 _ 10 a-16 
 
 Divide : 
 
 36. a-b by a^ + bK 41. J -{- 2 ah~^ + b'^ hy a^ + b~k 
 
 37. a — 6 by a^ — b^. 42. x^ — 2 + x~^ by a;^ — a;"^. 
 
 38. a + 6 by a^ + fti 43. 3-4a;-^ + a;-2 by o;-^ — 3. 
 
 39. a^ + b^ by a^ + b^. 44. 4 x^i/-^ — 5 ?/ + a? V by iK^ — 2/^. 
 
 40. x—1 by a;^ + a;^ + 1. 45. a^— W by a^ + b^. 
 
 Simplify the following : 
 
 46. {a^y. 49. (-a^)». 52. (8-*)*. 
 
 47. (a~^)^ 50. {-a?)\ 53. (16"^)^. 
 
 48. (a-'')2. 51. {-a^)-\ 54. (- ^V)"^. 
 
 Expand by the binomial formula : 
 
 55. {a^-b^y. 57. (a-'- b^y. 59. (a~* + i)^ 
 
 56. (a* + 6^^ 58. (a;"2-y^y. 60. {1 — x^y. 
 
 Simplify the following : 
 
 61. \xi. 64. Va~^6 ^ 67. {■^^a'h^)'^. 
 
 62. Va"*. 65. V^.T%-3. 68. (^m'hr^)k 
 
 63. Va"^. 66. ^aFb". 69. (4aj2»?/-V)i 
 Extract the square root of 
 
 70. a.-^ + 2 a;^ + 3 a; + 4 a;^ + 3 + 2 a;~2 4- x-\ 
 
 71. ar^ + ?/ + 4 2;~2— 2 a;?/^ + 4 .tz"' — 4 y^z-^ 
 
 72. a + 4 &* + 9 c^ - 4 a^^^ + 6 ah^ - 12 ^^c*. 
 
THEORY OF EXPONENTS 249 
 
 Extract the cube root of 
 
 73. a2 + 6a^-f 12a* + 8. 
 
 74. a-^ah^ + ^ah^ -h\ 
 
 75. ^x-^ — \2x~iy+&x'^y--f. 
 
 76. x^~&x + in x^ -20 + 15 x'^ - 6 a;-i + x~^. 
 
 77. Factor 4a;~^ — 9y~^, and express the result with positive 
 exponents. 
 
 Solution 
 
 § 128, 4 x-2 - 9 y-2 = (2 a;-i + 3 y-^) (2 x"! - 3 y-i) 
 
 _/2 3\/2_3\ 
 
 ~U 2//U y) 
 
 Factor the following and express the results with positive 
 exponents : 
 
 78. a-^ — h-\ 83. a^ - a;"*. 
 
 79. 9-x-\ 84. o? + 2 + a\ 
 
 80. 16 -a-*, j 85. h'-S + lQh-\ 
 
 81. 27-6-3. 86. 12-x-^-x-\ 
 
 82. h-'^ + y-^ 87. 2 - 3 a;"' - 2 a;-3. 
 
 Solve the following equations : 
 
 88. x^ = 2. 96. a."^ = 6. 
 
 97. a;~^ = 16. 
 
 98. 25 a;-* = 1. 
 
 99. x2=:243. 
 
 100. a;^ + 32 = 0. 
 
 101. x^ + a^ = 0. 
 94. ^a;^ = 25. 102. a;^ - 64 = 0. 
 96. 2a;"^ = ^. 103. a;~^ + 27 = 0. 
 
 89. 
 
 3^-^ = 8. 
 
 90. 
 
 x^ — 4. 
 
 91. 
 
 x^ = 16. 
 
 92. 
 
 ix^ = 9. 
 
 93. 
 
 a;-^ = 5. 
 
RADICALS 
 
 253. 1. What is indicated by V«? by a;^^ by -^9 by ah 
 
 2. Indicate in two ways the square root of 25 ; of 36 ; of 2 ; 
 of 3. 
 
 3. Which of these indicated roots can be obtained exactly ? 
 Which cannot be obtained exactly ? 
 
 254. An indicated root of an expression is called a Radical. 
 The root may be indicated by a radical sign or by a fractional 
 
 exponent. 
 
 \/5^, (5a)i, \/a^, («a;6)3, Va^ + 2ab + b^, and {a"^ + 2 ab + b^)^ are 
 radicals. 
 
 In the discussion and treatment of radicals only principal roots 
 will be considered. 
 
 255. The Degree of a radical is indicated by the index of the 
 root or by the denominator of the fractional exponent. 
 
 Va + X and {b + xy are radicals of the second degree. 
 
 256. When the indicated root of a rational number cannot be 
 exactly obtained, the expression is called a Surd. 
 
 V2 is a Surd, since 2 is rational but has no rational square root. 
 V 1 + V3 is not a surd, because 1 + VS is not rational. 
 
 Radicals may be either rational or irrational, but surds are 
 always irrational. 
 
 Both Vi and VS are radicals but only V3 is a surd. 
 
 257. An indicated even root of a negative number is called an 
 Imaginary Number ; as V— 4, V — a. 
 
 All other numbers, whether rational or irrational, are called 
 
 Heal Numbers; as V25, VS, a^, 4. 
 
 250 
 

 RADICALS 251 
 
 258. A surd may contain a rational factor, that is, a factor 
 which is a perfect power of the same degree as the radical. The 
 rational factor may be removed and written as the coefficient of the 
 irrational factor. 
 
 In VS = V4 X 2 and \/54 = \/'21 x 2, the rational factors are V4 and 
 \^; that is, VS = 2V2 and v^54 = 3v^. 
 
 259. A surd that has a rational coefficient is called a Mixed 
 Surd. 
 
 2V2, aVic^, and (a — b) Va + b are mixed surds. 
 
 260. A surd that has no rational coefficient except unity is 
 called an Entire Surd. 
 
 VS, VU, and \/cfi + a;^ are entire surds. 
 
 261. A radical is in its simplest form when the expression 
 under the radical sign is integral, contains no factor that is 
 a power of the same degree as the radical, and is not itself a 
 perfect power whose exponent is a factor of the index of the 
 radical. 
 
 V? is in its simplest form ; but Vf is not in its simplest form, because 
 I is not integral in form ; VS is not in its simplest form, because the square 
 root of 4, a factor of 8, may be extracted ; \/25, or 25^, is not in its simplest 
 form, because 25^ =(52)^ = 5^ = 5^, or \/5. 
 
 REDUCTION OF RADICALS 
 
 262. To reduce a radical to its simplest form when it has a rational 
 
 factor. 
 
 Examples 
 
 1. Reduce V20a^ to its simplest form. 
 
 process 
 
 V20a« = V4a« x5 = V4^« X VS = 2a^V5 
 
 Explanation. — Since the highest factor of 20 (fi that is a perfect square 
 is 4a^, V^Oo^ is separated into two factors, a rational factor ^4 a**, and an 
 irrational factor V5. y/20a^ =\/4a*' x V5, § 229, Prin. 3. Extracting the 
 square root of 4 a^ and prefixing.the root to the irrational factor as a coeflBcient, 
 the result is 2 a^VS. 
 
252 RADICALS 
 
 2. Keduce V^- 864 to its simplest form. 
 
 PROCESS 
 
 ^37864=^/- 216 x 4 =\/^^216 x ^/4 = - 6^ 
 
 RuLK. ^- Separate the radical into two factors one of which is 
 its highest rational factor. Extract the required root of the rational 
 factor, multiply the result by the coefficient, if any, of the given 
 radical, and place the product as the coefficient of the irrational 
 factor. 
 
 Simplify the following : 
 
 3. 
 
 V12. 
 
 4. 
 
 V75. 
 
 5. 
 
 ^16. 
 
 6. 
 
 V128. 
 
 7. 
 
 ^250. 
 
 8. 
 
 ^32. 
 
 9. 
 
 V600. 
 
 10. 
 
 V500. 
 
 11. 
 
 a/160. 
 
 12. 
 
 ^3000. 
 
 13. 
 
 ^81. 
 
 14. 
 
 v'189. 
 
 15. V162. 27. V243a^ri' 
 
 16. Vl8a2. 28. V128a%^ 
 
 17. V256. 29. V405 aY. 
 
 18. ^Wc\ 30. V375a;Y. 
 
 19. V50a. 31. (245 a V')^- 
 
 20. a/640. 32. (135a;V)l 
 
 21. V84. 33. (a^ + Sa^)*. 
 
 22. a/72. 34. (16x-16)^. 
 
 23. a/192. 35. Vl8 re - 9. 
 
 24. V800. 36. ^W-^^2^. 
 
 25. V3645. 37. a/8-20 61 
 
 26. a/735. 38. 5 V4 a- + 4. 
 
 39. -\/bx^-10xy + b y-. 41. (3 am- + 6 am + 3a)i 
 
 40. a/4 a^ - 24 a^o; + 36 ax\ 42. (x*?/ - 3 aj^^^ + 3 a^^/^ _ a.y4)i 
 
 / a 
 
 43. Reduce a/ to its simplest form. 
 
 \2f 
 
 PROCESS 
 
 ^/Z =^r«^5IZ ^ Jj^ X V2^ = -^ a/2^ 
 
RADICALS 253 
 
 Explanation. — Since a radical is not in its simplest form when the ex- 
 pression under the radical sign is fractional, the denominator is to be removed ; 
 and since the radical is of the second degree, the denominator must be made 
 a perfect square. The smallest factor that will accomplish this is 2 y. Multi- 
 plying the terms of the fraction by this factor, the largest rational factor 
 
 of the resulting radical is found to be \/-^, which is equal to -^- There- 
 
 \ 4 y4 2 y'^ 
 
 fore, the irrational factor is v'2 y, and its coefficient is — -• 
 
 ' "' 2y^ ' 
 
 Simplify the following : 
 
 "■ ^i- ' 62. yf^. 66 - ' 2 
 
 45. 
 
 vi- 
 
 46. 
 
 vi. 
 
 47. 
 
 Vi. 
 
 48. 
 
 V|. 
 
 49. 
 
 V|. 
 
 50. 
 
 ^A- 
 
 51. 
 
 n 
 
 3f 
 
 54. ^'£ 58. 
 
 Mil 
 
 3x 
 
 60. 
 
 50 a^y 
 
 61 2.V W^^ 3 (g + hf sj a + b 
 
 x-2y^ 2y ' a-b \l (a - bf 
 
 263. To reduce a radical to its simplest form when the expression 
 under the radical sign is a perfect power of a degree corresponding to 
 eome factor of the index of the root. 
 
 Examples 
 
 1. Reduce v9a^ to its simplest form, 
 
 PROCESS 
 
 -^/9^^ = V(3^y = (3 a)^ = (3 a)^ = ^3^ 
 5$. Keduce V64a*'6'^ to its simplest form. 
 
 PROCESS 
 
 ^/6iaFb^' = ^/^^^W* = 6 (2 ab)'^ = b(2 ab)^ = b ^la^ 
 
254 RADICALS 
 
 Simplify the following : 
 
 3. a/36. 7. a/1600. 11. ^'9 d'U'c\ 
 
 4. a/25. 8. A/27a8. 12. a/121 aV. 
 
 5. A/ii4. 9. A^343. 13. Va%^c*d\ 
 
 6. a/81. 10. ^289. 14. -^{x'-2xy-\-y^. 
 
 264. To reduce a mixed surd to an entire surd. 
 
 Examples 
 1. Express 2 a a/5 h as an entire surd. 
 
 PROCESS 
 
 2a a/66 = V4^ a/56 =A/4a2x5&=A/20^ 
 
 Rule. — Raise the coefficient to a power corresponding to the 
 index of the given radical, and introduce tJte result under the radical 
 sign as a factor. 
 
 Express as entire surds : 
 
 2. 2 a/2. 6. 3 a/3. 10. ^V2. 14. i^/^. 
 
 3. 3V5. 7. 4 a/5. 11. fV^. 15. fA/ffa^. 
 
 4. 5a^. 8 I a/8. 12. ^Vbc. 16. I^H. 
 
 5. 3 a/2. 9 a2A/6. 13. fV|. 17. |^. 
 
 18. ^±^^/^H5. 19. ^^±iJl §_. 20. l(a-6)l 
 
 265. To reduce radicals of different degrees to equivalent radicals 
 of the same degree. 
 
 1. Express a^ by an equivalent radical with an exponent in 
 higher terms. 
 
 2. What is the degree of the radical x^? Express x* as a 
 radical of the 12th degree. Express x^ as a radical of the 12th 
 degree. Express 6^ and 6* as radicals of the same degree. 
 
 Examples 
 
 1. Reduce a^3, a/2, and a/4 to equivalent radicals of the 
 same degree. 
 
RADICALS 255 
 
 PROCESS 
 
 </3 = 3^ = 3A=^3^ = ^27 
 V2 = 2i=2A = ^2"«=^^ 
 ^ = 4* = 4^^ = ^^4* = ^256 
 
 Rule. — Express the given radicals with fractional exponents 
 having a common denominator. 
 
 Raise each number to the power indicated by the numerator of its 
 fractional exponent, and indicate the root expressed by the common 
 
 denominator. 
 
 I 
 
 Reduce to equivalent radicals of the same degree : 
 
 2. V2 and ^/3. 9. Va&, -^ab", and ^/2. 
 
 3. V5 and ^s/G. 10. Va, V&, Vx, and Vy. 
 
 4. -y/T and VIO. 11. ^/a + b and Vic + t/. 
 
 5. -v^, V2, and ^5. 12. V|, ^v^S^, and 2V5. 
 
 6. V4, V2, and V3. 13. -\/x, y/xy, and ■y/x^y'^. 
 
 7. "\/l3, VS, and V4. 14. (a + &)Va— & and y/a — b. 
 
 8. V3, ^5, and ^/J^. 15. Va+&, \/a2+62, and Va^. 
 
 ADDITION AND SUBTRACTION OF RADICALS 
 
 266. Radical terms that, in their simplest forms, are of the 
 >ame degree and have the same number under the radical sign 
 are called Similar Radicals. 
 
 267. Principle. — Only similar radicals can be united into one 
 term by addition or subtraction. 
 
 Examples 
 
 1. Find the sum of V50, 2^/8, and 6VJ. 
 
 Explanation. — To ascertain whether or not the given 
 -v/50 = 5 V2 expressions are similar radicals, each may be reduced to 
 
 o 6/q 9/9 **^ simplest form. Since, in their simplest form, they are 
 
 _ of the same degree and have the same number under the 
 6 V ^ = 3 V 2 radical sign, they are similar, and their sum is the sum 
 Sum — 1 a/2 ^^ ^^^ coeflScients prefixed to the common radical factor. 
 
266 RADICALS 
 
 Find the sum of 
 ' 2. ^^50, Vis, and V98. 
 
 3. V27, Vl2, and V75. 
 
 4. V20, V80, and ViS. 
 
 5. V28, V63, and VTOO. 
 
 10. V|, Vi2l, Vl, and ViJ. 
 
 11. V|, VT5, |V3, and VT2. 
 
 12. ^% iV3, 1^9, and Vl47. 
 
 13. ViO, V28, -v^25, and VT75. 
 
 14. a/375, a/44, V192, and a/99 
 
 Simplify : 
 
 15. a/245- a/405 +V45. 
 
 16. V 12 + 3 V75 - 2 v/27. 
 
 17. 5V72 + 3a/18-a/50. 
 
 18. Vr28 + v/686-A/54. 
 
 19. A/IT2-V341+ a/448. 
 
 20. Vl35- a/625 4- a/320. 
 
 21. vi+^i+<m- _ 
 
 22. V864- a/4000 + a/32 
 
 6. a/250, a/T6, and A^Si. 
 
 7. A/i28, a/686, and Vi- 
 
 8. a/135, a/320, and ^625. 
 9. 
 
 500, a/108, and V- 32. 
 
 23. A/l28a;+V375a;-A/54x. 
 
 y 
 
 25. 
 
 V^i-v 
 
 16 aar^ , /4 aa^ 
 
 A/(a + &)2c-A/(a-&)«c, 
 
 27 
 
 28. 6Vff + 4^|f-8VfH- 
 
 29. </'- 96 a-^ + 2 a/31? - a/5^- + a/40^. 
 
 30. Vabx - V«^&^ + a/8 a^'fis^^. 
 
 31. A/3a? + 30«2^75a._V3a^-6a^ + 3a;. 
 
 32. a/5 a^ + 30 a* + 45 a^ -^baP- 40 a* + 80 a\ 
 
 33. A/50+V9-4Vi+A/^=24+A/27-V64. 
 
 34. a/| + 6V|- ^18+ a/36- A/if + ^125-2 VS- 
 
RADICALS 257 
 
 MULTIPLICATION OP RADICALS 
 
 268. 1. What is the exponent of a in d^ x a^? in a^ x a^? 
 in a^ X a^? in a* x a* ? in a^ x a» ? in Va x -v^a ? 
 
 2. When the fractional exponents indicate different roots, what 
 must be done before the radicals can be multiplied together ? 
 
 Examples 
 PEOCESSEs. — 1. V7xV5=V35 
 
 2. 5V3x2Vi5 = 10V45 = 10x3V5 = 30V5 
 
 3. 2 V3 X 3^ = 2a/27 X 3-^^ = 6\/i08 
 
 Rule. — If necessa'ry, reduce the radicals to the same degree. 
 
 Multiply the coefficients together for the coefficient of the product 
 and the factors under the radical sign for the radical factor of the 
 product, and simplify tlie result. 
 
 Multiply : 
 
 4. V2 by Vs. ' 13. 2V6 by Vl8. 
 
 5. V2 by V6. 14. 2^3 by 3^46. 
 
 6. V3 by Vis. 15. 2V6 by 3V6. 
 
 7. V3 by V48. 16. 3V3 by 2^6. 
 
 8. 2V5 by 3VI0. 17. </5 by ViO. 
 
 9. 3V20 by 2V2. 18. 2^250 by V2. 
 
 10. V2 by 3V3. 19. 2V24 by VlS. 
 
 11. V2 by 2V5. 20. V28 by 3V7. 
 
 12. V3 by 3V3. 21. 2V2 by V512. 
 
 Find the value of 
 
 22. Va6 X Vbc x Vcd x Vda. 
 
 23. VaV X Vl2^ X V75^. 
 
 24. V2^ X Voftc X ^/4a^. 
 
 ALG. 17 
 
258 RADICALS 
 
 85. Vm7i X ^Ttihx X ^mn*. 
 
 26. V2 axy x Va;?/ x -\/a?xy. 
 
 27. ^x~^y X Va;~V x -Vx~^y^. 
 
 28. Va^=^ X \/a262 x </(a - 6)-2. 
 
 29. V|xVfxV|. 32. ^rx^fxVf. 
 
 30. VixV|xV|. 33. ^jx^xV|. 
 
 31. ^1 X ^f X VJ. 34. . \/| X V| X ^|. 
 35. Multiply 2V2 + 3V3 by 5V2-2V3. 
 
 Solution 
 
 2V2 + 3V3 
 5V'2-2>/3 
 
 20 + 15V6 
 - 4\/6-18 
 
 20+ iiVe-is 
 
 = 2+ll\/6. 
 
 36. V5+V3 by V5-V3. 
 
 37. V7 + V2 by V7 - V2. 
 
 38. V6-V5 by V6-V5. 
 
 39. 5-V5 by 1 +V5. 
 
 40. 4V7 + 1 by 4V7-1. 
 
 41. 2V2+V3 by 4V2+V3. 
 
 42. 2V3 4-3V5 by 3V3 + 2V6. 
 
 43. Sa+VS by 2a — V5. 
 
 44. 2V6-3V5 by 4V3-ViO. 
 
 45. a"- a6V2 + b^ by a" + a6V2 + b\ 
 
 46. a: — Vx^ + yz by Va; + V.y2. 
 
 47. xy/x — a; Vi/ + yVx — 2/V2/ by Va; + Vy. 
 
 Multiply ; 
 
RADICALS 259 
 
 DIVISION OP RADICALS 
 
 269. 1. What is the exponent of a in a^-j-a^? in a^-^a^? 
 in a^-i-a^? in cC^-i-a^? in ^a^-^a2 
 
 2. When the fractional exponents indicate different roots, what 
 must be done before one radical can be divided by another ? 
 
 Examples 
 PROCESSES. — 1 . V60 -7- Vl2 = V5 
 
 2. ^/2H-V2=^4-^8=^| = ^||=^^/32 
 
 Rule. — If necessary, reduce the radicals to the same degree. 
 
 To the quotient of the coefficients annex the quotient of the radical 
 factors under the common radical sign, and reduce the result to its 
 simplest form. 
 
 Find the quotient of 
 
 
 
 4. V50-hV8. 
 
 12. 
 
 2v^-r-V8. 
 
 5. V72--2V6. 
 
 13. 
 
 -\/ax -^y/xy. 
 
 6. 4V5-J-V40. 
 
 14. 
 
 ^/2ab^^</a*b\ 
 
 7. 6V7H-V126. 
 
 15. 
 
 -\/a^3^ -j- V2 ax. 
 
 8. ^4--V2. 
 
 16. 
 
 ^9a'b^-^^3ab. 
 
 9. 7^/135-^/9. 
 
 17. 
 
 \/4 a^y^ -7- \/2 xy. 
 
 10. 7V75H-5V28. 
 
 18. 
 
 Va — 6 -T- Va + 6. 
 
 11. ■v/l6^v/32. 
 
 19. 
 
 s^^Vf. 
 
 20. Divide VIS-VS by V3. 
 
 21. Divide a/6-2V3 + 4 by V2. 
 
 22. Divide V2 + 2 + ^ V42 by ^ V6. 
 
 23. Divide 5V2 + 5V3 by Vl04-Vi5. 
 
 24. Divide 5 + 5V30 + 36 by V5 + 2V6. 
 
260 
 
 RADICALS 
 
 INVOLUTION AND EVOLUTION OP RADICALS 
 
 270. In finding powers and roots of radicals it is frequently 
 convenient to use fractional exponents. 
 
 Examples 
 
 1. What is the cube of 2V'ttx^? 
 
 Solution. (2Vax^Y = 2^(ax^)^ = 8 a^x^ = SVcM = 8 aX^Vax. 
 
 2. What is the square of 3 V^ ? 
 
 Solution. (3v^)2 = 9(a;5)t = 9x^ = 9v^ = Qxv^x^. 
 
 3. What is the cube of V2 + 1 ? 
 
 Solution 
 (\/2 + l)8=(V^)3 + 8(V2)2.'l +3V'2.12 + 1» 
 = 2V2 + 6 + 3V2 + 1 
 = 7 + 5V2. 
 In such cases expand by the binomial formula. 
 
 Square : 
 
 4. 3Va6. 
 
 5. 2^^^. 
 
 6. x^/2^. 
 
 7. nWlb. 
 
 8. a-y/a^b. 
 
 Expand : 
 
 19. (2+V6)== 
 
 20. (2-|-V2)2 
 
 Cube: 
 
 9. 2V5. 
 
 10. 3V2. 
 
 11. 2^/^. 
 
 12. \/a^. 
 
 13. ^4^«. 
 
 Involve as indicated; 
 
 14. (-2V2a6)*. 
 
 15. (-■y/2y/xy. 
 
 16. (-V2\/'a^)*. 
 
 17. (-2V^^)«. 
 
 18. (-3aV)«. 
 
 22. (2-Vsy. 25. (Vx±iy. 
 
 23. (V7-^/6)l 26. (Va-Vbf 
 21. (2+V5)3 24. (2V2-V3)2 27. Vx±lf. 
 28. What is the cube root of — 27 Vox? 
 
 Solution. -^ -27Vax.= i- 27)\ax)^ =-SVax. 
 
RADICALS 261 
 
 29. What is the fourth root of -y/^x ? 
 Solution. ^V2x =[(2x)^]i =(2a;)^ =\/2«. 
 
 Find the square root of Find the cube root of 
 
 30. V2. 33. -y/^. 36. ^2x. 39. - 27 V^. 
 
 31. ^/5. 34. \/x^\ 37. V7V. 40. -64a/^ 
 
 32. ^'^. 35. VaV. 38. v/SmV. 41. — Vo^. 
 
 Simplify the following indicated roots : 
 
 42. V-v/4aV. 44. (V8aV)i 
 
 43. 'VVa^a^. 45. (vVy»)"*". 
 
 46. 
 
 
 RATIONALIZATION 
 
 1 ^72 
 
 271. How may the fraction be reduced to the form -^— ? 
 
 •^ _ V2 2 
 
 A to ^^ -^ to 2V5^ -^ to ^? 
 V2 2 V5 5 vx a; 
 
 272. The process of rendering a surd expression rational is 
 called Rationalization. 
 
 The factor by which a surd expression is multiplied to render 
 it rational is called the Rationalizing Factor. * 
 
 The denominator of — - is rendered rational by multiplying it by Vs. 
 
 ^ V3 ♦ 
 2 2v3 /— 
 Thus, — = . v3 is a rationalizing factor for the denominator. 
 
 \/3 3 
 
 273. A binomial, one or both of whose terms are surds, is called 
 a Binomial Surd. 
 
 2 + V3, Vx + Vy, and a* — 6' are binomial surds. 
 
 274. Two binomial surds of the second degree whose product 
 is rational are called Conjugate Surds. 
 
 Vo + Vb and Va — Vb are conjugate surds ; also a + y/b and a — Vb. 
 Conjugate surds differ only in the sign of one of the terms. 
 
262 RADICALS 
 
 275. Principle. — A binomial surd of the second degree may he 
 rationalized by miUtiplying it by its conjugate. 
 
 Thus, the product of a + Vfc and a —Vb is a^ — b. 
 
 Examples 
 
 1. Find the simplest rationalizing factor for V3a;. 
 Solution. -k/Sx x VS x = ^9 a;"^ = 3 x. 
 
 .•. y/Wx is the simplest rationalizing factor. 
 
 2. Find the simplest rationalizing factor for ■v^4 a. 
 Solution. \/4 o x v^2 a^ = y/8 a^ = 2 a. 
 
 ,-. V2 a^ is the simplest raiionalizing factor. 
 
 Find the simplest rationalizing factor for 
 
 3. V3. 5. Vox, 7. V9^. 9, \/4. 
 
 4. V6. 6. vTa. 8. Vs. 10. ^/oF. 
 
 11. Find the simplest rationalizing factor for Vo + V3. 
 Solution. ( V5 + \/3)(V5 -•\/3) = (\/5)2_( V3)2 = 5 - 3 = 2, 
 
 .•. \/5 — y/3 is the simplest rationalizing factor. 
 
 Explanation. — Since (^5)2 = 5 and (\/3)2 = 3, Vd + VS may be 
 rationalized by multiplying it by some factor that will give the square of 
 each term, but no other terms. Since the product of the sum and difference 
 of two numbers is equal to the difference of their squares, the simplest 
 rationalizing factor for V5 + VS is its conjugate, VS — \/3 (Prin.). 
 
 12. Find the simplest rationalizing factor for Va +-v^. 
 
 Solution. — By § 111, Va + \/b^, or a^ + fit, is exactly contained in the 
 sum of any like odd powers of a^ and fet, and also in the difference of any 
 like even powers of a^ and ftt. Since in raising ai and fcl to the same power 
 the exponents | and f are multiplied by the index of the power, the lowest 
 like powers of a^ and bs that are rational numbers are the sixth powers, 
 which are even powers. Hence, the rational expression of lowest degree in 
 which ai + fit is exactly contained is (a^)^ - (6^)*, or a^ — b*. 
 
 Dividing a' — 6* by a^ + bh or by Va + y/h^, the rationalizing factor is 
 found to be a2 — a^ftt + a^b^ - ab^ + a^b^ — b^. 
 
RADICALS 263 
 
 Find the simplest rationalizing factor for 
 
 13. V3 + V2. 16. Va + V^. 19. -s/a^ + ^• 
 
 14. V3-V2. 17. a-2V6. 20. Va--^. 
 
 15. 2 + V3. 18. 3V^ + 2?/. 21. y/x+-\/f. 
 
 3 
 22. Rationalize the denominator of 
 
 Vl2 
 
 Solution 
 3 _ 3x\/3 ^3\/3^3V3^V3 q,, i^^ 
 VT2 VB X V3 V36 6 2' 
 
 Rationalize the denominators : 
 
 23. 1-. 26. ^. 29. -i^. 32. ^ 
 
 V3 V6 ^x' ^ax" 
 
 24. A. 27. -1=. 30. ^. 33. V«+^. 
 
 V5 V'4 V2 aa; Va — h 
 
 25. J|. 28. 2^. 31. #. 34. ^^^H?. 
 
 ^3 V6y ^12 a/^+2 
 
 35. Rationalize the denominator of _ ~ _ » 
 
 V7 + V3 
 
 Solution 
 V7 - V3 _ C V7 - >/3) (\/7 - V3) ^ 7 - 2 V21 + 3 _ 10 - 2V21 _ 5 - V2T 
 V7+V3~(\/7+V3)(>/7-V3) 7-3 4 2 ^* 
 
 Rationalize the denominators : 
 
 36. ^^ + A 38. ^^-^. 40. 1--^^ 
 
 V3 - V2 V2 + V3 V8 - V7 
 
 37. ^. 39. ^-^^ . 41. V|±V5. 
 
 V5 - V3 2 - V2 Va; - V^ 
 
 42. 4V2 + 6V3 ^^ a;-V^^^=T 
 3V3-2V2 ' a; + V^^31 
 
 43. V^ + 1 — 2 . ^g aVa— Va; + 1 
 Va; + 1 + 2 * Vo^ + V^Tl 
 
264 RADICALS 
 
 46. ^x + y-^/x-y ^^ \/a? + a-\-l — l 
 
 -s/x + y + Va. — y Va'^ + a-\-l+l ' . 
 
 5 
 
 48. Find the approximate value of 
 
 V3 
 
 SOLCTION 
 
 6 _ by/S _ 6 X 1.73205 _ ^ ggg^g 
 V3 3 3 
 
 In solving the following examples, the work will be lessened by observing 
 that V2 = 1.41421, V3 = 1.73205, and \/5 = 2.23607. 
 
 Find the approximate value of 
 
 49. A.. 52. J^. 55. 8l^. 
 
 V2 V45 2-V3 
 
 50. J-. 53. ^. 56. 1±^. 
 V5 V50 2 - V2 
 
 Ri 6 -. 1 fi7 5 4-2V5 . 
 
 51. 54. — ♦ o7. — • 
 
 V8 V125 6 - 2 V6 
 
 58. Simplify — = ^ ^• 
 
 V2+V3+V5 
 
 Solution 
 Since (V2 + V3)-V5 =(V2 -\/5) + V3, 
 
 y/2-VS-V5 ^ (V2-V5)-VS ^, (V2-V6) + V3 
 y/2+V3+y/l (V2 + \/3) + \/5 (\/2+\/3)-V^ 
 _ 2 - 2 VlO + 5-3 _ 4-2v/l0 
 2 + 2V6 + 3-5 2v'6 
 
 _ 2-\/T0 _ 2v^-2VT5 _ y/e-VIS 
 ~ V6 ~ 6 3 * 
 
 E-ationalize the denominators : 
 
 59. V2-V5-V7. 61 
 
 V2+V5+V7 V2+V3+V5 
 
 V3+V2 g2 2V2-3V3 + 4V 
 
 V34.V2-V6' ' V2+V3-V5 
 
RADICALS 265 
 
 276. To find the square root of a binomial surd. 
 
 (V2 + Ve)" = 2 + 2 Vi2 + 6 = 8 + 2 Vl2; 
 (V4 -H V3)2 = 4 + 2 Vl2 + 3 = 7 + 2 Vl2 ; 
 or (2 + V3)2 = 4+4V3 +3 = 7 + 4 V3. 
 
 1. Since (V2 + V6)^ = 8 +2 Vl2, what is the square root of 
 8 + 2 Vi2 ? 
 
 2. How does the product of the terms of the square root of 
 8 + 2 Vl2 compare with the irrational term 2 Vl2 ? 
 
 3. How does the sum of the squares of the terms of the square 
 root compare with the rational term 8 ? 
 
 4. How, then, may the square root of 8 + 2 Vl2 be found from 
 the termsof 8 + 2V12? 
 
 5. How may the terms of the square root of 7 + 2 Vl2, or the 
 equivalent expression 7 + 4 V3, be found ? After the irrational 
 term is divided by 2, what two factors of the result are selected 
 for the terms of the root ? 
 
 277. A surd of the second degree is called a Quadratic Surd. 
 
 Vx, 4 Va;, Vx + Vy, and 3 + 2 y/b are quadratic surds. 
 
 278. Principlk. — The terms of the square root of a quadratic 
 binomial siird that is a perfect square may be obtained by dividing 
 the irrational term by 2 and then separating the quotient into two 
 factors, the sum of whose squares is the rational term. 
 
 Examples 
 1. Find the square root of 14 + 8 V3. 
 
 Solution 
 
 Since, if 14 + 8 VS is the square of a binomial quadratic surd, the irra- 
 tional term 8 V3 is twice the product of the terms of the root (Prin.), 4 \/3, 
 or V48, IS the product of the terms of the binomial surd. Since the two 
 factors of V48, the sum of whose squares is 14, are VQ and Vs, the required 
 square root is equal to VB + VS. 
 
 .-. Vl4 + 8 \/3 = V6 + V8. 
 
 ALG. — 17 
 
266 RADICALS 
 
 2. Find the square root of 11 — 6 V2. 
 
 Solution 
 
 Vll - 6v/2 = Vll - 2>/l8 =(3 - v^) 
 
 = 3 - \/2. 
 
 Find the square root of each of the following : 
 
 3. 12 + 2V35. 11. 12 + 4V6. 
 
 4. 16-2V60. 12. 11+4V7. 
 
 5. 15 + 2V26. 13. 12-6V3. 
 
 6. 16-2V55. 14. 17 + 12V2. 
 
 7. 114-2V30. 15. 15-6V6. 
 
 8. 7-2V10. 16. 18 + 6V5. 
 
 9. 3-2V2. 17. a' + b + 2aVb. 
 10. 6-I-2V5. 18. 2a-2Va2-6'. 
 
 PROPERTIES OP QUADRATIC SURDS 
 
 279. The square root of a rational number cannot be partly 
 raiional and partly a quadratic surd. 
 
 For, if possible, let Vy = Vb ± m. 
 
 By squaring, j/ = 6 ± 2 mVft + m', 
 
 and Vb=± y-'^'-^ ; 
 
 2 m 
 
 that is, a surd is equal to a rational number, which is impossible. Therefore, 
 Vy cannot be equal to Vb ± m. 
 
 280. In any equation containing rational numbers and quadratic 
 
 surds, as a -{- y/b = x -\- -Vy, the rational parts are equal, and also 
 the irrational parts. 
 
 Let a + Vfe = X + Vy. (1) 
 
 Since a and x are both rational, if possible, let 
 
 a = X ± »n. (2) 
 
 Then, x±m + \/6 = x + Vy, (3) 
 
 and Vy=Vb±m. (4) 
 
RADICALS j 267 
 
 Since, § 279, equation (4) is impossible, a = x ±m is impossible ;. that is, 
 a is neither greater nor less than x. 
 
 Therefore, a = x, and, Eq. (1), Vb=y/y. 
 
 Hence, if a + Vb = x + Vy, a = x, and Vb = VM 
 
 281. If Va+V6 =Vx-\- Vy, then Va — V& = V'a; — ^/y, when 
 a, b, X, and y are rational and a > V6. 
 
 For, squaring, a + ^^1) = x + 2Vxy + y. 
 
 Therefore, § 280, a = z + y, and Vb-=2 Vxy. 
 
 Hence, a—-\/b = x + y — 2 Vxy. 
 
 Whence, v a — Vb = Vx — y/y. 
 
 Examples 
 1. Find the square root of 21 + 6ViO. 
 
 Solution 
 Let y/x-\-Vy=: V2I + 6VIO. (1) 
 
 Then, § 281, \/x-Vy = V2I -6v/l0. (2) 
 
 Multiplying, x-y= \/441 - 360 = Vsl, 
 
 or x~y = 9. (3) 
 
 Squaring (1), x + 2Vxy + y = 21 + OVlO. 
 
 Therefore, § 280 a; + y = 21. (4) 
 
 Solving (4) and (3), x = 15, ?/ = 6. 
 
 .'. Vx = W5, Vy = V6. 
 Hence, the square root of 21 + 6 VTo is -v/lS + V6. 
 
 Find the square root of 
 
 /2. 25 + 10V6. 8. 16 + 6V7. 14. 2+V3. 
 
 3. 19 + 6V2. 9. 21-8V5. 15. 6+V35. 
 
 4. 45 + 3OV2. ^0. 47 - 12 ViT. V 16. 1 + |V2. 
 
 5. 35-14V6. 11. 56 4-32V3. 17. 2 -I- f VO. 
 
 6. 11 + 6V2. vJ2. 35 - 12 V6. 18. 30 4- 20V2. 
 
 7. 24-8V5. 13. 56-12V3. 19. 18 - 6 V5. 
 
268 RADICALS 
 
 RADICAL EQUATIONS 
 
 282. An equation involving an irrational root of an unknown 
 number is called an Irrational, or Radical Equation. 
 
 X* = 3, Va; + 1 = Vx — 4 + 1, and y/z — 1 = 2 are radical equations. 
 
 283. An equation containing quadratic surds involving x may 
 be rationalized with respect to x by writing all the terms in the 
 first member and multiplying both members by the proper ration- 
 alizing factor. 
 
 To rationalize Vx — .3 = 2. (1) 
 
 Transposing, Vaj - 3 -2 = 0. (2) 
 
 Multiplying by the conjugate surd, 
 
 (Vx-3-2)(Vx~- 3 + 2)=0, (3) 
 
 or X - 3 - 4 = 0. (4) 
 
 The rationalization of such equations, however, is more con- 
 veniently accomplished by the process of squaring. 
 
 Thus, by squaring (1), (4) may be obtained in the form 
 
 X - 3 = 4. (5) 
 
 It is evident, then, that squaring an equation is equivalent to 
 multiplying both members by the same unknown expression, an 
 operation likely to introduce roots (§ 196). The roots introduced, 
 if any, when an equation is rationalized are those of the equation 
 or equations formed by placing each rationalizing factor equal to 
 zero. 
 
 Thus, in squaring the equation Vx — 3 = 2, the root of Vx — 3 = — 2 is 
 introduced. But if Vi is taken to mean either + 2 or — 2, this equation has 
 the same root as the given equation and no root has been introduced. 
 
 Repeated squaring is often necessary to free an equation of 
 quadratic surds involving x. This corresponds to repeated ration- 
 alization with respect to particular surds. 
 
 Thus, Vx - 5 = - Vx + 5. 
 
 Squaring, x - 10 Vx + 25 = + (x -f 6) = x + 5. 
 
 Simplifying, y/x = 2. 
 
 Squaring, x = 4. 
 
 Or, transposing in the given equation, 
 
 Vx - 6 + Vx + 6 = 0, 
 
RA DICALS 269 
 
 Rationalizing with respect to Vx + 5, 
 
 (Vx- 5 + \/x + 5)(\/S-5— Vx + 5) = 0. 
 
 Expanding, x — lOVx + 25 — (x + 5) = 0. 
 
 Simplifying, Vx — 2 = 0. 
 
 Rationalizing, ( Vx — 2) ( Vx + 2) = 0, 
 
 or X — 4 = 0. 
 
 In squaring the first time, the factor Vx— 5— Vx+5 is introduced; and 
 in squaring the second time the factor Vx+2, which is ^ of the product 
 ( Vx + 5+ Vx + 5)( Vx + 5 — Vx + 5), or 10 Vx + 20, is introduced. 
 
 284. It follows from the preceding discussion that : 
 If a radical equation is rationalized by multijylying by a rational- 
 izing factor or by squaring, the resulting equation has all the roots 
 of the given equation. 
 
 Whether the given radical equation has all the roots of the 
 rational equation depends upon the method agreed upon of 
 verifying radical equations. 
 
 As illustrated above, each of the equations 
 
 Vx - 5 + Vx + 5 = 0, (1) 
 
 v^ _ 5 _ Vx + 5 = 0, (2) 
 
 Vx + 5 + Vx+5 = 0, (3) 
 
 and Vx + 5 - Vx + 5 = (4) 
 
 is rationalized by finding the product of them all, which is x— 4=0. Hence, 
 the equation x — 4 = has the roots of the four equations ; that is, each 
 equation has the root x = 4, or has no root. 
 
 If X = 4, Vx = V4 and Vx + 5 = VO. If Vi is either + 2 or - 2 and V9 
 is + 3 or — 3, the equations are verified as follows : 
 
 (1) becomes 2-5 + 3 = 0, 
 
 (2) becomes 2-5-(-3)=0, 
 
 (3) becomes -2 + 5+(-3)=0, 
 
 (4) becomes -2 + 5-3 = 0. 
 
 To prevent confusion in making numerical substitutions, it is 
 customary to regard only the positive or principal square root in 
 expressions like Vi, V9, V5, etc. (See § 225.) 
 
 For example, by common agreement + V9 means +(+3), or +3, — V9 
 means —(+3), or —3; +V6 means the positive square root of 5; etc. 
 With this understanding equations (2), (3), and (4) cannot be verified for 
 X = 4, and since they have no other root, they may in this sense be regarded 
 as impossible equations. 
 
270 RADICALS 
 
 When the equations given in this section have been freed from 
 the radical signs, the resulting equations will be found to be 
 simple equations. Other varieties of radical equations are treated 
 subsequently. 
 
 Examples 
 
 1, Given V2 cc + 4 = 10, to find the value of x. 
 
 Solution 
 y/2z + 4 = 10. 
 Transposing, y/2x = 6. 
 
 Squaring, 2 a; = 36. 
 
 .-.a: =18. 
 
 2. Given Va; — 7 + Va; = 7, to find the value of x. 
 
 Solution 
 
 X- 7 f yJx = 7. 
 
 Transposing, Va; — 7 = 7— y/x. 
 
 Squaring, a; — 7 = 49 — HVx + X. 
 
 Transposing and combining, 14 Vx = 56. 
 Dividing by 14, Vx = 4. 
 
 Squaring, x = 16. 
 
 3. Given \14 + Vl + Va; + 8 = 4, to find the value of x. 
 
 Solution 
 
 \14+Vl+Vx + 8 = 4. 
 
 Squaring, 14 + Vl + Vx + 8 = 16. 
 
 Transposing, etc., Vl + \/x + 8 = 16 - 14 = 2. 
 
 Squaring, . 1 + Vx + 8 = 4. 
 
 Transposing, etc., Vz + 8 = 4 — 1 = 3. 
 
 Squaring, z + 8 = 9. 
 
 .-. a; = 9-8 = l. 
 
 Verification. \i4 4. Vl + vT+8 = Vi4 + Vl + 3 
 
 = V14T2 = 4. 
 
RADICALS 271 
 
 4. Given 2y^-_& = 
 2 Vcw; + 6 
 
 : 3^ - 2 ^^ to find the value ( 
 3 Voic + 36 
 
 
 Solution 
 
 
 
 2Vax-6 
 
 3V«x-26 
 
 
 2 Vox +6 
 
 3Vax + 36 
 
 2 7) 
 
 -1 56 
 
 
 2Vax + 6 
 
 3Vax + 36 
 
 Canceling, 
 
 26 
 
 66 
 
 2Va« + 6 
 
 3\/ax + 36 
 
 Dividing by — 6, 
 
 2 
 
 5 
 
 2\/ax + 6 
 
 3Vax + 36 
 
 Clearing of fractions, etc. 
 
 ,, 6Vax- 10%/ax = 
 
 = 56-66. 
 
 
 .% a/ox: 
 
 _6 
 4 
 
 Squaring, etc., 
 
 X - 
 
 62 
 16 a 
 
 Suggestions. — 1. When the equation is free from fractions, 
 transpose so that the radical term, if there is hut one, or the more 
 complex radical term, if there is more than one, mxiy constitute one 
 member of the equation ; then raise each member to a power of the 
 same degree as that radical. Simplify the result. If the equation 
 is not freed from radicals by the first involution, proceed again as 
 at first. 
 
 2. It is sometimes convenient to rationalize denominators before 
 clearing of fractions or involving. 
 
 Solve the following equations : 
 
 5. Va; + 11 = 4. 12. l + 2Vi = 7-V». 
 
 6. Va; + 5 = 3. VlS. Va; + 16 - V^ = 2. 
 
 7. Va;-a2=6. 1^14. \/2» - V2 a; - 15 = 1. 
 
 8. ^a; - 1 = 2. 15. Var« + a; + l = 2-a;. 
 A- \/ar^^ = a. 16. 3V^^^ = 3a;-3. 
 
 10. Vx-\-h = a. 17. Vaf + 2=Va; + 32. 
 
 11. 1 + V« = 5. 18. 5 — Va; + 5 = Va. 
 
272 RADICALS 
 
 19. 2Vx — x = x-8Vx. 25. V3 a; - 5 + V3 a; + 7 = 6. 
 
 20. V4a^+6a;-10=2a:+4. 
 
 21. Va^ — 5a;-f 7 + 2 = a;. 
 
 26. Vl6a; + 3+Vl6x4-8 = 6. 
 
 27. V9 ic + 8 + V9^ -4 = 0. 
 
 22. 4-V4-8a; + 9a^ = 3a;. 28. ^1 + a; V^^+H = 1 + ». 
 
 23. V2(l-a;)(3-2a;)-l=2a;. 29. a/t + 3 V5¥^^36 -4 = 0. 
 
 24. V2a;-l+V2» + 4 = 5. 30. 2a;+A/4a:2- Vl6a^-7=1. 
 31. V7 +Vl +V4 +>/l +2 V^ = 3. 
 
 32. ^ = V3a; + 2+V3a;-l. 
 
 V3a; + 2 
 
 33. 
 
 34. 
 
 35. 
 
 36. 
 
 37. 
 
 V2x + 9 ^ V2a;4-20 
 V2a - 7 V2^ - 12 
 
 V^ + 18 32 
 
 v^ + 2 V^ + 6 
 
 4-1. 
 
 ■y/x — 1 Va; — 3 
 
 Va; + 5 Va; — 1 
 
 y/x — 6 _ -y/x — 8 
 V» — 1 Va; — 5 
 
 ■y/x — 3 _ Va; — 4 
 
 Va; + 1 Va;— 2 
 
 38. 
 
 39. 
 
 ^40. 
 
 41. 
 
 42. 
 
 ^ + 6 V2x + 2 
 
 V2a; + 4 V2a; + 1 
 
 y/nic -f V2 a; + 3 ^ 8 
 VlT^-V2a;-|-3 3 
 
 2V2x + 4 3 Va; + 1 + 9 
 
 2V2a;-4 3VST1-9 
 A/V5«+11 ^iV5x-lQ 
 
 V4a; + 3 + 2Va;-l 
 
 V4a; + 3-2Vx^^ 
 
 = 5. 
 
 43. 
 
 44. 
 
 45. 
 
 Vaj 4- 1 — Va; — I 1 
 
 Va; + 1 + Va; — I 2 
 
 _^-3_^V^+V3_^2V3. 
 V^-V3 2 
 
 Vi9a;-f-V2a;-|-ll 
 
 = 2*. 
 
 Vl9a;-V2a;-Hll 
 46. 2V^-V4a;-22-V2 = 0. 
 
RADICALS 
 
 273 
 
 47. l + V(3-5a;)2 + 16 = 2(3-a;). 
 
 48. -\/x + yx — Va^ — x= y/a. 
 
 49. Va+Va;— (a — 6)^ = a + 6. 
 
 50. Vwiw — a; — Va; Vwn - 
 
 51. aV» — ftVS = a- + 6^ 
 
 1 = Vwin Vl — X. 
 -2 ah. 
 
 62. V5 ax — 9 a' + a = VSaa;. 
 6 a 
 
 53. -\/x -j- 3 a = 
 
 V» + 3 a 
 
 — v». 
 
 54. V2^-V2 
 
 a; 
 
 7 = 
 
 V2a;-7 
 55. \/2a; + Vl0a; + l=V2^ + l 
 
 56. 
 
 Va; + g + Va; — a _ 2 4. Va^*^ — «^ 
 Va; + a — Va; — a ** 
 
 57. V» + V2^ + V3^ = Va. 
 
 Solution 
 Vx + y/2x + V3« = Va. 
 Factoring, VicCl + V2 + V3) = Va. 
 
 Multiplying by 1+ a^ - V3, 
 
 v^(l +2V2 + 2-3) = V5(l +V2-V3). 
 Vx • 2 V2 = Va(l + V2 - V3). 
 Squaring, 8 a; = a (1 + V2 - V3)2. 
 
 X = ^ (1 + V2 - V3)3. 
 
 ■^58. V2^ + V3^ + V5 a; = Vw. 
 1^59. V2^ + V3x — VS^ = Vc. 
 60 
 
 V^^^4-V2(a;-a)= A|3a; + aV2. 
 61. V^^n+V2a;-2 = V3a;-3 + V2. 
 "3 + V4a;-6 = V2a + V^. 
 
 .^62. V2^ 
 
 ALG. — lb 
 
274 REVIEW 
 
 REVIEW 
 
 Reduce the following to their simplest forms : 
 
 Qa^ — 7 0^ — 5x ^ X — y y + x 4a^^ 
 
 9a^ — 25x x + y y — X ai* — y* 
 
 2 8x' + 18x-5 g V2-V3 V2 4-V3 
 
 • 12a^ + 5a;-2* " V2-|-V3 V2-V3" 
 
 g^ aV — aVx -I- a;^ a;+V^+y xVx+yVy 
 
 2 a V6 + 6 
 
 a 
 
 10. —r^ =-1 + 
 
 2+V5 V243 1-V2a; 1 + V2a; l-2a; 
 
 13. 
 
 a; 4- Va^ — a^ a; — Vx^ 
 
 Vic^ — a^ a; + Var* 
 
 14. 
 
 Va + 1 + Va — 1 Va 4- 1 — Vc 
 
 Va + 1— Va — 1 Va + l+Va— 1 
 
 a^-2ax-3x' 6 a' + 7 aa; + 2 ar' 
 So? + 6 ax + 2 7? a'-Aax + Saf' 
 
 16 a'-6 ^ gg - 4 g V6 + 46 
 
 d'~2aVb + b a^ + 2 a V6 + & 
 
 Va6 
 
 "■ i'-tH^4} »• ^K^'^^/S^a 
 
 1 + g + g^ ' / a Va; \ /_ot Va; ' 
 
 ift l+V« + a „„ VVa; « 7 VV^ « 
 
 1 — Vg + a / g Va;\ / a Va;' 
 
 1 — Vg vVa; '^ / \ Va; * 
 
 
REVIEW 275 
 
 Expand : 
 
 21. (a^-by. 25. (a-2+a-y. 31. (a-Vb)*. 
 
 22. C^a-Sby. ''• ^"" + '^'- 32. (Vx+V^)«. 
 
 27. (a^-6*)«. 33. (V2-V3)*. 
 
 ^^' (^"2)* 28. (a*-6-i)*. 34. (V5-2)«. 
 
 29. (a-*-6-i)« 35. (^4-^)«. 
 
 24. (aa: + -j- 3^j^ (ai + 6^)«. 36. (V2-^/2)«. 
 
 Extract the square root of 
 
 37. ^ + 3ar^-ar'-^ + |- 
 
 38. ^+42/2+:^-2a^+f-.V«. 
 4 lb 4 
 
 39. a-4-12aV + 54a& + 108aM + 816* 
 
 40. 1 + 2-y/x — x — 2 xVx + of. 
 
 41. a + 46 + 9c — 4Va6 + 6Vac — I2V6C. 
 
 42. cc^ — 4 xVxy + 6xy — 4: yVxy + 3/^ 
 
 43. a^-12 aV + 60 x^y - 160 a;?/^ + 240 tcV 
 
 - 192 ic*?/2 + 64 f. 
 Find the square root of 
 
 44. 81234169. 48. 56 + 14 Vl5. 
 
 45. 64064016. 49. 47 - 12 Vi5. 
 
 46. .00022801. 50. 62 + 20V6. 
 
 47. .1 to four places. 51. 51 - 36V2. 
 
 Extract the cube root of 
 
 52. a:3_9a.+ 27a;-i-27a;-3. 
 
 64. cc^ -f- 3 ar* V^ — 5 xVx -f 3 Va; — 1. 
 
276 REVIEW 
 
 55. Find the cube root of 2 V2 - 6^2 + 3 V2 ^/4 - 2. 
 
 56. Find the cube root of (a + 12 & + 3 c)Va 
 
 -(6a + 86 + 6 c)V6 - (3 a + 12 & + c)Vc + 12Va6c. 
 
 57. Extract the cube root of 510,082,399. 
 
 58. Extract the cube root of 1,042,590,744. 
 
 59. Extract the cube root of 2 to three decimal places. 
 
 60. Find the first four terms of VI +x — x'. 
 
 61. Find the first three terms of ^1 + a^. 
 
 62. Find the fourth root of 
 
 a« - 4 a^VoFi + 6 a^b-^ - 4 ab-'Vab^' + 6-^ 
 
 63. Find the sixth root of 
 
 8 - 48 Va + 120 a - 160 aVa + 120 a^ - 48 aWa + 8 a". 
 
 If a" X a" = «"*+" for all values of m and n, show that 
 
 64. a- = l. 67. (a&)o = l. 
 
 65. J = V^^ = {Va)\ ^^- ^«^'>' = 
 
 69. f^]=: 
 
 66. 2a-i = ?i!?. 
 
 a V& 
 
 Find the values of the following : 
 
 70. 16^ 73. (a*x*)i. 76. Qi^)'^. 
 
 71. 27*. 74. (byyK 77. (36)-f 
 
 79. For what values of n is (a — 6)" = (6 — a)" ? 
 Simplify and express with positive exponents : 
 
 80. (36 a-3 ^ 25 a-T2. 83. (VaFx^^ -h^/a^^i 
 
 81. (8 a»a* X 64 a-*a;-T* 84. (Va^*^ Va26)-i 
 
 82. (ah^y ^ (ah^y. 85. (Va-;-va)^\/a. 
 
REVIEW 
 
 277 
 
 86 
 
 a + b 2 a^b 
 
 a^ — b^ a^ + b^ a^ — b^ 
 
 1 + a-'b . / I + ab-' + a'b-' ^ 1 + a ^&^ 
 • 1 _ a-'b ' [l - ab-' + d'b-^ 1 - a'^b'^j 
 
 Solve the following equations : 
 
 a; + l_a7 — 1_3 — 5a; 
 x—1 x + 1 1 — a^ 
 
 88. 
 
 fto 7-2a; 2^-1 ,,_ 5^-61 17 + 3a; 
 89. -^^ 5" + ^"^" "• 
 
 2 a; 
 
 30 
 
 90. 
 
 91. 
 
 92. 
 
 4 a; -17 3|-22a; 
 9 33 
 
 3x — 5y 2x 
 
 x" 
 
 =^-s^-s- 
 
 KM+H 
 
 8.v-9 ^y I 7 
 12 2 12 
 
 lP^a;-^-24U0. 
 3 V 8 
 
 93. j 
 
 I (a; 
 
 3a; + l=2y, 
 
 (a: + 5)0/ + 7) = (a;+l)(i/-9) + 112. 
 
 r a; - 2/ = 3, 
 (a; + l)(a; + 2)-(a;-2)(a;+l) = ll2/ + 2. 
 
 Simplify and express with positive exponents : 
 
 94. 
 
 96. 
 
 96. 
 
 2 7 ' sy 
 
 3 a-* + 2 a; 
 
 \a-\a\a^^Y\K 
 
 a.-2«/j 
 
 \xy' 
 
 b 
 
 97^ f a; ^ ^ . a;-yT 
 La;-*y-i " (a^)"'J 
 
 98. Ka^&^)^-^(a"^"^l*- 
 
 g 4- & a — b 
 
 a* _ &J a3 + 6^ 
 
 b + a 
 ab 
 
 99 
 
 100. 
 
QUADRATIC EQUATIONS 
 
 285. 1. What is the value of x in the equation 3cc = 24? 
 What kind of. an equation is it? 
 
 2. What powers of x are found in the equation a;- + 2 x = 3 ? 
 Which is the higher power ? 
 
 3. What is the value of x in the equation a;- = 9 ? How many- 
 values has X ? How do they compare numerically ? 
 
 4. Factor a^ — 5a; + 6 = 0, and so find the values of x. How 
 many values has x ? 
 
 286. An integral equation that contains the square of the 
 unknown number, but no higher power, is called a Quadratic 
 Equation,* or an equation of the second degree. 
 
 It is evident, therefore, that quadratic equations may be of 
 two kinds — those which contain only the second power of the 
 unknown number, and those which contain both the second and 
 first powers. 
 
 x^ = 15 and aoii^ + bx = c are quadratic equations. 
 
 PURE QUADRATICS 
 
 287. An equation that contains only the second power of the 
 unknown number is called a Pure Quadratic. 
 
 ax^ = b and ax^ — cx^ = be are pure quadratics. 
 
 Pure Quadratics are also called Incomplete Quadratics, because 
 they lack the first power of the unknown number. 
 
 288. Since pure quadratics contain only the second power of 
 the unknown number, they may be reduced to the general form 
 axi^ = b, in which a represents the coefficient of x^, and b the sum 
 of the terms that do not involve a^. 
 
 278 
 
QUADRATIC EQUATIONS 279 
 
 289. Principles. — 1. If the square root of each member of a 
 quadratic equation is extracted and the second member of the result- 
 imj equation is given the sign ± , the resulting equation is equivalent 
 to the given equation. 
 
 2. Every pure quadratic equation has two roots numerically equal, 
 but having opposite signs. 
 
 For the equation A- = B^, or A^ - B^ = 0, or (A - B)(A + B)= 0, is 
 equivalent to the two equations 
 
 A- B = and A + B = 0, 
 
 or A =+ B and A = — B; 
 
 that is, to the two equations A = ± B. 
 
 Examples 
 
 1. Given 10 XT = 99 — or, to find the value of x. 
 
 Solution 
 10x2 = 99-x2. 
 Transposing, etc., 11 x^ = 99. 
 
 Dividing by 11, x^ = 9. 
 
 Extracting the square root of each member. Ax. 7, 
 
 x = ±3. 
 
 Strictly speaking, the last equation should be ± x = ± 3, which stands for 
 the equations +x=+3, +x = — 3, — x=: — 3, and — x = + 3. But since 
 the last two equations may be derived from the first two, the first two express 
 all the values of x. For convenience, the two expressions, x = + 3 and 
 X = — 3, are written x = ± 3. 
 
 Consequently, in extracting the square roots of the members of an equa- 
 tion, it will be sufficient to write the ambiguous sign before the root of one 
 member. 
 
 2. Find the roots of the equation 3 a^ = — 15. 
 
 Solution 
 3x2 = -15. 
 
 Dividing by 3, x^ = — 5. 
 
 Extracting the square root, x = ± V — 6. 
 
280 QUADRATIC EQUATIONS 
 
 Solve the following equations : 
 
 3. 7«^-25 = 5a^ + 73. 14. -^ ^^^ = 0. 
 
 a-\-h X 
 
 5. {a-xy={Sx+a){x-a). ^^' ^^2 "*" ^TJI^ " ^^• 
 
 6. aa:2=(a-6)(a2-62)-6ar'. « a; a6 
 
 lb. — I — = — • 
 
 7. a^x^+2ax'={a^-iy-a?. x a x 
 
 8. (a; + 2)2-4(x + 2) = 4. 17. V^F+8 -^ = x. 
 
 10. -^: 1- ^ = 2|. 19 jg+g . a; — q ^ g^' + ft" 
 
 1-a; l + a; • x+b x-b x" - b^' 
 
 , , a; , a^ — 15 x ^^ — - 6 
 
 12 5 a; 5 
 
 12. ^±-? + £zi3 = 4_ 21. ^^ -Var' + 12 = a;. 
 
 a:-3x + 3 Var^ + 12 
 
 13 ''^~2 _i_ ^ + 2 _ _ ;[ 22 ^"^^ , a; — a 2a 
 
 x+1 x—1 x—a x+a 1—a 
 
 x+7 x—1 7 
 
 23. 
 
 a^-7a; a^ + 7x x^-7S 
 
 a' 
 
 24. X + Var^ — d^- 
 
 Va^ — a? 
 
 25. V25 - 6a; + V25 + 6aj = 8. 
 
 26. 2^±V^^^4 
 2a;-V4ar^-l 
 
 27. Vl+a; _^_ Vl - a; ^^ 
 1 4- VH- a; 1 + Vl — a; 
 
 V^Ti+V^^^ '^ 
 
 1 11 
 
 28. 
 
 29. 
 
 1 + Vl - » VI + a; + 1 » 
 
QUADRATIC EQUATIONS 281 
 
 2 2 
 
 30. -I =a?. 
 
 X + V2 — a^ a: — V2 — or^ 
 
 31. 
 
 Vo; -f 2 g — Va; — 2 a _ x 
 Va;-2a + Va; + 2a 2 a 
 
 32. J^^=^+\/^^ = - 
 ^a; + a ^a; — a 
 
 Problems 
 
 1. If 25 is added to the square of a certain number, the sum 
 is equal to the square of 13. What is the number ? 
 
 2. What number is that whose square is equal to the differ- 
 ence of the squares of 25 and 20 ? 
 
 3. If a certain number is increased by 5 and also decreased 
 by 5, the product of these results will be 75. What is the 
 number ? 
 
 4". Two numbers are as 3 to 4, and the sum of their squares is 
 equal to the square of 15. What are the numbers ? 
 
 5. Two numbers are as 4 to 3, and the sum of their squares is 
 400. What are the numbers ? 
 
 6. A gentleman has two square rooms whose sides are as 
 2 to 3. He finds that it takes 9 square yards more than twice as 
 much carpeting for the larger room as for the smaller. What is 
 the length of a side of each room ? 
 
 7. A man who owns a field 80 rods square sells one fourth of it. 
 If the part he sells is also a square, how long is each of its sides ? 
 
 8. A man had a rectangular field whose width was f of its 
 length. He built a fence across it so that one of the two parts 
 thus formed was a square. If the square field contained 10 acres, 
 what were the dimensions of the original field ? 
 
 9. How many rods of fence will inclose a square gardei 
 whose area is 2^ acres ? 
 
 10. The sum of two numbers is 10, and their product is 21 
 What are the numbers ? 
 
 Suggestion. — Represent the numbers by 6 + « and 5 — x. 
 
282 QUADRATIC EQUATIONS 
 
 11. The sum of two numbers is 16, and their product is 55. 
 What are the numbers ? 
 
 12. The sum of two numbers is 26, and their product is 69. 
 What are the numbers ? 
 
 13. The sum of two numbers is 5, and their product is — 14. 
 What are the numbers ? 
 
 14. Factor a^ + lTa+eO by the method suggested in the 
 preceding problems. 
 
 Suggestion. + 60 is the product of the arithmetical terms, and + 17 is 
 their algebraic sum. 
 
 15. Separate a? + 2 a — 2 into two factors. 
 
 16. Separate or^ — 2 a; — 1 into tv/o factors. 
 
 17. Divide 24 into two parts whose product is 143. 
 
 18. The length of a ten-acre field was 4 times its width. What 
 were its dimensions ? 
 
 19. The sum of the squares of two numbers is 394, and the 
 difference of their squares is 56. What are the numbers ? 
 
 20. A man has two square fields that together contain 51^ 
 acres. If the side of one is as much longer than 50 rods as that 
 of the other is shorter than 50 rods, what are the dimensions of 
 each field ? 
 
 AFFECTED QUADRATICS 
 
 290. A quadratic equation that contains both the second and 
 the first powers of one unknown number is called an Affected 
 Quadratic. 
 
 a;2 + 3 X = 10, 4 x^ _ x + 1 = 0, and ax^+bx-\- c = are affected quadratics. 
 
 Affected Quadratics are also called Complete Quadratics. 
 
 291. Since affected quadratic equations contain both the second 
 and first powers of the unknown number, they may always be 
 reduced to the general form of ax^ -|- 6a; + c = 0, in which a, 6, 
 and c may represent any numbers whatever. 
 
 The term c is sometimes called the absolute term. 
 
QUADRATIC EQUATIONS 283 
 
 292. To solve affected quadratics by factoring. 
 
 Reduce the equations to the form ay? + 6a; + c = 0, and solve by 
 the methods of § 141. 
 
 Solve the following equations by factoring : 
 
 1. i»2_5a; + 6 = 0. 7. 10a;2_27a; + 5 = o. 
 
 2. x'-5x = 24:. 8. Q{x' + l)=^lSx. 
 
 3. ar' - 1 = 3(a; + 1). 9. x^ — {a — h)x = ab. 
 
 4. 2a;2-7a; + 3 = 0. 10. 2 ^ -^ ax-2 a^ = i). 
 
 5. 2 0^^-2; -3 = 0. 11. d{h^ + a?)=\Ohx. 
 
 6. 3iK2_2a;_8 = 0. 12. x^-2ax + {a + l){a-l)=0. 
 
 13. Solve the equation x^ + 100 x + 2491 = 0. 
 
 Solution 
 Since, § 99, 100 is tlie sum of the arithmetical terms of the factors of 
 «2 + 100 s; 4- 2491, and their product is 2491, 50 + p and 50 —p may repre- 
 sent the two factors of 2491 whose sum is 100. 
 Then, (50 + p) (50 -p)= 2491. 
 
 Expanding, 2600-^2 = 2491. 
 
 Transposing, etc., p2 _ 9. 
 
 p = ± 3. 
 50+p = 53, 50'-p = 47. 
 Therefore, § 130, x^ + 100 a; + 2491 = (a; + 53) (x + 47) = 0, 
 
 and a; = — 53 or — 47. 
 
 Since p = — 3 gives no new values of 50 + p and 50 - p^ the negative value 
 of p may be disregarded. 
 
 14. Solve the equation a^ _|_ 3 a; _ 208 = 0. 
 
 
 x2 -f 3 a; - 
 
 208 = 
 
 :0. 
 
 
 
 Let 
 
 (i+p)(f 
 
 -p) = 
 
 : - 208. 
 
 
 
 Expanding, 
 
 1 
 
 -Jj2 = 
 
 : - 208. 
 
 
 
 Solving, 
 
 
 p = 
 
 :±¥. 
 
 
 
 Factormg the given equation. 
 
 = 16, f- 
 
 p = - 
 
 13. 
 
 
 (x + 16)(a:- 
 
 -13) = 
 
 :0. 
 
 
 
 
 
 ' X = 
 
 : - 1« or 
 
 + 13. 
 
 
284 QUADRATIC EQUATIONS 
 
 Solve the following equations : 
 
 15. a^ + 10a; + 21 = 0. 28. x^^ a; -756 = 0. 
 
 16. ar2 + 12a;-28 = 0. 29. a;^- a; -506 = 0. 
 
 17. a^- 20 a; + 51 = 0. 30. ar^ + 2 a;- 168 = 0. 
 
 18. x^^ 60 a; + 891 = 0. 31. a;=^ + 6 a; - 135 = 0. 
 
 19. a;2 - 44 a; + 403 = 0. 32. x^ + 3 a;- 154 = 0. 
 
 20. a^ + 20 a; -629 = 0. 33. a^ + 5a; + 2 = 0. 
 
 21. a.2 - 30 .'>; - 2275 = 0. 34. a;2 + a;-10 = 0. 
 
 22. a;2 + 24 a; + 119 = 0. 35. a^ - x- 1 = 0. 
 
 23. a^ + 2 a; -323 = 0. 36. x^- 2a;- 4^0. 
 
 24. a;2_ga;-475 = 0. 37. ar^- 3a; -9 = 0. 
 
 25. ar^ + 8a;-768 = 0. 38. a;2 + 4a; + 8 = 0. 
 
 26. a;2 + 3a;-418 = 0. 39. x^-\-6x-]-U = 0. 
 
 27. a;2 4. 5 a; _ 546 = 0. 40. a;2 + 8a; = -25. 
 
 293. First method of completing the square. 
 
 1. What is the square of a; + 3 ? ofa; + 5? of a; + 10? 
 
 2. If a;^ + 20 a; are the first two terms of the square of a bino- 
 mial, what is the first term of the binomial ? 
 
 3. Since 20 a; is twice the product of the two terms of the 
 binomial, and the first term of the binomial is x, how may the 
 second term of the binomial be found ? 
 
 4. Since the second term of the binomial is 10, what must be 
 added to a^-^ + 20 a; to complete the square of the binomial ? 
 
 5. What term must be added to a;^ + 6a; to complete the 
 square of some binomial ? How is the term found ? 
 
 6. What term must be added to a;^ + 8a; to complete the 
 square ? to ar^ — 10 a; ? to a;^ — 14 a; ? 
 
 7. What must be added to both members of the equation 
 a;* — 12 a; = 13 to make the first member a perfect square ? 
 
QUADRATIC EQUATIONS 285 
 
 Examples 
 
 1 . Solve the equation a^ — 6 a? = 40. 
 
 PROCESS Explanation. — Completing the square in the 
 
 first member by adding to each member the square 
 0^ - 6 a; = 40 ^f ^alf the coefficient of a;, x^ - 6 a: + 9 = 49. 
 a^— 6a; + 9 = 49 Extracting the square root of each member, 
 
 a;-3 = ±7 a:-3=±7. 
 
 Using first the upper sign of ± 7, the simple equa- 
 
 ^ = +'+<J^10 tion X — 3 = + 7 gives x = 10. Next using the lower 
 
 a; = — 7-j-3 = _4 sign of ± 7, the simple equation x — 3 = — 7 gives 
 
 X = — 4. 
 
 Since each of the values 10 and — 4 satisfies the given equation when 
 
 substituted for x, x = 10 or — 4. 
 
 2. Solve the equation a;^ — 5 x = 14. 
 
 Solution 
 
 x2-5x = 14. 
 Completing the square, x^ — 5x + (|)2 = 14 + ^ = 5^. 
 
 Extracting the square root, a; — f = ± f . 
 
 Taking the upper sign x = f + f = 7. 
 
 Taking the lower sign, x = f — |= — 2. 
 
 3. Solve the equation 10 a:^ + 19 a; = 15. 
 
 Solution 
 
 10x2 + 19x= 15. 
 Dividing by coefficient of x^ x^ + ]^x= \^. 
 
 Completing the square, x'' + ^x+ (ig)2 = j^ + |f^ = |^. 
 
 Extracting the square root, a; + ^ = ± f^. 
 
 Taking the upper sign, x = - ^| + f^ = f. 
 
 Taking the lower sign, ^ =■ — \i — ih — ~h 
 
 Rule. — Transpose so that the terms containing x? and x are in 
 one member of the equation and the known terms in the other, and 
 make the coefficient of x^ unity by dividing both members of the 
 equation by the coefficient of a^. 
 
 Add to each member of the equation the square of half the coeffi- 
 cient of X, and extract the square root of each member. 
 
 Solve the two simple equations thus obtained. 
 
286 QUADRATIC EQUATTONS 
 
 Find the values of x in the following equations ; 
 
 4. x'~2x=U3. 
 
 5. x^ + 2x=im. 
 
 6. a^- 4 a; =117. 
 
 7. a^- 6 a; = 160. 
 
 8. a;2-8a:=180. 
 
 9. a;2 + 2a; = 120. 
 
 10. a^ + 6a; = 187. 
 
 11. a;2-12a; = 189. 
 
 12. a.-2 4- 10 X = 171. 
 
 13. a^- 22 a; = 48. 
 
 14. a^ + 30a;=31. 
 
 26. 
 
 27. 
 
 28. 
 
 a; + 2 X 2x 
 
 29. a^ + (m + n) (m — n) = 2 mx. 
 
 294. Other methods of completing the square. 
 
 By the previous method, when x^ had a coefficient, the equation 
 was divided by that coefficient so that the ter^n containing ar might 
 always be a perfect square. The same result may be secured in 
 other ways. 
 
 Thus, if the term containing x^ is 3 x?, it may be made a perfect 
 square by multiplying by 3 ; if 8 x^, by multiplying by 2 ; if ax^, 
 by multiplying by a. 
 
 In the completed square a^x^ + 2 ahx + h"^, it is evident that the 
 third term, h^, is the square of the quotient obtained by dividing 
 the second term by twice the square root of the first. 
 
 15. 
 
 a^ + 3a; = 10. 
 
 16 
 
 ar'-3a! = 180. 
 
 17. 
 
 a^ + 15 a; = 54. 
 
 18. 
 
 x'-x= 930. 
 
 19. 
 
 a^ + 13 a; = 140. 
 
 20. 
 
 a^ - 11 a; + 28 = 0. 
 
 21. 
 
 hx'-3x-2 = (i. 
 
 22. 
 
 6a^-5a;-6 = 0. 
 
 23. 
 
 2a;2 + 9a; = 35. 
 
 24. 
 
 3 a^ - 7 X = 10. 
 
 25. 
 
 4a.-2-19a; = 5. 
 
 1 3 _10 
 
 « -1- 1 a; - 1 3 
 
 x" 3 a; - 5 
 
 _a; + 2 
 
 a;-2 2 
 
 5 
 
 1 a;-2_a;-7 
 
QUADRATIC EQUATIONS 287 
 
 Examples 
 
 1. Solve the equation 5 a^ + 12 a; = 9. 
 
 Solution 
 5 a;2 + 12 a; = 9. 
 Multiplying by 5, 25 x^ + 60 « = 45. 
 
 Completing the square, 25 x2 + 60 a; + 36 = 81. 
 Extracting the square root, 5 x + 6 = ± 9. 
 
 5 a; = - 6 ± 9. 
 .-. a; = f or - 3 
 
 Explanation. — Since the coefficient of x^ is not a perfect square, it may 
 be made a perfect square by multiplying the members of the equation by 5. 
 
 Since, if to 25 x- + 60 x there were added such a term as would make the 
 trinomial a perfect square, 60 x would be equal to twice the product of the 
 square root of 25 x^ and the square root of this third term, the square root 
 of the third term is obtained by dividing 60 x by 2 -\/25 x^ ; that is, by 10 x. 
 60 X H- 10 X = 6, and &^, or 36, added to both members completes the square 
 of (5 X + 6). 
 
 2. Solve the equation 8a^ — 10x = 3. 
 
 Solution 
 8x2-10x = 3. 
 Multiplying by 2, 16 x* - 20 x = 6. 
 
 Completing the square, 16 x^ - 20 x + (|)2 = 6 + ^ = ^. 
 Extracting the square root, 4 x — | = ± |. 
 
 4x = |±| = 6 or -1. 
 .-. X = f or — \. 
 
 General Rule. — Transpose so that the terms containing x^ and x 
 are in one member of the equation and the known terms in the other. 
 
 If the term containing the second power of the unknown number is 
 not a perfect square, make it such by multiplying or dividing the 
 members of the equation by some number. 
 
 Add to each member of the equation the square of the quotient 
 detained by dividing the term containing the first power of the 
 unknown number by twice the square root of the term, containing the 
 second power. 
 
 Extract the square root of each member, and solve the two simple 
 equations. 
 
288 QUADRATIC EQUATIONS 
 
 Solve the following equations : 
 
 
 3. 2a^-5x = 4:2. 
 
 8. 3aj2 + 4cc = 95. 
 
 
 4. 6ar'-5a; + l = 0. 
 
 9. 7a^4-2a; = 32. 
 
 
 5. 4a^-12a; = 27. 
 
 10. 8a^-18ic = 5. 
 
 
 6. 83^^ + 200; = 48. 
 
 11. 6a^ + 5a; = 4. 
 
 
 7. lSx^ + 6x = 4:. 
 
 12. 5a^ + 6x = S. 
 
 
 13. Solve the equation 
 
 ax^ + bx + c = 0. 
 Solution 
 
 
 
 ax2 + bx + c = 0. 
 
 (1.^ 
 
 Transposing c, 
 
 ax^ + bx = — c. 
 
 (21 
 
 Multiplying by a, 
 
 a'^x'^ + abx = — ac. 
 
 (3) 
 
 Completing the square, 
 
 7)2 ;)2 
 a^x^ + abx + — = ~-ac. 
 4 4 
 
 (4) 
 
 Multiplying by 4, 
 
 4 a^x^ + iabx + b'^ - b^ - 4 ac. 
 
 (5) 
 
 Extracting the square root, 
 
 2ax + b = ± Vft^ - 4 ac. 
 
 (6) 
 
 
 . „ - 6 ± V62 - 4 ac 
 
 (7) 
 
 2a 
 
 It is evident that (5) can be obtained by multiplying (2) by 4 a and add- 
 ing b^ to both members. Hence, when a quadratic has the general form of 
 (1), if the absolute term is transposed to the second member, as in (2), the 
 square may be completed and fractions avoided by multiplying by 4 times the 
 coefficient ofx^ and adding to each member the square of the coefficient ofx in 
 the given equation. 
 
 This is called the Hindoo method of completing the square. 
 
 Solve the following equations by the Hindoo method: 
 
 14. 
 
 2x2 + 3a; = 27. 
 
 21. 
 
 4a;2-a;-3 = 0. 
 
 15. 
 
 2a^ + 5x = l. 
 
 22. 
 
 5a^_2a;-16 = 0. 
 
 16. 
 
 2x' + lx = -Q. 
 
 23. 
 
 3a^ + 7a;_ 110 = 0. 
 
 17. 
 
 3 ar^ - 5 a; = 2. " 
 
 24. 
 
 2cc2-5a;-150 = 0. 
 
 18. 
 
 4ar'-15aj = 4. 
 
 25. 
 
 3a^ + a;-200 = 0. 
 
 19. 
 
 5ot?-lx = -2. 
 
 26. 
 
 15a^_7a;-2 = 0. 
 
 20. 
 
 6a^ + 5a; = -l. 
 
 27. 
 
 7ar'-20a;-32 = 0. 
 
QUADRATIC EQUATIONS 289 
 
 295. To solve quadratics by a formula. 
 
 Since every quadratic can be reduced to the general form 
 aoi? -f 6a; + c = 0, in which a, b, and c represent any numbers 
 whatever, and since the roots of this equation are 
 
 Ex. 13, § 294, X = -&±V&^-4ac ^ 
 
 2a 
 
 the vahies of the unknown number in any affected quadratic 
 equation may be found by substituting the coefficient of or^ for a, 
 the coefficient of x for 6, and the absolute term for c. 
 
 Examples 
 
 1. Solve the equation 6 ar* — x — 15 = 0, 
 
 Solution. — Since a = 6, 6 = — 1, and c = — 15, by the above formula, 
 
 ^^ l±V(-l)-^-4 x6(-^15) 
 2x6 
 
 12 3 2 
 
 Solve by the above formula : 
 
 2. 2x'-ir5x + 2 = 0. 10. 2ar' + 3a;-l = 0. 
 
 3. 3ar^ + llx + 6 = 0. 11. ^^ + 2x-4. = 0. 
 
 4. Q3?-lx-\-2 = {). 12. a2_5a; = _3. 
 
 5. 4r' + 4a;-15 = 0. 13. S7?-&x = -2. 
 
 6. 2«* + 3a;-9 = 0. 14. 4ar'- 3a:- 2 = 0. 
 
 7. 2x2 + 3a; + l=0. 15. ar'- 6a; + 10 = 0. 
 
 8. 3a;*-13a; = 10. 16. a? + 4.x + 12 = 0. 
 
 9. 7a;^ + 9a; = 10. 17. ar«-8a; = -20. 
 
 Solve by any method : 
 
 18. a;2_6a;4-5 = 0. 22. a*-12x = 64. 
 
 19. ar»-8a; + 7 = 0. 23. 18 ar' + 6 a; = 4. 
 
 20. 2a;2-5a; = 42. 24. ar^-a;-72 = 0. 
 
 21. 7ar' + 2x = 32. 25. 4 ar* - 12 a; = 27. 
 
 ALG. — 19 
 
290 QUADRATIC EQUATIONS 
 
 26. x' = Sx + 10. gg 1 + x x-1 
 
 27. a^- 30 = 13a;. 
 
 28. a:2_i2a; = 28. 39. 
 
 29. a^-50x-=159. 
 
 30. a^ + 8a; = 84. 
 
 33. 
 
 9(x - 1) 6 
 
 40. 
 
 31. x + ^-^ = 0. "• 
 
 X 2 
 
 32. ^-H?=:?^. 42. 
 9 3 4 
 
 a^-2 . 43. 
 
 a; 
 
 -3 
 
 a;— 2 
 
 5 
 
 
 X 
 
 a; — 5 
 a; 
 
 3 
 
 X 
 
 -5 
 
 2 
 
 X 
 X 
 
 
 a; 4- 12 
 x + ij 
 
 = 7. 
 
 X 
 
 X 
 
 + 4 , ^ x + 3 
 -2' ^-x-3' 
 
 
 4 a; 
 
 a; + 3 
 
 - 4 
 
 X 
 
 -1 
 
 X 
 
 
 X 
 
 + 2^ 
 
 1 a; + 2 
 
 2 2 a; 
 
 X 
 
 5 a; 
 
 a; + 6 
 a;-f3 
 
 = 3. 
 
 X 
 X 
 
 + 2 
 
 -7 
 
 a; + 5 
 a; — 5 
 
 = 1. 
 
 X 
 X 
 
 tI- 
 
 a; + 2 
 x-2 
 
 23 
 "lO" 
 
 2 
 
 a; + l 
 
 5 X 
 
 -8 
 
 1 
 
 -2a; 
 
 7 
 
 2 
 
 34. =-. 44. 
 
 a^-2x + l 4 
 
 35. ^_?_?^28. 45. 
 4 3 
 
 36. —-^ + —^ = 4. 46. 
 2x + l aJ-3 
 
 37. '^^-1=^^. 47. 
 
 a;-|-2 x-2 
 
 Literal Equations 
 
 1. Solve the equation a;'' — -a; a; + l = 0. 
 
 b a 
 
 ' Solution 
 
 a;2_«a;-^a; + l=0. 
 b a 
 
 Factoring, fx-'^(x--\ = 0. 
 
 Therefore, x = " or -. 
 
 b a 
 
QUADRATIC EQUATIONS 291 
 
 Solve by any of the preceding methods : 
 2. x^ — ax = ah — hx. Q. ^x — 2ax = 3? — \0a. 
 
 Z. <x?-\-ax = ac + cx. 7. a:^ + 3 6a; = 5 ccc + 15 6c. 
 
 4. ar^ = (m — n) a; + mn. 8. 2 a6a; — a^ = 14 a6 — 7 a;. 
 
 5. ar'-3 6a;=2ttaj — 6a6. 9. Qx^-ir^ax = 2hx + ah. 
 
 10. oca? — 6ca; — 6d -f adx = 0. 
 
 11. a:^ + 4 ma; + 3 7ia; + 12 mn = 0. 
 
 12. x^ — 2ax = a^. „_ x . a 5 
 
 Ab. --\ -• 
 
 13. ar' + 4 6a; = 6l a x ^ 
 
 14. a;2 = 4aa,_2a2. 27. ^ + ^ = 1- 
 
 4 3 3 
 
 15. x^ — ax + a^ = 0. 
 
 28 _1^__^ = 3 
 
 16. ^2 = 6a; + 61 ' a;-a a; + a 
 
 17. a;2^^^^^^0. ^ 2a 
 
 29. ^ — • 
 
 18. a;'-2a; + a = 0. a;-a a; 
 
 19. 4aa;-a;2^3(ij2 ^^ 1 ^^ aa; - 4 
 
 aa; + 4 16 
 
 20. ar — a = l + aa;. 
 
 21. x" - b' = a" - bx. 31. ar' + ^a; = ^^. 
 
 6 6 
 
 22. 2162-4 6a; = a;2. 
 
 „„ „ , o (2a2 + l)a; 
 
 23. 5aa; + 6a^ = 6ar^. ^2. ^ + 2 = ^ ^ ^ ■ 
 
 24. a;^-l = 4aa;-al ^^ ^ 2a;^4(a6-l) 
 
 25. a;2 4- 6^ = 4(a2 + 6a;). * a6 a6 
 
 34. a.-^ — 2 (a — 6)a; = 4 a6. 
 
 35. a;2 + 2(a + 8)a; = -32a. 
 
 36. a;^ + a; + 6a; + 6 = ax + a. 
 
 37. 2aa; — a+2 6a; — 6 = 2a;2 — a;. 
 
 38. a^ + A(a — l)x = 8a — 4:0^. 
 
 39. a(a; — 2 a + 6) + a(a; + a — 6) = a;^ — (a — 6)1 
 
292 QUADRATIC EQUATIONS 
 
 4n ^'^ + ^' . a — 2 a; _ 8 
 
 41. 
 
 2a — a; a + 2x 3 
 
 1 1 ^ 3 + a^ 
 
 a — a; a + a; a? — x^ 
 
 42. -1_+ 1 1 
 
 43. 
 
 a; + a aj + d a — 6 
 a^ + 1 a + h c 
 
 X c a + b 
 
 .. 2x — a , o 4a 
 44. \- S = 
 
 b 2x 
 
 45. 6ig I ^^ a(x + 2b) 
 a — X a + 6 
 
 46. Va + a; — Va — .a; = V2 x. 
 
 47. Va; — a + V^ — x = Vft — a. 
 
 48. Va;^- 6- = Va; + 6 Va + 6. 
 
 49. Va — X + V6 — a; = Va + 6 — 2 a;. 
 50. Solve and verify Va; + 1 + Va; — 2 — V2 a; — 5 = 0. 
 
 Solution 
 
 Vx+1 + V'a;-2 - V2x- 6 = 0. 
 
 Transposing, Vx + 1 + Va; — 2 = V2 x — 5. 
 
 Squaring, x + 1 + 2\/x"^ — x — 2 + x — 2 = 2x — 5. 
 
 Simplifying, Vx^ — x — 2 = — 2. 
 
 Squaring, x^ — x — 2 = 4, 
 
 Solving, X = — 2 or 3. 
 
 Verification. — Substituting — 2 for x in the given equation, 
 V^^+ \/34- V^^ = 0; 
 that is, VZrT + 2\/^n"- 3V^n" = 0. 
 
 Therefore, — 2 is a root of the given equation. 
 Substituting 3 for x in the given equation, 
 
 Vi + Vi - Vl = 0, 
 
 which is not true according to the convention adopted in the discussion in 
 § 284. Hence, 3 is not to be regarded as a root of the given equation. 
 
QUADRATIC EQUATIONS 293 
 
 Note. — In the previous verification, when only positive square roots are 
 taken, the second value obtained for x does not satisfy the given equation, 
 yet this indicates no error in the process of rationalizing, for the equation 
 can be verified by admitting negative square roots. But, as explained 
 in § 283, it is more convenient to regard 3 as the root of the equation 
 Vx + 1 — Vx — 2 — y/2x — b = 0, which has for its first member one of the 
 rationalizing factors of the given equation. 
 
 Solve and verify : 
 
 51. 8V^ — 8aj = f. 53. a; — 1 -I- Va; + 5 = 0. 
 
 52. 3a;+V^ = 5V4^- 54. cc — 5— Va; — 3 = 0. 
 
 55. V4 a; + 17 + Va; + 1 —4 = 0. 
 
 56. Va; — 1 + V2 a; — 1 — Vo « = 0. 
 
 57. V2a;-7-V2^+Va;-7=0. 
 
 58. Va; + 3+V4a; + l-Vl0a; + 4 = 0. 
 
 59. V6~+x + Va; - VlO — 4a; = 0. 
 
 60. V4 aj - 3 - V2 a; -I- 2 = Va; - 6. 
 
 61. V2 a; -f 3 - Va; + 1 = V5 a; - 14. 
 
 62. V3a;-5+Va; — 9=V4a;-4. 
 
 63. Va; + a^ - Va; - 2 a^ = V2a; - 5 aK 
 
 Problems 
 
 297. 1 . The sum of two numbers is 8, and their product is 15. 
 Find the numbers. 
 
 Solution 
 
 Let X = one number. 
 
 Then, 8 — x = the other. 
 
 Since their product is 15, 8 x — x^ = 15. 
 
 Solving, X = 3 or 5, 
 
 and 8 — X = 5 or 3. 
 
 Therefore, the numbers are 3 and 5. 
 
294 
 
 QUADRATIC EQUATIONS 
 
 2. A party hired a coach for $12. In consequence of the 
 faihire of three of them to pay, each of the others had to pay 
 20 cents more. How many persons were in the party ? 
 
 Let 
 Then, 
 
 and 
 
 Therefore, 
 
 X 
 
 x-3 
 
 12 
 x-3 
 
 Solution 
 the number of persons, 
 the number who paid, 
 
 the number of dollars each paid, 
 
 12 
 
 — = the number of dollars each should have paid. 
 
 X 
 
 12 1 ^ 12 
 x — 3 5 X 
 
 X = 15 or — 12. 
 
 Solving, 
 
 The second value of x is evidently inadmissible. Hence, the number of 
 persons in the party was 16. 
 
 3. A cistern can be filled by two pipes in 24 minutes. If it 
 takes the smaller pipe 20 minutes longer to fill the cistern than 
 the larger pipe, in what time can the cistern be filled by each 
 pipe ? 
 
 Let 
 Then, 
 
 Since 
 
 and 
 
 and 
 
 Solving, 
 
 SOLCTION 
 
 X = the number of minutes required by the larger pipe. 
 X + 20 = the number of minutes required by the smaller. 
 
 - = the part which the larger pipe fills in one minute, 
 = the part which the smaller pipe fills in one minute, 
 
 X4-20 
 
 ^^ = the part which both pipes fill in one minute, 
 
 1 ^J_ 
 
 24* 
 
 1 + 
 
 X x + 20 
 
 X = 40 or - 12. 
 
 Hence, the larger pipe can fill the cistern in 40 minutes, and the smaller 
 in 60 minutes. 
 
 4. Divide 20 into two parts whose product is 96. 
 
 5. Divide 14 into two parts whose product is 45. 
 
 6. A man purchased a flock of sheep for $75. If he had 
 paid the same sum for a flock containing 3 more, they would 
 have cost $ 1.25 less per head. How many did he purchase ? 
 
QUADRATIC EQUATIONS 295 
 
 7. A rectangular garden is 12 rods longer than it is wide, 
 and contains 1 acre. What are its dimensions ? 
 
 8. If each side of a square field were lengthened 4 rods, the 
 area would be increased 136 square rods more than \ of it. What 
 are the dimensions of the field ? 
 
 9. A rectangular lot is 8 rods longer than it is wide. What 
 are its dimensions, if it contains 1\ acres ? 
 
 10. In a column of 600 soldiers each file contained 3 men more 
 than 9 times as many men as each rank. How broad and how 
 deep was the column ? 
 
 11. A party had a dinner that cost $60. If there had been 
 5 persons more, the share of each would have been $ 1 less. 
 How many persons were there in the party ? 
 
 12. A man worked a certain number of days for $30. If he 
 had received $1 per day less than he did, he would have been 
 obliged to work 5 days longer to earn the same sum. How many 
 days did he work ? 
 
 13. Find two consecutive numbers the sum of whose squares 
 is 61. 
 
 14. Find two consecutive numbers the sum of whose recipro- 
 cals is 1^^. 
 
 15. A picture that was 8 inches by 12 inches was placed in a 
 frame of uniform width. If the area of the frame was equal to 
 that of the picture, what was the width of the frame ? 
 
 16. A merchant purchased a quantity of flour for $ 100, and 
 retailed it at a gain of $ 1 per barrel. After he had sold $ 100 
 worth of it, he had 5 barrels of it left. How many barrels did 
 he buy, and at what price ? 
 
 17. A merchant sold a hunting coat for $ 11, and gained a 
 per cent equal to the number of dollars the coat cost him. What 
 was his per cent of gain ? 
 
 18. A railway train traveled 5 miles an hour slower than 
 usual and was one hour late in making a run of 280 miles. How 
 many miles per hour did it travel ? 
 
296 QUADRATIC EQUATIONS 
 
 19. A rectangular park 56 rods long and 16 rods wide was 
 surrounded by a street of uniform width containing 4 acres. 
 What was the width of the street ? 
 
 "20. A boatman rowed 8 miles up a stream and back in 3 hours. 
 If the velocity of the current was 2 miles an hour, what was his 
 rate of rowing in still water? 
 
 21. A man who owned a lot 56 rods long and 28 rods wide 
 constructed a road around it, thereb}^ decreasing the area of the 
 lot 2 acres. What was the width of the road ? 
 
 22. A man bought two lots of cloth and paid 96 shillings for 
 each. There were 20 yards in all, and the number of shillings 
 per yard paid for each was the same as the number of yards of 
 the other. How many yards of each did he buy ? 
 
 23. Find the price of eggs, when 2 less for 30 cents raises the 
 price 2 cents per dozen. 
 
 24. A and B started at the same time and traveled toward a 
 place 75 miles distant. A traveled one mile an hour faster than 
 B and reached the place 2^ hours before B. At what rate did 
 each travel ? 
 
 25. A person drew a quantity of wine from a cask filled with 
 81 gallons of pure wine, and replaced it with water. He then 
 drew from the mixture as many gallons as he drew before of pure 
 wine, when it was found that the cask contained only 64 gallons 
 of pure wine. How many gallons did he draw each time ? 
 
 26. The circumference of the fore wheel of a coach is 5 feet 
 less than that of the hind wheel. If the fore wheel makes 150 
 more revolutions than the hind wheel in going a mile, what is 
 the circumference of each wheel ? 
 
 27. Two pipes running together can fill a cistern in 24 hours. 
 The larger pipe can fill the cistern in 2 hours less time than the 
 smaller. How many hours will it take each pipe alone to fill 
 the cistern ? 
 
 28. It took a number of men as many days to dig a ditch as 
 there were men. If there had been 6 more men, they would have 
 done the work in 8 days. How many men were there ? 
 
QUADRATIC EQUATIONS 297 
 
 EQUATIONS IN THE QUADRATIC FORM 
 
 298. An equation that contains but two powers of an unknown 
 number or expression, the exponent of one power being twice 
 that of the other, is in the Quadratic Form. 
 
 Equations in the quadratic form can be reduced to the general 
 form aa^" + 6a;" + c = 0, in which n represents any number. 
 
 Examples 
 
 1. Given a;* + 6 ar' — 40 = 0, to find the values of x. 
 
 Solution 
 5C* + 6 x2 - 40 = 0. 
 Factoring, (x^ - 4) (a;2 + lO) = 0. 
 
 .-. x^-4 = or x2 + 10 = 0, 
 
 and x = ±2 or ± V— 10. 
 
 2. Given x^ — x* = 6, to find the values of x. 
 
 First Solution 
 x^ -x^ = 6. 
 Completing the square, x* — x' + (i^)^ = ^^. 
 
 Extracting the square 
 
 root, x^ - ^ = ± f . 
 
 
 .-. xi = 3 or - 2. 
 
 Raising to the fourth j 
 
 power, X = 81 or 16. 
 
 
 Second Solution 
 
 
 x2 - xi = 6. 
 
 Let X* = p, then, 
 
 x^ - i)2, and p"^- p = 6. 
 
 
 .-. p2_p_6 = 0. 
 
 Factoring, 
 
 (P-3)(.i> + 2) = 0. 
 
 
 .: p = S or - 2 ; 
 
 that is, 
 
 JF* = 3 or - 2. 
 
 Whence, 
 
 « = 81 or 16. 
 
298 QUADRATIC EQUATIONS 
 
 3. Solve the equation a; — 4 o;^ + 3 x^ = 0. 
 
 Solution 
 
 Let x^ — p, then, x» = p"^, and x = p\ 
 
 Then, p^ -ip^ + Zp=iQ. 
 
 Factoring, p(p2 _ 4p + 3) = o. 
 
 Whence, J? = 0, 
 
 or p2_4p + 3_0. 
 
 Factoring, (p - 1) (p - 3) = 0. 
 
 Whence, J> = 1 or p = 3. 
 
 That is, x^ = 0, 1, or 3. 
 
 .-. x = 0, 1, or 27. 
 
 4. Given a:^ — 7 a; 4- Va^ — 7 a; + 18 = 24, to find the value of x. 
 
 Solution 
 
 x2 _ 7 X + Vx2 - 7 X + 18 = 24. (1) 
 
 Adding 18, x2 - 7 x + 18 + Vx2 - 7 x + 18 = 42. (2) 
 
 Put p for (x2 - 7 X + 18)2 and p^ for (x2 - 7 x + 18). (3) 
 
 Then, p^+p-^2-0. (4) 
 
 Solving, p = 6 or - 7. (6) 
 
 That is, Vx2 - 7 X + 18 = 6, (6) 
 
 or Vx2 _ 7 X + 18 = - 7. (7) 
 
 Squaring (6), ^2 - 7 x + 18 = 36. 
 
 Solving, -, ' X = 9 or - 2. 
 
 Squaring (7), x2 - 7 x + 18 = 49. 
 
 Solving, X = 1 ± |\/T73. 
 
 Hence, x = 9, - 2, or J ± JvTTS. 
 
 5. Solve the equation a^ — 9a^ + 8 = 0. 
 
 Solution 
 
 x6- 9x3 + 8 = 0. (1) 
 
 Factoring, (x^ - l)(x3 - 8)= 0. (2) 
 
 Therefore, x^ - 1 = 0, (8) 
 
 or 358 _ 8 = 0. (4) 
 
QUADRATIC EQUATIONS 299 
 
 If the values of x ate found by ti'ansposing the known terms in (3) and 
 (4) and extracting the cube root of each member, only one value of x will 
 be obtained from each equation. But if the equations are factored, three 
 values of x are obtained. 
 
 Factoring, (x - 1) (x2 + x + 1) = 0, (5) 
 
 and (x-2)(x2 + 2x + 4) = 0. (6) 
 
 Writing each factor equal to zero, and solving : 
 
 From Eq. (5), x = 1. - i + ^V^S, - ^ - ^V^^. (7) 
 
 From Eq. (6), x = 2, - 1+ V^ - 1 - V^^. (8) 
 
 Since the values of x in (7) are obtained by factoring x* — 1 = 0, they 
 may be regarded as the three cube roots of the number 1. Also, the values 
 of X in (8) may be regarded as the three cube roots of the number 8 (§ 225). 
 
 6. Solve the equation a;* + 4a^— 8a; + 3 = 0. 
 
 Solution 
 
 Extracting the square root of the first member as far as possible, 
 
 x* + 4x8-8x + 3 I X- + 2 X - 2 
 
 X* 
 
 2x2 + 2x 1 
 
 4x3 
 
 4 x3 + 4 x2 
 
 2x2 + 4X' 
 
 -2| -4x2-8x + 3 
 -4x2-8x + 4 
 
 - 1 
 
 Since the first member lacks 1 of being a perfect square, the square may 
 be completed by adding 1 to each member, giving the following equation : 
 
 X* + 4 x8 - 8 X + 4 = 1. 
 Extracting the square root, x2 + 2x — 2=±1. 
 
 .-. x2 + 2 X - 3 = 0, and x2 + 2 X - 1 = 0. 
 Solving, X = 1, - 3, - 1 ± V2. 
 
 Solve the following equations : 
 
 7. a^-13a^ + 36 = 0. 11. 5x* + Qa?-ll = 0. 
 
 8. ic* - 25 a^ + 144 = 0. 12. 2.'c* - 8 ar'- 90 = 0. 
 
 9. a;*-18a^ + 32 = 0. 13. a;* - 5^^ + 6= 0. 
 10. 3a;* + 5ar'-8 = 0. 14. sc^ -|- 3 a;^ - 28 = 0. 
 
300 QUADRATIC EQUATIONS 
 
 15. x^-Sx^ = -2. 25. x-^x^ + 2x^ = 0. 
 
 16. xi — x^ = Q. 26. 5x = xVx + QVx. 
 
 17. a; + 2V^ = 3. 27. 3 a; = rc^/^ -f- 2^^. 
 
 18. a;* -2x^ = 3. 28. a;"^ - 3 - 4 x^ = 0. 
 
 19. ar» + 8a;^-9 = 0. 29. a;"^- 6x5 = 1. 
 
 20. xi-\-xi-2 = 0. 30. x-^-^x^ = 2x-^. 
 
 21. a/x + 3Vx = 30. 31. x + 2xi = '6x^. 
 
 22. aa^» + 6a;" + c = 0. 32. 2 x + Vx = 15 a; V^. 
 
 23. a;^-4aj-5a;^ = 0. 33. Vx + 5 + 6x"^ = 0. 
 
 24. a;^ — ar^ — 12 a;^ = 0. 34. x* = 8 a; + 7 a^ Vx. 
 
 35. (x-3)2 + 2(a;-3) = 3. 
 
 36. (ar^ + l)2 + 4(a^ + l) = 45. 
 
 37. (a^-4)2-3(a^-4) = 10. 
 
 38. {a?-2xy-2{x'-2x) = S. 
 
 39. (x2 - x)2 - (x^ -x)- 132 = 0. 
 
 40. X—5 + 2Va; — 5 = 8. 
 
 41. ^2 - 3 x + 6 +2 Va^ - 3 a; + 6 = 24. 
 
 42. x'-bx^ 2y/x? - 5 a; - 2 = 10. 
 
 43. x2 — X - Va^ — a; + 4 — 8=0. 
 
 44. ar' — 5 x + 5 Var' — 5 x + 1 = 49. 
 
 45. a;+10 = 2Va; + 10-}-5. 49. a; + 2 Vx + 3 = 21. 
 
 46. a; - 3 = 21 - 4 Vx - 3. 50. 2 a; - 3 V2 a; + 5 = - 5. 
 
 47. 2a;-6V2x-l = 8. 51. ar^ + xVx - 72 = 0. 
 
 48. x = 11 — 3 Vx + 7. 52. x~^ — 5 a;"^ 4- 4 = 0. 
 
 53. /'12_;^y_^8/12_j\ 33^ 
 
QUADRATIC EQUATIONS 301 
 
 = 8. 
 
 ■V- 
 
 66. /'l^Y-4^1--' 
 
 58. (x-a)^-3a^(x-a)^ + 2a^ = 0. 
 
 59. Find the three cube roots of — 1. 
 
 60. Find the three cube roots of — 8. 
 
 61. Find the four fourth roots of 1. 
 
 Solve the following equations : 
 
 62. ic« - 28 a^ + 27 = 0. 65. a;* + 2 ic^ _ g. ^ 30. 
 
 c 66. X* — 4a^ + 8x = -3. 
 
 63. 2:^-^ = 7. 
 
 a^ 67. ic* - 2 ar' + aj = 132. 
 
 64. a^-16 = 0. 68. a.-^ - 6 a;^ _|_ 27 a; = 10. 
 
 69. x* + 2a^-\-5x' + 4:X-60 = 0. 
 
 70. a^ + 6a^ + 7a;2_6a,_8 = 0. 
 
 71. a;*- 6 ar'H- 15 a,-2-18a; + 8 = 0. 
 
 72. a;*-10«8 + 35a^-50a; + 24 = 0. 
 
 73. 16 x* - 8 a^ - 31 ar^ + 8 a; +15 = 0. 
 
 74. 4a;*-4a:3-7a^ + 4a; + 3 = 0. 
 a^ a; + l 7 
 
 75. 
 
 a; + 1 a^ 12 
 
 Suggestion. — Since the second term is the reciprocal of the first, put p 
 
 for the first term and — for the second. 
 P 
 
 Then, p-l = i 
 
 76. ^+2 ^2. 78. §±l + 2i^=f. 
 
 2a^ + a; ar^ + 4aj + 2 5 
 
 77 a^ + 1 I 4 ^5 0,24.1 4(a;-l) ^21 
 
 ■ 4 "^ar*-;-! 2* ' a;-l o^ _,_ ^ 5 
 
302 QUADRATIC EQUATIONS 
 
 80. _^_ + ^:^ = -l^. 83. x'-2x-^- ^ = 11. 
 
 ar — 1 X 6 ar — 2a; — 6 
 
 81. x' + x + l--—^ - = % 84. a^-a; + -— -^ = 7. 
 
 ar + a; + l d ar — a; — 4 
 
 82. ;>?-3x+ I =1. 85. a;^-2a;+ ^ =4. 
 
 ar — 3a; + 2 ar — 2x + l 
 
 86. 1—-^ - g 3 = 0. 
 
 SIMULTANEOUS EQUATIONS INVOLVING 
 QUADRATICS 
 
 299. An equation whose terms are homogeneous with respect 
 to the unknown numbers is called a Homogeneous Equation. 
 
 x2 — a;y + y2 — 6^ jg3 ^ y3 — 12, and ax + ?/ = 10 are homogeneous equations. 
 
 300. An equation that is not affected by interchanging the 
 unknown numbers involved in it is called a Symmetrical Equation. 
 
 2x^ + xy + 2y^ = 32, x^ + y^ = 28 are symmetrical equations. 
 
 301. Many simultaneous equations involving quadratics may 
 be solved by the rules for quadratics, if they belong to one of the 
 following classes : 
 
 1. When one is simple and the other quadratic. 
 
 2. When the unknown numbers are involved in a similar manner 
 in each equation. 
 
 3. WJien each equation is homogeneous and qxiadratic. 
 
 I. Simple and quadratic. 
 1. Solve the equations 
 
 x-\-y = l, 
 3a^ + 2/2^43^ 
 
 Solution 
 
 x+y =1. (1) 
 
 3 «2 + 8/2 = 43. (2) 
 
 From (1), y = 7 -X. (3) 
 
 Substituting in (2), 3a;2 + (7 - x)2 = 43. (4) 
 
X = 3 or J. 
 
 (6) 
 
 y = i. 
 
 (6) 
 
 y = ¥- 
 
 <7) 
 
 ( when X = 3, j/ = 4, 
 
 
 QUADRATIC EQUATIONS 303 
 
 Solving, 
 
 Substituting 3 for x in (3), 
 
 Substituting ^ for x in (3), 
 
 That is, X and y each have two values 
 
 [ when X = ^, ?/ = ^^• 
 
 Equations of this class may be solved by finding the value of 
 one unknown number in terms of the other in one equation, and 
 then substituting it in the other. 
 
 Solve the following equations : 
 
 V + 2/2_20, (x = 6-y, 
 
 ^' ' x = 2y. " [x' + f = 72. 
 
 10x + y = 3xy, (xy{x-2y) = 10, 
 
 y — x = 2. 1 an/ = 10. 
 
 x' + xy = 12, ^ (3x(y + l) = 12, 
 
 x — y = 2. \Sx = 2y. 
 
 II. Unknown numbers similarly involved. 
 
 {ic 4- w = 7 
 xy = 10. 
 
 Solution 
 
 x + y = 7. (1) 
 
 xy = 10. (2) 
 
 Squaring (1), z^ + 2xy-{-y^ = 49. (3) 
 
 Multiplying (2) by 4, 4 xy = 40. (4) 
 
 Subtracting (4) from (3), x^-2xy + y^ = 9. (5) 
 
 Extracting the square root, x — y = ± S. (6) 
 
 From (1) + (6), x = 5 or 2. 
 
 From (1) - (6), y = 2 or 6. 
 
 Such equations may be solved by the method illustrated in 
 example 1, but the method given above of finding the value of 
 
304 QUADRATIC EQUATIONS 
 
 X — y, so that it may be combined with the value of x-{-y to dis- 
 cover the vahies of x and y, is preferable. 
 
 f x^ 4- y^ = 25 
 9. Solve the equations \ ' 
 
 [x + y = 7. 
 
 Solution 
 
 x2 + 2/2 = 25. 
 x + y = 7. 
 Squaring (2), x^ + 2xy + y^ = 49. 
 
 Subtracting (1) from (3), 2xy = 24. 
 
 Subtracting (4) from (1), x^ — 2 xy + y^ = 1. 
 Extracting the square root, x — y = ±1. 
 
 From (2) + (6), x = 4 or .3. 
 
 From (2) -(6), . 2/ = 3 or 4. 
 
 10. Solve the equations 
 
 a!* + 2/^ = 97, 
 x + y = l. 
 
 Solution 
 
 x*-\-y* = 97. 
 
 x + y = l. 
 4th power of (2), 
 
 x^ + ^afiy + 6 x2j/2 4. 4 xyS + y* = 1. 
 
 Subtracting (1 ) , 4 x^y + 6 xV + 4 ajys := _ gg. 
 
 Dividing by 2, 2 x^y + 3 x^y^ + 2 xyS = _ 48. 
 
 2 xy X square of (2), 2 x^ + 4 x^y'^ + 2xy^ = 2 xy. 
 
 Subtracting (5) from (6), xV -2xy = 48. 
 
 Solving for xy, xy = — 6 or 8. 
 
 Equations (2) and (8) give two pairs of simultaneous equations, 
 
 x+y=l ^ fx+y=l 
 and 
 
 xy = — 6 [ xy = 8 
 
 Solving these by previous methods, 
 
 x = 3, or -2, or ^ + |V-31, or ^ _ i-sA-Si. 
 y=-2, or 3, or ^ - ^V-Sl, or J + ^V- 31. 
 
QUADRATIC EQUATIONS 305 
 
 Solve the following equations : 
 
 r^ + f = m, j^ rx^ + t/^ = i3, 
 
 "• \xy = 7. ' \x + y + xy = ll. 
 
 P + ^ = 8, 15. P^^^' = ''' 
 
 lx + y = 9, ^, |a^ + .^,^ + ,-21, 
 
 .a^ + 2/3^243. ■ \x' + xy + f = 7- 
 
 13 
 
 III. Homogeneous quadratics. 
 
 (x^-xy + y' = 21, 
 17. Solve the equations |^2_2a^ = _15. 
 
 
 SOLCTIOK 
 
 
 x^-xy + y^ = 21. 
 
 
 y^-2xy =- 16. 
 
 Assume 
 
 x = vy. 
 
 Substituting in (1), 
 
 «2j/2 _ vy2 + y-2 = 21. 
 
 Substituting in (2), 
 
 y^ — 2vy^ = — 15. 
 
 
 21 
 
 Solving (4) for y% 
 
 ^ v'^-V + l 
 
 
 „2_ 15 , 
 
 Solving (5) for y^, 
 
 ^ -2t;-l 
 
 
 15 21 
 
 Comparing the values oiy, 2v-i v^-v + 1 
 
 Clearing, etc., 
 
 5 1)2 - 19 w + 12 = 0. 
 
 Factoring, 
 
 («-3)(5v-4) = 0. 
 
 
 .-. t) = 3 or |. 
 
 Substituting 3 for v in (7) or in (6) , y = ±y/S } 
 and since x = vy, * = ±3v3 
 
 Substituting | for v in (7) or in (6), y = ± 5 
 and since x = vy, a; = ± 4 
 
 ra; = +3\/3, or -3v/3, or +4, or -4. 
 
 Hence, | y = + VS, or - V3, or + 6, or - 6 
 
 ALG.— 20 
 
 (1) 
 
 (2) 
 (3) 
 (4) 
 (5) 
 
 (6) 
 (7) 
 
 (8) 
 
 (9) 
 (10) 
 
 (11) 
 
306 QUADRATIC EQUATIONS 
 
 Solve the following equations : 
 
 [x'-dxy^-S. ' \u?-Sxy + 2y' = ^. 
 
 ^^ {a? + xy = 12, {x^-xy + f = 21, 
 
 ' \xy + 2y^ = 5. ' \^+2f=:21. 
 
 r a;2 + 2 2/2 = 44, f 2 0-^ - 3 a:?/ + 2/ = 100, 
 
 i a;?/ — 2/^ = 8. ' \y? — y^z= 75. 
 
 302. Many simultaneous equations that belong to one or more 
 of the preceding classes, and many that belong to none of them, 
 may be readily solved by special devices. 
 
 24. Solve the equations \ 
 
 f x' + xy = 12, 
 
 Solution 
 
 
 a;2 + xy = 12. 
 
 (1) 
 
 xy + ?/2 = 4. 
 
 (2) 
 
 Adding, x'^ + 2xy + y'^ = 16. 
 
 (3) 
 
 .-. a; + 2/ = + 4 or — 4. 
 
 (4) 
 
 Subtracting (2) from (1), x"^ - y^ = g. 
 
 (5) 
 
 Dividing (5) by (4), x — y = + 2 or — 2. 
 
 (6) 
 
 Combining (4) and (6), a; = 3 or — H ; y = 1 or — 1. 
 
 
 25. Solve the equations 
 
 x^ + a;y + .V* = 48, 
 x^ — xy + y^ = 12. 
 
 Solution 
 
 x4 + xV- + 2/^ = 48. (1) 
 
 x^ - x?/ + 2/'^ = 12. (2) 
 
 Dividing (1) by (2), x'^ + xy + y^ = 4. (3) 
 
 From (3) - (2), xy = - 4. (4) 
 
 By adding (4) to (3), and subtracting (4) from (2), the values of (x + y)^ 
 and (x — y)^ may be found and the solution readily completed. 
 
QUADRATIC EQUATIONS 307 
 
 X^-y =26, 
 
 26. Solve the equations 
 
 x-y = 2. 
 
 Solution 
 
 x8 - y» = 26. (1) 
 
 x-y = 2. (2) 
 
 Dividing (1) by (2), z^ + xy + y^ ^ IS. (3) 
 
 Squaring (2), x^ -2xy + y^ rz4. (4) 
 
 Subtracting, dividing by 3, xy = 3. (5) 
 
 By adding (5) to (3), the value of (x + yy may be obtained and the 
 solution completed as in previous examples. 
 
 c . . ■ ix* + y* = 82 
 
 27. Solve the equations i 
 
 I x — v = 2. 
 
 x* + y* = 82, 
 
 y 
 
 Solution 
 
 a;* + y* = 82. (1) 
 
 x-y = 2. (2) 
 
 Assume x = u + v, (3) 
 
 and y = u — V. (4) 
 
 Substituting these values in (1), 
 
 M* + 4 uH + 6 uH^ + 4 uv^ + V* 
 
 + M* - 4 M8tj + 6 mV _ 4 Mt;3 4- w* = 82, (5) 
 
 and in (2), 2v = 2. (6) 
 
 Dividing (5) by 2, m* + 6 mV + ^4 = 41. (7) 
 
 Dividing (6) by 2, v = 1. (8) 
 
 Substituting 1 for i; in (7) and solving, « = ± 2 or ± V— 10. (9) 
 Hence, from (3) and (4), x = 3 or — 1 or 1 ± V— 10, 
 
 and y = lor— 3or— 1± V— 10. 
 
 ( x^ + y^ + x-^y = 14:, 
 28. Solve the equations \ 
 
 [xy = 3. 
 
 Solution 
 
 x^ + y^ + x+y = U. 
 xy = 3. 
 Adding twice the second equation to the first, 
 
 x2 + 2 acy + j/2 4. X + y = 20. 
 
308 
 
 QUADRATIC EQUATIONS 
 
 Completing the square, {x+ yY -\- x ->r y + (^)2 = 20J. 
 Extracting the square root, x + 2/ + i=±f. 
 
 .•. X + y = 4 or — 6. 
 The soUition may be completed by solving the equations, 
 
 'x + y = 4 
 xy = 3 
 
 and 
 
 x+y =-6 
 x?/ = 3 
 
 The student will doubtless discover many other methods for 
 solving simultaneous equations. All the preceding solutions are 
 but illustrations of devices that are important only because they 
 are often applicable. 
 
 Solve the following simultaneous equations : 
 
 29. 
 
 30. 
 
 31. 
 
 32. 
 
 33. 
 
 34. 
 
 35. 
 
 36. 
 
 37. 
 
 x — y = 5. 
 
 3^ + 2/3 = 28, 
 
 [x + y = 4:. 
 
 n + x = y, 
 1 ic2 + 2/2 = 61. 
 
 r l+x = y, 
 
 a^ + ?/2 = 40, 
 xy = 12. 
 
 x(x + y) = x, 
 y(x-y)=-l. 
 
 a^-{-3xy-f = A3, 
 x + 2y = 10. 
 
 (x^ + xy-{-f = 19, 
 
 ( x^ -^ 3 xy = y"" + 23, 
 [x + 3y = 9. 
 
 38. 
 
 39. 
 
 2x^-\-xy-5y^ = 20, 
 2x-3y = l. 
 
 ^ + — = 21, 
 
 y y 
 
 x-y = 2. 
 
 40. -I 
 
 [3^ 
 
 {xy 
 
 xy = 48, 
 -/ = 12. 
 
 41. ■! 
 
 { 21? + xy = —Q, 
 
 42. 
 
 43. 
 
 44. 
 
 45. -! 
 
 46. 
 
 [xy -\-y^ = 15. 
 
 4 ajt/ = 96 — x?y^, 
 X + ?/ = 6. 
 
 r x'-xy = 8, 
 \xy + f = 12. 
 
 a? — xy =■ Q, 
 :(? + y^ = 61. 
 
 {'j? + xy = ll, 
 {xy — y^ = 12. 
 
 \2x-y = 2, 
 \2x^ + f = ^. 
 
47. 
 
 48. 
 
 49. 
 
 50. 
 
 51. 
 
 52. 
 
 53. 
 
 54. 
 
 55. 
 
 66. 
 
 57. 
 
 58. 
 
 QUADRATIC EQUATIONS 
 
 I 5 0^2 + 4 «2 = 41. 
 
 309 
 
 0^2 + 4 y2 = 41 
 
 2xy-f = 12, 
 Sxy + oo? = 104. 
 
 1 a^ + / = 106. 
 
 ' a^ + a:y + 2/* = 84, 
 , X — V^ + 2/ = 6. 
 
 ' 4:0? — 2 xy ->ry' = 13, 
 .8 0^ + ^ = 65. 
 ■6ar^ + 6t/2^13 3.y^ 
 .a:2_,/2^20. 
 
 a^ + 2^^-3(0^ + 2/)= 8, 
 
 a; 4- 2/ + 2^ = 11- 
 
 3xy + 2x + y = 25, 
 
 9a;_4 ?/ 
 
 2/ a; 
 
 0^ + x?/ = 40, 
 
 27 + 2 2/2 = 3 0^. 
 
 ' 0^2 ^ a;?/ = 24, 
 . xy^ -\- X = 56. 
 
 a;* + 2/* = 82, 
 x + y = 4. 
 
 ic* - y^ = 369, 
 
 ar^ - 2/2 = 9. 
 
 f ^ + 2/' - 78 = a; + 2/, 
 59. i 
 
 I «?/ + a; + 1/ = 39. 
 
 60 
 
 |x-2 + 2.'r2/ + 32/- = 43, 
 *' l2a^ + 3a;2/ + 42/'' = 62. 
 
 61. 
 
 62. 
 
 63. 
 
 64. 
 
 65. 
 
 x^ — xy + 2y^ = 46, 
 a? + xy + 3y^ = 111. 
 
 ar^ - 7 a;?/ + 12 2/2 = 0, 
 xy-\-3y = 2x + 21. 
 
 a^ - 2/3 = 37, 
 a^(2/-a;) = -12. 
 
 a; + 2/ = 25, 
 V^+V2/ = 7. 
 
 a-3 + 2/3^225 2/, 
 3^2 - 2/2 = 75. 
 
 f a^ + aw + 2 2/2 = 11, 
 
 * l2a:2_^5^2^22. 
 
 a^ + 2/^ = 3 a:2/ + 5, 
 .x* + y* = 2. 
 
 67. 
 
 68. 
 
 a;^ 4- 2/ = ' 
 
 x-y 
 
 x^y — an/2 _ Q 
 
 1 (a; - 2/) (a^ - 2/^0 = ^• 
 I a^ + 2/ = a; - 2/^ + 42, 
 ' U.v = 20. 
 
 ^^ I a; + y + 2 V a; + 2/ = 24, 
 I a; — 2/ + 3 Va; — y = 10. 
 
 (a^ + 2/' + 6V^+7 = 55, 
 
310 QUADRATIC EQUATIONS 
 
 x-{- y _ 2 X a? — a?y _ 3 
 73. I y x + y f~x^y S' 
 [ a^ = 2/2 + 16. 
 
 a? — xy = a? -\- h^, 
 
 74. I , ^ 
 
 y xy — y =1 2 ah, 
 
 ^^ {x-2y = 2{a + h), 
 [xy + 2f = 2b(b-a). 
 
 \x'y + xy^ = 2a(a^-by 
 
 i a — X ,b + y__ ay + bx 
 X y a^ — b^' 
 
 a^ + 2/2 = 2(a' + 62). 
 
 Problems 
 
 303. 1. The sum of two numbers is 12, and their product is 
 32. What are the numbers ? 
 
 2. The sum of two numbers is 17, and the sum of their 
 squares is 157. What are the numbers ? 
 
 3. The difference of two numbers is 1, and the difference of 
 their cubes 91. What are the numbers ? 
 
 4. The sum of two numbers is 82, and the sum of their square 
 roots is 10. What are the numbers ? 
 
 5. It takes 52 rods of fence to inclose a rectangular garden 
 containing 1 acre. How long and how wide is the garden ? 
 
 6. The sum of the squares of the terms of a fraction is 89, 
 and the fraction is |^ larger than its reciprocal. What is the 
 fraction ? 
 
 7. Find two numbers such that their product is 8 greater 
 than twice their sum and 48 less than the sum of their squares. 
 
 8. If 63 is subtracted from a certain number expressed by 
 two digits, its digits will be transposed; and if the number is 
 multiplied by the sum of its digits, the product will be 729. 
 What is the number ? 
 
QUADRATIC EQUATIONS 311 
 
 9. A man expended $ 6 for canvas. Had it cost 4 cents less 
 per yard, he would have received 5 yards more. How many 
 yards did he buy, and at what price per yard ? 
 
 10. If the difference of two numbers multiplied by the 
 greater is 160, and multiplied by the less is 96, what are the 
 numbers ? 
 
 11. A rectangular flower garden containing 54 square rods 
 was enlarged to twice its former size by making an addition 
 of 1\ rods on all sides. What were the original dimensions of 
 the garden? 
 
 12. If it requires 200 rods of fence to inclose a rectangular 
 field of 15 acres, what are its dimensions ? 
 
 13. A rectangular field contains 20 acres. If its length were 
 20 rods less and its width 8 rods less, its area would be 8 acres 
 less. What are its dimensions ? 
 
 14. A man found that he could buy 16 more sheep than cows 
 for $ 100, and that the cost of 3 cows was $ 15 greater than the 
 cost of 4 times as many sheep. What was the price of each ? 
 
 15. Eight persons contributed $30 to pay for a set of books. 
 One half of the amount was contributed by women, and the other 
 half by men, each man giving $ 2 more than each woman. What 
 did each woman and what did each man contribute ? 
 
 16. A man loaned $ 1000 in two unequal sums at such rates 
 that both sums yielded the same annual interest. The larger 
 sum at the higher rate of interest would have yielded $36 
 annually, the smaller sum at the lower rate, $16 annually. 
 What sums did he invest, and at what rates of interest? 
 
 17. If 2 is added to the numerator and subtracted from the 
 denominator of a certain fraction, the result will be the recipro- 
 cal of the fraction; if 3 is subtracted from the numerator and 
 added to the denominator, the result will be ^ of the original 
 fraction. What is the fraction ? 
 
 18. The product of two numbers is 59 greater than their sum, 
 and the sum of their squares is 170. What are the numbers ? 
 
312 QUADRATIC EQUATIONS 
 
 19. A purse contained $50.50 in gold and silver coins. If 
 there were fifteen coins, and if each gold coin was worth as many 
 dollars as there were silver coins and each silver coin was worth 
 as many cents as there were gold coins, how many coins of each 
 kind were there ? 
 
 20. Two men working together can complete a piece of work 
 in 6f days. If it would take one man 3 days longer than the 
 other to do the work alone, in how many days can each man do 
 the work alone ? 
 
 21. The fore wheel of a carriage makes 12 revolutions more 
 than the hind wheel in going 240 yards. If the circumference 
 of each wheel were one yard more than it is, the fore wheel 
 would make 8 revolutions more than the hind wheel in going 240 
 yards. What is the circumference of each wheel ? 
 
 22. A sum of money on interest for 1 year at a certain per 
 cent amounted to $11130. If the rate had been 1% less and 
 the principal $ 100 more, the amount would have been the same. 
 Find the principal and rate. 
 
 23. A number multiplied by another composed of the same 
 two digits, but reversed, gives a product of 4032. If the first 
 divided by the second is equal to If, what are the numbers ? 
 
 24. The town A is on a lake and 12 miles from B, which is 
 4 miles from the opposite shore. A man rows across the lake 
 and walks to B in 3 hours. In returning he walks at the same 
 rate as before, but rows 2 miles an hour less than before. If it 
 takes him 5 hours to return, find his rates of walking and rowing. 
 
 25. A, B, and C started together to ride a certain distance. A 
 and C rode the whole distance at uniform rates, A two miles an 
 hour faster than C. B rode with C for 20 miles, and then, by 
 increasing his speed two miles an hour, reached his destination 
 40 minutes earlier than C and 20 minutes later than A. Find 
 the distance, and the rate at which each traveled. s 
 
 26. Find two numbers such that their product is equal to the 
 difference of their squares, and the difference of their cubes is 
 equal to the sum of their squares. 
 
QUADRATIC EQUATIONS 313 
 
 PROPERTIES OP QUADRATICS 
 
 304. Every quadratic equation may be reduced to the form 
 
 aa^ + bx + c = 0, (1) 
 
 in which a is positive and b and c are positive or negative. 
 Denote tlie roots by rj and rg. Then, § 294, Ex. 13, 
 
 ft + Vy-4ac^^^^_-&-V&^-4ac^ ^2) 
 
 2a 2a 
 
 In the following discussion of the nature of the roots of a quadratic equa- 
 tion, the student should keep in mind the distinctions between rational and 
 irrational, real and imaginary. For example, 2 and Vi are rational, and real 
 also ; \/2 and Vs are irrational, but real ; V— 2 and V— 5 are irrational, 
 and also imaginary. 
 
 1. Suppose that &^ — 4 ac is positive. 
 
 Then, V6^ — 4 ac is a positive real number and — V6'' — 4 ac is 
 a negative real number. Hence, the roots are real and unequal. 
 
 If 6^ — 4 ac is a perfect square, the roots are rational ; otherwise 
 they are irrational. 
 
 2. Suppose that 6^ — 4 ac = 0. 
 
 Then, Vb^ — 4 etc = and the roots are real and equal. 
 
 3. Suppose that b^ — 4ac is negative. 
 
 Then, Vft^ — 4ac and — V6^ — Aac are imaginary, and conse- 
 quently both roots are imaginary. 
 
 Principles. — 1. In any quadratic equation ax^ -\-bx + c = 0, if 
 &^ — 4 ac is positive, the roots are real and unequal; if b^—4 ac=0, 
 the roots are real and equal ; if b^ — 4 ac is negative, both roots are 
 imaginary. 
 
 2. If W — A:ac is a perfect square or is equal to zero, the roots are 
 rational; othenvise they are irrational. 
 
 When the roots are real, their signs are found by comparing the 
 values of b and c. 
 
 If c is positive, — & is numerically greater than ± V6^ — 4 ac, 
 whence both roots have the sign of — 6 ; if c is negative, — & is 
 numerically less than ± V&^ — 4 ac, whence r^ is positive and r^ is 
 
314 QUADRATIC EQUATIONS 
 
 negative. The root haviug the sign opposite to that of h is the 
 greater numerically. Hence, 
 
 Principle 3. — If c is positive, both roots have the sign opposite 
 to that of b; if c is negative, the roots have opposite signs, and the 
 numerically greater root has the sign opposite to that of b. 
 
 The following are special cases: 
 
 1. If c = 0, (1) reduces to the form ax^ + bx = 0, which has two roots, — - 
 and 0. " 
 
 2. If 6 = 0, (1) becomes the pure quadratic equation ax^ + c = 0, whose 
 roots are numerically equal with opposite signs. 
 
 3. If c = and b = 0, (1) becomes ax'^ = 0, which has two zero roots. 
 
 4. If a = or if a = and 6 = 0, (1) ceases to be a quadratic equation. 
 But if these coefficients differ from zero, however little, (1) is still a quad- 
 ratic equation and has two roots. To discover the nature of the roots in 
 these cases, rationalize the numerators in (2). 
 
 Then, n = ^ /^ and rg- ^ ^' (3) 
 
 -b - V62 - 4 ac -b+ Vb^ - 4 ac 
 
 Suppose that a is very small as compared with b and c. 
 Then, the denominator of ri is very nearly equal to — 6 — VP, or to — 2 b, 
 and the denominator of ^2 is very small. 
 
 Hence, the smaller a is the less will the first root differ from — - , the 
 
 b 
 root of the simple equation bx + c = 0, and the greater will be the numerical 
 value of the second root. 
 
 Suppose that a and b are very small as compared with c. 
 
 Then, both denominators in (3) are very small as compared with the 
 numerators. 
 
 Hence, the smaller a and b are the greater will both roots be in numerical 
 value. 
 
 Examples 
 
 1. What is the nature of the roots of a^ — 7a; — 8 = 0? 
 
 Solution. — Since 6^ _ 4 ^^^ = 49 + 32 = 81 = 9^, the roots are real and 
 unequal (Prin. 1), and rational (Prin, 2). Since c is negative, the roots have 
 opposite signs and, b being negative, the positive root is the greater numeri- 
 cally (Prin. 3). 
 
 2. What is the nature of the roots of Sx^ + Tyx + S^O? 
 
 Solution. — Since b^ — iac = 25 — 36= — 11, both roots are imaginary 
 (Prin. 1). 
 
QUADRATIC EQUATIONS 315 
 
 Find the nature of the roots of the following equations : 
 
 3. 
 
 ar^ — 5 a; — 75 = 0. 
 
 9. 
 
 x^ + x-2 = 0. 
 
 4. 
 
 X- -\-bx-\-Q = 0. 
 
 10. 
 
 4ar-4a;+i = 0. 
 
 5. 
 
 a;^ + 7 a; - 30 = 0. 
 
 11. 
 
 4a^-f6a;-4 = 0. 
 
 6. 
 
 a;- — 3 a; + 5 = 0. 
 
 12. 
 
 2 0^-9x4-4 = 0. 
 
 7. 
 
 a^ + 3a;-5 = 0. 
 
 13. 
 
 4a^4-16a;-f7 = 0. 
 
 8. 
 
 ar + X + 2 = 0. 
 
 14. 
 
 9a^-M2a; + 4 = 0. 
 
 305. Formation of quadratic equations. 
 
 Any quadratic equation, as aa? + hx + c = 0, may be reduced, 
 by dividing both members by the coefficient of ar, to the form 
 7? -\- px -\- q = Q, whose roots are 
 
 -P+^P'-Al and r,^-P-^P'-AR. 
 2 2 
 
 Adding the roots, i\ + rg = — — - = —p- 
 
 Li 
 
 Multiplying the roots, r^r^ = ^ ~ ^^ ~ — ^i = q. Hence, 
 
 4 
 
 306. Principle. — The sum of the roots of a quadratic equa- 
 tion having the form a? -\- jox -\- q = is equal to the coefficient of x 
 with its sign changed, and their product is equal to the absolute term. 
 
 Substituting — (r^ + r^ for p, and r^rg for gina:;^-|-j9a;-f-g' = 0, 
 
 x^ — (ri + r.2)x + ri>2 = 0. 
 
 Expanding, of — r^x — r^ -\- r-^r^. = 0. 
 
 Factoring, (x — rj) (a; — rj) = 0. 
 
 Hence, to form a quadratic equation when the roots are given : 
 
 Rule. — Subtract each root from x and place the product of the 
 remainders equal to zero. 
 
 Examples 
 
 1. Form an equation whose roots are —5 and 2. 
 
 Solution, (ac + .5) (x - 2) = 0, or x^ + 3 x - 10 = 0. 
 
 Or, since the sum of the roots with their signs changed is +5 — 2, or 3, and 
 the product of the roots is —10 (Prin.), the equation is x^ + 3 x — 10 = 0. 
 
316 GENERAL REVIEW 
 
 Form equations whose roots are 
 
 2. 
 
 6,4. 
 
 
 8. 
 
 a, — 3 a. 
 
 14. 
 
 3+V2, 3-V2. 
 
 3. 
 
 5, -3. 
 
 
 9. 
 
 a + 2, a -2. 
 
 15. 
 
 _2-V5, -2+V5. 
 
 4. 
 
 3, -i. 
 
 
 10. 
 
 6 + 1, &-1. 
 
 16. 
 
 2 ± 3V|. 
 
 5. 
 
 if 
 
 
 11. 
 
 a + h, a— b. 
 
 17. 
 
 -i(3±V6). 
 
 6. 
 
 -2, - 
 
 h 
 
 12. 
 
 Va — Vft, V&. 
 
 18. 
 
 i(-l±V2). 
 
 7. 
 
 -h- 
 
 t- 
 
 13. 
 
 i(«iV&). 
 
 19. 
 
 a(2±2V5). 
 
 GENERAL REVIEW 
 
 307. 1 . Add X -\/y + y Vic + v'a^, x^y^ — -y/x^y — Va^, Va^ 
 — Vic?/^ — V^, and y Vic — ic V4^ — V9 a-t/. . 
 
 2. From the sum of 2a + 36 — 3^/ and 2 1/ — a — 3 6 sub- 
 tract (a — 6 — y) — (a + & + y). 
 
 3. What number must be added to a to give b — a? 
 
 4. If a = 2 and 6 = 3, find the value of 
 
 a + 26 b a? — o?b 
 
 a a— b 2a— b 
 
 5. What number must be subtracted from a — 6 to give 
 b-a + c? 
 
 6. Simplify a — {6 — a-[a — 6— (2a + 6) + (2a-6)-a]-6}. 
 
 7. A grocer sold m pounds of sugar at a cents a pound, and 
 a pounds of tea at 6 cents a pound. If the sugar cost him b cents 
 a pound and the tea m cents a pound, what was his gain by the 
 transaction ? 
 
 8. Multiply a^ + 2a^b + 2 ab^ + 6^ by a^-2 a^b + 2 aft^ - 6^ 
 
 9. Multiply a;"-^ — 2 2/"-^ by 2a; + 2/^. 
 
 10. Multiply xVx-i-xVy + y^/x + y^y by y/x — Vy- 
 
 a a+b a a+b 
 
 11. Multiply 2x^-5y'^ by 2 a^^ + 5 y~^. 
 
 12. Expand (of - y"") (x" + y") (ct^" + y^). 
 
 13. Divide x* — y*hy x — y. 
 
GENERAL REVIEW 317 
 
 14. Divide ic* — 3 a^ — 20 by a; — 2, using detached coefficients. 
 
 15. Prove that x^—b^ is divisible by x + 6. 
 
 16. Divide {a-\-b) + x by (d + 6)* + ici 
 
 17. Factor 9x^-12 a; + 4. 
 
 18. Factor 9ar' + 9a;4-2. 
 
 19. Factor a^ — 3 a; + 2. 
 
 20. Prove that a; — a is a factor of a;" + 3 ax"'^ — 4 a". 
 
 21. Separate a^^ — 1 into six rational factors. 
 
 22. Factor 4 (od + bcf -(a^-b^-(^+ cpf. 
 
 23. Find the H. C. D. of a^ - /, a^ + 2 xy + y% and y^ + xy. 
 
 24. Find the H. C. D. of 3a^-a;-2 and 6x2 + a;-2. 
 
 25. Find the H. C. D. of 4 x' - 11 a:^ + 11 x - 12, 
 
 2 X* 4- a:^ — 4 a:^ + 7 a; — 15, and 2 a;* + a^ — a: — 12. 
 
 26. Find the L. C. M. of 4 a^ftx, 6 aWf, and 2 oxy. 
 
 27. Find the L. C. M. of a:^ — y^, x-\-y, and xy — y^. 
 
 28. Reduce — — — to its lowest terms. 
 
 3ar^-4a;-4 
 
 29. Reduce —-^ „ to its lowest terms. 
 
 x^ - 2 aj^* + 1 
 
 ^2 yi g2 2 be 
 
 30. Reduce — to its lowest terms. 
 
 a2-62 + c2 + 2ac 
 
 31. Simplify -^ - -^- + ^ 
 
 X + 1 1 — X X? — 1 
 
 32. Simplify -^±|- + -£::i|- + 4^. 
 ^ •^2a;-22/ 2a; + 22/ y^-a;* 
 
 ^^" ^°^P ^ ^ (a - 6) (6 _ c) ~ (c - 6)(c - a) "^ (c- a)(a- 6)* 
 
 34. Simplity(a + l)(a' + i)(a-l^ 
 
 35. Simplify = = — 
 
 X-\-- X 
 
 X X 
 
8 
 
 
 GENERAL REVIEW 
 
 
 
 36 
 
 Simplify 
 
 [ " 11 M 
 
 1 + 1 «' + ! 
 
 X 
 
 H- 
 
 X 
 
 - x~ 
 
 1 
 
 
 1-1 
 
 I X 
 
 x-1 
 
 37. 
 
 Prove that ^x£ = ^. 
 h q hq 
 
 
 
 38. 
 
 Prove tlial 
 
 a m an 
 b n bm 
 
 
 
 
 
 39. Divide ^-^ by -1— J-. 
 
 V^ V^ A/y Vx 
 
 40. Raise a — & to the seventh power. 
 
 41. Expand (2 a 4- 3 b)*. 
 
 42. Expand ( V^ + -^f. 
 
 43. Square Va + 6 — c. 
 
 44. Extract the square root of a^+2 aVab-\-3 ab+2 bVc^+b\ 
 
 45. Extract the cube root of 8 a^ - 36 a^b + 54 ab^ - 27 b^ 
 
 46. Extract the square root of a + & to four terms. 
 
 47. Find the sixth root of 4826809. 
 
 48. Reduce Vf to its simplest form. 
 
 49. Reduce V25 a* to its simplest form. 
 
 50. Find the approximate value of • 
 
 51. Multiply 2 + a/8 by 1 - V2. 
 
 52. Simplify ^ + ^ A 
 
 V6 + 2 
 
 53. Prove that oiP = l. 
 
 54. Prove that ax'" = — • 
 
 ar 
 
 55. Prove that x^ = i^c? ; also that x^ = (-^f. 
 
 56. Findthevalueof 125^; of (^—Vl 
 
 ' U2; 
 
GENERAL REVIEW 
 
 319 
 
 Solve the following equations : 
 
 59. 
 60. 
 
 61. 
 62. 
 
 68. 
 
 69. 
 
 70. 
 
 71. 
 
 72. 
 
 73. 
 
 74. 
 
 57. 
 
 58. 
 
 + 
 
 + 
 
 a — b 
 
 X 
 
 1 
 
 X 
 
 b 
 
 + 
 
 a-\-b 
 
 1 
 
 X 
 
 c 
 
 + 
 
 a + b 
 
 = 0. 
 
 mx^ — nx = mn. 
 
 X* + - = 
 
 2 2 
 
 (1 + xy+ (1 - xy= 242. 
 1 + a; 
 
 1 +a;+Vl + a^ 
 
 67. 
 
 r a; + 2/ = 8, 
 
 ^ + z = 4, 
 [ z -{- X = 6. 
 
 fl 1 ..^ 
 
 - + - = 10, 
 \ X y 
 
 \'- + ' = 10. 
 
 [x y 
 
 {2x-\-^y + z = % 
 \x + 2y + Zz = 13, 
 [Sx-\- y + 2z = ll. 
 
 (ax + y + z = 2(a + l), 
 I x + ay + z = 3a + l, 
 [ a; + 2/ -f az at a* -j- 3. 
 
 ■ ar^ + a^ = 24, 
 .y^ + xy = 12. 
 
 r a^ + 3 07/ = 7, 
 la^ + 42/2^13^ 
 
 a^?/ + a^ = 6, 
 a;3 + / = 9. 
 
 63. 
 64. 
 
 65. 
 
 66. 
 
 3. 
 
 75. 
 
 y/x — 9 = V^ — 1. 
 
 ar^ + Va^ + 16 = 14. 
 
 x + ar^ + (1 + a; + a^)2 = 55. 
 1 + a; 
 
 76. 
 
 77. 
 
 78. 
 
 79. 
 
 80. 
 
 81. 
 
 82. 
 
 83. 
 
 84. 
 
 1 — a; + Vl +a^ 
 ' a^ + a; = 26 — ?/2 - 2/, 
 an/ = 8. 
 
 |V^=12, 
 
 {x-\-y — Va; + y = 20. 
 [a:2-2/=' = 7, 
 
 a^ - 2/* = 175. 
 
 a^ - a^2 = - 6, 
 
 a; — a^ = 9. 
 
 xy=zx + y, 
 
 ar' + 2/2 ^ 8. 
 
 ar*?/" — 4 a;^/ = 5, 
 
 x-2 + 42/2 = 29. 
 
 2a^ + 2/ = 9a!2/, 
 
 a; + 2/ = 3. 
 
 a^ + a;?/ + / = 189, 
 
 X + Va^ -f 2/ = 21. 
 ■ a;^ + 2/' = 4, 
 , iK^ + 2/ = 16. 
 j V^ - V2/ = f (» - y), 
 
320 GENERAL REVIEW 
 
 85. A and B hired a carriage for themselves and four friends. 
 If all had paid, A and B would each have had 4 dollars less to 
 pay. What was the cost of hiring the carriage ? 
 
 86. What number is that to which if 12 is added and from 
 ^ of the smn 12 is subtracted, the remainder is 12 ? 
 
 87. A grocer has two kinds of sirup worth 50 and 80 cents 
 per gallon respectively. How many gallons of each must he take 
 to make a mixture of 45 gallons worth 60 cents a gallon ? 
 
 88. How many dimes and how many quarters must be taken 
 so that 18 coins are worth $ 3 ? 
 
 89. In a certain weight of gunpowder the saltpeter was 5 
 pounds more than half the weight, the sulphur 2 pounds less 
 than a fifth, and the charcoal 1 pound more than a tenth. Find 
 the number of pounds of each. 
 
 90. How far down a river whose current runs 3 miles an hour 
 can a steamboat go and return in 8 hours, if its rate of sailing in 
 still water is 12 miles an hour ? 
 
 91. A woman being asked what she paid for her eggs, replied, 
 " Six dozen cost as many cents as I can buy eggs for 32 cents." 
 What was the price per dozen ? 
 
 92. A gentleman had not room in his stables for 8 of his 
 horses, so he built an additional stable ^ the size of the other, 
 and then had room for 8 horses more than he had. How many 
 horses had he ? 
 
 93. In a mass of copper, lead, and tin, the copper was 5 
 pounds less than half the whole in weight, and the lead and tin 
 each 5 pounds more than \ of the remainder. Find the weight 
 of each. 
 
 94. At what time between 4 and 5 o'clock do the hands of 
 a clock make a straight line ? 
 
 95. A person who can walk n miles an hour has a hours at his 
 disposal. How far may he ride in a coach that travels m miles 
 an hour and return on foot within the. allotted time ? 
 
GENERAL REVIEW 321 
 
 96. A merchant sold half a car load more than half his grain ; 
 then he sold half a car load more than half the remainder, when 
 he found that if he could sell half a car load more than half of 
 what he still had, he would have none left. How many car 
 loads of grain had he ? 
 
 97. Four years ago A's age was \ of B's, and 4 years hence 
 it will be I of B's age. What is the age of each ? 
 
 98. A person being asked the time of day replied that the 
 time past noon was f of the time to midnight. What was the 
 time of day ? 
 
 99. If 3 is added to each term of a certain fraction, the value 
 of the fraction will be |^ ; if 3 is subtracted from each term, the 
 value will be |. What is the fraction ? 
 
 100. A boatfaian rows such a distance down a stream that it 
 takes him 4 hours to return. If it takes him 2 hours to row 
 down and the current is 2 miles an hour, what is his rate of 
 rowing in still water ? 
 
 101. A man received $ 2.50 per day for every day he worked, 
 and he agreed to forfeit $ 1.50 for every day he was idle. If he 
 worked 3 times as many days as he was idle and received $ 24, 
 how many days did he work ? 
 
 102. A jeweler has two silver cups, and a cover worth $ 1.50. 
 The first cup with the cover on it is worth 1^ times as much as 
 the second cup, and the second cup with the cover on it is worth 
 \\ as much as the first cup. Find the value of each cup. 
 
 103. Some smugglers discovered a cave that would exactly 
 hold their cargo, which consisted of 13 bales of cotton and 33 
 casks of wine. While they were unloading, a revenue cutter 
 hove in sight, and they sailed away with 9 casks and 5 bales, 
 leaving the cave two thirds full. How many bales, or how many 
 casks, would the cave hold ? 
 
 104. Twenty-eight tons of goods are to be transported in carts 
 and wagons, and it is found that it will require 15 carts and 12 
 wagons, or else 24 carts and 8 wagons. How much can each cart 
 and each wagon carry ? 
 
 ALG. — 21 
 
322 GENERAL REVIEW 
 
 105. There is a number whose three digits are the same. If 7 
 times the sum of the digits is subtracted from the number, the 
 remainder is 180. What is the number ? 
 
 106. A and B can do a piece of work in m days, B and C in w 
 days, A and C in p days. In what time can all together do it ? 
 How long will it take each alone to do it ? 
 
 107. Two passengers together have 400 pounds of baggage 
 and are charged, for the excess above the weight allowed free, 40 
 and 60 cents respectively. If the baggage had belonged to one 
 of them, he would have been charged $ 1.50. How much baggage 
 is one passenger allowed without charge ? 
 
 108. Divide 20 into two parts such that the sum of the two 
 fractions formed by dividing each part by the other is 4^. 
 
 109. It takes 1000 square tiles of a certain size to pave a hall, 
 or 1440 square tiles whose dimensions are one inch less. Find 
 the area of the hall floor. 
 
 110. The sum of two numbers is 16, and the difference of their 
 squares is 128. What are the numbers ? 
 
 111. Find two numbers such that their sum, their product, and 
 the difference of their squares are all equal. 
 
 112. Divide 25 into two parts such that the difference of their 
 square roots is 1. 
 
 113. The difference of two numbers is 6, and their product 
 is equal to twice the cube of the less number. What are the 
 numbers ? 
 
 114. It took a number of men as many days to pave a side- 
 walk as there were men. Had there been 3 men more, the work 
 would have been done in 4 days. How many men were there ? 
 
 115. The product of two numbers is 8, and the sum of their 
 squares is 14 greater than the sum of the numbers. What are 
 the numbers ? 
 
 116. A rectangular lawn 50 feet long and 40 feet wide has a 
 walk of uniform width around it. If the area of the walk is 64 
 square yards, what is its width ? 
 
GENERAL REVIEW 323 
 
 117. A merchant sold goods for 56 dollars and gained as many 
 hundredths of the cost as there were dollars in the cost. Find 
 the cost of the goods. 
 
 118. A person swimming in a stream that runs 1^ miles per 
 hour finds that it takes him 3 times as long to swim a certain 
 distance up the stream as it does to swim the same distance down. 
 What is his rate of swimming in still water ? 
 
 119. A drover bought some oxen for $ 900. After 5 had died, 
 he sold the rest at a profit of $ 20 each and thereby gained $ 350. 
 How many oxen did he buy ? 
 
 120. A detachment from an army was marching in regular 
 column with 5 men more in depth than in front. On approaching 
 the enemy, the front was increased by 845 men, and the whole 
 was thus drawn up in 5 lines. Find the number of men in the 
 detachment. 
 
 121. A round iron bar weighed 36 pounds. If it had been 
 1 foot longer and of uniform diameter, each foot of it would have 
 weighed \ a pound less. Find the length of the iron bar and its 
 weight per foot. 
 
 122. A farmer has two cubical granaries. The side of one is 
 3 yards longer than the side of the other, and the difference of 
 tlieir solid contents is 117 cubic yards. What is the length of 
 the side of each ? 
 
 123. Two workmen, A and B, were employed at different 
 wages. At the end of a certain number of days A received $ 30, 
 but B, who had been idle two days in the meantime, received 
 only $ 19,20. If B had worked the whole time, and A had been 
 idle two days, they would have received equal sums. Find the 
 number of days, and the daily wages of each. 
 
 124. By traveling 5 miles an hour less than its usual rate a 
 train was 50 minutes late in running 300 miles. Find the usual 
 rate of speed and the time usually required to make the trip. 
 
 125. Find two numbers such that their sum, their product, and 
 the sum of their squares are all equal. 
 
324 GENERAL REVIEW 
 
 126. A merchant bought two lots of tea, paying for both $34. 
 One lot was 20 pounds more than the other, and the number of 
 cents paid per pound was in each case equal to the number of 
 pounds bought. How many pounds of each did he buy ? 
 
 127. A and B hired a pasture into which A put 4 horses, and 
 B as many as cost him 18 shillings per week. Afterward B put 
 in 2 additional horses, and found that he must pay 20 shillings 
 per week. How much was paid for the pasture per week ? 
 
 128. By lowering the selling price of apples 1 cent a dozen, 
 an apple woman finds that she can sell 60 more than she used to 
 sell for 60 cents. At what price per dozen did she sell them at 
 first? 
 
 129. A and B are two stations 300 miles apart. Two trains 
 start at the same time, one from A, the other from B, and travel 
 to the opposite station. If the first train reaches B 9 hours after 
 the trains meet, and the second train reaches A 4 hours after 
 they meet, when do they meet, and what is the rate of each train ? 
 
 130. If a carriage wheel 14| feet in circumference takes one 
 second longer to revolve, the rate of traveling will be 2| miles 
 less per hour. How fast is the carriage traveling ? 
 
 131. A railway train, after traveling 2 hours, was detained 1 
 hour by an accident. It then proceeded at f of its former rate, 
 and arrived 7f hours behind time. If the accident had occurred 
 50 miles farther on, the train would have arrived 6^ hours behind 
 time. What was the whole distance traveled by the train ? 
 
 132. A person rents a certain number of acres of land for 
 $ 200. He retains 5 acres for his own use and sublets the rest 
 at $ 1 an acre more than he gave. If he receives $ 10 more than 
 he pays for the whole, how many acres does he rent, and at what 
 rate per acre ? 
 
 133. A and B left Chicago and walked in the same direction 
 at uniform rates. B started 2 hours after A and overtook him 
 at the 30th milestone. Had each traveled half a mile more per 
 hour, B would have overtaken A at the 42d milestone. At what 
 rate did each travel ? 
 
RATIO AND PROPOKTION 
 
 308. 1. What is the relation of 10 a; to 5 a; ? of 3 a; to 12 a; ? of 
 8a to 2a? ofl4mto7m? of 4a to 8a? of 26 to 66? 
 
 2. In finding the relation, or ratio, of 10 a to 5 a, which is the 
 dividend, the number that precedes, or the number that follows f 
 Which is the divisor ? 
 
 3. What is the ratio of a to 6 ? Since 6 may not be exactly- 
 contained in a, how may the ratio be expressed ? 
 
 4. Since the ratio of two numbers may be expressed in the 
 form of a fraction, what operations may be performed upon the 
 terms of a ratio without changing the ratio ? 
 
 309. The relation of two numbers that is expressed by the 
 quotient of the first divided by the second is called their Ratio. 
 
 310. The Sign of Ratio is a colon (:). 
 
 A ratio is also expressed in the form of a fraction. 
 
 The ratio of a to 6 is written a : 6 or -• 
 
 h 
 The colon is sometimes regarded as derived from the sign of division by 
 omitting the line. 
 
 311. The first term of a ratio is called the Antecedent. 
 It corresponds to a dividend, or numerator. 
 
 312. The second term of a ratio is called the Consequent. 
 It corresponds to a divisor, or denominator. 
 
 313. The antecedent and consequent form a Couplet. 
 
 In the ratio a : 6, or -, a 
 b 
 
 terms a and b form a couplet. 
 
 In the ratio a : 6, or -, a is the antecedent, h the consequent, and the 
 b 
 
 325 
 
326 RATIO AND PROPORTION 
 
 314. The ratio of the reciprocals of two numbers is called the 
 Reciprocal, or Inverse Ratio of the numbers. 
 
 It may be expressed by interchanging the terms of the ratio 
 of the numbers. 
 
 The inverse ratio of a to 6 is - : - . Since --=-- = -, the inverse ratio of 
 ^ ah aba 
 
 a to b may be written - , or 6 : a. 
 a 
 
 315. The ratio of the squares of two numbers is called the 
 Duplicate ratio ; the ratio of their cubes, the Triplicate ratio ; the 
 ratio of their square roots, the Subduplicate ratio; the ratio of 
 their cube roots, the Subtriplicate ratio of the numbers. 
 
 The duplicate ratio of a to 6 is a^ : b^ ; the triplicate ratio, a^ : b^; the 
 subduplicate ratio, Va : y/b ; the subtriplicate ratio, \/a : y/b. 
 
 316. Principle. — Multiplying or dividing both terms of a ratio 
 by the same number does not change the ratio. 
 
 Examples 
 
 1. What is the ratio of 8 m to 4 m ? of 4 m to 8 m ? 
 
 2. Express the ratio of 6 : 9 in its lowest terras; 12x:lQy\ 
 am : bm ; 20 ab: 10 be ; (m + w) : ('^n^ — n^)- 
 
 3. Which is the greater ratio, 2:3 or 3:4? 4:9 or 2:5? 
 
 4. What is the ratio oi^to\? | to ^ ?' | to f ? 
 Suggestion. — When fractions have a common denominator, they have 
 
 the ratio of their numerators. 
 
 5. Reduce a : b and x:y to ratios having the same consequent. 
 
 6. When the antecedent is 6 a; and the ratio is ^-, what is the 
 consequent ? 
 
 317. It is evident from § 316, that the ratio of two rational 
 fractions may be expressed by the ratio of two integers. 
 
 Eor example, — ; _ may be reduced to the form — xny: -xny, 
 , ?i V n y 
 
 or my : bn. ^ ^ 
 
 But the ratio of two numbers, when one is rational and the 
 other irrational or when they are dissimilar surds, cannot be 
 expressed by the ratio of two integers. 
 
 Thus, the ratio V2 : 3 cannot be expressed by any two integers- 
 
RATIO AND PROPORTION 327 
 
 318. If the ratio of two numbers can be expressed by the ratio 
 of two integers, the numbers are called Commensurable Numbers, 
 and their ratio a Commensurable Ratio. 
 
 319. If the ratio of two numbers cannot be expressed by the 
 ratio of two integers, the numbers are called Incommensurable 
 Numbers, and their ratio an Incommensurable Ratio. 
 
 • r \/2 1.414213+ 
 The ratio v2 : 3 = -— = — — cannot be expressed by any two 
 
 o o 
 
 integers, because there is no number that, used as a common measure, will 
 be contained in both V2 and 3 an integral number of times. Hence, v^ and 
 3 are incommensurable, and \/2 : 3 is an incommensurable ratio. 
 
 It is evident that by continuing the process of extracting the square root 
 of 2, the ratio V2 : 3 may be expressed by two integers to any desired de- 
 gree of approximation, but never with absolute accuracy. 
 
 320. 1. What two numbers have the same relation to each 
 other as 2 to 3 ? 
 
 2. Name several couplets that express the same ratio as 2:5. 
 How may it be indicated that the ratio of 2 to 5 is the same as 
 that of 2 a to 5 a ? 
 
 3. What number has the same ratio to 12 a that 5 6 has to 3 6 ? 
 
 4. What number has the same ratio to 10 a that 10 a has to 2 ? 
 How does this number times 2 compare with 10 a? 
 
 321. An equality of ratios is called a Proportion. 
 
 3 : 10 = 6 : 20 and a :x = b : y are proportions. 
 
 The double colon (: ;) is often used instead of the sign of 
 equality. 
 
 The double colon has been supposed to represent the extremities of the 
 lines that form the sign of equality. 
 
 The proportion a : b = c : d, ov a : b : : c : d, is read, the ratio of a 
 to b is equal to the ratio of c to d, or a is to 6 as c is to d. 
 
 322. The antecedents and consequents of the ratios that form 
 a proportion are called the Antecedents and Consequents, respec- 
 tively, of the proportion. 
 
 In a : b = r. : d, the antecedents of the proportion are a and c, and the 
 consequents are b and d. 
 
328 RATIO AND PROPORTION 
 
 323. The first and fourth terms of a proportion axe called the 
 Extremes of the proportion. 
 
 In the proportion a : 6 = c : (Z, the extremes are a and d. 
 
 324. The second and third terms of a proportion are called the 
 Means of the proportion. 
 
 In the proportion a -.h = c -.d, the means are h and c. 
 
 325. The terms of a proportion are also called Proportionals. 
 In the proportion a: b = b : c, b is called a Mean Proportional 
 between a and c, and c is called a Third Proportional to a and b. 
 
 In the proportion a : & = c : d, d is called a Fourth Proportional 
 to a, b, and c. 
 
 Since a proportion is an equality of ratios each of which may 
 be expressed as a fraction, a proportion may be expressed as an 
 equation each member of which is a fraction. Hence, it follows 
 that: 
 
 326. General Principle. — The changes that may be made 
 upon a proportion without destroying the equality of its ratios are 
 based upon the changes that may be made upon the members of an 
 equation without destroying their equality and upon the terms of 
 a fraction without altering the value of the fraction. 
 
 PRINCIPLES OP PROPORTION 
 
 327. 1. Let any four numbers form a proportion, as a : 6 = c : d. 
 
 2. Express the proportion as a fractional equation. 
 
 3. If this equation is cleared of fractious, what terms of the 
 proportion does the first member contain ? the second member ? 
 
 Principle 1. — In any proportion the product of the extremes 
 is equal to the product of the means. 
 
 If a : b = c : d, then, ad = fee. 
 
 Since a mean proportional serves as both means of a proportion, 
 if a:b = b: c, b^ = ac, ov b = ^/ac. Hence, 
 
 The mean proportional between two numbers is equal to the square 
 root of their product. 
 
RATIO AND PROPORTION 329 
 
 Principle 1 may be established as follows : 
 Let a -.b = c :d represent any proportion. 
 
 Then, §310, J = |. 
 
 a 
 
 Clearing of fractions, ad = he. 
 
 Therefore, the product of the extremes is equal to the product of the 
 means. 
 
 NuMEKicAL Illustration 
 
 2 : 5 = 6 : 15. 
 
 2 X 15 = 5 X 6. 
 
 30 = 30. 
 
 328. 1. Transform the proportion a:6 = c:d in accordance 
 with Prin. 1. 
 
 2. Since ad = he, how may the value of a be found ? the value 
 of d ? What terms of the proportion are a and d ? 
 
 3. How, then, may either extreme of a proportion be found ? 
 How may either mean be found ? 
 
 Principle 2. —JE^iY^er extreme of a proportion is equal to the 
 product of the means divided by the other extreme. Either mean is 
 equal to the product of the extremes divided by the other mean. 
 
 If a:b = c:d, then, a = — , b = —, etc. 
 
 d c 
 
 Demonstrate Prin. 2, and give numerical illustrations. 
 
 329. 1. If ad = be, what will be the resulting proportion when 
 both members are divided by bd and reduced ? 
 
 2. What do the factors of ad, the first member of the equation, 
 form in the proportion ? What do the factors of be form ? 
 
 Principle 3. — If the produet of two numbers is equal to the 
 product of two other numbers, one pair of them may be made the 
 extremes and the other pair the means of a proportion. 
 
 If ad = be, then, a : b = c : d, or b : a = d : c, etc. 
 
 Demonstrate Prin. 3, and give numerical illustrations. 
 
830 RATIO AND PROPORTION 
 
 330. 1. Change the proportion a:b = c:d into an integral 
 equation by Prin. 1. 
 
 2. Divide the members of this equation by cd and reduce. 
 
 3. What terms of the given proportion now form the first 
 couplet ? the second couplet ? 
 
 Principle 4. — If four numbers are in proportion, the ratio 
 of the antecedents is equal to the ratio of the consequents ; that is, 
 the numbers are in proportion by Alternation. 
 
 If a : 6 = c : d, then, a •.c = h -.d. 
 
 Demonstrate Prin. 4, and give numerical illustrations. 
 
 331. 1. Change the proportion a:b = c: d into an integral 
 equation by Prin. 1. 
 
 2. Divide the members of this equation, be = ad, by ac, and 
 reduce. 
 
 3. What change has taken place in the order of the terms of 
 each couplet ? 
 
 Principle 5. — If four numbers are in proportion, the ratio 
 of the second to the first is equal to the ratio of the fourth to the 
 third ; that is, the numbers are in proportion by Inversion. 
 
 K a:h = c:d, then, h : a = d : c. 
 
 Demonstrate Prin. 5, and give numerical illustrations. 
 
 332. 1. Express the proportion a:b = c:d as a fractional 
 equation. 
 
 2. Add 1 to each member and reduce the mixed numbers to 
 fractional form. Write in the form of a proportion. 
 
 3. How may the terms of this proportion be formed from the 
 terms of the given proportion ? 
 
 4. Since, when a:b = c:d,b:a = d:c, if the changes just 
 indicated are made in the second proportion, how may the terms 
 of the resulting proportion be obtained from the terms of the 
 original proportion? 
 
RATIO AND PROPORTION 331 
 
 Principle 6. — If four numbers are in proportion, the sum 
 of the terms of the first ratio is to either term of the first ratio as 
 the sum of the terms of the second ratio is to the corresponding 
 term of the second ratio ; that is, the numbers are in proportion by 
 Composition. 
 
 If a lb = c :d, then, a-{-b:b=:c + d:d and a + b : a = c + d : c. 
 
 Demonstrate Prin. 6, and give numerical illustrations. 
 
 333. 1. Express the proportion a:b = c:d as a fractional 
 equation. 
 
 2. Subtract 1 from each member, and reduce the mixed num- 
 bers to fractional form. Write in the form of a proportion. 
 
 3. How may the terms of this proportion be formed from the 
 terms of the given proportion ? 
 
 4. Since, Avhen a : b — c : d, b : a = d : c, if the changes just 
 indicated are made in the second proportion, how may the terms 
 of the resulting proportion be obtained from the terms of the 
 original proportion ? 
 
 Principle 7. — If four numbers are in j^^'oportion, the differ- 
 ence between the terms of the first ratio is to either term of the 
 first ratio as the difference between the terms of the second ratio is 
 to the corresponding term of the second ratio; that is, the numbers 
 are in proportion by Division. 
 
 If a : 6 = c : (?, then, a — b:b = c — d:d and a — b •.a = c — d-.c. 
 
 Demonstrate Prin. 7, and give numerical illustrations. 
 
 334. 1. Change the proportion a:b = c:d according to Prin. 
 6, and also according to Prin. 7, using the same consequents in 
 each transformation. Express in fractional form. 
 
 2. Divide the first equation by the second. 
 
 3. How may the terms of the resulting proportion be formed 
 from the terms of the given proportion ? 
 
 Principle 8. — If four numbers are in proportion, the sum of 
 the terms of the first ratio is to their difference as the sum of the 
 
332 RATIO AND PROPORTION 
 
 terms of the second ratio is to their difference ; that is, the numbers 
 are in proportion by Composition and Division. 
 
 If a : b=c : d, a-\-b : a — b — c+d : c — d and a + b : b — a=c+d : d—c. 
 
 Demonstrate Prin. 8, and give numerical illustrations. 
 
 335. 1. Express the proportion a:b = c:d as a fractional 
 equation. 
 
 2. Raise each member to the nth power. 
 
 3. Also express the nth. root of each member. 
 
 4. How may these proportions be formed from the given 
 proportion ? 
 
 Principle 9. — If four numbers are in proportion, their like 
 
 powers, and also their like roots, tvill be in proportion. 
 
 11 11 
 If a :b = c :d, then, a" : 6" = c» : d" and a" : 6" = c" : d". 
 
 Demonstrate Prin. 9, and give numerical illustrations. 
 
 336. 1. Express a : 6 = c : d as a fractional equation. 
 
 2. What may be done to a fraction without changing its value ? 
 
 3. Multiply the terms of the first fraction by m and the terms 
 of the second fraction by n. Write as a proportion. 
 
 4. How may the terms of this proportion be formed from the 
 terms of the given proportion ? 
 
 5. Take the given proportion by alternation and multiply the 
 terms of the first couplet by m and those of the second couplet 
 by w. 
 
 6. How may the terms of this proportion be formed from the 
 terms of the given proportion ? 
 
 7. How may the given proportion be formed from the propor- 
 tions ma : mb = nc: nd and ma : nb = mc : nd ? 
 
 Principle 10. — In a proportion, if both terms of a couplet, or 
 both antecedents, or both consequents are multiplied or divided by 
 the same number, the resulting four numbers form a proportion. 
 
 If a : b = c : d, then, ma : mb — nc: nd and ma : nb = mc : nd ; also, If 
 ma : mb = nc : nd, or if ma : nb = mc: nd, then, a -.b = c: d. 
 
 Demonstrate Prin. 10, and give numerical illustrations. 
 
RATIO AND PROPORTION 333 
 
 337. 1. Express the proportions a:b = c:d and x:y = z:w 
 as fractional equations. 
 
 2. How may two equations be combined (Ax. 4 and 5) ? Com- 
 bine these two equations and write the results as proportions. 
 
 3. How may these proportions be formed from the given 
 proportions ? 
 
 Principle 11. — The products, and also the quotients, of corre- 
 sponding terms of two proportions form a proportion. 
 
 K a -.b = c :d and x : y = z : lo, ax :by = cz : dw, also _:__ = _:_. 
 
 X y z w 
 
 Demonstrate Prin. 11, and give numerical illustrations. 
 
 338. If a:h = c:d and c:d = e:f how does the ratio a : b 
 compare in value with the ratio e : f? 
 
 Principle 12. — If two proportions have a common couplet, the 
 other tivo couplets will form a proportion. 
 
 If a -.6 = :d and c :d = e if, then, a : ft = e : /. 
 
 Demonstrate Prin. 12, and give numerical illustrations (Ax. 1). 
 
 339. A proportion that consists of three or more equal ratios 
 is called a Multiple Proportion. 
 
 2:4 = 3:6 = 5: 10 and a .b = c :d = e -.f are multiple proportions. 
 
 340. A multiple proportion in which each consequent serves 
 also as the antecedent of the following ratio is called a Continued 
 Proportion. 
 
 2:4 = 4:8 = 8: 16 and a :b = b :c = c: d are continued proportions. 
 
 341. 1. Form a multiple proportion, as 
 
 2 : 4 = 3 : 6 = 5 : 10 = 10 : 20. 
 
 2. How does the ratio of the sum of the antecedents to the sum 
 of the consequents compare with the first ratio ? with the second 
 ratio ? with the ratio of any antecedent to its consequent ? 
 
 3. Investigate other multiple proportions. 
 
334 RATIO AND PROPORTION 
 
 Principle 13. — In any multiple proportion the sum of the ante- 
 cedents is to the sum of the consequents as any antecedent is to its 
 consequent. 
 
 Principle 13 may be established as follows : 
 
 Let a :b = c:d = e :/= g : h. 
 
 It is to be proved that 
 
 a + c + e + g:b-{-d + f+h = a:b, or c:d, etc. 
 Denoting each of the equal ratios by r, 
 
 l = r, | = r, l=r, l = r. (1) 
 
 b d f h ^ ^ 
 
 Hence, a = br, c = dr, e—fr, g = hr. (2) 
 
 Adding equations (2), 
 
 a + c + e + g=(b + d + f+h)r. (3) 
 
 Dividing hy (b + d + f+ h), 
 
 a + c+e + g ^^ 
 b+d+f+h 
 
 Replacing r by any of the equal ratios, 
 
 a + c + e + g ^a_c ^^^ 
 b + d+f+h b d' 
 
 That is, a + c + e + g :b + d+f+ h = a:b, OT c-.d, eta. 
 
 Examples 
 342. 1. In the proportion 3 : 5 = x : 55, what is the value of a;? 
 
 First Solution 
 3 : 5 = X : 56. 
 
 Prin. 2, x= ^^ = 33. 
 
 5 
 
 Second Solution 
 3 : 5 = a; : 55. 
 Prin. 10, 3 : 1 = a; : 11, 
 
 Prin. 1, x= 33. 
 
 Find the value of x in each of the following proportions : 
 
 2. 2:3 = 6 :a;. 4. l:a; = a;:9. 
 
 3. 5:« = 4:3. 5. 8:5 = x:10. 
 
RATIO AND PROPORTION 335 
 
 6. « + 1 . a; = 8 : 6. 8. a; + 2 : a; = 10 : 6. 
 
 7. a;: a;- 1 = 15: 12. 9. a; + 2 : a;- 2 = 3 : 1. 
 I 10. If a; + 5 : a; — 5 = 5 : 3, find the value of x. 
 
 11. What two numbers are mean proportionals between 1 
 and 25? 
 
 12. Show that a mean proportional between any two numbers 
 has the sign ± . 
 
 When a : b = c : d, prove that the following proportions are 
 true by deriving them from a:b = c: d: 
 
 13. d:b = c.a. 16. a^ : 6V = 1 : d^ 
 
 ,.„,11 __ & d 
 
 14. c:a = -:— 17. ma : - = mc : ~' 
 
 b a 2 2 
 
 15. b^ : d^ ^ a^ : c^. 18. ac : bd = c^ : d". 
 
 19. Vod : V& = Vc : 1. 
 
 20. a + b : c -\- d = a — b : c — d. p 
 
 21. a" + a% + ab"^ -(- b^ : a^ = <? ->^ <?d -\-cd^ ■\-d^:<?. ^ ' 
 
 22. 2a + 3c:2a-3c = 86 + 12d:86-12rf. 
 
 23. Solve the equation ^' + ^ = — ■ 
 
 ^ oa; 20 
 
 Solution 
 
 gg + x2 _ 41 
 ax 20' 
 
 Dividing by 2, «i±^ = il. 
 
 2 ax 40 
 
 Regarding this equation as a proportion, 
 
 by composition and division, - + ^ «^ + ^g _ 81^ 
 a^ — 2 ax 4- x'^ 1 
 
 - Extracting the square root, '^ = — 
 
 a — X 1 
 
 composition and division, — = — . 
 2x 8 
 
 .-. x = |a. 
 
336 RATIO AND PROPORTION 
 
 24. Solve the equation 
 
 Vaj + T+VcB Va; + 7 - ^^ 
 
 4+Va 
 
 Solution 
 
 ■■\x 
 
 y/x+1 +^/x _ y/x+l -v'x 
 4 + Va; 4 — y/x 
 
 By alternation, Prin. 4, ^^^^^+^ = ^ + ^^. 
 v^a; + 7 - Vx 4 - Vx 
 
 By composition and division, Prin. 8, 
 
 2Vx+7 _ 8 
 2Vx 2Vx 
 Since the consequents are equal, the antecedents are equal. 
 Therefore, 2\/x + 7 = 8. 
 
 Whence, reducing, x = 9. 
 
 25. Given V^' + 11 + 2^ V 2^+14 + 2|_ ^^ ^^^ ^ 
 Vaj + 11 - 2 V2 aj + 14 - 2| 
 
 Solution 
 Vx + 11 + 2 _ ^2 X + 14 + 
 
 v'x+11-2 V2x-fl4 
 By composition and division, Prin. 8, 
 
 2Vx + ll _ 2V2x + 14 
 4 Y 
 
 Dividing both terms of each ratio by 2, Prin. 10, 
 
 Vx + 11 _ \/2x+ 14 
 
 Dividing the consequents by f , Prin. 10, 
 
 Vx + ll _ V2 X+14 
 3 4 
 
 By alternation, Prin. 4, ^x, + U _ 3_ 
 
 V2 X + 14 4 
 
 Squaring, and applying Prin. 7, 
 
 x + ll_ 9 
 
 x + 3 7 
 Solving, X = 25. 
 
RATIO AND PROPORTION 337 
 
 Solve by the principles of proportion ; 
 
 ViB — Vm ^ 
 
 V^-fV2a _2 
 
 26. V^±V^ = ^. 29. 
 
 27. V|W2a^2. 3^_ 
 
 Va; 
 
 + 6 + Va; - 
 
 -&_ 
 
 - d. 
 
 Vx 
 
 + 6- 
 
 4-Va 
 
 Vx- 
 
 1 
 
 a 
 
 IX — 
 
 
 Va 
 
 — X 
 
 
 Vet — Va 
 Voa; — & 
 
 — a; 
 3V( 
 
 26 
 
 28. ^ + V^^13. 3^ 
 
 a; — Va; — 1 ^ ^aa; + h 3 Vaa; + 56 
 
 „- ^ ■ Va; + 2 + Vi + 1 3 , r; , 
 
 32. Given ^ ^ ^ = -, to find x. 
 
 Va; + 2 + Va; — 1 1 
 
 _„ ^ . Va 4- Va + X V6 4- Va; — 6 . n •, 
 
 33. Given ' ^ = ^ to find a;. 
 
 Va — Va + a; V6 — Va; — 6 
 
 /^- a — V2 ax — x^ a — b . n ■, 
 34. Given — = , to find x. 
 
 a+V2aa;-a;2 a + b 
 
 „,, />.• Va; + 1 + Va; — 2 Va; — 3 + -v'a; — 4 . ^ i 
 
 35. Given ^ ' = ' , to find x. 
 
 Va; + 1 — Va; — 2 Va; — 3 — Va; — 4 
 
 36. Divide $ 35 between two men so that their shares shall be 
 in the ratio of 3 to 4. 
 
 37. Two numbers are in the ratio of 3 to 2, and if each is 
 increased by 4, the sums will be in the ratio of 4 to 3. What are 
 the numbers ? 
 
 38. Divide 16 into two parts such that their product is to the 
 sum of their squares as 3 is to 10. 
 
 39. Divide 25 into two parts such that the greater increased 
 by 1 is to the less decreased by 1 as 4 is to 1. 
 
 40. The sum of two numbers is 4, and the square of their sum 
 is to the sum of their squares as 8 is to 5. What are the numbers ? 
 
 41. A dealer had two casks of wine. From the larger he drew 
 34 gallons, and from the smaller 8 gallons, after which their con- 
 tents were as 5 to 4. When half the original contents of each 
 cask had been drawn, he put 8 gallons into the larger and 6 into 
 the smaller. If the ratio of their contents was then 5 to 3, what 
 was the capacity of each ? 
 
 ALG. —22 
 
VARIATION 
 
 343. One quantity or number is said to vary directly as another, 
 or simply to vary as another, when they depend on each other in 
 such a manner that if one is changed the other is changed in the 
 same ratio. 
 
 Thus, if a man earns a certain sum per day, the amount of wages he earns 
 varies as the number of days he works. 
 
 344. The Sign of Variation is cc. It is read ' varies as.' 
 
 345. The expression xc^y means that if x is doubled, y is 
 doubled, or if x is divided by a number, y is divided by the same 
 number, etc. ; that is, that the ratio of cc to 3/ is always the same, 
 or constant. 
 
 If the constant ratio is represented by k, then when ccocy, 
 - = A;, or x = ky. Hence, 
 
 y 
 
 If x varies as y, x is equal to y multiplied by a constant. 
 
 346. One quantity or number varies inversely as another when 
 it varies as the reciprocal of the other. 
 
 Thus, the time required to do a certain piece of work varies inversely as 
 the number of men employed. For, if it takes 10 men 4 days to do a piece of 
 work, it will take 5 men 8 days, or 4 men 10 days, to do it. 
 
 1 . 1 X 
 
 In a; oc -, if the constant ratio of cc to - is Zc, t = ^» or aw = Tc. 
 
 Hence, y 
 
 If X varies inversely as y, their product is constant. 
 
 347. One quantity or number varies jointly as two others when 
 it varies as their product. 
 
 888 
 
VARIATION 339 
 
 Thus, the amount of money a man earns varies jointly as the number of 
 days he works and the sum he receives per day. For, if he should work 
 three times as many days, and receive twice as many dollars per day, he 
 would receive six times as much money. 
 
 In a: oc yz, if the constant ratio of x to yz is A:, 
 
 — ='k, or X — Jcyz. Hence, 
 yz 
 
 If X varies jointly as y and z, x is equal to their prodrict multiplied 
 by a constant. 
 
 348. One quantity or number varies directly as a second and 
 inversely as a third when it varies jointly as the second and the 
 reciprocal of the third. 
 
 Thus, the time required to dig a ditch varies directly as the length of the 
 ditch and inversely as the number of men employed. For, if the ditch were 
 10 times as long and 5 times as many men were employed, it would take 
 twice as long to dig it. 
 
 1 V 
 
 In xccy ' -, or xcc~, if Z; is the constant ratio, 
 z z ' 
 
 x-^'- = ]c, or x = k-- Hence, 
 
 z z ' 
 
 If x varies directly as y and inversely as z, x is equal to ^ multi- 
 plied by a constant. 
 
 349. If X varies as y when z is constant, and x varies as z when 
 y is constant, then x varies as yz lohen both y and z are variable. 
 
 Proof. — Since the variation of x depends on the variations of y and z, 
 suppose the latter variations to take place in succession, each in turn pro- 
 ducing a corresponding variation in x. 
 
 While z remains constant, let y change to yi* thus causing x to change 
 to x'. 
 
 Then, ^=y-. (1) 
 
 a;' 2/1 
 
 Now while y keeps the value yi, let z change to «i, thus causing x' to 
 change to x\. 
 
 Then, *! = £. (2) 
 
 Xi Zi 
 
 * In algebraic notation Xi, X2, Xs, etc., are read 'a; sub one,' 'a; sub two,' 
 'X sub three,' etc. 
 
340 
 
 Multiplying (1) by (2), 
 
 VARIATION 
 
 X 
 
 _ yz 
 
 
 Xi' 
 
 yizi 
 
 
 :. X : 
 
 Xl 
 
 yizi 
 
 yz 
 
 kyz, 
 
 ^1 
 
 where k is the constant 
 
 yizi 
 Hence, x cc yz. 
 
 Thus, the area of a triangle varies as the base when the altitude is con- 
 stant, varies as the altitude when the base is constant, and varies as the 
 product of the base and altitude when both vary. 
 
 Similarly, if x varies as each of three or more numbers, y, z, 
 V, ••• when the others are constant, when all vary x varies as their 
 product. 
 
 That is, xccyzv ••-. 
 
 Thus, the volume of a parallelopiped varies as the length, if the width and 
 thickness are constant ; as the width, if the length and thickness are con- 
 stant ; as the thickness, if the length and width are constant ; as the product 
 of any two dimensions, if the other is constant ; or as the product of the 
 three dimeusions, if all vary. 
 
 ExAMPIiES 
 
 350. 1. li X varies inversely as y, and x = 6 when y = S, what 
 is the value of x when y — 12? 
 
 Solution 
 Since x x -, let k be the constant ratio of x to — 
 
 y y 
 
 Then, § 346, xy = k. (1) 
 
 Hence, when x = 6 and y = 8, 
 
 A; = 6 X 8, or 48. (2) 
 
 Since k is constant. A; = 48 when y = 12. 
 
 Hence, Eq. (1) becomes 12 a; = 48. 
 
 Therefore, when y — \2, x = i. 
 
 2. The volume of a cone varies jointly as its altitude and the 
 square of the aiameter of its base. When the altitude is 15 and 
 the diameter of the base is 10, the volume is 392.7. What is the 
 volume, when the altitude is 5 and the diameter of the base 
 is 20? 
 
VARIATION 341 
 
 • ' Solution 
 
 Let V, H, and D denote the volume, altitude, and diameter of the base, 
 respectively, of any cone, and v the volume of a cone whose altitude is 
 
 5 and the diameter of whose base is 20. 
 
 Since VxHD^, or V=kHD^, 
 
 and V= 392.7 when H=lo and D = 10, 
 
 392.7 = A; X 15 X 100. (1) 
 
 Also, since V becomes v when H=b and D = 20, 
 
 v = kx 5x 400. (2) 
 
 Dividing (2) by (1), -^^ = ^ ^ ^^ = 1 (3) 
 
 .-. -y = f of 392.7 = 523.6. 
 
 3. It xccy and yccz, prove that xacz. 
 
 Proof 
 
 Since xccy and y oc 2, let m represent the constant ratio of x to y, and 
 n the constant ratio of y to 2. 
 
 Then, §345, x = my, (1) 
 
 and y = m. (2) 
 
 Substituting n^ for y in (1), x = mm. (3) 
 
 Hence, since mn is constant, xcc z. 
 
 4. The circumference of a circle varies as its diameter. If 
 the circumference of a circle whose diameter is 1 foot is 3.1416 
 feet, what is the circumference of a circle whose diameter is 
 100 feet? 
 
 5. The area of a circle varies as the square of its diameter. 
 If the area of a circle whose diameter is 10 feet is 78.54 square 
 feet, what is the area of a circle whose diameter is 20 feet ? 
 
 6. The distance a body falls from rest varies as the square 
 of the time of falling. If a stone falls 64.32 feet in 2 seconds, 
 how far will it fall in 5 seconds ? 
 
 7. The area of a triangle varies jointly as its base and alti- 
 tude. The area of a triangle whose base is 12 inches and altitude 
 
 6 inches is 36 square inches. What is the area of a triangle 
 whose base is 8 inches and altitude 10 inches? What is the 
 constant ratio ? 
 
342 VA RIA TION 
 
 8. A wrought iron bar 1 square inch in cross section -and 
 1 yard long weighs 10 pounds. If the weight of a uniform bar 
 of given material varies jointly as its length and the area of its 
 cross section, what is the weight of a wrought iron bar 36 feet 
 long, 4 inches wide, and 4 inches thick ? 
 
 9. The weight of a beam varies jointly as the length, the area 
 of the cross section, and the material of which it is composed. 
 If wood is Y^<j as heavy as wrought iron, what is the weight of a 
 wooden beam 24 feet long, 12 inches wide, and 12 inches thick ? 
 
 10. What is the weight of a brick 2 in. x 4 in. x 8 in., if the 
 material is \ as heavy as wrought iron ? 
 
 11. If 10 men can do a piece of work in 20 days, how long will 
 it take 25 men to do it ? 
 
 12. If a men can do a piece of work in b days, how many men 
 will be required to do it in c days ? 
 
 13. The distances, from the fulcrum of a lever, of two weights 
 that balance each other vary inversely as the weights. If two 
 boys weighing 80 and 90 pounds, respectively, are balanced on the 
 ends of a board 8|^ feet long, how much of the board has each ? 
 
 14. A water carrier carries two buckets of water suspended 
 from the ends of a 4-foot stick that rests on his shoulder. If 
 one bucket weighs 60 pounds and the other 100 pounds, and they 
 balance each other, what point of the stick rests on his shoulder ? 
 
 15. The weight of a body near the earth varies inversely as 
 the square of its distance from the center of the earth. If the 
 radius of the earth is 4000 miles, what would be the weight of a 
 4-lb. brick at the distance of 4000 miles from the earth's surface ? 
 
 16. A boy wishes to ascertain the height of a tower. He 
 knows that it is 31 feet 6 inches from his window to the pave- 
 ment below, and that the distance through which a body falls 
 varies as the square of the time of falling. He drops a marble 
 from his window and finds that it strikes the pavement in 1.4 
 seconds. Then he throws a stone to the top of the tower and 
 observes that it takes just 3 seconds for it to descend. What 
 is the height of the tower ? 
 
VARIATION 343 
 
 17. A horse tied with a rope 45 feet long to a stake in the 
 center of a pasture eats all the grass within reach in 1^ days. If 
 his rope were 15 feet longer, how many days would it take hira to 
 eat all the grass within reach ? 
 
 18. The illumination from a source of light varies inversely 
 as the square of the distance. How far must a screen that is 10 
 feet from a lantern be moved so as to receive one fourth as much 
 light ? 
 
 19. The number of times a pendulum oscillates in a given 
 time" varies inversely as the square root of its length. If a pen- 
 dulum 39.1 inches long oscillates once a second, what is the 
 length of a pendulum that oscillates twice a second? 
 
 20. How long must a pendulum be to oscillate once in three 
 seconds ? 
 
 II 
 
 21. If XCC-, and if x = 2 when y = 12, and 2 = 2, what is the 
 
 value of X when ?/ = 84 and z = l ? 
 
 22. If XCC-, and if aj = 60 when y = 24, and 2 = 2, what is the 
 value of y when cc = 20 and 2 = 7? 
 
 23. If X varies jointly as y and z and inversely as the square 
 of w, and if a; = 30 when ?/ = 3, 2 = 5, and to = 4, what is the 
 value of X expressed in terms of y, z, and w? 
 
 24. If ic«- and ycc-, prove that xccz. 
 
 25. li xccy and zccy, prove that (x ± z)xy. 
 
 26. Three spheres of lead whose radii are 6, 8, and 10 in., re- 
 spectively, are united into one. Find the radius of the resulting 
 sphere, if the volume of a sphere varies as the cube of its radius. 
 
 27. The volume of a cone varies jointly as its altitude and the 
 square of the diameter of its base. The altitudes of three cones, 
 8, P, and E, are 30 ft., 10 ft., and 5 ft., respectively. The 
 diameter of the base of P is 5 ft. and that of E is 10 ft. If the 
 volume of S is equivalent to that of P and E combined, what is 
 the diameter of the base of ^S ? 
 
PROGRESSIONS 
 
 351. 1. How does each of the numbers 2, 4, 6, 8, 10, 12, ••• 
 
 compare with the number that follows it ? How may any term 
 after the first be obtained from the preceding term ? 
 
 2. How may any term of 2, 2^, 3, Z\, ••• after the first be ob- 
 tained from the preceding term ? 
 
 3. Write a series of six terms beginning with a and increasing 
 by a constant number d. 
 
 4. How may any term, after the first, of the series 3, 6, 12, 
 24, 48, •••be obtained from the preceding term ? 
 
 5. How may any term, after the first, of the series 1, ^, \, \, ••• 
 be obtained from the preceding term ? 
 
 6. Write a series of six terms beginning with a and increasing 
 by a constant multiplier r. 
 
 352. A succession of numbers, each of which after the first is 
 derived from the preceding number or numbers according to some 
 fixed law, is called a Series. 
 
 353. The successive numbers are called the Terms of the series. 
 The first and last terms of a series are the Extremes, the inter- 
 vening terms the Means. 
 
 In the series a, a + d, a + 2 d, c + 3 (^, a + 4 d, the terms a and a + id 
 are the extremes and the other terms are the means. 
 
 354. A series consisting of a limited number of terms is 
 called a Finite Series. 
 
 355. A series consisting of an unlimited number of terms is 
 called an Infinite Series. 
 
 344 
 
PROGRESSIONS 345 
 
 ARITHMETICAL PROGRESSION 
 
 356. A series each term of which after the first is den ^ed from 
 the preceding by the addition of a constant number is called an 
 Arithmetical Series, or an Arithmetical Progression. 
 
 357. The number added to any term to produce the next is 
 called the Common Difference. 
 
 2, 4, 6, 8, ••• and 15, 12, 9, 6, ••• are arithmetical progressions. In the 
 first, the common difference is 2 and the series is ascending ; in the second, 
 the common difference is — 3 and the series is descending. 
 
 A. P. is an abbreviation of the word.s Arithmetical Progression. 
 
 358. To find the nth, or last term. 
 
 1. In the arithmetical progression a;, cc + 2, cc + 4, a; + 6, what 
 is the common difference ? How many times does it enter into 
 the second term ? into the third term ? into the fourth term ? 
 
 2. From the first term of the series a, a + d, a + 2 d, a + ^ d, '•' 
 how is the second term formed ? the third term ? the fourth 
 term ? the fifth term ? the nth term, or any term ? 
 
 3. What is the nth. term of the series a, a — d, a — 2 d, ••■? 
 
 369. When a represents the first term of an A.P., d the com- 
 mon difference, I the nth, or last term, and n the number of terms, 
 
 l = a + (n~l)d. (I) 
 
 Examples 
 
 1. What is the 10th term of the series 3, 6, 9, ••• ? 
 
 PROCESS Explanation. —Since the series 3, 6, 9, ••• is an 
 
 Z = a -f (w — l)d A. P. the common difference of whose terms is 3, sub- 
 
 ^ _ 3 ^ (10 1)3 stituting 3 for «, 3 for d, and 10 for n in the formula 
 
 7 ^ oA for the last term, the last term is found to be 30. 
 
 2. Find the 20th term of the series 7, 11, 15, •••. 
 
 3. Find the 16th term of the series 2, 7, 12, •••. 
 
 4. Find the 24th term of the series 1, 16, 31, ••♦. 
 
346 PROGRESSIONS 
 
 5. Find the 18th term of the series 1, 8, 15, •••. 
 
 6. Find the 13th term of the series — 3, 1, 5, •••. 
 
 7. Find the 49th term of the series 1, 1|, If, •••. 
 
 8. Find the 15th term of the series 45, 43, 41, •••. 
 Suggestion. — The common difference is — 2. 
 
 9. Find the 10th term of the series 5, 1, — 3, •••. 
 
 10. Find the 16th term of the series a, 3 a, 5 a, •••. 
 
 11. Find the 12th term of the series a — b, a + 6, a + 3b, •••. 
 
 12. Find the 7th term of the series x ~ 3y, x — 2 y, •••. 
 
 13. A body falls 16^1^ feet the first second, 3 times as far the 
 second second, 5 times as far the third second, etc. How far will 
 it fall during the 10th second ? 
 
 360. To find the sum of n terms of a series. 
 
 1. Express 5 terms of the series a, a -\- d, a -{-2d, •••. 
 
 2. How may the term before the last term be obtained from 
 the last term ? If I represents the last term and d the common 
 difference, what will be the term next to the last ? the second 
 term from the last ? the third term from the last ? 
 
 3. How, then, may the series a, a + d, ••• be written in reverse 
 order, if the last term is I? 
 
 361. Let a represent the first term of an A.P., d the common 
 diiference, / the last term, n the number of terms, and s the sum 
 of the terftis. 
 
 Writing the sum of n terms in the usual order and then in the 
 reverse order, and adding the two equal series, 
 
 s = tt + (a + d) + (a + 2 d) + (a + 3 d) H +1 
 
 s = I -\- (I - d) + (I - 2 d) + (I - 3 d) + -■ + a. 
 
 2 s = (a + + (a + + (a + + (a + + ••• + (« + 0- 
 .\2s = n(a-\-l). 
 
 s='^{a + l),ovn(^'^y (II) 
 
PROGRESSIONS 
 
 347 
 
 Examples 
 
 1. What is the sum of 20 terms of the series 2, 5, 8, ••• ? 
 
 PROCESS 
 
 2 = a + (w - 1) d = ^ + (20 - 1) X 3 = 59 
 
 a-\-l 
 
 2 y V 2 ; 
 
 -±^^= 
 
 610 
 
 Explanation, — Since the last term is not given, it is found by the pre- 
 vious case and substituted for I in the formula for the sum. 
 
 2. What 
 
 3. What 
 
 4. What 
 
 5. What 
 
 6. WTiat 
 
 7. What 
 
 8. What 
 
 9. What 
 10. What 
 
 s the sum of 16 terms of the series 1, 5, 9, ••• ? 
 s the sum of 10 terms of the series — 2, 0, L', ••• ? 
 s the sum of 6 terms of the series 1, 3^, 6, ••• ? 
 s the sum of 8 terms of the series a, 3 a, 5 a, ••• ? 
 s the sum of n terms of the series 1, 7, 13, ••• ? 
 s the sum of a terms of the series x, x -\- 2 a, ••• ? 
 s the sum of 7 terms of the series 4, 11, 18, ••• ? 
 s the sum of 10 terms of the series 1, — 1, — 3, ••• 
 s the sum of 10 terms of the series 1, |, 0, ••• ? 
 
 11. How many times does a common clock strike in 12 hours ? 
 
 12. A body falls 16^ feet the first second, 3 times as far the 
 second second, 5 times as far the third second, etc. Hbw far will 
 it fall in 10 seconds ? 
 
 13. Thirty flower pots are arranged in a straight line 4 feet 
 apart. How far must a lady walk who, after watering each 
 plant, returns to a well 4 feet from the first plant and in line 
 with the plants, assiuning that she starts at the well ? 
 
 14. A boy took a 30-day job on the following terms : he was 
 to receive 5 cents the first day, 10 cents the second day, 15 cents 
 the third day, etc. How much was he paid for the thirtieth day, 
 and what was the whole amount of his earnings ? 
 
348 PROGRESSIONS 
 
 362. The two fundamental formulae, 
 
 (I) l = a + (7i-l)d and (II) s=-(a + 0» 
 
 Li 
 
 contain five elements, a, d, I, n, and s. Since these formulae are 
 independent simultaneous equations, if they contain but two 
 unknown elements they may be solved. Hence, if any three of 
 the five elements are knoivn, the other two may be found. 
 
 Examples 
 
 1. The last term of an A. P. is 58, the common difference is 3, 
 and the sum of the series is 260. Find the number of terms and 
 the first term. 
 
 Solution 
 
 Substituting 58 for Z, 3 for d, and 260 for s in both (I) and (II), 
 
 (I) becomes 58 = a + (n - 1)3. (1) 
 
 (II) becomes 260 = - (a + 58) . (2) 
 
 (1) X n, 58 n = na + 3 w^ - 3 n. (3) 
 
 (2) X 2, 520 = na + 58 n. (4) 
 
 (3) - (4) , 58 n - 520 = 3 n^ - 61 n. (5) 
 
 3n2 - 119 n + 520 = 0. 
 
 (n-5)(3n- 104) = 0. 
 
 .•. « = 5, the number of terms. 
 
 Substituting in (1), a = 46, the first term. 
 
 Since the number of terms must be expressed by a positive integer, frac- 
 tional or negative values of n are rejected. 
 
 2. How many terms are there in the A. P. 2, 5, 8, •••, if the 
 sum is 610 ? 
 
 Solution 
 
 Since a, d, and s are given, and n, but not I, is required, n may be found 
 by eliminating I from (I) and (II) and solving the resulting equation. 
 
 From (I) and (II), l = a+(n-l)d = iLniHi. 
 
 Substituting 2 for c, 3 for d, and 610 for s, and solving, 
 
 n = 20. 
 
PROGRESSIONS 349 
 
 t3. How many terms are there in the series 2, 6, 10, ••• 66 ? 
 4. What is the sum of the series 1, 6, 11, ••• 61 ? 
 
 5. How many terms are there in the series — 1, 2, 5, •••, if 
 the sum is 221 ? 
 
 6. Determine the series 2, 9, 16, ••• 86. 
 
 7. Determine the series — 10, — 8^, — 7, •••to 10 terms. 
 
 8. The sum of the series •••22, 27, 32, ••• is 714. If there 
 ire 17 terms, what are the first and last terms ? 
 
 9. If s = 113§, a = ^, and d = 2, find n. 
 
 10. What is the sum of the series - 16, — 11, — 6, ••• 34 ? 
 
 11. What is the sum of the series ••• —1, 3, 7, ••• 23, if the 
 number of terms is 16 ? 
 
 12. What are the extremes of the series ••• 8, 10, 12, •••, if s = 
 300, and n = 20 ? 
 
 13. How many terms are there in the series 1, 5, 9, ••• V^ 
 
 14. What is the sum of an A. P. whose extremes are x and y, 
 if the number of terms is 6 ? 
 
 363. To insert arithmetical means. 
 
 Examples 
 
 1. Insert 5 arithmetical means between 1 and 31. 
 
 Solution. — Since there are 5 means, there must be 7 terms. Hence, in 
 I = a -\- {n — \) d, Z = 31, a — \, n = 7, and d is unknown. 
 
 Solving, d = 5. 
 
 Or, since there are 6 means, there must be 6 terms after the first. 
 
 Hence, d = 5ijzJ = 5. 
 
 6 
 
 .-. 1, 6, 11, 16, 21, 26, 31, is the series. 
 
 2. Insert 9 arithmetical means between 1 and 6. 
 
 3. Insert 10 arithmetical means between 24 and 2. 
 
 4. Insert 7 arithmetical means between 10 and — 14. 
 
 5. Insert 6 arithmetical means between — 1 and 2. 
 
350 PROGRESSIONS 
 
 6. Insert 14 arithmetical means between 15 and 20. 
 
 7. Insert 3 arithmetical means between a — h and a + 6. 
 
 8. Deduce the formula for the common difference when m 
 arithmetical means are to be inserted between a and I. Find the 
 first mean. 
 
 9. What is the arithmetical mean between 2 and 6 ? between 
 10 and 20 ? between — 3 and 5 ? between a and 6 ? 
 
 364. Principle. — The arithmetical mean between two numbers 
 is equal to half their sum. 
 
 The above principle may be established as follows : 
 
 Let a and b represent any two numbers, and A their arithmetical mean. 
 
 It is to be proved that A = ^ "*" - 
 
 Since the two numbers and their arithmetical mean form the arithmetical 
 progression a, A, b, 
 
 § 356, A — a = b — A, 
 
 2A = a + b. 
 
 .: A = «i-^. 
 2 
 
 Examples 
 Find the arithmetical mean between 
 
 1- fandi ^ ^+Jand^^^. 
 
 X — y X -{- y 
 
 2. a + 6 and a — b. 
 
 5. 1 - a; and N-*- ~ ^) • 
 
 3. (a + 6)2 and (a - 6)1 1 + x 
 
 Problems 
 
 365. Problems in Arithmetical Progression involving two 
 unknown elements commonly suggest series of the form 
 
 X, x + y, x + 2y, x + 3y, etc. 
 
 Frequently, however, the solution of problems is more readily 
 accomplished by representing the series as follows : 
 
 1. When there are three terms, the series may be written 
 
 x-y, X, x + y. 
 
PR0GRE»8I0NS 351 
 
 2. When there are five terms, the series may be written, 
 
 a; - 2 2/, x - y, x, x + y, x + 2 y. 
 
 3. "When there are foxir terms, the series may be written, 
 
 X - 3 y, X - 2/, X + J/, X + 3 y. 
 
 The sum of the terms of a series represented as above evidently contains 
 but one unknown number. 
 
 1. The sum of three numbers in arithmetical progression is 
 30, and the sum of their squares is 462. What are the numbers ? 
 
 Solution 
 Let the series be as — y, x, x + y. 
 
 Then, (x - y) + x + (x + y) = 30, (1) 
 
 and (X - i/)2 + x2 + (X + y)^ = 462. (2) 
 
 From (1), 3x = 30. (3) 
 
 X = 10. (4) 
 
 From (2), 3 x^ + 2 y2 = 462. (5) 
 
 Substituting 10 for X, 2y2 = 162. • (6) 
 
 Solving, y = ± 9. 
 
 Forming the series from x = 10 and y = ± 9, the terms are 
 1, 10, 19, or 19, 10, 1. 
 
 2. The sum of three numbers in arithmetical progression is 
 18, and their product is 120. What are the numbers ? 
 
 3. The sum of three numbers in arithmetical progression is 
 21, and the sum of their squares is 155. What are the numbers ? 
 
 4. There are three numbers in arithmetical progression the 
 sum of whose squares is 93. If the third is 4 times as large as 
 the first, what are the numbers ? 
 
 5. The product of the extremes of an arithmetical progression 
 of 3 terms is 4 less than the square of the mean. What are 
 the numbers, if their sum is 24 ? 
 
 6. The sum of four numbers in arithmetical progression is 14, 
 and the product of the means is 12. What are the numbers ? 
 
 7. The sum of seven numbers in arithmetical progression is 
 98, and the sum of their squares is 1484. What are the numbers ? 
 
852 PROGRESSIONS 
 
 8. The sum of five numbers in arithmetical progression is 15, 
 and the product of the extremes is 3 less than the product of the 
 terms next to the extremes. What are the numbers ? 
 
 9. A number is expressed by three digits in arithmetical pro- 
 gression. If the number is divided by tlie sum of its digits, the 
 quotient is 20^ ; and if the number is increased by 594, the result 
 is the number with its digits in the reverse order. What is 
 the number? 
 
 10. Find the sum of the odd numbers from 1 to 100. 
 
 11. The product of the extremes of an arithmetical progression 
 of 10 terms is 70, and the sum of the series is 95. What are the 
 extremes ? 
 
 12. Fifty-five logs are to be piled so that the top layer shall 
 consist of 1 log, the next layer of 2 logs, the next layer of 3 logs, 
 etc. How many logs must be placed in the bottom layer ? 
 
 13. It cost Mr. Smith f 19.00 to have a well dug. If the cost 
 of digging was $1.50 for the first yard, f 1.75 for the second, 
 $ 2.00 for the third, etc., how deep was the well ? 
 
 14. The product of the extremes of an arithmetical progression 
 of 15 terms is 93, and the sum of the first and last means is 34. 
 What is the progression ? 
 
 15.- How many arithmetical means must be inserted between 
 5 and 37, so that the ratio of the first mean to the last mean 
 may be ^j ? 
 
 16. How many arithmetical means must be inserted between 
 4 and 25, so that the sum of the series may be 116 ? 
 
 17. Prove that the equimultiples of the terms of an arith- 
 metical progression are in arithmetical progression. 
 
 18. Prove that the difference of the squares of consecutive 
 integers are in arithmetical progression, and that the common 
 difference is 2. 
 
 19. Prove that the sum of n consecutive odd integers, beginning 
 with 1, is n\ 
 
PROGRESSIONS 353 
 
 GEOMETRICAL PROGRESSION 
 
 366. A series of numbers each of which after the first is derived 
 by multiplying the preceding number by some constant multiplier 
 is called a Geometrical Series, or a Geometrical Progression. 
 
 2, 4, 8, 16, 32 and a*, a^, a^, a are geometrical progressions. 
 
 In the first series the constant multiplier is 2 ; in the second it is -• 
 
 a 
 
 G. P. is an abbreviation of the words Geometrical Progression. 
 
 367. The constant multiplier is called the Ratio. 
 
 It is evident that the terms of a geometrical progression 
 increase or decrease numerically according as the ratio is numer- 
 ically greater or less than 1. 
 
 368. To find the nth, or last term. 
 
 1. In the geometrical progression 3, 6, 12, 24, what is the 
 ratio of 6 to 3 '/ of 12 to 6 ? of 24 to 12 ? 
 
 2. In the geometrical progression a, ar, ar^, ar', ••• what is the 
 ratio ? How many times does the ratio enter as a factor into the 
 second term ? into the third term ? into the fourth term ? 
 
 369. When a represents the first term of a G. V., r the ratio, 
 and I the last or nth term, 
 
 I = ar^'-K (I) 
 
 Examples 
 1. Find the 9th term of the series 1, 3, 9, •••. 
 
 PROCESS 
 
 Explanation. — In this example a = 1, r = 3, and 
 
 I =zl X 3^ Substituting these values in the formula for I, the 
 
 , r.,-ni last term is 6561. 
 
 ( = 6561 
 
 2. Find the 10th term of the series 1, 2, 4, •••. 
 
 3. Find the 8th term of the series \, ^, 1, •••. 
 '4. Find the 9th term of the series 6, 12, 24, .... 
 
 ALG, — 23 
 
354 PROGRESSIONS 
 
 5. Find the 11th term of the series ^, 1, 2, •••. 
 
 6. Find the 7th term of the series 2, 6, 18, •••. 
 
 7. Find the 6th term of the series 4, 20, 100, .... 
 
 8. Find the 6th term of the series 6, 18, 54, •••. 
 
 9. Find the 10th term of the series 1, ^, \, •••. 
 
 10. Find the 10th term of the series 1, |, ^, •••. 
 
 11. Find the 8th term of the series ^, ^, ^, •••. 
 
 12. Find the 11th term of the series a^%, o}%\ ■-. 
 
 13. Find the nth term of the series 2, V2, 1, •••. 
 
 14. A man worked for 25 cents the first day, 50 cents the 
 second day, f 1 the third day, and so on for 10 days. How much 
 did he receive the tenth day ? 
 
 15. If a man begins business with a capital of $200 and 
 doubles it every year for 6 years, how much will he have at 
 the end of the sixth year? 
 
 16. If the population of the United States is 76 millions in 
 1900 and doubles itself every 25 years, what will it be in the 
 year 2000? 
 
 17. A man's salary was raised \ every year for 5 years. If 
 his salary was $ 512 the first year, what was it the sixth year ? 
 
 18. The population of a city at a certain time was 20,736, and 
 increased in geometrical progression 25% each decade. What 
 was the population at the end of 40 years ? 
 
 19. A man who wanted 10 bushels of wheat thought $1 a 
 bushel too high a price. But he agreed to pay 2 cents for the 
 first bushel, 6 cents for the second, 18 cents for the third, and 
 so on. What did the last bushel cost him ? 
 
 20. From a grain of corn there grew a stalk that produced 
 an ear of 150 grains. These grains were planted, and each pro- 
 duced an ear of 150 grains. This process was repeated until 
 there were 4 harvestings. If 75 ears of corn make 1 bushel, how 
 many bushels were there the fourth year ? 
 
PROGRESSIONS 355 
 
 370. To find the sum of a finite series. 
 
 Let a represent the first term, I the nth. term, or the last term, 
 r the ratio, n the number of terms, and s the sum of the terms. • 
 
 Then, s = a + ar + a/-^ + a?-* -| \- ar'^'K (1) 
 
 (l)xr, rs — ar + ar^ + a7-^ -{- \- ar"~^ -\- ar\ (2) 
 
 (2)-(l), s(r-l)=ar"-a. 
 
 (11) 
 
 
 s = 
 
 r - 
 
 — a 
 
 a(?-" 
 r — 
 
 -1). 
 1 
 
 But, since ar 
 
 "-» = / 
 
 ar 
 
 " = rl. 
 
 
 
 Substituting 
 
 rl for 
 
 ar" 
 
 in (II), 
 
 
 
 
 
 
 rl — a 
 
 a 
 
 -rl 
 
 r-r 
 
 or ^^^-L'. (Ill) 
 
 Examples 
 
 1. Find the sum of 6 terms of the series 3, 9, 27, •••. 
 
 PROCESS 
 f^iyM ^ Explanation. ; — Since the first term a, the 
 
 * — . _ ^ ratio r, and the number of terms n, are given, 
 
 and fonnula II gives the sum in terms of a, r, 
 ^ _ o X o — o _ -j^Qg2 ^.'^'i "> formula II is used. 
 
 2. Find the sum of 8 terms of the series 1, 2, 4, •••. 
 
 3. Find the sum of 8 terms of the series 1, ^, \, •••. 
 
 4. Find the sum of 10 terms of the series 1, 1^, 2\, •••. 
 
 5. Find the sum of 7 terms of the series 2, — |, f, •••. 
 
 • 6. Find the sum of 12 terms of the series — ^, \, ~ j, •••. 
 
 7. Find the sum of 7 terms of the series 1, 2 a:, 4 a^, •••. 
 
 8. Find the sum of 7 terms of the series 1, — 2 a;, 4 a^, •••. 
 
 9. Find the sum of n terms of the series 1, a^, x*, •••. 
 10. Find the sum of n terms of the series 1, 2, 4, •••. 
 
356 PROGRESSIONS 
 
 11. Find the sum of n terms of the series 1, \, \, •••. 
 
 12. The extremes of a geometrical series are 1 and 729, and the 
 ratio is 3. What is the sum of the series ? 
 
 13. What is the sum of the series 3, 6, 12, •••, 192 ? 
 
 14. What is the sum of the series 7, - 14, 28, •-, - 224 ? 
 
 371. To find the sum of an infinite geometrical series. 
 
 If the ratio r is numerically less than 1, it is evident that the 
 successive terms of a geometrical series become numerically less 
 and less. Hence, in an iniinite decreasing geometrical series, the 
 nth term I, or ar^~^, can be made less than any assignable number, 
 though not absolutely equal to zero. 
 
 (Ill) may be written s = — — ^ 
 
 1-r 
 
 Since, by taking enough terms, I and consequently rl can be 
 made less than any assignable number, the second fraction may 
 be neglected. 
 
 Hence, the formula for the sum of an infinite decreasing geomet- 
 »:ical series is 
 
 a 
 
 Examples 
 
 1. Find the sum of the series 1, y^^, y^^, •••. 
 
 Solution 
 Substituting 1 for a and ^^ for r in (IV) , 
 
 s = — 1— = — = — 
 
 2. Find the value of .185185185 .... 
 
 Solution 
 
 Since .185185185 ••• = .185 + .000185 + .000000185 + ..., a = .185 and 
 J- = .001. 
 
 Substituting in (I V) , . 185186185 • . . = s = '^^'^ = — • 
 ^ ^ 1 - .001 27 
 
PROGRESSIONS 357 
 
 Find the value of 
 
 3. 1 + 1 + 1+.... 6. .407407.... 
 
 4. 3 + f + y3^+.... 7. .363636. ... 
 
 5. i_^_i_| . 8. 1.94444 =^ 
 
 372. To insert geometrical means between two terms. 
 
 Examples * •' 
 
 ft 
 
 1. Insert 3 geometrical means between 2 and 162. 
 
 PROCESS Explanation. — Since there are three means, there are 
 
 ; -- (ifn-\ five terms, and n — 1 = 4. Solving for r and neglecting 
 
 1 fio _ 9 ^ imaginary values, r = ± 3. 
 
 IbZ — Z r* Therefore, the series is either 2, 6, 18, 54, 162, or 2, -6, 
 
 r = ± 3 18,-54, 162. 
 
 2. Insert 3 geometrical means between 1 and 625. 
 
 3. Insert 5 geometrical means between 4^ and -fy^. 
 
 4. Insert 4 geometrical means between ^^ and |-|. 
 
 5. Insert 4 geometrical means between 5120 and 5. 
 
 6. Insert 4 geometrical means between 4V2 and 1, 
 
 7. Insert 5 geometrical means between r/ and h^. 
 
 8. Insert 6 geometrical means between — 2 and ^ V2. 
 
 9. Insert 4 geometrical means between x and — y. 
 
 373. Principle. — The geometrical mean between two numbers 
 is equal to the square root of their product. 
 
 The above principle may be established as follows : 
 
 Let a and b represent any two numbers, and G their geometrical mean. 
 It is to be proved that G = y/ab. 
 
 Since the two numbers and their geometrical mean form the geometrical 
 progression a, G, b, 
 
 § 366. ^ = A, 
 
 ^ ^ a G' 
 
 '♦ f '• G^^ab. 
 
 .: G = \/ab. 
 
 V'-N 
 
358 PROGRESSTOIVS 
 
 Find the geometrical meaxi between 
 
 1. 8 and 50. 4. (a + bf and (a - 6)*. 
 
 2. 4^ and 34. ox , ,? 
 2 ^ ^ a^ + ab . ab + b^ 
 ^ , , 1 o 5. — — ' and — -• 
 
 3. 11^ and f. a'-ab ab-b^ 
 
 6. 25 a^ - 10 a; + 1 and x- + 10 a- + 25. 
 
 374. Since formulae I and II, or III, which is equivalent to 
 II, are two independent simultaneous equati)|ns containing five 
 elements, if three elements are known, the other two may be found 
 by elimination. 
 
 Problems 
 376. 1. Given r, I, and s, to find a. 
 
 2. The ratio of a geometrical progression is 5, the last term is 
 625, and the sum is 775. What is the first term ? 
 
 3. The ratio of a geometrical progression is ^, the sum is \, 
 and the series is infinite. What is the first term ? 
 
 4. Find I in terms of a, r, and s. 
 
 5. Find the last term of the series 5, 10, 20, •••, the sum of 
 whose terms is 155. 
 
 6. If 1+ 1^ + ^-1 = 1|. (1 + y^), what is the last term, 
 
 , and the number of terms ? 
 
 7. Deduce the formula for r in terms of a, I, and s. 
 
 8. If the sum of the geometrical progression 32 ••• 243 is 665, 
 what is the ratio ? Write the series. 
 
 9. The sum of a geometrical progression is 700 greater than 
 the first term and 525 greater than the last term. What is the 
 ratio ? If the first term is 81, what is the progression? 
 
 10. Deduce the formula for r in terms of a, n, and I. 
 
 11. The first term of a geometrical progression is 3, the last 
 term is 729, and the number of terms is 6. What is the ratio ? 
 Write the series. 
 
 12. Find I in terms of r, n, and s. 
 
PROGRESSIONS 359 
 
 13. The sum of the 12 terms of a geometrical progression whose 
 ratio is 2 is 4095. What is the 12th term ? 
 
 14. The velocity of a sled at the bottom of a hill is 100 feet 
 per second. How far will it go on the level, if its velocity 
 decreases each second \ of that of the previous second ? 
 
 15. From a cask of vinegar ^ was drawn off and the cask was 
 filled by pouring in water. Show that if this is done 6 times, 
 the contents of the cask will be more than y^^ water. 
 
 16. A ball thrown vertically into the air 100 feet falls and 
 rebounds 40 feet the first time, 16 feet the second time, and so 
 on. What is the whole distance through which the ball will 
 have passed when it finally comes to rest ? 
 
 17. Show that the amount of $ 1 for 1, 2, 3, 4, 5 years at com- 
 pound interest varies in geometrical progression. 
 
 18. Show that equimultiples of numbers in geometrical pro- 
 gression are also in geometrical progression. 
 
 19. The sum of three numbers in geometrical progression is 
 19, and the sum of their squares is 133. WTiat are the numbers ? 
 
 Suggestion. — When there are but three terms in the series tliey may be 
 represented by x^, xy, y^, or by x, Vary, y. 
 
 20. The product of three numbers in geometrical progression 
 is 8, and the sum of their squares is 21. What are the three 
 numbers ? 
 
 21. If 4 is a geometrical mean between two numbers whose 
 sum is 10, what are the numbers ? 
 
 22. The product of three numbers in geometrical progression 
 is 64, and the sum of their cubes is 584. What are the numbers ? 
 
 23. The sum of the first and second of four numbers in geo- 
 metrical progression is 15, and th& sum of the third and fourth 
 is 60. What are the numbers ? 
 
 Suggestion. — Four unknown numbers in geometrical progression may 
 
 a;2 w2 
 
 be represented by — , x, y, — • 
 y 25 
 
 24. The sum of the first and third of three numbers in geo- 
 metrical progression is 130, and their product is 625. What are 
 the numbers ? 
 
360 PROGRESSIONS 
 
 25. Divide $700 among three persons so that the first shall 
 receive $ 300 more than the third, and the share of the second 
 shall be a geometrical mean between the shares of the first and 
 third. 
 
 26. If a, h, and c are in geometrical progression, show that 
 their reciprocals also are in geometrical progression. 
 
 27. The difference between two numbers is 24, and their 
 arithmetical mean exceeds their geometrical mean by 6. "What 
 are the numbers ? 
 
 HARMONICAL PROGRESSION 
 
 376. 1. Examine the series 1, \, \, \, •••. Has it a constant 
 difference ? Has it a constant ratio ? 
 
 2. Take the reciprocal of each term. What kind of a series 
 is thus formed? How, then, may the series 1, \, ^, -^, •••, be 
 described ? 
 
 377. A series the reciprocals of whose terms form an arith- 
 metical progression is called a Harmonical Series, or a Harmonical 
 Progression. 
 
 3, \i 1, f, f) h ■•• is a harmonical progression, because \, f, 1, f, |, 2, ••■ 
 *he reciprocals of its terms form an arithmetical progression. 
 
 H. P. is an abbreviation for the words Harmonical Progression. 
 
 378. Problems in harmonical progression are commonly solved 
 by taking the reciprocals of the terms and employing the prin- 
 ciples of arithmetical progression. There is no general method, 
 however, for finding the sum of the terms of a harmonical pro- 
 gression. 
 
 379. Principle 1. — The harmonical mean between two numbers 
 is equal to twice their product divided by their sum. 
 
 The above principle may be established as follows : 
 £et H represent the harmonical mean between a and 6. 
 It is to be proved that H= ^ 
 
 a + 6 
 
 § 377, -, — , - are in arithmetical progression. 
 
 a H b 
 
PROGRESSIONS 361 
 
 Hence, § 356, 
 
 1 1 _ 1 _1. 
 b H H a 
 
 Clearing of fractions, 
 
 aH-ab^ah- bH. 
 
 Transposing, 
 
 aH+bH=2ab. 
 
 
 .-. if_2«^ 
 
 §364, 
 
 ^- 2 
 
 §373, 
 
 G'=\/^. 
 
 § 379, 
 
 a + 6* 
 
 Multiplying (1) by (3), 
 
 AH^ah. 
 
 Taking the square root, 
 
 yjAH=\'ab. 
 
 From (2) and (5), Ax. 1, 
 
 a = y/AH. 
 
 a + h 
 
 380. Principle 2. — The geometrical mean between two numbers 
 is also the geometrical mean between their arithmetical and harmoni- 
 cal means. 
 
 The above principle may be established as follows : 
 
 (1) 
 
 (2) 
 (3) 
 
 (4) 
 (6) 
 
 Hence, § 373, G is the geometrical mean between A and H. 
 
 Examples 
 
 1. Find the 12th term of the H. P. 6, 3, 2, .... 
 
 Solution. — The reciprocals of the terms form the arithmetical progression 
 
 f) 5> 2» • 
 in which a = J and d = |. 
 
 Substituting ^ for a, J for d, and 12 for n in (I) , 
 
 §359, Z = ^ + (12 -1)1 = 2. 
 
 Therefore, § 377, the 12th terra of the given harmonical progression is J. 
 
 2. Find the 10th term of the harmonical series ^, \, ^, •••. 
 
 3. Insert 6 harmonical means between 1^ and 12. 
 
 4. Insert 2 harmonical means between 2 and 5. 
 
 5. Insert 7 harmonical means between 12^ and 2^. 
 
362 • PROGRESSIONS 
 
 6. Insert 3 hannonical means between b and a. 
 
 7. Find the nth term of the H. P. |, i, j\, .... 
 
 8. The 3d and 4th terms of a H. P. are 2i and 1|-, Write the 
 first 6 terms. 
 
 Find the harraonical mean between 
 
 9. 2 and 3. 13. a — c and a + c. 
 
 10. I and I. 14. 1 — Va and 1 + \^. 
 
 11. 21 and 1^. 15. o and -. 
 
 a 
 
 12. 21 and 10. ^g V6 and V3. 
 
 17. The 5th term of a harmonical progression is -^j, and the 
 ] 1th term is 2^- What is the first term ? 
 
 18. The arithmetical mean between two numbers is 5, and 
 their harmonical mean is 3^. What are the numbers ? 
 
 19. If one number exceeds another by 2, and their arithmetical 
 mean exceeds their harmonical mean by ^, what are the numbers ? 
 
 20. If a, b, and c are in harmonical progression, prove that 
 a — b:b — c = a:c. 
 
 21. If a, &, c, and d are in harmonical progression, prove that 
 ab: cd = b — a:d~c. 
 
 22. If 6 is the harmonical mean between a and c, prove that 
 
 1 111 
 
 + i -=- + 
 
 6 — a b — c a c 
 
 23. When b — a:c — b = a:x, prove that x = a, if a, b, and c 
 are in arithmetical progression ; that x = b, if a, b, and c are in 
 geometrical progression; and that x — c, if a, b, and c are in 
 harmonical progression. 
 
 24. The harmonical mean between two numbers is 5^ and their 
 arithmetical mean is 6^. What is their geometrical mean ? 
 
 25. Prove that x + xy, 2xy, and xy -\- xy^ are in harmonical 
 progression. 
 
 26. If b -\- c, c + a, and a -\-b are in harmonical progression, 
 prove that a?, b'\ c- are in arithmetical progression. 
 
IMAGINARY AND COMPLEX NUMBERS 
 
 381. 1. If from V— 25 the rational factor V25 is removed, 
 what irrational factor remains ? 
 
 2. Simplify V— 25, V— 16, V— a^. What common part, or 
 unit, have the indicated square roots of negative numbers ? 
 
 3. What is the square of Vi? of V5? of V9? of y/2x? 
 What is the effect of squaring a radical of the second degree ? 
 
 What, then, is the square of V— 4 ? of V— a? of V— 1 ? 
 
 382. Up to this point the only numbers whose nature has been 
 discussed have been numbers that differ from arithmetical num- 
 bers in having a sign, + or — , to indicate quality or direction. 
 These numbers are called real numbers, and may be briefly 
 described as numbers ivhose squares are x>ositive. 
 
 There are numbers, however, whose squares are negative. They 
 constitute one class of imaginary numbers, defined in § 257. In 
 this chapter only imaginary numbers of the second degree are 
 treated. 
 
 Let — a be any negative real number. 
 
 Then, V— a will represent an imaginary number. 
 
 Since + V— a = + V« • V— 1 = + V— 1 • Va 
 
 and — V— a = — ^a • V— 1 = — V— 1 • Va, 
 
 the positive imaginary unit is + V— 1, and the negative imagi- 
 nary unit is —V— 1. 
 
 Since the square of the square root of a number is the number 
 itself, 
 
 This relation is sufficient to explain operations with imaginary 
 numbers. 
 
 363 
 
-x ^n 
 
 '^ 
 
 364 IMAGINARY AND COMPLEX NUMBERS 
 
 383. Relation between the units +1, — 1, V— 1, and — V— 1. 
 
 In the accompanying figure the length of any radius of the circle represents 
 the arithmetical unit 1. The line drawn from to A, called the line OA, 
 ^ represents the positive unit + 1, and 
 
 the line OA" represents the nega- 
 tive unit — 1. Every real number 
 lies somewhere on the line X'X, 
 which is supposed to extend indefi- 
 nitely in both directions from 0. 
 
 ' -Y X'Xis called the axis of real num- 
 
 bers. 
 
 The direction of any line drawn 
 from 0, as OB, that is, the quality 
 or direction sign of the number 
 represented by that line, is deter- 
 ^ mined with reference to the fixed 
 
 line OA by finding what part of a revolution is required to swing the line 
 from the position OA to the required position. By common consent revo- 
 lution of the line OA is performed in a direction opposite to that of the hands 
 of a clock, as shown by the arrows. OB is reached after ^ of a revolution, 
 OA' after \ of a revolution, OA" after J of a revolution, etc. 
 
 Since OA", or —1, represents ^ of a revolution of OA, the square of OA", 
 or (— 1)2, represents 1 revolution of OA, which produces OA, or + 1. 
 Hence, OA", or — 1, represents the square root of + 1, or (+ 1)^. 
 
 Similarly, since OA' represents J of ^ of a revolution of OA, and OA" 
 represents ^ of a revolution of OA, OA' represents the square root of OA", 
 or of - 1 ; that is, OA' = V^. 
 
 If OA" is swung J of a revolution from the position OA" to the position 
 OA'", OA" will be multiplied by V— 1 just as OA is multiplied by V— 1 to 
 produce OA'. Hence, the result OA'" = — 1 • V— 1 = — V— 1. 
 
 + 1, represented by OA, and — 1, represented by OA", are the units for 
 real numbers, that is, are real units. Just as the real number + a is repre- 
 sented by a line a units long extending from toward X, and the real num- 
 ber — a by a line a units long extending from O in the opposite direction, so 
 the imaginary number + av^^l, or (-|-V— 1) x a, is represented by a line 
 a units long extending from toward Y, and the imaginary number — a V — 1 , 
 or ( — V— 1) X a, by a line a units long extending from in the opposite 
 direction, toward Y'. Hence, +V— 1 and — V— 1 are the units for imagi- 
 nary numbers, that is, they are, imaginary units ; + aV— 1 is called a posi- 
 tive imaginary number and — a V— 1 a negative imaginary number. 
 
 Y' Y is called the axis of imaginary numbers. If Y' Y is taken as the 
 axis of real numbers, then X'X becomes the axis of imaginary numbers. 
 Hence, it is seen that imaginary numbers have as much reality as real num- 
 bers. Imaginary uumbers were named before their nature was understood. 
 
IMAGINARY AND COMPLEX NUMBERS 
 
 365 
 
 384. In the graphical illustration of the relation between real 
 and imaginary numbers it was assumed that -f 1 • V— 1 = V— 1, 
 or V— 1 • 1 ; that is, the Commutative Law for multiplication was 
 assumed to apply when imaginary numbers were involved. It is 
 evident, if the discussion of number is to proceed, that in any 
 operation imaginary numbeijs must obey all the laws of real num- 
 bers except those which determine the quality of the result ; and 
 that the quality of the result is determined, as far as the imaginary 
 numbers are concerned, by the relation (V— 1)*' = — 1. 
 
 385. Powers of V^^. 
 
 (V^)^ =-1; 
 
 and so on. Hence, if n = or a positive integer, 
 
 ( V^rT)*n+l ^ ^ V"3T ; ( V^T)^"+2 = - 1 ; 
 
 (V^^)*"+^ = -V^^; ( V^^)*"+* = + 1. 
 
 Hence, any even ])oiver o/ V— 1 is real and any odd power is 
 imoffinary. 
 
 386. Operations involving imaginary numbers. 
 
 Examples 
 Find the value of 
 
 2. (V^^)^ 4. {-y/^^y\ 6. (V^^)^'. 8. (-V^^ 
 
 9. Add V— a* and V— 16 a*. 
 
 Solution 
 
 V-a* + V- 16 a* = a^y/^^ + 4 aV^ = 5 aV^^. 
 
366 IMAGINARY AND COMPLEX NUMBERS 
 
 Simplify the following : 
 
 10. V^4-f-V-49. 13. V^n^ + 4V^^. 
 
 11. V^l)+V-64. 14. 5V^18-V^72. 
 
 12. 2V^^ + 3V^^. 15. 3V^^-V^80. 
 
 16. ( V'^^ + SV^b) + ( V^^ - 3 V^^). 
 
 17. (V— 9a;y — V— xy) — (V— 4a;?/ + V— x?/). 
 
 18. V-a^ + V-4ic2-V-ar* + 3a;V- 
 
 X. 
 
 19. V- 16 - 3V^^ + V- 18 + V- 50 + V- 25. 
 
 20. V^8 + aV^2-V-98-5V^=='2^^. 
 
 21. V- 16 aV + V- aV _ ^_ g ^2^ 
 
 22. V1-5-3V1-10 + 2V5-30. 
 
 23. Multiply 3V-10 by 2V- 3. 
 
 PROCESS 
 
 3 V^IO X 2V^^ = 3 VlO V^^ X 2V8V^^ 
 
 = 6V10 X 8 X (- 1) 
 
 = - 6 V80 = - 24 V5. 
 
 ExPLANATiox. — In order to determine the sign of the product, each imagi- 
 nary number is reduced to the form bV — 1. The numbers are then multiplied 
 together as ordinary radicals, observing, however, that V— 1 x V— 1 = — 1. 
 
 24. Multiply V^2 + 3 V^^ by 4 V^^ - V^3. 
 
 First Solution Second Solution 
 
 (V2 + 3V3)(4V2-\/3)(-l) -4V4- 12V6 « 
 
 = (8 + 12V6->/6-9)(-l) +3V9+ V6 
 
 = 1-11^^ ~l _ IIV6 
 
 Multiply : 
 
 25. 3V^5 by 2V^15. 28. SV^l by V^^. 
 
 26. 4V- 27- by V-12. 29. V^l25 by V^^^IOS. 
 
 27. 2V^^ by 5V^^. 30. V^^^^ by V^^^30. 
 
IMAGINARY AND COMPLEX NUMBERS 
 
 367 
 
 31. v^::6+V^^ by V^6-V^^. 
 
 32. V— o6 + V — a by V— ab —^—a. 
 
 33. V —xy + V— a; by V— xy + V— a;. 
 
 34. V^r^-V^=T2by V^^-V"^75. 
 
 35. V— a +V— 6 +V— c by V— a + V— 6 — V— c. 
 
 36. Divide V— 12 by V^^. 
 
 Solution 
 
 37. Divide Vl2 by 
 
 
 V3 
 
 : = V4 = 2. 
 
 Solution 
 
 \/l2 V4 
 
 V-3 VsV^l 
 
 38. Divide 5 by (V- 1/. 
 
 V3T v^T 
 
 1 
 
 Solution 
 5 _ 5 (+ 1) _ 5(V^^)^ 
 
 (>/^=n:)8 (v^8 (>/^)3 
 
 2\/-l. 
 
 dV^n.. 
 
 Divide : 
 
 39. V-18 by V^^. 
 
 40. V27 by V^^. 
 
 41. 14V^^ by 2V^7. 
 
 42. — V-a- by V- 6^. 
 
 43. 1 by V^n!. 
 
 44. (V^)*-V^ by V^. 
 
 45. V^+(V^)' by V^. 
 
 46. V8 - 3 Vl4 by V^^. 
 
 47. Vi2+V3 by V^. • 
 
 57. 
 
 48. - 2 by V^^. 
 
 49. (V^^)' by iV^=n:. 
 
 50. (V^^y by (V^^^ 
 
 51. V4a6 by V— 6c. 
 
 52. V-20- V"^ by 2V^. 
 53. 
 54. 
 
 V^16-V-6 by 2V-2. 
 (V3T)H by _ i^ZIl, 
 
 55. (V^^^)'" by (V^^)-l 
 
 56. V-a'+bV^l by V— a6. 
 4 by V^2 . V^^ • V^^. 
 
368 IMAGINARY AND COMPLEX NUMBERS 
 
 387. For brevity V — 1 is often written i. 
 
 Including all intermediate fractional and incommensurable 
 values, the scale of real numbers may be written 
 
 .■.._3 2 1...0... + 1... + 2..- + 3.-. (1) 
 
 and the scale of imaginary numbers, composed of real multiples 
 of + i and — i, may be written 
 
 Zi 2 i i ...0 ... + i--' + 2i ... + 3/..- (2) 
 
 Since the square of every real number except is positive and 
 the square of every imaginary number except i, or 0, is negative, 
 the scales (1) and (2) have no number in common except 0. Hence, 
 
 An imaginary number cannot he equal to a real number nor cancel 
 any part of a real number. 
 
 388. The algebraic sum of a real number and an imaginary 
 number is called a Complex Number. 
 
 2 + 3 V — 1, or 2 + 3 i, and a + h V — 1, or a + hi, are complex numbers. 
 a^ + 2aby/ — \ —b^ is a complex number, since a^ + 2 a6 V — 1 — 62 — 
 (a2 - 62) + 2 ah y/"^^. 
 
 389. Two complex numbers that differ only in the signs of 
 their imaginary terms are called Conjugate Complex Numbers. 
 
 a + 6 V — 1 and a — 6 V — 1, or a -\- hi and a — 6i, are conjugate com- 
 plex numbers. 
 
 390. Tlie sum and product of two conjugate complex numbers are 
 both real. 
 
 Let a + hV — 1 and a — b V— 1 be conjugate complex numbers. 
 
 Their sum is 2 a. 
 
 Since (V — 1)2 = — 1, their product is, 
 
 §97, c2-(6V^^)2 = a2-(-62) 
 
 = a- + 62. 
 
 391. If two complex numbers are equal, their real parts are equal 
 and also their imaginary parts. 
 
 Let a + b V^^ = x + y V- 1. 
 
 Then, a - x = (y - b) V^, 
 
 which, § 387, is impossible unless a = x and y = b. 
 
IMAGINARY AND COMPLEX NUMBERS 
 
 369 
 
 392. //"a + ftV — 1 = 0, a and h being real, then a = and 6 = 0. 
 
 For, squaring, a^^ 4- 2 o6 V — 1 — 6^ = o, 
 
 a2 _ 52 ^ _ 2 a6 V^H, 
 which, § 387, is true only when a = and 6 = 0. 
 
 393. Graphical representation of a complex number. 
 
 The sum of 3 positive real units and 2 positive imaginary units is found by 
 counting 3 units along OX in the positive direction from and from that 
 point, Z), measuring 2 units upward at 
 right angles to OX in the direction of 
 the axis of imaginary numbers. The 
 line OP represents, by its length and 
 direction, the combined effect or sum 
 of the directed lines OD and DP, that 
 is, the complex number 3 + 2 1. 
 
 The same result may be obtained 
 by counting 2 units along OY up- 
 ward from O and from the end of 
 the second division measuring 3 units 
 toward the right at right angles to 
 Y in the direction of the axis of real numbers, 
 sents either 3 + 2 i or 2 i + 3. 
 
 Similarly, the line OP' represents by its length and direction 2\ — \i or 
 — J I -f 2 J, and the line OP' represents — J + i or i — J. 
 
 Represent the following numbers graphically : 
 
 1. 3 + 4 1. 3. 5 + 2i. 5. 1-2. 
 
 2. 2-3?". 4. b-2i. 6. 4i-l. 
 
 Hence, the line OP repre- 
 
 394. Relation of complex numbers to real and imaginary numbers. 
 
 Let a and. 6 represent any real numbers. 
 
 In the figure of § 393 let P represent any point a units dis- 
 tant from Y' Y and h units distant from X'X. 
 
 Then OP, or the complex number a + h V — 1, represents any 
 number whatever. 
 
 If P lies on the axis of real numbers, 6 = and the complex 
 number a -f- 6 V — 1 = a, a real number. 
 
 If P lies on the axis of imaginary numbers, a = and the com- 
 plex number a + 6 V — 1 = 6 V — 1, an imaginary number. 
 
 If P lies in both axes, a = and 6 = 0, and the complex num- 
 ber a + h V— 1 = 0. 
 
 ALG. — 24 
 
370 
 
 IMAGINARY AND COMPLEX NUMBERS 
 
 395. Operations involving complex numbers. 
 
 Examples 
 1. Add 3 - 2V^i: and 2 + SV^T. 
 
 Solution 
 
 3 - 2\/:r"i + 2 + 5V^^ = (3 + 2) + (- 2\/^^ 4- 5 V^n:) 
 
 = 5 + 3V^. 
 
 ExPLANATiox. — Since, § 387, the imaginary terms cannot unite with the 
 real terms, the simplest form of the sum is obtained by uniting the real and 
 the imaginary terms separately and indicating the algebraic sum of the results. 
 
 Simplify 
 
 the following : 
 
 
 
 
 2. 
 
 (5+V= 
 
 ■4) + (V-9- 
 
 3). 
 
 
 3. 
 
 (2-V= 
 
 .16) + (3+V- 
 
 -4). 
 
 
 4. 
 
 (3-V^ 
 
 r8) + (4+V^ 
 
 - 18). 
 
 
 5. 
 
 (V-20 
 
 -V16) + (V^ 
 
 -45+V4). 
 
 
 6. 
 
 (4+V= 
 
 •25)-(2+V- 
 
 -4). 
 
 
 7. 
 
 (3-2V 
 
 _5)-(2-3V-5). 
 
 
 8. 
 
 (2-2V 
 
 -l+3)-(Vl6-V-16), 
 
 
 9. 
 
 V-49- 
 
 _2-3V-4 
 
 -V-1 + 6. 
 
 10. Expand (a + &V— l)(a + &V— 1). 
 
 Solution 
 § 91, (a + 6\/^n)(a+ &v'^=n')= a^ + 2 aftV^H 4.(6v'3i)2. 
 
 §384, =a2 + 2a6V=^- 62. 
 
 11. Expand (Vo — V- 3)^ 
 
 Solution 
 
 (V5 
 
 3)2 = 5_2V^rT5 4.(_3) 
 = (5_3)-2V- 16 
 = 2-2V-15. 
 
 Expand the following : 
 
 12. (2 + 3V^T)(l+V^^). 
 
 13. (5 _ v^n;)(i - 2V^i). 
 
 14. (V2+V^:^)(V8-V^^). 
 
 15. {2 + Zif. 
 
 16. (2-3i/. 
 
 17. (a - Uf. 
 
IMAGINARY AND COMPLEX NUMBERS 
 
 371 
 
 18. Show that (1 -f.V^^)(l +V^^)(1 +V^^)=-8. 
 
 19. Show that (- 1 + V^=^)(- 1 +V^^)(- 1 -I- V^^) 
 
 20. Show that (- i + iV^(-Hi V^3)(-|+i-V^) 
 
 21. Divide 8 + V^^ by 3 + 2V^^. 
 
 First Solution 
 
 = 1. 
 
 3+2\/-l 
 
 2-V-l 
 
 8+aA^ = 6+ V-1 + 2 
 6 + 4^^^! 
 
 -3\/^n: + 2 
 
 - SV^T + 2 
 
 The real term of the dividend may always be separated into two parts, 
 of which will exactly contain the real term of the divisor. 
 
 Second Solution 
 3 + 2-^/^:^1 (3 + 2V":n)(3-2\/^n;) 9 + 4 
 
 = 2-^-1. 
 
 26. 16 + 4 V^r5 by 3 - V - 
 
 27. a^ + 6^ by a - feV^^. 
 
 28. a + hi by ai + h. 
 
 29. (1 + 0' by 1 - i. 
 
 Divide : 
 ^22. 3 by 1 -V^^. 
 
 23. 2 by 1 +V- 1. 
 
 24. 4+V4by2-V^^. 
 
 25. 9 + V^^ by 3 - V- 
 30. Find by inspection the square root of 3 + 2V— 10. 
 
 Solution 
 3 + 2V:n[0=(5-2)+2V5- -2 = 5 + 2V5. -2 +(-2). 
 
 .-. V3 + 2>/^^=V5 + 2V5. -2+(-2) = V5+ V^. 
 Find by inspection the square root of 
 /31. 4 + 2V^=^2T. 33. 6-2V^=^. 35. 12V^n:-5. 
 32. 1+2V^6. 34. 9-f 2V^^2. /36. h'' + 2ahV^^- 
 
 yi. Verify that — 1 + V— 1 and — 1 — V— 1 are roots of 
 equation a^ + 2ic + 2 = 0. 
 
 38. Expand (| + ^V^r3)3. 
 
 the 
 
372. IMAGINARY AND COMPLEX NUMBERS 
 
 396. If Va -\-bi = \/x + iy/y, then Va — hi = y/x — i^/y, when 
 a and b are real and x and y are positive. 
 
 For suppose that Va + hi = Vx + iVy. 
 
 Squaring, a + hi = 3? + 2 iy/xy —^ 
 
 .'. § 391, a = 35 — y ana~Br= 2 i\/xy. 
 
 a — bi =:x — 2 iVxy — 
 = (Vx — iVyy. 
 Hence, Va — bi = Vx — i Vy. 
 
 Examples 
 1. Find the square root of — 5 — 12 V— 1. 
 
 Solution 
 
 Let Vx-iVy = V- 5- 12V^^. (1) 
 
 Then, § 396, Vx + iVy - V- 5 + 12 V^^. (2) 
 
 Multiplying, x -\- y = V25 + 144, 
 
 or X + 2/ = 13. (3) 
 
 Squaring (1), x — 2 iy/xy — ?/ = — 5 — 12 V— 1. 
 
 Therefore, § 391, a; - ?/ = - 5, (4) 
 
 Solving (3) and (4), x = 4, y = 9. 
 
 .-. Vx = 2, iy/y = 3 V— 1. 
 Hence, by (1), V- 5 - 12V^ = 2 - 3V^. 
 This result may be verified by squaring 2 — 3 V— 1. 
 
 2. Find the square root of — 5 + 12V— 1. 
 
 Solution 
 By the preceding example, v — 5 — 12 V— 1 =2— 3V— 1. 
 
 Therefore, § 396, V-5 + l^yT^ = 2 + 3 V^^. 
 
 Find the square root of 
 
 
INEQUALITIES 
 
 397. One number is said to be greater than another when the 
 remainder obtained by subtracting the second from the first is 
 positive, and to be less than another when the remainder obtained 
 by subtracting the second from the first is negative. 
 
 If a — ft is a positive number, a is greater than 6 ; but if a — ft is a negative 
 number, a is less than ft. 
 
 Any negative number is regarded as less than ; and, of two 
 negative numbers, that more remote from is the lessi "' 
 
 Thus, — 1 is less than ; — 2 is less than — 1 ; — 3 is less than — 2 ; etc. 
 
 An algebraic expression indicating that one number is greater 
 or less than another is called an Inequality. 
 
 398. The Sign of Inequality is > or <. 
 
 It IS" placed between two unequal numbers with the opening 
 toward the greater. 
 
 Thus, a is greater than ft is written « > ft ; a is less than ft is written a<,b. 
 
 The expressions on the left and right, respectively, of the sign 
 of inequality are termed the Jirst and second members of the 
 inequality. 
 
 399. When the first members of two inequalities are each 
 greater or each less than the corresponding second members, the 
 inequalities are said to subsist in the same sense. 
 
 When the first member is greater in one inequality and less in 
 another, the inequalities are said to subsist in a contrary sense. 
 
 a; > ffl and 2/ > ft subsist in the same sense, also a; < 3 and y < 4 ; but x > 6 
 and y <ia subsist in a contrary sense. 
 
 373 
 
374 INEQUALITIES 
 
 400. 1. If 2 is added to each member of the inequality 8 > 5. 
 how will the two inequalities subsist ? How will they subsist, if 
 2 is subtracted from each member ? 
 
 2. Investigate other inequalities. 
 
 Principle 1. — If the same number or equal numbers are added 
 to or subtracted from both members of an inequality, the reciting 
 inequality will subsist in the same sense. 
 
 Let a > 6. 
 
 Then, § 397, a -- b = p, a. positive number. 
 
 Adding c — c = 0, Ax. 2, a + c — (b + c) = p. 
 Therefore, a + c>b + c. 
 
 401. 1. What is the effect of adding 2 to each member of the 
 inequality x — 2>y? What is the effect of subtracting 2 from 
 each member of the inequality a + 2 > 6 ? 
 
 2. If a term is transposed from one member of an inequality 
 to the other, what must be done to its sign ? 
 
 Principle 2. — A term may be transposed from one member 
 of an inequality to the other, provided its sign is changed. 
 
 Let a — b>c — d. 
 
 Adding b to each side, Prin. 1, a>b + c — d. 
 
 Adding — c to each side, Prin. 1, a — c > 6 — (?. 
 
 402. Principle 3. — If the signs of all the terms of an inequality 
 are changed, the resulting inequality will subsist in a contrary sense. 
 
 Let a — 6 > c — d. 
 
 Transposing every term, Prin. 2, 
 
 — c + d> — a + b; 
 that is, — a + b< — c + d. 
 
 403. 1. If both members of the inequality 10 > 8 are multi- 
 plied by 2, how will the two inequalities subsist ? How will they 
 subsist, if both members are divided by 2 ? 
 
 2. How will they subsist in each case, if the multiplier or 
 divisor is — 2? 
 
INEQUALITIES 375 
 
 Principle 4. — If both members of an inequality are multiplied 
 or divided by the same number, the resulting inequality ivill subsist 
 in the same sense if the multiplier or divisor is positive, but in the 
 contrary sense if the multiplier or divisor is negative. 
 
 Let a~>b. 
 
 Then, § 397, a — b = p, a. positive number. 
 
 Multiplying by m, ma — mb — mp. 
 
 If fli is positive, mp is positive, 
 
 and ma > mb. 
 
 If m is negative, mp is negative, 
 
 and mb > ma, or ma < mb. 
 
 Putting — for m, the principle holds also for division. 
 m 
 
 404. 1. If the corresponding members of the inequalities 6 > 5 
 and 4 > 2 are added, how will the resulting inequality subsist ? 
 How, if — 5 > — 6 and — 2 > — 4 are added ? 
 
 Principle 5. — If the corresponding members of any number of 
 inequalities subsisting in the same sense are added together, the 
 resulting inequality ivill subsist in the same sense. 
 
 Let a > b, c'>d, e >/, etc. 
 
 Then, § 397, a — b, c — d, e — f, etc., are positive numbers. 
 
 Hence, their sum, a + c + e+ ••• — (b ■\- d-\- f+ •••), is a positive number; 
 that Is, a + c + e +•••>/> + (?+/+ ••.. 
 
 405. Principle 6. — If each member of an inequality is sub- 
 tracted from the corresponding members ofa7i equation, the resulting 
 inequality will subsist in a contrary sense. 
 
 Let a > 6 and let c be any number. 
 
 Then, § 397, a — 6 is a positive number. 
 
 Since a number is diminished by subtracting a positive number from it, 
 
 c — (a — ft) < c. 
 
 Transposing, c — a<c — b. 
 
 That is, if each member of the inequality « > ft is subtracted from the cor- 
 responding member of the equation c — c, the result is an inequality subsisting 
 in a contrary sense. 
 
376 INEQUALITIES 
 
 406. It is evident that the difference of two inequalities sub- 
 sisting m the same sense, or the sum of two inequalities subsist- 
 ing in a contrary sense, or the product, or the quotient of two 
 inequalities, member by member, may have its first member 
 greater than, equal to, or less than its second. 
 
 For example, take the inequality 12 > 6. 
 
 Subtracting 8 > 2, or adding — 8 < — 2, the result is 4 = 4. 
 
 Subtracting 8 > 1, or adding — 8 < — 1, the result is 4 < 5. 
 
 Multiplying and dividing by 3 > 2, the results are 36 > 12 and 4 > 3. 
 
 Multiplying by — 2 > — 4, the result is — 24 = — 24. Dividing by 4 > 2, 
 the result is 3 = 3. 
 
 Multiplying and dividing by — 2 > — 3, the results are — 24 < — 18 and 
 - 6 < - 2. 
 
 Examples 
 
 407. 1. Find one limit of x in the inequality 3 a; — 10 > 11. 
 
 Solution 
 
 3a;-10>ll. 
 
 Prin. 2, 3x>21. 
 
 Prin, 4, x>7. 
 
 Therefore, the inferior limit of x is 7 ; that is, x may have any value greater 
 than 7. 
 
 2. Find the limits of x in the simultaneous inequalities 
 3 a; -f- 5 < 38 and 4 a; < 7 ic - 18. 
 
 Solution 
 
 3a;-f5<38. (1) 
 
 4x<7x-18. (2) 
 
 Transposing in (1), Prin. 2, 3x<33. 
 
 .-.Prin. 4, a;<ll. 
 
 Transposing in (2), Prin. 2, — 3x < — 18. 
 
 Changing signs, Prin. 3, 3 x > 18. 
 
 .-.Prin. 4, x>6. 
 
 The inferior limit of x is 6, and the superior limit is 11 ; that is, the given 
 inequalities are satisfied simultaneously by any value of x between 6 and 11. 
 
INEQUALITIES 377 
 
 3. Find the limits of x and ?/in3x— ?/> — 14 and x-\-2y = 0. 
 
 Solution 
 3a;-y>-14. (1) 
 
 X + 2 «/ = 0. (2) 
 
 Multiplying (1) by 2, Prin. 4, 
 
 6 X - 2 y > - 28. (3) 
 
 Adding (2) and (3), 7 x > - 28. 
 
 .-. x>-4. (4) 
 
 Multiplying (2) by 3, 3 x + 6 y = 0. (6) 
 
 Subtracting (5) from (1), Prin. 1, 
 
 -7 2/>-14. 
 Dividing by — 7, Prin. 4, y < 2. 
 
 That is, X is greater than — 4, and y is less than 2. 
 
 Find the limits of x in the following : 
 
 4. 6a;-5>13. {4.x-ll>\x, 
 
 5. oa;-l<6a; + 4. ^* l20-2a;>10. 
 
 6. 3a;-|a;<30. r3-4a;<7, 
 
 7. 4a; + l<6a;-ll. * l5a; + 10<20. 
 
 10. a; + — -f — >25 and <30. 
 3 6 
 
 11. Find the limits of a; in ar^ + 3 cc > 28. 
 
 Solution 
 
 x2 + 3 X > 28. 
 Transposing, Prin. 2, x2 + 3 x - 28 > 0. 
 
 Factormg, (x - 4) (x + 7 ) > 0. 
 
 That is, (x — 4)(x + 7) is positive. 
 
 If (x — 4) (x + 7) is positive, either both factors are positive or both are 
 negative. Both factors are positive, when x > 4 ; both factors are negative, 
 ■vhen X < - 7. 
 
 Therefore, x can have all values except 4 and — 7 and intermediate values. 
 
378 INEQUALITIES 
 
 Find the limits of x in the following : 
 
 12. a^ + 3a:>10. 16. a^>9a;-18. 
 
 13. a;2 + 8ar>20. 17. cc^ + 40a; > 3(4a;- 25). 
 
 14. y? -{-5x> 24. 18. x^ ->rhx>ax + ah. 
 
 15. (a; - 2) (3 - a;) > 0. 19. (a; - 3) (5 - a;) > 0. 
 
 Find the limits of x and y in the following, and, if possible, 
 one positive integral value for each unknown number : 
 
 2x-3y<2, jy = 3a; + 4, 
 
 ■ 2a; + 52/ = 25. ' \25<2y + Zx. 
 
 3x + 2?/ = 42, fy — a;>9, 
 
 3x ^>lb. [20^15-^- 
 
 x + ?/ = 10, f X > w + 4, 
 
 22.-1 25. 1 ^^^ ' 
 
 4x<3t/. la;-2^ = 8. 
 
 If a, b, and c are positive and unequal, 
 
 26. Which is the greater, — — — or — i ? 
 
 ^ ' a + 26 a + 3& 
 
 27. Prove that a^ + b->2 ah. 
 
 Suggestion. — Whether a — 6 is positive or negative, (o — ft)* is positive. 
 
 28. Prove that a^ + W + c?>ah + etc -f he. 
 
 29. Prove that a^ + 6^ > a^6 + a6l 
 
 30. Which is the greater, ^^d^ or ^^!±_^ ? 
 
 a + 6 a- + 6^ 
 
 31. Prove that 1 > 1, except when 2 a = 3 6. 
 
 3& 4a 
 
 32. Prove that (a — 2 &) (4 & — a) < fe^, except when a = 3 6. 
 
 33. Prove that a^ + ft'' 4- c^ > 3 a&c. 
 
 34. Prove that the sum of any positive real number, except 1, 
 and its reciprocal is always greater than 2. 
 
VARIABLES AND LIMITS 
 
 408. A number that has the same value throughout a discussion 
 is called a Constant. 
 
 Arithmetical numbers are constants. A literal number, as a or x, is con- 
 stant in a discussion, if it keeps the same value throughout that discussion. 
 
 409. A number that under the conditions imposed upon it may 
 have a series of different values is called a Variable. 
 
 Variables are usually represented by x, y, z, etc. 
 
 The numbers .3, .33, .333, .3333, . . . are successive values of a variable 
 approaching the constant |. 
 
 The commensurable numbers 1, 1.4, 1.41, 1.414, 1.4142, . . . are succes- 
 sive values of a variable approaching the constant V2. 
 
 410. An expression whose value depends upon the value of a 
 variable is called a Function of that variable. 
 
 2 a; + 1 is a function of x, because, if successive values are given to x, 
 2 X + 1 will take successive values. For example, if x = 0, 2 x + 1 = 1 ; if 
 
 x = l, 2x-|-l = 3; ifx = 2, 2x + l = 5, etc. x^ — 2 x and are also 
 
 functions of x. ~^ 
 
 x^ + 2 X2 — 5 z2 is a function of x and z. 
 
 Every function of a single variable is a variable. 
 
 The variable to which different values may be given at pleasure 
 or according to some law is called the Independent Variable, and 
 the function of the independent variable is called the Dependent 
 Vanable. 
 
 In the first illustration x is the independent variable and 2 x -|- 1, the func- 
 tion of X, is the dependent variable. 
 
 411. To discuss functions of a variable it is necessary to sup- 
 pose that the variable takes its successive values according to 
 some definite law of change. 
 
 379 
 
380 VARIABLES AND LIMITS 
 
 Where a variable takes a series of values that approach nearer 
 and nearer a given constant, so that by taking a sufficient number 
 of steps the difference between the variable and the constant can 
 be made numerically less than any conceivable number however 
 small, the constant is called the Limit of the variable, and the 
 variable is said to approach its limit. 
 
 The variable .3, .33, .333, .3333, . . ., whose increase at each step is y^ of 
 the previous increase, approaches J as a limit. For .3 differs from ^ by less 
 than yJj, .33 by less than y^j, .333 by less than ys\^, etc., and by taking a 
 sufficient number of steps it is possible to obtain a value of the variable differ- 
 ing from ^ by less than any number that can be conceived of, however small. 
 
 412. The difference between a variable and its limit is a vari- 
 able whose limit is zero. 
 
 As .3, .33, .333, . . . approaches its limit ^, the variable difference | — .3, 
 ^ — .33, ^ — .-363, . . ., or ^, 5^5, ^^Viy • • • approaches the limit zero. 
 
 A variable may approach a constant without approaching it as 
 a limit. 
 
 The variable 6.6, 6.66, 6.666, . . ., in approaching 6§ as a limit, approaches 
 7 also, but not as a limit. 
 
 A variable in approaching its limit may be always less than its 
 limit, or always greater, or sometimes greater and sometimes less. 
 
 The variable .3, .33, .333, ... is always less than its limit |. 
 The variable, .7, .67, .667, ... is always greater than its limit 1 — |. 
 / The sum of the first n terms of the geometrical series 1, — J, J, — 5, ^^, 
 — 3^2, ... is a variable whose successive values 1, ^, |, f, |^, f^, . . . are 
 alternately greater and less than the limit |. 
 
 A variable may increase or decrease according to its law of 
 change and become numerically greater or less than any assign- 
 able number. 
 
 The variable 1, —2, 4, — 8, 16, . . . may become numerically greater than 
 any number that can be assigned. The variable 1, ^, \, \, ^^, . . • may 
 become numerically less than any number that can be assigned. 
 
 413. A variable that may become numerically greater than any 
 assignable number is said to be Infinite. 
 
 The symbol of an infinite number is 00. 
 
 414. A variable that may become numerically less than any 
 assignable number is said to be Infinitesimal. 
 
 An infinitesimal is a variable whose limit is zero. 
 
VARIABLES AND LIMITS 381 
 
 The character is used as a symbol for an infinitesimal number 
 as well as for absolute zero, the result obtained by subtracting a 
 number from itself. 
 
 415. A number that cannot become either infinite or infinitesi- 
 mal is said to be Finite. / 
 
 416. Interpretation of -• 
 
 If the numerator of the fraction - is constant while the de- 
 
 X 
 
 nominator decreases regularly until it becomes numerically less 
 than any assignable number, the quotient will increase regularly 
 and become numerically greater than any assignable number. 
 
 .-. ^ = 00. That is, 
 
 If a finite number is divided by an infinitesimal number, the 
 quotient tvill be an infinite number. 
 
 417. Interpretation of — 
 
 00 
 
 If the numerator of the fraction - is constant while the de- 
 
 X 
 
 nominator increases regularly until it becomes numerically greater 
 than any assignable number, the quotient will decrease regularly 
 and become numerically less than any assignable number. 
 
 .-. - = 0. That is, 
 
 00 
 
 If a finite munber is divided by an infinite number, the quotient 
 will be an infinitesimal number'. 
 
 418. The symbol = is read 'approaches as a limit.' 
 The abbreviation lim. is used for limit. 
 
 X = a is read 'x approaches a as its limit.' The expression lira, x = a ia 
 equivalent to x = a, and is read ' the limit of a; is a.' 
 
 Though 00 represents a variable that may transcend all finite 
 values, it is convenient to use the expression a; = oo to indicate 
 that X increases numerically without limit. 
 
 Thus, as « = 00, - :i 0. 
 
 • ■ X 
 
 419. Principle 1, — A variable cannot approach two unequal, 
 limits at the same time. 
 
382 VARIABLES AND LIMITS 
 
 For in approaching the more remote as a limit the variable would reach a 
 value intermediate between the two, and thereafter in approaching one as 
 a limit it would recede from the other, which, therefore, could not be a limit. 
 
 420. Principle 2. — If two variables are always equal and each 
 approaches a limit, their limits are equal. 
 
 For in their values the two variables are but one. 
 
 Hence, Prin. 1, the limit of their common values is their common limit. 
 
 421. 1. Since the limit of .333 •••is i, what is the limit of 
 2 + .333 •••, or 2.333 ••• ? of 4 + .333, or 4.333 ... ? of 5.333 ... ? 
 of 2 - .333 ..., or 1.666 ... ? of .333 .13, or .203 ... ? 
 
 2. What is the limit of the algebraic sum of a constant and a 
 variable ? 
 
 Principle 3. — TTie limit of the algebraic sum of a constant and 
 a variable is the algebraic sum of the constant and the limit of the 
 variable. 
 
 422. 1. Since the limit of .333 ... is i, what is the limit of 
 .666...? of .111 ... ? 
 
 2. Since the limit of 1 + i + i -f ... is 2, what is the limit of 
 
 •^ 4- 3 _i_ 3 I , . , ? of J -4- L -I- 1 4- . . . '.> 
 
 3. How may the limit of the product of a variable and a con- 
 stant be obtained from the limit of the variable ? 
 
 Principle 4. — The limit of the product of a variable and a finite 
 constant is equal to the ^product of the constant and the limit of the 
 variable. 
 
 The above principle may be established as follows : 
 
 Case 1. — When the limit of x is 0. 
 
 Let k be any finite constant. 
 
 It is to be proved that kx ~ 0. 
 
 However small any number, as q, may be, since a: = 0, x may be made 
 numerically less than q -^ k. 
 
 Hence, kx may be made numerically less than q ; that is, kx may be made 
 numerically less than any number Jiowever small. 
 
 Therefore, §411, kx = 0. 
 
 Case 2. — Whe7i the limit of x is not 0. 
 
 Let k be any finite constant, a the limit of x, and y the variable that must 
 be added to x to produce a. 
 
[VARIABLES AND LIMITS 383 
 
 It is to be proved that kx = ka. 
 
 Since x + 2/ = a, x = a — y; 
 
 .-. kx = ka — ky. 
 
 Prin. 2, lim. {kx) = lim. {ka — ky) 
 
 Prin. 3, = ka — lim. {ky). 
 
 But since, § 412, y = 0, by Case 1, lim. ky = 0. 
 
 Hence, lim. {kx) — ka — = ka ; 
 
 that is, kx = ka. 
 
 Note. — The principle holds for the limit of a variable divided by a con- 
 stant, since dividing by k is equivalent to multiplying by -• 
 
 423. Principle 5. — The limit of the variable sum of any finite 
 number of variables is equal to the sum of their limits. 
 
 The above principle may be established as follows : 
 
 Let a; = a, 2/ = 6, 2 = c, etc., to any finite number of variables, as n. 
 
 It is to be proved that lim. {x -{■ y + z -\- •••) =: a + 6 + c + ■••. 
 
 Let t?i, ij2i ^3, ••• be the variable differences between x, y, z ••• and their 
 respective limits. 
 
 Then, a; + y + 2 + ••• = (a + 6 + c + •••) - {vx + V2 + »3 + •••)> 
 and lim. (x + y + z + ■.•) = lim. [(« + 6 + c + •••) — {vx + V2 + «3 ^ — )] 
 
 Prin. 3, = a + 6 + c + ••• — lim. (»i + W2 + »3 + •••)• 
 
 Since, §412, »i = 0, ^2 ^ 0, v^ = 0, etc., however small any number, as 
 g, may be, each of the n variables, vi, V2, vg, etc., may be made less than 
 q H- n, and hence their sum may be made less than q. 
 
 Therefore, §411, lim. {vi + V2 + vs + ••■) = 0. 
 
 Hence, lim. {x + y + z + ••■) = a + b + c + •••. 
 
 424. Principle 6. — The limit of the variable product of two 
 or more variables is equal to the product of their limits. 
 
 The above principle may be established as follows : 
 
 Given, x = a and y = b. 
 
 It is to be proved that lim. {xy) = ah. 
 Let t7i = a — « and V2= b — y. 
 
 Then, xy = {a — vi) {b — v^), 
 
 and lim. {xy) = lim. [ab — bvi — av^ + v^v^l 
 
 Prin. 5, Prin. 4, =ab — b lim. vt — a lim. v^ + lim. {viVa) 
 
 § 412, =ab + lim. {viv^). 
 
384 VARIABLES AND LIMITS 
 
 Since when vi and v^ are near their common limit 0, their product is much 
 less than either vi or v^, 
 
 lim. (viVz) = 0. 
 Hence, lim. {xy) = ab + = ab. 
 
 Similarly, the principle may be established for any number of variables. 
 
 425. Principle 7. — The limit of the variable quotient of two 
 variables is equal to tJte quotient of their limits, provided the limit 
 of the divisor is not 0. 
 
 The above principle may be established as follows: 
 
 V 
 Let X = - : and let Mm. z be not 0. 
 
 2 
 
 It is to be proved that lim. x = ^ . 
 
 lira, z 
 
 Since x = -, v = xz. 
 
 z 
 Prin. 6, lim. y = lim. a; • lim. z. 
 
 Therefore, lim. x = ^-i!^^. 
 
 lim. z 
 
 The principle has no meaning when ^ = 0, since lim. y cannot be divided 
 byO. 
 
 426. When by causing a variable x to approach sufficiently 
 near to a it is possible to make the value of a given function of x 
 approach as near as we please to a finite constant I, I is called the 
 limit of the function when x = a. 
 
 Suppose that .1, .11, .111, .1111, ... are successive values of x approaching 
 I as a limit. Then, the corresponding values of 1 — 2a;, a function of x, are 
 .8, .78, .778, .7778, values of a variable approaching ^ as a limit. As a; = J, 
 the function of x, 1 — 2x, = | ; for by causing x to approach sufficiently 
 near to \ it is possible to make 1 — 2 x approach as near as we please to |. 
 
 The expression 
 
 lim. [function of x\^a 
 
 is read ' limit of function of x as a; approaches a as a limit.' 
 
 Thus, lim. (1 — 2x)x=\ — I indicates that as x approaches its limit |, 
 1 — 2 X approaches the limit |. 
 
 427. In finding the limiting values of the functions given in 
 the following examples, the student is expected to apply the 
 principles that have been established above. 
 
 Finding the limiting value of a function of a; as x = a is called 
 evaluating the function for x = a. 
 
VARIABLES AND LIMITS 
 
 385 
 
 Examples 
 If x = a, y = 2, and z = 0, find the limit of 
 
 1. x + y + z. A. x-\y+ax-y^. 
 
 2. axy-x^. 5. {x + y)x -{x - y)z. 
 
 o ^ y. 
 
 2 a 
 
 ^ a+x+z ^ x+y 
 x — y a 
 
 Find the value of 
 
 7. Lim.r^-^^ + »' 
 X — 2 
 
 x=3 
 
 8. Lim. 
 
 9. Lim. 
 10. Lim. 
 
 f a^ + 4 a; + 1 " 
 L X* + a^ 4- 1 _ 
 
 V - 3 a; + 4' 
 
 aj + e 
 
 " 1 + a;" " 
 1 4- a; 
 
 j;=-l 
 
 £=2 
 
 z=\ 
 
 428. Indeterminate forms -, — , x qo, qo x 0, oo — oo. 
 
 00 
 
 For all values of a and x, 
 a 
 
 X _a X 
 l~l 'i 
 
 X 
 
 = a. 
 
 If a; = 00, (1) becomes - = x oo = a. 
 
 (1) 
 
 If a; = 0, (1) becomes - = oo x = a. 
 
 00 
 
 Since a denotes any number whatever, -, — , Ox oo, and 
 
 00 
 
 00 X are symbols for indeterminate numbers. 
 
 If k is any constant, oo, + A; = 0C2 ; ••. 002 — ooj = A;. 
 Hence, 00 — 00 is a symbol for an indeterminate number. 
 
 ADV. ALG. — 25 
 
886 VARIABLES AND LIMITS 
 
 429. Since every function of a single variable is a variable, it 
 is evident that the preceding principles apply to functions of a 
 variable. Thus, to apply to functions of a variable, Prin. 5, 6, and 
 7 may be stated as follows : 
 
 Tlie lirfiit of the sum of a finite number of functions of x is equal 
 to the sum of their limits. 
 
 The limit of the product of a finite number of functions of x is 
 eqxial to the product of their limits. 
 
 Tlie limit of the quotient of two fxmctions of x is equal to the 
 quotient of their limits, provided the limit of the divisor is not zero. 
 
 These principles fail to give a limit whenever the result 
 obtained involves one of the indeterminate forms, 
 
 00 — 00, X 00, QO X 0, -, ^• 
 ' ' ' Q' CO 
 
 430. The preceding principles of limits lead to the conclusion 
 that the limit of a function is found by substituting the limits of 
 the variables for the variables, except when such a substitution 
 gives an indeterminate form (§ 429). 
 
 Thus, if lim. x = 5 and lim. ?/ = 2, the limit of 4 a; — 3 y is found by substi- 
 tuting 5 for X and 2 for y in the function ix — 3y. But if substitution is 
 
 employed directly to evaluate the functions 7^, - x y, and 
 
 X — I x{x — \)y 
 
 (a;2 — 1) -^ (x — 1) when x = 1 and y = 0, these functions take the forms 
 
 00 — 00 , 00 X 0, and 0-4-0, respectively. 
 
 When the method of evaluation by substitution in the given 
 function fails, the evaluation of the function is performed by the 
 aid of Prin. 2. 
 
 Thus, to evaluate (a-2 — 1) ^(x — 1) when x = 1, find another function of 
 X, as X + 1, equal to the given function (x^ — 1)-t-(x— 1) for all values 
 assumed by x while approaching the limit 1. 
 
 If X takes the successive values 
 
 2, I, I, I, Ih 11. •••. approaching 1, 
 then both functions (x^ — 1) -r- (x — 1) and x + 1, take the successive values 
 3, i, I, V' il. fl» •••> approaching 2. 
 Since the two functions are equal for all values of x as x approaches its 
 limit 1, by Prin. 2 they have the same limit. This limit is lim. (x + l)xii, 
 which by substituting lim. x for x is found to be 1 + 1, or 2. 
 
VARIABLES AND LIMITS 
 
 387 
 
 Examples 
 
 Find the value of 
 
 Lim. 
 
 2. Lim. 
 
 fx" - 81 
 
 
 _a^-4 
 
 x=2 
 
 "1 - ar'" 
 
 
 J--X 
 
 x = l 
 
 5. Lim. 
 
 6. Li 
 
 m. 
 
 3. Lim. , 
 
 4. Lim.r?ii^" 
 
 
 xp^— 6 
 
 a^ — 5 a; + 6" 
 ar'-4 
 
 -/:=2 
 
 3a^ + 5ar'-2a;-l' 
 
 a^ — 8 x^ + 7 
 
 x=l 
 
 7. Find the limiting value of — — — ~ ^t_ when x = 
 
 and also when a; = oo. 
 
 Solution. — As x approaches the limit 0, the first three terms of the 
 numerator and also of the denominator become infinitely small as compared 
 with the fourth, and, consequently, may be neglected. Hence, when x = 0, 
 the fraction approaches the limiting value |. 
 
 As X = 00, that is, as x becomes indefinitely greater, the last three terms 
 of the numerator and also of the denominator become infinitely small as 
 compared with the first, and, consequently, may be neglected. Hence, when 
 
 X = cc, the fraction approaches the limiting value — ^, or — 
 
 Find the limiting values of the following when x = and when 
 X = 00 : 
 
 8. 
 
 10. 
 
 11. 
 
 1+:^ 
 
 2 + X* + x« 
 
 1 - X-' 
 
 - X* - 2 x« 
 
 5x3- 
 
 aj2 -1- 4 X + 2 
 
 2 x^ 4- 3 .x^ - X -f- 1 
 
 4:X* 
 
 -3a^ + x + l 
 
 2x*- 
 
 X^ — X^ -{- X -{- 1 
 
 2x*- 
 
 3ar^ + 2x^-2 
 
 2x2 + x + l 
 
 12. 
 
 13. 
 
 14. 
 
 15. 
 
 5x + 10 
 
 x2 + 2x + 2 
 
 3x-4 
 
 x2 - X - 8 
 
 2x2-4x + l 
 
 2x-l 
 
 4ar^ + 5a^ + 2x 
 
 2x3 + x-f 1 
 
INCOMMENSURABLE NUMBERS 
 
 431. Though an incommensurable number (§ 227) cannot be 
 expressed by any integer or by any fraction with integral terms, 
 a commensurable number can always be found to differ from any 
 incommensurable number by less than any number that may be 
 assigned, however small. 
 
 For example, though V2 cannot be expressed by a decimal that 
 terminates, commensurable numbers can be found to approximate 
 to the true value of V2. For by the process of evolution a com- 
 mensurable number can be found to differ from V2 by less, 
 successively, than .1, .01, .001, .0001, •••, the difference finally 
 becoming smaller than any number, however small, that can be 
 assigned. 
 
 To generalize : let p and q be variables each of which may take 
 any integral value whatever. Then, any commensurable number 
 
 P 
 
 may be exactly represented by -, and since q may be made as 
 
 large as we please and p may be given any integral value what- 
 
 P 
 ever, any incommensurable number may be represented by - to 
 
 any required degree of approximation. 
 
 Thus, any incommensurable number may be included between 
 
 p p -\- \. 1 
 
 - and , each of which differs from it by less than -; and 
 
 q q ' . . 5 . 
 
 since q may be made as large as we please, the limit of this 
 
 difference is zero. 
 
 Hence, an incommensurable number may he regarded as the limit 
 of a variable commensurable mimher. 
 
 Thus, \/2 is the limit of the series 1 + ^ + t^ + rsW + "••' ^^^^ ^"'" °^ " 
 terms of which is 1, 1.4, 1.41, 1.414, •••, as n takes the successive values 1, 2, 
 3, 4, .... 
 
 888 
 
INCOMMENSURABLE NUMBERS 389 
 
 432. By § 429, the principle of § 420 may be stated as follows : 
 
 If two functions of the same variable or variables are equal for all 
 values of the variables, the limits of the two functions, as the variables 
 approach their resjyective limits, are equal. 
 
 Thus, X + y and y -\- x are equal for all values of x and y, and if x and y 
 approach limits, lim. ( x+ ?/) = lim. (y + x). 
 
 433. Commutative, Associative, and Distributive Laws. 
 
 Let a, b, and c be incommensurable constants, the limits, § 431, 
 of the commensurable variables, x, y, and z, respectively. 
 
 1. Covimutative Law for Addition and Subtraction. 
 
 § 55, x + y := y +x. 
 
 Hence, § 432, lim. (x -\- y) = lim. (y + x) ; 
 
 that is, § 423, a+b = b + a. 
 
 Similarly, a — 6 = — & + a. 
 
 2. Commutative Law for Multiplication and Division. 
 § 82, xy = yx. 
 
 Hence, § 432, lim. (xy) = lim. (yx) ; 
 
 that is, § 424, ab = ba. 
 
 Similarly, § 106, a^ - = -. a. 
 
 3. Distributive Law for Multiplication and Divisipn. 
 §§ 85, 87, z being either a monomial or a polynomial, 
 
 (x-\-y)z = xz+ yz. 
 Hence, § 432, lim. \fx + y)z'] = lim. (xz + yz) ; 
 that is, §§ 424, 423, (a + &)c = ac + be. 
 
 Similarly, § 106, (a + 6)i = a • i + 6 • - ; 
 
 c c c 
 
 that is, (a-{-b)-i-c = a^c-\-b-^c. 
 
 4. Associative Laiv. 
 
 The Associative Law follows from the Commutative Law, sub- 
 ject to the rules for signs given in § 68, 2 and § 104, 2. 
 
390 INCOMMENSURABLE NUMBERS 
 
 434. Meaning of a" when n is incommensurable. 
 
 Since an incommensurable number is neither even nor odd, if a 
 is negative or imaginary, the rule for the sign of a power given in 
 § 218, Prin, 1, cannot be used to find the sign of (a)" when n is 
 incommensurable. In the following discussion, then, a will 
 denote a positive real number. 
 
 Since V3 lies between corresponding terms of the two series 
 17 173 1732 p 
 
 ^' 10' 100' 1000' 
 
 — ) 
 
 18 174 1733 p + 1 
 
 and 2, ^^, ^^^, ^^^, -., -^-, •.., 
 
 it may be inferred that a/^ lies between corresponding terms of 
 
 a*, a^', a^^, a^^, •••, o«, •••, 
 
 p+i 
 
 and a^, a^^, a^^^, a^^^^, ••-, a « , •••, 
 
 1 
 
 provided a' is restricted to indicate the principal 5th root. 
 
 Passing to the general case, in a" let n be the incommensurable 
 
 limit of the variable exponent x. 
 
 p p _|_ 1 
 
 Then two values of x, as — and , can be found to include 
 
 '9 Q ' 
 
 n between them ; and by taking q sufficiently great the difference 
 
 between the values of x, and, consequently, the difference between 
 
 n and either value of x, may be made as small as we please. 
 
 It will now be proved that the difference between the j^oivers 
 p+i p 
 
 a ' and a', and, consequently, the difference between a" and either 
 power, can be made as small as we please by taking the difference 
 between the corresponding exponents sufficiently small; that is, that 
 
 p+\ p 
 lim.(a « — a^\q^oo — 0. 
 1. When a > 1. . 
 
 1 1 I 
 Since a > 1, a»> 1 and a»— 1 is positive. Let a»— 1 = z. 
 1 
 .-. a»=F= 1+2!, and a= (1 + zy = 1 + qz -\ , 
 
 every term of which is positive, z being positive. 
 
 Therefore, a>l + qz, whence z < , 
 
 which approaches as q- = 00. ^ 
 
INCOMMENSURABLE NUMBERS 391 
 
 1 
 Hence, when a > 1, lim. (a« — l)^=oo = 0. (1) 
 
 Since a^ <,a when q>p, a« remains finite as q = cc. Hence, 
 
 p 1 p+i p 
 
 § 424 and (1), lim. [a«(a«- l)^q=oo = 0, or lim. (a « — a«)g=oo = 0. 
 
 p+i p 
 
 2. When a = 1, a « - a«= 0. 
 
 3. When a is positive and < 1, 
 
 In this case -> 1 and f - ) — 1, or a «— 1, is positive. 
 a \aj 
 
 Hence, by (1), lim. (a"'- l)^^* = 0. (2) 
 
 p+} 
 Since a is positive and < 1, a « cannot become greater than 1 
 however great q becomes. Therefore, § 424 and (2), 
 p+i _i p+i p 
 
 lim. [— a « (a *— l)]^=oc = 0, or lim. (a « — a«)5ix = 0. 
 
 It has been proved that lim. (a « — a«)gi » = for all positive real 
 
 values of the base a. This formula is true when - is substituted 
 
 1 .« 
 
 for a, since - is positive and real when a is positive and real, 
 a 
 
 p+i _p 
 Therefore, lim. (a « — a ^)q^!» = 0, 
 
 in which the values of x are negative. Hence, 
 
 When a is any positive real base, the difference between two 
 
 p+i p p+i 
 
 values of a' that include a" between them, as a ^ and a«, or a « 
 
 _p 
 and a ', can be made as small as we please by taking the difference 
 
 between the corresponding commensurable exponents sufficiently 
 
 small. 
 
 Therefore, when n is incommensurable and a is positive and 
 
 real, a" is defined as the limit of a^ when x = n, 
 
 435. When the change in a function of a variable correspond- 
 ing to an infinitesimal change in the variable is an infinitesimal, 
 the function is called a Continuous Function of the variable, or the 
 function is said to vary continuously with the variable. 
 
 a* is a continuous function of x when a is positive and real. 
 
392 
 
 INCOMMENSURABLE NUMBERS 
 
 436. Proofs of the laws of exponents. 
 
 Let m and n be incommensurable constant exponents, and a 
 any positive real number. Then, m and n may be defined by the 
 relations x = m and y = n, and by § 434, a" and a" may be defined 
 by the relations a' = a™ and a" = a", x and y being variable com- 
 mensurable numbers. 
 
 I. § 240, 
 
 a' ■a'' = a'+v. 
 
 (1) 
 
 §432, 
 
 lim. {a' • a") = lim. a*+». 
 
 (2) 
 
 §424, 
 
 lim. (ct" . W) = (lim. a^) (lim. a") 
 
 
 by definition. 
 
 = a" • a". 
 
 (3) 
 
 By definition, 
 
 lim. a'+'' = a^^". 
 
 (4) 
 
 By (3) and (4), 
 
 a" • a" = a^^". 
 
 (5) 
 
 II. By law I, 
 
 a""" • a" = a"'""''+" = a"*. 
 
 
 
 a"-" = a" ^ a". 
 
 (6) 
 
 III. § 249, 
 
 (a')'' = a^''. 
 
 (7) 
 
 By definition, 
 
 (a')" - (a^)" = when y = n, 
 
 (8) 
 
 and 
 
 (a*)" — (a")" = when x = m. 
 
 (9) 
 
 (8) + (9), §423, 
 
 (ay - (a"*)" = ; 
 
 
 that is, when x = m and y = n, lim. (a')* = (a")". 
 
 Again, by definition, lim. a'" = a"™ ^"^"^ 
 
 ft'''"' * ^''"' ' 
 
 = a"". 
 Since, § 432, by (7), lim. (a'^y = lim. a'*, 
 by (10) and (11), (a™)" = a"". 
 
 IV. § 250, Vaf = a'^, 
 
 1 1 
 
 or . (ay = a''. 
 
 Hence, by the preceding proof, 
 
 V. § 251, (aby = a'b'. 
 
 § 432, lim. (aby = lim. (a'J'). 
 
 By definition and § 424, (aby = a^'b". 
 
 (10) 
 
 (11) 
 
INTERPRETATION OF RESULTS 
 
 437. When the roots obtained by solving an equation satisfy 
 the equation, the solution has been properly performed ; but the 
 results found in solving a ])7-oblem may sometimes be at variance 
 with some condition of the problem. Consequently, the inter 
 pretation of results becomes important. 
 
 POSITIVE RESULTS 
 
 438. Since the numbers sought in the solution of a problem 
 are arithmetical rather than algebraic, when positive results are 
 obtained, it is not likely that they will conflict with the condi- 
 tions of the problem. Sometimes, however, even a positive result 
 violates one or more of the conditions of a problem. 
 
 In such cases the problem is impossible. 
 
 439. 1. A club consisting of 25 members raised the sum of 
 $ 13 by assessing the men 80 cents each and the women 40 cents 
 each. How many men were there, and how many women? 
 
 
 Solution 
 
 Let 
 
 X = the number of men. 
 
 Then, 
 
 2b — X — the number of women ; 
 
 
 |a; + §(25-a;)=13; 
 
 whence, 
 
 x = 7J, 
 
 and 
 
 26-x = 17^. 
 
 Though the numbers found will satisfy the equation, yet since 
 the number of men and the number of women cannot be frac- 
 tional, the problem is impossible. 
 
 393 
 
394 INTERPRETATION OF RESULTS 
 
 2. The second digit of a number expressed by two digits is 
 twice the first, and 4 times the first digit is 9 greater than the 
 second. What is the number ? 
 
 
 
 Solution 
 
 Let 
 
 X = the first digit. 
 
 Then, 
 
 2 « = the second digit ; 
 
 .•. 
 
 4a; = 2a; + 9; 
 
 whence, 
 
 a; = 4|, the first digit, 
 
 and 
 
 2 X = 9, the second digit. 
 
 While these numbers satisfy the equation, they fail to satisfy 
 the implied condition that the digits must be integers. 
 Hence, the problem is impossible. 
 
 NEGATIVE RESULTS 
 
 440. A few examples will suggest the methods to be employed 
 in the interpretation of negative results. 
 
 1. If A is 40 years old and B is 30, in how many years will A 
 
 be twice as old as B ? 
 
 Solution 
 
 Let X = the number of years after which A will be 
 
 twice as old as B. 
 
 Then, 40 + a: = 2 (30 + a:) ; 
 
 whence, x = — 20. 
 
 Though the result is algebraically correct, inasmuch as — 20 
 substituted for x satisfies the equation, nevertheless it is arith- 
 metically absurd. Hence, the conditions of the problem are 
 inconsistent with each other. Had the result been + 20, the 
 statement that A would be twice as old as B in 20 years would 
 have been arithmetically reasonable. However, since —x=+ 20, 
 the equation will give a result arithmetically reasonable, if — a; is 
 substituted for x ; that is, if x is taken to represent the number of 
 years since A was twice as old as B. 
 
 The conditions of the problem should, therefore, be modified as 
 follows : If A is 40 years old and B is 30, how many years ago 
 was A twice as old as B ? 
 
INTERPRETATION OF RESULTS 396 
 
 2. How much money has A, if \ of his money is 5 dollars more 
 than ^ of it ? 
 
 Solution 
 
 Let X = the number of dollars A has. 
 
 Then, 
 
 5-? = 5. 
 4 3 
 
 Solving X = - 60. 
 
 While the result — 60 satisfies the equation, it violates the sup- 
 position, made in the problem, that A has some money. 
 If — ic is substituted for x, the equation becomes 
 
 = 5 : whence, a; = 60. 
 
 3 4' ' 
 
 The problem when modified to express conditions arithmetically 
 reasonable will be : How much money has A, if \ of it is 5 dollars 
 less than ^ of it ; or, if — 60 dollars is interpreted as 60 dollars in 
 debt : How much money does A owe, if \ of what he owes is 5 
 dollars more than ^ of it ? 
 
 441. From the above discussions we may infer: 
 
 1. A negative result indicates that some quantity in the problem 
 has been ajyplied in the wrong direction. 
 
 2. A jxtssible problem analogous to the given j^roblem may be 
 formed by changing the absurd, conditions to their opposites. 
 
 Problems 
 
 442. Interpret arithmetically the negative results obtained by 
 solving the following : 
 
 1. If A is 40 years old and B is 25, in how many years will B 
 be half as old as A ? 
 
 2. Find the numbers whose sum is 6 and difference 10. 
 
 3. What fraction is equal to f if 1 is added to its numerator, 
 or to I, if 1 is added to its denominator ? 
 
 4. A boy bought some apples for 24 cents. Had he received 
 4 more for that sum, the cost of each would have been 1 cent less 
 How many did he buy ? 
 
396 INTERPRETATION OF RESULTS 
 
 5. A man worked 7 days, during which he had his son with 
 him 3 days, and received 22 shillings. He afterwards worked 
 5 days, during which he had his son with him 1 day, and received 
 18 shillings. What were the daily wages of each ? 
 
 ZERO RESULTS 
 
 443. When the result obtained by solving a problem is zero, it 
 may sometimes indicate that the problem is impossible, and some- 
 times it may be the proper answer to the question. 
 
 1. A dealer had two kinds of tea worth 75 and 60 cents per 
 pound, respectively. How many pounds of each must he take to 
 make a mixture of 45 pounds worth $27? 
 
 Solution 
 Let X = the number of pounds of the better kind. 
 
 Then, 45 — x = the number of pounds of the poorer kind ; 
 
 .-. I X + 1(45 - x) = 27 ; 
 whence, x = 0. 
 
 This result means that no such mixture can be made. In fact, 
 45 pounds of the poorer tea is worth $ 27. 
 
 2. A is 48 years old, and B is 16 years old. After how many 
 years will A be 3 times as old as B ? 
 
 Solution 
 Let X = the required number of years. 
 
 Then, 48 + x = 3(16 + x). 
 
 Solving, X = 0. 
 
 This result indicates that A is now 3 times as old as B. 
 
 3. What number is equal to the square of itself ? 
 
 Solution 
 Let X = the number. 
 
 Then, x = x^, 
 
 x(;x - 1) =0; 
 
 .-. X = 1 or 0. 
 These results indicate that no number except 1 is equal to the 
 square of itself. 
 
INTERPRETATION OF RESULTS 397 
 
 INDETERMINATE RESULTS 
 
 444. 1. A lady being asked her age replied, "If from 3 times 
 my age you take 4 years and divide the difference by 2, you will 
 have twice my age less half of my age 4 years hence." What was 
 her age ? 
 
 Solution 
 Let X = the number of years. 
 
 Then, 3^^ = 2x-^, ^ (1) 
 
 3a:-4 = 4x-x-4, (2) 
 
 (3-3)x = 0; (3) 
 
 .-. - = ?• (4) 
 
 Since (2) may be reduced to the identity 3a; — 4 = 3x — 4, it 
 may be satisfied by any value of a; whatever. This relation, 
 § 428, is expressed by (4). Hence, the problem is indeterminate. 
 
 Problems 
 
 445. 1. If twice a certain number is subtracted from the 
 square of the number, the result will be 1 less than the square 
 of a number 1 less. What is the number ? 
 
 2. A father is 30 years older than his son, and the sum of their 
 ages is 30 years less than twice their father's age. What is the 
 son's age ? 
 
 3. The sum of the first and third of three consecutive integers 
 is equal to twice the second. What are the integers ? 
 
 4. A bought 400 sheep in two flocks, paying $1.50 per head 
 for the first flock and $ 2 per head for the second. He lost 30 
 of the first flock and 56 of the second, but by selling the rest of 
 the first flock at $ 2 per head and the rest of the second at $ 2.50 
 per head, he neither lost nor gained. How many sheep were 
 there in each flock originally ? 
 
 5. A and B receive the same monthly salary. A is employed 
 10 months in the year and his annual expenses are f 600. B is 
 employed 8 months in the year and his annual expenses are $ 480. 
 If A saves as much money in 4 years as B saves in 5 years, what 
 is the monthly salary of each ? 
 
398 INTERPRETATION OF RESULTS 
 
 INFINITE RESULTS 
 
 446. An infinite result indicates that the problem is impossible. 
 
 1. If a man's yearly income is a dollars and his yearly ex- 
 penses are a dollars, in how many years will he have saved 6 
 
 dollars ? 
 
 Solution 
 
 Let X = the required number of years. 
 
 Then, x = — ^ = -, or «. 
 
 a — a 
 
 That is, he will never have saved h dollars in this way. 
 
 2. A reservoir is fitted with three pipes. One pipe can fill the 
 reservoir in 15 hours, the second can fill it in | of that time, and 
 the third pipe can empty it in 6 hours. If the reservoir is full 
 and the three pipes are opened, in what time will it be emptied ? 
 
 
 
 Solution 
 
 Let 
 
 
 X ■= the required number of hours. 
 
 Then, 
 
 16 10 
 
 1_1 
 6 X 
 
 Solving, 
 
 
 60 
 x = — , or oo. 
 
 
 That is, the reservoir will 7iever be emptied under these con- 
 ditions. 
 
 3. What numb 
 
 3r add 
 
 Led to both terms 
 
 make the fraction 
 
 equal 
 
 tol? 
 
 Solution 
 
 Let 
 
 
 X = the number. 
 
 Then, 
 
 
 1 + a;_j 
 
 2 + x 
 
 Clearing, 
 
 
 1 + a; = 2 + x. 
 
 Solving, 
 
 
 x-lor--^; 
 ' 
 
 that is, 
 
 
 X =+ CO or — a 
 
 Consequently, there is no such number ; but the larger the 
 number in numerical value, the nearer will the resulting fraction 
 approach the value 1. 
 
INTERPRETATION OF RESULTS 399 
 
 THE PROBLEM OF THE COURIERS 
 
 447. Two couriers, A and B, travel on the same road in the 
 direction from X to F at the rates of m and n miles an hour, 
 respectively. At a certain time, say 12 o'clock, A is at P, and B 
 is at Q, a miles from F. Find when and where they are together. 
 
 X. I F 
 
 P Q 
 
 Solution 
 
 Suppose that time reckoned from 12 o'clock toward a later time is positive, 
 and toward an earlier time, negative ; also, that distances measured from P 
 toward the right are positive, and toward the left, negative. 
 
 Let X represent the number of hours from 12 o'clock, and y the number 
 of miles from P, when A and B are together. Then, they will be together 
 y — a miles from Q. 
 
 Since A travels mx miles and B travels nx miles before they are together, 
 
 y = mx, (1) 
 
 and y — a = nx. (2) 
 
 Solving (1) and (2), 
 
 X = — - — , the required time. (3) 
 
 n 
 ma 
 
 , the required distance. (4) 
 
 DiscnssiON 
 
 1. When a>0 andm'^n. 
 
 When a > and m > n, the numerator and denominator in (.3) and also in 
 (4) are positive ; hence, x and y are positive. 
 
 That is, A overtakes B some time after 12 o'clock, somewhere at the right 
 of P. 
 
 2. When a > and m<n. 
 
 When a > and m < ?i, both x and y are negative. 
 
 That is, at 12 o'clock B is ahead of A and gaining on him, and they were 
 together some time before 12 o'clock and somewhere at the left of P. 
 
 3. When a > and m= n. 
 
 When a > and m = n, x and y are positive and infinitely great. 
 That is, at 12 o'clock B is ahead of A and traveling at the same rate ; 
 consequently, he will never be overtaken by A. 
 
400 INDETERMINATE EQUATIONS 
 
 4. When a = and m ^ n 07- m <^ n. 
 
 When a = and m> n or m <, u, x — and y = 0. 
 
 Ifm>n,x = +0 and y = -f- 0. That is, at 12 o'clock A and B are to- 
 gether, and A is passing B. 
 
 If m < w, X = - and y = - 0. That is, at 12 o'clock A and B are to- 
 gether, and B is passing A. 
 
 5. When a = and m = n. 
 
 When a = and m = n, x = - and « = - • 
 ^0 
 
 That is, A and B are together at 12 o'clock, and since they travel at the 
 
 same rate they will be together at all times. 
 
 olOic 
 
 INDETERMINATE EQUATIONS 
 
 448. While a problem that presents more unknown literal num- 
 bers than independent equations involving them is in general 
 indeterminate (§ 214), yet frequently by the introduction of a 
 condition or conditions not leading to equations, the number of 
 values of the unknown numbers may be limited and these values 
 algebraically determined. A common condition is that the results 
 shall be positive integers. 
 
 1. Solve the equation 5x -\-3y = 35 in positive integers. 
 
 Solution 
 
 Since x and y are positive integers, 5 x must be equal to 5 or a multiple of 
 5, and 3 y must be equal to 3 or a multiple of 3. Since the sura of these 
 multiples is 35, if the multiples of 5 are subtracted from 35, one or more of 
 the remainders will be a multiple of 3, if the problem is possible. 
 
 The only multiples of 5 that subtracted from 35 leave multiples of 3 are 
 6 and 4 times 5. 
 
 .•. X = 1 or 4 ; whence, y = 10 or 5. 
 
 Or, since x must be a positive integer and by transposition 5x = 35 — Sy, 
 the values of x must be 1, 2, 3, 4, 5, or 6, if the equation is possible. Sub- 
 stituting these values of x in the given equation and rejecting all those that 
 give negative or fractional values for y, the positive integral values are found 
 to be X = 1 or 4, and y = 10 or 6. 
 
INDETERMINATE EQUATIONS 401 
 
 2. Solve the equation 5 a; + 8 ?/ = 107 in positive integers. 
 
 
 Solution 
 
 
 
 
 
 bx + %y 
 
 = 107. 
 
 
 (1) 
 
 Dividing by 5, x + y +^: 
 
 5 
 
 = 21 + |. 
 5 
 
 
 (2) 
 
 Collecting integral and fractional terms, 
 
 
 
 
 
 x + y-2\: 
 
 _2-3j/ 
 5 
 
 
 (3) 
 
 Since x -\- y 
 
 - 21 is integral, ^~^^ = 
 5 
 
 = an integer. 
 
 
 (4) 
 
 If 2-3y_ 
 
 2-5 
 w, an integer, then, y = -— 
 
 — , which is in 
 
 the 
 
 fractional 
 
 5 ' o , , . 3 
 
 form. To avoid this, the coefl&cient of y in the number placed equal to w 
 
 2 — 3 u 
 
 should be made equal to unity. Since " is equal to an integer, any 
 
 5 
 multiple of it is equal to an integer. Since 5 is contained in 3 times — 3 y, 
 — 2 J/ times with a remainder of y, multiplying (4) by 3, 
 
 6- 9m 
 
 ^ = an mteger 
 
 = l_22/ + i±i^. (6) 
 
 5 
 
 Then, let -JtJM. = x^^ an integer. (6) 
 
 Solving for y, y = bw — \. (7) 
 
 Substituting in (3), z = 23 - 8 w. (8) 
 
 Equations (7) and (8) are called the general solution of the given equation 
 in integers. 
 
 To make y and x positive integers, it is evident from (7) that we must 
 take ro > ; and from (8) that we must take mj < 3. 
 
 Since w is an integer greater than and less than 3, w = 1 or 2, 
 
 When to = 1, x = 15, «/ = 4 ; 
 
 when 10 = 2, a; = 7, y = 9. 
 
 3. Determine whether the equation 10 a; + 15 = 53 may be 
 satisfied by integral values of x and y. 
 
 Solution. — Dividing by 5, 2 x + 3 ?/ = \5. 
 
 If X and y are integers, the first member is integral. 
 
 Since the first member is equal to the fraction ^^/, it cannot be an integer 
 Hence, x and y cannot be integers at the same time ; that is, the equation 
 is not satisfied by integral values of x and y. 
 
 ADV. ALG. — 26 
 
402 INDETERMINATE EQUATIONS 
 
 Solve the following equations in positive integers: 
 
 4. 5x-f32/ = 49. 8. 12a; + 5y = 61. 
 
 b. 3x + 2y = b. 9. Q>x + ly = 72. 
 
 6. 2a; + 7^ = 48. 10. 5x + 9i/ = 75. 
 
 7. 8a; + 5.v = 80. 11. 6a; + 9?/ = 100. 
 
 Find the least integral values of x and y in the following: 
 
 12. 2x = 9 + 3?/. 14. 7a;- 2^ = 6. 
 
 13. 5i/ = 2a;-|-7. 15. 5a;-3?/ = l. 
 
 f, 1 1 • fa; + ?/4-2 = 6l 
 
 16. Solve the equations i ,, \ va. positive integers. 
 
 {2x-\-y -z = l ] ^ ^ 
 
 Solution 
 x + y + z = 6. (1) 
 
 2 X + y - z = 7. (2) 
 
 Adding, 3 x + 2 y = 13. (3) 
 
 Solving (3) for positive integers, x = 3 and y = 2. 
 Substituting in (1), 2 = 1. 
 
 Solve the following equations in positive integers: 
 
 [ a; + ?/ + 2; = 8, f 3 a; + 2 ?/ = 17, 
 
 17. •! ^ ' 19. I ^ ' 
 I a;-?/ + 2 2 = 6. 1^ + 22 = 14. 
 
 18. |2- + 3, + . = 15, ^^^ iy + z = l, 
 l3x + i/-2 = 8. l3a;-2 = 7. 
 
 21. Separate 100 into two parts one of which is a multiple 
 of 11, and the other a multiple of 6. 
 
 22. In what v/ays may a weight of 19 pounds be weighed with 
 5-pound and 2-pound weights ? 
 
 23. A man has $.300 that he wishes to expend for cows and 
 sheep. If cows cost $ 45 apiece and sheep $ 6 apiece, how many 
 can he buy of each ? 
 
 24. If 9 apples and 5 oranges together cost 52 cents, what is 
 the cost of one of each ? 
 
INDETERMINATE EQUATIONS 403 
 
 25. A grocer sold two packages of sugar for f 1.25. One pack- 
 age contained a certain number of pounds of 7-cent sugar, the 
 other a certain number of pounds of 5-cent sugar. How many 
 pounds were there in each package ? 
 
 26. A man sold 9 animals — sheep, hogs, and cows — for $100. 
 If he received $ 3 for a sheep, $ 6 for a hog, and $ 35 for a cow, 
 how many of each did he sell ? 
 
 27. A woman expended 93 cents for 14 yards of cloth, some at 
 
 5, some at 7, and the rest at 10 cents a yard. How many whole 
 yards of each did she buy ? 
 
 28. Divide 74 into three parts that shall give integral quotients 
 when divided by 5, 6, and 7, respectively, the sum of which 
 quotients shall be 12. 
 
 29. A purse contained 30 coins, consisting of half-dollars, 
 quarters, and dimes. How many coins of each kind were there, 
 if their aggregate value was f 6.50 ? 
 
 30. A man bought 100 animals for $99. There were pigs, 
 sheep, and ducks. If he paid $ 6 for a pig, $ 4 for a sheep, and 
 50 cents for a duck, how many of each did he buy ? 
 
 31. What is the least number that will contain 25 with a 
 remainder of 1, and 33 with a remainder of 2 ? 
 
 32. Find the least number that divided by 10 and by 11 will 
 leave remainders of 3 and 6, respectively. 
 
 33. What is the least number that will contain 2, 3, 4, 5, and 
 
 6, each with a remainder of 1, and 7 without a remainder ? 
 
 34. A man selling eggs to a grocer took them out of his basket 
 4 at a time and there was 1 egg over. The grocer put them into 
 a box 5 at a time and there were 3 over. Both lost the count ; 
 but knowing that there were between 6 and 7 dozen eggs, the 
 grocer paid for 6^ dozen. How many eggs did he lose ? 
 
 35. Four boys have a pile of marbles. A throws away 1 and 
 takes \ of the remainder ; B throws away 1 and takes \ of the 
 remainder; C throws away 1 and takes \ of the remainder; D 
 throws away 1, and each boy takes \ of the remainder. At least 
 how many marbles must have been in the pile, and how many 
 does each boy now have ? 
 
MATHEMATICAL INDUCTION 
 
 449. Induction is generally understood as the process of infer- 
 ring a general principle from particular instances. But mathe- 
 matical induction diifers in some respects from ordinary induction. 
 Mathematical Induction is the process of proving a general prin- 
 ciple by means of a known fact together with a conditional 
 principle. The known fact is that the principle under consider- 
 ation is true for as many of the iirst consecutive cases as we 
 examine, and the conditional principle is that if it is true for the 
 wth case it holds true for the (w + l)th case. 
 
 This method of proof is illustrated by the following examples : 
 
 1. In the series of odd numbers 1, 3, 5, 7, •••, the first odd 
 number = 2(1) — 1 ; the second odd number = 2(2) — 1 ; the 
 third odd number = 2(3) — 1. From these particular instances 
 it may be inferred that the «th odd number = 2n — 1, and it is 
 now stated as a principle to be investigated that in the series 
 1, 3, 5, 7, •••, any odd number is 1 less than twice the number of 
 the term. 
 
 Supposing that the wth odd number is 2n — 1, by the law of 
 the given series the (n + l)th odd number is (2w — 1) -f 2. 
 
 But, §§ 56, 85, (2n - 1) + 2 = (2n + 2) - 1 = 2(71 + 1) - 1 ; 
 that is, the principle holds for the {n -\- l)th odd number on con- 
 dition that it holds for the nth odd number. 
 
 Therefore, since the principle is true for the third odd number 
 it holds for the fourth ; since it is true for the fourth it holds for 
 the fifth ; and so on. Hence, the principle is true generally'. 
 
 2. By trial x — y, x- — y-, ar' — i/^, x* — y*, and x^ — i/^ are each 
 found to be divisible by x —y. From these special cases it may 
 be inferred, and stated as a principle to be proved or disproved, 
 
 404 
 
MATHEMATICAL INDUCTION 405 
 
 that the difference of any like powers of two numbers is divisible 
 by the difference of the numbers. But the proved fact that x — y, 
 or — y"^, ••• , ar* — ?/* are each divisible hj x — y is no warrant for 
 accepting without further inquiry the statement that ic" — y^, for 
 example, is divisible by x — y. It is first necessary to know 
 whether a^ — y^ can be expressed in terms of these numbers 
 known to be exactly divisible by x — y. Since x^ — y^ = x^ — x^y 
 + x^y — if = afix — y) -\- y{x' — y^), each term of which has been 
 proved to be divisible hy x — y, x^ — y^ is divisible hy x — y; and 
 this illustrates the general proof. 
 
 For since a;"+' — ?/""^^ = a;"+' — x^y -\- x"// — y"+^ 
 = x\x-y)-]-y{x" -y"), 
 
 which is divisible by x —.y on condition that ic" — ?/" is divisible 
 by a; — y, it follows that a;"+^ — ?/"+' is divisible hy x — y on condi- 
 tion that x" — I/" is divisible hy x — y. 
 
 These two things then are proved : 
 
 First, the difference of the same jwwers of any two numbers, up 
 to and including the fifth powers, is divisible by the difference of the 
 numbers. 
 
 Second, if the difference of the nth powers of any two numbers is 
 divisible by the difference of the numbers, the difference of the next 
 higher jwivers is divisible by the difference of the numbers. 
 
 Therefore, since by actual division a^ — y^ has been proved to 
 be divisible by x — y, x^ — y^ is divisible hy x — y; since, as just 
 proved, x^ — y*^ is divisible hy x — y, x^ — y'' is divisible by a; — y ; 
 since, as just proved, x^ — y^ is divisible by x—y, x? — y^ is 
 divisible hy x — y; etc. 
 
 Hence, for all positive integral values of n, however great, 
 »" — y" is divisible hy x — y. 
 
 3. Let it be required to derive a formula for squaring any 
 polynomial. 
 
 By actual multiplication, rearranging terms, 
 (a + 6f = a2 + 62^_2a6; (1) 
 
 (a + & -f c)2 = a^ + 62 + c^ + 2 a& + 2 ac + 2 6c; (2) 
 
 (a + 6 + c + («)' = a^ + 6^ + c^ + d2 
 
 + 2 a6 + 2 ac + 2 ad + 2 6c + 2 6d + 2 cc?. (3) 
 
406 MATHEMATICAL INDUCTION 
 
 It is proved, then, that the square of a polynomial of n terms, 
 provided that n is not greater than 4, is equal to the sum of the 
 squares of the terms, plus twice the product of each term by 
 each term that follows it. It may be inferred that this principle 
 is of general application, but without further proof it does not 
 follow that it holds for polynomials of more than four terms. 
 ' Suppose, however, that the principle is true for a polynomial 
 of n terms, n being any positive integer. If this is so, then, 
 
 (Oi + 0-2 + 03+ •■• + (/„)- = af + a| + a| + ••• + < 
 
 + 2 «i(a2 + rtg + ••• + a„)+ 2 a^{a^ + 04 + ••• + a„) 
 + ...+2a„_ia„ (4) 
 
 is'the formula for the square of a polynomial of n terms. 
 
 Since {a^ + a, + 013+ \- a„+i)- = [a^+ia^+a^ + \- a^^-^J, 
 
 by (1), (ai + a, + 03 ••• + a„+i)' 
 
 = a\+ {a-i + 03 H + a„+i)- + 2 a,(a2 + 0,3 + ••• + a„+i) ; 
 
 if (4) is true, 
 
 = of + [a| + a| + ••• + a^^i + 2 a^ia^ + ••• + a„+i) 
 + 203(04+ ••• +o„+i)+ ••• +2a„a„+i] 
 
 + 201(02 + 03+ h a„+i); 
 
 rearranging, 
 
 = af + o^ + o? + ••• + o^+i + 2 ai(o2 + O3 + ••• a„+i) 
 + 2 02 (03 + ••• + a„+i) + 2 03(04 + ••• + o„^i) 
 + ••• +2a„o„+i. (5) 
 
 But (5) expresses the same law for forming the square of a 
 polynomial of » + l terms that (4) expresses for forming the square 
 of a polynomial of n terms. Hence, if the principle is true for 
 polynomials of n terms, it is true for polynomials of n + 1 terms. 
 
 By actual multiplication the principle has been proved true for 
 polynomials of two, three, and four terms, respectively. There- 
 fore, being true for polynomials of four terms, the principle holds 
 true for polynomials of five terms \ being true for polynomials of 
 
MATHEMATICAL INDUCTION 407 
 
 live terms, it holds true for polynomials of six terms ; and so on 
 indefinitely. 
 
 Hence, the principle is universally true. 
 
 4. Let it be required, to prove by mathematical induction that 
 12 + 22 + 3^ 4- ... +v? = i n(n + 1) (2n + 1). (1) 
 
 Supposing that (1) is true, then, Ax. 2, 
 
 124.22+32+ ... +/iH.(w+l)'=f<'* + IK2«+l)4-(n + l)0 
 § 85, = \{n -H l)[»i(2 % + 1) + 6(n + 1)] 
 
 = i(H + l)(2n2 + 7n+6) 
 = i(w + l)(n + 2)(2H+3) 
 §§56,85, =K«+1)(^^+1)(2*"^+1)- (2) 
 
 By comparing (2) with (1) it is seen that the sum of the 
 squares of the first {n + 1) integers has the same form with 
 respect to {n + 1) that the sum of the squares of the first n in- 
 tegers has with respect to n ; that is, it has been proved that if 
 the formula is true for n terms, it is true for n + 1 terms. 
 
 It can be verified by actual trial that the formula is true for 
 1, 2, 3, 4, ... terms, as far as we please. Supposing that the 
 verification stops with w = 5. By the above proof, since the 
 formula is true for five terms it holds true for six terms ; being 
 true for six terms it holds true for seven terms ; and so on indefi- 
 nitely. Hence, the principle expressed by (1) is true for all 
 integral values of n. 
 
 Examples 
 Prove by mathematical induction that : 
 
 1. 12 + 32 + 52+ ... to«terms=-(2n-l)(2n + l). 
 
 o 
 
 2. 1 . 3 + 2 . 4 + 3 . 5 + ... to w terms = -(2 ^2 + 9 w + 7). 
 
 6 
 
 3. 1.3-f3.5-f5.7-f...tow terms = - (4 ^2 + 6 w - 1). 
 
 o 
 
 4. a -f ar -\- ai^ + ••• to w terms = \ '-• 
 
 1 — r 
 
 5. ic2" — y2n is divisible hj x + y, n being a positive integer. 
 
THE BINOMIAL THEOREM 
 
 450. The Binomial Theorem derives a lormula by means of 
 which any power of a binomial may be expanded into a series, 
 whether the index of the power is positive or negative, integral 
 or fractional. 
 
 POSITIVE INTEGRAL EXPONENTS 
 451. The powers of (a + x), expanded in § 221, may be written 
 
 2 • 1 
 
 (a + x)- = a- 4- 2 ax + ^ — - a^. 
 
 (a + xy = a' + 3a'x + 'p-^ax^ + - 
 ^ 1-2 1 
 
 (a + xy = a'' 4- 5 a*x + ^^ aV + J 
 
 5.4 
 1.2' 
 
 ■of. 
 
 3 1.2.3-4 
 
 3-2.1 
 
 3-4.5 
 
 ■a^. 
 
 If the law of development revealed in the above is assumed to 
 apply to the expansion of any power of any binomial, as the wth 
 power of (a + x), the result is 
 
 ^ 1-2 1-2-3 ^ ^ 
 
 From formula (1) it is evident that in any term, 
 
 1. The exponent of x is 1 less than the number of the term. 
 Hence, the exponent of x in the (r + l)th term is r. 
 
 2. The exponent of a is n minus the exponent of x. 
 Hence, the exponent of a in the (r + l)th term is w — r. 
 
 408 
 
BINOMIAL THEOREM 409 
 
 3. The namber of factors in the numerator and in the denomi- 
 nator of the coefficient is 1 less than the number of the term. 
 
 Hence, the coefficient of the (?* -f l)th term has r factors in the 
 numerator and r factors in the denominator. 
 
 Therefore, the (r + l)th, or general term, is 
 
 n(n~l){n-2)-- tor factors ^„_,^^ ^2) 
 
 1 • 2 • 3 • • • to >• factors 
 
 Since, when there are two factors in the numerator, the last is 
 w — 1, when there are three factors, n — 2, when there are four 
 factors, n — 3, etc. ; when there are r factors, the last is n— (r— 1), 
 or w — r + 1. Hence, (2) may be written 
 
 n{n-V){n- 2) -" jn-r + l) ,^. 
 
 Hence, the full form of (1) is 
 
 This is called the Binomial Formula. 
 
 452. Since it has already been proved, by actual multiplica- 
 tion, that the binomial formula is true for the second, third, 
 fourth, and fifth powers of a binomial, it remains to discover 
 Avhether it is true for powers higher than the fifth. 
 
 If the binomial theorem, when assumed to be true for the nth 
 power, can be j)roved to be true for the {n + l)th power, it will 
 then have been proved to be true for the sixth power, since it is 
 known to be true for the fifth power ; also for the seventh power, 
 being true for the sixth power ; and in like manner for each suc- 
 ceeding power. 
 
 It then remains to prove that if (I) is true for the rtth power, 
 it will hold true for the (n -f l)th power. 
 
 To find the expansion of (a + x)"+* (I) may be multiplied by 
 a-faj. But since the (r-f l)th term of the product will be the 
 algebraic sum of a times the (r4-l)th term of {a+xy and x times 
 the rth term of (a+x)", (I) should be prepared for multiplication 
 
410 
 
 BINOMIAL THEOREM 
 
 by writing also the Hh term, obtained from the (r + l)th term byj 
 substituting r for r -f 1, or r — 1 for r. Then (I) is written 
 
 (a + xy = a" + na^-'x + ^^^"^^ a" V + ... 
 
 n{n-l){n-2)"'{n-r + 2,) +^ ^ 
 ■^ 1.2.3."(r-l) 
 
 ^ n(n-l)(n-2).-.(n-9- + 2)(n-r + l) ^n-r^ , , ^n 
 1 -2 -3 ••• (r — l)r 
 
 Multiplying both members by a + x, 
 
 (a + a;)"+i = a"+' + n 
 + 1 
 
 a"a; + 
 
 n (yi — 1) 
 
 1-2 
 
 + n 
 
 a"-'^x^ + 
 
 + 
 
 n (n — 1) (w — 2) ••• (n — r + 2) (n — r + 1) 
 
 + 
 
 1.2.3-(r-l)r 
 
 n(n — l)(n — 2) ••• (n — r + 2) 
 
 1.2.3..- (r- 1) 
 
 
 ,n+l 
 
 .n+1 
 
 = a"+^ + (n + l)a"a; + [V^^]_J^ + 
 
 Ai-r+l \ A^(n-l)(n-2).-.(n-r + 2) \ „_,^i ^_^ 
 "^V r "^ A l-2.3...(r-l) ; 
 
 = a»+i + (ri + 1) a»a; + ^^^-tl)^ a" ia;2 + . . . 
 
 J. * ^ 
 
 +(^)( "^'' - 1^.^; .-3^!. (;^' 1-) "" ^ '> -'"-^+ ■■■+' 
 
 That is, 
 
 (a + «)«+i = a"+i + (n + 1) a"a; + (^^-±l)l^a»-V + ... 
 
 (r.-fl)n(n-l)-.(n-r + 2)^„_,^t^ ... +a;»+i. (II) 
 
 Since upon comparison it may be seen that (II) and (I) have 
 the same form, ri + 1 in one taking the place of n in the other, 
 (II) and (I) must express the same law of formation. 
 
 Therefore, if the formula is true for the wth power, it holds 
 true for the (n + l)th power. 
 
BINOMIAL THEOREM 411 
 
 Hence, the binomial formula is true for any positive integral 
 exponent. 
 
 This method of proof is a proof by Mathematical Induction. 
 
 463. If — a; is substituted for x in (I), the terms that contain 
 the odd powers of — x will be negative, and the terms that con- 
 tain the even powers will be positive. Therefore, 
 
 (a-xY = a" - wa'-^x + ^<^ -^ l^ju-i^s _ gH) 
 
 1 ■ a 
 
 If a = 1, (I) becomes 
 
 454. From (I) it is seen that the last factor in the numerator 
 of the coefficient is n for the 2d term, n — 1 for the 3d term, 
 n — 2 for the 4th term, n — (n — 2), or 2, for the rith term, and 
 n — {n — 1), or 1, for the {n + l)th term ; and that the coefficient 
 of the (w + 2)th term, and that of each succeeding term, contains 
 the factor n — n, or 0, and therefore reduces to 0. Hence, 
 
 Wlien n is a positive integer, the series formed by expanding 
 {a + x)" is finite and has w -f 1 terms. 
 
 455. By formula (I), when n is a positive integer, 
 
 / . \n n , «-i , n(n — l) -_o'i , , n(w — 1) ••• 2 • 1 _ 
 
 (a + X)" = a" + «a" 'a; + -^^ ^a" -or + ••• -f -'^ ^. cc". 
 
 1-2 1 • 2 ••• (« — 1)71 
 
 / . \n ^ > n-1 , Mn — 1) „_2 •> , . n(n — 1) ••• 2 • 1 _ 
 
 (x 4- a)" = af -f nx^^a + -^^ ^aj" V + ••• + -^^ ^- a", 
 
 1-2 1 • 2 ••• (n — V)n 
 
 A comparison of the two series shows that : 
 . The coefficients of the latter half of the expansion of (a + a;)", when 
 n is a positive integer, are the same as those of the first half, written 
 in the reverse order. 
 
 Examples 
 1. Expand (3 a - 2 by. 
 
 Solution. — Substituting 3 a for a, 2 6 for x, and 4 for n in (III), 
 (3a-26)* = (3a)*-4(3«)3(26)+i^(.3a)2(26)2-l^|-l|(3a)(26)« 
 
 1.2.3.4^ ^ 
 = 81 a* - 216 a% + 216 a%^ - 96 afts + 16 6*. 
 
412 BINOMIAL THEOREM 
 
 2. Expand f^ + bx\\ 
 
 Solution 
 
 Since (f + ''-y=[^l+2x)J=|(l + 2.)S, 
 
 (1 + 2 a;)5 may be expanded by (IV), and the result multiplied by -• 
 
 0J4 
 
 3. 
 4. 
 
 cpand : 
 
 (b - 7iy. 
 
 (i+a-y. 
 
 10. 
 
 MJ- 
 
 15. 
 
 {''-:-j- 
 
 5. 
 
 (2-3aj)«. 
 
 11. 
 
 fa _ x\\ 
 \x a) 
 
 16. 
 
 n~l 1 
 
 (^x~^-xy. 
 
 6. 
 
 (x^ - xf. 
 
 
 \^ yJ 
 
 17. 
 
 {ax "^ —b^xf. 
 
 7. 
 
 (x + x-y. 
 
 12. 
 
 18. 
 
 f^~a VbY 
 \^/b Va'J 
 
 
 
 
 
 
 8 
 
 (2 a + V^)^. 
 
 13. 
 
 (\/a^ + </F«)3. 
 
 
 1 — / — 
 
 9 
 
 (a + aVa)*. 
 
 14. 
 
 (2V2-^3f. 
 
 19. 
 
 m^u 
 
 456. To find any term. 
 
 Any term of the expansion of a power of a binomial may be 
 obtained by substitution in (2) or (3), § 451. 
 
 In the expansion of a power of the difference of two numbers, as (a — x)", 
 since the exponent of x in the (r + l)th term is r, the sign of the general 
 term is + if ?• is even, and — if r is odd. 
 
 Examples 
 1. Find the 12th term of (a - 6)". 
 
 Solution 
 
 12th term = 14 • 13 • 12 . 11 ■ 10 ■ ■ 8 • 7 ■ 6 . 5 ■ 4 3 
 
 1.2.3.4.5.6.7.8.9.10.11 ^ ^\i, 
 
 = _ 14 • 13 -12 ^3511 = _ 364 a%n_ U^ 
 Or 
 
 By § 4.55, since there are 15 terms, the coefBcient of the 12th terra, or the 
 4th term from the end, is equal to that of the 4th term from the beginning. 
 
 .-. 12th term = - ^jJ^jJ3 oSju _ _ 364 38511, 
 1 • ^ • o 
 
BINOMIAL THEOREM 413 
 
 2. In the expansion of (ic^+2iB)", find the term containing x^. 
 Solution .— Since (x^ + 2 x)" = fa^^f 1 + - ) 1 = a;''^( 1 + - ) , every term 
 
 of the series expanded from M + - j will be multiplied by x"^^^. 
 
 Hence, the term sought is that which contains | - J , or — ; that is, the 
 (7 + l)th, or 8th term. ^^' ^' 
 
 8th term = x'^'^ W-^^-^-'^ f-V = 42240 xi*. 
 
 1.2.3.4 vx; 
 
 3. Find the 4th term of (a + 2)^". 
 
 4. Find the 4th term of {x - 3?/)'2. 
 
 5. Find the 8th term of {x + yf. 
 
 6. Find the 5th term of {x-1 yf\ 
 1. Find the 3d term of {a- - a-y. 
 
 8. Find the 20th term of (1 + xf\ 
 
 9. Find the 16th term of (1-2 x)"^. 
 
 10. Find the middle term of {a + 3 h)\ 
 
 11. Find the 6th term of (x-\-^°- 
 
 12. Find the middle term off ^ — -] • 
 
 
 
 13. Find the two middle terms of f - - 
 
 aj 
 
 
 lA ^^'^n^^ fVio r>r«oflRr>ionf r>-f n^ in i-\\o a•v^ 
 
 riancinn r\f 1 
 
 n^ _1_ «^5 
 
 457. The formula given for the expansion of (a + x)" is true, 
 tinder certain conditions, for all commensurable values of «, 
 whether they are positive, negative, integral, or fractional, and 
 the student will, therefore, be able to expand such expressions ; 
 but the proof for negative and fractional exponents and the dis- 
 cussion of the conditions under which the expansion for these 
 exponents gives the true valvie of (a + aj)" will be deferred. (See 
 pages 520-523.) 
 
 In the expansion of (o + x)", if n is negative or fractional, 
 
 none of the binomial coefficients, ^^ ~ — L — — ~ >\' ~ -^i g^p 
 
 1 • 2 ' 1-2.3 
 
 can become ; consequently, when such exponents are given, the 
 series developed can have no end. 
 
414 BINOMIAL THEOREM 
 
 Examples 
 
 1. Expand (1 — y)-'^ and find its (r + l)th term. 
 Solution. — Substituting 1 for a, y for x, and — 1 for n in (III), 
 
 (1 - 2/)-l = 1-1 -(- 1) l-2y + ^11=^ 1-3 y2 _ ^iI:i|lIzLlll-4y8 + ... 
 
 = 1 + 2/ + 2/- -I- 2/^ + •••• 
 Tlie (r + l)th term is evidently y^. 
 
 Since (1 — y)"^ = , the above expansion of (1 — j/)"^ may be verified 
 
 by division. ~ ^ 
 
 2. Expand (a -|- x)^ to five terms and find the 10th term. 
 
 Expand to four terms : 
 
 3. {a + h)^. 8. VIT^. 13. (l+a;)l 
 
 4. (a-|-6)"i 9. V(9 - xf. 14. (l-ha)-\ 
 
 5. (a -6)1 10. {a^-x-^)i. 15. (1 _ a)-i. 
 
 6. </{a - &)•''. 11. (a^-x*)-«. 16. (1 - a;)-^ 
 
 1 
 
 7. ,^ 12. ( ,- 3/- )- 17. (l-xV 
 
 ^(a-6/ VVa-Va;/ ^ ^ 
 
 18. Find the (r + l)th term of {a + x)^. 
 
 19. Show that 
 
 {\-x-x^-' = 1 + 0; + 2.t2 + 3ar'+ 5a;* + 8ar'+ 13.T«-1- 21 x" + 
 
 20. Eind the square root of 24 to three decimal places. 
 Solution. V24 =(24)2 = (25 - 1)^= (25)^(1 - ^.)^= 5(1 - ^V)^- 
 
 = 5 - .1 - .001 - .00002 = 4.89898 - = 4.899, nearly. 
 
 Find the values of the following to three decimal places : 
 
 21. Vs. 23. V26. 25. V9. 
 
 22. VT7. 24. -s/25. 26. </M. 
 
LOGARITHMS 
 
 458. 1. What power of 3 is 9 ? 27? 81? 243? 729? 
 
 2. What power of 5 is 25 ? 125? 625? 3125? 5? 1? i? 
 
 3. Express 100 as a power of 10 ; 1000 as a power of 10 ; 10,000 
 as a power of 10 ; 10 as a power of 10 ; 1 as a power of 10. 
 
 459. The exponent of the power to which a fixed number called 
 the Base must be raised in order to produce a given number is 
 called the Logarithm of the given number. 
 
 When 10 is the base, the logarithm of 100 is 2, for 100 = 10^ ; the logarithm 
 of 1000 is 3, for 1000 = 10^ ; the logarithm of 10,000 is 4, for 10,000 = 10*. 
 
 460. When a is the base, x the exponent, and m the given 
 number, x is the logarithm of the number m to the base a. 
 
 It is written log„ m = x. 
 
 When the base is 10, it is not indicated. 
 
 Thus, the logarithm of 100 to the base 10 is 2. It is written log 100 = 2. 
 
 461. Logarithms may be computed with any arithmetical 
 number except unity as a base, but the base of the Common or 
 Briggs System of logarithms is 10. 
 
 Since 10° = 1, the logarithm of 1 is 0. 
 Since 10' = 10, the logarithm of 10 is 1. 
 Since 10^ = 100, the logarithm of 100 is 2. 
 Since 10^ = 1000, the logarithm of 1000 is 3. 
 Since 10~' = ^, the logarithm of .1 is — 1. 
 Since 10~^ = y^^, the logarithm of .01 is — 2. 
 
 462. It is evident, then, that the logarithm of any number 
 between 1 and 10 is a number greater than and less than 1. 
 For example, the logarithm of 4 is approximately 0.6021. Again, 
 the logarithm of any number between 10 and 100 is a number 
 greater than 1 and less than 2. For example, the logarithm of 
 50 is approximately 1.6990. 
 
 416 
 
416 ^ LOGARITHMS 
 
 Most logarithms are incommensurable uurabers. All the laws 
 established for commensurable exponents apply also to incom- 
 mensurable exponents. The proofs for incommensurable expo- 
 nents are given in § 436. 
 
 463. The integral part of a logarithm is called the Characteris- 
 tic; the fractional or decimal part, the Mantissa. 
 
 In log 60 = 1.6990, the characteristic is 1 and the niarcissa .6990. 
 
 464. The following examples will illustrate the cJiaracteristic 
 and mantissa, and their significance : 
 
 log 458/ = 3.6609 ; that is, 4580 = lO'^'^. 
 
 that is, 458.0 = lO^*^. 
 
 that is, 45.80 = 10^*^. 
 
 that is, 4.580 =10"'^. 
 
 that is, .4580 = 10-i+«*». 
 
 that is, .0458 ^ 10-2+ ««»; 
 that is, .00458 = 10-''+««». 
 
 log458A =2.6609 
 log45.8i =1.6609 
 log4.58J> =0.6609 
 log .4580 =1.6609 
 log .0458 = 2.6609 
 log .00458 = 3.6609 
 
 From the above examples it is evident that : 
 
 465. Principles. — 1. The characteristic of the logarithm of a 
 mimber greater than 1 is positive and 1 less than the number of 
 digits in its integral part. i 
 
 2. The characteristic of the logarithm of a decimal is negative and 
 numerically 1 greater than the number of ciphers immediately follow- 
 ing the decimal point. 
 
 466. To avoid writing a negative characteristic before a positive 
 mantissa, it is customary to add 10 or some multiple of 10 to the 
 negative characteristic, and to indicate that the number added is 
 to be subtracted from the whole logarithm. 
 
 Thus, f.6609 is written_9.6609 - 10 ; 2.3010 is written 8.3010 - 10 ; 14.9031 
 is written 6.9031 - 20 ; 28.8062 is written 2.8062 - 30 ; etc. 
 
 467. It is evident, also, from the examples, that in the loga- 
 rithms of numbers expressed by the same figures in the same 
 order, the decimal parts, or mantissas, are the same, and that the 
 logarithms differ only in their characteristics. Hence, tables of 
 logarithms contain only the mantissas. 
 
LOGARITHMS 417 
 
 468. The table of logarithms on the two following pages gives 
 the decimal parts, or mantissas, correct to four places, for the 
 common logarithms of all numbers from 1 to 1000. 
 
 469. To find the logarithm ofjajumfeer. - ' 
 
 Examples 
 
 1. Find the logarithm of 765. >/ 
 
 Solution. — In the following table the letter N designates a vertical 
 column of numbers from 10 to 99 inclusive, and also a horizontal row of 
 figures 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The first two figures of 765 appear as the 
 number 76 in the vertical colunm marked N on page 419, and the third figure 
 5 in the horizontal row marked N. 
 
 In the same horizontal row as 76 are found the mantissas of the logarithms 
 of the numbers 760, 761, 762, 763, 764, 765, etc. The mantissa of the loga- 
 rithm of 765 is found in this row under 5, the third figure of 765. It is 8837 
 and means .8837. 
 
 By Prin. 1, the characteristic of the logarithm of 765 is 2. 
 
 Hence, the logarithm of 765 is 2.8837. 
 
 2. Find the logarithm of 4. ^^^ 
 
 SoLiTiox. — Although the numbers in the table ap^^^B begin with 100, 
 the table really includes all numbers from 1 to 1000, siiSBKmbers expressed 
 by less than three figures may be expressed by three figures by adding deci- 
 mal ciphers. Since 4 = 4.00, and since, § 467, the mantissa of the logarithm 
 of 4.00 is the same as that of 400, which is .6021, the mantissa of the loga- 
 rithm of 4 is .6021. 
 
 By Prin. 1, the characteristic of the logarithm of 4 is 0. 
 
 Therefore, the logarithm of 4 is 0.6021. 
 
 Verify the following from the table : 
 
 3. log 10 =L0000. 9. log .2 =9.8010-10. 
 
 4. log 100 = 2.0000. 10. log 542 =2.7340. 
 
 5. log 110 = 2.0414. 11. log 345 =2.5378. 
 
 6. log 2 =0.3010. 12. log 5.07 = 0.7050. 
 
 7. log 20 =1.3010. 13. log 78.5 = 1.8949. 
 
 8. log 200 = 2.3010. 14. log .981 = 9.9917 - 10. 
 
 ADV. Al.O.— 27 
 
418 
 
 LOGARITHMS 
 
 Table of Common Logarithms 
 
 N 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 lO 
 
 0000 
 
 0043 
 
 0086 
 
 0128 
 
 0170 
 
 0212 
 
 0253 
 
 0294 
 
 0334 
 
 0374 
 
 II 
 
 0414 
 
 0453 
 
 0492 
 
 0531 
 
 0569 
 
 0607 
 
 0645 
 
 0682 
 
 0719 
 
 0755 
 
 12 
 
 0792 
 
 0828 
 
 0864 
 
 0899 
 
 0934 
 
 0969 
 
 1004 
 
 1038 
 
 1072 
 
 1106 
 
 13 
 
 "39 
 
 "73 
 
 1206 
 
 1235 
 
 1271 
 
 1303 
 
 1335 
 
 1367 
 
 1399 
 
 1430 
 
 14 
 
 1461 
 
 1492 
 
 1523 
 
 i£^3 
 
 1584 
 
 1614 
 
 1644 
 
 1673 
 
 1703 
 
 1732 
 
 15 
 
 1761 
 
 1790 
 
 1818 
 
 184^ 
 
 1875 
 
 1903 
 
 1931 
 
 1959 
 
 1987 
 
 2014 
 
 16 
 
 2041 
 
 2068 
 
 2095 
 
 2122 
 
 2148 
 
 2175 
 
 2201 
 
 2227 
 
 2253 
 
 2279 
 
 17 
 
 2304 
 
 2330 
 
 2355 
 
 2380 
 
 2405 
 
 2430 
 
 2455 
 
 2480 
 
 2504 
 
 2529 
 
 18 
 
 2553 
 
 2577 
 
 2601 
 
 2625 
 
 2648 
 
 2672 
 
 2695 
 
 2718 
 
 2742 
 
 2765 
 
 19 
 
 2788 
 
 2810 
 
 2833 
 
 2856 
 
 2878 
 
 2900 
 
 2923 
 
 2945 
 
 2967 
 
 2989 
 
 20 
 
 3010 
 
 3032 
 
 3054 
 
 3075 
 
 3096 
 
 3118 
 
 3139 
 
 3160 
 
 3181 
 
 3201 
 
 21 
 
 3222 
 
 3243 
 
 3263 
 
 3284 
 
 3304 
 
 3324 
 
 3345 
 
 3365 
 
 3385 
 
 3404 
 
 22 
 
 3424 
 
 3444 
 
 3464 
 
 3483 
 
 3502 
 
 3522 
 
 3541 
 
 3560 
 
 3579 
 
 3598 
 
 23 
 
 3617 
 
 3636 
 
 3655 
 
 3674 
 
 3692 
 
 37" 
 
 3729 
 
 3747 
 
 3766 
 
 3784 
 
 24 
 
 3802 
 
 3820 
 
 3838 
 
 3856 
 
 3874 
 
 3892 
 
 
 3927 
 
 3945 
 
 3962 
 
 25 
 
 3979 
 
 3997 
 
 4014 
 
 4031 
 
 4048 
 
 4065 
 
 4082 
 
 4099 
 
 4116 
 
 4133 
 
 26 
 
 4150 
 
 4166 
 
 4183 
 
 4200 
 
 4216 
 
 4232 
 
 4249 
 
 4265 
 
 4281 
 
 4298 
 
 27 
 
 4314 
 
 4330 
 
 4346 
 
 4362 
 
 4378 
 
 4393 
 
 4409 
 
 4425 
 
 4440 
 
 4456 
 
 28 
 
 4472 
 
 4487 
 
 4502 
 
 4518 
 
 4533 
 
 4548 
 
 4564 
 
 4579 
 
 4594 
 
 4609 
 
 29 
 
 4624 
 
 4639 
 
 4654 
 
 4669 
 
 4683 
 
 4698 
 
 4713 
 
 4728 
 
 4742 
 
 4757 
 
 30 
 
 4771 
 
 4786 
 
 4800 
 
 4814 
 
 4829 
 
 4843 
 
 4857 
 
 4871 
 
 4886 
 
 4900 
 
 31 
 
 4914 
 
 4928 
 
 4942 
 
 4955 
 
 4969 
 
 4983 
 
 4997 
 
 501 1 
 
 5024 
 
 5038 
 
 32 
 
 5051 
 
 5065 
 
 5079 
 
 5092 
 
 5105 
 
 5"9 
 
 5132 
 
 5145 
 
 5159 
 
 5172 
 
 33 
 
 5185 
 
 5198 
 
 5211 
 
 5224 
 
 5237 
 
 5250 
 
 5263 
 
 5276 
 
 5289 
 
 5302 
 
 34 
 
 5315 
 
 5328 
 
 5340 
 
 5353 
 
 5366 
 
 5378 
 
 5391 
 
 5403 
 
 5416 
 
 5428 
 
 35 
 
 5441 
 
 5453 
 
 5465 
 
 5478 
 
 5490 
 
 5502 
 
 S5H 
 
 5527 
 
 5539 
 
 5551 
 
 36 
 
 55^3 
 
 5575 
 
 5587 
 
 5599 
 
 5611 
 
 5623 
 
 5635 
 
 5647 
 
 5658 
 
 5670 
 
 37 
 
 5682 
 
 5694 
 
 5705 
 
 5717 
 
 5729 
 
 5740 
 
 5752 
 
 5763 
 
 5775 
 
 5786 
 
 38 
 
 5798 
 
 5809 
 
 5821 
 
 5832 
 
 5843 
 
 5855 
 
 5866 
 
 5877 
 
 5888 
 
 5899 
 
 39 
 
 5911 
 
 5922 
 
 5933 
 
 5944 
 
 5955 
 
 5966 
 
 5977 
 
 5988 
 
 5999 
 
 6010 
 
 40 
 
 6021 
 
 6031 
 
 6042 
 
 6053 
 
 6064 
 
 6075 
 
 6085 
 
 6096 
 
 6107 
 
 6117 
 
 41 
 
 6128 
 
 6138 
 
 6149 
 
 6160 
 
 6170 
 
 6180 
 
 61^1 
 
 6201 
 
 6212 
 
 6222 
 
 42 
 
 6232 
 
 6243 
 
 6253 
 
 6263 
 
 6274 
 
 6284 
 
 6294 
 
 6304 
 
 6314 
 
 6325 
 
 43 
 
 6335 
 
 6345 
 
 6355 
 
 6365 
 
 6375 
 
 6385 
 
 6393 
 
 6405 
 
 6415 
 
 6425 
 
 44 
 
 6435 
 
 6444 
 
 6454 
 
 6464 
 
 6474 
 
 64041. 
 
 6493 
 
 6503 
 
 6513 
 
 6522 
 
 45 
 
 6532 
 
 6542 
 
 6551 
 
 6561 
 
 6571 
 
 6580 
 
 6590 
 
 6599 
 
 6605 
 
 6618 
 
 46 
 
 6628 
 
 6637 
 
 6646 
 
 6656 
 
 6665 
 
 6675 
 
 6684 
 
 6693 
 
 6702 "6712 1 
 
 47 
 
 6721 
 
 6730 
 
 673^1 
 
 674t> 
 
 6758 
 
 6767 
 
 6776 
 
 6785 
 
 6794 
 
 6803 
 
 48 
 
 6812 
 
 6821 
 
 6830 
 
 6839 
 
 6848 
 
 6857 
 
 6866 
 
 6875 
 
 6884 
 
 6893 
 
 49 
 
 6902 
 
 6911 
 
 6920 
 
 6928 
 
 6937 
 
 6946 
 
 6955 
 
 6964 
 
 6972 
 
 6981 
 
 50 
 
 6990 
 
 6998 
 
 7007 
 
 7016 
 
 7024 
 
 7033 
 
 7042 
 
 7050 
 
 7059 
 
 '067 
 
 51 
 
 7076 
 
 7084 
 
 7093 
 
 7101 
 
 7110 
 
 7118 
 
 7126 
 
 7>35 
 
 7 '43 
 
 7152 
 
 52 
 
 7160 
 
 7168 
 
 7177 
 
 7185 
 
 7193 
 
 7202 
 
 7210 
 
 TJlS 
 
 • 7326 
 
 J^ZS 
 
 53 
 
 7243 
 
 7251 
 
 7259 
 
 7267 
 
 7275 
 
 7284 
 
 7292 
 
 
 
 ') 
 
 54 
 
 7324 
 
 7332 
 
 7340 
 
 7348 
 
 7356 
 
 7364 
 
 7372 
 
 7 
 
 r®" 
 
 9 
 
 N 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 ~5~ 
 
 6 
 
y 
 
 'y 
 
 . r 
 
 V 
 
 LOGARITHMS 
 
 419 
 
 Table of Common Logarithms 
 
 N 
 
 
 
 1 
 
 2 
 
 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 55 
 
 7404 
 
 7412 
 
 7419 
 
 7427 
 
 7435 
 
 7443 
 
 7451 
 
 7459 
 
 7466 
 
 7474 
 
 56 
 
 7482 
 
 7490 
 
 7497 
 
 7505 
 
 7513 
 
 7520 
 
 7528 
 
 7536 
 
 7543 
 
 7551 
 
 57 
 
 7559 
 
 7566 
 
 7574 
 
 7582 
 
 7589 
 
 7597 
 
 7604 
 
 7612 
 
 7619 
 
 7627 
 
 58 
 
 7634 
 
 7642 
 
 7649 
 
 7657 
 
 7664 
 
 7672 
 
 7679 
 
 7686 
 
 7694 
 
 7701 
 
 59 
 
 7709 
 
 7716 
 
 7723 
 
 7731 
 
 7738 
 
 7745 
 
 7752 
 
 7760 
 
 7767 
 
 7774 
 
 60 
 
 7782 
 
 7789 
 
 7796 
 
 7803 
 
 7810 
 
 7818 
 
 7825 
 
 7832 
 
 7839 
 
 7846 
 
 61 
 
 7853 
 
 7860 
 
 7868 
 
 7875 
 
 7882 
 
 7889 
 
 7896 
 
 7903 
 
 7910 
 
 7917 
 
 62 
 
 7924 
 
 7931 
 
 7938 
 
 7945 
 
 7952 
 
 7959 
 
 7966 
 
 7973 
 
 7980 
 
 7987 
 
 63 
 
 7993 
 
 8000 
 
 8007 
 
 8014 
 
 8021 
 
 8028 
 
 8035 
 
 8041 
 
 8048 
 
 8055 
 
 64 
 
 8062 
 
 8069 
 
 8075 
 
 8082 
 
 8089 
 
 8096 
 
 8102 
 
 8109 
 
 8116 
 
 8122 
 
 65 
 
 8129 
 
 8136 
 
 8142 
 
 8149 
 
 8156 
 
 8162 
 
 8169 
 
 8176 
 
 8182 
 
 8189 
 
 66 
 
 8195 
 
 8202 
 
 8209 
 
 8215 
 
 8222 
 
 8228 
 
 8235 
 
 8241 
 
 8248 
 
 8254 
 
 67 
 
 8261 
 
 8267 
 
 8274 
 
 8280 
 
 8287 
 
 8293 
 
 8299 
 
 8306 
 
 8312 
 
 8319 
 
 68 
 
 8325 
 
 8331 
 
 8338 
 
 8344 
 
 8351 
 
 8357 
 
 8363 
 
 8370 
 
 8376 
 
 8382 
 
 69 
 
 8388 
 
 8395 
 
 8401 
 
 8407 
 
 8414 
 
 8420 
 
 8426 
 
 84^ 
 
 8439 
 
 8445 
 
 70 
 
 8451 
 
 8457 
 
 8463 
 
 8470 
 
 8476 
 
 8482 
 
 8488 
 
 8494 
 
 8500 
 
 8506 
 
 7X 
 
 8513 
 
 8519 
 
 8525 
 
 8531 
 
 8537 
 
 8543 
 
 8549 
 
 8555 
 
 8561 
 
 8567 
 
 72 
 
 8573 
 
 8579 
 
 8585 
 
 8591 
 
 8597 
 
 8603 
 
 8609 
 
 8615 
 
 8621 
 
 8627 
 
 73 
 
 8633 
 
 8639 
 
 8645 
 
 8651 
 
 8657 
 
 8663 
 
 8669 
 
 8675 
 
 8681 
 
 8686 
 
 74 
 
 8692 
 
 8698 
 
 8704 
 
 8710 
 
 8716 
 
 8722 
 
 8727 
 
 8733 
 
 8739 
 
 8745 
 
 75 
 
 8751 
 
 8756 
 
 8762 
 
 8168 
 8fc5 
 
 8774 
 
 8779 
 
 8785 
 
 8791 
 
 8797 
 
 8802 
 
 76 
 
 8808. 
 
 8814 
 
 8820 
 
 8831 
 
 -8837 
 
 8842 
 
 8848 
 
 8854 
 
 8859 
 
 77 
 
 8865 
 
 8871 
 
 8876 
 
 8882 
 
 8887 
 
 8893 
 
 8899 
 
 8904 
 
 8910 
 
 8915 
 
 78 
 
 8921 
 
 8927 
 
 8932 
 
 8938 
 
 8943 
 
 8949 
 
 8954 
 
 8960 
 
 8965 
 
 8971 
 
 79 
 
 8976 
 
 8982 
 
 8987 
 
 8993 
 
 8998 
 
 9004 
 
 9009 
 
 9015 
 
 9020 
 
 9025 
 
 80 
 
 9031 
 
 9036 
 
 9042 
 
 9047 
 
 9053 
 
 9058 
 
 9063 
 
 9069 
 
 9074 
 
 9079 
 
 81 
 
 9085 
 
 9090 
 
 9096 
 
 9101 
 
 9106 
 
 9112 
 
 9117 
 
 9122 
 
 9128 
 
 9133 
 
 82 
 
 9138 
 
 9143 
 
 9149 
 
 9154 
 
 9159 
 
 9165 
 
 9170 
 
 9175 
 
 9180 
 
 9186 
 
 83 
 
 9191 
 
 9196 
 
 9201 
 
 9206 
 
 9212 
 
 9217 
 
 9222 
 
 9227 
 
 9232 
 
 9238 
 
 84 
 
 9243 
 
 9248 
 
 9253 
 
 9258 
 
 9263 
 
 9269 
 
 9274 
 
 9279 
 
 9284 
 
 9289 
 
 85 
 
 9294 
 
 9299 
 
 9304 
 
 9309 
 
 9315 
 
 9320 
 
 9325 
 
 9330 
 
 9335 
 
 9340 
 
 86 
 
 9345 
 
 9350 
 
 9355 
 
 9360 
 
 9365 
 
 9370 
 
 9375 
 
 9380 
 
 9385 
 
 9390 
 
 87 
 
 9395 
 
 9400 
 
 9405 
 
 9410 
 
 9415 
 
 942e. 
 
 9425 
 
 9430 
 
 9435 
 
 9440 
 
 88 
 
 9445 
 
 9450 
 
 9455 
 
 9460 
 
 9465 
 
 9469 
 
 9474 
 
 9479 
 
 9484 
 
 9489 
 
 89 
 
 9494 
 
 9499 
 
 9504 
 
 9509 
 
 9513 
 
 9518 
 
 9523 
 
 9528 
 
 9533 
 
 9538 
 
 90 
 
 .9 = 
 
 i?ll^ 
 
 9557 
 
 9562 
 
 9566 
 
 9571 
 
 9576 
 
 9581 
 
 9586 
 
 91' 
 
 c 
 
 9605 
 
 9609 
 
 9614 
 
 9619 
 
 9624 
 
 9628 
 
 9633 
 
 92 
 
 9 
 
 ! 9647 ' 
 
 9652 
 
 9657 
 
 9661 
 
 9666 
 
 9671 
 
 9675 
 
 9680 
 
 93 
 
 c 
 
 9601 
 
 9699 
 
 9703 
 
 9708 
 
 9713 
 
 9717 
 
 9722 
 
 9727 
 
 94 < 
 
 97J5 
 
 gjy> 
 
 9754 
 
 9759 
 
 9763 
 
 9768 
 
 9773 
 
 95 
 
 9791 
 
 9795 
 
 9800 
 
 9805 
 
 9809 
 
 9814 
 
 9818 
 
 96 
 
 : 9836 
 
 9841 
 
 9845 
 
 9850 
 
 9854 
 
 9859 
 
 9863 
 
 97 
 
 7 9^1 
 
 9886 
 
 9890 
 
 9894 
 
 9899 
 
 9903 
 
 9908 
 
 98 
 
 I i 9926 
 
 99*30 
 
 9934 
 
 9939 
 
 9943 
 
 9948 
 
 9952 
 
 99 
 
 N| 
 
 1 
 
 '5 
 
 9969 
 
 9974 
 
 9978 
 
 9983 
 
 9987 
 
 9991 
 
 9996 
 
 f 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
420 LOGARITHMS 
 
 15. Find the logarithm of 6253. 
 
 Solution. — Since the table contains the mantissas not only of the loga- 
 rithms of numbers expressed by three figures, but also of logarithms expressed 
 by four figures when the last figure is 0, the mantissa of the logarithm of 625 is 
 first found, since, § 467, it is also the mantissa of the logarithm of 6250. It 
 is found to be .7959. 
 
 The next greater mantissa found in the table is .7966, which is the man- 
 tissa of the logarithm of 626 or of 6260. Since the numbers 6250 and 6260 
 differ by 10, and the mantissas of their logarithms differ by 7 ten-thou- 
 sandths, it may be assumed as sufficiently accurate that each increase of 1 
 unit, as 6250 increases to 6260, produces a corresponding increase of .1 of 7 
 ten-thousandths in the mantissa of the logarithm. Consequently, 3 added 
 to 6250 will add .3 of 7 ten-thousandths, or 2 ten-thousandths, to the man- 
 tissa of the logarithm of 6250 for the mantissa of the logarithm of 6253. 
 
 Hence, the mantissa of the logarithm of 6253 is .7959 -f .0002, or .7961. 
 
 Since the number is an integer expressed by 4 digits, the characteristic is 
 3 (Prin. 1). 
 
 Therefore, the logarithm of 6253 is 3.7961. 
 
 Note. — The difference between two successive mantissas in the table is 
 called the Tabular Difference. 
 
 Find the logarithm of 
 
 16. 1054. 20. 21.09. 24. .09095. 
 
 17. 1272. 21. 3.060. 25. .10125. 
 
 18. .0165. 22. 441.1. 26. 54.675. 
 
 19. 1906. 23. .7854. 27. .09885. 
 
 470. To find a number whose logarithm is given. 
 
 Examples 
 1. Find the number whose logarithm is 0.9472. 
 
 Solution. — The two mantissas adjacent to the given mantissa are .9469 
 and .9474, corresponding to the numbers 8.85 and 8.86, since the given 
 characteristic is 0. The given mantissa is 3 ten-thousandths greater than 
 the mantissa of the logarithm of 8.85, and the mantissa of the logarithm of 
 8.86 is 5 ten-thousandths greater than the mantissa of the logarithm of 8.85. 
 
 Since the numbers 8.85 and 8.86 differ by 1 one-hundredth, and the man- 
 tissas of their logarithms differ by 5 ten-thousandths, it may be assumed as 
 sufficiently accurate that each increase of 1 ten-thousandth in the mantissa 
 is produced by an increase of | of 1 one-hundredth in the number. Conse- 
 
LOGARITHMS 421 
 
 quently, an increase of 3 ten-thousandths in the mantissa is produced by an 
 increase of f of 1 one-hundredth, or .00(3, in the number. 
 Hence, tlie number whose logarithm is 0.9472 is 8.856. 
 
 2. Find the number whose logarithm is 9.4180 — 10. 
 
 Solution. — Given mantissa, .4180.- 
 
 Mantissa next less, .4166 ; ^gures corresponding, 261. 
 
 Difference, 14 
 
 Tabular difference, 17)14(.8 
 
 Hence, the figures corresponding to the given mantissa are 2618. 
 Since the characteristic is 9 — 10, or —J, the number is a decimal with no 
 ciphers immediately following the decimal point. 
 
 Hence, the number whose logarithm is 9.4180 — 10 is .2618. 
 
 Find the number corresponding to. ' ^ ' i 
 
 3. 0.3010. 8. 3.9545. 13. 9.3685-10. ' 
 
 4. 1.6021. 9. 0.8794. 14. 8.9932-10. 
 6. 2.9031. 10. 2.9371. 15. 8.9535-10. 
 
 6. 1.6669. 11. 0.8294. 16. 7.7168-10. 
 
 7. 2.7971. 12. 1.9039. 17. 6.7016-10. 
 
 471. Multiplication by logarithms. 
 
 • Since logarithms are the exponents of the powers to which a 
 constant number is to be raised, it follows that : 
 
 472. Prixciple. — The logarithm of the product of two or more 
 numbers is equal to the sum of their logarithms; that is, 
 
 To any base, log (win) = log m -j- log n. 
 
 The above principle may be established as follows : 
 Let logo m = X and logo n = y, a being any base. 
 
 It is to be proved that loga (jwn) = x + y. 
 § 459, • a' = w, 
 
 and *«» = n. 
 
 Multiplying, §§ 240, 436, ' 0*+* = m». 
 Hence, § 460, loga (inn) = z + y 
 
 = log„ m + loga n. 
 
422 LOGARITHMS 
 
 Examples 
 1. Multiply .0381 by 77. 
 
 Solution 
 Prin., log (.0381 x 77) = log .0381 + log 77. 
 
 log .0381= 8.5809 -10 
 log. 77 = 1.8865 
 Sum of logs = 10.4674 - 10 
 = 0.4674 
 0.4674 = log 2.934. 
 .-..0381 X 77 = 2.934.' 
 
 Note. — Three figures of a number corresponding to a logarithm may be 
 found from this table with absolute accuracy, and in most cases the fourth 
 will be correct. In finding the logarithms of numbers or the numbers corre- 
 sponding to logarithms, allowance should be made for the figures after the 
 fourth, whenever they express .5 or more. 
 
 Multiply : 
 
 2. 3.8 by 56. 6. 2.26 by 85. 10. 1.414 by 2.829. 
 
 3. 72 by 39. 7. 7.25 by 240. 11. 42.37 by .236. 
 
 4. 8.5 by 6.2. 8. 3272 by 75. 12. 2912 by .7281. 
 
 5. 1.64 by 35. 9. .892 by .805. 13. 289 by .7854. 
 
 473. Division by logarithms. 
 
 Since the logarithms of two numbers to a common base repre- 
 sent exponents of the same number, it follo.ws that : 
 
 474. Principle. — Tlie logarithm of the quotient of two numbers 
 is equal to the logarithm of the dividend min^s the logarithm of the 
 divisor; that is, 
 
 To any base, log (m -f- w) = log m — log n. 
 
 The above principle may be established as follows: 
 
 Let loga m = x and logo n = y, a being any base. 
 
 It is to be proved that loga(TO -;- n) = x — y. 
 § 459, a'' = m and as = n. 
 
 Dividing, §§ 248, 436, a^-y -m-i-n. 
 Hence, § 460, loga(m -=- n) = a; — y 
 
 = logo m - logo n. 
 
LOGARITHMS 423 
 
 Examples 
 
 1. Divide .00468 by 75. 
 
 Solution 
 
 Prin., log (.00468 -- 75) = log .00468 - log 76. 
 
 log .00468 = 7.6702 - 10 
 
 log 75 = 1.8751 
 
 Difference of logs = 5.7951 - 10 
 
 5.7951 - 10 = log .00006239. 
 
 .-. .00468 -f- 75 = .00006239. 
 
 2. Divide 12.4 by 16. 
 
 Solution 
 
 Prin., log (12.4 -^ 16) = log 12.4 - log 16. 
 
 log 12.4 = 1.0934 = 11.0934 - 10 
 
 log 16 = 1.2041 
 
 Difference of logs = 9.8893 - 10 
 
 9.8893 - 10 = log .775. 
 
 .-. 12.4 -- 16 = .lib. 
 
 Suggestion. — The positive part of the logarithm of the dividend may 
 be made to exceed that of the divisor, if necessary, by adding 10 — 10 or 
 20 - 20, etc. 
 
 Divide : 
 
 3. 3025 by 55. 8. 10 by 3.14. 13. 1 by 40. 
 
 4. 4096 by 32. 9. .0911 by .7854. 14. 1 by 75. 
 
 ' 5. 3249 by 57. 10. 2.816 by 22.5. 15. 200 by .5236. 
 
 6. .2601 by .68. 11. 4 by .00521. 16. 300 by 17.32. 
 
 7. 3950 by .250. 12. 26 by .06771. 17. .220 by .3183. 
 
 475. Extended operations in multiplication and division. 
 
 Since dividing by a number is equivalent to multiplying by its 
 reciprocal, for every operation of division an operation of multi- 
 plication may be substituted. In extended operations in multipli- 
 cation and division with the aid of logarithms, the latter method 
 of dividing is the more convenient. 
 
424 LOGARITHMS 
 
 476. The logarithm of the reciprocal of a number is called the 
 Cologarithm of the number. 
 
 The cologarithm of 100 is equal to the logarithm of j^, which is — 2. 
 The cologarithm of 100 is — 2 is abbreviated to colog 100 = — 2. 
 
 477. Since the logarithm of 1 is and the logarithm of a 
 quotient is obtained by subtracting the logarithm of the divisor 
 from that of the dividend, it is evident that the cologarithm of a 
 number is minus the logarithm of the number, or the logarithm 
 of the number with the sign of the logarithm changed ; that is, if 
 log„ m = x, then, colog„ m = —x. 
 
 Since subtracting a number is equivalent to adding it with its 
 sign changed, it follows that : 
 
 478. Principle. — Instead of subtracting the logarithm of the 
 divisor from the logarithm of the dividend, the cologarithm of the 
 divisor may be added to the logarithm of the dividend; that is, 
 
 To any base, log (m ~n) = log m + colog n. 
 
 Examples 
 
 , ^. . ,, 1 r .063 X 58.5 X 799 , , .,, 
 
 1. Find the value of ^ ^^ ^ ^ ^ — — by logarithms. 
 
 458 X 15.6 X .029 "^ ^ 
 
 Solution 
 .068 X 58.5 X 799 ^ ^gg ^ ,3 , ^ 799 ^ J- x J- x ^. 
 458 X 15.6 X .029 458 15.6 .029 
 
 log .06.3= 8.7993-10 
 
 log 58.5= 1.7672 
 
 log 799 = 2.9025 
 
 colog 458 = 7.3391 - 10 
 
 colog 15.6= 8.8069-10 
 
 colog .029= 1.5376 
 
 log of result = 31.1526 - 30 
 = 1.1526. 
 .*. result = 14.21. 
 
LOGARITHMS 425 
 
 Find the value of 
 
 110 X 3.1 X .653 g 15 X .37 x 26.16 
 
 33 X 7.854 X 1.7* 
 6000 X 5 X 29 
 
 11 X 8 X .18 X 6.67 
 
 78 
 
 x52x 
 
 1605 
 
 338 
 
 X 767 
 
 x431 
 
 .5 
 
 x.315 
 
 x428 
 
 .7854 X 25000 x 81.7 
 ^ 3.516 X 485 X 65 ^ 
 
 3.33 X 17 X 18 X 73* * .317 x .973 x 43.7 
 
 479. Involution by logarithms. 
 
 Since logarithms are simply exponents, it follows that: 
 
 480. Principle. — Tlie logarithm of a poiver of a number is 
 equal to the logarithm of the number multiplied by the index of the 
 power ; that is, 
 
 To any base, log m" = n log m. 
 
 The above principle may be established as follows : 
 
 Let loga m = 05, and let n be any number, a being any base. 
 
 It 5s to be proved that loga m" = nx- 
 
 § 459, a^ = m. 
 
 Raising each member to the nth power, 
 
 Ax. 6 and §§ 249, 436, a*" = m». 
 
 Hence, § 460, loga w» = nx = n log, m. 
 
 Examples 
 1. Find the value of .25^. 
 
 Solution 
 Prin,, log .252 = 2 log .25. 
 
 log .26 = 9.3979 - 10. 
 2 log .25 = 18.7958 - 20 
 = 8.7958 - 10. 
 8.7958 -10 = log .06249. 
 .-. .252 = .06249. 
 
 Note. — By actual multiplication it is found that .252 = .0625, whereas 
 the result obtained by the use of the table is .06249. 
 
 Also, by multiplication, 182 _ 324^ whereas by the use of the table it is 
 found to be 324.1. Such inaccuracies must be expected when a four-place 
 table is used. 
 
426 LOGARITHMS 
 
 Find by logarithms the value of 
 
 2. V. 7. .782 12. 4.073. 17. {^)\ 
 
 3. 11=*. 8. 8.051 13. .5433. 18. {\f. 
 
 4. 47='. 9. 8.332. 14. 1\ 19. (tS^)*. 
 
 5. 4.92. 10. 6.613. 15 1 02*. 20. (A^/. 
 
 6. 5.22. 11 JI42 16. 1.7383. 21. {^\^)\ 
 
 481. Evolution by logarithms. 
 
 Since logarithms are simply exponents, it follows that: 
 
 482. Principle. — The logarithm of the root of a number is equal 
 to the logarithm of the number divided by the index of the require.-^ 
 root; that is, 
 
 To any base, log Vm = ^^S^- 
 
 n 
 
 The above principle may be established as follows: 
 
 Let logo m = X, and let n be any number, a being any base. 
 
 It is to be proved that loga y/m = x ^ n. 
 
 § 459, a"^ = m. 
 
 Taking the ?ith root of each member, 
 
 Ax. 7 and §§ 250, 436, a^-" = Vm. 
 
 Hence, § 460, log„ v^m = x ^ n = \9SsJB. 
 
 n 
 
 Examples 
 
 1. Find the square root of .1296 by logarithms. 
 
 Solution 
 
 Prin., log \/.1296 = J log .1296. 
 
 log.l296 = 9.1126 -10. 
 
 2 )19.1126-20 
 9.5.563 - 10 
 
 9.5663 -10 = log. 360. 
 .♦. V.1296 = .36. 
 
LOGARITHMS 427 
 
 Find by logarithms the value of 
 
 2. 225I 9. 133li 16. V2. 23. V2. 
 
 3. I2.25I 10. 1024^^ 17. V3. 24. ^:027. 
 
 4. .2025i 11. .67242. I8. V5. 25. V30|. 
 
 5. 324^ 12. 5.929I 19. V6. 26. Vm 
 
 6. .512I 13. .46241 20. ^/2. 27. Vi52. 
 
 7. .llSli 14. 1.464li 21. v4. 28. v'j032. 
 
 8. 3.375I 15. .000321 22. \/3. 29. V^025. 
 
 Simplify the following : 
 
 30. ^ 35. IM^^. 
 
 15 X 3.1416 11 
 
 31. 100^ 36. J.434X96* 
 
 4 
 
 48x64x11 >'64xl500 
 
 52=^ x 300 .32 X 5000 x 18 
 
 * ' 12 X .31225 X 400000' ' 3.14 X .1222 x s" 
 
 ,, / 400 38 11x2.63x4.263 
 \55x 3.1416 * 48x3.263 
 
 34. 50x?l. 39. /^ 
 
 /35 
 
 81.63 \ 1 06^* 
 
 40. 2^ x (i)^ X ^f X VI. 
 
 483. Solution of exponential equations. 
 
 Exponential equations, or equations that involve unknown 
 exponents, are solved by the aid of the principle that, in any 
 system, equal numbers have equal logarithms. 
 
 In simple cases the solution of such equations may be per- 
 formed by inspection, but in general it is necessary to use a table 
 of logarithms. 
 
 Examples 
 
 1. Find the value of x in the equation 2"" = 32 V2, 
 
428 LOGARITHMS 
 
 Solution 
 
 2« = 32V2 = 2622 = 2^; 
 therefore, log (2^) = log (2^), 
 
 or a; log 2 = ^^ log 2, 
 
 .-. X = V- 
 
 2. Find the value of x in the equation 2"= = 48. 
 
 Solution 
 Taking the logarithm of each member, 
 
 X log 2 = log 48. 
 
 log 2 
 1.6812 
 
 = 5.58+. 
 
 0.3010 
 
 3. Find the value of x in the equation 3^^ — 20 • 3^ + 99 = 0. 
 
 Solution 
 Factoring the given equation, 
 
 (3'^-9)(3»'-ll)=0. 
 
 .-. 3^ = 9 or 11. 
 
 Solving the equation 3»= = 9 by inspection, since 9 = 3*, 
 
 x-2. 
 
 Taking the logarithm of each member of 3* = 11, 
 
 xlog3 = log 11. 
 
 .^^logli^ 1,0414 ^2.18+. 
 log 3 0.4771 
 
 Therefore, the value of x is either 2 or 2. 18+. 
 
 4. Given x^ = y^ and x" = y^, to find x and y. 
 
 Solution 
 
 Raising the members of the first equation to the arth power, and those of 
 the second equation to the 3d power, 
 
 and x^y = y^'. 
 
 Hence, by inspection, 2x — Sy. 
 
 Squaring, since 4 x^ = 4 y^, 4 x^ = 9 y^ = 4 y», 
 .: y = or |, 
 and X = or ^. 
 
LOGARITHMS 
 
 429 
 
 5. Given 3^ = 27/ and 2" = y, to find x and y. 
 
 Solution 
 
 3^ = 2 y. 
 
 2^ = ?/. 
 
 (1.5)^ = 2. 
 
 /.x log 1.5 = log 2. 
 
 ^ log 2 ^ 0.3010 
 *~ log 1.5 0.1761* 
 
 log X ^ 0.2328 ; 
 
 X = 1.709. 
 
 log y — X log 2. 
 
 /. log log j^ = log X + log log 2 
 
 = 0.2328 + 1.4786 
 
 = 1.7114. 
 
 logy = 0.5145; 
 
 y = 3.270. 
 
 Dividing (1) by (2), 
 
 Hence, by tables, 
 
 By logarithms, 
 ♦whence, by tables, 
 From (2), 
 
 by (4) and tables. 
 
 Hence, by tables, 
 whence, 
 
 (1) 
 (2) 
 
 (3) 
 
 (4) 
 (5) 
 
 Solve the following ; 
 
 6. 
 7. 
 8. 
 9. 
 10. 
 
 3' = 81. 
 4- = 10. 
 2^ = 80., 
 
 2'"= 512, 
 
 12. 3-'^ - 36-3^ + 243 = 0. 
 
 13. log log a; = log 2. 
 
 14 
 
 11. (2^ = 256. 
 
 15. 
 16. 
 
 I 4^ = 20 y. 
 2^= 512. 
 5^'= 625. 
 
 17. 
 
 18. 
 
 19. 
 
 r 2^+^ = 6, 
 1 2^+1 = 3^. 
 
 4^+" =32, 
 22x-» ^ 4_ 
 
 [x =l+logy. 
 
 484. Logarithms applied to the solution of problems in compound 
 interest and annuities. 
 
 1. What is the amount of f 1 for 1 year at 6% ? By what 
 must the amount for 1 year be multiplied to find the amount for 
 2 years at 6% compound interest? 
 
 2. By what must the amount for 2 'years be multiplied to 
 obtain the amount for 3 years, compound interest? By what 
 must the amount for 3 years be multiplied to obtain the amount 
 for 4 years, compound interest? 
 
430 LOGARITHMS 
 
 3. Since the amount of any principal at 6% compound interest 
 for 1 year is 1.06 times the principal ; for two years, 1.06 x 1.06, 
 or 1.06^ times the principal ; for 3 years, 1.06 x 1.06 x 1.06, or 
 1.06^ times the principal, etc., what will be the amount {A) of any 
 principal (P) for n years at any rate per cent (r) ? 
 
 Formula. A = P(l + ?•)". 
 
 Expressing the formula by logarithms, 
 
 log ^ = log P + ?i log(l + r). (1) 
 
 .-. log P = log ^ - 71 log(l + r) ; (2) 
 
 also log(l + r)= ^°g^-^"g-^ . 
 
 and lo g^-logP 
 
 (3) 
 (4) 
 
 Examples 
 
 1. What is the amount of $475 for 10 years at 6% compound 
 interest ? 
 
 Solution 
 
 A = P(l + r)«. 
 log475 = 2.6767 
 log'l.Oeio = 0.2530 
 \ogA = 2.9297 
 .-.^ = $850.60. 
 
 2. What will be the amount of $ 225 loaned for 5 years at 8% 
 compound interest ? 
 
 3. Find the amount of $700 loaned for 5 years at 6% com- 
 pound interest. 
 
 4. Find the amount of $400 for 10 years at 3% compound 
 interest. 
 
 5. Find the amount of $1200 for 20 years at 4% compound 
 interest. 
 
 6. What principal will amount to $1000 in 10 years at 5% 
 compound interest? 
 
LOGARITHMS 431 
 
 7. What sum of money invested at 4 % compound interest, 
 payable semiannually, will amount to $ 743 in 10 years ? 
 
 8. What principal loaned at 4% compound interest will 
 amount to $ 1500 in 10 years ? 
 
 9. What sum of money invested at 4 % compound interest 
 from a child's birth until he is 21 years old will yield f 1000 ? 
 
 10. In what time will $800 amount to f 1834.50, if put at 
 compound interest at 5 ^J) ? 
 
 11. What is the rate per cent when $ 300 loaned at compound 
 interest for 6 years amounts to $ 402? 
 
 12. A man agreed to loan $ 1000 at 6 % compound interest for 
 a time long enough for the principal to double itself. How long 
 was the money at interest ? 
 
 485. A sum of money to be paid periodically for a given num- 
 ber of years, during the life of a person, or forever, is called an 
 Annuity. 
 
 The payments may be made once a year, or twice, or four times 
 a year, etc. 
 
 Interest is allowed upon deferred payments. 
 
 486. To find the amount of an annuity left unpaid for a given 
 number of years, compound interest being allowed. 
 
 1. If an annuity of a dollars is not paid at the end of the first 
 year, how much is then due ? 
 
 2. Upon what sum will compound interest be computed for 
 the second year ? What will be the amount of that sum, if the 
 rate is r ? What will be the whole sum due at the end of the 
 second year ? Ans. a 4- a(l + r). 
 
 3. Upon what sum will compound interest be computed for 
 the third year? What will be the amount of that sum at the 
 given rate ? Ans. a (1 + r) + a (1 + r)^. 
 
 What will be the whole sum due at the end of the third year ? 
 
 Ans. a + a (1 + r) + a (1 + rf. 
 
432 LOGARITHMS 
 
 4. What will be the whole sum due at the end of the fourth 
 year ? What will be the whole sum due at the end of the nth. 
 year? 
 
 487. Let a represent the annuity, n the number of years, ?• the 
 rate, and A the whole amount due at the end of the nth year. 
 
 Then, 
 
 A= a + a(l + r) + a(l -{- rf -\- a(l + ry -\ \-a(l-\- r)"-^ 
 
 = a\l + (l + r) + (1 + r)2+ (1 +r)3 + ... + (1 + r)"-'|. 
 
 Since the terms of A form a geometrical progression in which 
 1 4- »* is the ratio, § 370, the sum of the series is 
 
 A = ^[{l + ry-ll 
 
 488. Sometimes annuities, drawing interest, are not payable 
 until after a certain number of years. It is often necessary, 
 therefore, to find the present value of such annuities. 
 
 489. A sum that will amount to the value of an annuity, if 
 put at interest at the given rate for the given time, is called the 
 Present Value of the annuity. 
 
 490. 1. If P denotes the present value of an annuity due in n 
 years, allowing r % compound interest, to what sum will P be 
 equal in that time at the given rate ? Ans. P(l + r)". 
 
 2. Since the amount of the present value put at interest for 
 the given time at the given rate is equal to the amount of the 
 annuity for the same time and rate, equate the two sums and find 
 the value of P. 
 
 P(l + r)" = ?[(l + r)"-l]. 
 
 r (1 + r)» 
 
LOGARITHMS 433 
 
 Examples 
 
 1. What will be the amount of an annuity of $100 remaining 
 unpaid for 10 years at 6 % compound interest ? 
 
 Solution. ^ = - [(1 + r)» - 1]. 
 
 r 
 
 By logarithms, 1.06" = 1.7904 
 
 .-. 1.0610 - 1 = .7904 
 log 100 = 2.0000 
 
 log .7904 = 9.8978 - 10 
 
 colog .06 = 1.2218 
 .-. logyl = 3.1196 
 
 Hence, -4 = f 1317, the amount of the annuity. 
 
 2. What is the present value of an annuity of f 100 to continue 
 10 years at 6 % compound interest ? 
 
 Solution. p ^<1 AA+l)!^!. 
 
 r (1 + r)" 
 
 By logarithms, 1 .06^ = 1. 7904 
 
 .-. 1.0610 _ 1 ^ .7904 
 
 log 100 = 2.0000 
 
 log .7904 = 9.8978 - 10 
 
 colog .06 = 1.2218 
 
 colog 1.0610 ^ 9.7470 - 10 
 .'. log P = 2.8666 
 
 Hence, P = $ 735.50, the p. v. of the annuity. 
 
 3. To what sum will an annuity of $25 amount in 20 years at 
 4 ffo compound interest ? 
 
 4. What is the present value of an annuity of f 300 for 5 years 
 at 4 % compound interest ? 
 
 5. What will be the amount of an annuity of f 17.76 remain- 
 ing unpaid for 25 years at 3|^ % compound interest ? 
 
 6. What is the present value of an annuity of f 1000 to con- 
 tinue 20 years, allowing compound interest at 4i % ? 
 
 7. What annuity will amount to f 1000 in 10 years at 5% 
 compound interest ? 
 
 ADV. ALG. — 28 
 
PERMUTATIONS AND COMBINATIONS 
 
 491. All the different orders in which it is possible to arrange 
 a given number of things, taking either some or all of them at 
 a time, are called the Permutations of the things. 
 
 Thus, the permutations of the letters a and h are ah, ha ; the permuta- 
 tions of three letters, two at a time, are ah, ac, ha, be, ca, cb. 
 
 492. All the different selections that it is possible to make 
 from a given number of things, taking either some or all of them 
 at a time, without regard to the order in which they are placed, 
 are called the Combinations of the things. 
 
 Thus, while the permutations of three letters, two at a time, are ah and 
 ba, be and cb, and ca and ae, their comhinations, two at a time, are ah (or -7. 
 ha),< be (or cb), and ac (or ca) ; again, the six permutations of three letters 
 among themselves, viz., ahc, acb, bca, bac, cab, and cha, form but one com- 
 bination, abc (or acb, or bca, or bac, or cab, or cha). 
 
 It is evident that there can be but one combination of any number of 
 things taken all at a time. 
 
 493. Notation. — The symbol for the number of permutations 
 of n different things, taken r at a time, is P"^ ; of n different 
 things, taken w at a time, or all together, P^. 
 
 . The symbol for the number of combinations of n different 
 things, taken r at a time, is C" ; of n different things, taken n 
 at a time, or all together, is O^. 
 
 494. The product of the successive natural numbers from 1 
 to n, or from n to 1, inclusive, is called factorial n, written \n. 
 
 [5 = 1x2x3x4x5, or 5x4x3x2x1; 
 
 [w = 1 . 2 . 3 ..• (n - 2)(n - l)n, or w(n - l)(n - 2)(n - 3) ••• 3 • 2 • 1. 
 
 in is sometimes written n ! 
 
 434 
 
PERMUTATIONS AND COMBINATIONS 435 
 
 495. To find the number of permutations of n different things taken 
 V* at a time. 
 
 Since the permutations of the letters a, b, and c, taken 2 at a 
 time, are ab and ac, ba and be, ca and cb, formed by writing after 
 each of the letters a, b, and c, each of the other letters in turn, the 
 number of permutations of 3 different things taken 2 at a time is 
 3x2. 
 
 The number of permutations of n letters taken 2 at a time may 
 be found by associating with each of the n letters each of the 
 n — 1 other letters. Consequently, the number of permutations 
 of n different things taken 2 at a time is n{n — 1). 
 
 Since the number of permutations of n letters 2 at a time is 
 n(n — 1), if the letters are taken 3 at a time there will be n — 2 
 letters each of which may be associated with each of the n{n — 1) 
 permutations of letters taken 2 at a time. Hence, the number of 
 permutations of n different things taken 3 at a time is 
 
 n (n - 1) (m— 2). 
 
 Principle 1. — The number of permutations ofn different things 
 taken r at a time is equal to the continued product of the natural 
 numbers from n to n — (r — 1) inclusive. The number of factors 
 is r. That is, 
 
 P" = ?i(n — 1) (w — 2) ••• to r factors 
 
 = w(w-l)(n-2) ...(ji-r + l). (I) 
 
 Multiplying and dividing the second member of (I) by 
 
 (w — r) (w — r -^ i) (n — r — 2) ••• 2 • 1 ; that is, by \n — r , 
 
 \n 
 P? = -l-- (II) 
 
 I n — r 
 
 It will usually be more convenient to employ formula (I) in 
 solving numerical examples; but when simply algebraic results 
 are desired, formula (II) will be preferable. 
 
 496. When r = n, that is, when the things are taken all 
 together, the last, or nth, factor in (I) is 1. Consequently, 
 
 Principle 2. — The number of permutations ofn different things 
 taken all at a time is equal to \n. That is, 
 
 Pj = w (w — 1) (71 — 2) ••• to n factors = | w. (HI) 
 
436 PERMUTATIONS AND COMBINATIONS 
 
 Examples 
 
 1. Three boys enter a car in which there are 5 empty seats. 
 In how many ways may they choose seats ? 
 
 Solution. — Since the first boy may choose any one of 5 seats ; and since, 
 after he has chosen one of them, for each seat that he may choose, the second 
 boy may choose any one of the 4 seats remaining, the greatest possible num- 
 ber of ways in which two of the boys may be seated is 5 x 4. 
 
 Again, since after each clioice of seats made by two of the boys there will 
 be left to the third boy a choice of one of the 3 seats remaining, the number 
 of ways in which all may choose seats is 5 x 4 x 3, or 60. 
 
 Or, by (I), P;? = n{n - l)(n - 2) ... (?i - r + 1), 
 
 P| = 5 X 4 X 3 = GO. 
 
 2. How many numbers between 100 and 1000 can be expressed 
 by the figures 1, 3, 5 ? 
 
 Solution. — Since the numbei's lie between 100 and 1000, each must be 
 expressed by three figures. Hence, the number of numbers between 100 and 
 1000 that can be expressed by the figures 1, 3, and 5 is the same as the num- 
 ber of permutations 'of these 3 figures taken 3 at a time. 
 
 Since, Prin. 2, . P| = (3 = 3.2.1=:6, 
 
 there are six such numbers. They are 135, 153, 351, 315, 513, and 531. 
 
 3. How many permutations can be made of the letters in the 
 word Albany, each beginning with capital A ? 
 
 Solution. — Since A is to be prefixed to each permutation of the 5 other 
 letters, the required number is 
 
 P| = 5x4x3x2xl = 120. 
 
 4. In how many orders may 4 persons sit on a bench ? 
 
 5. How many permutations may be made of the letters in the 
 word steam f 
 
 6. If 10 athletes run a race, in how many ways may the first 
 and second prizes be awarded ? 
 
 7. In how many different orders may the colors violet, indigo, 
 blue, green, yellow, orange, and red be arranged ? 
 
 %t There are five routes to the top of a mountain. In how 
 many ways may a person go up and return by a different way ? 
 
^ 
 
 PERMUTATIONS AND COMBINATIONS 437 
 
 497. To find the number of combinations of n different things 
 taken /• at a time. 
 
 Since two letters, as a and h, have two permutations, ab and ba, 
 but form only one combination, the number of combinations of n 
 letters taken 2 at a time is one half the number of permutations 
 of n letters taken 2 at a time. 
 
 Since three letters taken 3 at a time have 3x2 permutations, 
 but form only one combination, the number of combinations of n 
 letters taken 3 at a time is obtained by dividing the number of 
 permutations of n letters taken 3 at a time by 3 x 2. 
 
 Since four letters taken 4 at a time have |4 permutations but 
 form only one combination, to obtain the number of combinations 
 of n letters taken 4 at a time, the number of permutations of n 
 letters taken 4 at a time must be divided by [4. 
 
 Hence it follows that : 
 
 Principle 3. — The numbei' of combinations of n different things 
 taken r at a time is equal to the number of permutations of n dif- 
 ferent things taken r at a time, divided by the number of permuta- 
 tions of r different things taken all togetlier. That is, 
 
 ri^ — p" _t. p*- _ '>^{n — 1) (n — 2) • • • to r factors 
 ?* (/ — 1) (r — 2) ••• to r factors 
 
 ^ n (n - 1) (n - 2) »• (w - r + 1) (IV) 
 1 .2 .3 ..• r 
 
 or, by (II), 
 
 (V) 
 
 498. Since for every combination of r things out of n different 
 things there is left a combination of n — r things, it follows that : 
 
 Prixciple 4. — Tlie number of combinations of n different things 
 is the same when taken n — r at a time as when taken r at a time. 
 That is, 
 
 (fn-r^Crr^ . ;- • • (VI) 
 
 ., [r \n — r 
 
 
 1 
 
 n 
 
 
 \n 
 
 — r 
 
 
 
 \n 
 
 
 \r 
 
 \n — r 
 
438 PERMUTATIONS AND COMBINATIONS 
 
 The above principle may be established as follows : 
 By(V). c: = --}^— (1) 
 
 Substituting n — r for r, cZ-r = 
 
 Since the second members of (1) and (2) are identical, cZ-r = Or 
 
 \r\n — r 
 
 \n — r\n —{n- 
 
 -r) 
 
 \n — r\r 
 
 
 (2) 
 
 The above principle is useful in abridging numerical computations. 
 Thus, the number of combinations of 18 things taken 16 at a time is com- 
 puted by Prin. 3 as follows : 
 
 ^18 _ 18 • 17 . 16 • 15 • 14 • 13 • 12 • 11 • 10 • 9- 8- 7 • 6 • 5 • 4 ■ 3 ^ jgg 
 ^® 1 . 2 . 3 • 4 . 5 • 6 . 7 . 8 . 9 . 10 . 11 . 12 . 13 • 14 . 15 • 16 
 
 By Prin. 4, the computation is abridged as follows : 
 
 1-2 
 
 Examples 
 
 1. A man has 6 friends and wishes to invite 4 of them to 
 dinner. In how many ways may he select his guests? 
 
 Solution. — Since each party, or combination, of 4 guests could be 
 arranged, or permuted, in [4 ways, the number of combinations must be — 
 of the number of permutations of 6 things taken 4 at a time. I— 
 
 Hence, the number of ways is 
 
 1x2x3x4 'o^ 
 
 2. A man and his wife wish to invite 11 of their friends, 6 men -^ 
 and 5 women, to dinner, but find that they can entertain only 8 
 guests. In how many ways may they invite 4 men and 4 women ? 
 
 SoLDTioN. — As in the previous example, 4 men may be selected from 6 
 men in 15 ways, and in a similar manner 4 women may be selected from 6 
 women in 5 ways. ^ 
 
 Since, when any set of 4 men has been invited, the party of 8 may be com- 
 pleted by inviting any one of 5 sets of 4 women, the whole number of differ- 
 ent parties that it is possible to invite is 15 x 5, or 75. That is, 
 
 r.6,,r.^_ 6-5.4.3 5.4.3-2 _ 
 
PERMUTATIONS AND COMBINATIONS 439 
 
 3. In how many ways may a baseball nine be selected from 
 12 candidates? 
 
 4. Find the value of C'^; of (7>|; of C^. 
 
 5. How many different combinations of 5 cards can be formed 
 from 52 cards ? 
 
 6. Which is the greater, C'^ or C'^ ? C't or G^l ? 
 
 7. From 11 Republicans and 10 Democrats how many different 
 committees can be selected composed of 6 Republicans and 5 
 Democrats ? 
 
 8. A man forgets the combination of figures and letters by 
 which his safe is opened. If they are arranged on the circum- 
 ferences of three wheels, one bearing the numbers to 9 inclu- 
 sive, another the letters A to M inclusive, and the third the letters 
 N to 2 inclusive, what is the greatest number of trials he may 
 have to make to open the safe ? 
 
 9. From 6 consonants and 4 vowels how many words may be 
 formed each consisting of 4 consonants and 2 vowels, if any 
 arrangement of the letters is considered a word ? 
 
 Solution. — The number of combinations is Ct x C\; and since by per- 
 muting the letters of each combination |^ words can be formed, the number 
 of words is C\x G\x [6. 
 
 10. In an omnibus that will seat 8 persons on a side there are 
 seated 4 persons, 3 on one side and 1 on the other. In how many 
 ways may 12 more persons be seated ? 
 
 Solution. — Since 5 persons must take seats on one side and 7 persons on 
 the other, 12 persons are to be divided into two classes, 6 and 7. The number 
 of these combinations, formula (V), is 
 
 C'/,orC?,=|i. 
 
 Since each combination of 5 may have [5 permutations of the 5 that 
 compose it, and each combination of 7 may have [7 permutations each of 
 which may be associated with each of the [5 permutations, the required 
 number of ways is C^i >j P5 x P7, 
 
 (12 
 or J=x|5x 7=112. 
 
 Or, since there are 12 persons to be seated in 12 seats, the number of ways 
 
 i8Pii = |12. 
 
 
 \* A 
 
 ^1 ; • 'A .> 
 
440 PERMUTATIONS AND COMBINATIONS 
 
 11. Out of 20 consonants and 5 vowels how many words con- 
 taining 3 consonants and 3 vowels can be formed, if any arrange- 
 ment of the letters is considered a word ? 
 
 12. How many different sums can be paid with a cent, a half- 
 dime, a dime, a quarter, and a dollar ? 
 
 13. From 5 boys and 5 girls how many committees of 6 can be 
 selected so as to contain at least 2 boys ? 
 
 14. A company of a soldiers is joined by another company of 
 h soldiers. In how many ways is it possible to leave c of them to 
 garrison the fort, dividing the rest into two scouting parties, one 
 of m, the other of n soldiers ? 
 
 15. If Cs = 2 Cl, find the number of things. 
 
 Solution.— By formula (V), CI = ,^,^ and CI = — -t^^ 
 
 |2 |w — 2 
 
 Since 01 = 2, CI 
 
 2> 
 
 \n-2 
 
 (V), 
 
 CI: 
 
 |5|u- 
 
 6 
 
 \n 
 
 2[n 
 
 \2\n- 
 1 
 
 
 [5\n- 
 1 
 
 -_5 
 
 2 
 
 |6|n- 
 
 -6 
 
 \n-2 
 = \5\n- 
 
 
 .:{n^ 
 
 -2: 
 
 5. 
 
 or (n-2)(n-3)(n-4)=5 x4x3x2xl, 
 
 |w-5 
 
 or (n-2)(n-3)(tt-4) =6 X 5 X 4. 
 
 .-. n = 8. 
 16. If 3 C^ = 2 C"t\ find n, Cl and (7"t\ 
 
 499. To find the number of circular permutations of n different 
 things taken n at a time. 
 
 Suppose four letters a, b, c, d, placed in a fixed position around 
 a circle in the order abed. Since the arrange- 
 ment may be read abed, bcda, cdab, or dabc, 
 without changing the direction in which the 
 letters are read, it is evident that each circu- 
 lar permutation of 4 letters taken all together 
 takes the place of 4 permutations of the 
 letters all together. 
 
 That is, the number of circular permuta- 
 tions of 4 things taken all together is one fourth of the number of 
 permutations of 4 things taken all together. 
 
PERMUTATIONS AND COMBINATIONS 44l 
 
 The whole number of permutations of n things taken all 
 together is [n. But if the n things are arranged around a circle, 
 n of these permutations may be obtained from any circular per- 
 mutation without disturbing the relative positions of the things. 
 
 Hence, 
 
 Principle 5. — The number of circular permutations of n things 
 
 taken all together is equal to -th of the whole number of their per- 
 
 n 
 
 mutations taken all together. That is, 
 
 In 
 i^* (circular) = = = \ n -1 . (VII) 
 
 n 
 
 Examples 
 
 1. In how many orders may 6 persons seat themselves around 
 a table ? 
 
 2. In how many orders may 4 gentlemen and their wives seat 
 themselves around a table ? 
 
 3. In how many orders may 4 gentlemen and their wives seat 
 themselves around a table so that each gentleman sits opposite 
 his wife ? 
 
 4. In how many orders may 4 gentlemen and their wives seat 
 themselves around a table so that each gentleman sits opposite 
 a lady ? 
 
 5. In how many ways may the colors violet, indigo, blue, green, 
 yellow, orange, and red be arranged on a disk, the colors radiating 
 from the center ? 
 
 500. To find the number of permutations of n things taken n at 
 a time when they are not all different. 
 
 If, in the permutation (a, b, c, d, e, f g), the letters h, d, and g 
 are permuted while the other letters remain fixed in position, the 
 resulting number of permutations will be the same as the number 
 of permutations of b, d, and g. If b, d, and g are different things, 
 the number of permutations resulting will be [3 ; but if 6, d, and g 
 become alike, there will be but 1 permutation. 
 
 That is, the number of permutations of any number of things 
 
442 PERMUTATIONS AND COMBINATIONS 
 
 when three of them are alike is equal to the number of permuta- 
 tions of the things, considered as all different, divided by |3 ; if 4 
 of the things are alike, by [4 ; ii p oi the things are alike, by [p. 
 Hence, it follows that : 
 
 Principle 6. — The number of permutations of n things, taken 
 
 \n 
 all together, when p of them are alike, is — • 
 
 [P 
 Since, if q of the remaining n—p different things become alike, 
 but different from the p like things, the number of permutations 
 must be divided by [g ; if r others become alike, by [r ; etc. : it 
 follows that : 
 
 Principle 7. — Tfie number of permutations of n things, taken 
 all together, when p of them are of one kind, q of another, r of 
 
 another, etc., is ■ — -= 
 
 \p\q\r-- 
 
 Examples 
 
 1. How many permutations may be made with the letters of 
 the word Mississippi taken all together ? 
 
 Solution. — The number is — = 34650. 
 
 [4 [4 [2 
 
 2. How many permutations may be made with the letters of 
 ^ch of the following words, all at a time in each case : zoology, 
 coefficient, ecclesiastical, divisibility ? 
 
 3. How many permutations may be made with the letters rep- 
 resented in the product a*b^c^ written out in full ? 
 
 501. To find the total number of combinational n different things. 
 
 The number of combinations of n different things taken succes- 
 sively 1, 2, 3, ••• w at a time is called the total number of combi- 
 nations of n things. 
 
 The total number of combinations of 2 things is 
 
 Cf+ 01=2 + 1 = 3, or 22-1. 
 The total number of combinations of 3 things is 
 
 Of + Ci + C'i = 3 + 3 -H 1 = 7, or 23 - 1. 
 
PERMUTATIONS AND COMBINATIONS 443 
 
 The total mimber of combinations of 4 things is 
 
 Ct + q + C| + C^4 = 4 + 6 + 4 + 1 = 15, or 2* - 1. 
 
 Principle 8. — The total number of combinations of n different 
 'things is 2" — 1. 
 
 The above principle may be established as follows: 
 
 § 452, when n is a positive integer, 
 
 (1 + x)» = l + nx + »(»-!) a;2 + ... + n(n - l)(n - 2) ... 1 ^„ 
 1-2 1 ■ 2 • 3 .•• n 
 
 Ifx = l, 2n = 1 + n + "(" - ^) + .- + '<"" - ^)('^ - ^) - ^ 
 
 1.2 1 .2.3...ri 
 
 Prin. 3, =i+Gl + C';+ ... + C',: = 1 + C;.^. 
 
 '^ total - "^ 
 
 Examples 
 
 1. How many different sums can be paid with a cent, a 3-cent 
 piece, a half-dime, a dime, a quarter, and a dollar ? 
 
 Solution. — Total 0^ = 2^-1 = 63. 
 
 2. A man has 10 friends. In how many ways may he invite 
 one or more of them to dinner ? 
 
 3. How many different quantities can be weighed by weights 
 of 1 oz., 1 lb., i lb., 5 lb., and 10 lb. ? ^ 
 
 4. How many signals can be made with 7 flags ? 
 
 5. By permuting the letters of the word counter, how many 
 permutatiatis ^i be formed 
 
 (a) ending in er ? 
 
 (b) with n as the middle letter ? 
 
 (c) without changing the position of any vowel ? 
 
 (d) beginning with a consonant ? 
 
 6. How many numbers can be formed with the digits 1, 2, 3, 
 4, 3, 2, 1, so that the odd digits always occupy the odd places? 
 
 7. If the number of permutations of n different things taken 
 5 at a time is equal to 24 times the number of permutations of 
 the same number of things taken 2 at a time, find n. 
 
444 PERMUTATIONS AND COMBINATIONS 
 
 8. How many permutations can be made of the letters of the 
 word international, taken all together ? 
 
 9. A man has five coats, six vests, and eight pairs of trousers. 
 In how many different suits may he appear ? 
 
 10. In how many ways may a committee of 5 men and 3 
 women be chosen from 10 men and 5 women ? | 
 
 11. Mr. Alexander has 5 pairs of gloves. In how many ways 
 can he select a right-hand glove and a left-hand glove without 
 selecting a pair ? 
 
 12. In how many ways can a party of 12 young people form a 
 ring? 
 
 13. In how many ways can 5 ladies and 5 gentlemen be seated 
 at a round table so that no two ladies sit together ? 
 
 14. A man belongs to a club of 16 members and every day he 
 invites three members to dine with him. For how many days 
 can he invite a different party each day ? 
 
 15. In how many ways may 12 persons seat themselves at two 
 round tables, 6 persons at each table ? 
 
 16. In how many ways may 8 books be arranged on a shelf so 
 that two particular books will not be together ? 
 
 17. In how many ways may a committee of 5 be appointed 
 from 12 boys and 10 girls so that there will be more boys than 
 girls in the committee ? 
 
 18. A boat's crew consists of 8 men, of whom 3 can row only 
 on the port side and 2 only on the starboard side. In how many 
 ways can the crew be seated ? 
 
 19. In how many different numbers less than 1000 does the 
 digit 9 appear ? 
 
 20. From a detachment of 20 soldiers 4 are detailed each night 
 for picket duty. For how many successive nights can a different 
 picket be formed and how many times will each soldier be on 
 duty during this interval ? 
 
PROBABILITY 
 
 502. Suppose that a person draws one ball at random from a 
 bag containing 3 white balls and 2 black balls, and it is required 
 to find his chance of drawing a white ball. 
 
 Since this event can happen in 3 ways, namely, by drawing 
 any one of the 3 Avhite balls, and can fail in 2 ways, namely, by 
 drawing either of the 2 blacft balls, it is seen that the ratio of 
 the number of ways favorable to drawing a white ball to the 
 whole number of ways, favorable and unfavorable, is 3:5. This 
 is expressed by saying that the chance, or prohability, of drawing 
 a white ball is f . 
 
 Similarly, the probability of drawing a black ball is |. 
 
 If one of the black balls is removed before drawing, the proba- 
 bility of drawing a white ball is increased to f and the probability 
 of drawing a black ball is diminished to \. If both black balls 
 are removed, the probability of drawing a white ball amounts to 
 certainty, expressed by |, or 1, and the probability of drawing a 
 black ball becomes zero, the occurrence being impossible. 
 
 503. If an event can happen in a ways and fail in b ways, and 
 all of these ways are equally likely to occur, the chance, or proba- 
 bility, that it will happen is r, and the probability that it 
 
 1 Ct ~p" 
 
 will fail is r^- 
 
 a -\- b 
 
 It a = b, the chances are said to be even. 
 
 504. The ratio of the probability that an event will happen to 
 the probability that it will fail is called the odds in favor of the 
 event, and the reciprocal ratio is called the odds against the 
 event. 
 
 445 
 
446 PROBABILITY ^ 
 
 \ 
 
 Thus, if a person draws one ball at random from a bag containing 3 white 
 balls and 2 black balls, the odds are 3 to 2 in favor of his drawing a white 
 ball and 2 to 3 against his drawing a white ball. 
 
 505. Principles. — 1. If the probability that an event will hap- 
 pen is equal to p, the probability that it will fail is equal to 1 — p. 
 
 2. If an event is certain to happen, its probability is 1 ; if an event 
 is certain to fail, its probability is zero. 
 
 The above principles may be established as follows : 
 Suppose that an event can happen in a ways and fail in h ways. 
 
 1. Then, by the definition of probability, the probability that the event 
 
 will happen is and the probability that it will fail is . Since 
 
 the sum of these probabilities is 1, if the probability that the event will hap- 
 pen is equal to p, the probability that it will fail is equal to 1 — p. 
 
 2. If an event is certain to happen, the probability that it will fail is 
 equal to zero ; that is, 1 — p = 0, or p = 1 . 
 
 If the event is certain to fail, the probability that it will fail, as has just 
 been proved, is equal to 1 ; that is, 1 — p = 1, orp = 0. 
 
 Problems 
 
 1. If one ball is drawn at random from a bag containing 3 
 white balls, 4 black balls, and 5 red balls, what is the probability 
 that it is red *? What are the odds against its being red ? 
 
 Solution 
 
 Since 5 of the 12 balls are red, the probability of drawing a red ball is 
 5 in 12, or ^j. 
 
 Since 7 of the 12 balls are not red, the probability that the ball drawn is 
 not red is ^^, and the odds against drawing a red ball are ^^ to ^, or 7 to 5. 
 
 2. If three coins are tossed, what is the probability (a) that all 
 
 fall heads ? (6) that a certain two fall heads and the third tail ? 
 
 (c) that any two fall heads and one tail ? (d) that at least two 
 
 fall heads ? 
 
 Solution 
 
 For convenience call the coins A, B, and C. 
 
PROBABILITY - 447 
 
 (a) Since the probability that any coin falls head is J, and since, whether 
 A falls head or tail, B may fall either head or tail, the probability that both 
 A and B fall heads is ^ x ^, or \. Again, since when both A and B fall 
 heads C may fall either head or tail, the probability that all fall heads is 
 
 (b) Since the probability that a certain two, as A and B, fall heads is 
 ^ X ^, or \, and since when A and B have fallen heads C may fall either head 
 or tail, the probability that a certain two fall heads and the third tail is | x |, 
 orf 
 
 (c) Since A and B, or A and C, or B and C may fall heads while in each 
 case the third coin has even chances of falling head or tail, the probability 
 that some two coins will fall head and the third coin tail is 3 times as great 
 as the probability that a certain two will fall heads and the third tail. 
 Hence, by (fc) the probability of this event Ls |. 
 
 (d) The probability that at least two coins fall heads is greater than the 
 probability that two fall heads and one tail, for the former includes, while 
 the latter excludes, the probability that all fall heads. Since by (c) the 
 probability that two fall heads and one tail is |, and by (a) the probability 
 that all fall heads is |, the probability that at least two coins fall heads is 
 I + i, or ^• 
 
 3. From a bag containing 8 white balls and 4 black balls, 4 
 balls are drawn at random. What is the probability that 2 are 
 white and 2 are black ? 
 
 First Solotion. — Suppose that the balls are drawn one at a time. The 
 probability that the first drawn is white is /^ ; if the first is white, the proba- 
 bility that the second is white is j'^j^ ; when two white balls have been drawn, 
 the probability that the third is black is ^ ; and when 2 white balls and 1 
 black ball have been drawn, the probability that the next is black is |. 
 
 Hence, the probability of drawing 2 white balls and 2 black balls in the 
 
 14 
 order WWBB is /^ • t't • i^u • I ) ^^^ similarly for each of the -=- orders in 
 which 2 white balls and 2 black balls can be drawn. I— L- 
 
 (8-7-4-3)C4-3.2 -1) ^56 
 '■ ^"(12.11 •10.9)(1.2. 1-2) 165' 
 
 Second Solutiov. — From 12 balls 4 balls can be selected in C^ ways. 
 Again, from 8 white balls 2 white balls can be selected in G\ ways, and from 
 4 black balls 2 black balls can be selected in C\ ways. 
 
 Therefore, out of C^ ways of selecting 4 balls from 12, C\ • C\ lead to" 
 selecting 2 white balls and 2 black balls. 
 
 . D = ^il_^2= (R-7)(4.3)( 1.2.3.4) ^ ^6 
 ' C^ (1.2)(1.2)(12.11.10.9) 165 
 
448 PROBABILITY 
 
 J 
 
 4. A and B and eight other persons seat themselves at random 
 around a circular table. What is the probability that A and B 
 sit together ? 
 
 First Solution 
 
 Since A may sit with each of the 9 other persons in 2 ways, on the right 
 or on the left, but whenever he sits on the right of one person he sits on the 
 left of another, the whole number of ways in which he can sit with different 
 persons is 9. Now on 2 occasions A sits with B, once on B's right and once 
 on B's left. Hence, if the ten persons seat themselves at random, the prob- 
 ability that A and B will sit together is \. 
 
 Second Solution 
 
 If A and B and the other 8 persons seat themselves at random, §499, the 
 whole number of ways in which they may sit around the circular table is [9. 
 
 To find how many of these ways lead to seating A and B together, sup- 
 pose that A and B first seat themselves side by side while the remaining 
 persons take at random the 8 remaining seats. A and B may sit together in 
 two ways, A on the right or on the left of B, and with each of these ways the 
 other persons may be seated in any one of F% ways. Hence, 2 T% ways lead 
 to seating A and B together. 
 
 Since there are 2 Pg, or 2 [8, ways that lead to seating A and B together 
 and |9 ways in all, the probability that A and B sit together is 
 
 2J|_2J8_2 
 |9~9|8~9" 
 
 5. In a hat are placed 25 tickets numbered from 1 to 25. 
 Three tickets, each entitling the holder of their duplicates to a 
 prize, are drawn from the hat. What is the probability that A, 
 wno holds a duplicate of number 10 and also of number li, will 
 receive at least one prize ? 
 
 Solution 
 The whole number of ways in which 3 tickets can be drawn is 
 
 -,25 _ 25 • 24 • 2.3 
 '^ 1.2.3 
 
 = 2300. 
 
 To find how many of the 2300 possible ways of drawing 3 tickets involve 
 the drawing of any particular ticket, as ticket number 10, suppose that all 
 the groups of 3 tickets in which ticket number 10 is found are formed by first 
 selecting this ticket and then associating with it every possible combination 
 of the remaining 24 tickets taken 2 at a time. From this it is evident that 
 any particular ticket is found in C^ of the 2300 possible groups of 3 tickets. 
 
PROBABILITY 449 
 
 Hence, the probability that A will draw a prize with the duplicate of num- 
 ber 10 is 
 
 O^ ^ 12 • 23 ^ 3 
 2300 2300 25' 
 
 Since he has the same chance with the duplicate of number 11, his chance 
 of drawing one prize is f^. 
 
 In like manner, since any two particular tickets, as numbers 10 and 11, are 
 found in Cf of the 2300 possible groups of 3 tickets, the probability that A 
 will draw two prizes is 
 
 Cf ^ 1 
 2300 100* 
 
 Since the probability that A will draw at least one prize includes not only 
 the probability that he will draw a prize with one or the other of the two 
 duplicates, but also the probability that he will draw prizes with both of them, 
 the probability that he will draw at least one prize is /^ + ^^g, or \. 
 
 6. From a bag containing 4 white balls and 8 red balls one ball 
 is taken at random. What is the probability that it will be white ? 
 What are the odds against drawing a white ball ? 
 
 7. A die has 2 white sides and 4 black sides. What is the 
 probability that a white side will be thrown ? 
 
 8. When two dice are thrown, what is the chance of throwing 
 double sixes ? of throwing a 5 and a 6 ? 
 
 9. If two cards are drawn at random from a pack of 52 cards, 
 4 of which are kings, what is the probability that both are kings ? 
 
 10. A committee of four is chosen by lot .from 5 Democrats and 
 4 Republicans. What is the probability that it will bo composed 
 of 2 Democrats and 2 Republicans ? 
 
 11. From a bag containing 5 black balls, 6 red balls, and 7 
 white balls, 4 balls are drawn at random. What is the probar 
 bility that all are black ? that 2 are black and 2 are white ? 
 
 12. What is the probability that 6 can be thrown in a single 
 throw of two dice ? 
 
 13. What is the chance of throwing over 20 with four dice ? 
 
 14. From an urn containing 40 tickets, numbered from 1 to 40, 
 four tickets are drawn. If each ticket drawn entitles the holder 
 of the duplicate to a prize and A holds duplicates of two tickets, 
 what is the chance of A's obtaining at least one prize ? 
 
 ADV. ALG. — 29 
 
450 PROBABILITY 
 
 15. A" has three shares in a lottery that ofPors 2 prizes to 18 
 blanks. B has two shares in a lottery that offers 5 prizes to 25 
 blanks. Compare their chances of drawing at least one prize. 
 
 ■ Suggestion. — Since out of C^ ways of drawing 3 tickets from 20 there 
 are C 3* ways of drawing three blanks, the probability that A will draw .3 
 
 blanks is -^ = ^^ ' ^^ ' ^^' ■= — ■ Hence, Prin. 2, the probability that A will 
 (>|> 20 . 19 • 18 95 ^ 
 
 fail to draw .3 blanks, and therefore draw at least one prize, is 1 — f f , or ||, 
 
 16. A has three tickets in a lottery containing 10 prizes and 90 
 blanks. B has three tickets in another lottery containing 5 prizes 
 and 45 blanks. Show that B has a better chance than A of win- 
 ning at least one prize. 
 
 506. When two or more events are regarded, in the occurrence 
 or non-occurrence of some or all of them, as joining to make one 
 event, the resulting event is called a Compound Event. 
 
 Thus, if an ace is thrown with each of two dice, these two events may be 
 regarded as joining to produce a compound event, namely, throwing a pair of 
 aces. • 
 
 Again, if the event of throwing an ace with one die is joined to the event 
 of failing to throw an ace with the other die, the occurrence and non-occur- 
 rence of the event of throwing an ace with a single die are joined to produce 
 a compound event, namely, throwing only one ace with two dice. 
 
 507. Events that compose a compound event are said to be 
 Independent or Dependent according as the occurrence of one does 
 not or does affect the probability of the occurrence of the other or 
 others. 
 
 Thus, in the event of throwing double sixes with two dice, the component 
 events are independent. But in the compound event that six shall be thrown 
 with one die on the second trial, the component events are dependent, since 
 the success of the second trial depends partially upon the failure of the first 
 trial. 
 
 508. Events that form a compound event by joining the occur- 
 rence of each event to the non-occurrence of each of the others 
 are called Mutually Exclusive Events. 
 
 The compound event of throwing over 9 with two dice is composed of three 
 mutually exclusive events. For when 10 is thrown, neither 11 nor 12 can be 
 
PROBABILITY 451 
 
 thrown ; when 11 is thrown, neither 10 nor 12 can be thrown ; and when 12 
 is thrown, neither 10 nor 11 can be thrown. 
 
 509. Principles. — 3. Tlie probability of a compound event com- 
 posed of independent events is equal to the product of the probabilities 
 of the independent events. 
 
 4. If p is the probability that an event will happen in one trial, 
 the iwobability that it ivill happen n times in succession is p". 
 
 5. The probability that all of two or more independent events will 
 fail is (1 -_pi) (1 -p^ •••. 
 
 Principle 3 may be established as follows : 
 
 Let P be the probability of a compound event composed of events whose 
 probabilities are pi, p2i Pz-, •••• 
 
 It is to be proved that P = piP2Ps •••• 
 
 Suppose that the first event can happen in a ways and fail in b ways, all 
 equally likely ; and that the second can happen in c ways and fail in 
 d ways, all equally likely. 
 
 Since the events are independent, not only may both happen together, but 
 either may happen while the other fails, or both may fail together. Hence, 
 the probability that both will happen together is the ratio of the number of 
 ways they can happen together to the whole number of ways they can 
 happen or fail. 
 
 Since each of the a cases in which the first event happens may be asso- 
 ciated with each of the c cases in which the second event happens, there are 
 ac cases in which both can happen together. 
 
 Similarly, there are ad cases in which the first happens while the second 
 fails, be cases in which the first fails while the second happens, and bd cases 
 in which both fail together. 
 
 Hence, if the probability that both will happen together is p, 
 
 ac + ad + be + bd 
 
 n. c 
 
 ■PiPa- 
 
 a + b c + 
 
 If there is a third event, whose probability of happening is ps, the con- 
 currence of the three events may be regarded as the concurrence of two 
 events by considering the concurrence of the first two as a single event whose 
 probability of happening is p = pip-2. Hence, the probability of the three 
 events happening together is pps, or pip-iPs- 
 
 Since this method can be extended to the consideration of any number of 
 independent events, P = piPiPs •••• 
 
 The proofs of Principles 4 and 5 are left for the student. 
 
452 PROBABILITY 
 
 Problems 
 
 1. The probability that A will die within 20 years is -^, and 
 that his wife will die during that interval is -^^. What is the 
 probability that at the end of 20 years (1) both will be dead? 
 (2) both will be living ? (3) A will be living and his wife dead ? 
 (4"^ A will be dead and his wife living ? 
 
 Solution. — 1. Prin. 3, P = ^^ x i\ = ^V 
 
 2. Prin. 5, i'^ (1 - A)(l - A)= A- 
 
 3. Prin. 5, Prin. 3, p = (1 _ jl) ^3^ = .^zi^, 
 
 4. Prin. ii, Prin. 3, P = (1 - t\) A = \. 
 
 2. A crosses the United States, passing through five cities 
 where he has friends. If the odds are 3 to 1 against his meeting 
 a friend in any one of the five cities, what is the probability that 
 he will meet a friend in each city ? 
 
 510. Principle 6. — The prohahiUty of a compound event com- 
 posed of a series of events, dejyendent or independent, is equal to the 
 probability that the first happens, multijilied by the probability that, 
 after the first has happened, the second happens, etc. 
 
 For if the events are independent, considering them to happen in any 
 order does not modify the probability of any of them ; and if they are 
 dependent, after each has happened and so has modified the probability of 
 the following event, the modified event may be regarded as independent of 
 the preceding event. 
 
 Problems 
 
 1. From a bag containing 3 white balls and 5 black balls A 
 draws a ball, and after A has replaced the ball B draws a ball. 
 What is the probability that A draws a white ball and B a black 
 ball ? What would have been the probability of A drawing a 
 white ball, and B a black ball, had A not replaced the ball he 
 drew? 
 
 Solution 
 
 Since by replacing the ball A removes the effect of his drawing upon the 
 probability that B draws a black ball, the events under the first supposition 
 are independent, but under the second they are dependent. 
 
 In the first case, since the separate probabilities are | and f, the joint 
 probability is | x f , or ^|. 
 
PROBABILITY 453 
 
 In the second case, supposing that A draws a white ball, B's cliance of 
 drawing a black ball out of 2 white balls and 5 black balls is f. Tliat is, the 
 joint probability of A drawing a white ball and B a black ball would have 
 been f x f , or \\. 
 
 2. A starts on an ocean voyage. The probability that the ship 
 will encounter a storm is |. If she does, the probability that she 
 will spring a leak is y'^, but 10 to 1 if she springs a leak her 
 engines will be able to pump her out. If they fail, the proba- 
 bility that the compartments will keep her afloat is f. If she 
 sinks, it is an even chance that any one passenger will be saved 
 by passing boats. What is the probability that A will be lost 
 at sea ? 
 
 511. Principle 7. — TJie probability of a compound event com- 
 posed of two or more mutually exclusive events is equal to the sum 
 of their separate p)Tobai)ilities. 
 
 For if one event can happen a times out of n, another h times out of ra, 
 another c times out of n, etc., if whenever one event happens the others fail, 
 such a union of occurrence and non-occurrence can happen a + 6 + c + ••• 
 times out of n times. 
 
 Hence, the probability of such a compound event is 
 
 or« + ^+^+...; 
 n n n 
 
 that is, when it is impossible for more than one of a series of events to 
 happen at the same time, the probability that some one of them will happen 
 is the sum of the probabilities of the individual events. 
 
 512. Let p be the probability that an event will happen in one 
 trial, and let it be required to find : (a) the probability that it 
 will happen exactly r times in n trials ; (6) the probability that 
 it will happen at least r times in n trials. 
 
 (a) By Prin, 4, the probability that the event will happen any 
 particular r times is p'', and since, Prin. 1, the probability that 
 the event will fail in any one trial is 1—p, the probability that 
 it will fail the remaining n — r times is (1 — />)"~^ 
 
 Hence, Prin. 3, the probability of the compound event, that the 
 event will happen in any parficidar r trials and fail in the remain- 
 ing n — r trials, is ^""(l — j?)""''. 
 
454 
 
 PROBABILITY 
 
 Since out of n trials, exactly r successful trials can be made in 
 as many ways as there are combinations of r things out of n 
 different things, when each excludes the others the probability 
 that the event will happen exactly r times out of n is C" times 
 as great as the probability that the event will happen in any 
 particular r trials and fail in the remaining n — r trials. 
 
 Hence, the probability that the event will happen exactly r times in 
 n trials is 
 
 CrP' {1 - py-'. (A) 
 
 (6) Happening at least r times in n trials includes the following 
 mutually exclusive events, namely, happening exactly n, n — 1, 
 '•', r times in n trials. Substituting these values successively for 
 r in C"^' (1 — pY'", and adding, Prin. 7, the probability that the 
 event will happen at least r times in n trials is 
 
 p" + Cfp"-^ (1 - i?) + C?p"-2 (1 - pY 
 
 + ■■' + c^p^(1-pY-% 
 
 or 
 
 j9" + np"-*(l —p) + 
 
 n(n-l) „_2^. _ 
 
 \1 
 
 p"-\i-py 
 
 + • • • to w — r + 1 terms. 
 
 0') 
 
 Miscellaneous Problems 
 1. Show that the probabilities of throwing 2, 3, 4, •••, 12, re- 
 spectively, in one throw of 2 dice are 3V, ^V s%, ^, j\, ^^, i\^ ^% 
 
 3% 3F» -sV respectively. 
 
 Solution 
 
 Since a die has six faces numbered from 1 to 6, the probability of throw- 
 ing any particular number from 1 to 6 with one die is J, and the probability 
 of throwing the same number with two dice, or of throwing any two desig- 
 nated numbers, one with each die, is J x J, or ■^. 
 
 2, 3, 4, •••, 12 can be thrown as follows: 
 
 2 = 
 
 3 = 
 
 4 = 
 
 and so on. 
 
 1+1, 
 1+2 1 
 2 -f 1 I ' 
 
 14-3 
 
 2 + 2 
 
 3 + 1 
 
 12: 
 11 
 
 10 = 
 
 + 
 
 {5 + 6 
 I 6 + 5 
 
 f4 + 6 
 5 + 6 
 6+4 
 
I 
 
 PROBABILITY 455 
 
 Hence, 2 and 12 can be thrown each in 1 way ; 3 and 11, each in 2 ways ; 
 4 and 10, eacli in 3 ways ; 5 and 9, each in 4 ways ; (5 and 8, each in 5 ways ; 
 and 7, in 6 ways. Therefore, since the probability of throwing two dice in 
 any designated way is 5*5, if po, p.i, •••, denote the probabilities of throwing 
 2, 3, •••, respectively, in one throw, 
 
 P-2 = j\, Ps = /s, Pi = t%, P5 = 3*S^ P6 = A' P^ =sV 
 
 Ps = A> P9 = Ai i'lo = 3^S' Pn = A' P12 = ih- 
 
 2. What is the probability of throwing an ace (a) six times in 
 six trials ? (6) at least once in six trials ? 
 
 Solution 
 
 (a) Since the probability of throwing an ace in a single trial is |, the 
 probability of throwing an ace six times in succession is (J)^ = jsiss- 
 
 The same result riiay be obtained by substituting ^ for p, 6 for n, and 6 
 for r in formula (A). 
 
 (6) If the ace is not thrown at least once in six trials, then the person 
 who throws the die must fail to throw the ace six times in succession. 
 
 Since the probability of failing to throw an ace is 1 — ^, or |, with each 
 trial, the probability of failing to throw an ace six times in succession is(|)^ 
 
 Hence, the probability of throwing an ace at least once in six trials is 
 1 ~(l)^i ^^ mH' which is slightly less than |. 
 
 The same result may be obtained by substituting ^ for p, | for I — p, 
 6 for n, and 1 for r in formula (B). Thus, if the probability of throwing at 
 least one ace in 6 trials is P, 
 
 /iy/5y 6-5-4-3-2 nW5\s 
 VOy Uj 1-2.3.4.5U/UJ 
 
 6.5.4.3/lW5\* , 6.5.4.3.2 
 
 1.2.3.4 
 
 = a + §)«-(!)« = 1 -(!)« = HHi- 
 
 3. If A and B take turns in the order named in drawing one 
 ball from a bag containing 2 white balls and 4 black balls, what 
 chance has each of being the first to draw a white ball ? If A 
 and B are drawing for a prize of $ 25, how much less is B's chance 
 worth than A's ? 
 
 Solution 
 
 First trial. — The probability that A will draw a white ball is |, or ^, and 
 the probability that he will draw a black ball and so give B an opportunity 
 to draw from the remaining 2 white balls and 3 black balls is 1 — |, or |. 
 
456 PROBABILITY 
 
 Second trial. — The probability that B will have an opportunity to draw 
 and will draw a white ball is § x §, or ^, and the probability that he will 
 draw a black ball and so give A an opportunity to draw from the remaining 
 2 white balls and 2 black balls is | (1 - |), or f 
 
 Third trial. — The probability that A will have an opportunity to draw 
 and will draw a white ball is § x ^, or \, and the probability that he will 
 draw a black ball and thus give B an opportunity to draw from the remain- 
 ing 2 white balls and 1 black ball is |(1 — ^), or ^. 
 
 Fcnirth trial. — The probability that B will have an opportunity to draw 
 and will draw a white ball is | x |, or t^^, and the probability that he will 
 draw a black ball and thus give A an opportunity to draw from the remain- 
 ing 2 white balls is J (1 — |), or ^. 
 
 Fifth trial. — The probability that A will have an opportunity to draw 
 and will draw a white ball is jij x f, or j^^, and the probability that B will 
 have another opportunity to draw is JI5 x S, or 0. 
 
 Hence, A's chance for the prize is ^ + i + 1*5, or |, and B's chance is 
 ^ + j^, or |. Since the prize is worth f 25, A's chance is worth | of $ 25, 
 or $ 15, and B's chance is worth f of $25, or $ 10, that is, $5 less than A's. 
 
 Note. — A's expectation is $15 and B's expectation is $10. 
 
 4. What is the probability of throwing more than 10 in a 
 single throw with two dice ? 
 
 5. Compare the probabilities of throwing 3 with one die and 
 6 with two dice. 
 
 6. The probabilities of three events are |, |, and \, respec- 
 tively. What is the probability that one, but not more than one 
 of them, Avill happen ? 
 
 7. The odds in favor of A's solving a certain problem are 2 
 to 1, and the odds in favor of B's solving it are 3 to 2. If both 
 attempt the solution, what is the probability that it will be solved ? 
 AVhat is the probability that both will solve it ? 
 
 8. If 4 coins are tossed, what is the probability that at least 
 one will fall head ? that one and only one will fall head ? 
 
 9. Which is the greater, the probability of throwing at least 
 one ace in six trials, or the probability of throwing at least one 
 head of a coin in two trials ? 
 
 10. Show that the probability of throwing an ace with a single 
 die is ^ in two trials and ^^ in thi-ee trials. 
 
PROBABILITY 457 
 
 11. What is the chance that a backgammon player who has 
 two throws with two dice will make exactly 10 ? 
 
 12. The door of A's house has a spring lock that fails to lock 
 the door 2 times out of 5. A comes home, and in the darkness 
 cannot distinguish between 6 keys, one of which fits the lock. 
 What is the probability that he will get in without a key ? with 
 the first key he uses ? with the first two keys ? with the first 
 three keys ? 
 
 13. What is the chance of throwing just 30 in three throws of 
 two dice ? 
 
 14. Find the probability of throwing double fives with two dice 
 exactly 3 times out of 4. 
 
 15. A, B, and C take turns, in the order named, in drawing 
 one ball from a bag containing 2 black balls and 4 balls of other 
 colors, agreeing that the first to draw a black ball shall receive 
 $ 150. What is the value of the expectation of each ? 
 
 16. From a bag containing 10 5-dollar bills and 20 2-dollar 
 bills, I have the privilege of drawing two bills at random. What 
 is the value of my expectation ? 
 
 17. A, B, and C toss a coin, in the order named, until one wins 
 by tossing head. Find the chance that each has of winning. 
 
 18. A and B throw a die alternately, in the order named, until 
 one wins by throwing an ace. What chance has each of winning ? 
 
 19. A and B toss a dollar alternately, in the order named, agree- 
 ing that the dollar shall go to the one who first throws head. 
 What is the value of the expectation of each ? If the dollar is 
 their joint property, each owning half, what is the probable gain 
 or loss of each ? 
 
 20. From a bag containing 10 balls numbered 1, 2, 3, •••, 10, 
 five balls are drawn at random, each ball being replaced before 
 the next is drawn. Find the probability of drawing the ball 
 numbered 4 exactly three times ; at least twice. 
 
 21. A machinist works 300 days a year. If the odds are 1000 
 to 1 against his meeting with an accident on any particular work 
 day, show that the odds against his escaping injury for 5 years 
 are about 3 to 1. (Use logarithms.) 
 
SIMPLE CONTINUED FRACTIONS 
 
 513. An expression of the form a-\ -r 
 
 Continued Fraction. c -\ 
 
 e + 
 
 It is usually written in the more compact form 
 
 is called a 
 
 a + 
 
 c+ e-t- 
 
 514. A continued fraction in which each numerator is 1 and 
 each denominator is a positive integer is called a Simple Con- 
 tinued Fraction. 
 
 The integral part a may be either a positive integer or zero. 
 
 In this chapter only simple continued fractions will be discussed, and for 
 the sake of brevity these will be called continued fractions. 
 
 515. A continued fraction is said to be terminating or infinite 
 according as the number of denominators is finite or indefinitely 
 great. 
 
 A terminating continued fraction may be reduced to a simple 
 fraction as in § 184. 
 
 516. In the continued fraction 
 111 
 
 ttj + 
 
 02+ «3+ «4-| 
 
 which will be used as the general symbol of a continued fraction, 
 Ui is called the first convergent, ai-\ — the second convergent, 
 tti -1 the third convergent, and so on. 
 
 «£+ Os 
 
 468 
 
SIMPLE CONTINUED FRACTIONS 459 
 
 When ttj is lacking, is regarded as the first convergent. 
 «i) <*2j %) ••■ s-re called partial quotients. 
 
 Og -\ , a^-\ , ••• are called complete quotients. 
 
 a3 + '" «4-l 
 
 517. Let - be a fraction whose terms are integers greater 
 h 
 than 1. 
 
 Dividing a by 6, let a^ be the quotient and 5i the remainder. 
 
 r^, a 5. 1 
 
 Then, - = a, +- = a, + j- 
 
 Dividing h by 6i, let ttj be the quotient and h^ the remainder. 
 
 r^i a 1 1 
 
 Then, r = «iH r = '^'i-l r' 
 
 62 1 
 
 h. 
 
 Continuing this process exactly as in finding the H. C. D. of a 
 and h, the result is a continued fraction. Since a and h are posi- 
 tive integers, the process must finally come to an end when the 
 H. C. D., or if none exists, then unity itself, is obtained as a 
 divisor. Hence, 
 
 A fraction tchose terms are integers greater than 1 may be con- 
 vetied into a terminating continued fraction. 
 
 Examples 
 
 1. Keduce ^^ to a continued fraction. 
 
 Solution. 24J 1 51!^ = ai 
 
 144 
 
 7_|24[3 = a2 
 21 
 3_|7t2 = a8 
 6 
 
 . 1^ = 6+ — — - lJ3[3 = a4 
 
 * ■ 24 3+2+3' 
 
 Reduce to continued fractions : 
 
 2. H- 3. H- 4. If 5. ■^. 6. H- 7. IH- 
 
46a SIMPLE CONTINUED FRACTIONS 
 
 518. In «, + A^ -4-^, let ^, ^^ ^^ ••, ^-, ... 
 
 be the successive convergents reduced to simple fractions. Then, 
 Pi_'h. "Pi _ <^2<^i + 1 . V^ _ a^iSL^i + 1) + ^1 _ asPi + Pi . 
 
 91 ~" 1 ' 92 ~ a2 • 1 ' 93 ~ «3a2 + 1 " ^392 + 9i ' 
 
 j94 ^ a4[a3(a2ai + 1) + «i] + (a2«i+l) _ ^4^3 + Pi 
 g* ~ ai{a3fi2 + 1) + a, ~~ «493 + 92 
 
 From an examination of the third and fourth convergents it is 
 seen that they are formed according to a certain law. If the wth 
 convergent obeys this law, 
 
 Pn _ anPn-l + Pn-2 /-^N 
 
 9»~«n9n-l + 9n-2' 
 
 It remains to be shown that if (1) is true for the wth convergent, 
 it will be true for the (n + l)th convergent. 
 
 Since the (n + l)th convergent differs from the nth only in 
 
 having a„ + instead of a,„ 
 
 ^n+l 
 
 » , {^-~^n )Pn-^+Pn-2 
 
 Fn+l \ <^n + l/ 
 
 9:;7i~7 r^; ' 
 
 («n + ^)9«-l + 9«-2 
 
 multiplying both terms by a„_^i and rearranging, 
 
 ^ «n+l(<^n A.-1 + Pn-2) + Pn-1 
 a„+l(«n9n-l + 9»-2)+ 9«-l 
 
 "n+l9n T^ 9n-l 
 
 which has the same form as (1), n + 1 taking the place of n. 
 
 Hence, if (1) is true, (2) is true ; that is, if the law of forma- 
 tion of successive convergents revealed in the third and fourth 
 convergents is true for any particular convergent, it holds true for 
 the next convergent. 
 
 Therefore, since the law is true for the fourth convergent, it 
 holds for the fifth ; and being true for the fifth, it holds for the 
 sixth ; and so on. 
 
 Hence, formula (1) expresses the law of formation of the suc- 
 cessive convergents to a continued fraction, each convergent after 
 
SIMPLE CONTINUED FRACTIONS 461 
 
 the second being obtained from the two convergents immediately 
 preceding it. 
 
 This formula is called the recurrence formula. 
 
 Examples 
 
 1. Find the first six convergents to 
 
 3 1 1 1 1 1 
 
 ^2+6+ 2+ 6+ 2+...' 
 
 Solution. — The first two convergents, 3 and |, are found by inspection, 
 and the first convergent is written | because its denominator is needed in 
 forming the denominator of the third convergent. By the recurrence formula, 
 
 3d convergent = ^AJ1±1 = ^ ■ 4th convergent = 2(45)_+I = 21 
 6(2)4-1 13 2(13)4-2 28 
 
 5th convergent = ^Jm±^ = 627 ^^^ eonvergent = 2(627)+97 ^ 1351 . 
 6(28)4-13 181 2(181)4-28 390 
 
 Hence, the first six convergents are 3, |, f^, ||, ff|, -VW- 
 Note. — In applying the recurrence formula, if the first convergent is 0, 
 it is written ^. 
 
 Find the first five convergents to 
 
 2. 2+ — — -^^ ^ 4. 0-f— ^ 
 
 3. 1-f 
 
 1 + 3+5+7 + .. . 2+1 + 3+2 + 
 
 1 1 1 1 g 111 1 
 
 2+3+4+5 + ...* ' 1+1 + 1+2 + ...* 
 
 519. To illustrate the principles proved in the succeeding ar- 
 ticles the student may calculate and compare the convergents of 
 any continued fraction, as 
 
 ■^ 1 1 1 1 1 1 
 + 1 + 2 + 3+ 1 + 2 + 3 + ...' 
 The convergents of this continued fraction are 1, 2, ^, i^, ^|, 
 1^, 105^ ...^ or 1, 2, 1.66, 1.7, 1.6923, 1.6944, 1.6942, .••, approxi- 
 mately. 
 
 It is observed that : 
 
 (1) If any tvro consecutive convergents are chosen, the even 
 convergent is always greater than the odd one. 
 
 (2) The even convergents form a descending series and the odd 
 convergents form an ascending series. 
 
 (3) The difference between consecutive convergents grows less 
 
462 
 
 SIMPLE CONTINUED FRACTIONS 
 
 1 
 
 and less, and the even and odd convergents seem to approach a 
 common limit. Later the student will be able to show that this 
 limit is ^(6 + V6), an incommensurable number and the actual 
 value of the continued fraction. 
 
 The accompanying diagram shows the relative size of the 
 successive convergents, represented by the dotted lines. 
 
 520. Every terminating continued fraction has a single definite 
 value, which is the same as that of its last convergent. 
 
 To show that every infinite continued fraction has a single 
 definite value, it is necessary to prove that: 
 
 Principle 1. — T7ie difference between any two consecutive con- 
 vergents, as the (n-\-l)th and the nth, is numerically equal to the 
 recijyrocal of the product of their denominators. This difference 
 is negative when the minuend is an odd convergent, and positive 
 when the minuend is an even convergent ; that is, 
 
 Pn+i j)„^(-l)"+^ 
 
 220 - 221 - 1 
 
 Thus, in § 519, 
 
 aud 
 
 9n+l9» ' 
 
 
 a _ £i = 22 _ 
 35 Qi 13 
 
 17 
 10 
 
 i>6 _ P6 = ^ _ 
 
 ge 25 36 
 
 22 
 13 
 
 hi9«-P«9«+i = (-!)' 
 
 (1) 
 
 13-10 
 
 793 - 792 
 
 36 . 13 
 
 13-10 
 
 1 
 36-13" 
 
SIMPLE CONTINUED FRACTIONS 463 
 
 The above principle may be established as follows : 
 
 Suppose that the principle is true for the nXh and (m — l)tli convergents. 
 
 Then, ^-^ = ^", or p„g„_i - p„_,g,, = (- 1)» (2) 
 
 Since by the recurrence formula j}„+i=a„+ip„+p,i_i and g'„+i = a„+ig'„+g„_i, 
 
 Pn+\qH-Pnqn+l-((tn+lPn+Pn-l)qn-Pn(an+iqn + qn-l) 
 
 canceling, = Pn-iqn - Pnqn-i 
 
 by (2), =-(-!)»=(- 1)(- !)«=(- l)«+i. 
 
 Dividing by ,„,„,., f-_|.= (^\ 
 
 qn+l ?n Qn+iqn 
 
 Therefore, if (2) is true, (1) is true ; that is, if the difference between the 
 
 nth and the (n — l)th convergents is found by Prin. 1, the difference between 
 
 the (n + l)th and nth convergents is found by Prin. 1. 
 
 T, 4. • e -iQ Pi Pi a2ai+l ai 1 
 
 But since, § ol8, - — ■^— — 
 
 gi 02 1 02 • 1 
 
 the principle is true for the first two convergents. Therefore, it holds for 
 the second and third ; and being true for the second and third, it holds 
 for the third and fourth ; and so on. Hence, the principle is true universally. 
 
 Corollary.* — The convergents to a continued fraction are 
 fractions in their loivest terms. 
 
 For in (1), if Pn and g„, or p„+i and qn+ii had a common factor, this 
 factor would be a factor of Pn+iqn —Pnqn+h or of ± 1, which is impossible. 
 
 521. Principle 2. — TJie even convergents to a continued frac- 
 tion form a descending series and the odd convergents an ascending 
 series. 
 
 The above principle may be established as follows : 
 
 By Prin. 1, P^ _ Pn ^ i^iy^ ^ _ tl^ ^ (^ 
 
 g»+l ?n Qn+iqn 9n+l9n 
 
 and ^»_^^:.^l21". (2) 
 
 qn qn-1 qnqn-1 
 
 Adding (1) and (2), g^^ _ ^i^ (gn.i- g«-i)(- 1)" . (3^ 
 
 * ^ ^ qn+l qn-1 qn+\qnqn-l 
 
 By the recurrence formula, g„+i >q„> g„-i, and since they are all positive, 
 the sign of (3) depends only upon the sign of ( — 1)". 
 
 If n is odd, the convergents in the first member of (.3) are even con- 
 vergents and, (— 1)» being negative, their difference is negative; that is, 
 the even convergents form a descending series. 
 
 * A corollary to a principle is a subordinate principle easily deduced from 
 the main principle or from some part of its proof. 
 
464 SIMPLE CONTINUED FRACTIONS 
 
 If n is even, the convergents in tlie first member of (3) ai-e odd con- 
 vergents and, (— 1)" being positive, their difference is positive ; that is, the 
 odd convergents form an ascending series. 
 
 522. Principle 3. — Every infinite continued fraction has a 
 single finite value, which is the common limit of the terms of the 
 descending series formed by its even convergents and the terms of 
 the ascending series formed by its odd convergents, as the number 
 of convergents increases without limit. 
 
 The above principle may be established as follows : 
 
 By Prin. 1 any even convergent is greater tlian the preceding odd con- 
 vergent, which by Prin. 2 is greater than any preceding odd convergent. 
 Therefore, every even convergent is greater than every odd convergent that 
 precedes it. 
 
 By Prin. 2 any even convergent is greater tlian any following even con- 
 vergent, and by Prin. 1 each of the latter is greater than either of the adja- 
 cent odd convergents. Therefore, every even convergent is greater than 
 every odd convergent that follows it. 
 
 Hence, every even convergent is greater than every odd convergent. 
 
 It is evident from the law of formation of the denominators of successive 
 convergents that they form an ascending series, in which qi <q2<Q3 < •••• 
 
 Therefore, in i^^, (^ii^, .., t^^, (-1)""\ .., 
 
 9251 QsQ-i qnqn-i qn+iqn 
 
 whose terms, Prin. 1, are the differences of successive convergents, one even 
 and the other odd, as n increases without limit the nth difference approaches 
 zero as a limit. 
 
 Since every even convergent is greater than every odd convergent, and as 
 the number of convergents taken increases without limit, the difference 
 between successive convergents approaches zero as a limit, the terms of the 
 descending series of even convergents and the terms of the ascending series of 
 odd convergents approach a common limit, which is the value of the continued 
 fraction itself. 
 
 Corollary. — The value of an infinite continued fraction lies 
 between any two consecutive convergents. 
 
 523. Principle 4. — Ayiy convergent to an infinite continued 
 fraction is nearer the continued fraction than any preceding 
 convergent. 
 
 The above principle may be established as follows : 
 
 Let X represent the value of the continued fraction. 
 
 To show that the rath convergent is nearer to x than the (n — l)th con 
 vergent, first find x in terms of these convergents. 
 
I SIMPLE CONTINUED FRACTIONS 465 
 
 By the recurrence formula the (n + l)th convergent involves the nth 
 and the {n — l)lh convergents and the (n + \)i\\ partial quotient, an+\. Let 
 k represent the complete {n + l)th quotient. Since when the complete quo- 
 tient is substituted for the partial quotient the result is the fraction itself, 
 
 kPn + Pn-l 
 
 Therefore,* x ~ 
 
 kqn + qn-i 
 
 Pn-l k(Pnqn-l ~ Pn-lQn) k 
 
 Qn 1 qn-i{lcq„ + qn-i) qn-i(kq„ + g„_i) 
 
 and ^~a; = 
 
 Pn ^ JPn?n-l~Pn-l9n 
 
 <?n qn{kq„ + g„_i) q„(kqn + q„-i) 
 
 Since q„>qn-i and A;>1, the difference between the nth convergent 
 and X is less than the difference between the (n — l)th convergent and x ; 
 that is, any convergent is nearer to the continued fraction than the next pre- 
 ceding convergent, and is nearer, therefore, than any preceding convergent. 
 
 524. Principle 5. — Any convergent to a continued fraction is 
 nearer to the value of the continued fraction than any other fraction 
 whose denominator is smaller than that of the convergent. 
 
 The above principle may be established as follows : 
 
 Let X be the value of a continued fraction, ^ its nth convergent, and ? 
 
 qn b 
 
 any other fraction whose terms are positive integers, and whose denominator 
 is less than q„. 
 
 It is to be proved that ^ is nearer to x than _• 
 
 a Qn Pn ^ a 
 
 If possible, let t be nearer to x than — ' Then, Prin. 4, t is nearer to 
 
 X than ^5zl. Therefore, since by Prin. 3, Cor. , x lies between the nth and 
 
 the (n — l)th convergents, t lies between them. 
 
 Hence, « ^<^',.^\ 
 
 b Qn \ qn ?n-l 
 
 or, Prin. 1, aq.^-bpn-^ _J_^ 
 
 bqn-l qnQn-l ^ ' 
 
 Since g„> 6, making the denominators in (1) alike by substituting qn for 
 6 in the first denominator decreases the first fraction. 
 
 Hence, in order that - may be nearer to x than — , we must have 
 
 ag„_i~ &;?„ i<l, 
 
 which is impossible, because a, qn-\i b, and p„_i are integers. 
 
 * The sign ~, read ' the difference between,' indicates that the less num- 
 ber is to be subtracted from the greater. 
 
 ADV. ALG. — 30 
 
466 SIMPLE CONTINUED FRACTIONS 
 
 Hence, since 6 < <?„, t is not nearer to x tlian — ; that is, the ntli conver- 
 
 gent is nearer to x than any other fraction whose denominator is smaller 
 than that of the nth convergent. 
 
 525. Limit of error of the nth convergent. 
 
 Since, Prin. 3, Cor., the value of a continued fraction lies between 
 the values of any two consecutive convergents, the error made by 
 taking the nth convergent instead of the whole fraction is less 
 than the difference between the nth. convergent and the next con- 
 vergent. By Prin. 1 this difference is numerically equal to the 
 reciprocal of the product of their denominators. Hence, 
 
 Principle 6. — The error made by taking the nth convergent of 
 an infinite continued fraction is less than unity divided by the prod- 
 uct of the denominators of the nth and (n -f l)th convergents. 
 
 Since error < , 
 
 <lnQn+l 
 
 and = — ; ■ r< 
 
 Qngn+i 9«(an+i7» + 9„-i) a„+i9„ 
 the error corresponding to the nth. convergent is small when a„+i, 
 the next partial quotient, is large. Hence, 
 
 Corollary. — Any convergent that immediately precedes a large 
 partial quotient is a near approximation to the value of the continued 
 fraction. 
 
 Examples 
 
 111111 1 
 
 1. Compute the value of 10 -|- 
 
 1 + 4+2 + 4+1 + 20+1 + — 
 
 correct to the nearest fourth decimal place. 
 
 e ^ . 10 11 54 119 530 649 ( ) 
 
 Solution. — Convergents, — , — , — , — , — •, -—, -Hr;, •••• 
 ^ 1 1 5 11 49 60 1249 
 
 Since the error must be less than \ of .0001, or less than 25^5, by 
 Prin. 6 the first convergent that is sufficiently accurate is -^, whose denomi- 
 nator multiplied by the next denominator exceeds 20000. 
 
 Hence, continued fraction = ^ = 10.8167, to the nearest ten-thousandth. 
 
 2. Find the value of tt = 3 + ^ -^ -^ ^ ^ to the 
 nearest sixth decimal place. " ' "" 
 
 3. If 2.20462 pounds make 1 kilogram, find six commensurable 
 fractions that represent approximately the ratio of a kilogram to 
 a pound, and estimate the error for each fraction. 
 
SIMPLE CONTINUED FRACTIONS 467 
 
 526. To convert a quadratic surd into a continued fraction. 
 
 Let it be required to convert 2 + Vll into a continued fraction. 
 The largest integer in 2 + Vll is 2 + 3, or 5. 
 
 2 + Vll = 5+Vll-3 = 5 + ^^^~^ 
 
 2 
 rationalizing the numerator, = 5 H =r 
 
 VlH-3 
 
 reducing the numerator to 1, = 5 H -:=n (1) 
 
 Vll + 3 ' 
 
 2 
 Similarly, Vn+3 = 3 + Vn^ = 3 + ^; (2) 
 
 1 
 
 also, — 3—='+— I— =^+ vnTi - ^'^ 
 
 2 
 The last denominator in (3) may be transformed by (2), then 
 the last denominator in the result may be transformed by (3), and 
 so on. Substituting the results (2), (3), and so on in (1), 
 
 or 
 
 2+vn=5+-^^ — - 
 
 
 
 3+ 1 
 • Vll+3 
 
 1 
 
 
 3+-^ 
 6+— 4— 
 
 vn+3 
 
 2 
 
 
 
 
 ^+ 1 
 
 
 3+1 , 
 
 6+ 1 
 
 
 6+1 , 
 
 3+ ^1 
 Vll+3 
 
 1 
 
 «-6 + ... 
 
 o 1 ^/TT-^ \ ^ ^ ^ 
 
 1 
 
 
 ^^^^^-^^3+6+3+6+...' 
 
 
 in which the partial quotients, 3 and 6, recur. Such a continued 
 fraction is called a periodic, or a recurrimj continued fraction. 
 
 By the method just exemplified, any positive quadratic surd 
 may be converted into a periodic continued fraction. 
 
468 SIMPLE CONTINUED FRACTIONS 
 
 527. To convert a periodic continued fraction into a quadratic surd. 
 
 1. Let it be required to convert 6 + - — - — into a 
 
 * o+6+3+6-f--" 
 
 ♦ 
 
 quadratic surd. The recurring part is marked by asterisks. 
 
 1 1 
 
 Then, a; = 6 H z z — = 6 + 
 
 3 + i^" 3fl (2) 
 
 G + o + • • • X 
 
 Simplifying, etc., a^ — 6a; — 2 = 0, a quadratic equation whose 
 roots, 3 + Vil and 3 — VH, are quadratic surds, one positive and 
 one negative. Since the given continued fraction is positive, its 
 value X is the positive root, 3 + VH. 
 
 Note. — The second member of (2), being a terminating continued frac- 
 tion, may be simplified by finding its third convergent. 
 
 2. To convert 5 + into a quadratic surd, 
 
 3+ 6+ 3+ 6h 
 
 let x = 5-\- - — 
 
 3+ 6+ 3+ 6+--- 
 1 hen, X — 5 = ^ 
 
 3+g 1 1 3+6 + (x-5) 
 
 "^3+6 + ... 
 Simplifying, etc., ar^ — 4 a; — 7 = 0, whence a; = 2 + VH. 
 
 3. To convert 5 -\ — - — into a quadratic surd, 
 
 4+ 6+ 3+ 6+"- 
 « * 
 
 first find the value of the complete quotient whose integral part 
 
 is the first recurring partial quotient 6. 
 
 Byl, 6 + ^^;^ = «^ = 3+VlT. (1) 
 
 Next, put y = 5 + — -^^ — = 5 + — -- (2) 
 
 '^ ^ ^4+6+3+6 + --- 4+a; ^^ 
 
 * * 
 
 Substituting (1) in (2), ?/ = -f (40 - Vll). 
 
 Hence, every periodic continued fraction is equal to a quadratic 
 surd and is a positive root of a quadratic equation with rational 
 coefficients. 
 
SIMPLE CONTINUED FRACTIONS 469 
 
 The proof 
 
 is as follows : 
 
 
 
 Let 
 
 2/ = 6i+-^ 
 
 ^2 + • 
 
 111 11 
 
 • bn + ffll + a2 + ••• Or + Ol + — 
 
 
 
 --^.. 
 
 1 1 
 
 • 6„ + X 
 
 (1) 
 
 In (1), 
 
 a-2 + • 
 
 = aiH 
 
 (2) 
 
 Since x equals the (»• + l)th convergent of (2), by the recurrence formula, 
 
 xqr + qr-1 
 
 Clearing (3) , qrX^ -(Pr- qr-i)x -Pr-i = 0, 
 
 a quadratic in x whose positive real root (§ 304, I) is 
 
 2qr ^ ^ 
 
 Since ai recurs, ai must be equal to or greater than 1 . 
 • ■• Pr^qr^ qr-i, Pr — qr-1 IS a positive integer, and x is real. 
 To prove that the radical in (4) is a quadratic surd, write 
 
 VCiJr - qy-l)'^ + 4 Pr-\qr = ^(Pr + qr-l)'^ + 4:(P>-i<lr - Prqr-l) 
 
 Prin. 1, formula (1), = VCp-- + Qr-i)'^ ± 4. 
 
 Since no positive integral square increased or diminished by 4 is a perfect 
 square, the radical is a quadratic surd. 
 
 Hence, x is a quadratic surd. 
 
 Therefore, if the value of x is substituted for x in (1) and the result is 
 simplified by repeated rationalization of denominators and reduction to higher 
 terms, y will finally be reduced to the form of a quadratic surd. 
 
 Examples 
 
 Reduce each of the following to a periodic continued fraction 
 and find a near approximation (§ 525) to its value : 
 
 1. Vl5. 3. V7. 5. V'2i. 7. VlO. 
 
 2. Vi26. 4. 2 4-V7. 6. Vl9. 8. 1 : VlO. 
 
 Reduce the following to quadratic surds : 
 9.2 + ^—^. 11.14-^ ^ ^ 
 
 8 + 2 + ... 3 + 2 + 2 + ... 
 
 « » # 
 
 10. 1+-1-Tx-^^ 12.3+-^ 111 
 
 10 + 10 + ... 1+1+1+1+6+1 + 
 
THEORY OF NUMBERS 
 
 SCALES OF NOTATION 
 
 528. The theory of numbers treats of the properties of positive 
 integers. In this chapter the word number will be used in the 
 sense of positive integer, and the terms integer, even number, odd 
 number, power, root, etc., will be employed in the same sense as 
 in arithmetic, unless the contrary is stated. 
 
 529. The Arabic system of notation is called the decimal 
 system, because ten units of any order make one unit of the next 
 higher order. 10 is called the radix of the scale of notation. 
 
 Any number in the decimal system may be expressed in a scale 
 of descending powers of 10, the exponent of the highest power 
 being one less than the number of digits of the number. 
 
 Thus, 256473 = 200000 + 50000 + 6000 + 400 + 70 + 3 
 
 = 2(10)5 + 5(10)* + 6(10)8 + 4(10)2 + 7(-io) + 3. 
 
 530. Any number greater than 1 may be used as the radix of a 
 scale of notation. 
 
 The meaning of the expression 256473 in the decimal scale has just been 
 shown. But in the scale whose radix is 8, 
 
 256473 = 2(8)5 + 5(^)4 + 6(8)8 + 4(8)2 ^ 7(8) + 3. 
 
 In the binary scale the radix is 2 ; in the ternary, 3 ; in the 
 qimternaiy, 4 ; in the quinary, 5 ; in the senary, 6 ; in the septen- 
 ary, 7 ; in the octary, 8 ; in the nonary, 9 ; in the denary, or deci- 
 tnal, 10 ; in the undenary, 11 ; in the duodenary, or duodecimal, 12 ; 
 and so on. 
 
 The general symbol for any radix is r. 
 
 470 
 
I THEORY OF NUMBERS 471 
 
 531. Principle. — Any integer N may he expressed in the scale 
 ivhose radix is r in one and only one loay. 
 
 The above principle may be established as follows : 
 
 Divide N by r, and let the quotient be Ni and the remainder s. 
 
 Then, N = Niv + s. (1) 
 
 Divide Ni by r, and let the quotient be N^ and the remainder q. 
 
 Then, Ni = N^r + g, 
 
 whence, by (1), iV = N^r"^ + qr -\- s. (2) 
 
 Continuing the process, a quotient less than r is finally obtained. Suppose 
 this occurs after r — 1 divisions, and denote the last quotient and remainder 
 by a and 6, respectively. 
 
 Then, iV" = ar»-i + &r»-2 + \-qr+s. 
 
 The coefficients a, ft, •••, 5', s are the digits of the number, and since each is 
 a remainder of some division by r, each is less than r. Also, since there can 
 be but one remainder less than r for each division, there is but one way of 
 expressing N in the scale of r. 
 
 Corollaries. — 1. The exponent of the highest power of the 
 radix is one less than the number of digits. 
 
 2. In any scale the number of different characters, including 0, 
 that are required to express all numbers is equal to the radix. 
 
 For example, in the duodecimal scale the ten Arabic figures and two other 
 characters, t for ten and e for eleven , are required ; in the quinary scale only 
 the figures 0, 1, 2, 3, 4 are required. 
 
 532. To change from the decimal to another scale. 
 
 Examples 
 1. Change 3010 from the decimal to the senary scale. 
 
 Explanation. — Since 6 units of any order make one 
 
 PROCESS ^"^^^ ^^ ^^^ next higher order, 3010 units of the first 
 
 order make 501 units of the second order and 4 units of 
 
 6 [3010 the first order ; 501 units of the second order make 83 
 
 6 [501 ... 4 units of the third order and 3 units of the second order ; 
 
 6 [83 ...3 and so on. 
 
 ^ M g p. Hence, 3010 when expressed in the senary scale is the 
 
 ^~ "' sum of 2 units of the fifth order, 1 unit of the fourth 
 
 Z ••• i. order, 5 units of the third order, 3 units of the second 
 
 order, and 4 units of the first order. 
 
 For convenience the radix of the scale is indicated by a subscript figure, 
 
 which may be omitted in the decimal scale. Thus, 3010 = 215346. 
 
_6 
 83 
 
 472 THEORY OF NUMBERS 
 
 2. Express in the octary scale 50, 128, and 5283. 
 
 3. Express in the quinary scale 12, 342, and 6627. 
 
 4. Express in the duodecimal scale 15, 100, and 6053. 
 
 5. Express in the binary scale the numbers from 1 to 10. 
 
 533. To change from any scale to the decimal scale. 
 
 Examples 
 
 1. Change 215346 to the decimal scale. 
 
 PROCESS 
 
 21534 Explanation. — Since each unit of any order is equal to 6 
 
 6 units of tiie next lower order, 2 units of the fifth order are equal 
 
 To to 12 units of the fourth order, and adding 1, the number of 
 
 units of the fourth order in 215346, the whole number of units 
 
 of the fourth order is 13. 
 
 Continuing in a similar way to reduce the units of each 
 
 6 order to units of the next lower order, the whole number of 
 
 601 units of the lowest order is found to be 3010, which is expressed 
 
 g in the decimal scale. 
 
 3010 
 
 Change the following to the decimal scale : 
 
 2. 42;. 3. 66548. 4. UQ^- 5. 6e4i2. 6. IIIIOOI2. 
 
 534. Arithmetical processes in any scale. 
 
 The processes are performed in the same manner as in the deci> 
 mal scale. The student must simply bear in mind each time the 
 number of units of each order required to make one of the next 
 higher order. 
 
 When the process is complicated or when different scales are involved, j.11 
 the numbers may first be reduced to the decimal scale. 
 
 Examples 
 Perform the operations indicated : 
 
 1. 38,2 + 45i2 + e6i2. 3. 42415-33235. 5. I3047--257. 
 
 2. IOI2 + IIO2 + lOllOlo. 4. 4e5823i2 x 15i2. 6. Vi32i;. 
 
 7. Multiply 2I12 by 13^ and express the product in the decimal 
 scale. 
 
THEORY OF NUMBERS , 473 
 
 8. Show that in any scale the expression 121 represents a per- 
 fect square. 
 
 9. In what scale is 5 times 6 expressed by 36 ? 
 
 10. In what scale is \ of the number 100 equal to the number 30 ? 
 
 11. Which of the weights of 1, 2, 4, 8, ••• pounds must be taken 
 to aggregate 75 pounds, if not more than one of each is used ? 
 
 PRINCIPLES OF DIVISIBILITY 
 
 535. In this subject ' divisible ' is used for ' exactly divisible.' 
 Also, since there is no remainder when zero is divided by any 
 number, zero is regarded as exactly divisible by any number. 
 
 536. If the number JV^= ar"^ + &/•"-- -\ \-pr + qr + .s is 
 
 divided by r, the remainder is s, the last digit ; if iV is divided by 
 9-^, the remainder is the number that is expressed by the last two 
 digits ; if A^ is divided by ?-^, the remainder is the number that is 
 expressed by the last three digits ; and so on. 
 
 Likewise, if N is divided by a factor of r, the remainder is the 
 same as that obtained by dividing the last digit by that factor ; if 
 N is divided by a factor of r^, the remainder is the same as that 
 obtained by dividing qr + s, the number expressed by the last two 
 digits, by that factor ; and so on. Hence, 
 
 Principle 1. — (a) If a number in the scale of r is divided hy 
 y™, the remainder is the number that is expressed by the last m digits 
 of the given riumber. 
 
 (b) If the number is divided by a factor of r™, the remainder is 
 the same as that obtained by dividing by that factor the number that 
 is expressed by the last m digits. 
 
 Corollaries. — 1. In the decimal system a number ending in m 
 ciphers is divisible by 10"*. 
 
 2. In the decimal system 
 
 (a) Every number ending in or 5 is divisible by 5. 
 
 (b) Every mimber ending in 00, 25, 50, or 75 is divisible by 25. 
 
 (c) Every number ending in three ciphers or in three digits that 
 express a multiple of 125 is divisible by 125. And so on. 
 
 3. In the decimal system 
 
 (a) Every mimber ending in 0, 2, 4, 6, or 8 is divisible by 2. 
 
474 , THEORY OF NUMBERS 
 
 (b) Every number ending in two ciphers or in two digits that 
 express a multiple of 4 is divisible by 4. 
 
 (c) Eve^-y number ending in three ciphers or in three digits that 
 express a multiple of 8 is divisible by 8. And so on. 
 
 Note, — If the radix is odd, the last digit of a number does not indicate 
 whether the number is even or odd. Thus, llg = 6 ; I25 = 7 ; II25 = 32. 
 
 537. From a?-»-^ -f-6r"-2 -\ ypt^ -\-qr +s 
 
 subtracting a 4-_6 + f-p + q -j- s 
 
 the result = a^-"-^ - l) + 6(r»-2-l) -\ \-p{r^ - 1) + qO'-I), 
 
 which, § 111, is divisible by r — 1. Hence, 
 
 Principle 2. — The difference betiveen any number and the sum 
 of its digits is divisible by the radix less one. 
 
 Corollary. — In the decimal system the difference between any 
 number and the sum of its digits is divisible by 9. 
 
 538. Since the difference between any number and the sum of 
 its digits is a multiple of r — 1, r being the radix, and since the 
 number that must be added to this multiple of r — 1 to produce 
 the given number is the sum of the digits, it follows that : 
 
 Principle 3. — (a) TJie remainder obtained by dividing a mim- 
 ber in the scale ofrbyr — 1 is the same as that obtained by dividing 
 the sum of the digits by r — 1. 
 
 (b) TJie remainder obtained by dividing the number by any factor 
 ofr — 1 is the same as that obtained by dividing the sum of the digits 
 by that factor. 
 
 Corollaries. — 1. A number in the scale of r is divisible by 
 r — 1, if the sum of its digits is divisible by r — 1; or by any factor 
 of r — 1, if the sum of the digits is divisible by that factor. 
 
 2. In the decimal system 
 
 (a) Any number is divisible by 9, if the sum of its digits is divisible 
 by 9. 
 
 (b) Any number is divisible by 3, if the sum of its digits is divisible 
 by 3. 
 
 (c) Any even number is divisible by 6, if the sum of its digits is 
 divisible by 3. 
 
 The test called casting out the nines is an application of Prin. 3. For 
 applications of this test, see the author's Standard Arithmetic, page 400. 
 
THEORY OF NUMBERS 475 
 
 539. Let N = a?-""^ -\ \-hr* + hi^ +pr^ + qr + s. 
 
 Subtracting the coefficients of the even powers of r and also the 
 coefficients of the odd powers of r with their signs changed, and 
 calling the subtrahend d, 
 
 N=s + qr +pi^ -{-kr^ -{-hr^ -\ 
 
 d= s— q +j3 — k + h 
 
 N-d= 5(r + l) + ^(7-2-l)+ fc(j'3 + 1)4- /i(r* _!)+.. . 
 
 By § 111, this result is divisible by r + 1. 
 
 Then, let JV— d = m (r + 1), m being an integer. 
 
 .-. ]Sf=m(r + l)-\-d. 
 
 Therefore, if d is divisible by r + 1, ^ is also. Hence, 
 
 Prijj ciPLE 4. — A number in the scale of r is divisible by r -\-l, 
 if the difference between the su7ns of its alternate digits is divisible 
 by r + 1. 
 
 Corollary. — In the decimal system ayiy number is divisible by 
 
 11, if the difference bettoeen the sums of its alternate digits is divisible 
 by 11. 
 
 Examples 
 
 Which of the numbers 2, 4, 8, 16, 5, 10, 15, 20, 25, 100, 3, 6, 9, 
 
 12, 11, 22, 33 are divisors of the following ? 
 
 1. 1000. 4. 6250. 7. 522. 10. 54945. 
 
 2. 1001. 5. 1620. 8. 275. 11. 10240. 
 
 3. 1012. 6. 2673. 9. 729. 12. 80080. 
 
 PRIME AND COMPOSITE NUMBERS 
 
 540. Suppose that x is prime to a but is a factor of ab. 
 
 Since - can be converted into a continued fraction that termi- 
 
 X 
 
 nates with the nth convergent, if the preceding convergent is -, 
 
 § 520, (1), « ^-P = 1, or aq-^px^ 1. 
 
 X q qx 
 
 .'. abq-^bpx — b. 
 
 Since ab and x are each divisible by x, the difference between 
 ab • q and x • bp, which is b, is divisible by x. Hence, 
 
476 THEORY OF NUMBERS 
 
 Principle 1. — Any factor of ah that is prime to a must be a 
 factor of b. 
 
 Corollaries. — 1. If x is prime to a, b, c, •••, it is prime to 
 their product. 
 
 2. If a', b', c', ••• a»-e pnme to each of the numbers a, b, c, •••, 
 then a'b'c' ••• is prime to abc •••. 
 
 3. Powers of different prime numbers or of mimbers prime to 
 each other are prime to each other. 
 
 541. By definition, every composite number can be resolved 
 into at least two factors, each greater than 1 and less than the 
 number. If any of these factors is composite, it may be resolved 
 into factors ; and if this process is continued far enough, the given 
 number will finally be resolved into prime factors. 
 
 To discover whether a composite number has more than one set 
 of prime factors, suppose N= abc •••, in which a,b,c, ••• are prime 
 numbers; and if possible, suppose N=ABC • ••, in which A,B,C,-" 
 are prime numbers. 
 
 Then, abc-- = ABC--. 
 
 Since A must divide the product abc •••, and since a, b, c, ••• 
 are prime numbers, A must be equal to some one of the factors 
 a, b, c, ••-. Similarly, each of the factors B, C, ••• must be equal 
 to some one of the factors a, b, c, •••; that is, each factor in 
 ABC '■■ corresponds to an equal factor in abc •••. Hence, 
 
 Principle 2. — A composite number may be resolved into one 
 and only one set of prime factors. 
 
 Corollary. — Every composite number has at least one factor 
 equal to or less than its square root. 
 
 Therefore, if a number has no divisor equal to or less than its 
 square root, the number is prime. 
 
 542. By § 497, IV, the number of combinations of n different 
 
 ,, . . 1 i. i.- • w(^ — l)(w — 2) ••• to r factors 
 things taken r at a time is ^ ^-\ — ~-^ 
 
 Since the number of combinations of n different things taken r 
 itt a time must be a whole number. 
 
 Principle 3. — The product of any r consecutive integers is divisi- 
 ble by jr. 
 
THEORY OF NUMBERS 477 
 
 n 1 Tr-\i w(w — l)(n — 2) '•• (71 — r + 1) 
 Corollaries. — 1. IJ N = -^ ^ f ^^ ' — ^, evei-y 
 
 \r 
 
 jn'inie factor of the numerator that is greater than r is a factor 
 
 of N. 
 
 2. If n is a positive integral exponent, tlie binomial coefficients 
 
 n n(n — 1) n(n — 1) (n — 2) . , 
 
 -, -!^ ^, — ^^ '-^ ^, •••, are integers. 
 
 1'1.2' 1.2.3'' ^ 
 
 l^XAMPLES 
 
 1. Show that ji^ — w is an even number when n is integral. 
 
 Proof 
 
 The integer n must be even or else odd. 
 
 Every even number has the form 2p, p being an integer, and every odd 
 number has the form 2 p + 1 or the equivalent form 2(p — 1)+ 1, or2p — 1. 
 
 If n is even, let n = 2 p. 
 
 Then, n^ — n = n(n — 1) = 2p(2p — 1), an even number. 
 
 If n is odd, let 7i = 2p + 1. 
 
 Then, n^ — n = n(n — 1) = (2p + l){2p), an even number. 
 
 Hence, whether n is even or odd, n^ — n is even. 
 
 Or, again, by Prin. 3 the product of any two consecutive integers is 
 divisible by [2. Hence, n^ — n, or n(n — 1), is even. 
 
 2. Show that every perfect cube belongs to one of the forms 
 
 7n,7n-\-l,7n — l. 
 
 Proof 
 
 Every integer must belong to one of the forms 7 p, 7 p ± I, 7 p ±2, 
 7p ±3 ; for 7p + 4 = 7(p +1) — 3, which has the same form as 7 p — S, 
 and 7p — 4: = 7(p — 1) +3, which has the same form as 7p + 3 ; and simi- 
 larly, 7 p ± 5 and 7p ±() may be reduced to the forms 7 p ±2, and 7 p ± 1, 
 
 respectively. 
 
 (7p)8 = 348 p8 = 7(49p3):= 7 „. 
 
 (7/)±l)3 = 343p3 J. U7p^ + 2lp± 1 
 
 = 7(49 p3 ± 21 p2 + 3 J,) J. 1 =, 7 ,j ± 1, 
 
 (7j9 ± 2)3 = 343 1)3 ± 294 p2 + 84i9 ± 8 
 
 = 7(49 j93 J. 42p^ + \2p± 1)± 1 = 7n± 1. 
 
 (7p ± 3)3 = 343 p3 ± 441 p2 + is9p ± 27 
 
 = 7(49p8±63p2 + 27p±4)T 1 = 7 n T 1. 
 
 Caution. — Not every number belonging to one of these forms is a perfect 
 cube. For example, 7(6) + 1, or 43, is not a perfect cube. 
 
478 ''W-;^va^^^ \^^F NUMBERS 
 
 3. Find what rau ,3 of x will make a^ — 3x + 5 a 
 
 perfect square. 
 
 Solution 
 
 Let x^-3x + b =(.x - 7n)2. 
 
 Solving, * = f*'-=^. 
 
 2 n?. — 3 
 
 which is rational for all rational values of m. 
 
 Thus, substituting •••, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, ... for m, 
 
 x=..., -1, -1,1,1,1,4, - 1, I, Jgi, .... 
 Any of these values substituted renders x^ — 3 a; + 6 a perfect square. 
 
 4. Find which of the numbers 431, 211, 253, 301, 623, 323 are 
 prime and factor the others. 
 
 5. Of the first ten numbers beginning with 1001, which are 
 composite ? 
 
 Prove the following : 
 
 6. The sum of any number of even numbers is even. 
 
 7. The sum or difference of two odd numbers is even. 
 
 8. The difference between the squares of two odd numbers is 
 divisible by 8. 
 
 9. The difference between a number and its cube is divisible 
 by 6. 
 
 10. The number w' — 5 n^ -f- 4 n is divisible by 120. 
 
 11. Every perfect square has one of the forms 3n,3n + l. 
 
 12. Every perfect square has one of the forms 4 n, 4 n + 1. 
 
 13. Every prime number except 2 and 3 belongs to one of the 
 forms 6 ?i + 1) 6 n — 1. 
 
 14. What rational values of x will make a^ + 5 a; + 2 a perfect 
 square ? 
 
 15. What values of x will rationalize -\/ax + & ? 
 
 Suggestion. — Put y/ax + b = m. 
 
 16. What is the least multiplier that will make 33957 a perfect 
 square ? a perfect cube ? 
 
 Suggestion. — Put the number into the form aPb^(f ••• and examine the 
 exponents of each factor. 
 
(1) 
 
 DETERMIN rp' 
 
 ^king constitueD*-" *^^'^ xaiau\&T l-.uo^* 
 
 DETERMINANTS 
 
 543. Solving the simultaneous independent equations 
 
 we have x = -^^ ^-i, y — -^-^ ^-i* 
 
 aib.2 — a>J)i a-Pi — ci-P\ 
 
 Comparing the values of x and y it is observed that : 
 
 1. They have the same denominator. 
 
 2. The numerator of the value of x may be formed from the 
 denominator by replacing the coefficients of x by the correspond- 
 ing known terms k^ and k^. 
 
 3. The numerator of the value of y may be formed from the 
 denominator by replacing the coefficients of y by the correspond- 
 ing known terms k^ and k^. 
 
 The common denominator Uih^ — api is called the determinant 
 of the system. 
 
 A convenient symbol for Oiftg — «2&i> suggested by the arrange- 
 ment in (1) of the coefficients of x and y in two columns and two 
 rows, is 
 
 called a determinant of the second order. ^ 
 
 ai&2 — ^2^1 is called the developed form, or the development, of 
 this determinant. 
 
 0162 ^'Hd — a^i are called its constituents. 
 
 Oj, a2, 61, &2 are called its elements. 
 
 Note. — Some authors employ the terms eZemen^ and constituent with the 
 meanings here.given to constituent and element, respectively. 
 
 479 
 
480 TTIEORY 6f numbers 
 
 544. To deveiop 5 • ^^.Wne'seL'^^^ make a^ — Sa- 
 
 By definition, I "* '^ \= a^b^ ■— a^b^- 
 
 \(h fh\ 
 
 The second member may be written h-2ai — bia^i or — bia2 + Oib^, etc. 
 
 1. The positive term, afii or fta^D is obtained by multiplying the 
 element aj in the Jirst column and first row by the element 62 ii^ 
 the next column and next row ; or by multiplying the element 63 in 
 the second column and second row by the element ay in the preced- 
 ing column and preceding row. 
 
 The selection of an element from any column or row before the 
 selection of an element from a preceding column or row consti- 
 tutes an inversion. 
 
 Then, the positive term formed in the first way presents no 
 inversions, but formed in the second way presents two inversions, 
 namely, the selection of an element from the second column 
 before that of an element from the preceding column, and the 
 selection of an element from the second row before that of an 
 element from the preceding row. 
 
 In either case the positive term of the development presents an even 
 number of inversions. 
 
 2. The negative term, — a2&i, or — ftia,? is obtained by multi- 
 plying the element a.^ in the first column and second row by the 
 element b^ in the second column and first row, and making the 
 product negative ; or by selecting the elements in the reverse 
 order and making the product negative. In the first way there is 
 an inversion of rows, in the second way, an inversion of columns. 
 
 In either case the negative term of the developinent presents an odd 
 number of inversions. 
 
 545. Any square array of n^ elements arranged in n columns 
 and n rows represents a determinant of the nth order. 
 
 In harmony with the principles of the preceding article a deter- 
 minant of any order is now defined as a square array of numbers 
 that, by common agreement, represents the algebraic sum of all 
 the products, or constituents, that can be formed by taking one 
 element, but not more than one, from each column and from each 
 
DETERMINANTS 
 
 481 
 
 row, making constituents that present an eveii number of inver- 
 sions positive and constituents that present an odd number of 
 inversions negative. 
 
 546. Development of any determinant. 
 
 I f\ ^1 Cl 
 
 Let a^ h-2 c.2 be a determinant of the third order. 
 
 By the defiiaition of a determinant, each constituent of this 
 determinant contains three elements as factors, one and only one 
 taken from each column and from each row. 
 
 Hence, the constituents involving ai are Uih/ig and — aib^Ci, the 
 latter being negative because it presents one inversion. There- 
 fore, the sum of the constituents involving a^ is 
 
 ai02Cs — ftiOgCo, or aj 
 
 '-^3 ^3 
 
 which may be obtained from the given determinant by canceling 
 
 or deleting the elements that cannot be associated with %, 
 
 Oi -&J — Ci 
 
 thus : (jf2 &2 Ca 
 
 . (tg O3 C3 
 
 The determinant of the next lower order 
 
 by which aj is 
 
 — ^2 
 
 multiplied is called the minor of the element a^. When the minor 
 is given the proper sign, in this case + , it is called the co-factor 
 of the element. 
 
 Similarly, the sum of the constituents involving a^ is 
 
 61 Ci 
 derived by deleting the elements that cannot be associated with a^, 
 
 thus : «2 ^ ^ 
 
 <p3 ^3 ^3 
 and giving o^ the sign — , because in each constituent ag is chosen 
 before an element of the preceding row. 
 
 In tM.. case, since 
 
 02 is D^^^'■^l^.'t■. 
 
 = a2X — 
 
 , the co-factor of 
 
 31 
 
482 
 
 DETERMINANTS 
 
 Similarly, the sum of the constituents involving a^ is 
 
 b.2 C2 
 
 Since each constituent of the given determinant must involve 
 either a^, a.^, or a^, we have found all the constituents. Hence, 
 
 «i 
 
 The same result is obtained by nsing any column or any row 
 of elements as the first column is used above. 
 
 For example, selecting the elements of the second column. 
 
 61 Ci 
 
 
 
 
 
 
 
 
 
 
 b2 
 
 Ci 
 
 
 &i 
 
 Cl 
 
 
 h 
 
 Ci 
 
 62 ~£s 
 
 = ai 
 
 
 - ttz 
 
 
 
 + 013 
 
 
 
 
 \0s 
 
 C3 
 
 
 &3 
 
 C3 
 
 
 h 
 
 C2 
 
 63 C3 
 
 
 
 
 
 
 
 
 
 Oj 
 
 6x 
 
 Cl 
 
 
 
 
 
 
 
 
 
 
 
 
 
 a2 
 
 C2 
 
 
 ttl 
 
 Cl 
 
 , «! 
 
 Cl 
 
 «2 
 
 62 
 
 C2 
 
 = -6, 
 
 
 
 + 62 
 
 
 
 -&3 
 
 
 
 
 
 
 ttg 
 
 C3 
 
 
 ttg 
 
 C3 
 
 ^2 
 
 C2 
 
 «3 
 
 &3 
 
 C3 
 
 
 
 
 
 
 
 
 
 = — ttoftiCg + agftjCa + aiftgCs — a3&2Ci — ai&3C2 + a A^d 
 which is the former result differently arranged. 
 
 The above discussion applies to a determinant of any order. 
 Hence, 
 
 The development of a determinant of any order is equal to the 
 algebraic sum of the products of the elements of any cohimn or 
 row and their respective co-factors. 
 
 547. The minors corresponding to the elements aj, Oa^, •••, 
 61, &2) •••) are denoted by A^, A^, •••, B^, B^, •••. 
 
 548. Number of constituents. 
 
 Since the co-factors of each of the n elements in any selected 
 column or row of a determinant of the ?ith order are determinants 
 of the (w — l)th order, a determinant of the nth order has n times 
 as many constituents as a determinant of the (n — l)th order; 
 this, in turn, has {n — 1) times as many constituents as a deter- 
 minant of the (n — 2)th order ; and so on, until a determinant of 
 the 2d order is reached, which has 2 constituents. Hence, 
 
 A determinant of the yith order has n(n — l)(n — 2) ••• 2, or [w, 
 constituents. 
 
DE TERMINA NTS 
 
 483 
 
 1. Develop the determinant 
 
 Examples 
 
 6 9 8 
 
 10 11 12 
 
 14 15 16 
 
 Solution. — Multiplying the elements of the first column by their 
 co-factors, and adding, the given determinant is reduced to 
 
 111191 IQ 9t \ IQ 9i\ 
 
 15 lel-^His lel+i^lii i2h-^-2^« + 280 = i6. 
 
 12 3 4 
 5 6 9 8 
 9 10 11 12 
 3 14 15 16 
 Solution. — Proceeding as in Ex. 1, the given determinant is reduced to 
 
 2. Develop the determinant 
 
 6 9 
 
 8 
 
 
 10 11 
 
 12 
 
 -5 
 
 14 15 
 
 16 
 
 
 2 3 
 
 4 
 
 
 10 11 
 
 12 
 
 + 9 
 
 14 15 
 
 16 
 
 
 2 
 
 3 
 
 4 
 
 
 6 
 
 9 
 
 8 
 
 -3 
 
 4 
 
 15 
 
 16 
 
 
 10 11 12 
 
 Since by Ex. 1 the first determinant is equal to 16, the given determinant 
 = 16-5. 2I!? !5|-5(-10)L! .!l-5.14l ^ 41 
 
 15 16 
 9 8 
 
 15 16 
 
 + ^'2li5 lel +^(-«)|i5 lel + ^-i^ 
 
 11 12 
 
 3 41 
 
 3.2 
 
 ,..41 134 
 
 .Sf—e^ —3-10 
 
 ^ ^ 11 12 ^^"98 
 
 11 12 
 
 = 16 + 40 - 600 + 560 + 432 + 648 - 1512 - 120 - 144 + 360 
 = - 320. 
 
 Develop the following determinants : 
 
 4 
 
 9 
 
 2 
 
 3 
 
 5 
 
 7 
 
 8 
 
 1 
 
 6 
 
 1 
 
 7 
 
 1 
 
 3 
 
 3 
 
 3 
 
 5 
 
 1 
 
 5 
 
 3 
 
 2 
 
 
 
 
 
 6 
 
 4 
 
 1 
 
 1 
 
 1 
 
 2 
 
 2 
 
 3 
 
 4 
 
 3 
 
 2 
 
 2 
 
 5. 
 
 3 2 1 
 
 4 8 3 
 
 5 2 1 
 
 3 4 2 5 
 
 3 12 
 
 12 1 
 
 2 2 7 
 
 1111 
 
 4 2 7 
 
 3 2 7 
 
 2 2 1 
 
 2 
 
 3 2 2 1 
 2 12 1 
 13 5 1 
 
 * For economy of space the sign of a negative element may be written 
 above the element. 
 
484 
 
 DETERMINANTS 
 
 ^ 
 
 10. Express ain? — 2 a — hmn -\-2hc-\-mx — ncx as a determi- 
 nant. 
 
 Solution 
 
 Since there are 6, or [3, constituents, it is likely that the required determi- 
 nant is of the third order, and that the terms — 2 a and mx have a factor 1 
 or — 1 unexpressed. 
 
 avP' — 2a — bmn + 2bc + mx — ncx 
 
 = a(n • H — 2 • 1)— 6(m • n — 2 • c)+ x(m • 1 — n • c) 
 
 (a2_62)_c2. 
 1- (x^ + 1). 
 m^ — (n^ — n). 
 
 Express as determinants ; 
 
 
 
 
 
 
 
 
 
 
 a 
 
 m 
 
 c 
 
 n 
 
 1 
 
 
 m 
 
 c 
 
 
 m 
 
 c 
 
 
 
 
 
 
 
 -b 
 
 
 
 + X 
 
 
 
 ^ 
 
 b 
 
 n 
 
 1 
 
 2 
 
 n 
 
 
 2 
 
 n 
 
 
 n 
 
 1 
 
 
 X 
 
 2 
 
 n 
 
 11. 
 
 12. 
 13. 
 20. 
 21. 
 22. 
 
 14. 
 15. 
 16. 
 
 a + X. 
 b' + l. 
 
 25 - 21. 
 
 42 + 33. 
 
 ab — cd. 
 
 ^liVi - Vs) + XiiVs - Vi) + x^iVx - Ih)- 
 a^ — abc — abc + 6^ -f c^x — abx. 
 abc — axy — acx -f- xyz -\- abx — bh. 
 
 17. 
 18. 
 19. 
 
 23. a 
 
 3 a 
 
 4 2 
 
 5 a 
 
 + c 
 
 REDUCTION OF DETERMINANTS 
 549. A determinant that is equal to zero is called a vanishing 
 determinant. 
 
 550. 1. How does 
 
 compare in form and value with 
 
 5 2 
 4 3 
 
 9 
 
 
 7 
 2 
 
 4 
 1 
 
 with 
 
 L 
 
 7 2 
 4 1 
 
 9 
 
 
 6 
 8 
 
 9 
 4 
 
 w 
 
 :th 
 
 6 8 
 9 4 
 
 2 8 1 
 
 
 2 3 7 
 
 
 Ol 
 
 6l 
 
 Ci 
 
 
 tti ttg ttj 
 
 3 5 6 
 
 with 
 
 8 5 3 
 
 9 
 
 tta 
 
 b. 
 
 C2 
 
 with 
 
 bi &2 6j 
 
 7 3 
 
 4 
 
 
 
 1 
 
 6 4 
 
 
 
 ttg 
 
 &3 
 
 Cs 
 
 
 
 Ci 
 
 Cg Cj 
 
 2. How is the value of a determinant affected by changing the 
 rows into columns and the columns into rows ? 
 
 Principle 1. — Tlie value of a determinant is not changed by 
 changing its columns into rows and its rows into columns, provided 
 that their order of succession is not changed. 
 
DETERMINANTS 
 
 485 
 
 The above principle may be established as follows : 
 
 Since the 1st, 2(i, •••, nth columns become the 1st, 2d, •••, wth rows, re- 
 spectively, and vice versa, the relative position of the elements is not changed. 
 
 Therefore, each element of any column or row has the same co-factor as 
 before the reduction. Hence, the value of the determinant is not changed. 
 
 Corollary. — Wluitever is true of the columns of a determinant 
 is true of its rows, and vice versa. 
 
 551. 
 
 1. What is the value of 
 3 
 ^ of 
 
 of 
 
 3 
 
 2 
 
 5 
 
 
 
 
 
 
 
 4 
 
 1 
 
 6 
 
 2. What is the value of a determinant if all the elements of 
 one column or row are zeros ? 
 
 Principle 2. — A determinant that has one or more columns or 
 rows of zeros is equal to zero. 
 
 For since each constituent must have for a factor an element of the 
 column or row whose elements are zeros, each constituent is equal to zero. 
 
 552. 1. 
 
 9 4 I 
 
 15 8 r 
 
 How does 
 with 
 
 12 
 
 compare in form and value with 
 3 
 
 with 
 
 2. What is the effect of multiplying or dividing all the elements 
 in a column or row by the same number ? 
 
 Prixciple 3. — Multiplying or dividing all the elements in a 
 column or row of a determinant by the same number multiplies or 
 divides the determinant by that number. (§ 85, § 104, 3.) 
 
 Corollary. — Changing the signs of all the elements in one 
 column or row changes the sign of the determinant. 
 
 553. 1. How are 
 
 and 
 
 formed from 
 
 How do they compare with the latter in value ? 
 
 2. Show that 
 
 3 
 
 7 
 10 
 
 3 
 
 7 
 10 
 
 3 
 
 7 
 10 
 
 Principle 4. — The interchange of any two columns or of any 
 two rows of a determinant changes the sign of the determinant. 
 
486 
 
 DETERMINANTS 
 
 The above principle may be established as follows : 
 
 Let Z) be a determinant of the nth order and D' a determinant formed by 
 interchanging any two columns of D. 
 
 It is to be proved tliat D' =— D. 
 
 By the definition of a determinant, § 545, the elements forming each con- 
 stituent may be selected from the columns in any order we please, taking one 
 but not more than one fi-om each column and row, provided each constituent 
 so formed is given the proper sign showing the even or odd number of inver- 
 sions of the established order of columns and rows. 
 
 Then, let the last two elements of each constituent be chosen from the two 
 columns to be interchanged in the order in which these columns stand, giv- 
 ing the result the proper sign. By this method, when the columns have been 
 interchanged, each constituent will have one more inversion than before, 
 namely, the inversion in the order of the last two columns. 
 
 Hence, the sign of each constituent will be changed by interchanging the 
 two columns, and by the Distributive Law this changes the sign of D ; that 
 is, D' =-D. 
 
 554. By changing places successively with each of the preced- 
 ing columns, any column may be made the leading column, pro- 
 vided, Priu. 4, that when the number of columns supplanted by the 
 advancing column is odd the sign of the determinant is changed. 
 
 Since the same is true of the advance of any row to the position 
 of leading row, any element may be brought to the position of 
 leading element by a proper number of advances of its column and 
 row, provided that the sign of the determinant is changed when 
 the sum of the number of columns and the number of rows preced- 
 ing the column and row in which the element stands is odd. 
 
 Therefore, since the co-factor of the leading element is always 
 positive, the sign of the co-factor of any element is -\- when the com- 
 bined number of cohtmns and rows preceding the column and row of 
 the element is even, and — when this number is odd. 
 
 12 3 
 Thus, in the determinant 4 5 
 7 8 
 negative ; of 5, positive ; of 6, negative ; of 7, positive ; etc. 
 
 555. The preceding principle suggests a device for developing 
 a determinant of the third order. 
 
 the co-factor of 4 is negative ; of 2, 
 
 In 
 
 ttj bi 
 a^ &2 
 
 draw diagonals, thus 
 
DETERMINANTS 
 
 487 
 
 The constituents ai^gCs ^-nd — agftoCi, whose elements lie on the 
 diagonals, are called the principal diagonal and the secondary diago- 
 nal, respectively. In a determinant of the third order the principal 
 diagonal is posiYive and the secondary diagonal is negative. 
 
 To find the other positive and negative constituents, by two 
 interchanges of columns, and again by two, we have 
 
 Oi 
 
 61 
 
 Ci 
 
 
 h 
 
 Ci 
 
 Oi 
 
 
 Ci 
 
 «! 
 
 61 
 
 a^ 
 
 h 
 
 C2 
 
 = 
 
 h 
 
 C2 
 
 ^2 
 
 = 
 
 C2 
 
 02 
 
 &2 
 
 ttg 
 
 h 
 
 C3 
 
 
 h 
 
 C3 
 
 ag 
 
 
 C3 
 
 ttg 
 
 63 
 
 The principal diagonals of these determinants are the three 
 positive constituents of the given determinant and the secondary 
 diagonals are the three negative constituents. 
 
 The three equal determinants and their diagonals are written i-n 
 the form + 
 
 in which the principal diagonals are positive and the secondary di- 
 agonals are negative. 
 
 Caution. — This device does not apply to determinants of a higher order 
 than the third. 
 
 556. 1. What is the value of 
 
 7 
 of 
 
 ? of 
 
 10 
 6 
 4 
 
 2. Form other determinants each with two columns or rows 
 alike or differing by a constant multiplier. What value has each,? 
 
 Principle 5. — If the corresponding elements in any tivo columns 
 or rows of a determinant are the same, or if the elements in one col- 
 umn or row are equimultiples of the corresponding elements in the 
 other, the determinant is equal to zero. 
 
 The above principle may be established as follows ; 
 
 1. Let D be a determinant having two identical columns or rows. 
 
 By Prin. 4, if these two cohimns or rows are interchanged the sign of the 
 determinant will be changed, giving —D. But since the two columns or 
 rows are identical, interchanging them does not change the determinant. 
 
 Hence, D = - D. But D = - D only wlien Z) = 0. Therefore, Z) = 0. 
 
488 
 
 DE TERM IN A NTS 
 
 2. Let the elements in one column or row be m times the corresponding 
 elements in another column or row, and, Prin. 3, let the determinant be 
 represented by mD. 
 
 Then, as in 1, mD = — niD, 
 
 which is true only when Z) = 0, for m is not equal to zero. 
 
 557. To what determinant of the second order is 
 
 ai 
 
 6i 
 
 Ci 
 
 0^2 
 
 &2 
 
 C2 
 
 as 
 
 bs 
 
 Cs 
 
 eqital if ag = and ag = ? if Z>i = and Cj = ? if &2 = f> and 
 Co — O? if all the elements but one in any column or row are 
 equal to zero ? 
 
 Principle 6. — If all the elements but one in any column or row 
 of a determinant are equal to zero, the determinant is equxil to a 
 single determinant of the next lower order, namely, the product of 
 the element and its co-factor. 
 
 For each of the co-factors corresponding to the other elements in that 
 column or row has the coefficient 0, and so becomes 0. 
 
 558. By Prin. 6, any determinant may be written as the minor 
 of the element 1 or — 1 of a determinant of the next higher order 
 equal to the given determinant, provided that the other elements 
 in the same column or row as 1 or — 1 are zeros. 
 
 Thus, 
 
 b 
 d ~ 
 
 1 * * 
 a 6 
 
 1 or 
 
 a b * 
 c d * 
 
 , or 
 
 * a b 
 i 
 
 c d 
 
 
 1 
 
 
 * c d 
 
 m which each asterisk stands for any finite number. 
 
 559. 1. IfZ> = 
 
 , showthati)=(5a-3c) + (56-3d). 
 3 d as determinants, each having 
 
 a -\-h 3 
 c + d 5 
 
 2. Write 5 a — 3 c and 5 h ■ 
 the same second column as 2). 
 
 3. Into what two determinants, then, may D be resolved ? 
 
 Principle 7. — If each element of any column or row of a deter- 
 minant is compound, the determinant may he written as the algebraic 
 sum of two or more determinants. (§ 85.) 
 
 560. It follows from Prin. 7 that if two or more determinants 
 differ only in the elements of one column or row, they may be 
 united into a single determinant. 
 
DE TERMINANTS 
 
 ^89 
 
 Thus, 
 
 5 2 3 
 
 
 -3 2 3 
 
 
 5-3 2 3 
 
 
 2 2 3 
 
 2 4 3 
 
 + 
 
 1 4 3 
 
 = 
 
 2 + 143 
 
 = 
 
 3 4 3 
 
 1 5 4 
 
 
 -15 4 
 
 
 1-1 5 4 
 
 
 5 4 
 
 561. 1. Separate 
 
 2. Separate 
 
 5-3 3 10 
 6-4 4 20 
 
 8-7 7 30 
 is the value of the second determinant ? 
 
 5-3 3 
 6-4 4 
 
 -7 7 
 
 10- 
 
 -12 
 
 20- 
 
 -16 
 
 30- 
 
 -28 
 
 into two determinants. What 
 
 into four determinants. What 
 
 5 
 
 3 10 
 
 6 
 
 4 20 
 
 8 
 
 7 30 
 
 is the value of each ? Then, what simpler form has 
 
 Principle 8. — If the elements of any column of a determinant 
 are increased or diminished by the corresponding elements or by 
 equimultiples of the corresp)onding elements of any other column, the 
 value of the determinant is not changed. 
 
 Tlie same is true of any ttvo rows. 
 
 The above principle may be established as follows : 
 
 ffli 6i ••• ki 
 02 b-z ••• k^ 
 
 Let 
 
 bn 
 
 kfi 
 
 be any determinant, 
 
 and let m be any positive or negative number. 
 
 1. Suppose that the elements of the second column are multiplied by m 
 and added to the corresponding elements of the first column. 
 Then, by Prin. 7, the resulting determinant is resolved thus : 
 
 Oi 4- mbi bi • 
 0(2 + fnbfi bz • 
 
 . ki 
 
 • kz 
 
 = 
 
 «! bi ••• ki 
 
 az bz •■' kz 
 
 an bn ••• *„ 
 
 D + = D. 
 
 + 
 
 mbi bi • 
 mbz bz • 
 
 mbn bn ■ 
 
 ■ ki 
 
 ■ kz 
 
 a„ + mbn bn • 
 
 .5, 
 
 •• kn 
 
 ' • ^n 
 
 Prin 
 
 2. Let the modified column be any column after the first, say the ?*th. 
 Then, by r interchanges of columns the modified column may be made 
 
 the leading column, and the determinant may be resolved as in 1, into B + O 
 or — D + 0, according as the number of columns preceding the rth is even 
 or odd. In either case by restoring the leading column to its original posi- 
 tion the result obtained will be Z), the given determinant. 
 
 3. A similar proof may be given for modifying any row. 
 
490 
 
 DETERMINANTS 
 
 1. Evaluate the determinant 
 
 Examples 
 
 2 3 4 1 
 
 4 2 12 
 112 3 
 
 5 3 10 
 
 Solution 
 
 2 
 
 3 
 
 4 1 
 
 
 4 
 
 1 
 
 2 
 
 1 
 
 1 2 
 
 2 8 
 
 = 
 
 5 
 
 
 
 3 10 
 
 
 1 2 10 
 
 6 3 8 
 
 112 3 
 
 5 3 10 
 
 
 
 
 
 
 
 # 
 
 1 2 10 
 
 
 1 2 IT) 
 
 
 _ 
 
 6 3 8 
 
 _ 
 
 9 52 
 
 
 
 9 62 
 
 
 5 3 10 
 
 
 7 40 
 
 
 7 40 
 
 = 4. 
 
 Explanation. — The aim is to reduce each determinant in turn to a de- 
 terminant of the next lower order (Prin. 6) by adding such multiples of the 
 elements of some column or row to the corresponding elements of one or 
 more other columns or rows (Prin. 8) that all of the elements but one in some 
 row or column of the resulting determinant shall be zeros. The column or 
 row, multiples of whose elements are added (or subtracted), may be called 
 an operating column or row and is marked with an asterisk. 
 
 Thus, selecting the third row for an operator, we subtract 3 times the 
 operator from the first row, obtaining T 2 10 ; and add 2 times the operator 
 to the second row, obtaining 6 3 8. The operator itself must be btought 
 down unchanged, in order that the parts added or subtracted may be vanish- 
 ing determinants. 
 
 Since all the elements except —1 in the second column of the resulting 
 determinant are zeros, this determinant (Prin. 6) is equal to —1 times its co- 
 factor, which is negative, because, § 554, the element — 1 is preceded by ele- 
 ments in an odd number of columns and rows. Hence, — 1 times this negative 
 co-factor gives a positive determinant of the third order. 
 
 Continuing this process, the result obtained is 4. 
 
 2. Show that 
 
 Prin. 8 and 5, 
 
 14 7 
 
 2 5 8 
 
 3 6 9 
 
 is a vanishing determinant. 
 Solution 
 
 1 4 7 
 
 
 1 3 6 
 
 2 5 8 
 
 = 
 
 2 3 6 
 
 3 6 9 
 
 
 3 3 6 
 
 3. Evaluate the determinant 
 
 3 
 5 
 
 9 11 
 5 3 
 
 = 0. 
 
 4 
 
 7 
 
 2 3 
 
 5 
 
 9 
 
 13 
 
 2 
 6 
 
DETERMINANTS 
 
 491 
 
 Solution 
 
 12 3 4 5 
 
 2 4 5 7 9 
 
 3 5 9 11 13 
 2 4 5 3 2 
 12 3 5 6 
 
 
 
 
 
 
 1 
 
 
 
 1 
 
 
 
 1 
 
 
 1 
 
 
 
 1 
 
 2 
 
 = 
 
 
 
 1 
 
 5 
 
 8 
 
 
 
 
 
 
 1 
 
 1 
 
 
 111 
 
 
 1 1 1 
 
 10 12 
 
 _ 
 
 1 5 8 
 
 15 8 
 
 
 1 1 
 
 11 
 
 
 * 
 
 1 
 
 
 4 7 
 
 1 4 7 
 
 = — 
 
 1 1 
 
 1 1 
 
 
 
 = -3. 
 
 Evaluate the following ; 
 
 5. 
 
 8 
 
 4 
 
 6 
 
 2 
 
 2 
 
 4 
 
 2 
 
 3 
 
 4 
 
 4 2 12 
 
 2 3 2 5 
 
 3 2 12 
 
 5 6 4 9 
 
 5 2 7 5 
 
 6 3 14 
 
 4 2 13 
 6 3 2 5 
 
 4 12 1 
 
 5 2 3 1 
 
 2 112 
 
 3 2 4 6 
 
 10. 
 
 2 4 4 6 1 
 
 2 5*312 
 
 3 112 1 
 2 1112 
 5 2 2 3 1 
 
 2 12 3 3 
 12 2 2 4 
 
 3 2 13 2 
 
 2 3 4 2 
 
 3 3 2 3 
 
 a 1 1 1 
 1 a 1 1 
 1 1 a 1 
 Ilia 
 
 562. To factor a determinant. 
 
 1. Factor Z) = 
 
 Examples 
 
 X b b 
 
 X a c 
 
 -y a c 
 
 Solution. — If a — c, the second and third columns are identical and, 
 Prin. 5, the determinant vanishes. Hence, by the Factor Theorem, § 136, 
 a — c is a factor of D. 
 
492 
 
 DETERMINANTS 
 
 Again, \l x= — y, the second and third rows are identical and P ^0. 
 Hence, x + y is a factor of D. 
 
 Since every constituent of D is of the third degree and (a — c) (x + y) is 
 of the second degree, D must have another factor of tlie first degree. Substi- 
 
 a 
 
 tuting for 6, D is equal to x times the co-factor 
 
 , VFhich is equal to 0; 
 
 that is, Z> = 0. Hence, the other factor of Z) is 6 — 0, or 6 ; or it may be —h. 
 It remains to find wliether D — b {a — c) {x + y) or — 6 (a — c) (x + 2/). 
 The secondary diagonal of Z) is + aby and this is the only constituent of 
 D involving a, 6, and y. Since the sign of aby is + in ?> (a — c) (v + y) 
 but — in — 6 (« — c) (x + 2/), Z) = 6 (a — c) (x + y). 
 
 Factor the following determinants by inspection : 
 
 X 
 
 1 
 
 h 
 
 y 
 
 1 
 
 a 
 
 X 
 
 1 
 
 a 
 
 3. 
 
 a^ a 1 
 
 b^ b 1 
 c2 c 1 
 
 2 2 5 
 
 2x5 
 X S 5 
 
 563. Solution of simultaneous simple equations. 
 
 It has been shown that in a system of two simultaneous simple 
 equations of the form ax -{-by = c, either unknown number is equal 
 to a fraction whose denominator is the determinant of the system and 
 whose numerator is the determinant of the system with the known 
 terms substituted for the corresponding coefficients of that unknown 
 number. 
 
 By trial, the principle is found to hold for the solution of three 
 simultaneous simple equations. 
 
 j- axx + byy + Ci^ = Ai, 
 
 Thus, given j a^x + h^y + C'lZ = ki^ 
 
 \ a?,x + bzy + csz = ^3. 
 
 Solving by the ordinary process of elimination, then rearranging and 
 grouping terms, 
 
 ^__ k-i (&2C3 - &3C2) - ki (ftiC3 - fesPi) + ^'3 JbxC-i - &2C1) 
 «i (62C3 — 63C2) - ai {b\Gz - bsci) + as (61C2 — boCi) 
 
 ki 
 
 bi 
 
 Cl 
 
 ki 
 
 &2 
 
 C2 
 
 ks 
 
 63 
 
 Cs 
 
 «! 
 
 61 
 
 Cl 
 
 az 
 
 &2 
 
 C2 
 
 as 
 
 63 
 
 C3 
 
 Similarly for the values of y and ». 
 
DETERMINANTS 
 
 493 
 
 The principle will now be proved to be general ; 
 
 Let 
 
 aix + biy + ciz + ••• = Ai, 
 a2X + b-iV + c-iZ + ■■■ =k2, 
 
 (1) 
 
 i a„x + ft»y + c„2 H = An 
 
 be a system of n simple equations. Let D represent the determinant of the 
 system, Di the determinant of the system with the known terms fci, k^^, ••• 
 substituted for the corresponding coefficients of x, and ^i, A^^ ••• the co-fac- 
 tors of «!, 02, •••• 
 
 Then, 
 
 D 
 
 ax 
 
 hi 
 
 Ci ••• 
 
 
 ki 
 
 bi 
 
 Ci .. 
 
 <H 
 
 h» 
 
 C2 ••• 
 
 , and Dx = 
 
 k.2 
 
 62 
 
 C2 •• 
 
 a„ 
 
 bn 
 
 C„ ••• 
 
 Kn 
 
 b„ 
 
 c„ •• 
 
 (2) 
 
 Multiplying the first equation of the system by Ai, the second by A2, etc., 
 and adding the resulting equations, 
 
 (aiAi + aoAz + — f- a„^„)x 
 
 = kiAi + koA^ + ••• + k„A„ 
 
 (3) 
 
 + ibiAi + 62^2 + 1- b„An)y 
 
 + ■ . 
 
 Since the coefficient of x in (3) is the sum of the products of the elements 
 in one column of D and their co-factors, the coefficient of x is equal to D, 
 and the second member of (3) is equal to Dx- The coefficient of y in (3) 
 differs from that of x only in having the elements of the second column of D 
 repeated in the first column, bi, bz, •■• replacing ai, ao, •••, thus : 
 
 61 61 Ci 
 
 62 &2 C2 
 
 bn b„ Cn ■ 
 
 By Prin. 5, this determinant is equal to zero, and in like manner the coeffi- 
 cients of the other unknown numbers vanish. Hence, (3) becomes 
 
 Dx = Dx 
 
 whence, x = — 2- 
 D 
 
 Similarly, Dy = Dy-, whence, y = 
 
 So for each unknown number. 
 
 Examples 
 
 Solve the following by determinants : 
 
 2x-t5y = 9, 
 3x + 2y = S. 
 
 Sx + 2y = 12, 
 4:X + 3y = 17. 
 
 D 
 
 (4) 
 
 1. 
 
 2 a; + 7 ?/ = 30, 
 a; + 4 ^ = 17. 
 
 \bx-\- y = 12, 
 
 [2x-^Sy = 10, 
 
494 
 
 DE TERMINA NTS 
 
 8. 
 
 9. 
 
 10. 
 
 4a;-3?/ = 8, 
 x + 4y = 21. 
 
 3x-2y=-2, 
 2x-3y=-5. 
 
 ( ax + by = c, 
 [ mx + ny = d. 
 
 ax — by=r, 
 cx-\- dy = s. 
 
 2x + 5y + 2z = 27, 
 8x-\-6y + 3z = A6, 
 3x + 7y + 5z = 47. 
 
 9x + 2y-\-z = 25, 
 5x-\-y-\-z = 14:, 
 
 7x + Sy + 2z = 25. 
 
 ( (a + h)x —(a — b)y = 4: ab, 
 y(a-b)x+{a + b)y = 2a?~2b\ 
 
 {3x + 2y + 3z = 17, 
 
 12. I 2a; + ?/ + 2 2 = 10, 
 [5x-\-5y-{-z = 29l. 
 
 ' 2 a; + 3 2/ - 4 2 = 18, 
 
 13. • x + y-\-z = 12, 
 
 .5x — y — z = 12. 
 
 (x + 2y + z = 0, 
 
 14. ^ 2x + y + z = 2a — b, 
 x—y — 2z = 3b. 
 
 x-{-y = 2 a, 
 lb. \ y + z = 3a — bf 
 [z + x = 3a. 
 
 16. 
 
 17. 
 
 u — x + 2y — 3z=—5, 
 3u~x-[-y — 2z = 2, 
 2u + x-{-y — z = 9, 
 
 I —5u + 2x — 7y + z= — 12. 
 
 '2u + 3v — 4:X + y = 0, 
 
 u — v-\-x — y= —2, 
 
 7 ti + 2v — 3x + y = %, 
 I 5 t< + 8 i; - 10 a; + 3 ?/ = 3. 
 
 564. An equation in which every term is of the first degree in 
 some unknown number is called a Homogeneous Linear Equation. 
 
 ax = by., or ax — by = 0, is a homogeneous linear equation. 
 
 565. By § 563 the denominator of the value of each unknown 
 number in 
 
 [ a^x -{- b^y + c-^z -\ = 0, 
 
 agaj + da^ + C2Z+ ••• =0, (1) 
 
 . a„a; + 6„2/ + c„2 + 
 
 
 
I 
 
 DETERMINANTS 
 
 495 
 
 is the determinant of the system, and the numerator is the same 
 determinant with substituted for each cqelficient of the unknown 
 number. Therefore, the numerator in each case will have one 
 column composed entirely of zeros, and, Prin. 2, will be equal to 0. 
 
 , 
 
 .'. 0; = —, y = — , z = — , etc. 
 D ^ D D 
 
 Hence, each unknown number is equal to zero, except when 
 i> = 0, in which case each is indeterminate and the system is 
 indeterminate. 
 
 The case in which i) = is the case in which the equations in 
 (1) are not independent. For if it is possible to form any equa- 
 tion in (1) by combining multiples of two or more of the other 
 equations by addition or subtraction, it is possible to make two 
 rows of D identical by the same process. 
 
 Hence, if n homogeneous linear equations involving n unknown 
 numbers are independent, the unknown numbers are separately equal 
 to zero. 
 
 566. A system of w — 1 independent linear equations involving 
 n unknown numbers is indeterminate (§ 214, proof) ; but if the 
 equations are homogeneous, the ratio of any two unknown num- 
 bers may be found. 
 
 Thus, let a\X -\- h\y -f Ciz = 0, 
 
 and a2X + h^y -f C2« = 
 
 be given, to find the ratios of x to y, x to 0, and y to z. 
 
 X . ^ y 
 
 From (1), 
 From (2), 
 
 Solving, 
 also, from (5), 
 
 z z 
 
 02-4-62^ = - 
 
 z z 
 
 C2- 
 
 -Cl 
 
 61 
 
 
 a\ 
 
 -Cl 
 
 - C2 
 
 K 
 
 and ^ = 
 
 z 
 
 ai 
 
 - C2 
 
 «! 
 
 61 
 
 «i 
 
 fti 
 
 0,1 
 
 62 
 
 
 a2 
 
 62 
 
 
 -Cl fti 
 
 y 
 
 — C2 62 
 
 z 
 
 Ol — Cl 
 
 
 ai — C2 
 
 6i 
 
 Cl 
 
 62 
 
 C2 
 
 Cl 
 
 ax 
 
 C2 
 
 a% 
 
 (1) 
 
 (2) 
 
 (3) 
 (4) 
 
 (6) 
 
CONYEKGENCY OF SERIES 
 
 567. By division —i— = 1 + cc + 0^2 _|_ ^ _l ^ (i^ 
 
 an infinite series the sum of whose first n terms is 
 
 l+x + x^ + a^-\ H ^•" = ^—^^ 
 
 1 —X 
 
 Therefore, the diiference between the fraction from which the 
 series arose and the sum of the first n terms of the series is 
 
 1 1 — a;" x" 
 
 or 
 
 1—x 1 — x' 1 — x 
 
 As n increases without limit, this difference decreases numeri- 
 cally without limit if x is numerically less than 1, but increases 
 numerically without limit if a; = 1 or is numerically gi-eater than 1. 
 
 Hence, if aj'is numerically less than l,sthe sum of the first n 
 terms of the series approaches the value of the fraction as a limit, 
 but if ic = 1 or is numerically greater than 1, the sum of the first 
 n terms diverges farther and farther from the value of the fraction 
 as n increases, and approaches no fixed value as a limit. 
 
 li x= —\, the series becomes 
 
 1-1 + 1-1 + -, 
 
 the sum of whose first n terms oscillates between 1 and accord- 
 ing as n is odd or even, while the value of the fraction is ^. 
 It is evident from the above discussion that the series 
 
 1+X + X^ + 'J^+--- 
 
 stands for no definite fixed number unless x is numerically less 
 than 1, in which case the sum of the first n terms, as n is indefi- 
 nitely increased, converges toward the fixed number 
 
 J- — X 
 496 
 
CONVERGENCY OF SERIES 497 
 
 568. When the sum of the first n terms of an infinite series 
 approaches a constant number as a limit, as n is indefinitely 
 increased, the limit is called the sum of the series, and the series 
 is called a Convergent Series. 
 
 As n increases without limit, the sum of the first n terms of the series 
 l + | + ^ + ^ + --. approaches 2 as a limit. Hence, 2 is the sum of the series, 
 and the series is convergent. 
 
 The series 1 + a; + x^ + x^ + ••• is convergent for all values of x numeri- 
 cally less than 1. 
 
 569. A series that is convergent without regard to the signs of 
 its terms is called an Absolutely Convergent Series. 
 
 The series \-\ + \-\j^ ..., i + i + |+i + ..., _i_i_i_i 
 
 are absolutely convergent ; for even when the signs of all the terms are alike, 
 the sum of the first n terms remains finite however great n becomes. But 
 the series '^ — \ + \ — \ + •"■, though convergent, is not absolutely convergent, 
 since l + ^ + ^ + l^ + .-is not convergent, as will be shown later. 
 
 570. When the sum of the first n terms of an infinite series 
 can be made numerically greater than any finite number by tak- 
 ing n sufiiciently great, the series is called a Divergent Series. 
 
 The series 1 + a; + a:^ 4. x^ -f ... is divergent if x = 1 or any number numeri- 
 cally greater than 1, as 2 or —2 or 3, etc. 
 
 571. When the sum of the first n terms of an infinite series 
 oscillates between certain fixed values, as n is indefinitely in- 
 creased, the series is called an Oscillating Series. 
 
 The series \ + x -\- x- -{- x^ + •■• is oscillating if x = — 1, for the sum of the 
 first n terms oscillates between 1 and according as n is odd or even. 
 
 The series l-l-2-l + l-l-2-f-l-l-2-l-|-l-|-2-f-.. is oscillat- 
 ing, for the sum of the first n terms, as n takes the values 1, 2, 3, •••, is one 
 of six fixed values that recur in the order 1, 0, —2, —3, —2, 0. 
 
 572. Any infinite series may be represented by 
 
 Wi + W2 + «3 H \-u^-\ , 
 
 in which «i denotes the first term, Wg the second term, etc., and w„ 
 the nt\v term, or any term. Since n may be given the successive 
 values 1, 2, 3, •••, a series may be represented by its nth term. 
 
 ADV. ALG. 32 
 
498 CONVERGENCY OF SERIES 
 
 Theseries ? + ^ + ^ + ^ +...+ 2?-±I-f ... is represented by ^^^^: the 
 1 2 3 4 71 n 
 
 series 1-1 + 1- 1+ ... + (_i)n-ii + ... is represented by (-l)»-il. 
 1234 n n 
 
 The sum of the first n terms of a series is represented by S„. 
 If the series is convergent, its sum is represented by S, which is 
 defined by the relation 
 
 lim. (.S„)n=x = /S. 
 
 If the series is convergent, the series beginning with the 
 (w + l)th term is a convergent series and has a sum, repre- 
 sented by Rn, which is called the Residue and defined by the 
 
 relation 
 
 S==S^ + R,, or R, = S-S„. 
 
 573. From the above discussion and definitions it is seen that 
 the expression 
 
 Ui + u^ + Us-l h«„H 
 
 represents a definite number S only when the series is convergent. 
 Since a convergent series has a definite sum, it can be used in 
 demonstrations, but divergent and oscillating series can be so 
 used only under special restrictions. Hence, it is important to 
 discover principles by which the convergency of infinite series 
 may be tested. 
 
 ELEMENTARY PRINCIPLES 
 
 574. The ratio of the (n -\- l)th term of a series to the nth is 
 called its Ratio of Convergency. 
 
 In a geometrical series the ratio of convergency is constant, but 
 in general the ratio of convergency of a series is variable, increas- 
 ing or decreasing with the number of terms. 
 
 In 1 +-4--H 1 1 1 1- •••, the ratio of convergency is the con- 
 
 2 4 8 2"-i 2" 
 
 stant ^. In 1 + 3 + 9 + 27 + — h 3»-i + 3» + •••, the ratio of convergency is 
 the constant 3. 
 
 In 1 + 3 + 5 + 7H f-(2n-l) + (2n + l)H , the ratio of conver- 
 gency is the variable ratio ^ "^ , which decreases as n increases. 
 2 w — 1 
 
 575. The convergence or divergence of a series may often be 
 discovered by examining its ratio of convergency, or by com- 
 
CONVERGENCY OF SERIES 499 
 
 paring the series with an auxiliary series whose nature as con- 
 vergent or divergent is already known. 
 
 576. Principle 1. — The nature of a series as convergent, 
 divergent, or oscillating, is not changed by prefixing or removing 
 a finite number of terms. 
 
 For the sum of a finite number of terms is a finite or fixed number. 
 Hence, if the sum of the series is finite, or infinite, or oscillating, when this 
 sum has been increased or diminished by a fixed number, it will still be 
 finite, or infinite, or oscillating, as the case may be. 
 
 577. Principle 2. — A series of positive and negative numbers 
 is convergent, if the series formed by making the terms all positive or 
 all negative is convergent. 
 
 For the effect of changing the signs of some of the terms of a convergent 
 series of positive numbers or of negative numbers is to decrease the numeri- 
 cal value of the sum of the series. 
 
 578. Principle 3. — If each term of a given series, or each term 
 after any finite number of terms, is numerically less than the corre- 
 sponding term of an absolutely convergent series, the given series is 
 absolutely convergent ; if each term is numerically greater than the 
 corresponding term of a divergent series of numbers having the same 
 sign, the given series is divergent. 
 
 Thus, since l-}-i + J + ^+-" is an absolutely convergent series, 1 -f- 
 i + i + i+"' is a convergent series. But though 1— ^ + ^ — ;! + ••• is a 
 convergent series, it does not follow that l + |-f-f + f + '-' is a conver- 
 gent series, because the convergent series used as a standard of comparison 
 is not absolutely convergent. 
 
 Again, since l-f^ + l + f-t-^-t- "-is a divergent series, 1 + \+{\ + \) 
 + (i + J + 4 + i) + (i + ••• + A) + ••• is a divergent series ; that is, 1 + i 
 -f^-|-J + ^-|-^ + |^ + | + >" is a divergent series. 
 
 579. Principle 4. — If, for all values of n greater than a certain 
 finite value, the ratio of convergency of a given series of positive 
 numbers is less than the ratio of convergency of a convergent series 
 of positive numbers, the given senes is convergent; but if the ratio 
 of convergency is greater than that of a divergent series of positive 
 numbers, the given series is divergent. 
 
 Thus, in 1-1-1 H 1 \ f- ••• the ratio of convergency is a 
 
 variable having the successive values 1, i, ^, i, •••• It is observed that the 
 
500 CONVERGENCY OF SERIES 
 
 ratio of convergency of the given series is smaller, after the third term, than 
 that of the series I + ^ + \+ ^ + •■■, which is known to be convergent. 
 Hence, the given series is more rapidly convergent than the latter. 
 
 The series 1 + 2 • 2 + 3 • 2"^ + 4 • 2" + ••• is more rapidly divergent than 
 the series 1 + 2 + 2^ + 2^ + .... 
 
 580. Principle 5. — If the ratio of the corresponding terms of 
 two series is finite for all values of n, and if the terms of both series 
 are jwsitive, then both series are convergent or both are divergent. 
 
 The above principle may be established as follows : 
 
 Let Ml, M2, Ms, •••, M„, ■••, and vi, Vo, Vs, •••, v„, ••• be the corresponding 
 terms of two series beginning with the first or after any finite number of 
 terms. Let these terms be such positive numbers that the ratio m„ : v„ is 
 finite for all values of n. 
 
 Let — = n, — = r2, — = /"s, •••, »m being the greatest and r„ the least of 
 
 Vi V2 Vs 
 
 all the ratios. 
 
 Then, Mi + u^ + us + ••• = nui + r^vo + rzva + •••, 
 
 which is < Tmivi + v^ + vs + •••), but > Vp («i + 172 + ^3 + •••)• 
 
 .-. — lies between »•„ and r„, and hence is equal to a finite 
 
 Vi + V-2 + VS + - m PI 
 
 number, as C. 
 
 .-. Ml + M2 + t(3 + ••• =C{Vi + V2 + V3 + •••)• 
 
 Hence, if the sum of the terms of one series is finite, the sum of the terms 
 of the other is finite ; and if the sum of the terms of one series can become 
 infinite, the sum of the terms of the other can become infinite. 
 
 Hence, both series are convergent or both are divergent. 
 
 TESTS OF CONVERGENCY 
 
 581. First Test. — An infinite geometrical series is absolutely 
 convergent if its ratio is numerically less than 1, oscillating if its 
 ratio IS —1, divergent if its ratio is 1 or numerically greater than 1. 
 
 The above test may be established as follows : 
 
 By § 370, the sum of the geometrical series a, ar, ar^, ••• ar"-^ (1) 
 
 is 5„ = <1^U!0. (2) 
 
 1 — r 
 
 1. Let r be numerically less than 1. 
 
 Then, since as n increases indefinitely, r» = 0, (2) becomes 
 
 \im.Sn=S = -^^, 
 1 — r 
 
 a finite value, since r is a constant finite ratio and either positive or negative. 
 
 Hence, in this case the series is absolutely convergent. 
 
CONVERGENCY OF SERIES 501 
 
 2. Let r = - 1. 
 
 Then, (1) becomes a, — c, a, — a, •••, 
 
 the sum of the first n terms of which oscillates between a and 0. 
 Hence, in this case the series is oscillating. 
 
 3. Let r be equal to or numerically greater than 1. 
 
 Then, by taking n great enough, the second member of (2) may be made 
 numerically greater than any assignable number. 
 Hence, in this case the series is divergent. 
 
 582. Second Test. — A series is absolutely convergent if, as n 
 increases witJiout limit, the limit of the ratio of convergence is numeri- 
 cally less than 1, but divergent if the limit of the ratio of convergency 
 is greater than 1. 
 
 When the ratio of convergency approaches 1 as a limit, the test 
 fails and other tests must be applied. 
 
 The above test may be established as follows : 
 Let lim.'^^^ = Z. 
 
 1. When I is numerically less than 1. 
 
 Let r represent any fixed number numerically less than 1 but numerically 
 greater than I. Also let m be any finite positive integer. 
 Then, numerical values only being regarded, 
 
 Mm+l . „ Mm+2 . „ Mm+3 ^ „ Mm+4 . „ 
 Um Um+x Um+2 «m+3 
 
 whence M„+i < U^r, Mm+2 < "m+ir, Mm+a < Mm+2»', Mm+4 < ?<m+3r, •-, 
 
 or Um+1 < UmV, M„+2 < UmT^, U^+S < Mm»^, Mm+4 < ?<m»^, ••• ; 
 
 that is, on and after a finite number of terms each term of the given series is 
 numerically less than the corresponding term of an infinite geometrical series 
 whose ratio r is numerically less than 1. 
 
 Hence, by the first test and Prin. 3, the series is absolutely convergent. 
 
 2. When I is numerically greater than 1. 
 Then, as in 1 it can be shown that 
 
 nm+l>Umr, «m+2>Mm»^, «m+3 > "m?^, Mm+4 > Mm^*, ••• J 
 
 that is, on and after a finite number of terms each term of the given series is 
 numerically greater than the corresponding term of an infinite geometrical 
 series whose ratio r is numerically greater than L 
 
 Hence, by the first test and Prin. 3, the series is divergent. 
 
502 CONVERGENCY OF SERIES 
 
 The failure of the second test is illustrated as follows 
 
 In each of the series l + i + i + i+ h-H 1 and 1 _ ^ + ^ _ 
 
 11 
 
 i + ••• + (— 1)"~^- 4- (— 1)" h •••, the ratio of convergency is numeri- 
 
 n n + 1 
 
 cally equal to — - — Though this ratio is numerically less than 1 for finite 
 
 n + 1 
 values of n, it approaches 1 as a limit as ji is increased without limit. Since 
 the limit of the ratio of convergency is neither less nor greater than 1 
 numerically, the second test is not applicable. In fact the first series is 
 divergent, as illustrated under Prin. 3, and the second is convergent, by 
 the third test, which follows. 
 
 583. Third Test. — If the terms of a series are alternately 
 positive and negative, and each term is numerically greater than 
 the following term, the series is convergent, provided the limit of 
 the nth term is zero. 
 
 The above test may be established as follows : 
 
 Let U1-U2 + US- Ui-\ + (- 1)»-1m„ + M2» + U2n+1 
 
 be the series, and let mi > «2 > ms > • ■ • > "n > 0, so that un=0. 
 The sum of an even number of terms may be written 
 
 Ssn = («1 - «2) + ("3 - M4) + ••• + (M2»-1 - M2n), (1) 
 
 or San = Ui- («2 - «3) - (M4 - Ms) (M2n-2 - «2n-l) - M2». (2) 
 
 Since Mi>t*2>«3> •••, each group in (1) and (2) is a positive number. 
 Therefore, by (2), <S'2„ < mi ; and, by (1), the sums *S'2, St, Ss, •-, Szn 
 are positive and increase with n. 
 
 Hence, lim. (S2n)n^^ lies between and ui. 
 
 The sum of an odd number of terms can be written 
 S2n+i = S2n + W2n+i, a positive number, 
 
 or S2n+1 = «1 - (M2 - Us) - (M4 - Ms) (M2n - M2n+l), 
 
 which is less than ui. 
 
 Hence, 11m. (/S^2n+i)n==x> also lies between and «i. 
 
 Since the limit of the nth term is zero, 
 
 lim. (M2n+l)n=^ = ; 
 
 that is, lim. (52„+i - S2n)n^ = 0. 
 
 Hence, S2n and S2n+i approach a common limit and the series is convergent. 
 
 Note. — If the rath term approaches a constant, not zero, as a limit, say 
 lim. u„ = a, Sin+i and S2n approach limits that differ by o, and the series is 
 oscillating. Thus, |-| + |_f +...-|-(_i)«-i^+l + ... jg an oscillating series, 
 because lim. (*S'2»+i - /S2»)„=«, = lim. («2»f i)»i» = 1. 
 
CONVERGENCY OF SERIES 503 
 
 584. Effect of grouping or rearranging the terms of a convergent 
 series. 
 
 The sum S of an absolutely convergent series is not changed by 
 grouping or rearranging its terms in any manner, for by taking a 
 sufficient number of groups of the derived series all the terms of 
 Sn may be included, and the sum of m groups approaches S as S^ 
 approaches S. 
 
 Thus, §371, 1-^+1 _j + ^_^ij + ... = |j 
 
 also, § 371, (1 _ ^) + (i _ ^) + (3I5 _ ^^) + ... = I ; 
 
 also (1 + J _ I) + (^^ + 5»^ _ I) + (5^, + ^^._ ji^) + ... 
 
 = (5_n /A_1N /5_1\ .. 
 
 4 V 42 44^ ) 2V 22 2*^ } 
 
 §371, =Kil)-KI) = l- 
 
 But when a convergent series is not absolutely convergent, that 
 is, when its convergence depends partly upon the signs of its 
 terms, grouping or rearranging its terms may change the sum of 
 the series or even cause the series to become divergent. 
 
 Such series are said to be conditionally convergent 
 
 Thus, i_i + ^_i + ... + (_l)«-il4.... 
 
 is a conditionally convergent series. Denote the sum by S. 
 
 Then , 6^ = 1 - i + i - i + i - i + ^ - J + i - T^s + ^ij - ,35 + . . . ; 
 but (1 + I _ i) + (1 + I _ ^) + (J + ^ _ I) + ... 
 
 = S-t\S = \S. That is, by this method of grouping the sum 
 
 has been increased one half. 
 
 The sum of this series may be changed at will by suitably grouping and 
 reaiTanging the terms. 
 
 By grouping similarly the terms of the conditionally convergent series 
 1 — y/\ + V| — y/\ + •••, the resulting series can be shown to be divergent. 
 
504 CONVERGENCY OF SERIES 
 
 685. The standard auxiliary series for determining the con- 
 vergency of series is the geometrical series l-\-ar-\- a>•^-|-ar^^ . 
 
 The next in importance is the series ■::- + w\ 1 1- •••• 
 
 Ip 2^ 3* 4^ 
 
 The series :i- + i^+:^ + -j- + "- is absolutely convergent when p 
 
 is positive and greater than 1 ; otherwise it is divergent. 
 
 The proof is as follows: 
 
 1. Let p > 1. Then, the first term is equal to 1 ; the sum of the next two 
 
 terms < — | — , or — ; of the next four terms < — | i h — , or — ; 
 
 2p 2p 2p 4^ 4p 4p 4p 4p 
 
 of the next eight terms < — ; etc. Hence, after the first term, the terms of 
 
 the series, grouped as shown, are less than the corresponding terms of the 
 geometrical series 
 
 2p iP 8p 
 
 which is absolutely convergent by the first test. 
 
 Hence, by Prin. 3, the given series is absolutely convergent when jp > 1. 
 
 2. Let p — 1. Then, the series may be written 
 
 Since each of the grouped terms is greater than the corresponding term of 
 the divergent series 1 + | -f- (J + ^) + (i + | + | + |) + ..., by Prin. 3 the 
 given series is divergent when p = 1. 
 
 3. Let p<l. 
 
 It is evident that the series is more rapidly divergent than when p = \, 
 especially if p is negative. 
 
 Examples 
 1. Find the first four terms and represent the sum of the 
 
 series whose nth term is (— 1)""^— • Test the convergency of the 
 series. '— 
 
 Solution 
 
 Substituting 1, 2, 3, 4, ..., and m + 1 successively for n in (-l)»-i— , 
 
 Ira 
 
 [4 [4' ""* " ' \ n4-\ ' ' |ra + l 
 
 12 [3 [4 ^ ^ [n ^ [n±l 
 
CONVERGENCY OF SERIES 505 
 
 Ratio of convergency = ^^^^ = ^--. — ^ x 
 
 \n _i 
 
 M„ I w + 1 (- l)''-i tt + 1 n + l 
 
 which is numerically less than 1 and approaches as a limit as n increases 
 without limit. 
 
 Hence, 2d test, the series is absolutely convergent. 
 
 2. Apply the third test to the series of example 1. 
 
 3. Test the convergency of the series of example 1 by com- 
 parison with a geometrical series. 
 
 Solution 
 The given series may be written 
 
 i-l + J L_ + \ I + .... 
 
 2 23 2.3.4 2.3.4.5 2.3.4.5.6 
 The infinite geometrical series whose ratio is — J may be written 
 
 l_l + _i 1_ + --J 1 + .... 
 
 2 2.2 2.2.2 2.2.2.2 2.2-2.2.2 
 
 Since each term of the given series after the second is numerically less 
 than the corresponding term of an infinite decreasing geometrical series, and 
 since, § 581, every infinite decreasing geometrical series is absolutely con- 
 vergent, by Prin. 3, the given series is absolutely convergent. 
 
 4. Compare the series of example 1 with the auxiliary series 
 
 14-1 + 1 + 1 + 1+1 + =.. 
 li.'^2^ 3^ 4^ 5" 6^ • 
 
 Solution 
 Let p = 2. Then, the auxiliary series becomes 
 
 1 I 1 I 1 I 1 , 1 , 1 I 
 1.12.23.34.46.56.6 
 
 The series in Ex. lis 1-1 + 1-1 + 1-1 + ... 
 
 1 1.2 12. 3 1.2-3.4 1.2.3.4.5 1.2.3.4.5.6 
 
 each term of which after the third is numerically less than the corresponding 
 term of the auxiliary series. 
 
 Since. § 585, the auxiliary series is absolutely convergent when p =2, the 
 given series is absolutely convergent. 
 
606 CONVERGENCY OF SERIES 
 
 5. Discuss the convergency of the series 
 
 Solution 
 
 3x = M2 = (2 . 2 - l)x2-i ; 5 a;2 = M3 = (3 . 2 - l)x8-i ; 7 x^ = M4 = (4 • 2- l)x*-i. 
 
 Similarly, «„ =(n • 2 — l)a?*-i = (2 n — l)x»-i, and substituting n + 1 for 
 n, M„+i = [2(n + 1) - ljx(»+i)-i = (2 n + l)x» 
 
 . M„+i^ (2n+l)a^ ^ 271 + 1 3.^/1 ■ 2 \ 
 M„ (2n-l)x»-i 2n-l V 2n-lj' 
 
 •■■ ""■ ["^J-. = «"• [(' + 2^1)'=]^. = (' + ">"^ = "^- 
 
 Hence, 2d test, the series is absolutely convergent when x is numerically 
 less than 1, divergent when x is numerically greater than 1, and the test fails 
 when X is numerically equal to 1. 
 
 It is easy to see, however, that the series is divergent when x = 1, since 
 1 + 3 + 5 + 7+ •••is plainly divergent. Finally, when x = — 1, the series 
 becomes 1 — 3 + 5 — 7 + 9 — 11 + •••. The sum of 2 terms is —2; of 
 3 terms, + 3 ; of 4 terms, — 4 ; of 5 terms, + 5 ; of n terms, (— l)"-in. 
 Since by taking n large enough, the sum of the first n terms can be made 
 numerically greater than any assignable number, the series is divergent. 
 
 Hence, the series is absolutely convergent when x is numerically less than 
 1, but divergent in all other cases. 
 
 Write the first six terms of the series denoted by 
 
 Find the ratio of convergency of 
 
 10. 1+3 + 5 + 7 + ". . 12. l_f+5_7 + .... 
 
 11. l-^ + |_J^+.... 13. l_2x + 3ar^-4a;3+.... 
 
 Discuss the convergency of the following series : 
 
 15. 1+I + I + I+.... ^ 2 2^ 2 ^ 
 
 16. l--i + i-| + -. 20. ^^ + ^_^^ + ^^_^+.... 
 
 17. l+f + | + i+"-. ^ 2 . 2 3 • 2=^ 4 . 2^ 
 
 18. |+f + | + T'6 + -. ' 3 "^ 3=^ "^ 33 "^ " 
 
 1111 
 Vl + 1 V2 + I V3 + 1 V4 + 1 
 
CONVERGENCY OF SERIES 607 
 
 23. l-2ic + 3a^-4a^ + 5a;*-6ar*4-".. 
 
 24. l + 2a; + 4x'2+6ic3 + 8a;< + 10ar'+--. 
 
 25. l + ia; + |a^+iVa^ + rr^' + T^^ + -- 
 
 1-2 2-3 3.4^4.5 
 29. a + {a + d)r + {a + 2 d)!^ -\ \-{a + {n~ l)d]r"-^ + .-. 
 
 586. Convergency of the binomial series. 
 
 It has been proved in § 452 that when w is a positive integer 
 the series 
 
 a" + na"-^a: + ^^^^a--V + ''^^~^]^^3~^^ a»-V+-- (1) 
 
 terminates and is equal to (a + xy. It will be shown in a later 
 chapter that if (a 4- xy can be expanded into a series of ascend- 
 ing powers of x when n is negative or fractional, the series is 
 infinite and has the form of (1). It is important, then, to inquire 
 for what values of x, if any, this infinite series is convergent. 
 The (r + l)th and the rth terms, respectively, are 
 
 n{n-V) .- (n - r + 2)(n - r + 1) ^„-.^ 
 1.2...(r-l)r 
 
 ^^•^ 1.2...(r-l) ' 
 
 and, consequently, the ratio of convergency of the series is 
 
 /n-r + iy^^ /n + l_j\x 
 \ r Ja \ r Ja 
 
 in which n, x, and a are constants and r is a variable, expressing 
 
 the variable number of terms, 1, 2, 3, •••,00. Hence, as r increases 
 
 71 4- 1 
 without limit, approaches the limit zero, and 
 
 r 
 
 iim.[('i±i-iYl =(o-i)?=-5. 
 
a»ri 
 
 508 CONVERGENCY OF SERIES 
 
 Kow is numerically greater or less than 1, according as x 
 
 is numerically greater or less than a. 
 
 Hence, by the second test, the binomial series developed from 
 (a + xy is absolutely convergent ivhen x is numerically less than a, 
 and divergent when x is numerically greater than a. 
 
 The second test fails when x is numerically equal to a. 
 
 When x = a, (1) may be written 
 
 _ 1.2 1-2.3 J ^ ^ 
 
 When x = — a, (1) may be written 
 
 a.[l - n + ^^^ - " (" - W-'g- ^^ + ■■■]■ (3) 
 
 The nature of the series in (2) and (3) depends upon the value 
 of n. The full discussion of these cases is beyond the scope of 
 this book. 
 
 In ChrystaVs Algebra, Part II, page 131, it is shown that (2) is absolutely 
 convergent when n is positive, conditionally convergent when n lies between 
 and — 1, oscillating when n = — 1, and divergent when n< — 1 ; also that 
 (3) is convergent when n is positive, and divergent when n is negative. 
 
 587. Sum and difference of two convergent series. 
 
 Let ?7„ = rtj + ^2 + 1*3 + • ' • + Un 
 
 and Vn=Vi + V2 + Vs-] h v„ 
 
 be the sums of the first n terms of two convergent series whose 
 sums are CTand V, respectively. Then, §§ 568, 423, as n = oo, 
 
 Therefore, 
 
 (ui + Vi) + (u2 + V2) + (us + Vs)-\ 1- (m„ + v„) H 
 
 is a convergent series whose sum is U+V. 
 
 Hence, if the corresponding terms of ttvo convergent series are 
 added, the residting series is convergent, and its sum is equal to the 
 sum of the sums of the two given series. 
 
 The same principle applies when one convergent series is to be 
 subtracted from another, term by term, for this operation is equiv- 
 alent to adding the convergent series formed by changing the 
 signs of all the terms of the subtrahend. 
 
UNDETERMINED COEFFICIENTS 
 
 588, Coefficients assumed in the demonstration of a principle 
 or the sohition of a problem, whose values, not known at the 
 outset, are to be determined by subsequent processes, are called 
 Undetermined Coefficients. 
 
 To expand (x — l)(x + l)(x — 2) without actual multiplication, 
 
 put (x-l){x + l)(x-2) =x«-\- Ax^-\- Bx+ C, (V) 
 
 and determine A, B, and C by means of the fact that (1) is an identity and 
 must be true for all values of x. 
 
 By (1), when a: = 0, 2=0; 
 
 whenx = l, 0=\+A + B+C; 
 
 when x = -l, =- I + A - B + C ; etc. 
 
 By solving these three conditional equations, the coefficients A, B, and C, 
 undetermined in (1), are found to be ^ = — 2, £ = — 1, 0=2. 
 
 .: (x-l)(x + l)(«-2) =x3-2a;2_a; + 2. 
 
 589. Let po + Pix -f p^x^ + • • • + p,.-i«"-' + Pn^^ + ■-, (1) 
 called a power series in x, be convergent when x = a, and let it be 
 required to find toward what limit the sum S tends when x = 0. 
 
 The ratio of convergency of (1) when a; = a is -^-^ • a. 
 
 Pn-\ 
 
 Since (1) is convergent when x= a, the limit of the ratio of 
 convergency as w = oo cannot exceed 1 numerically, and conse- 
 quently must be numerically less than 1 when x<.a numerically, 
 say when x = b (not zero). 
 
 Hence, § 582, p^ + p^h -{■p^'' + - + p^.^b'^-' + p,J)" +— (2) 
 is absolutely convergent. Therefore, it is possible to find a fixed 
 number, as q, that is numerically greater than any term of (2). 
 
 609 
 
510 UNDETERMINED COEFFICIENTS 
 
 Take q>Pib, q>pj)^, q>Psb^, •••, q>pj>'', numerically. 
 Then, pi < |, i>2 < p, Pa < ^3' -> P» < |.» numerically. 
 Therefore, by (1), 
 PiX +p!^ -\-Ps^ + ••• < y + ^ + ^ + •••> numerically, 
 
 Since & is a value of x between a and 0, when a; = the series 
 in the second member of (3) is an infinite decreasing geometrical 
 series. 
 
 .-. PiX+p^+p^-\--", =S—po, <^-7 , or < -5^, numerically, 
 
 b—x b—x 
 
 which approaches zero as a limit as x = 0. 
 
 That is, as x = 0, S — po = 0, or S = Pq. Hence, 
 
 Principle. — If a power series in x is convergent for a value of 
 x, not zero, it is absolutely convergent for any smaller value of x, and 
 as x = the sum of the series approaches the first term as a limit. 
 
 590. JjQt A + Bx + Ca? + — =:^ A' -\- B'x + C'x^ + ••., (1) 
 
 for every value of x that makes both series convergent. 
 
 Then, for sxich values of a:, as w = oo, 
 lim. {A + Bx + Cx' + -) = lim. (A' + B'x + Ca^+ •..). 
 
 Therefore, § 589, 
 lim. {A + Bx + Cx" + •..)x=o = lim. (A' + B'x + C'x" + "•)x=0, 
 or A = A'. (2) 
 
 .-. by (1), Bx + Cx" + - = B'x + C'x" + ... , 
 or x(B+Cx + -)=x(B' + C'x +'.•). (3) 
 
 Since a; = but is not equal to zero, the members of (3) may 
 be divided by x. 
 
 .-. B+Cx + Dx'-\-- = B'+Cx-\-D'x'+'-, (4) 
 
 for all values of x that make the given series convergent. 
 
 From (4), by the reasoning applied to (1), 
 
 lim. (B+Cx + Dx'+ •••)x=o = lim. (B' + C'x + D'x^ + ...)a;=o ; 
 that is, B = B', whence x (C + Dx -\- —) = x(C + D'x + ..•). 
 
 Similarly, C=C', D = D', etc. Hence, 
 
UNDETERMINED COEFFICIENTS 511 
 
 Principle of Undetermined Coefficients. — If two series 
 arranged according to the ascending powers of x are equal for every 
 value of X that makes both series convergent, the coefficients of the 
 like powers of x are equal each to each. 
 
 Since lim. {A -f- Bx + Ca^ + •••)a;=o = ^ also when the series is 
 finite (§ 423), the Principle of Undetermined Coefficients applies 
 when one or both of the series are finite. 
 
 591. An algebraic expression is said to be developed when it 
 is transformed into a series whose sum, if it is finite, or the limit 
 of whose sum, if it is infinite, is equal to the given expression. 
 
 DEVELOPMENT OF FRACTIONS 
 Examples 
 
 592. 1. Develop the fraction - — — — ^« 
 
 ^ l + x + ar^ 
 
 Solution 
 
 The first term of the development is evidently 1 -r- 1, or 1 ; and since the 
 denominator is not exactly contained in the numerator, the development is 
 an infinite series beginning with 1 and proceeding according to ascending 
 powers of x. 
 
 To determine the coefBcients of the various powers, assume 
 
 ^ ^^'^ = \ + Ax + Bx"^ + Cx3 + Dx* + ... (1) 
 
 1 + X + x^ 
 
 true for all values of x that make the second member a convergent series. 
 Clearing of fractions and collecting terms, 
 
 l+2x=l + ^|x + -B 
 + 1 +^ 
 
 X2+ C 
 
 + B 
 
 + A 
 
 x^ + D 
 + C 
 + B 
 
 X* + ... 
 
 + ••• (2) 
 
 + - 
 
 + 1 
 
 The first member may be regarded as the infinite series, 
 
 1 +2x + 0x2 + 0x8 + ..., 
 
 which has a definite sum for all values of x, while the second member is an 
 infinite series having a definite sum for such values of x as make the series 
 assumed in (1) convergent. 
 
 Therefore, since (2) is true for all values of x that make the assumed 
 series convergent, by the principle of undetermined coefficients the coefficients 
 of like powers of x in (2) may be equated. 
 
512 UNDETERMINED COEFFICIENTS 
 
 Hence, A-\-\=2; .-. A = \; 
 
 JS + ^ + l = 0; .. S = -2; 
 
 C+5 + ^ = 0; .-. C=l; 
 
 Z) + C + B = ; .-. i) = 1 ; etc. 
 
 • 1 + 2 « . = ] + x _ 2 x2 + x3 + X* - .... 
 
 \ + X-\- X 
 
 The fraction may be developed also by division. 
 2. Develop the fraction — ^ + 2x' ^ 
 
 Solution 
 
 Q 
 
 Since the first term of the quotient is evidently — , or 2 aT", 
 
 surae 2 - a: + 2 x" ^ ^^.^ _|_ j^^.^ + (7 + ^a; + ^x2 + — . 
 
 x^ — 2 x^ 
 
 Clearing of fractions, 
 
 2-x + 2x2 = ^+ B\x+ C 
 -2a\ -IB 
 
 X2+ i?|X3+ JFiX* + ' 
 
 -2C -2D 
 
 Equating the coefficients of like powers of x, § 590, and solving, 
 ^ = 2;5 = 3; C=8;Z)=16;^ = 32; etc. 
 
 . .^ — X + 2 X _ o ,v.-2 I Q ,>.-l 1 Q 1 1ft« 1 00/>-2 1 
 
 X2 - 2 X3 
 
 
 
 
 Develop to five terms : 
 
 
 
 ^- 1-0." 
 
 10. 
 
 l-2a;-a^ 
 
 17. 
 
 1 l + 3a; 
 
 11. 
 
 x — o^ 
 
 18. 
 
 1 +a; 
 
 \ + 2x-y? 
 
 5 1 
 
 12. 
 
 \-x 
 
 19. 
 
 ' l-2a; 
 
 6 3 
 
 13. 
 
 1 
 
 20. 
 
 ^- 1-x 
 
 \~X — '3? 
 
 7 1 
 
 14. 
 
 2 + a;-2a^ 
 l-a; + 2a:2 
 
 21. 
 
 1 — ace 
 
 Q 2 + 3ar2 
 ' 1-2.'b2 
 
 15. 
 
 a^ + a^ 
 l-2a; + a^ 
 
 22. 
 
 4rc-3a^ 
 ■ l + 2a; 
 
 16. 
 
 l-2a; 
 
 ^- + -i? -ir x^ 
 
 23. 
 
 2-5a; 
 
 2x-Q^ 
 
 l + a; + a^ 
 
 a; + cc^ + ic* 
 
 1 
 
 a — ic 
 
 1 
 
 an- a? 
 
 a 
 
 1-a; 
 
 a 
 
 h — X 
 
 c 
 
 CIX 
 
UNDETERMINED COEFFICIENTS 513 
 
 DEVELOPMENT OF SURDS 
 Examples 
 
 693. 1. Develop the expression Va -|- a; by the use of unde- 
 termined coefficients. 
 
 Solution 
 
 Assume y/a + x = A + Bx + Cx"^ + Dx? + Ex^ + •■•. 
 
 Squaring, a-\-x- A^ + 2 ABx + B^\x^ + 2 AD\x^ + C'^lx* + •... 
 
 + 2AC\ -\-2Bc\->r2AE\ 
 + 2Bd\ 
 Equating the coefficients of like powers of x, § 590, 
 
 A'^ = a; .-. J = Va. 
 
 2AB=\; .: B = ^. 
 2a 
 
 B^ + 2AC = 0; .'. C= -— . 
 8a2 
 
 2AD + 2BC = 0; .: D = -^. 
 16 a8 
 
 C^ + 2AE + 2BD = 0; .: E=- -A^. 
 
 128 a* 
 
 .*. va + X = v« 1 -\ h • — ). 
 
 V 2a Ba-'^ 16a3 128 a* / 
 
 The given surd may also be developed by the extraction of the root indi- 
 cated or by the use of the binomial formula. But whatever the method of 
 development, the series obtained is equal to the surd only for such values of z 
 as make the series convergent. 
 
 Develop to five terms by undetermined coefficients : 
 
 2. VI — X. 8. VI + 
 
 X. 
 
 3. VI + 2 «. 9. vr+^T^. 
 
 4. Vl + 4 aj. 10. (1 + 4 a; + 6 ar^ 4- 4 a^ + a?*)i 
 
 5. V4-}-a;. 11. {l-3x-\-3x^-a^i 
 
 6. Vo^^. 12. (l+x)i 
 
 7. ^/a^-x^. 13. Vl + 2x + 3x^ + i3^ + .'.. 
 
 ADV. ALG. — 33 
 
514 UNDETERMINED COEFFICIENTS 
 
 PARTIAL FRACTIONS 
 
 594. To resolve a fraction into partial fractions is to separate it 
 into simpler fractions whose sum is equal to the given fraction. 
 
 It is necessary to consider only proper fractions. For if the 
 degree of the numerator is not lower than that of the denomi- 
 nator, the fraction may be reduced by division to the sum of an 
 integer and a proper fraction. 
 
 If the denominator of a fraction is composite, the fraction may 
 be resolved into an indefinite number of sets of partial fractions. 
 But the number of sets is limited to one by requiring that each 
 partial fraction shall be a proper fraction incapable of resolution 
 into simpler fractions. 
 
 1. To select the denominators of the partial fractions. 
 
 Since the resolution of a fraction into partial fractions is the 
 converse of the process of uniting fractions with different denomi- 
 nators into a single fraction having their lowest common denomi- 
 nator, it is evident that the denominators of the partial fractions 
 must comprise all the various prime factors of the given 
 denominator. 
 
 When the prime factors of the given denominator are all 
 different, it is sufficient to assume each of them for the denomi- 
 nator of a partial fraction. But when any prime factor is 
 repeated in the given denominator, say m times, there must be 
 a partial fraction having the wth power of this factor for a 
 denominator, for otherwise the L. C. M. of the denominators 
 would not contain the mth power of the factor. There may or 
 may not be denominators that are lower powers of the factor 
 than the mth, but provision should be made for such denomi- 
 nators. If they do not occur among the partial fractions, this 
 will be indicated by numerators that reduce to zero. 
 
 Two illustrations of the selection of denominators will be given. 
 
 Thus 5x^--Sx-2^ _ ^ ^ . ._^2^ . _ .^ (1) 
 
 and ^-26^0^+1^-- ^i_ +-J^ ^ ^« +_Z^_-. (2) 
 
UNDETERMINED COEFFICIENTS 515 
 
 In (1) no numerator can be equal to zero because each denomi- 
 nator is needed to produce the L. C. D., (a^ + 1) (a; + 3) {x + 4). 
 But in (2) it is possible that ^5 or N^ may be equal to zero, since 
 the L. C. D. would not be changed by omitting the corresponding 
 fractions. 
 
 2. To assume expressions for the numerators of the partial 
 fractions. 
 
 It has been agreed that each numerator shall be of lower degree 
 than the corresponding denominator, and that each partial frac- 
 tion shall be of such a form that it cannot be resolved into 
 simpler fractions. 
 
 Therefore, in (1) the numerators -^2 and ^3 must be numerals, 
 and the numerator N^ may be either a numeral or an expression of 
 the first degree in x. Hence, the numerators of the partial frac- 
 tions have the form 
 
 5cB2_3a.--24 _Ax + B ^ C . D ,^. 
 
 (a^ + 1) (a; + 3) (a; 4- 4) x' + l x + 3 x + 4. 
 
 T r . 5ar'-3a;-24 «-2, 3 4 
 
 In fact, — — = — \- 
 
 (ar^ -I- 1) (x 4- 3) (a; + 4) a.-^ + 1 ' a- + 3 a; -f 4 
 
 In (2) the numerators N^ and iV^ corresponding to powers of 
 {x -f 3), the prime factor in the denominator, must be of lower 
 degree than this prime factor, for otherwise either numerator 
 could be separated into two parts, one of them a multiple of 
 {x 4- 3), which would allow the fraction to be further decomposed. 
 
 Hence, the numerators of the partial fractions have the form 
 
 26a;4-18 _Ax + B , C . I) , E ... 
 
 "■" „ 1 o'/^ , o\2 ' /„ 1 o\s' V / 
 
 (a^ + l){x + Sy x^-fl a; -f 3 (.r + 3)- (a! -4- 3)3' 
 
 in which, as special cases, either A or B, but not both, may be equal 
 to zero, and either C or D or both, but not E, may be equal to zero. 
 
 In fact 26a^ + 18 ^ 1 1__ 6_, 
 
 ' (x' -\- 1) (x + Sy af + l (x + 3)- (x + sy 
 
 Rule. — Take each prime factor of the given fraction for a 
 denominator of a paTiial fraction, and ivhen any j^i'^'^ne factor 
 occurs m times in the given denominator, use also its second, third, 
 fourth, •••, mth powers as denominators. 
 
 Assmne for each numerator an expression tvith undetermined 
 
516 UNDETERMINED COEFFICIENTS 
 
 coefficients of a degree one lower than the prime factw of the cor- 
 responding denominator. 
 
 Suppose that the denominator of the given fraction is of the 
 nth degree in x. Then, by the above rule, there will be n unde- 
 termined coefficients ; and if the assumed identity is cleared of 
 fractions and the second member is then arranged according to 
 powers of x, the second member will be of the (ii — l)th degree 
 and have n terms. 
 
 Hence, by § 590, the two equal series obtained by clearing of 
 fractions will furnish n conditional equations from which the n 
 undetermined coefficients may be found. 
 
 Examples 
 
 3 
 1. Resolve into its partial fractions. 
 
 Solution 
 
 3 An 
 
 Assume that = — — 1 ^^— is an identity. 
 
 l-5x + 6a;2 l_3x l-2x 
 
 Clearing of fractions, 3 = yi — 2^x+B — 3 Bx. 
 
 That is, Z + a X = A + B - \2 A ^ '^ B) X. 
 
 .-. § 690, ^ + -B = 3 and 2 ^ + 3 J9 = 0. 
 
 Solving, ^ = 9 and JS = - 6. 
 
 3 9 6 
 
 Hence, 
 
 l-5a; + 6x-'' l-3a; l-2x 
 
 2. Resolve ; -— - into its partial fractions. 
 
 1 — 4a; + 4a^ ^ 
 
 Solution 
 
 Assume that :; — ^ ~ ^ ^^ , = —^ — + is an identity. 
 
 \-^x + 4:X^ \-2x (l-2a;)2 ^ 
 
 Reducing, b — Qx = A + B—2 Ax. 
 
 .: § 690, ^ + B = 5and - 2 A = - 6. 
 
 Solving, ^ = 3 and B = 2. 
 
 Hence, 5-6x ^_J__^ 2 
 
 1 - 4 X + 4 a;2 l - 2 a; (1 - 2 a;)2 
 
UNDETERMINED COEFFICIENTS 517 
 
 7 11 7; -I- 7 a^ 
 
 S. Resolve '—^ into its partial fractions. 
 
 (1 - x)\l +X + x") 
 
 Solution 
 It is evident that the fractions corresponding to the factor (1 — x)* will be 
 
 ^ ^ and ^ 
 
 1-x (i-xy-2 (i-x)^ 
 
 Since the factor (1 + x + x^) is quadratic and has no rational simple 
 factor, the numerator corresponding to the denominator 1 + x + x"'^ may 
 have two terms ; therefore, assume that 
 
 7-llx + 7x'2 _ A B C D-irEx .j. 
 
 (1 - X)S(1 + x + X2) 1-x (1 - X)2 (1 - X)3 1+ X + X^ 
 
 is an identity. 
 
 Then, 7-11x4-7x2 = ^(1 - x)2(l + x + x2) + ^(1 -x)(l + x + x2) 
 + C(H-x + x2) + (Z)+^x)(l-x)8 (2) 
 
 is an identity ; that is, is true for all values of x. 
 
 Since there are five coefficients, A, B, C, D, and E, to be determined, 
 and since (2) is true for all values of x, by giving x in succession each of 
 five different values, five independent equations involving the undetermined 
 coefficients may be formed, and from these equations the coefficients may 
 be determined. 
 
 Let X = 1 ; then, (1 — x), (1 — x)2, and (1 — x)^ reduce to 0, and the 
 identity (2) becomes 
 
 3 = 3C; .-. 0=1. (3) 
 
 Let X = ; then, 7 = A + B+C+B, 
 
 or, since C = 1, A + B-j- Dr=6. (4) 
 
 Let X = - 1 ; then, 2b = 4A + 2B+\ + SD-8E, 
 or, dividing by 2, 2A + B + 4D-4E-12. (5) 
 
 Let X = 2 ; then, 13 = 7A-7B + 7-D-2E, 
 
 or 1A-1B-D-2E = Q. (6) 
 
 Let X = - 2 ; then, 57 = 27 ^ + S + 3 + 27 D - 54 j&, 
 or, dividing by 0, 3A + B-\-'iD-QE = %. (7) 
 
 Solving the equations. (4), (5), (6), and (7), we have, together with (3), 
 ^ = 2, 5 = 0, C=l. Z) = 4, E-2. 
 
 Hence, 7 - 11 x + 7 x^ ^ _2_ ^ l ^ 4 + 2x . 
 
 (1 - x)8(l + X + x2) 1-X (1 - x)8 1 + X + xa 
 
518. UNDETERMINED COEFFICIENTS 
 
 X -\~ 2 x^ 
 
 4. Resolve -— l- — into its partial fractions. 
 
 1 — .^■^ 
 
 Suggestion. — Assume ?-±l^ = _^ + B + Cx ^ 
 1 - X^ 1 -X 1 + x + x^ 
 
 5. Resolve — , ^ ~ into its partial fractions. 
 
 ar — 1 
 
 Suggestion. — When the numerator is not of lower degree than the de- 
 nominator, the numerator should be divided by the denominator until the 
 remainder is of lower degree than the denominator. The fractional part of 
 the resulting mixed expression may then be resolved into partial fractions, 
 and these may be annexed to the integral part. 
 
 Resolve each of the following into its partial fractions : 
 
 — 6x 
 
 2x 
 
 S-Gx + x" 
 
 3-\-Ax 
 
 l_|-8a; + 16iK2 
 
 3a; 
 
 l + ar^ 
 
 l-x-ex" 
 
 x — x^ 
 
 l-2a! + 2a^ 
 
 12. 
 
 13. 
 
 8. " • 14. 
 
 15. 
 
 10. -± •^•"^^"^ , . 16. 
 
 (x - 5)3 
 
 x'-S 
 x^-l 
 
 2a^ + 9x+ll 
 
 a^ + 4a;-f4 
 
 2 - 6 .T -f 6 a^ 
 l-6a; + llar'-6a;3' 
 
 49 
 
 11 3a;-2 ^^ 1 + 2a; 4- 3a^ + ga:^ 
 
 (l-x)(l-2xy (2_3a;)2(3-a;) 
 
 3a;-2 
 (a; -3)=^ 
 
 REVERSION OF SERIES 
 
 595. To revert a convergent series of ascending powers of x is 
 to express the value of x by means of a series of ascending powers 
 of the sum of the given series. 
 
 Let it be required to revert the series 
 
 y = ax -{- ba^ -{- cx^ + dx* + • • •, (1) 
 
 in which x has' any value that will render the series convergent. 
 Assume x = Ay + By--{-Cf + Dy* -{ . (2) 
 
UNDE TERM I NED COEFFICIEN TS 
 
 519 
 
 Substituting this value T)f x in (1), and dropping all terms 
 involving a higher power of y than the fourth, 
 
 y = aAy + aB I y^ + aC 
 
 + bA'\ +2 bAB 
 + cA' 
 
 Since (1) is an identity, § 590, 
 aA = 1 ; .-. A = 
 
 aB + bA' = 0; .: B = 
 
 aC+2bAB + cA^ = 0; .: (7 = 
 
 f + aD 
 + bB^ 
 + 2bAC 
 -\-3cA'B 
 + dA* 
 
 y* + 
 
 bA^^ b 
 a a? 
 
 - 2 bAB - cA^ 
 a 
 
 2b'-ac 
 
 «5 
 
 aD+bB^ + 2 bAC + 3 cA'^B + dA* = 0; 
 
 T) __ _ g^d — 5 abc -f 5 b^ 
 ~ a' 
 
 Hence, 
 
 5 abc + 5 6^ 4 , 
 
 — r-^ y + 
 
 1 6 9 , 2 &^ — ac .J a'd 
 
 « = -2/ - ^r H ^ — y 
 
 Examples 
 
 1. Revert the series y = x 1 1- .-. 
 
 ^ 2 3 4 
 
 Solution. — Substituting in (8) 1 for a, — ^ for fe, | for c, 
 
 values of the undetermined coefficients in (2) are found ; 
 
 whence, x = y + ^y'^ + ly^ + ^y* -] . 
 
 Revert the series in the following equations : 
 
 2. y = x -i- x^ -\- x^ + X* -{- ■••. 
 
 3. y = x — 3x^ + 5x^ — 7x*+ •■: 
 
 4. y = x + 2x^+3a^ + Ax*-] . 
 
 5. y = 2a;4•3a^4-4a^ + oa^^ . 
 
 /y2 /yi3 n/A 
 
 6. y = x+ ^-}.5L+.^+.... 
 ^ 2624 
 
 7. y = X — 3 .T^ + 5 a^. 
 
 8. y = x-2a^ + 2ix^-2x* + "'. 
 
 (3) 
 
 ^ for d, the 
 
620 BINOMIAL THEOREM 
 
 9. Find the approximate value of x to four terms in the series 
 
 2~2 12 30 56 ■ 
 
 Solution 
 
 Reverting, x = 2(i) - K^)^ - Mh)' - i¥V(i)* - - 
 
 = 1 ~ i — ^ ~" j-hs — ••• 
 = .8189+. 
 
 Find the approximate value of x in the following : 
 
 10. J = a; + - + — + — +•••. 
 2 6 24 60 
 
 11. l = .-^V^-2^+.... 
 5 3 10 7 
 
 BINOMIAL THEOREM - FRACTIONAL AND 
 NEGATIVE EXPONENTS 
 
 596. It has been shown (§ 453, formula IV) that when n is 
 a positive integer, 
 
 (l + x)" = l+nx. + !i(^a- + ^(!L^il^^ (1) 
 
 It is yet to be proved that this formula is true when n is a 
 positive fraction, a negative integer, or a negative fraction. 
 
 I. When n is a positive fraction. 
 
 Let n = -, in which jp and q are positive integers. 
 
 Then, § 246, (1 + xy = ^(1 + a;)^ 
 
 §452, =Vl+F«+-. (2) 
 
 Assume, § 593, -\/\+px+ ••• = A + Bx + Gx' -{- ." (3) 
 
 where x may have any value that will make both series convergent. 
 Raising both members to the gth power, 
 
 1 +px + ... =\_A + iBx-{-Cx^ + •••)]' 
 
 = A^->rqA^-'^{Bx-\-Ca?+ •••) + •••. 
 
BINOMIAL THEOREM 521 
 
 Equating the coefficients of like powers of x in two terms, 
 
 1 = J." and p = qA^-^B) 
 
 P 
 whence, A = l, and B — — 
 
 Substituting these values in (2) and (3), 
 
 ^ q 
 
 That is, the formula is true for two terras, when n is a positive 
 fraction. 
 
 II. Wlien n is negative, and either integral or fractional. 
 
 1 
 
 (1 4- xy 
 1 
 
 § 244, (1 + x)-'^ 
 
 § 452 and Case I, 
 
 1 + nx + ••• 
 
 Dividing the numerator by the denominator, 
 
 (1 + x)-" — l — nx-] . 
 
 That is, the formula is true for two terms when n is negative 
 and either integral or fractional. 
 
 Therefore, the formula is true for tivo terras, whether the 
 exponent is positive or negative, integral or fractional, and the 
 coefficient of the second term is n. 
 
 597. To find the coefficients of terms after the second, assume 
 
 (1 + xy = l + nx + Ax" + Ba^+Cx*+—. (1) 
 
 Since only two terms of the expansion of the first member are 
 known, the coefficients A, B, etc., cannot be determined from (1) 
 in its present form. 
 
 To find the value of the undetermined coefficients A, B, etc., we 
 involve them in an identical equation, so that the coefficients of 
 like powers of the sarae variable raay be equated (§ 590). 
 
 Since in (1) x represents any number, put l-\-{x + z) for 1 + x. 
 
622 BINOMIAL THEOREM 
 
 Then, (1 -f-aj + z)» = [l + (x + z)]" 
 by (1), =l + n{x-\-z) + A{x + zy + B{x + zf-\-'" 
 
 = l+nx + Ax^ + B2(?-\ 
 
 + {:,i + 2Ax + ^Bx^ +-■)% + •■: (2) 
 Since in (1) x represents any number, and since 1 +{x-\-z) 
 
 = {l + x) + z^{l+x)(l+-^, put 1 + :^^ for 1+x. 
 Then, 
 
 (ij^x + zy = {\+xy(\+^y 
 
 by(i), ■ =a-i-.r(i+.^^+^^.^^^3+-) 
 
 -= (1 4- xy + ?i (1 + xy-h 
 
 + ^ (1 + xy-h"" + B{1+ xy-h^ + . . .. (3) 
 
 Therefore, from (8) and (2), Ax. 1, 
 (1 + xy + n (1 + xy-h + ^ (1 + a;)"- V ^B{l-\- a;)"" V + • . • 
 
 = 1+ na; + Aa? + Bx^ -\ \- {n + 2 Ax + ^ Bx^ -\ )z^ 
 
 when both members are convergent. 
 
 Equating the coefficients of z, 
 
 § 590, ?i(l + xy-^ = n + 2 ^a; + 3 Ba^ + — . 
 
 Multiplying each member by 1 + a;, 
 
 w(l + a;)» = 7t + (2^4-w)a;+(3^ + 2^)a:2_,_...^ 
 
 (1) X n, n(l-\-xy — n-\-'in?x-\-nAii?-\ . 
 
 Equating coefficients, 2 A + n = n^, 
 and 3B + 2.4 = w^; 
 
 whence, A = ^ ^ ~^ K 
 
 and .(n-l)(.-2X 
 
 1.2.3 
 
BINOMIAL THEOREM 523 
 
 In like manner any number of successive coefficients may be 
 found. Substituting these values in (1), 
 
 {l + xy = l + nx + '^^f^^ + ^^" - ^l%- ^> a^+ •>•. (4) 
 
 The expansion of (1 + a;)" is not the expansion for the most 
 general form of a binomial, since the first term is 1 ; but putting 
 
 X 
 
 for X 
 a 
 
 or 
 
 V ay a 1-2 a' 1-2.3 a^ '^^ 
 
 \ a j a\ 1-2 
 
 Multiplying by a", 
 
 (a + a:)" = a" + ««" 'a; + ^^^^^ ~/^ a"" V + • • • . (6) 
 
 J. * ^ 
 
 The binomial formula has thus been proved true when n is any 
 positive integer, any positive fraction, any negative integer, or 
 any negative fraction, that is, when n is any commensurable expo- 
 nent, provided the second member of the formula is convergent 
 when it is an infinite series. 
 
 598. It has been shown (§ 457) that the series developed from 
 (a+xy is infinite for fractional or negative values of n, and (§ 586) 
 convergent when x is numerically less than a. Hence, in this 
 case the true value of (a + »)" may be found to any required 
 degree of accuracy. The series is divergent when x is numerically 
 greater than a, but in this case the true value of (a + a;)" may be 
 found to any reqiiired degree of accuracy by expanding (a; + a)", 
 for the latter expansion then gives a convergent series. 
 
 Thus, VTOI is not found from (1 + 100)i = 1 + .50 - 2500 + •••, but from 
 (100 + \)^ = \{) -\- ^ - ^^ + ..., which approaches VIoT as a limit. 
 
 When x = — a, (a + a;)»=0" = ; when x = a, (a + a-)" = (2 a)", 
 the value of which may be found by separating 2 a into a binomial 
 whose first term is numerically greater than the second and ex- 
 panding by the binomial formula. Thus, (5 + 5)^ = 10^ = (9 + 1)^. 
 
 Exercises for practice will be found on page 414. 
 
EXPONENTIAL AND LOGARITHMIC SERIES 
 
 599. In a previous chapter the student learned the use and 
 advantage of common logarithms. But the principles upon which 
 the computation of the tables was based had not then been 
 established. Now, however, by the application of the principles 
 of convergence of series, he will be able, by means of two infinite 
 series, called the exponential and logarithmic series, respectively, 
 to derive a formula for computing logarithms in any system. 
 
 600. The exponential series. 
 
 The exponential series is the development in ascending powers 
 of X of the xth. power of a certain constant base. The series is 
 derived as follows: 
 
 By the binomial formula, if nx is commensurable and n is 
 numerically greater than 1, 
 
 f^ VV" _ ^ I *^^ _i_ nx{nx — 1) nx{nx — 1) {n x — 2) .^. 
 
 (2) 
 
 '\2 n^\^ 
 
 When x = \, (1) becomes 
 
 (l I ^Y = l I 1 I H^-1) I ri{n-l){n-2) 
 \ n) ^ ^ n^^ /^ w3[3 ^ 
 
 By (2) and (1), since (l + ^X ' = (l + ^^> 
 
 \ I 1 I K^-1) I n(n-l)(n-2) T 
 
 _ n'\2 «3|3 ^ J 
 
 _ w , , nx(nx — 1) nxjnx — l){nx — 2) ,o\ 
 
 ~ i?J2 n3|3 ^^ 
 
 In (3) it is permissible to let x have any finite value while n 
 increases numerically without limit. For whatever the value of 
 
 624 
 
I 
 
 EXPONENTIAL AND LOGARITHMIC SERIES 525 
 
 X, the law of variation of n as ?i = ± oo may be so taken that nx 
 is always commensurable. Accordingly, let n = ± oo . 
 
 Then, in (3), lini. 
 lim. 
 lim. 
 
 i(n - 1)' 
 
 , or lim. ( 1 — ) = 1 ; 
 
 ')>(n-l)(n-2)' 
 
 , or lim. ( 1 — 
 
 3« 
 
 = 1; 
 
 )ix(nx — 1) 
 
 , or lim. [x^ | = a^ ; 
 
 and so on. Hence, for all finite values of x, (3) becomes 
 
 usually written 
 
 \2 '13 
 ^^ = ^ + ^^1 + 1 + 
 
 (4) 
 
 (5) 
 
 y 
 
 In (5) the base e is a constant, 1 + 1+ — + — H , whose value, 
 
 I 
 
 as shown later, is 2.7182818 approximately ; and the exponent x is 
 a variable, and so may have any finite value. 
 
 Since in e'' the variable is an exponent, e"" is called the 
 exponential function of x, and the series developed from e"" is called 
 the exponential series. 
 
 601. To derive a formula applicable to any positive constant 
 base a, let log^ a = k. 
 
 Then, a = e*, 
 
 and a" = e'" = e('°Ke«)''. 
 
 ... by(5), a^ = l + (log.a)a. + (l5S^V^^^^S^V'-.. (6) 
 
 for all finite values of x. This is the Exponential Formula, or the 
 exponential series when the exponent of e is (log^ a)x. 
 
 602. In the exponential series (5) the ratio of convergency is 
 
 Iv*. I 1 Jb tie JO 
 
 ?<,. \r !?• — 1 r 
 in which the exponent x is fiiiite. 
 
 Therefore, as r increases without limit, the ratio, of convergency 
 approaches zero as a limit. Hence, § 582, the exponential series 
 is absolutely converger^ for all finite values of the variable involved. 
 
526 EXPONENTIAL AND LOGARITHMIC SERIES 
 
 603. The constant e is used as the base of a system of loga- 
 rithms called the Napierian System* or the natural system. Its 
 
 approximate value is computed from the series l + l + i- + l-| 
 
 as follows : 
 
 Adding, 
 
 l-fl 
 
 l-^[3 =(1-12) -3 = 
 
 1h-^ =(1-|3) -^4 = 
 
 1^[5 =(1^[4) -^5 = 
 
 1^16 =(1^[5) ^6 = 
 
 l-^[7 =(1-^(6) ^7 = 
 
 lH-|8 =(1^|7) -^8 = 
 
 1-|20 = (1^|9) -10 = 
 1 -[11 = (1-110)- 11 = 
 
 = 2.000000000 
 = .500000000 
 .166666667 
 .041666667 
 .008333333 
 .001388889 
 .000198413 
 .000024802 
 .000002756 
 .000000276 
 .000000025 
 
 e =2.7182818 to 7 places. 
 
 Examples 
 From the exponential series deduce the following ; 
 
 1. i(e^-e-=) = x +- + - + - + .... 
 ^\ J 1 3 [5 [7 
 
 2. i(e + e-i) = l + l-|.l + l+.... 
 2^ ^ ^ [2 [4 [6 
 
 3. ie = l+l + ^+i + .... 
 ' |3 [5 [7 
 
 4. 1=2 4 6 8 _ 
 
 e L§ L? IJ [9 
 
 5. i(e<'4-e-'^)=l-- + --^V— . 
 
 ^^ ^ |2 [4 |6 
 
 * Logarithms were invented by Baron Napier of Scotland in 1614. 
 
EXPONENTIAL AND LOGARITHMIC SERIES 527 
 
 604. The logarithmic series. 
 
 The logarithmic series is the expansion of log^ (1 + a;) in ascend- 
 ing powers of X. It is derived as follows : 
 
 By the exponential formula, when 1 + a? is the base and y the 
 exponent, 
 
 (l4-a.y = l + [log.(l+a.-)].v + ^^°^'^^^-^^^^''^V »-. (1) 
 
 By the binomial theorem, 
 
 (l + ^y = l+y. + yJl^^ + H=^!lii^^^+.... (2) 
 Equating the second members of (1) and (2), 
 
 1 + [log. (1 + .)], + p°g-(^+-)]y + ['°«-(i^+-)]y + ... 
 
 l + y. + y^^ + y'^-'^^^-^^ ^+.... (3) 
 
 Equation (2) is true when x is numerically less than 1 (§ 586), 
 and y is finite and commensurable (§ 597). Equation (1) also is 
 true under these conditions, for the exponent y is finite, and when 
 X is numerically less than 1, the base 1 + x is positive. 
 
 Therefore, since the two series have the same sum (1 + a;)* and 
 both are absolutely convergent for all finite commensurable values 
 of y, X being numerically less than 1, equation (3) is an identical 
 equation. Accordingly, equating the coefficients of y in the two 
 series, § 590, 
 
 loc (1 I x)-x \-^x'-\ (- ^)(-^) :^\ (-l)(-2)(-3) , 
 log,(l+a)-x + ^ 2^+ 1.2.3 ^ 1.2.3.4 "^ ^ ' 
 
 Simplifying the second member, 
 
 log,(l + a.)=a;-| + |-| + -. . (4) 
 
 This series is called the Logarithmic Series. It is absolutely 
 convergent when x is numerically less than 1 (§ 582), and con- 
 ditionally convergent when a; = 1 (§§ 583, 584), 
 
628 EXPONENTIAL AND LOGARITHMIC SERIES 
 
 605. To compute a table of natural logarithms. 
 
 Since the logarithmic series 
 
 log,(l4-a.)=a;-| + |-|V... (1) 
 
 is not convergent for values of x greater than 1, it cannot be used 
 vo find the natural logarithm of any number greater than 2. 
 
 Hence, it is necessary to modify (1) to make it available for 
 computing the natural logarithm of any positive number, however 
 great ; and for ease in computation it is desirable that the series 
 obtained be rapidly convergent. 
 
 Substituting — x for x in (1), 
 
 log.(l-a-)=-^-|-|-|*-.... (2) 
 
 Subtracting (2) from (1), § 587, 
 
 log. (1 + a^)- log. (1 - a;)= 2(^0^ + I + 1 + 
 
 or, §474, log.l±| = 2(^a. + | + |+...Y (3) 
 
 which is true when x is numerically less than 1 and when x = l. 
 
 Let w be a positive number whose natural logarithm is known, 
 and let m be a greater positive number whose natural logarithm 
 is to be computed. Then, since 
 
 is positive and less than 1, 
 
 m-\-n '^ 
 
 this value may be substituted for x in (3). 
 
 . m — n 
 -|.„ _m — n 1+x _ m + n_2m_m, 
 ~ m + n 1 — x . m — n 2n~ n^ 
 m-\-n 
 
 and in (3), since log. ^— ^ = log, — = log. m — log. n, 
 
 1 1 ■ of*'* — " . IM — nV 1/m — wV 1 ,.. 
 
 Iog,« = log,» + 2[_^^^ + -(^-^j +S(^ +■■■} W 
 
 This is the logarithmic formula for m > n > 0. 
 
EXPONENTIAL AND LOGARITHMIC SERIES 529 
 
 Since log^ 1 = 0, by substituting 1 for n and 2 for m, log, 2 may 
 be found ; then by substituting 2 for n and 3 for m, log^ 3 may be 
 found ; etc. Hence, a table of logarithms may be constructed by 
 substituting for n in (4) the successive values 1, 2, 3, 4, ••• and 
 for m values greater by 1 in each instance. 
 
 Substituting w -f 1 for m in (4) gives the more convenient 
 formula, 
 
 (5) 
 
 log,(»+l) = log.n+2[^ + g^jJ^ + g^^J;^ + . 
 
 which is true for all positive values of n. 
 
 Since the ratio of convergency of the series is always less than 
 
 -, the series is very rapidly convergent. 
 
 {2n-\-iy 
 
 Thus, when w = 1, the series is more rapidly convergent than the geo- 
 metrical series ^ + jV + 1J5 + ••• ; when n = 2, the series is more rapidly 
 convergent than the geometrical series ^ + yjj + yy^jj + ••• ; etc. 
 
 Examples 
 
 1. Find the natural logarithm of 2 to the nearest sixth decimal 
 place. 
 
 Solution. — Substituting 1 for n and for logj 1 in the formula for 
 log.rn-l-1), 
 
 ^ V3 3.38^5.36 7.37^ ) 
 
 _2 2 2 2 
 
 ~ 3 "^ 3T38 "^ 6T35 "'TTs^ "*" ■■"* 
 
 2 2 2 2 2 2 
 
 Since — = - h- 9, also — = — h 9, also — = — r- 9, etc., and since these 
 38 3 35 33 3' 38 
 
 quotients are to be divided by 1, 3, 5, 7, •••, respectively, the computation 
 
 may be arranged conveniently as follows : 
 
 3 I 2.00000000 
 
 Adding, 
 
 9 
 
 .66666667 -f- 1 = 0.66666667 
 
 9 
 
 .07407407 -- 3 = .02469136 
 
 9 
 
 .0082.3045 -- 5 = .00164609 
 
 9 
 
 .00001449-7 = .0001.3064 
 
 9 
 
 .00010101 -9 = .00001129 
 
 9 
 
 .00001129 s- 11 = .00000103 
 
 ADV. ALG. 3 
 
 .00000125 - 13 = .00000010 
 loge 2 = 0.69314718 
 4 
 
630 EXPONENTIAL AND LOGARITHMIC SERIES 
 
 The error committed in omitting the rest of the series is 
 2 2 2 
 
 15.3^6 17-3" 19.319 
 
 2/11 \ 2 1 
 
 which is less than {\ A 1 h-'-liOr x — ^— , 
 
 15.31H 9 92 /' 16-315 i_i' 
 
 9 
 X -, or 
 
 4. log, 5. 
 
 6. log, 7. 
 
 8. log. 9. 
 
 5. log, 6. 
 
 7. log, 8. 
 
 9. iog,m 
 
 15-316 8' 15-313.8' 
 
 2 
 ■which is much less than , , or .00000010, the last term found. 
 
 Hence, the part of the logarithm omitted does not affect the sixth decimal 
 place, and log, 2 = 0.693147 to the nearest sixth decimal place. 
 Compute to the neaxest sixth decimal place : 
 
 2. log, 3. 
 
 3. log, 4. 
 
 Suggestion. — The logarithm of a composite number is equal to the sum 
 of the logarithms of its factors. 
 
 Prove the following : 
 
 10. log.a-log,6 = _-+^(^^+-(^__J+..., 
 
 11. log.Va?^ = log.x-(^ + jL + _L.+ ...^,when,r>l. 
 
 606. The base e arises naturally in the process of finding a 
 formula for computing logarithms. In this and in other theoreti- 
 cal work, natural logarithms are used. But in numerical calcula- 
 tions, common or Briggs logarithms are the most convenient, 
 because the base of the common system is the same as the base of 
 our decimal system of notation. 
 
 Hence, the next problem is to discover how to change natural 
 logarithms to common logarithms. 
 
 607. To find the logarithm of a number to any given base. 
 
 Let iV be a number whose logarithm to the base a is sought. 
 By § 605, both log, N and log, a may be computed, and, there- 
 fore, are regarded as known numbers in this discussion. 
 
 Suppose that log, N=l, or N= e', (1) 
 
 and that log, a = k, or a = e*. (2) 
 
EXPONENTIAL AND LOGARITHMIC SERIES 531 
 
 Further, let r be the multiplier, as yet unknown, by which 
 log, N is multiplied to produce log„ N; that is, let 
 
 log„iNr=rlog, xV=rZ. (3) 
 
 By (3), N=ar\ 
 
 and by (2), N= {eY = e*''- (4) 
 
 By (4) and (1), e^' = e'; 
 
 whence, krl = I, and r = - , a constant. (5) 
 
 A; 
 
 The constant multiplier by which the logarithms in any system 
 are derived from the corresponding natural logarithms is called 
 the Modulus of the given system. The modulus of the natural 
 system is 1. 
 
 To find the modulus of any given system, let a be the given 
 base. 
 
 Since by (5), »• = 7^ and by (2), A; = log, a, 
 
 fc 
 
 the modulus is r = Hence, 
 
 log. a 
 
 Principle 1. — TTie modulus of any system of logarithms is equal 
 to the reciprocal of the natural logarithm of its base. 
 
 608. Let M denote the modulus of the common system. 
 Since by actual calculation log, 10 = 2.30258509- ••, 
 
 by Prin. 1, M = ,-^--r = = .43429448.... 
 
 ^ ' log, 10 2.30258509... 
 
 Hence, by the definition of the modulus of a system, 
 
 Principle 2. — TTie common logarithm of a number is equal to 
 its natural logarithm multiplied by .43429448.... 
 
 Therefore, by (4) and (5), § 605, when m > n > 0, 
 
 1 1 , o Tirf'^n — 71 , 1/m — ri\^ , 1/m — n\^ , 
 
 logio m = logio n+2M — ^ + oi T" + S — T" + 
 
 [_m + n o\m + ny o\m + nj 
 
 and logio (n + 1) 
 
 logio n+2 M 
 
 1 +!_,+ !__, + , 
 
 _2« + l 3(2 71 + 1)3 5(2 n + 1) 
 in which Jlf= .43429448.- 
 
 ,(1) 
 , (2) 
 
532 
 
 EXPONENTIAL AND LOGARITHMIC SERIES 
 
 Examples 
 1. Compute the common logarithm of 11 to. six decimal places. 
 Solution. — By (2), the formula for the common logarithm of n + 1, 
 
 logioll =logiolO + 
 
 (^ 
 
 1 
 
 21 "*" 3 • 218 
 
 + ■ 
 
 1 
 
 5.216 
 
 For convenience the computation is arranged as follows : 
 logio 10 = 1.00000000 
 
 + 
 
 21 
 
 21 
 
 21 
 
 .86858896 
 
 .04136138 H- 1 = .04136138 
 .00196959 
 
 .00009379 - 3 = .00003126 
 Adding, logio U = 1.04139264 
 
 In the above computation only the first two terms of the series are used. 
 
 Since the third term is less than — of the second term, or less than lir of 
 
 212 ' **i 
 
 .00003126, say less than .00000008, and the fourth term is less than ^\j of 
 the third term, and so on, it is evident that the terms after the second cannot 
 affect the sixth decimal place. 
 
 Hence, logio H = 1.041393, to the nearest sixth decimal place. 
 
 2. Compute the common logarithms of 2, 3, and 7, and verify 
 the following table : 
 
 Number 
 
 Logarithm 
 
 Number 
 
 Logarithm 
 
 Number 
 
 Logarithm 
 
 1 
 
 0.000000 
 
 5 
 
 0.698970 
 
 9 
 
 0.954243 
 
 2 
 
 0.301030 
 
 6 
 
 0.778151 
 
 10 
 
 1.000000 
 
 3 
 
 0.477121 
 
 7 
 
 0.845098 
 
 11 
 
 1.041393 
 
 4 
 
 0.602060 
 
 8 
 
 0.903090 
 
 12 
 
 1.079181 
 
 3. Find the four-place common logarithms of the composite 
 numbers from 14 to 25 inclusive. 
 
 4. Find the common logarithms of 225, 175, and .014 to four 
 places of decimals. 
 
 5. By using the above table find the four-place common 
 logarithms of .125, 46.2, 1.62, .0625, j\, 9|, 1.1, and yf^. 
 
 6. Show that logi, 18 = ^f^±-^{^. 
 
 ^^ 2 log, 2 -f log, 3 
 
 7. Given log. 2 = 0.69314718 •••, to find logj 7 and log2 8. 
 
SUMMATION OF SERIES 
 
 609. The summation of arithmetical and geometrical progres- 
 sions and of infinite decreasing geometrical series has been dis- 
 cussed in an earlier chapter. Other series may be summed by 
 somewhat similar devices, as will now be shown. 
 
 RECURRING SERIES 
 
 610. In any geometrical series, as a + ax-{- oaf + •••, each term 
 after the first is equal to x times the preceding term. In the 
 series 1 + 2 x— x^ + S jc^ — 19 x* + 62 a^ — • • • each term after the 
 third is equal to the algebraic sum of — 2 a; times the preceding 
 term and 3 a^ times the term preceding that. For example, 
 8x'=-2x(-a^)+3x^(2x); -19 x*:=-2 x(Saf)+3 x^(-a^ -, etc. 
 
 A series in which every term, after a certain term, is equal to 
 the algebraic sum of the products formed by multiplying the r 
 preceding terms, respectively, by r fixed multipliers is called a 
 Recurring Series of the rth order. 
 
 A geometrical series is a recurring series of the first order, and is the sim- 
 plest kind of a recurring series. The series l4-2x— ac^+Sx^— 19a;*-t-62a;S 
 
 is a recurring series of the second order. 
 
 611. If Un represents any term, after the third, of the series 
 
 l + 2x-a^ + Sa^-19x* + 62x^ , ^^ 
 
 then, ti„ = — 2x (w„_i) +3x^ (w„^2)- ""^^ 
 
 Transposing, etc., 1 («„) +2x (w„_i) — 3 a^ (u^-z) = 0. (1) 
 
 Equation (1) states the relation existing between any three 
 terms of the series, n being greater than 3 ; and the expression 
 
 1 4- 2 a; - 3 a^, 
 formed by taking the coefficients of w„, u„^i, and m„_2 in this rela- 
 tion, is called the Scale of Relation of the series. 
 
 633 
 
534 SUMMATION OF SERIES 
 
 612. To find the scale of relation. 
 
 Examples 
 
 1. Find the scale of relation of the series 
 
 1 + a, _|. 4a^ + lOa^ + 22a;* + 460?^ + .... 
 
 Solution. — Since the series is evidently not a recurring series of the 
 first order, try it for a recurring series of the second order, and assume 
 1 + Ax + Bx^ as the scale of relation. 
 
 If this supposition is correct, each term, after a certain term, increased 
 by Ax times the preceding term and Bx'^ times the term preceding that, must 
 be equal to zero. Therefore, beginning with one of the following equations 
 each succeeding equation must be satisfied by the same values of A and B. 
 4 x2 + Ax{x) + B3c\l) =0, or 4+ A+ B = 
 10x^ + Ax(ix^) +Bz^(x) =0, orlO+ ^A+ B = 
 22 3^ + Ax{10 x3) + Bx'^(i x2) = 0, or 22 + 10 ^ + 4 B = 
 46 «8 + ^a;(22 x^) + Bx%10 x^) =0, or 46 + 22 ^ + 10 £ = 0. 
 No values of A and B will satisfy all of these equations, but ^ = — 3 and 
 B=2 satisfy the last three, as may be discovered by solving any two of 
 them and trying the values obtained in the other. 
 Hence, the scale of relation is 1 — 3 x + 2 x^. 
 
 2. Find the scale of relation of the series 
 
 Solution. — Assuming that the series is a recurring series of the second 
 
 order whose scale of relation is 1 + Ax + Bx% it is found that no values of A 
 
 and B make the equations that may be formed on this supposition consistent. 
 
 Not even the last three equations thus obtained are consistent. Therefore, 
 
 the series is next tried for a recurring series of the third order. Assuming 
 
 1 4- Ax + Bx"^ + Cx^ for the scale of relation, we obtain, as in the preceding 
 
 example, 
 
 ^'3+ A-2B+ C = 0, -i-3A + SB+C = 0, 
 
 -3 + 3^+ B-2C = 0, 7-4^-3B + 3C = 0. 
 
 All of these, after the first, are satisfied by ^ = 1, S = 2, C = 1. 
 
 Hence, the scale of relation is 1 + x + 2 x^ + x^. 
 
 Find the scale of relation of 
 
 3. l-\-2x-\-2x^ + 2a^ + 2x*-^'". 
 
 4. l + 2x + Sa^ + ^a^ + 5a^-\-6T^ + "'. 
 
 6. l-\-x-2x^-2c(^ + 7x* + 7a^-20a^ . 
 
 7. l + x-\-x'-ii^~5x*-llx!'-15a^-9a^-\-—. 
 
SUMMATION OF SERIES 
 
 535 
 
 613. To find the sum or generating function. 
 
 By dividing the numerator by the denominator, or by employ- 
 ing the method of undetermined coefl&cients (§ 592), certain frac- 
 tions may be developed into series of ascending powers of a;. 
 
 The present problem is, given the series, to find the generath^g 
 fraction, or generating function. 
 
 Let it be required to find the sum or generating function of 
 
 l+a; + 4a^ + 10a;3_,_22x* + 46ar'-|-94a/'-f .••. 
 Multiplying the series by the scale of relation, 1 — 3 ic -f- 2 a^, 
 
 14. a; + 4 ar^ + 10 ar^ + 22 »* -f 46 ar' + 94 a:« + ••• 
 1-3 a; + 2 x" 
 
 1 -f 1 a; + 4 
 
 a^ + 10 ar* + 22 
 
 a^ + 46 
 
 a^+ 94 
 
 a:« + ... 
 
 -3 -3 
 
 - 12 - 30 
 
 -66 
 
 -138 
 
 _ ... 
 
 2 
 
 + 2+8 
 
 + 20 
 
 + 44 
 
 + ••• 
 
 l-2x + 3aj2+ Oaf'+ Oa;*+ Oa^+ Oa;«+... 
 
 Since the series multiplied by the scale of relation gives l — 2x 
 + 3x2 + 0a^ + 0x*+ ..., or l-2a; + 3a^, if l-2a; + 3ar' is 
 divided by the scale of relation, the series will be obtained. 
 
 That is, — — — r is the generating function of the series. 
 
 ' l-3a; + 2a;2 ^ ^ 
 
 The generating function is also the sum of the series, if the 
 series is convergent. 
 
 Similarly, any recurring series is shimmed by multiplying the" 
 series by its scale of relation and indicating that the result is to 
 be divided by the scale of relation. 
 
 Suppose, for example, that the scale of relation of ao + aix 4- a2X^ + asx^ 
 + •■■ \^\ -\- px + qx'^ and that every term after the second is formed by means 
 of the relation 
 
 1 (a„x") + i3a;(a„_ia;"-i) + qx^(^an-2X'^-^) = 0, or a„ + pa„_i + ga„_2 = 0. 
 Multiplying the sum of the first n terms by I + px + qx^, 
 
 ao+ ai X + a2 x^ + •■• + a„_] x» 1 
 1 + p x + q x^ 
 
 ao + ai I X + 02 
 paol +pai 
 + qao 
 
 x^ + 
 
 + 
 + 
 
 + a,,-! 
 
 + Pan2 
 
 + qon-s 
 
 Xn-1 
 
 + i)an-i[x» 
 
 -f quo -r ••• -^ (/w»-3 -T qan-21 -t- t/«n-lX"^' 
 
 ao + (oi + pao) X + x^ + \- Oa;»-i + (pon-i + gan-2)x» + qa„-ix"+^ 
 
536 SUMMATION OF SERIES 
 
 Hence, the sum of the first n terms is 
 
 " 1 + pa; + 3x^ 1 + i)« + 3*2 
 
 If the series is convergent, the second fraction in the above value of 8n 
 approaches zero as a limit when n is increased without limit, and the true 
 sum is 
 
 o _ gp + («! + pao)x 
 ~ \ + px + qx^ 
 
 Examples 
 Find the sum or generating function of 
 
 1. l+2.'c + 2«2_^2ar' + 2a;^+.... 
 
 2. l-\-x + 23^ + ^u?-\-nx*-\ . 
 
 3. X - Z:x? + 1 x" - 11 X* + 41:(^ . 
 
 4. l-^x + a? + Za? + lx*-\-^l T- + ■-. 
 
 5. l+a; + 2a;2-f-4a^-|-7a;* + 13ic^ + 24a^ + 44a;^+.... 
 
 6. 2~^x + x' + la?-12a*-a^ + S2a^-^2x^ . 
 
 614. To find the general term. 
 
 When the generating function may be separated into partial 
 fractions with denominators of the first degree, the series may be 
 separated into two or more geometrical series, and the general 
 term of the given series may be expressed as the sum of the gen- 
 -eral terms of the component geometrical series. 
 
 Examples 
 1. Find the general term of 1 + 2 a; + 6 a^ + 22 or* + 86 a;* -}- -., 
 
 Solution. — The generating function is found to be TL?.? 
 
 l-5a; + 4a;2 
 Separating this into its partial fractions, § 594, 
 
 l_5x + 4a;2 i _ g; l-4a; 3 
 
 \\-x] S\\-4txl 
 
 Since — — = 1 + a; + x^ + ... + x' + ••. 
 
 1 — X 
 
 and — =-^ = 1 4- 4x + 16x2 + ... + 4»-x'- + 
 
 1 — 4x 
 
 the (r + l)th, or general, term of the given series is 
 
 f x-- + I (4'xO, or A (2 + 4r)xr. 
 
SUMMATION OF SERIES 537 
 
 Find the general term and the twelfth term of 
 2. l-2a;+7i«2-20a;3+61a;* . 
 
 4. l + 5a; + 19ar' + 65x-3 + 211a;^ + —. 
 
 DIFFERENCE SERIES 
 
 615. If each term of a series is subtracted from the following 
 term, the successive differences form another series called the 
 Jirst order of differences ; if each term of the first order of dif- 
 ferences is subtracted from the following term, a second order of 
 differences is formed ; and so on. 
 
 By continuing this process an order of equal differences ma} 
 sometimes be obtained. 
 
 A series whose mth order of differences is a series of equal 
 numbers is called a Difference Series of the /nth order, m being 
 finite. 
 
 Thus, in the series 1, 8, 27, 64, 126, 216, •••, the successive orders of dif- 
 ferences are ■ , „ -,,. r.» ^, ^, 
 
 1st order, 7, 19, 37, 61, 91, ••• 
 
 2d order, 12, 18, 24, 30, ... 
 3d order, 6, 6, 6, ••• 
 
 Hence, 1, 8, 27, 64, 125, 216, ••• is a difference series of the third order. 
 
 Similarly, 7, 19, 37, 61, 91, ••• is a difference series of the second order 
 whose first order of differences is 12, 18, 24, 30, .•• ; and 12, 18, 24, 30, ••• ia 
 a difference series of the first order, also called an equidifferent series, or an 
 arithmetical series. 
 
 616. To find the nth term of a difference series. 
 
 Let Gi, a^, eta, ••• denote the terms of a difference series, and let 
 
 di, c?2, d^, ... denote t\ve first terms, respectively, of the first, second, 
 
 third, .•• orders of differences. 
 
 It is required to find the nth term of the series, or a„, in terms 
 
 of tti, di, ^2, ^3, '••, that is, in terms of the first term of the series' 
 
 and the first term of each order of differences. 
 Series, a^, a^, a^, a^, -•• 
 
 1st differences, ag — Oi, ag — ag, a^ — as, ... 
 2d differences, ag — 2 ag + ai, a4 — 2 a^ -f- a2, ••• 
 3d differences, a4 — 3 ag + 3 a2 — «!, • • • ; etc. 
 
538 SUMMATION OF SERIES 
 
 Since di = a2 — Oi, 
 
 transposing, 02 = 01 + ^1. (1) 
 
 Since ^2 = ag — 2 ag + ctj = Og — 2 (oj + dj) + Oj , 
 transposing, etc., ag = aj + 2 dj + dg. (2) 
 
 Since d3=a4— Sag+Saa— ai=a4— 3(ai+2di+d2)+3(ai+di) — a^, 
 transposing, etc., a4 = ttj + 3 dj + 3 dg + dg. (3) 
 
 If, as suggested by (1), (2), and (3), tiie coefficients in the 
 formula for the nth term are the same as the coefficients in 
 the expansion of the (n — l)th power of a binomial, then, the 
 formula for the nth. term is 
 
 a.=a.+ (.-l)d.+ (''-V<''-^> c;,+ (''-^)C-^>(''-«) .i,+ .... (4) 
 
 Assuming that this formula is true for the nth term of a dif- 
 ference series, the nth term of the first order of differences, which 
 is a difference series having dg, dg, ••• for the first terms of its 
 orders of differences, respectively, is 
 
 «„+! -a„ = d, + (n- l)d2-f- ^''~^^i''~^^ dg + .... (5) 
 
 Adding (4) and (5), 
 a„+i = (h + [(n - 1) + 1] d, + ^^[(n - 2) + 2] d,, 
 
 , , , n(n — 1) ■, , n(n — l)(n — 2), , ,„. 
 
 = Oi + wdi + ^ ,„ ^ d2 + -!^ -^ ^dg + .... (6) 
 
 Since (6) has the same form as (4), n + 1 simply taking the 
 place of n, if the assumed law is true for the nth term it holds 
 for the (n + l)th term. This law is true for the fourth and pre- 
 ceding terms, as shown in (1), (2), and (3). Hence, it holds for 
 the fifth term, and being true for the fifth term it holds for the 
 sixth, and so on. 
 
 Hence, (4) is the formula for any term. 
 
SUMMATION OF SERIES 539 
 
 617. To find the sum of n terms of a difference series. 
 
 To find the sum of n terms of the series Oi, a,, a^, •••, a„, form 
 an auxiliary series of which the given series is the first order of 
 differences, as 
 
 0, a^, Oi + a2, «i + ^2 + Osj •", «i + Oa + ag H h a„. 
 
 Then, the sum of the first n terms of the given series is the 
 same as the (n + l)th term of the auxiliary series, which may be 
 found by substituting for a^ a^ for dj, dj for d^, etc., in (6), the 
 formula for the (n + l)th term. 
 
 Denoting the sum of the first n terms by >S„, 
 
 o , n(n — V) J , n(n — V) (n — 2^ , , /rr\ 
 
 ^„=wai + -A-^ — !-d^ + ^ -^ '-d.,+ '-. (7) 
 
 Examples 
 
 1. Find the 10th term and the sum of 10 terms of the series 1, 
 
 2, 6, 15, 31, .... 
 
 Solution 
 
 Series, 1, 2, 6, 15, 31, ... 
 
 Ist differences, 1, 4, 9, 16, ••• 
 
 2d differences, 3, 5, 7, ••• 
 
 3d differences, 2, 2, ••• 
 
 Therefore, by (4), aio = 1 + 9(1) + 1^ (3) + \^^ (2) = 286 ; 
 
 and by (7), Sxo = 10(1) + yfj (1) + yfl^y (3) + ^^.'^^.g^.' J (2) =^ 836. 
 
 Note. — Since each term of the fourth and of each succeeding order of 
 differences is 0, all the terms after the fourth of (4) and (7) vanish. 
 
 Sum to 12 terms and find the 12th term : 
 
 2. P, 2^ 3^ 43, .... 5. 2, 8, 18, 33, 54, .... 
 
 3. l^ 2*, 3^ 4^ .... e. 2, 7, 14, 23, 34, .... 
 
 4. 1, 8, 21, 40, 65, .... 7. l^ 2% 3', 4% 5', 6', 7', .... 
 
 618. Piles of spherical shot. 
 
 1. If the base of a pile of spherical shot is a triangle with n 
 shot on a side, the next course has w — 1 shot on a side, the next 
 n — 2, and so on ; and if the pile is complete, the top course con- 
 sists of a single shot. Calling this the first course, the second 
 
640 SUMMATION OF SERIES 
 
 course has 2 shot on a side and contains 2 + 1 shot ; the third 
 has 3 shot on a side and contains 3 + 2 + 1 shot ; the fourth 
 has 4 shot on a side and contains 4 + 3 + 2 + 1 shot ; and, in 
 general, the nth course has n shot on a side and contains 
 
 w + (ri - 1) + (n - 2) H h 2 + 1 shot, or |n(7i + 1) shot. 
 
 Hence, when a pile of spherical shot has the form of a triangu- 
 lar pyramid the number of shot is 
 
 1 + 3 + 6 + 10 + 15 + - + i n(7i + 1). 
 
 Summing as a difference series, 
 
 series, 1, 3, 6, 10, 15, ••• 
 
 1st differences, 2, 3, 4, 5, ••• 
 
 2d differences, 1, 1, 1, ••• 
 
 .-. §617,(7), ^„ = n.l+K^-^).2+^(^-^X^-^).l, 
 
 \2 [3 
 
 or S, = in(n + l){n + 2). (1) 
 
 2. If the base is a square, the top shot rests upon 4 shot ; these 
 rest upon 9 shot ; and in general, the nth course has n shot on a 
 side, and contains n.^ shot. 
 
 Hence, when a pile of spherical shot has the form of a square 
 pyramid, the number of shot is 
 
 1 + 4 + 9 + 16 + .-.+n2. 
 
 Summing as in 1, S„ = ^ n(n + 1) (2 n + 1). (2) 
 
 3. If the base is a rectangle n shot in width and m-\-n shot in 
 length, the top course is a row of m + 1 shot ; the second course 
 contains 2 (m + 2) shot ; the third, 3 (m + 3) shot ; and so on. 
 
 Hence, the number of shot in a wedge-shaped pile with a rec- 
 tangular base is 
 
 1 (m + 1) + 2(m + 2) + 3(m + 3) + ... + n(m + n). 
 Summing as a difference series, 
 series, (m+1), 2 m + 4, 3m + 9, 4m + 16, ••• 
 
 1st differences, m + 3, m + 5, m + 7, ••• 
 2d differences, 2, 2, 
 
 .-.§617, ^„ = n(m + l)+^<^^(m + 3) + ^^^-^)^^^g).2, 
 
 or /S„=^n(n + l)(3m + 2n + l). (3) 
 
SUMMATION OF SERIES 541 
 
 Problems 
 
 1. How many cannon shot are there in a triangular pyrami- 
 dal pile whose bottom course has 10 shot on a side ? 
 
 2. The base of a pyramid of round 10-inch shot is 10 feet 
 square. How many shot does it contain? 
 
 3. How many shot are there in a wedge-shaped pile whose 
 bottom course is a rectangle 9 shot in length and 7 shot in width ? 
 
 4. How many 8-inch shot can be piled on a rectangular plot 
 of ground 12 feet long and 10 feet wide ? 
 
 5. Find the number of balls required to complete a trian- 
 gular pyramid having 7 balls in each side of the top layer. 
 
 6. Find the number of balls required to complete a square 
 pyramid having 100 balls in the top layer. 
 
 7. Find the number of courses of shot in a triangular pyrami- 
 dal pile containing 165 shot. 
 
 8 . How many balls are there in an incomplete wedge-shaped pile 
 having 77 shot in the bottom layer and 21 shot in the upper layer ? 
 
 9. A wedge-shaped pile of shot consisting of 10 layers con- 
 tains 605 shot. How many shot are in the top row? 
 
 10. The number of shot in a square pyramid is f^f of the num- 
 ber in a triangular pyramid having the same number of shot in 
 each side of the base. How many shot are there in each pile ? 
 
 11. A fruit seller had two pyramids of oranges composed of 
 the same number of layers. One was a square pyramid and the 
 other a triangular pyramid, and the square pyramid contained 84 
 more oranges than the triangular pyramid. How many oranges 
 were there in each ? 
 
 12. A fruit seller has a triangular pyramid of oranges, and 
 wishes to make of them four square pyramids. How many 
 oranges must he use for the base of each of the square pyramids, 
 if the triangular pyramid has 2n layers ? 
 
 619. Interpolation. 
 
 620. The process of inserting between two terms of a series 
 numbers that obey the law of the series is called Interpolation. 
 
642 
 
 SUMMATION OF SERIES 
 
 621. When the law of the series is known, numbers may be 
 interpolated between any two terms by applying the law. 
 
 Thus, in the arithmetical series 1, 9, 17, 25, •••, three numbers may be 
 inserted, or interpolated, between 17 and 25 by substituting 3|, 3^, and 3| 
 successively for n in the formula I = a +(n — l)d = \ +{n — l)%. The 
 three numbers obtained, namely, 19, 21, and 23, may be regarded as the 
 (3J)th, (3|)th, and (3|)th terms of the series, respectively. 
 
 When the law of the series is not known, intermediate terms 
 may be found approximately by treating the series as a difference 
 series, and giving n the proper values in the formula for the 
 nth term. 
 
 Examples 
 
 1. Given log 20 = 1.3010, log 21 = 1.3222, log 22 = 1.3424, 
 log 23 = 1.3617, log 24 = 1.3802 ; to find log 22.5 to the nearest 
 ten-thousandth. 
 
 Solution 
 
 Series, 1.3010, 1.3222, 1.3424, 1.3617, 1.3802, ... 
 
 1st difierences, .0212, .0202, .0193, .0185, — 
 
 2d differences, - .0010, - .0009, - .0008, ... 
 
 3d differences, .0001, .0001, — 
 
 Hence, oi = 1.3010, di = .0212, c?2 =- .001, and dz = .0001 ; also, since 
 22.5 lies halfway between 22 and 23, whose logarithms are the third and 
 fourth terms of the series, n = 3.5. 
 
 Substituting these values in the formula for the nth term, § 616, (4), 
 
 log 22.5 = 1.3010 4- .053 - .001875 + .00003126 = 1.3522. 
 
 Note. — In the above example the third differences are not absolutely 
 equal to each other, as may be seen by taking the given logarithms to more 
 than four places of decimals, and, therefore, the fourth differences are not 
 equal to zero. But the fourth differences and all succeeding them are so 
 small that they may be disregarded when log 22.5 is required to only four 
 decimal places. 
 
 2. In the following table the surfaces and volumes of spheres 
 whose diameters are 1, 2, •.-, 5 are given. Find the surfaces and 
 volumes of the spheres whose diameters are 3.1, 3.2, 3.7. 
 
 Diam. 
 
 1 
 
 2 
 
 3 
 
 4 5 
 
 Surf. 
 
 3.1416 
 
 12.5664 
 
 28.2744 
 
 50.2656 78.5400 
 
 Vol. 
 
 .5236 
 
 4.1888 
 
 14.1372 
 
 33.5104 65.4500 
 
SUMMATION OF SERIES 543 
 
 3. Given ^350 = 7.0472987, ^/35i = 7.0540041, ^/352 = 
 7.0606967, ^353 = 7.0673767, ^354 = 7.0740440, to find 
 the cube root of 351.6. 
 
 4. Given 7!-^=. 00 1379310, yi^=. 001369863, ,4^=. 001360544, 
 7-^^= .001351351, y^ = .001342282, to find the reciprocal of 736. 
 
 SERIES WHOSE ^TH TERMS ARE FUNCTIONS OF N 
 
 622. The sum of the first n terms of a series is represented, in 
 general, by S^- But if the nth term of the series is a function of 
 n, the sum of the first n terms may be represented by prefixing 
 Sn to the nth term. 
 
 Thus, SnU^ means the sum of the first n squares; 65(211 + 1) means. 
 14-3 + 5 + 7 + 9, the sum of the first 5 odd numbers. 
 
 623. Summation by undetermined coefficients. 
 
 When the nth term of a series has the form a + 1)71 + en- + •••, 
 the series may be summed to n terms by employing undetermined 
 coefficients. 
 
 Examples 
 
 1. Find the sum of the first n squares. 
 
 Solution. — Since each term of 1^ + 2^ + 3^ + ... + rfi is a function of n, 
 assume S„.n'^ =^ A+ Bn + Cn^ + Dn^ + En*^ + ••• (1) 
 
 and ^„+i?i2 = ^ + 5(?i + 1) + C(n + 1)2 + i)(n+ 1)3 + ^(74 + 1)4 + .... (2) 
 
 Subtracting (1) from (2), 
 (n + l)2=J5+C(2n + l)+Z)(3n2+3„+i)^.^(4,i34.6„2+4„ + i) + ..._ ^3^ 
 
 Since the first member involves no power of n higher than the second, E 
 and all succeeding undetermined coefficients vanish, and 
 
 n2 + 2 n + 1 = ^ + (7(2 n + 1) + Z)(3 n2 + 3 n + 1) 
 = (B + a + i)) + (2 C + 3 D) M + 3 Z)n2 . 
 
 .-.§590, B + C + D = l, 2C + 3Z> = 2, and 3D = 1. 
 
 Solving, D = \, C=i, B = \. 
 
 .: 12 + 22 J.32 + ... + n2 = ^+ ^n + Jn2 + |n8 
 
 = ^ + iM(n + l)(2n + l). 
 Since Sxn^ = 1, I2 = ^ + J . 1 . 2 • 3, whence ^ = 0. 
 
 .-. 12 + 22 + 32+ ... + n2 = ^n(n + l)(2n + l). 
 
644 SUMMATION OF SERIES 
 
 Show by employing undetermined coefl&cients.that: 
 
 2. 1 + 2 + 3 -}-••. -fw=iri(« + 1). 
 
 3. l« + 23 + 33 + ... + w3 = ^w2(« + l)2. 
 
 4. 1^ + 2^ + 3* + ... + n* = ^\n(n -\-l){2n + l)(3n' + 3n- 1). 
 
 5. 1* + 3- + 5- + . . . + (2 n - 1)2 = ^ w (2 n + 1) (2 w - 1). 
 
 624. Compound series. 
 
 Some series may be summed readily by separating each term 
 Into two terms and adding the component series thus obtained. 
 
 Examples 
 1. Find the sum to n terms and to infinity of the series 
 2 2 2 
 
 1.22.33.4 
 
 Solution. — Separating the general term into two fractions having tne 
 same numerators as the terms of the series, 
 
 2 ^ 2 _ _2 
 
 n(n + 1) n n -\-\ 
 
 Since this is true for each term of the series, 
 
 -■=(!-i)-(i-i)-(i-l)— (!-;r^) 
 
 2 3 n — \ n 
 
 2 2 2 2 2 
 
 = 2- 
 
 2 3 n-\ n n+l 
 
 2 2n 
 
 n+ 1 n + 1 
 As n increases without limit, this sum approaches 2 as a limit. 
 
 Find the sum to n terms and to infinity of 
 
 2. J_ + J_ + J- + .... 5. J-+J- + J- + 
 
 1.3 2.4^3.5 3.5 5.7 7.9 
 
 3. _^+-^+ -^ + .... 6. -?- + ^ + -?- + 
 
 1.4 2.5 3.6 2.6 3.6 4.7 
 
 1.5^3.7^5.9 1.3 2.4 3.5 
 
FUNCTIONS OF A SINGLE VARIABLE 
 
 625. The general object of this chapter is the comparison of 
 the corresponding values of x and certain functions of x. For this 
 purpose X is regarded as a variable increasing continuously from 
 an indefinitely large negative value, denoted by — go, to an in- 
 definitely large positive value, denoted by + co. 
 
 TABULATION OP FUNCTIONS 
 
 626. By giving the independent variable x any values we 
 choose, the corresponding values of any given function of x, as 
 i(a^ — 6 x), may be obtained. The manner in which the function 
 changes with the independent variable may then be exhibited in 
 a table of correspondiiig values. For example. 
 
 when X = 
 
 -2 
 
 -1 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 8 
 
 \(x'-Qx) = 
 
 8 
 
 31 
 
 
 
 -^ 
 
 -4 
 
 — 4.1 
 
 -4 
 
 -^ 
 
 
 
 ^ 8 
 
 Since a;^ — 6 x = x{x — 6), if x> 6 and increases without limit, \{x'^—Q x) 
 will be positive and increase without limit. This is denoted by annexing to 
 the above table the corresponding values a; = + oo, ^{x^ — 6 x) = + oo. Again, 
 if X is negative and decreavses without limit, |(x2 — 6 x) will be positive and 
 increase without limit. This is denoted by prefixing to the table the cor- 
 responding values X = — 00, ^(x^ — 6 x) = -f- oo. 
 
 627. The symbol for any given function of x is f(x), read 
 ' function of .r.' Other functions of a; in a discussion may be 
 represented, if desired, by F(x), <f>(x), f'{x), etc., read ' large F 
 function of x,' ' phi function of a;,' '/-prime function of a?,' etc. 
 
 Since f{x) is a variable, depending upon the value assigned to 
 X, fix) is often represented by the variable y. 
 
 ADV. ALG. — 36 
 
 545 
 
546 FUNCTIONS OF A SINGLE VARIABLE 
 
 Values of f(x) corresponding to particular values assumed for 
 X, as 3, 0, — 1, a, are usually indicated thus : /(3), /(O), /(— 1), 
 
 If/Cx) = x2 - 6 x+ 0, /(0)= 9 ; /(1)= 1 -6 + 9 = 4 , /(2)=4-12 + 9 = l ; 
 /(3) = 9 - 18 + 9 = ; and in general, /(a) = a^ - 6 a + 9. 
 
 628. A convenient method of substitution. 
 
 Let it be required to find the value of a^-\- Aa^ + 7 x + G when 
 x = 2. The process of substituting 2 for x is similar to the 
 arithmetical process of reducing a compound denominate number 
 to a number of lower denominatioii. 
 
 Process Explanation. — Since a; = 2, x^ = 2 a;2. 
 
 Writing 2 x^ under 4 x- and adding, the 
 
 ar4-4ar-(- t x -\- 6 | 2 sum of the first two terms of /(x) is equal 
 
 2 a^ + 12 .r 4- .'58 to 6 a;2. Since x = 2, Gx"^ = 12x. Writ- 
 
 ar' -I- 6 ar 4- 19 r I 44 — ^2^ ^^^ ^^ ^ under 7 x and adding, the sum of 
 
 the first three terms of /(x) is equal to 
 19 X. Since x = 2, 19 x = 38. Adding 38 to 6, the remaining term of /(x), 
 44 is obtained for the value of /(x) when x = 2 ; that is, /(2) = 44. 
 In practice only the detached coefficients are used, thus : 
 
 1 + 4+ 7+ 6(_2 
 
 2 + 12 + 38 
 1 + 6 + 19 I 44 .-. /(2)=44. 
 
 Examples 
 Tabulate the following functions of x for integral values of x 
 between — 5 and 5 and for a; = ± co : 
 
 1. 2 a; — 1. 2. ar' — a^ + a; - 1. 3. .r' - 6 a;^ + 11 x — 6. 
 
 GRAPHICAL REPRESENTATION OF FUNCTIONS 
 
 629. Heretofore, functions of x have been given a purely- 
 numerical interpretation. By representing the values of f(x) by- 
 distances, it is possible to give a pictorial or graphical representa- 
 tion of the function passing through all its values as x increases 
 continuously from — cc to + oo. The function ^(x? — 6x) will be 
 used for purposes of illustration. 
 
 630. Since the values of x and ^(a^ — 6 a;) tabulated in § 626 are 
 real, they may be represented graphically by distances from a zero 
 
FUNCTIONS OF A SINGLE VARIABLE 
 
 547 
 
 point on a horizontal line, positive distances being laid off toward 
 the right and negative distances toward the left. But to avoid 
 confusing the values of x and f(x), it is customary to lay off the 
 values of the function on or parallel to a second line crossing the 
 first at right angles at the zero point, positive values of /(a;) being 
 represented by distances measured upward and negative values 
 by distances measured downward from the first line. 
 
 The first line is called the JIT-axis, and the second the K-axis, 
 the function of x being denoted by y. 
 
 Fig. 1 is a graphical represen- 
 tation of the manner in which 
 the function \(3r — 6 x) changes 
 with the independent variable x. 
 See the table of corresponding 
 values, § 626. 
 
 The point Pj, which is 2 units 
 to the left of the zero point and 
 at the same time 8 units above 
 it, represents the state of the 
 variable and the function when 
 x = — 2 and f(x) = 8. Similarly, 
 P.2 represents the next state of 
 the variable and the function, Pg 
 the next, and so on. 
 
 The lengths of the vertical 
 dotted lines represent the values of y, or f(x). They are called 
 ordinates. For example, y^ is the ordinate of Pj and is equal to 
 /(-2),or8. 
 
 The corresponding horizontal distances are called abscissas. 
 For example, — 2 is the abscissa of Pj. 
 
 The abscissa and ordinate of a point are called its coordinates. 
 
 The point whose abscissa is a and ordinate h is denoted by the 
 symbol (a, h). The abscissa is always written first. 
 
 Thus, the point Pi is represented by (—2, 8). 
 
 If the successive values of x between any two finite values are 
 taken sufficiently near together, the corresponding values of y 
 — i(^ — 6 a;) may be made to differ from each other as little as 
 we please ; that is, y varies continuously with x, or is a continuous 
 
 Fig. 1. 
 
548 
 
 FUNCTIONS OF A SINGLE VARIABLE 
 
 X 
 
FUNCTIONS OF A SINGLE VARIABLE 549 
 
 function of x (§ 435). Hence, the assemblage of all the points rep- 
 resenting corresponding values of x and y form a continuous line. 
 This line, the coordinates of every point of which satisfy 
 the equation y = ^(af — 6 x), is called the Locus of the equation 
 y = i(a^ _ 6 X), or the Graph of the function ^(x^ — 6x). Within 
 the limits of the drawing it is represented by the curved line 
 
 The graphs of several functions of x are shown in Fig. 2. 
 
 631. Roots of /(jr) = 0. 
 
 Consider the functions of x plotted in Fig. 2. 
 
 1. Since whenever the graph of a function crosses or touches 
 the axis of x the ordinate of that point is equal to zero, the func- 
 tion is equal to zero for the value of x denoted by the correspond- 
 ing abscissa. For example, a^ -1- 9 a^ + 23 a; -f- 15 is equal to zero 
 when a; = — 5 or — 3 or — 1, as shown by its graph crossing the 
 axis of X at these points; that is, the graph of the function 
 a^-f9a:^-f-23a/' + 15 shows that —5,-3 and — 1 are real roots 
 of the equation ar* + 9 a^ + 23 x + 15 = 0. 
 
 Again, a^ — 8a; + 16=0 when x = 4, as shown by its graph 
 touching the axis of x at this point ; that is, 4 is a real root of 
 the equation a^ — 8 a; + 16 = 0. 
 
 Hence, the real roots of f{x) = are represented graphically by 
 the abscissas of the points where the graph of f{x) crosses or touches 
 the axis of x. 
 
 2. Since, for all values of x, a^ — 8 a; + 20 is 4 greater than 
 x^ — 8 a; -f 16 and the latter is 4 greater than a^ — 8 a; + 12, the 
 graphs of these functions are exactly alike except in position. 
 
 Suppose, then, that the graph of a^ — 8 a; + 12, which exhibits 
 two unequal real roots, 2 and 6, is moved vertically to coincide 
 with the graph of .r^ — 8 a; -f- 16. It is evident that the two un- 
 equal roots become more and more nearly equal as the graph is 
 moved upward, and become equal when the graphs coincide. 
 
 Hence, if the graph off(x) touches the axis of x but does not cross 
 it or terminate in it, the abscissa of the point of tangency represents 
 two equal real roots off(x) = 0. 
 
 3. Now suppose that the graph is moved vertically once more 
 toward the graph of a;^ — 8 a; + 20. The graph ceases to touch 
 
550 FUNCTIONS OF A SINGLE VARIABLE 
 
 the axis of x, and we infer that there is no real value of x which, 
 substituted for x, can reduce the function to zero. 
 
 This inference accords with the fact that the roots of a^ — 8 a; 
 + 20 = are imaginary (§ 304, Prin. 1). 
 
 Hence, if the graph of f(x) has no point lying in the axis of x, all 
 the roots of f(x) = are imaginary. 
 
 Examples 
 
 Plot the following functions of x, and discover from the graphs 
 as much as possible concerning the roots of f(x) = : 
 
 In the following examples the student should substitute values of x close 
 enough to each other to determine the form of the graph with some accuracy. 
 
 Cross-section paper is very useful in this work. 
 
 It is often convenient to plot the values of f(x) to a smaller scale than 
 that employed for the values of x. 
 
 1. 2x + l. 4. 1-a^. 7. a^-ex' + llx-G. 
 
 2. xF — 5x + 4:. 6. a^-2x~l. 8. »" — 7 a; + 6. 
 
 3. a^_6a;-l-9. 6. x--^4. 9. a-^ - 9 a^ + 22 ar' - 32. 
 
 DERIVED FUNCTIONS 
 
 632. Another method of comparing the variation of f(x) with 
 that of X consists in determining for what values of x, if any, 
 f(x) increases or decreases when x increases continuously, and 
 in finding how fast f(x) is changing as compared with the corre- 
 sponding changes in x. 
 
 633. Let f(x) denote any function of x, and suppose that x is 
 increased by a positive number h from the value a to the value 
 a-{- h. Then h is called an increment of the variable. 
 
 Since the increment given to x produces a change in the value 
 of the function from /(a) to /(a + h), the change in the function, 
 f(a + h) —f(a), is called an increment of the function. 
 
 In general, /(a; + ^) —f(x) represents any increment of /(x). 
 
 When f(x) is increasing, the successive increments of the func- 
 tion are positive, and when f{x) is decreasing, they are negative. 
 
FUNCTIONS OF A SINGLE VARIABLE 551 
 
 634. When equal increments of x produce equal increments in 
 f{x), as in the case of the functions 2 x and 2 a; + 8, Fig. 2, the 
 rate of change of f{x) is uniform and may be obtained by finding 
 the ratio of any increment of fix), however large or small, to the 
 corresponding increment of x. When f{x) does not vary uni- 
 formly with X, the rate of change of f{x) with respect to x at any 
 instant during a small interval is obtained approximately on the 
 supposition that fix) varies uniformly with x during that interval. 
 
 The smaller the interval, then, or the smaller the value of h, 
 
 the more nearly will the ratio — ^ — represent the true 
 
 rate of change of f(x) as x changes from a to a + ^. 
 
 Hence, the limit of this ratio as /i = represents the rate of 
 change of f(x) at the instant when x = a. 
 
 The same is true for any value of x. Hence, 
 
 -f(x + h)~f(xy 
 
 lim. 
 
 h 
 
 h=0 
 
 represents the rate of change of f(x) at any instant as compared 
 with that of x. 
 
 635. The limit of the ratio of the increment of f{x) to the cor- 
 responding increment of x as the increment of x approaches the 
 limit zero is called the Derivative of the function with respect to jr, 
 or the First Derived Function of jr. 
 
 The formula for the derivative of f(x) with respect to x is 
 
 ■f(x + h)-f(xy 
 
 lim. 
 
 h 
 
 n 
 
 Jh=0. 
 
 This is represented by — f{x). 
 ax 
 
 Thus, — (x^ + 2 x) means the derivative with respect to a; of x^ -^ 2 x. 
 dx , n 
 
 Other notations are Dif(x), ^- w, -^, etc. 
 •'^ ^' dx dx 
 
 636. Since the derivative with respect to x of f(x) is usually 
 another function of x, which in turn may have a derivative with 
 respect to x, which may have still another, and so on, the first 
 derivative will be denoted by/'(x), the second hj f"(x), the third 
 by f'"(x), and so on. 
 
 Successive derivatives are sometimes denoted by /i(x), /'(x), fz{x), etc. 
 
652 FUNCTIONS OF A SINGLE VARIABLE 
 
 These derived functions are called the first derived function of x, 
 or the derivative of f{x) with respect to x, the second derived 
 function of x, and so on. 
 
 Examples 
 
 1. Find the derived functions of a^ — 8 a; + 12. 
 Solution. — Let x^ - 8 a; + 12 = fix) . 
 
 Then, f(x+ h)-fix)= x'^+2 hx + h^- (8 x+8h) + 12- (x'^-8x+12) 
 
 = 2hx-8h + h\ 
 
 ... fct^^-^ = 2x-8 + A. f\\\\ 
 
 TaMng the limit as /i = 0, 
 
 f'{x) = 2x — 8, the Jirst derived function. 
 Again, /'(x + /i)-/'(x)= [2(x + A)- 8]-(2x - 8) 
 = 2h. 
 . / '(x + /0-/'(x) _^ 
 •• h -^• 
 
 Taking the limit as A = 0, 
 
 /"(x)= 2, the second derived function. 
 
 Since the second derived function is a constant, its increment, and there- 
 fore its derivative, is zero. Hence, all derivatives of x^ — 8 x -f 12 after 
 the second vanish. 
 
 Find the successive derived functions of 
 
 2. 2x. ^. x- — 8x. 6. 3 — 2x — x^. 
 
 3. 2x4-1. 5. a.-2-8a; + 16. 7. a? + 9x^ + 23x-{- 15. 
 
 637. A function that involves only integral powers of x and is 
 integral with respect to x is called a rational integral function of x. 
 
 3 1 
 
 x^ + 1 + x"2 is rational but not integral with respect to x ; x^ + | x^ is 
 integral but not rational with respect to x ; x^ -f 5 x -t- 6, x^ -}- i x^ -t- 3 x, and 
 a;2 -I- ■v/2 X + 1 are both rational and integral with respect to x, that is, are 
 rational integral fimctions of x. 
 
 638. The general form of a rational integral function of x is 
 
 aaf + 6x"-' + ca;»-2 -| [-kx + l, 
 
 in which w is a positive integer, and a, b, c, •••, k, I do not 
 involve x. But it is more convenient to use the form 
 
 af -\-px^-^ + 9cc»-2 -\ \-sx + t, 
 
 which may be obtained by dividing the general form by a. 
 
FUNCTIONS OF A SINGLE VARIABLE 553 
 
 639. Derivative of a rational integral function of jr. 
 
 Let f{x) — X" + 2Jcc"-i + qx"-- H \-sx + t (1) 
 
 be a rational integral function of x. 
 Give X an increment h. Then, 
 
 f(x+h) = (x+h)'^+pix-\rhY-'+q(x+hy-'+ ••• +s(x+h)+t. (2) 
 Expanding and arranging to ascending powers of h, 
 
 f(x+h)= x" + px''-'^+ qx^-^-l \-sx-^t] 
 
 -{-h\nx''-'^ + (n—l)px"-^+{n—2)qx"-^-{ \-sl 
 
 ^^ln(n~l)x^-^+(n-l)(n-2)px''-^+-l 
 
 + ^"|n(n-l)(n-2)...l| 
 
 \n 
 
 ■ G^) 
 
 Subtracting f(x) from the first member and its value given in 
 (1) from the second, dividing the result by h, and taking the limit 
 of the result as /i = 0, the terms in (3) involving h^, h^, •■-, h" 
 vanish, and we have 
 
 lim.[M=^ 
 
 =nx^-'^-{-(7i—l)px''-^-j-(n—2)qx''-^-] \-s; 
 
 ^=0 
 
 that is, — (xf" + »a;"*^ + ga;"-^ -i h sx + 
 
 dx ' 
 
 = naf-^ + (w — \)px''-^ + (w — 2) qx""-^ -\ \- s. (4) 
 
 Hence, the derivative of a rational integral function of x is 
 
 obtained by midtiplying each term of f(x) by the exponent of x 
 
 in that term and diminishing the exponent of x by unity. 
 
 Thus, — (x8 + 5 a;2 + 7 3c _ 4) = 3 a;2 + 10 a; + 7. 
 
 dx 
 
 640. Denoting the successive derived functions of x by fix), 
 
 f'V), -', 
 
 f(x) = ic" -{- p«"-^ + qxT-'^ -\ \-sx + t, 
 
 f'{x) — na;"-^ + (n — l)pa;" ^ + (n — 2) qx''-^ -\ \-s, 
 
 f"(x) =n(n- 1) iC»-2 + (w _ 1) (n _ 2)px''-^ -j , 
 
 f"<(x) =n(n-l){n-2) x^'^ + • • •, 
 
 /"(x) =w(n-l)(n-2)...2.1.ir« = |w. 
 
654 FUNCTIONS OF A SINGLE VARIABLE 
 
 Hence, (3) may be written 
 f{x + h) =f(x) + hf'(x) + f^f"(x) +^f'\x) + -. + h\ (5) 
 
 Formula (5) gives a process of substituting x + h for x in a 
 rational integral function of x. It is a special case of Taylor's 
 Formula. 
 
 641. Continuity of a rational integral function of x. 
 
 By (5), § 640, if f{x) is a rational integral function of x, 
 
 fix + h) -f{x) = hf'ix) + |/"(a;) +^r'{x) + ■•• + h\ 
 
 The first member is the increment of f{x) produced by giving 
 x the increment h. To determine whether the given function is 
 continuous, it is only necessary to find whether an infinitesimal 
 change in the variable produces an infinitesimal change in the 
 function. Then, let h = 0. 
 
 Since the second member consists of a finite number of terms 
 and the coefficient of each power of h is finite for all values of x, 
 as /i = the second member approaches zero as a limit. 
 
 Hence, every rational integral function of x is continuous for all 
 finite values of x. 
 
 642. Derivative of the product of two or more functions. 
 
 Let y and z be rational integral functions of a; and let y^ and 
 z' be particular values of y and z corresponding to a; = a. 
 
 To find the derivative of yz with respect to x, suppose that, as 
 X increases from a to o + h,y changes from y' to y' + A; and z from 
 z' to z' + I If /i = 0, then, § 641, A; = and Z = 0. 
 
 When x = a, — yz = lim. 
 dx 
 
 \y' + k){z' + l) 
 
 ^1 
 
 J/t = 
 
 simplifying, = lim. 
 
 h 
 ■ kz' + (y' + k)r 
 h 
 Then, since y' and z' are constants, and as h = 0, y' -\- k = y', 
 
 h 
 
 — yz = z' Jim. 
 dx' 
 
 h = 
 
 '1 
 
 Jh = 
 + y'lim. - 
 
 §635, =z'^y + y'^z. 
 
 dx dx 
 
i 
 
 FUNCTIONS OF A SINGLE VARIABLE 565 
 
 Since this is true for each particular value of x, 
 
 d cl , d 
 
 —yz = z--y + y—z. 
 dx dx dx 
 
 Similarly, — v(yz) =yz — v -\-v — yz 
 dx dx dx 
 
 d , d , d 
 = yz—v + zv—y + vy—z, 
 dx dx dx 
 
 and so on for any number of variable factors. Hence, 
 
 The derivative toith respect to x of the product of any number of 
 
 rational integral functions of x is equal to the sum of the products 
 
 obtained by multiplying the derivative of ea/ih function by the product 
 
 of all the other functions. 
 
 Thus, — (x - a)(x - b)(x -c) = (x- 6)(x - c)— (x - a) 
 dx dx 
 
 + (x - c) (x - «) — (x - 6) 
 dx 
 
 + (x-a)(x-b) — (x- c) 
 dx 
 
 = (x — 6) (x — c) + (x — c) (x — a) 
 + (x — a) (x — b). 
 
 Examples 
 
 1. Find the derivative with respect to x of (x—l)(x—2)(x—3). 
 Test the result by comparing it with the derivative with respect 
 to X of the expanded product. 
 
 2. Find the first and second derived functions of (x — l)^(a; + 2) 
 and express each in the form (x — l)"^(x). 
 
 MAXIMA AND MINIMA 
 
 643. In tracing the graph of 3 — 2x— a^, Fig. 2, conceive x to 
 increase continuously from — oo to +qo. 
 
 As X increases continuously up to —1, the function 3 — 2 a; — o;^ 
 increases continuously up to 4, and as x increases continuously 
 from —1 to -f CO, the function decreases continously from 4 to 
 — CO. Hence, 3 — 2 a; — x^ is said to be an increasing function of x 
 for values of x less than —1 and a decreasing function of x for 
 values of x greater than —1. 
 
656 FUNCTIONS OF A SINGLE VARIABLE 
 
 .644. Any value at which a continuous function of a variable 
 ceases to be an increasing function and begins to be a decreasing 
 function is called a maximum of the function. For example, 4 is 
 a maximum oi 3 — 2 x — x^. 
 
 Any value at which a continuous function of a variable ceases 
 to be a decreasing function and begins to be an increasing function 
 is called a minimum of the function. For example, — 4 is a 
 minimum of x*^ — 8 a; + 12. 
 
 The maxima and minima of a function are called its turning 
 values, or critical values. 
 
 Cautiox. — A maximum of a function is not necessarily its gi-eatest value 
 nor a minimum its least value, as may be seen in the graph of x^ + 9 x^ + 23 a; 
 + 15. (Fig. 2.) 
 
 645. The test for turning values is as follows : 
 
 If f(a) is greater than either f(a + h) or f{a — h) however small 
 h may be taken, then f(a) is a maximiim of f{x). 
 
 If f(a) is less than either f(a + h) or f(a — h) however small h 
 may be taken, then f(a) is a minimum of f(x). 
 
 Thus, 3 — 2 a; — a;2 is greater for x = — 1 than for any other values of x in 
 the vicinity of — 1, however little they may differ from — 1. Hence, /(— 1), 
 or 4, is a maximum of 3 — 2 x — x^. 
 
 646. Let ax^ + bx -{- c be any quadratic function of x having 
 real coefficients. Placing the function equal to y and solving 
 for X, 
 
 b , 1 
 
 a; = - -^ ± -^ V4 ay - (4 oc - 6=^) . 
 
 2 a 2 a 
 
 Considering only real values of x, 4 ai/ — (4 ac — 6^) must be 
 
 positive or else equal to zero ; that is, 
 
 4 ay - (4 ac- 62)^0. 
 
 Therefore, §§ 401, 403, 
 
 if a is positive, y>:c , 
 
 ~ 4a 
 
 b^ 
 but if a is negative, 2/ ^ c — - — • 
 
 In the first case y, or ax^ + bx -\- c, may be as great as we 
 
 please, but cannot be less than c — -— , which is therefore a 
 
 . . 4a 
 
 minimum. 
 
FUNCTIONS OF A SINGLE VARIABLE 557 
 
 In the second case the function may be as small as we please, 
 
 but cannot be greater than c , which is therefore a maximum. 
 
 Hence, ^ ^ 
 
 Principle 1. — Every quadratic function of the form aa? -\-hx-\-c 
 has one and only one critical value, c — (6^ -h 4 a), which is a mini- 
 mum or a maximum according as a is positive or negative. TJie 
 coiTesponding value of x is —h -i-2 a. 
 
 Thus, x^ — 8 X + 12 has only one critical value, 12 — (64 ^4), or — 4, 
 and this is a minimum because the coefficient of x^ is positive ; the corre- 
 sponding value of X is 8 -T- 2, or 4. Again, 3— 2x— x^ has only one critical 
 value, 3— (4-: — 4), or 4, and this is a maximum because the coefficient of x^ 
 is negative ; the corresponding value of x is 2 h- — 2, or — 1. 
 
 647. Let a be a value of x that makes f{x) a maximum, and let 
 h be a small positive number, as small as we please. 
 
 Then, however small h is, if /(.^•) is continuous, by definition 
 
 /(a) >f{a ± h). (1) 
 
 Therefore, § 400, if k is any constant, 
 
 /(a) + k >f(a ± h) + k; that is, 
 
 Principle 2, — If f(a) is a maximum of f(x), f(a) + k is a 
 maximum of f(x) -\- k ; similarly, if f(a) is a minimum of f{x), 
 f(a) +k is a minimum off(x) -\- k. 
 
 Thus, in Fig. 2, is a minimum of x^ — 8 x + 16, 4 is a minimum of 
 x2 — 8 X + 16 + 4, or x2 — 8 X + 20, and — 4 is a minimum of x^— 8 x + 16— 4, 
 or x2 - 8 X + 12. 
 
 Again, if f(a) is a maximum of f(x), by (1) and the principles 
 of inequalities, 
 §402, _/(a)<_Xa±/0; 
 
 also, § 403, • m/(a) > mf(a ± h) ; 
 
 also, § 403, provided /(a) is not equal to zero, dividing (1) by 
 
 > - — , or < Hence, 
 
 f{a±h) f{a) f(a) f(a±h) 
 
 Principles. — 3. If f(a) is a maximum of f(x), — f{a) is a ' 
 minimum of — f{x). 
 
 4. Iffia) is a maximum of fix), mf(a) is a maximum ofmf(x). 
 
658 FUNCTIONS OF A SINGLE VARIABLE 
 
 5. If f(a) is a maximum of fix), is a minimum of • 
 
 JJK) •''"'^'Aa) ''fix) 
 
 Thus, 4 is a maximum of 3 — 2 x — x^ and — 4 is a minimum of x^ + 2 x — 3, 
 as may be discovered by applying Prin. 1 or by plotting their graphs ; also, 
 
 8 is a maximum of 2(3 — 2 x — x^) and \ is a^^inimum of 
 
 The same principles apply to minima of functions. 
 
 648. Let f{x) — uv, in which u and v are variables whose sum 
 is constant, say ?< + v = 2 A". 
 
 Since (w — vf is positive except when u = v, from the identity 
 4 MV = (m -}- ■v)^ — {u — vY, or 4 Mv = 4 Tc^ — (u — vy, 
 4 uv ^ 4 /c^ ; .-. uv ^ l<^. Hence, 
 
 Principle 6. — If f(x) is the product of tivo variable factors 
 whose sum is constant, the sqxiare of half this sum is a maximum 
 
 offi^y 
 
 Since 3 — 2 x — x^ = (3 + x)(l — x), the sum of the variable factors is 4, a 
 constant ; then, (|)2, or 4, is a maximum of the function. (See Fig. 2.) 
 
 Again, to find the critical value of x^ — 8 x + 12, 
 put x2 - 8 X + 12 = (x - 6)(x - 2) =/(x). 
 
 Then, (x - 6)(- x + 2) =-/(x). 
 
 Therefore, Prin. 6, [|(x — 6 — x + 2)]2, or 4, is a maximum of — /(x). 
 
 Hence, Prin. 3, — 4 is a minimum of /(x). 
 
 In geometrical language Prin. 6 is stated as follows : Of all rectangles icith 
 equal perimeters, the square is the greatest. 
 
 649. Let f{x) = u -\- v, in which u and v are variables whose 
 product is constant, say uv = Tc^. 
 
 Since (u + vy = 4: uv + (tt — vy, and since (u — vy is positive 
 except when u = v, in which case (u + v)^ = 4 uv, 
 
 (u + v)2 ^ 4 fc2 ; that is, [f{x)y ^ 4 1^. (1) 
 
 Since both members of (1) are positive, f(x) is numerically 
 greater than ± 2 A; or else equal to ± 2 k. When f(x) is positive, 
 f(x) ^ 2 A; ; and when f{x) is negative, f{x) -^ ~2k. Hence, 
 
 Principle 7. — If f{x) is the sum oftivo variables whose product 
 is constant, twice the positive square root of this product is a mini- 
 mum of fix) and twice the negative square root is a maximum. 
 
 If /(x) = = X + -, since x • - = 9, a constant, 6 is a minimum and 
 
 XXX 
 
 — 6 a maximum of the function. 
 
FUNCTIONS OF A SINGLE VARIABLE 
 
 559 
 
 6. 
 
 8. 
 
 1 + 
 
 X 
 
 x + ^+2. 
 
 X 
 
 xj\ x) 
 
 Examples 
 Find the critical values of the following functions : 
 
 1. rc^— o X + 6. 
 
 2. x^+x — 30. 
 
 3. 3 ar- 4 a; — 15. 7. a; + 
 
 4. —a? — 4. 
 
 5. (a;-3)(7-a;). 
 
 change of sign of 
 
 650. Critical values of continuous functions 
 th^ first derived function. 
 
 Since the first derived function of f{x) is the limit of the ratio 
 of the increment of f(x) to the increment of « as 7i = 0, ^ being a 
 positive increment, it follows that the function is increasing when 
 f'{x) is positive and decreasing when fix) is negative, and vice 
 versa. 
 
 Hence, if the derivative off(x) changes sign as x increases through 
 a, f{a) is a critical value of the function. 
 
 The value of a is found by solving the equation /'(x) = 0. For example, 
 the value of x that renders a;^ _ g x + 12 a minimum is 
 found by solving 2 a; — 8 = 0, giving x = 4, whence, 
 minimum =/(4) = 16 - 32 + 12 = - 4. (See Fig. 2.) 
 
 It does not always follow, however, that the values 
 of X that satisfy the equation f'{x) = correspond to 
 critical values of /(x). For the derivative may become 
 equal to zero without changing sign as in the accom- 
 panying graph of x^ — 1. 
 
 661. By (5), § 640, when x = a, 
 
 fCa+ h) -f(a) = hf'ia) +|/"(a) +||/"'(a) + - 
 
 and f(a - h) -f(a) = - hf'(a) +^V"(a) -||/"'(«) + 
 
 Suppose that y(a) is a critical value oif(x). Then, § 650, 
 f(a-\-h)-f(a) 
 
 ^|/"(«) +!>"(«) + 
 
 and f(a - h) -f{a) =|/"(«) - ^/'"(a) + 
 
560 FUNCTIONS OF A SINGLE VARIABLE 
 
 If h is taken sufficiently small, the terms in W, h*, •-■ may be 
 neglected. Hence, iij{a) is a critical value oif(x),f"(a) has the 
 same sign as /(a + h) — J{a) or f(a — h) —f(a). But by the test 
 of a critical value, § 645, f{a + h) —f{a) and f{a — h) —f(a) 
 are both negative when f(ci) is a maximum and both positive when 
 f(a) is a minimum. 
 
 Hence, if the first derived function of f(x) changes sign when 
 X = ci, /(«) is a maximum or a minimum of f(x) according as the 
 second derived function is negative or positive for x = a. 
 
 652. Discontinuous Functions. 
 
 653. When the increment of a function of a variable corie- 
 sponding to an infinitesimal increment of the variable is not 
 infinitesimal, the function is called a Discontinuous Function. 
 
 Let h be any increment of x in fix). 
 
 Then, f(x) is discontinuous for x = a, if the difference between 
 f{a + h) and f{a — h) cannot be made as small as we please by 
 taking h sufficiently small. 
 
 The function x-\-- is a discontinuous 
 
 X 
 
 function as may be seen from its graph. 
 For as x changes from a very small nega- 
 l tive number to a very small positive num- 
 ber, the function changes from a very large 
 negative number to a very large positive 
 Y' number; and the smaller the increment of 
 
 ic as a; passes through zero, the greater 
 is the increment of the function. Hence, the function is discon- 
 tinuous for a; = 0. 
 
 For all other values of x, however, the function is continuous. 
 The graph has two parts called branches, one in the third quad- 
 rant corresponding to negative values of x, the other in the first 
 quadrant corresponding to positive values of x. 
 
 654. Let h and k be corresponding increments of x and/(a;). 
 
 If the ratio - is always finite as A = 0, fc may be made as small 
 
 as we please by taking h sufficiently small ; that is, the function 
 is continuous if its derivative is finite for all values of x. 
 
 X^ 
 
FUNCTIONS OF A SINGLE VARIABLE 561 
 
 If the limit of this ratio can become infinite for some value of x, 
 an infinitesimal change in x may produce a finite or infinite 
 change in f(x) ; that is, the function is discontinuous for x = a if 
 the first derivative increases or decreases without limit as » = a. 
 
 Thus, X + i is discontinuous for x = ; for the first derivative of x -f x-'^ 
 
 X 
 
 is 1 — a;-^, or 1 — -- , which increases or decreases without limit as » = 0. 
 
 Examples 
 
 1. Discuss the function a^ + 9 a^ + 23 x + 15. (See Fig. 2.) 
 Discussion. — Since the function is rational and integral with respect tox, 
 it is continuous. Tabulating the values of x and f{x)^ 
 
 X = 
 
 — 00 
 
 -6 
 
 -5 
 
 -4 
 
 -3 
 
 -2 
 
 -1 
 
 
 
 1 
 
 2 
 
 + 00 
 
 /(a^)= 
 
 — 00 
 
 -15 
 
 
 
 3 
 
 
 
 -3 
 
 
 
 15 
 
 48 
 
 105 
 
 + 00 
 
 This shows that the curve begins in the third quadrant, crosses the axis 
 of X three times, giving the three real roots — 5, — 3, and — 1, crosses the 
 axis of y at the point (0, 15), and extends indefinitely upward in the first 
 quadrant ; also that the function is an increasing function up to some value 
 in the interval between x = — 5 and x = — 3, then a decreasing function to 
 some value in the interval between x = — 3 and x = — 1, and thereafter an 
 increasing function. To find the turning values, put /'(x)= 0. 
 
 Then, /'(x)= 3x2+ 18x + 23 = 0. 
 
 Solving, X = — 3 ± I V3 = — 4.155 or — 1.845, approximately. 
 Substituting these values in the second derived function 6 x + 18, — 4.155 
 makes /"(x) negative and — 1.845 makes /"(x) positive. 
 
 Hence, § 051, /(— 4.155) is a maximum and /(— 1.845) is a minimum. 
 The graph may now be plotted completely. 
 
 Discuss and plot the following functions : 
 
 2. a?-6x^ + llx-Q. 5. ar'-7a; + 6. 
 
 3. «2+l. 6. ic* — 5a^ + 4. 
 
 4. a^-10a; + 25. 7. x + 4x-\ 
 
 8. Show that first degree functions of x have no critical values 
 and that their graphs are straight lines. 
 
 9. Plot the graphs of a^ + a^ — 5x-\-3 and its derived func- 
 tions. What root has f(x) = when f{x) = has two equal 
 roots? 
 
 ADV. AlO. — 36 
 
THEORY OF EQUATIONS 
 
 655. Changing an equation to another whose roots are the 
 same as the roots of the given equation is called Reducing the 
 equation. 
 
 656. Changing an equation to another whose roots have known 
 relations to the roots of the given equation is called Transforming 
 the equation. 
 
 657. Every simple equation may be reduced to the form 
 ax-\-b = 0', every quadratic equation to the form aa^+6a;+c=0; 
 every equation of the third degree, or cubic equation, to the form 
 aa^ -\-bx^ + cx + d = 0; every equation of the fourth degree, or 
 biquadratic equation, to the form ax* + bxP -{-cx^ + dx + e = 0; etc. 
 
 Hence, these equations are called the general equations of the 
 1st, 2d, 3d, 4th, ••• degrees in x. 
 
 The general equation of the nth degree in x has the form 
 
 a.T" + bx"-^ + cx^-- H [-kx-\-l = 0. 
 
 658. Dividing both members of the general equation of the 
 nth degree in x by the coefficient of a;", the resulting equation 
 takes the form 
 
 X" -{-px"-^ + qx''-^ -\ \-sx + t = 0, 
 
 in which the coefficient of the highest power of x is 1. 
 
 This equation is called the reduced equation of the nth degree 
 in X. The first member is a rational integral function of x 
 (§ 638), and in this chapter the equation will often be written in 
 the brief form J{x) = 0. 
 
 The absolute term t may be regarded as the coefficient of x°. 
 
 The coefficients p, q, •••, s, t may be positive or negative or 
 equal to zero, integral or fractional, rational or irrational, real 
 
 562 
 
THEORY OF EQUATIONS 563 
 
 or imaginary. But unless the contrary is stated, it will be un- 
 derstood that they are rational and real. 
 
 If no coefficient is equal to zero, the equation is complete; if 
 any coefficient is equal to zero, the equation is incomplete. 
 
 659. Divisibility of ^(jr) = 0. 
 
 Let the first member of /(a;) = be divided hy x — a until the 
 remainder no longer involves x. 
 
 Denote the quotient by Q and the remainder by R. 
 
 Then, f{x) = (x-a)Q + B. (1) 
 
 Since M does not involve x, R has the same value whatever 
 value X has. To find R, then, let x = a. 
 
 When a is substituted for x, f{x) becomes /(a) by the defini- 
 tion of /(a) ; and the second member of (1) becomes • Q + jR, 
 
 ox R. 
 
 .-. f(a) = R; that is, i?=/(a). Hence, 
 
 Principle 1. — If f(x) is divided hy x—a, the remainder is 
 f(a). Consequently, 
 
 Corollary 1. — If f(a) = 0, that is, if a is a root of the equa- 
 tion f(x) = 0, f(x) is divisible by x — a (Factor Theorem), and con- 
 versely, if f(x) is divisible by x — a, then f{a) = and a is a root 
 of the equation f(x) = 0. 
 
 Principle 1 is sometimes called the Remainder Theorem. The 
 Factor Theorem, § 136, is a special case of the Remainder 
 Theorem. 
 
 If the coefficients of x"- +^xb""^ + ga;""- -| \-sx-\-tz=0 are all 
 
 integers, the first member cannot be divisible by x — a unless a 
 is a factor of t. Hence, 
 
 Corollary 2. — If f(x) has integral coefficients, every integral 
 root of f(x) = must be a factor of the absolute term. 
 
 660. Number of roots. 
 
 Since by Prin. 1, Cor. 1, to every root of the equation f(x) = 
 there corresponds a factor of f(x) of the form x — a, and, con- 
 versely, to every factor of this form there corresponds a root of 
 the equation f(x) — 0, it may be inferred that : 
 
 Principle 2. — Every equation of the nth degree in x has n and 
 only n roots. 
 
564 THEORY OF EQUATIONS 
 
 This principle cannot be proved until it has been proved that every equa- 
 tion has at least one root — a fact too difficult of proof for this work. But 
 assuming that every equation has at least one root, Prin. 2 may be estab- 
 lished conditionally as follows : 
 
 If fix) = x» + px"-! + gx» 2 ^ ... 4. sx 4- « = has a root, let it be a. 
 
 Dividing both members by a; — a, an equation of the (n — l)th degree is 
 obtained. If this has a root, let it be h. Dividing both members by a; — 6, 
 an equation of the (n — 2)th degree is obtained. By continuing this process 
 /(x) is resolved into n factors of the form x — a. Denoting these by x — a, 
 
 f{x)={x - a)(x- b) •'• (x - k). 
 
 Since the substitution of any one of the n numbers a, 6, •••, k will reduce 
 /(x) to ?ero, the equation f(x) = has these n numbers for roots. And 
 since the substitution of any other number than one of these will not reduce 
 /(x) to zero, /(x) = has no other roots. 
 
 Note. — In counting the roots, equal roots are regarded as different roots. 
 
 661. If a is a root of the equation f(x) = 0, then, by Prin. 1, 
 Cor. 1, fix) is divisible by a; — a. Removing the root a by divid- 
 ing both members of f{x) = by a? — a is called depressing the 
 equation fix) = to an equation of the next lower degree, and 
 this equation is called the depressed equation. 
 
 662. Horner's method of synthetic division. 
 
 Horner's method of synthetic division is an abridgment of the 
 method of division by detached coefficients. 
 
 In dividing 5.'c'+8x*+12a^+24a^-5x+7 by.ar'+2a^+3ic+5, 
 
 the process of division by detached coefficients is modified as 
 
 follows : 
 
 Dividend 
 Divisor 
 
 with last three 
 signs changed 
 
 Quotient 5 — 2 + l| 3+ 2 + 2 Remainder 
 
 Explanation. — By using the first term of the divisor as a trial divisor 
 and changing the sign of each term of the divisor ailer the first, it is possible 
 to substitute for the process of subtracting the partial products that of adding 
 the products obtained by multiplying the terms of the divisor after the first, 
 icith their signs changed, by the successive terms of the quotient. 
 
 Thus, dividing the coefficient of the first term of the dividend by the coeffi- 
 cient of the first term of the divisor, 5 is obtained for the coefficient of the 
 
 1 
 
 5+ 8 + 12 + 24- 5 + 7 
 
 2 
 
 -10-15-25 
 
 3 
 
 + 4+ 6 + 10 
 
 5 
 
 - 2- 3-5 
 
THEORY OF EQUATIONS ' 565 
 
 first term of the quotient. Multiplying the coefficients of the other terms of 
 the divisor with their signs changed by 5 gives the second horizontal line of 
 the process, — 10 — 15 — 25, which is added to the dividend. At present, 
 however, only the term — 10 is added. 
 
 Adding + 8 and — 10, the terms in the second column, and dividing the 
 sum by 1, the coefficient of the first term of the divisor, — 2 is obtained for 
 the coefficient of the next term of the quotient. Multiplying the coefficients 
 of the last three terms of the modified divisor by this term of the quotient 
 gives the third horizontal line of the process. 
 
 Adding + 12, — 15, and + 4, the terms in the third column, and dividing 
 the sum by the coefficient of the first term of the divisor, + 1 is obtained for 
 the coefficient of the next term of the quotient. Multiplying the coefficients 
 of the last three terms of the modified divisor by this term of the quotient 
 gives the fourth horizontal line of the process. 
 
 Adding + 24, — 25, + 6, and — 2, the terms in the fourth column, the 
 sum is 3, which stands for 3 a;2. Since this term is of lower degree than the 
 first term of the divisor, 5 x^ — 2 a; + 1 is the entire quotient, and 3 or? is 
 the first term of the remainder. Adding the last two columns, the entire 
 remainder is 3 x'-' + 2 x + 2. 
 
 663. When the divisor is a binomial of the form x-^a, the 
 first term is not written, and the second term with its sign changed 
 is written at the right of the dividend. Also, since each partial 
 product consists of but one term, all the partial products are 
 written in the same line under the dividend. For example, the 
 process of dividing a^ + 4a^ + 7a; + 6 by x — 2 is as follows : 
 
 -Dividend 1 + 4+ 7+ 6L2 
 
 Partial products 2 + 12 + 38 
 
 Quotient 1 + 6 + 19 I 44 Remainder 
 
 That is, the quotient is o:^ + 6 ic + 19 and the remainder 44. 
 
 It is seen that the process of dividing f(x) by x — 2 is identical 
 with the process of substituting 2 for x by detached coefficients 
 (§ 628), the remainder being the value of the function when x=2. 
 
 This illustrates the meaning of Prin. 1. 
 
 Examples 
 Divide by synthetic division : 
 
 1. ic* + a^ - 3 a^ - 17 a; - 30 by « - 3 ; by a^ + 2. 
 
 2. af + 4a^ — Bar' — a; + 2bya; — 1; bya; + l. 
 
566 THEORY OF EQUATIONS 
 
 3. 2 X* — 7 ic^ — 16 ar'^ — ic- +- 32 X — 10 by ic — 5 ; by a; — 1. 
 
 4. 2a;5_5^4_^4,p3_22a; + 21 by ar-2a; + 3. 
 
 664. To find the commensurable roots of f{x) = by trial. 
 
 Examples 
 1. Solve f{x) = a:^ - 8 x* +- 15 a.-3 + 20 0^ - 76 a; + 48 = 0. 
 
 SOLUTIOK 
 
 1 
 
 - 8 + 15 + 20 - 76 + 48 Li 
 
 1 _ 7 + 8 + 28-48 
 
 1- 
 
 - 7 + 8 + 28-48 
 2 _ 10 - 4 + 48 
 
 L2 
 
 1 
 
 -5- 2 + 24 
 -2 + 14-24 
 
 
 1-2 
 
 1-7 + 12 
 That is, a;2 - 7 a: + 12 = 0, or (x - 3)(x - 4) = 0. 
 
 .-. X = 1, 2, - 2, 3, 4. 
 
 Explanation. — By Prin. 1, Cor. 2, every integral root of the equation 
 
 must be a factor of 48. Substituting 1 for x the result is zero, and therefore 
 1 is a root of the equation ; also since the process of substituting 1 for x is 
 identical with that used to divide /(x) by x — 1, the remainder is zero, and 
 the depressed equation is x* — 7 x^ + 8 x'^ + 28 x — 48. This equation must 
 have the other roots of the given equation. 
 
 Similarly, removing the root 2 from the depressed equation, the next 
 depressed equation is x^ — 5 x^ — 2 x + 24 = 0. Removing the root — 2 from 
 this equation, the quadratic x^ — 7 x + 12 = is obtained, whose roots are 3 
 and 4. 
 
 Hence, the roots of /(x) = are 1,2, — 2, 3, and 4. 
 
 Solve by trial : 
 
 2. a;3-da^ + 23a;-15 = 0. 4. :^-7x + Q = Q>. 
 
 3. a;3_i0x^ + 29a;-20 = 0. 5. a;*-9x2 + 4x-f 12 = 0. 
 
 6. a;3-10x2_^33^_3e^0. 
 
 7. x-*-9a^4-21a:'+-a;-30 = 0. 
 
 8. 5x*~2x'-^ox'-lQx + 12 = Q. 
 
 9. 2a^-x*-12:»?-\-lx' + lQx-12==^{i. 
 
THEORY OF EQUATIONS 567 
 
 665. Newton's method of divisors. 
 
 The following method of limiting the number of factors of the 
 absolute term to be tried for roots is rarely expedient but is useful 
 as a check. 
 
 Let a be an integral root of the equation 
 
 y(x) = a;" + pyf-^ + qx""'^ + ••• + rx^ + sx + t = 0, 
 and suppose that all the coefficients are integers. 
 Then, a» + pa"-^ + Q-a""^ + ••• + ra- + sa-\-t = 0. 
 Transposing and dividing by a, 
 
 - = — s — ra — •••• — qa"-" —pa""'^ — a" 
 a 
 
 ,»-3 /r.«»-2 
 
 Hence, - is an integer. 
 a 
 
 Denoting the quotient by Qj, and transposing — s, 
 
 Qi + s = — ra— ••• — qa"~^ — pa" ~^ — a"~\ 
 Dividing again by a, 
 
 V^ "' ^ — — r — ••• — qa'*-* —pa"~^ — a"~^, an integer. 
 a 
 
 Denoting this quotient by Q^ and continuing the process, the 
 result of the nth division is 
 
 Qn-l + P = _ 1, 
 a 
 
 Let d be an exact divisor of the absolute term of f(x) = 0. 
 Then, by the above discussion we have the following : 
 
 Rule. — Divide tJie absolute term hjy d^and add the quotient to the 
 coefficient of x; divide the sum by d, and if the quotient is an integer, 
 add it to the coefficient of x? ; continue this process as long as the 
 quotients are integers or to n divisions. 
 
 If each quotient is integral and the nth quotient is — 1, then d is a 
 root of the equation ; 'oih&nmse d is not a root. 
 
 Thus, — 2 is a root ot x^ — 3 x^ — 16 x — 12 = 0, but 3 and 2 are not roots, 
 as shown by the following processes : 
 
 1_ 3-16-12 L 
 -1+5+ 9 
 
 -2 
 
 1 - 3 - 16 - 
 - 4 
 
 -12 L3 
 
 1 - 3 - 16 - 12 [2 
 -7-11- 6 
 
 + 2-10 
 
 -20 
 
 
 - 14 - 22 
 
568 
 
 THEORY OF EQUATIONS 
 
 666. Relation between the roots and coefficients of an equation. 
 
 Since the reduced equation of the nth degree in x is the product 
 of n equations of the first degree of the form a; — a = 0, an equa- 
 tion may be formed from its roots by subtracting each from x and 
 equating to zero the product of the n binomials thus formed. 
 
 Thus, suppose x = a, b, c, d. 
 
 Then, x — a = 0, x — b = 0, x — c = 0, x — d = 0. 
 
 .-. (x — a) (x — b)(x — c) (x — d)=0, 
 
 or 
 
 — a 
 
 a? -\- ah 
 
 y? — abc 
 
 - b 
 
 + ac 
 
 — abd 
 
 — c 
 
 -\-ad 
 
 — acd 
 
 -d 
 
 -f be 
 
 -bed 
 
 + bd 
 
 
 
 + cd 
 
 
 X + abed = 0. 
 
 Comparing the coefficients of the above equation with its roots 
 a, b, c, d, with their signs changed, the following rule for finding 
 the coefficients when the roots are given is obtained : 
 
 Rule. — Tlie coeffieient of the first term of fix) =0 is 1 ; the 
 coefficient of the second term is the sum of the roots with their signs 
 changed; the coefficient of the third term is the sum of the products 
 of the roots with their signs changed, taken two at a time; and, in 
 general, the coefficient of the (r + 1)^^ term is the sum of the products 
 of the roots with their signs changed, taken r at a time. 
 
 The absolute term is the product of all the roots with their signs 
 changed. 
 
 1. If the second term is wanting, the sum of the roots is 0. 
 
 2. If the absolute term is wanting, one root at least is 0. 
 
 Examples 
 Form the equations whose roots are : 
 
 1. 1, 2, 3. 3. - 1, - 3, 4. 
 
 2. 2, 3, 5. 4. 1, 1, - f 
 
 5. 2, -|±V5. 
 
 6. 1, -^±^y/^3. 
 
 667. It has been shown that the relations subsisting between 
 the roots and coefficients of an equation may be expressed by n 
 ■conditional equations, sufficient to determine the coefficients when 
 
THEORY OF EQUATIONS 569 
 
 the roots are known. It might be supposed, therefore, that when 
 the coefficients are known the roots may be found by aid of their 
 relations to the coefficients. It is not advantageous, however, to 
 endeavor to find the roots in this way, except when some particu- 
 lar relation between the roots is given in addition. 
 
 Thus, if the roots of x* + px- + qx + r = are to be found by means of 
 their relations to the coefficients p, q, and r, denoting the roots by a, 6, and c, 
 
 a + b + c = -p, 
 ab + ac+ be = q, 
 abc =— r. 
 
 Eliminating any two of the three unknown numbers, as b and c, it is found 
 that the solution of the system depends upon the solution of an equation of 
 the form 
 
 a^ + pcfi + 5'a + r = 0, 
 
 which is as difficult to solve as the given equation, being of the same form. 
 
 The following examples illustrate the method of solving an 
 equation by means of the general relations between its roots and 
 coefficients when the roots also have some given relation. 
 
 Examples 
 
 1. Solve the equation 48 a^ — 74 a?^ + 37 a; — 6 = 0, whose roots 
 are in geometrical progression. 
 
 Solution. — Represent 
 
 the roots by - , a, and ar. 
 r 
 
 
 
 Then, § 666, 
 
 a 74 
 
 --- a- ar = - — , 
 
 r 48 
 
 
 (1) 
 
 d 
 
 r 48 
 
 
 (2) 
 
 Dividing (2) by (1), 
 
 — a = — i, or a 
 
 = h 
 
 (3) 
 
 Substituting in (1), 
 
 r = 1 or f . 
 
 
 (4) 
 
 Testing these values in the third equation of condition by which the abso- 
 lute term is found from the roots, 
 
 (-»)(-„K-a0.or-a.=-A = ^(i)', 
 
 by (3) they satisfy this relation. 
 
 Hence, the roots are f , ^, f , when r = f ; 
 or f , J, §, when r = f . 
 
570 THEORY OF EQUATIONS 
 
 2. Solve 9a^— 7a; + 2 = 0, if one root is double another. 
 
 3. Solve 9 a^ — 27 a;^ + 23 a; — 5 = 0, whose roots are in A. P. 
 
 4. Solve the equation a^ — 12 a; + 16 = 0, which has two equal 
 roots. 
 
 5. Solve 2 a;* — 7 .T^ — 6 a^ + 44 a; — 40 = 0, which has three 
 equal roots. 
 
 6. Solve 72a;*+90a^-5a^- 40a;- 12 = 0, two of whose 
 roots are numerically equal but opposite in sign. 
 
 7. Solve 6a^ — lla^ + 6a; — 1 = 0, whose roots are in H.P, 
 
 8. Find the relation subsisting between the coefficients of the 
 equation aj* + pa? -{- qx-\-r = Q, when two roots are numerically 
 equal but opposite in sign, and by the aid of this relation solve 
 the equation a:^ — 5 a;^ — 4 a; + 20 = 0. 
 
 668. Fractional roots. 
 
 If possible, suppose that -, a rational fraction in its lowest 
 
 h 
 
 terms, is a root of the equatioirf* 
 
 f{x) = a;" + 29a;»-i + q^-^ -\ 1- sa; + ^ = 0, 
 
 whose coefficients 1, p, q, •••, s, t are integers. 
 
 Substituting this root for x and multiplying both members 
 
 by 6"-\ 
 
 — 4-i)a"~^ + qa^'-^b -\ h sa6»-2 + tb"-'^ = ; 
 
 b 
 
 whence, ^ = — (pa"-^ + qa^'-^b H + sab""-^ + tb"'^), 
 
 b 
 
 which is impossible, because the first member is a fraction in its 
 lowest terms and the second member is an integer. Hence, 
 
 Principle 3. — If the coefficient of the highest power of x in 
 f(x) = IS 1 and all the other coefficients are integers, no rational 
 fraction can be a root of the equation. 
 
 669. Imaginary roots and quadratic factors of f(x). 
 
 In the following pages every complex or pure imaginary root, that is, 
 every root involving an imaginary number, will be called an imaginary root. 
 
 Let a 4- 6 V— 1 and a —bV—1, in which a and b are real, b 
 is positive, and a is positive or zero, be substituted for x in an 
 equation f(x) = having positive integral exponents and real 
 coefficients. 
 
THEORY OF EQUATIONS 571 
 
 Theu, j\a + 6 V^l) = (a + 6 V^n[)" +i:'(a + 6 V^H^)""^ H 
 
 + s(a + 6V^^) + ^, (1) 
 
 and /(a - b V^^) = {a-h V^H!)" + p{a-b V^^)""' + ••• 
 
 + s(a-6V=^) + f. (2) 
 
 Since all the exponents are positive integers, (1) and (2) may 
 be fully expanded. Each will then consist of terms of three 
 kinds, those which do not involve 6, those which involve 
 even powers of 6V— 1, and those which involve odd powers 
 of 6 V — 1. Terms belonging to the first and second classes will 
 be real, and to the third class, imaginary. 
 
 Since the two complex numbers are alike except in the terms 
 involving h, the sum of the terms that do not involve h is the 
 same in (2) as in (1). Represent this sum by S. 
 
 Since any even power of — 6 V— 1 is equal to the same power 
 of 6 V— 1, the sum of the terms involving even powers of 6 V— 1 
 is the same in (2) as in (1). Represent this sum by P. 
 
 Since any odd power of — 6 V — 1 is numerically equal to the 
 same power of 6 V— 1 but opposite in sign, if the sum of those 
 terms that involve odd powers of 6 V— 1 in (1) is denoted by 
 Q V— 1, the sum of the corresponding terms in (2) is — Q V— 1. 
 
 Hence, /(a + b V^^) = S + P+Q V^ (3) 
 
 and /(a - b V^T) = S + P-Q V^l. (4) 
 
 Now suppose that one of the conjugate complex or imaginary 
 
 numbers, as a + 6V— 1, is a root of the equation f{x) = 0. 
 Then, S + P+Q V^^ = 0. (5) 
 
 Since the real number (S + P) can cancel no part of the 
 
 imaginary number Q V— 1, each must be equal to zero. 
 
 Therefore, S + P-Q V^^ = ; (6) 
 
 that is, a — & V— 1 also is a root of the equation. Hence, 
 
 Pbinciple 4. — If the equation f(x) = has real coefficients, 
 its imaginai-y roots, if any, occur in conjugate pairs. 
 
 CoROLLARiKS. — 1. To every pair of imaginary roots there corre- 
 sponds a quadratic factor of f(x) of the form (x — aY + h^. 
 
 2. If the equation f{x) =■ is of an odd degree, it has at least 
 one real root. 
 
572 THEORY OF EQUATIONS 
 
 Examples 
 
 1. One of the roots of the equation ic*— 2a^+10ar^— 8a;+24=0 
 isl+V— 5. Solve the equation. 
 
 Solution. — Since the coefficients are all real, the given root must enter 
 the equation with its conjugate in the quadratic factor 
 
 (x - 1 - \/^5) (x - 1 + V^), or x^ - 2 X + 6. 
 
 Removing this factor by synthetic division, the depressed equation is 
 x2 + 4 = 0, whose roots are 2 V— 1 and — 2 V— 1. 
 
 Hence, the roots of the given equation are 1 ± V— 5, ± 2^— 1. 
 
 Solve the following, having given the roots indicated : 
 
 2. 2ar'-lla^.+ 8a; + 7 = 0;3+V2. 
 
 3. 2a^-lla^ + 16a; + ll = 0;3-V^2. 
 
 4. a;<-2a^-ar= + 6a;-3 = 0; -i(3 + V^=^). 
 
 5. ic* + 3ar' + 5a^ + 4a; + 2=0; -1(1+^/"^). 
 
 6. 2ar>-a;^ + 4a^-2a;2 + 2a;-l = 0; V^ V^^. 
 
 TRANSFORMATION OF EQUATIONS 
 
 670. Frequently by transforming an equation its solution may 
 be made to depend upon that of an equation that is easier to solve. 
 
 For example, it is easier to substitute positive integral values 
 of the unknown number than negative or fractional values. 
 
 Hence, to find the negative roots of f{x) = by trial, first trans- 
 form to an equation whose roots are numerically equal to the roots 
 of fix) = but opposite in sign, and then search for the positive 
 roots of the transformed equation. 
 
 671. To change the signs of the roots. 
 
 Given f{x) = a;" + px"-^ + qx''-^ -\ \-sx-\-t = 0. (1) 
 
 Put — y for x. Then, the terms of (1) involving odd powers 
 will have their signs changed and 
 
 yn _ pyn-l _^ gyn-2 sy + t = (2) 
 
 or - y+py"-' - 92/"-' -\ sy -\-t = 0, (3) 
 
 according as n is even or odd. Changing all signs in (3), 
 
 yn _ pyn-i ^ gyn-2 _^gy_t=,0. (4) 
 
THEORY OF EQUATIONS 573 
 
 Since —y = x,y = — x and the roots of (2) and (4) are those of 
 (1) with their signs changed. (2) is obtained by changing the 
 signs of odd powers and (4) by changing the signs of even powers. 
 
 Principle 5. '— An equation in x may he transformed into another 
 having the same roots ivith opposite signs by changing the signs of 
 the terms involving either the odd or the even powers. 
 
 To change the signs of the roots ofar* + x3 + 2x + 3 = and of x^ + 2 x^ 
 + 4 a;2 + X + 5 = 0, and leave the highest power positive, change the signs 
 of odd powers in the first instance and of even powers in the second. The 
 transformed equations are, using y for the changed roots, 
 
 y4_y3_2y + 3 = and y^ + 2 y^ - ^y'^ + y - b = 0. 
 
 The absolute terras, 3 and 5, or 3 x" and 5 x°, are regarded as even powers. 
 
 672. To multiply the roots by a constant. 
 
 Given fix) — a;" + px""-^ + gx"-- H 1- sa; + < = 0, (1) 
 
 and m a constant by which each root is to be multiplied. 
 
 Let y = mx, (2) 
 
 whence, x =—• Substituting — for x in (1) and multiplying both 
 members of the resulting equation by m", 
 
 2/" 4- mpy""-^ + m^qy''~^ + • • • -f m'^'^sy + m^t = 0, (3) 
 
 whose roots by (2) are m tiines those of (1). Hence, 
 
 Principle 6. — An equation f(x) = may be transformed into 
 another whose roots are m times those of the given equation by multi- 
 plying the successive coefficients by 1, m, m^, m% •••, m", respectively. 
 
 To multiply the roots of x^ — 6 x^ + 11 x — 6 = by 2, the successive coef- 
 ficients are multiplied by 1, 2, 4, 8, respectively, and y or some other letter is 
 used for the unknown number in the transformed equation. Hence, the equa- 
 tion whose roots are twice those of the given equation is 
 ys _i2yi + Uy -48 = 0. 
 
 673. The principal use of the preceding transformation is to 
 clear equations of fractional coefficients and at the same time make 
 the coefficient of the highest power of x unity. 
 
 If x2 _ |a; -I- 3 = is cleared of fractional coefficients without changing the 
 roots f and 2, the coefficient of x^ in the resulting equation will not be 1, 
 But multiplying the roots by 2, giving y'^-ly + 12 = 0, the equation is cleared 
 of fractional coefficients and the coefficient of the highest power of the 
 unknown number is still 1. 
 
574 THEORY OF EQUATIONS 
 
 Examples 
 
 Transform the following equations to equations whose roots 
 are ten times the roots of the given equations : 
 
 1. x'+2x-+x-4.=0. 3. ar'-.9a-2+. 23a; -.015=0. 
 
 2. a^-Sar -4:X+12=0. 4. x*-.18a^+. 032 a; -.0015=0. 
 5. Transform 8 .r'* — 12 cc^ + 4 a;- — 10 .t — 3 = to an equation 
 
 with integral coefficients, the leading coefficient being 1. 
 
 
 
 
 
 Solution 
 
 
 
 
 Dividing by 
 
 8, 
 
 
 a;*- 
 
 -fa:3 + 
 
 ix'- 
 
 ■ix 
 
 -i-- 
 
 = 0. 
 
 Multiplying 
 
 the roots 
 
 ; by m 
 
 , Prin. 6, 
 
 
 
 
 
 
 
 2/4- 
 
 2 ^ 
 
 ^fy- 
 
 5m^ 
 
 2 
 
 y - 
 
 3to4 
 
 8 
 
 = 0, 
 
 whose coefficients are evidently integral when m = 2. 
 
 Substituting 2 for »n, y* - 3 yS + 2 j/'^ - 20 ?/ - 6 = 0, 
 an equation whose roots are double those of the given equation. 
 
 Transform the following equations to the reduced form/(a;)=0 
 having integral coefficients • 
 
 6. 4a;*+3a^-2a;+9=0. 9. 7?-\x'+j\x-J^=0. 
 
 7. 16 0^3-12 ar'+10a;-7=0. 10. (a;-lf(3.T-l)2=50. 
 
 8. 2a^— V-a^+-V- 2^-11= 0- H- 5a;^+.la^+.125x+.2=0. 
 
 12. On page 299 it is shown that the three cube roots of 1 are 
 1, ^ (— 1 ± V— 3). Prove by Prin. 6 that the three cube roots of 
 a? are a, ia(— 1 ± V— 3). 
 
 Find the three cube roots of 8, 27, — 8, and — 1. 
 
 674. To decrease the roots by a given number. 
 
 Given J{x) = .r" + p.T"-' + qx"-- + • ■ + sx + t = 0, (1) 
 
 and let it be required to find an equation of the same form whose 
 roots are h less. 
 
 First Method. Let y =:x — h, whence x = y + h. 
 
 Substituting (y + h) for a; in (1), 
 (?/ 4- /O" + PQ^ + hy-' + g(y + h)"-' + • • • 4- s(y + J>) + f = 0. (2) 
 
 Since y = x — h, (2) is the required transformed equation before 
 reduction to the same form as (1). 
 
THEORY OF EQUATIONS 
 
 575 
 
 Expanding the powers of {y -j- h) in (2) and arranging the result 
 in descending powers of y, 
 
 y'^ + nh y«-' + ^ — 11 /i2 
 
 + P 
 
 + (n- l)pli 
 
 + q 
 
 2/»-2.j 1_ /i« 
 
 + 
 + 
 
 
 + sh 
 + t 
 
 = 0, 
 
 (3) 
 
 which is the reduced form of the transformed equation. 
 
 Second Method. Except when n is small the expansion of the 
 powers of {y + h) is laborious. To discover an easier method of 
 finding the coefficients of 2/"~S y"~^ •••in the transformed equation 
 (3), represent these coefficients by P, Q, •••. 
 
 Then, ?/" + Py""^ + Q?/"-^ -\ \-Sy+T=0 (4) 
 
 is an equation obtained from (1) by substituting y + h for x, or 
 y for X — h. Therefore, if a; — ^ is substituted for y in (4), the 
 first member of the resulting equation must be identically equal 
 to the first member of (1) ; that is, 
 
 a;" + px"-^ 4- qx'"-^ -\ \- sx + t = {x — hy -\- P(x — h)"-^ 
 
 + Q{x - hy-' +...+S{x- h) + T (5) 
 
 is an identical equation. 
 
 From the form to which we have reduced f(x) in the second 
 member of (5), it is evident that T, the last coefficient of the 
 transformed equation, is the remainder when f{x) is divided by 
 X — h; S, the preceding coefficient, is the remainder when the 
 quotient is divided hy x — h; etc. 
 
 Hence, to transform an equation of the nth degree in x into 
 another whose roots are h less, 
 
 Rule. — Substitute y — h for x in the given equation and reduce 
 the resulting equoMon. 
 
 Or, divide f{x) hy x — h until the remainder does not involve x, 
 divide the quotient by x — h until the remainder does not involve x, 
 and so continue to n divisions. Tlie first coefficient of the given 
 equation together with the successive remainders taken in the reverse 
 order will be the coefficients of the transformed equation. 
 
676 THEORY OF EQUATIONS 
 
 Examples 
 1. Decrease the roots of 2 ar* - 19 a^ + 59 a; — 60 = by 2. 
 First Solution. — Let x - 2 = y. Then, x = y + 2. Substituting, 
 
 2(2/ + 2)3 -19(2/ + 2)2 + 59(2/ + 2) -60 = 0. (1) 
 
 Reducing, 2 j/S - 7 r/^ + 7 2/ - 2 = 0. (2) 
 
 Skcond Solution. — Dividing 2x3 — 19 x^ -). 59 a; — 60 by x — 2 as in the 
 following process, the quotient is 2 x^ - 15 x + 29 and the remainder is -2. 
 
 Dividing 2 x^- 15 x+29 by x-2, the quotient is 2 x- 11 and the remainder 
 is 7. 
 
 Dividing 2x — llbyx — 2, the quotient is 2 and the remainder is — 7. 
 
 Dividing 2 by x — 2, there is no quotient, and 2 itself is the remainder. 
 
 2 
 
 -19 
 
 59 
 
 -60 L2 
 
 
 4 
 
 -30 
 
 58 
 
 2 
 
 -15 
 
 29 
 
 -2 
 
 
 4 
 
 -22 
 
 
 2 
 
 -11 
 4 
 
 7 
 
 
 21 -7 
 
 Hence, the transformed equation is 2 2/' — 7 2^2 ^ 7 y _ 2. 
 In practice the first detached coefficient is written but once. 
 
 Transform the following to equations whose roots are 3 less : 
 
 2. a^ — 12 cc^ + 47 a; — 60 = 0. Prove by solving both equations. 
 
 3. a;<-4a^-19ar' + 46a; + 120 = 0. 
 
 4. 3a;*-19a^ + 21ar^+31a; + 12 = 0. 
 
 5. a;* - 2.75 x2 + .5 a; + 4.5 = 0. 
 
 675. To increase the roots by a given number. 
 
 Since increasing the roots of an equation by h may be regarded 
 as decreasing them by — h, the rule for increasing the roots by h 
 is the same as that given for decreasing the roots by h, except 
 that y+h and x+h take the place of y — h and x—li, respectively. 
 
 Examples 
 Tiansform the following to equations whose roots are 5 greater: 
 1. a;*+-9ar5 + 29a^ + 39a;+-18 = 0. 
 
THEORY OF EQUATIONS 511 
 
 2. ics + 20 a^ + 131 aj + 280 = 0. 
 
 ^ 4. or'^ + loa^-f 71 CC + 105 = 0. 
 
 676. To remove any assigned term except the first. 
 
 It is often desirable to transform an equation into another 
 lacking a certain term. This is done by decreasing the roots by 
 such a number h that the coefficient of the assigned term in the 
 transformed equation shall be zero. 
 
 Keferring to the formula for the equation whose roots are h 
 
 less than those of the given equation, § 674, (3), it is seen that 
 
 P 
 the second term is lacking when nh+p= 0, that is, when h= ; 
 
 that the third term is lacking when h has such a value that 
 n{n-l) j^2 + (« - l)ph + g = ; etc. 
 
 Si 
 
 Examples 
 
 1. Transform a;^ + 6a^ + 5x— 12=0 to an equation lacking 
 the second power of the unknown number. 
 
 Solution. — In this equation p = 6 and n = 3. Therefore, the value of h 
 tliat makes nA + j> = is — 2 ; that is, the roots of the given equation should 
 be decreased by — 2. 
 
 6 
 
 6 
 
 -12L 
 
 -2 
 
 -2 
 
 -8 
 
 6 
 
 
 4 
 
 -3 
 
 - 6 
 
 
 -2 
 
 -4 
 
 
 
 2 
 
 -7 
 
 
 -2 
 
 
 
 
 
 Hence, the transformed equation is 
 
 y3 4.0y2_7y_6 = 0, or«/3-7y-6=0. 
 2. Change 207* — 5a^ — a;4-4 = to the form o^ -f- oa; + ft = 
 with integral coefficients. 
 
 Solution. — The reduced form isa^ — ^x^ — ^x + 2 = 0. (1) 
 
 Since A = — - = 5, we first multiply the roots by 6 in order that h may 
 n 
 
 be an integer. 
 
 When y = 6 a;, (1) becomes y* - 15 y^ _ 18 y + 432 = 0. * (2) 
 
 In (2), p = - 15 and n = 3. .-. A = 5. 
 
 ADV. ALG. — 37 
 
678 THEORY OF EQUATIONS 
 
 Decreasing the roots of (2) by 6 as in Ex. 1, 
 
 when z = y-b, z^ - 93 ^ + 92 = 0. (8) 
 
 Since j/ = 6 a; and 2; = y — 6, « = 6a; — 5; that is, each root of (3) is 5 ]#ss 
 than 6 times the corresponding root of (1). 
 
 Change to the form a^ + ax + & = with integral coefficients, 
 and express the roots of the transformed equation in terms of 
 those of the given equation; 
 
 3. a^-12ic2_^43a._40 = 0. 6. a? + 3? -2 = 0. 
 
 4. a:3_3a;2_883,_240 = 0. 7. x' + ^ + x + 2 = Q. 
 
 5. a^-f 15a^ + 68a; + 96 = 0. 8. 2a^-a^-2a; + l = 0. 
 
 9. Transform a^-|-4a^ — 7a;^ — 22a;+24 = into an equation 
 lacking the third power of the unknown number. Solve. 
 
 677. To change the roots to their reciprocals. 
 
 Given f{x) = a;" + pa;"-^ + ^x"-^ -| \-ro^ + sx+t = 0. (1) 
 
 Let -=y. Then, a; = -. Substituting this value for x, 
 X y 
 
 yn yn-1 yn-i yl y 
 
 Clearing of fractions and arranging in descending powers of y, 
 
 ty^ + s2/»-i + 7^-2 + ... + 9/ +^^2/ + 1 =0 . (2) 
 
 Equation (2) is therefore the equation whose roots are the 
 reciprocals of the roots of (1). The transformed equation evi- 
 dently has the same coefficients as the given equation, but in the 
 reverse order. 
 
 DESCARTES' RULE OP SIGNS 
 
 678. In any series of algebraic numbers, every sequence of two 
 unlike signs is called a Variation of sign and every sequence of 
 two like signs is called a Permanence of sign. 
 
 In ac^ _ 2 X* + a;3 + 3 x2 — 5 a; — 4 there are three variations, H , }-, 
 
 -\ , and two permanences, + +, . In a;^ + 3 a;^ _ 5 x — 4, one variation 
 
 and two permanences occur. 
 
THEORY OF EQUATIONS 579 
 
 679. Let the signs of the first member of an equation in x be 
 
 + + + +0+- + +, 
 
 in which repr.esents the coefficient of each missing term. 
 Multiply by a; — a and so introduce a positive root. 
 
 + + + +0+- + + 
 
 + - . 
 
 + + + +0+- + 4- 
 
 -O-j-4- 0-+ 
 
 H ± + ±±- + - + ±- 
 
 Since the multiplicand has unlike signs, to each sign of the 
 multiplicand immediately following a like sign there corresponds 
 an ambiguous sign in the product. It will be observed, also, that 
 each ambiguous sign or set of ambiguous signs in the product 
 stands between two unlike signs, only interrupting a variation. 
 
 Therefore, if the ambiguous signs of the product and the corre- 
 sponding signs of the multiplicand are stricken out, the number of 
 variations will be unchanged in the multiplicand, and unchanged 
 or diminished in the product. 
 
 Hence, the least possible number of variations added by intro- 
 ducing the positive root a is found by comparing the remaining 
 signs of the multiplicand and product. 
 
 4-0--h0+--f 
 + + - + - + - 
 
 It is evident that each variation in the multiplicand corresponds 
 to at least one variation in the product, and that the product has 
 in addition one variation at the end, arising from multiplying the 
 last term of the multiplicand by x and then by — a. 
 
 Hence, each introduction of a positive root adds at least one varia- 
 tion of sign. 
 
 Let F(x) be the product of all the factors corresponding to the 
 negative and imaginary roots of f(x) = 0, and let a, b, •-• he the 
 positive roots. 
 
 Since the introduction of each positive root, when F(x) is 
 multiplied by (x — a), (x — b), •■•, causes an increase of at least one 
 in the number of variations of sign, in an equation f(x) = 0, the 
 number of positive roots may be equal to or less than the number 
 of variations, but cannot be greater than this number. 
 
680 THEORY OF EQUATIONS 
 
 Again, since X— x)=Q has the roots oif(x) = with their signs 
 changed, the negati-ve roots of f(x) = correspond to the positive 
 roots otf(^—x)= 0. Therefore, the number of negative roots of 
 f(x) = may be equal to, or less than, the number of variations 
 of sign in /( — . a;) = 0, but cannot be greater than this number. 
 Hence, 
 
 An equation f(x) = cannot have more positive roots than there 
 are variations of sign in f{x), nor more negative roots than there are 
 variations of sign inj\— x). 
 
 This is Descartes' rule of signs. 
 
 Note. — In applying Descartes' rule it will be understood that all zero 
 roots have been removed. Also af>, the power of x involved in the absolute 
 term, will be regarded as an even power of x. 
 
 680. The following may be deduced directly from Descartes' 
 rule: 
 
 1. If the signs of all the terms of f{x) are positive, the equation 
 f(x) = Jias no positive roots. 
 
 2. If all the terms of f(x) that involve even powers of x have the 
 same sign and all that involve odd powers of x have the opposite 
 sign, the equation f(x) = has no negaiive roots. 
 
 3. If the eqxiation f{x) = is complete, it cannot have more nega- 
 tive roots than there are permanences of sign in f{x). 
 
 The equation x^ + x* + a;^ + 5x + 4 = has no positive roots, for there are 
 no variations of sign in its first member. 
 
 The equation a* + x* — a;3_5x + 4=:0 has no negative roots, for there 
 are no variations of siga in /( — x) , or in x^ + x* + x^ + 5 x + 4. 
 
 The equation x* + 2x3 — Sx^— 2x + 2 = cannot have more than 2 
 negative roots, which is the number of permanences in /(x) ; for since the 
 
 equation is complete, each permanence of sign in + + \-, the signs of 
 
 /(x), corresponds to a variation of sign in H h +, the signs of /(— x). 
 
 681. Existence of imaginary roots. 
 
 It is often possible to detect the existence of imaginary roots 
 by applying Descartes' rule. For, first removing zero roots, if 
 any, if the sum of the greatest possible number of positive roots 
 and the greatest possible number of negative roots, computed by 
 Descartes' rule, is less than the whole number of roots, the number 
 of roots remaining must be the least possible number of imaginary 
 roots. 
 
THEORY OF EQUATIONS 581 
 
 Thus, «* + 2a; — 1=0 cannot have more than one positive root nor more 
 than one negative root. But the whole number of roots is four. Hence, 
 there are at least two imaginary roots. 
 
 Examples 
 Find the nature of the roots of the following : 
 
 1. a-*+4a^+6x2+5a;+2=0. 5. a;* - 4 ar'* + 27 = 0. 
 
 2. x-^- 3 ar'-x +3 = 0. 6. 2(?-dx-2 = 0. 
 
 3. a;*-10a;2_20.T-16 = 0. 7. ar'-l=0. 
 
 4. a;« + u;^ + a^ + X = 0. 8. a:^ + 1 = 0. 
 
 9. If the terms of f{x) are all positive and each involves an 
 even power of x, show that the roots of the equation f(x) = are 
 all imaginary. 
 
 10. Show that a;" — 1 = has two real roots when n is even, 
 one positive and one negative, but only one real root when n is 
 odd. 
 
 MULTIPLE ROOTS 
 
 682. If an equation has two roots each equal to a, a is called a 
 double root of the equation ; if an equation has three roots each 
 equal to a, a is called a triple root of the equation ; in general, if 
 an equation has m roots each equal to a, a is called a Multiple 
 Root whose order of multiplicity is m. 
 
 683. Let /(a;) = (a;-a)'»F(a;) = (1) 
 be an equation having m roots each equal to a. 
 
 Then, § 642, 
 
 f>(x)=^(x - arF(x) + (x - arfF(x) 
 
 = m{x — ay-'^Fix) + {x — a)'^F'(x) 
 = (x- a)'^-^lmF(x) + (x- a)F'{x)y (2) 
 
 By (2), f'(x) = has (m — 1) roots each equal to a. Hence, 
 Principle 7. — If a occurs m times as a root of an equation 
 f(x) = 0, it occurs (m— 1) times as a root of the first derived equation 
 f'(x)=0. 
 
 Thus, X* -6a;2 + 8a;- 3 = 0, or (x - l)^(x + S) = 0. has the root 1 
 occurring three times, and the derived equation 4 x^ — 12 a; + 8 = 0, or 
 4(x — l)"^(x + 2) = 0, has this root occurring twice. 
 
582 
 
 THEORY OF EQUATIONS 
 
 684. Let <f>{x) denote the H. C. D. of f(x) and f{x). 
 
 If f(x)=0 has no multiple roots, cl>(x) = 1 ; but if /(a;)=0 has 
 multiple roots corresponding to the factors (a; — a)"*, (x — by, •••, 
 then, Prin. 7, <t}{x) = (x—d)'^-\x — by-'^--. 
 
 Hence, if <f)(x) = can be solved, the multiple roots of f(x) = 
 may be found and removed. 
 
 Examples 
 
 1. Solve ic^-4a;« + 5ar'-6a;* + 32ar'-16a;-32 = 0. 
 
 Solution. — Let /(x) = a;^ - 4 a;6 + 5 a;5 _ 6 x* + 32 x'^ - 16 x - 32. 
 
 Then, /'(x) = 7 x" - 24 x» + 25 x* - 24 x" + 64 x - 16. 
 
 The H. C. D. of /(x) and /'(x) is found to be 0(x) = x^ - 3 x2 + 4. 
 
 Put x8 - 3 x2 + 4 = 0. 
 
 Solving by trial, the roots of 0(x) = are x = — 1, 2, 2. 
 
 Therefore, — 1 is a root of /'(x) = and 2 is a double root of /'(x) = 0. 
 
 Hence, Prin. 7, — 1 is a double root and 2 is a triple root of /(x) = 0. 
 
 Removing the corresponding factors (x + l)2(x — 2)^ from f(^x), the given 
 equation is depressed to the quadratic equation x^ + 4 = 0, whose roots are 
 2V'— 1 and — 2V— 1. Hence, the roots of the given equation are — 1, — 1, 
 2, 2, 2, 2V^, - 2V^^. 
 
 Solve the following equations : 
 
 2. x*-x'-3ar'-^5x-2=:0. 
 
 3. x^ + 6x* + 14:X^ + 20a^-\-24.x + 16 = 0. 
 
 4. a^-7x' + Ua^-2x'-15x + 9 = 0. 
 
 LOCATION OF REAL ROOTS 
 
 685. Let f(x) =x^ + 3x^ -llx- 5. Since, § 641, f(x) is con- 
 tinuous, if two numbers substituted for x 
 give results with opposite signs, f(x) passes 
 through zero and its graph crosses the 
 X-axis once or an odd number of times; 
 but if they give results with the same 
 sign, the graph does not cross the X-axis 
 during the interval or else crosses it an 
 even number of times. 
 
 For example, the given graph does not 
 cross the X-axis between — 4 and — 1, but crosses it once 
 
THEORY OF EQUATIONS 583 
 
 between — 1 and 0, twice between — 1 and 3, and three times 
 between — 5.2 and 3. 
 
 But each intersection gives a root of f{x) = 0. Hence, 
 
 Principle 8. — If two real numbers substituted for x in the first 
 member of the general equation f(x) = give results with opposite 
 signs, an odd number of real roots of the equation lie between the 
 numbers; but if they give results with the same sign, no roots or an 
 even number of roots lie between the numbers. 
 
 Corollary 1. — An equation of an odd degree has at least one 
 real root of a sign opposite to that of its absolute term. 
 
 For, if X = — 00, /(x) is negative ; if x = 0, /(x) has the sign of the abso- 
 lute term; if x=+oc, /(x) is positive. Hence, if the absolute term is 
 positive, /(x) changes sign for some value of x between — oo and ; but if 
 the absolute term is negative, /(x) changes sign for some value of x between 
 and + CO. 
 
 Corollary 2. — An equation of an even degree whose absolute 
 term is negative has at least two real roots, one positive and one 
 negative. 
 
 For, if X = — oo, /(x) is positive ; if x = 0, /(x) is negative ; if x = 4- oo, 
 /(x) is positive. Hence, /(x) changes sign for some value of x between 
 — CO and 0, and again for some value of x between and + oo. 
 
 Corollary 3. — Every equation whose absolute term is negative 
 has at least one real root. 
 
 686. Limits of the real roots. 
 
 687. If no positive real root of an equation can be greater 
 than a, then a is a Superior Limit of the positive real roots ; and 
 if no negative real root can be less than —b, then — 6 is an 
 Inferior Limit of the negative real roots. 
 
 688. Let f{x)=x''+px''-'^ + qx''-^-\ [-sx + t = (1) 
 
 be an equation having one or more of its roots positive. Then, 
 by Descartes' rule, some of the terms oif{x) iuust be negative. 
 
 To find a superior limit to the positive roots, let r denote the 
 number of positive terms preceding the first negative term, miss- 
 ing terms, if any, being counted ; and let — JV be the negative 
 coefficient of greatest absolute value. 
 
 For positive values of x, the sum of the terms beginning with 
 
584 THEORY OF EQUATIONS 
 
 the second and ending with the rth is positive, oi' if all are miss-' 
 ing, zero. Hence, dropping these terms will not increase f(x). 
 Again, making —N the coefficient of each term after the rth will 
 not increase f(x). 
 
 Therefore, every positive value of x that makes 
 
 a;" — iV(ic"-'' + tc"-'-^ H \-x + l) 
 
 positive makes f(x) positive, f{x) having the same or a greater 
 
 value. A^Ca^-'+i-n 
 
 x» —N(af-' + x"-'-^ H h a; + 1) = a;» — ^^^^ ±1, 
 
 X — 1 
 
 when a? > 1, > aj" 
 
 x — 1 
 
 Hence, when x>l, f(x) is positive if 
 
 X'" = or is positive : 
 
 x — 1 
 
 that is, if x" {x—1) — Nx'*~'"^^ = or is positive ; 
 that is, if af~'^(x — 1) — i\^= or is positive. 
 
 But when x>l, a;*-^ >(x — ly-^ and x"'^ (x — 1) > {x — Vf. 
 .: of-' {x-1)- N> (x - ly - N. 
 
 Hence, when a; > 1, f{x) is positive if 
 
 (a; — l)*" — ^= or is positive; 
 that is, if (x — ly is equal to or greater than N; that is, if x is 
 equal to or greater than ^/N^ 1. 
 
 Hence, for all values of x equal to or greater than -\/N-\-l,f(x) 
 is positive. 
 
 Since f{x)>0 for all values of x equal to or greater than 
 ■\/N+ 1, no real root of f(x) = can equal or exceed S/^+ 1 ; 
 that is, -\/N +1 is a superior limit of the positive real roots of 
 f{x)^0. 
 
 Thus, in a:* + «» - 7 a;2 - 13 x - 6 = 0, r = 2, and 2V = 13. Therefore, 
 VT3 + 1, or 4.6 + , is a superior limit to the positive roots. 
 
 When the superior limit found is not an integer, it is usually convenient 
 to take the next higher integer for a superior limit. For example, in the 
 illustration just given 6 may be taken for a superior limit. 
 
 689. To find an inferior limit to the negative roots of an equa- 
 tion, change the signs of all the roots by Prin. 5, and find a supe- 
 rior limit to the positive roots of the transformed equation. This 
 
THEORY OF EQUATIONS 585 
 
 number with its sign changed will be an inferior limit to the 
 negative roots of the given equation. 
 
 Examples 
 Find the integral limits to the real roots of 
 
 1. a:* + ar' + 7 a^- 8 a;-3 = 0. 
 
 2. ar'-3a;3-6a;2-14a;-8 = 0. 
 
 3. x« - 28 a^- 49 0-^ + 112.^ + 132 = 0. 
 
 4. a;«+ar^-4ic2 + 6ic-20 = 0. 
 
 5. a;'-a.-^ + 3a;'' + 76a;3-243ar' + 108 = 0. 
 
 690. The utility of the preceding principles and their failure 
 in certain cases may be illustrated in the solution of such an 
 equation asa^ — ar^ — 4ar + 3a; + 3 = 0. 
 
 Since no factor of the absolute term satisfies the equation, 
 Prin. Ij Cor. 2, the equation has no integral root. Also, Prin. 3, 
 the equation has no rational fraction for a root. Hence, the 
 equation has no commensurable root. 
 
 If the equation has a positive root, removing it would leave a 
 depressed equation of an odd degree with a negative absolute 
 terra. Since, Prin. 8, Cor. 1, such an equation has a positive root, 
 if the given equation has one positive root, it has two. Similarly, 
 if the given equation has one negative root, it has two. But we 
 cannot tell, without actual trial, whether the number of positive 
 roots or of negative roots is zero or two. 
 
 By § 688, if the equation has real roots, their superior limit is 
 5, and their inferior limit is — 3. " 
 
 Substituting - 2, - 1, 0, 1, 2, 3, 4, for x, 
 we obtain 5, - 2, 3, 2, 1, 30, 143, for f(x). 
 
 By Prin. 8, then, a negative root lies between — 2 and — 1, and 
 another between — 1 and 0. But without considerable labor in 
 substituting values between and 4, it cannot be determined 
 whether the other two roots are positive or imaginary. 
 
 When two roots of an equation are nearly equal or when there 
 are imaginary roots, the real roots are not found readily by trial. 
 In such cases the number of real and imaginary roots and the 
 situation of the real roots may be completely determined by 
 Sturm's theorem, which follows. 
 
686 
 
 THEORY OF EQUATIONS 
 
 691. Sturm's theorem. 
 
 Let f{x) = be an equation in x freed of multiple roots. 
 
 Then, f{x) and f\x) have no common factor involving x. 
 
 Divide J{x) by /'(x), as in the process of finding the H.C.D., 
 until the remainder is of lower degree than the divisor, introduc- 
 ing or rejecting positive monomial factors, if desired, hut not nega- 
 tive ones. Denote the remainder with its sign changed by F^jix). 
 Similarly, divide f'{x) by F^ix) and denote the remainder with its 
 sign changed by -^3(0;). Continuing in this way, since f(x) and 
 f'{x) have no common factor involving x, a remainder not involv- 
 ing x will finally be obtained. Denote this remainder with its sign 
 changed by F^(x). Then, J{x), f<{x), F^(x), F,(x), ■ ■ -, F^_^(x), fJx) 
 are called Sturm's functions. 
 Thus, let f(x) = a;3 + a.-2 - 2 X - 8, whence /'(a;) = 3 x2 + 2 x - 2. 
 3 + 2- 
 
 -13 
 
 3 + 15 
 
 -13-2 
 - 13 - 66 
 
 + 63 
 
 1 + 1- 
 
 2- 
 
 8 
 
 3 
 
 
 . 
 
 3 + 3- 
 
 U- 
 
 24 
 
 3 + 2- 
 
 2 
 
 
 1 - 
 
 4- 
 
 24 
 
 3 
 
 
 
 3- 
 
 12- 
 
 72 
 
 3 + 
 
 2- 
 
 2 
 
 + 14)- 
 
 14- 
 
 70 
 
 — 
 
 1 - 
 
 5 
 
 
 1 + 
 
 5 
 
 F^ix) = X + 6. 
 
 i?'3(x)=-63. 
 
 Hence, Sturm's functions are x3+ x^ — 2 x — 8, 3 x^ + 2 x — 2, x + 5, — 63. 
 
 Sturm's theorem may be stated as follows : 
 
 If a and b are two real numbers, b greater than a, then the number 
 of variations of sign obtained by writing f (a), f (a), F2(a), •••, F„(a) 
 in order, less the number of variations obtained by writing f(b), f'(b), 
 Fi(b), •••, FJb) in order, is equal to the number of real roots of 
 f{x) = that lie betioeen a and b. , 
 
 The proof is as follows : 
 
 1. Since fix) and f'{x) have no common factors, by § 148 no 
 two successive functions have a common factor; that is, no two 
 successive functions become zero for the same value of x. 
 
 Let Fr{x) be any of Stm-m's functions except f(x) and Fn(x). 
 
THEORY OF EQUATIONS 587 
 
 By the definition of Sturm's functions, if i^r-i(^) is divided by 
 Fr{x), tlie remainder is — F^^^{x). Represent the quotient by Q. 
 
 Then, § 102, 2^,_i(^) = QF^x) - F^^lx) 
 \iF,{x)=0, =-F,.^,(x); 
 
 that is, if any intermediate fundioyi becomes zero for any value of x, 
 the aOjax^ent functions have opposite signs for that value of x. 
 
 Let c' be a root of F^x) = 0, so that F^{c') — 0; and let c' — h 
 and c' + /t be so close to c' that no roots of the adjacent functions 
 lie between c' — h and c' + h. 
 
 Then, the adjacent functions have opposite signs for x =c' and 
 keep these signs while x increases from c' — ^ to c' + h. 
 
 Since both before and after x = c' the sign of F^x) is like that 
 of one of the adjacent functions and unlike that of the other, 
 ichen X is passing through a root of any intermediate function, 
 Sturm's functions maintain the same number of variations. 
 
 2. Let c be a root of f(x) = 0, so that /(c) = 0, and let h be a 
 positive number, as small as we please. 
 
 Then, /(c — h) denotes the value of f{x) just before, and /(c + h) 
 denotes the value of f{x) just after, x passes through the root c. 
 
 § 640, /f(c - h) = f{c)- hf\c) +,|V"(c)- -, 
 
 and /(c + h) = f{c)+ hf'(c) +~f"ic)+ .... 
 
 Since f(c) = 0, if h is taken sufficiently small the signs of 
 /(c — h) and f(c + h) will be the same as those of their numeri- 
 cally greatest terms, — hf'(c) and hf'(c), respectively. Then, h 
 being positive, f(c — h) has the sign opposite to that of /'(c), and 
 /(c -f h) has the same sign as f'(c). 
 
 That is, as x increases through a root of f(x) = 0, f[^x) changes 
 from a number having the sign opposite to that of f'(x) to a 
 number having the same sign as f'{x). Hence, as x increases 
 thi'ongh a root off(x) — 0, Sturm's functions lose one variation. 
 
 Therefore, since, as jx increases continuously from a to b, 
 Sturm's functions lose one variation every time x passes through 
 a root of ^f{x) = 0, but lose none when x passes through the roots 
 of the other functions, the number of variations lost between a and 
 b indicates the number of real roots between a and b. 
 
588 THEORY OF EQUATIONS 
 
 692. Since all the negative roots of an equation lie bet\ve»-.i 
 — 00 and 0, if — 00 is substituted for x in Sturm's functions and 
 then is substituted, the number of variations lost gives the 
 number of negative roots of the equation. 
 
 Similarly, if is substituted for x in Sturm's functions and 
 then 4- 00 is substituted, the number of variations lost gives the 
 number of positive roots of the equation. 
 
 Examples 
 
 1. Locate the roots of x-* — x^ — 4 a;^ + 3 a; -f- o = 0. 
 
 Solution. — In § OiH) it was found that one negative root of this equation 
 was situated between — 2 and — 1 and another between — 1 and 0. Tlie 
 nature and situation of the other two roots are found as follows : 
 
 Let f{x) = X* - x3 - 4 x2 + 3 X + 3. 
 
 Then, /'(x) = 4 x^ - 3 x^ - 8 x + 3. 
 
 Employing the process of finding the H.C.I), of /(x) and /' (x) , taking care 
 not to introduce or reject negative factors, it is discovered that/(x) and/'(x) 
 have no common factor. Hence, /(x) = has no multiple roots, and /(«) 
 and /'(x) are the first two of Sturm's functions. 
 
 Representing the successive remainders with their signs changed by i^2(x), 
 i?3(x), and Fi(x), Sturm's functions are: 
 
 /(x) = X* - x3 - 4 X--2 + 3 X + 3 
 /'(x) = 4x''-3x2-8x + 3 
 i?'2(x) = 35x2-28x-51 
 Fsix) = 11 X - 18 
 Fiix) = + 
 
 Substituting — co, 0, and + oo for x in these functions, 
 
 X 
 
 Kx) 
 
 /'(x) 
 
 F,(x) 
 
 Fs(x) 
 
 F^ix) 
 
 variations 
 
 — 00 
 
 + 
 
 - 
 
 + 
 
 — 
 
 + 
 
 4 
 
 
 
 + 
 
 + 
 
 - 
 
 - 
 
 + 
 
 2 
 
 + 00 
 
 + 
 
 + 
 
 + 
 
 + 
 
 + 
 
 
 
 Since two variations are lost between — oo and 0, by Sturm's theorem 
 there are two negative roots, as before ascertained ; and since two variations 
 are lost between and + co, the other two roots are real and positive. 
 
 By § 688, 5 is a superior limit of the positive roots. 
 
 Substitute in Sturm's functions 0, 1, 2, 3, •••, or as many of these numb«>i^« 
 as may be necessary to lose two variations. 
 

 
 
 THEORY 
 
 OF EQUATIONS 
 
 
 58£ 
 
 z 
 
 /(x) 
 
 f'C^) 
 
 Fi(x) 
 
 Fs(^) 
 
 F,(x) 
 
 variations 
 
 
 
 + 
 
 + 
 
 - 
 
 — 
 
 + 
 
 2 
 
 1 
 
 + 
 
 - 
 
 - 
 
 - 
 
 + 
 
 2 
 
 2 
 
 + 
 
 + 
 
 + 
 
 + 
 
 + 
 
 
 
 It is discovered tliat two variations are lost between 1 and 2 ; therefore, 
 there are two positive roots between 1 and 2. 
 
 Locate the real roots of the following equations and find the 
 number of imaginary roots : 
 
 2. 2ar^-15x'2 4-32a;-21 = 0. 
 
 3. 16ar'-4x-2-80a; + 75 = 0. 
 
 4. a:^-2ar^-7ar' + 10x + 10 = 0. 
 
 5. a;*-x-2_l0x-4 = 0. 
 
 693. Homer's method of solving numerical equations. 
 
 Commensurable roots expressed by two or more figures may be 
 found exactly by Horner's method, and real incommensurable roots 
 may be found to any degree of accuracy required. 
 
 1. Let it be required to find approximately the one positive 
 real root of the equation 
 
 f(x)=x' + 3x'-nx-5 = 0. (1) 
 
 Substituting 0, 1, 2, 3, ••• for x, the corresponding values of the 
 first member are /(O) = - 5, /(I) = - 12, /(2) = - 7, /(3) = 16, • • -. 
 
 Since f(x) is negative when x = 2 and positive when x = 3, the 
 one positive root must lie between 2 and 3 (Prin. 8). 
 
 Transforming (1) to an equation whose roots are 2 less, 
 
 3 
 
 -11 
 
 -5 
 
 2 
 
 10 
 
 -2 
 
 5 
 
 - 1 
 
 -7 
 
 2 
 
 14 
 
 
 7 
 
 13 
 
 
 2 
 
 
 
 9 
 
 tho first transformed equation is 
 
 f + 9f + 13y-7 
 
 (2) 
 
690 
 
 THEORY OF EQUATIONS 
 
 Since y = x — 2 and x lies between 2 and 3, the corresponding 
 value of y lies between and 1. Try .5. 
 
 19 13 -7 [^ 
 
 .5 4.75 8.875 
 
 9^5 17.75 ! 1.875 
 Since ^-f92/^ + 13y — 7 is equal to 1.875 when y = .5 and is 
 equal to — 7 when y = 0, the root of this equation must be less 
 than .5. Trying .4, it is found that the function does not change 
 sign. Hence, the root lies between .4 and .5. 
 
 Transforming (2) to an equation whose roots are .4 less, 
 
 9 
 
 13 
 
 .4 
 
 3.76 
 
 9.4 
 
 16.76 
 
 .4 
 
 3.92 
 
 9.8 
 
 1 20.68 
 
 .4 
 
 
 -7 
 6.704 
 
 Li 
 
 .296 
 
 ^ 
 
 .^1 
 
 10.2 
 
 the second transformed equation is 
 
 z« + 10.2 z^ + 20.68 z - .296 = 0, (3) 
 
 in which z = y — A = (x — 2) — A = x — 2.4. 
 
 Since the positive root of (2) lies between .4 and .5 and each 
 root of (3) is .4 less than the corresponding root of (2), the 
 positive root of (3) lies between and .1. Since this value of z 
 is a small fraction, in (3) the higher powers of z are small as 
 compared with the first power. Hence, the positive root of (3) is 
 very nearly the same as the root of 
 
 20.68 z - .296 = 0, or 20.68 z = .296. 
 
 Therefore, z = .01 +, and the next figure of the root is 1. 
 Transforming (3) to an equation whose roots are .01 less, 
 
 1.01 
 
 10.2 
 
 20.68 
 
 -.296 
 
 .01 
 
 - .1021 
 
 .207821 
 
 10.21 
 
 20.7821 
 
 - 088179 
 
 .01 
 
 .1022 
 
 
 10.22 
 
 20.8843 
 
 
 .01 
 
 
 
 lO-J^? 
 
THEORY OF EQUATIONS 
 
 591 
 
 the third transformed equation is 
 
 1^ + 10^23 v" + 20^8843 v - ^088179 = 0, 
 in which v = z — .01 = x— 2.41. 
 
 (4) 
 
 The next figure of the root is found by neglecting the higher 
 powers of v in (4) and solving the equation 
 
 20.8843 V - .088179 = 0, 
 
 which gives ^ = .004+. Hence, the root is 2.414 +. 
 
 The preceding transformations and the work of substituting 
 .004 for V in (4) to test the last figure may be written in one 
 process, as shown below. The coefficients of the successive trans- 
 formed equations are in heavy-face type. 
 
 f 
 
 ^ 
 
 3 -11 
 2 10 
 
 
 
 
 - 5 1 2.414 
 
 -2 
 
 5 
 2 
 
 -1 
 
 14 
 
 
 -7 
 6.704 
 
 7 
 2 
 
 13 
 
 3.76 
 
 
 
 -.296 
 
 .207821 
 
 9 
 
 .4 
 
 16.76 
 3.92 
 
 
 
 - .088179 
 
 .083700944 
 
 9.4 
 .4 
 
 9.8 
 .4 
 
 
 20.68 
 
 .1021 
 
 20.7821 
 
 .1022 
 
 
 
 - .004478056 
 
 10.2 
 
 .01 
 
 10.21 
 
 ■01 
 
 10.22 
 .01 
 
 
 20.8843 
 
 .040936 
 20.925236 
 
 
 
 10.23 
 
 .00 
 
 4 
 
 
 
 10.234 
 
 The broken lines mark the close of each transformation. 
 If more figures of the root are desired, the last transformation 
 may be completed and the process continued to any number of 
 
592 
 
 THEORY OF EQUATIONS 
 
 figures of the root by successive transformations in each of which 
 the root is decreased by tlie largest number that will not cause the 
 absolute term to change sign. 
 
 The solution of an equation f{x) = 0, as for example 
 
 ic3 + 3a^-llx-5 = 0, 
 
 (1) 
 
 whose positive root has just been found to three decimal places, 
 is often facilitated by plotting the graph oif{x). 
 
 Plotting corresponding values of x and f{x) in (1), the graph 
 given on page 582 is obtained. It shows that the equation has 
 two negative roots, and a positive root lying between 2 and 3. 
 
 The steps in Horner's process of approximation to the positive 
 root 2.414+ are illustrated graphically as follows : 
 
 In the accompanying figure the positive root is represented 
 graphically by the abscissa OP. Now if the F-axis is moved 
 parallel to itself 2 units toward the right to the position of 0' Y', 
 the abscissa of every point in the graph, including that of P, will be 
 decreased by 2. For example, the abscissa of P will become O'P. 
 
 Hence, moving the Y-axis 2 units toward the right has the effect 
 of decreasing the roots of the given equation by 2. 
 
THEORY OF EQUATIONS 593 
 
 The new axes, then, are O'X and 0' Y', the new equation, that 
 is, i\ie first transformed equation is 
 
 2/' + 92/^ + 132/-7 = 0, (2) 
 
 in which y = x — 2, and the curved line PiP^Pz • • • Pg, called the 
 graph of a:^ 4- 3 a*- — 11 cc — 5 with reference to the old axes, be- 
 comes the graph of i/* + 9 ?/^ + 13 y — 7 referred to the new axes. 
 
 Since y = x — 2, equation (2) has a root between and 1. 
 Accordingly, divide the portion of the X-axis between the second 
 and third divisions into tenths and plot the points Pi, P^, P^, P^, 
 Pg, corresponding to y — .1, .2, .3, .4, .5, respectively. It is found 
 that between .4 and .5 the ordinates change from negative to posi- 
 tive. Therefore, the graph crosses the X-axis between y= A and 
 y = .5, that is, between x = 2.4 and x = 2.5. 
 
 Hence, the F-axis may be moved .4 of a unit nearer to P than 
 has been done, without causing the origin to pass beyond P. 
 Making this transfer, then, the new axes are 0"Xand 0"Y", and 
 the second transformed equation is 
 
 z" + 10.2 z^ -f 20.68 z - .296 = 0, (3) 
 
 in which z = y — A = x — 2.4. 
 
 Since z = y — A, equation (3) has a root between and .10. 
 
 Dividing the portion of the X-axis between the divisions 2.4 
 and 2.5 into ten equal parts, each .01 of a unit in length, it is 
 found by substituting .01, .02, ••• for z in (3) that the first mem- 
 ber changes sign between z = .01 and z = .02, that is, the graph 
 crosses the X-axis between x = 2.41 and x = 2.42. 
 
 Hence, the F-axis may be moved .01 of a unit nearer to P, 
 without causing the origin to pass beyond P. This decreases the 
 roots of (3) by .01 and gives the third transformed equation 
 
 v" 4- 10.23^^-1- 20.8843 V -.088179 = 0, (4) 
 
 in which v = z — .01 = x — 2.41. 
 
 This process of approximation can be continued indefinitely. 
 
 Graphically considered, then, Horner's method of approximation 
 consists in moving the F-axis toward the right, first, the greatest 
 number of units possible without causing the origin to pass 
 beyond the intersection of the graph and the X-axis, then the 
 
 ADV. ALG. — 38 
 
594 
 
 THEORY OF EQUATIONS 
 
 1 
 
 ;ible 1 
 
 greatest possible number of tenths, then the greatest possible 
 number of hundredths, etc. The approximation to the root at 
 any stage is the total distance the origin has been moved. 
 
 2. The process on page 591 ilhistrates the general method of 
 extending a positive root to any number of figures after finding 
 enough figures to separate it from other roots, if any, that are 
 nearly equal to the root sought. 
 
 The following process illustrates certain modifications of 
 Homer's method for roots that are nearly equal, as explained 
 
 \5A 
 
 aelow. 
 
 Let a* 
 
 -12ar^ + 38 
 
 ,af 
 
 + 8 
 
 a; - 112 = 
 
 0. 
 
 
 1 
 
 -12 38 
 
 
 8 -112 
 
 
 6 -35 
 
 -7 3 
 
 
 16 
 
 115 
 
 
 23 
 
 3 
 
 
 6 -10 
 
 
 
 -35 
 
 - 2.9824 
 
 
 -2 -7 
 
 
 
 -12 
 
 1 .0176 
 
 
 5 
 
 15 
 
 
 
 4.544 
 - 7.456 
 
 
 
 3 
 
 8 
 
 
 
 
 5 
 
 3.36 
 
 
 
 5.952 
 
 
 y" 
 
 8 
 
 11.36 
 
 
 
 -1.504 
 
 
 
 .4 3.52 
 
 
 
 
 
 8.4 14.88 
 
 
 
 
 
 .4 
 
 3.68 
 
 
 
 
 
 8.8 
 
 18.56 
 
 
 
 
 
 .4 
 
 
 
 
 
 9.2 
 .4 
 
 
 
 
 
 
 
 
 2* 9.6 
 
 Explanation. — By Sturm's theorem it is found that the equation has a 
 negative root and three positive roots, one between 2 and 3, and two between 
 5 and 6. The above process is concerned with the roots between 5 and 6. 
 
 Decreasing the roots by 5, the first transformed equation is 
 
 yi + 8y^ + Sy'--l2y + S = 0, (1) 
 
 which has only two positive roots, both between and 1, the least positive 
 root having been passed, that is, changed to a negative root. This is indi- 
 cated by the change of sign of the absolute term from — to 4- and the loss 
 of one variation of sign. 
 
 Since the first transformed equation has ticn roots between and 1, the 
 rule that the next figure of the root is the greatest number of tenths by 
 
THEORY OF EQUATIONS 595 
 
 which the roots can be decreased without changing the sign of the absolute 
 term is not sufficient to guard against passing the roots, for it is possible, by 
 using too large a number, to pass both roots at the same time and still leave 
 the absolute term positive in consequence of tioo changes in sign. 
 
 The roots may be separated, however, by applying the rule that the equa- 
 tion loses one variation of sign for each positive root that is passed when 
 the roots are decreased by a number greater than the root in question. 
 
 If the roots of (1) are decreased by .4, as shown in the process, or by any 
 less number of tenths, the absolute term remains positive, and besides, no 
 variations are lost. Hence, both positive roots are greater than .4. But if 
 the roots are decreased by .5 or by any greater number of tenths, two varia- 
 tions are lost. Hence, both positive roots are less than .5. 
 
 That is, both positive roots of (1) lie between .4 and .5, and the corre- 
 sponding roots of the given equation lie between 5.4 and 5.5. 
 
 The second transformed equation 
 
 z* + 9.6 z^ + 18.56 22 _ 1.504 z + .0176 = (2) 
 
 has two roots between and .10. Trying .01, .02, .03, •••, it is found that 
 the absolute term changes sign and one variation is lost in passing from .01 
 to .02, and again in passing from .06 to .07. Hence, the roots of (2) are 
 .01+ and .06 + , and the corresponding roots of the given equation are 5.41 + 
 and 5.46 + . 
 
 The roots are now separated and the above process may be continued to 
 find either of them by the easier method given in 1. 
 
 In finding the less root 5.414 + , the absolute terms will continue to be 
 positive ; in finding the greater root 5.464 +, they will be negative beginning 
 with the third transformed equation. 
 
 Rule for Positive Roots. — Find the first figure of the root by 
 trial or by Sturm's theorem, and decrease the roots by this number. 
 
 If the root is separated from other roots, find each succeeding 
 figure of the root by decreasing the roots by the greatest number of the 
 next loioer order that will not cause the absolute term to change sign. 
 
 If the root is not seiJarated from other roots, first separate the 
 roots by Sturm's theorem ; or decrease the roots by the greatest 
 number of the next lower order that will not cause the equation to 
 lose more variations than there are positive roots that are less than 
 the root sought. 
 
 After several figures of the root have been found, the next figure may be 
 found u.'^ally by neglecting powers of the unknown number higher than the 
 first and solving the resulting simple equation ; or, if the coefficient of the 
 first power is small in comparison with the coefficients of the higher powers 
 or is equal to zero, by neglecting the powers higher than the second and 
 solving the resulting quadratic equation. 
 
 ADV. ALG. — 38 
 
596 
 
 THEORY OF EQUATIONS 
 
 694. To avoid decimals and to abbreviate the work when many 
 figures of the root are required, Horner's process is modified as 
 explained below. Compare with the process on page 591. 
 
 f 
 
 ^ 
 
 v^ 
 
 3 -11 
 2 10 
 
 
 
 
 
 — 
 
 5 12.41421 
 
 2 
 
 6 
 
 2 
 
 -1 
 
 14 
 
 
 — 
 
 7000 
 
 6704 
 
 7 
 2 
 
 1300 
 
 376 
 
 
 
 - 296000 
 
 207821 
 
 90 
 
 4 
 
 
 1676 
 392 
 
 
 
 - 88179000 
 
 83700944 
 
 94 
 4 
 
 206800 
 
 1021 
 
 
 
 
 - 4478O5600f) 
 
 4193648 
 
 98 
 4 
 
 
 207821 
 1022 
 
 
 
 - 2844080^ 
 
 209704 
 
 1020 
 
 1 
 
 1021 
 
 1 
 
 
 
 20884300 
 
 40936 
 
 20925236 
 
 40952 
 
 
 
 - 74704(ji^ 
 
 1022 
 1 
 
 
 2096618/^ft)() 
 
 205 
 
 
 
 10230 
 
 4 
 
 
 
 2096824 
 205 
 
 
 10234 
 4 
 
 10238 
 4 
 
 
 209702^p(f) 
 1 
 
 209704 
 1 
 
 
 
 i 
 
 
 20970^|!)p 
 
 
 ijD;80 
 
 
 
 
 
 
 Explanation. — 1. Decimals are avoided by multiplying the roots of each 
 transformed equation by 10, the numbers substituted for y^ z, v, ••• being 
 regarded as 4, 1, 4, ••• instead of .4, .01, .004, •••. 
 
 2. After several figures of the root have been obtained (in this case 
 beginning with the fourth transformed equation), nearly as many more 
 are accurately found by a contracted process. For since dividing both mem- 
 bers of an equation by the same known number does not change the roots, 
 
THEORY OF EQUATIONS 597 
 
 after the roots of the fourth transformed equation have been multiplied by 
 10, we may divide each coefficient by 1000 ; and if the decimal parts thus 
 formed are cut off, the accuracy of the root will not be affected for several 
 figures. Some allowance for the decimal cut off is made, when convenient. 
 Thus, 102.420x2 is nearer 205 than 204 and is taken as 205. 
 
 In practice the two operations of multiplying the roots by 10 and dividing 
 the equation by 1000 are performed simultaneously by omitting to annex 
 the ciphers and then cutting off the last figure of the coefficient of the first 
 power, the last two figures of the coefficient of the second power, and so on. 
 In the above process this practically destroys the leading coefficient and re- 
 duces the equation to the quadratic 102 vfl + 2096619 w — 4478056 = 0, next to 
 the quadratic u'^ + 209703 u — 284408 = 0, and finally to the simple equation 
 
 20970 t - 74704 = 0. 
 
 Solving this equation by dividing 74704 by 20970, t = 3.562+. 
 Hence, the root of the given equation is 2.4142135624- . 
 
 695. To find the value of a negative root, Horner's method 
 may be applied to find the corresponding positive root of that 
 transformed equation whose roots are numerically equal to the 
 roots of the given equation, but opposite in sign. The resulting 
 positive root with its sign changed will be the required negative 
 root. 
 
 696. Horner's method may be used to find the principal nth. 
 root of any number. For example, the principal fifth root of 2 
 may be found by solving the equation o^ — 2 = by Horner's 
 method. 
 
 Examples 
 
 1. Find by Horner's method the root lying between 1 and 2 of 
 the equation 4 ic^ — 3 a^ + .r — 14 = 0. Test the result by multi- 
 plying the roots by 4 and substituting 4 times the root found. 
 Find the imaginary roots. 
 
 2. The equation 8 ar' — 29 cc^ -f 29 a; — 21 = has a root between 
 2 and 3. Find all the roots. 
 
 3. The number and situation of the real roots of the equation 
 x'' — a^ — 4a^-f3a; + 3 = have been discussed in example 1, 
 page 588. Find all the roots to the third decimal place. 
 
 4. Find all the roots of .r* - 6 ar^ -f 5 a;^ -f- 14a; — 4 = to the 
 nearest third decimal place. 
 
 5. Find the fifth root of 330383.69407 by Horner's method. 
 
698 THEORY OF EQUATIONS 
 
 697. Newton's method of approximation. 
 
 If one and only one real root of the equation 
 
 f{x) = a;" +pa;''-^ + qx''-^ -\ \-sx + t = (1) 
 
 lies between a and b, and if a is less than this root by a number h 
 that is small in comparison with a, then a + /i is the root, and in 
 
 (a + hy + p{a + hy-'^ + q{a + /i)"-^ -| \- s{a-^h) + t = (2) 
 
 the terms involving h"^, h% h*, ••• may be neglected. 
 Then, approximately, 
 
 a" +pa^-^ +qa"-^ + \- sa + t 
 
 + [wa»-i + p{n — l)a"-2 + q{n — 2)a"-3 H h s]^ = 0; 
 
 that is, §639, /(a)+/'(a)fe = 0, or^ = -^. (3) 
 
 Thus, to approximate to the root of x^ — 2x— 5 = lying between 2 
 and 3, denote the root by 2 + ^ as in (2), neglect the terms involving h^ 
 and h? in (2 + hY — 2(2 + A) — 5 = 0, and solve the resulting equation 
 lOA — 1 = 0. The result is A = .1, approximately. 
 Or proceed as follows, using the result given in (3) : 
 f{x)=x^ - 2x - 5 and/''(a;)=: 3x2 - 2. 
 .-. /(2) = 8 - 4 - 5 = - 1 and /'(2) = 12 - 2 = 10. 
 Hence, by (3), h = — .1^ = .1, and the root is 2.1, approximately. 
 Next, using 2.1 for a, it is found that /(2.1)= .061 and /'(2.1)= 11.23. 
 
 Hence, by (3), the next addition to the root is — , or — .006, and the 
 
 ^ ^ 11.23 
 
 root is 2.1 — .005, or 2.095, approximately. 
 
 Note. — This method possesses but little practical value. 
 
 RECIPROCAL BQUATIOITS 
 
 698. When each root of an equation is the reciprocal of some 
 other root or of itself, the equation is called a Reciprocal, or Recur- 
 ring, Equation. 
 
 699. Since the only numbers that are reciprocals of themselves 
 are 1 and — 1, there are four kinds of reciprocal equations : 
 
 1. Those whose roots in pairs are reciprocals of each other. 
 
 2. The first kind with the root 1 in addition. 
 
 3. The first kind with the root — 1 in addition. 
 
 4. The first kind with the roots 1 and — 1 in addition. 
 
THEORY OF EQUATIONS 699 
 
 The four kinds may be illustrated, respectively, as follows: 
 
 1. 6xt + 5a;3-38x=^ + 5x+6 = ; roots, 2, |.-3, -\. 
 
 2. 2x3 - 7 x'^ + 7 a: - 2 = ; roots, 2, \, 1. 
 
 3. 4x3-13x2- 13x + 4 = 0; roots, 4, |, -1. 
 
 4. 4 X* - 17 x3 + 17 X - 4 = ; roots, 4, \, 1, - 1. 
 
 It is evident that equations of the second, third, and fourth 
 kinds may be depressed, by dividing by a? — 1, a; + 1, and x^ — 1, 
 respectively, to equations of the first kind. 
 
 Hence, the first kind is the standard reciprocal equation. 
 
 700. Let f{x) = mx" + pa;""^ + ga;«-2 h \-r2i?-\-sx+t=0 (1) 
 
 be a reciprocal equation. 
 
 Then, for all values of x that make f{x) = 0, 
 
 ^-\ (2) 
 
 Substituting this value of x in (1), multiplying both members 
 of the resulting equation by a;", reversing the order of the terms, 
 and changing signs throughout in case t is negative, 
 
 ± te" ± sa;"-^ ± rar"-- ± •• • ± ga;^ ± pa; ± m = 0, (3) 
 
 the upper signs being used when t is positive and the lower ones 
 when t is negative. 
 
 Since (1) and (3) have the same roots, their first members are 
 composed of the same factors and, consequently, are identical. 
 
 .-. m=zt, p = s, q=.r,-'-; or else m = — t, p = —s, q = —r, •••. 
 
 Hence, in a reciprocal equation, the coefficients of the terms equally 
 distant from the first and last terms of f(x) are equal when the last 
 term is positive, and numerically equal with opposite signs when the 
 la^t term is negative. 
 
 701. Let f{x) = ± ^a;" ± sof-' ± ra;"-=' + \-ra? + sx + t = 
 
 be a reciprocal equation, the lower signs corresponding to a 
 negative absolute term. 
 
 If n is odd, f(x) has an even number of terms that may be 
 grouped in pairs having equal or numerically -equal coefficients. 
 If the corresponding terms have like signs, by § 111, 4, each of 
 the groups t(x" + 1), sx(x^~^ + 1), rx^{x^~* + 1)> *" contains a; + 1 
 
600 THEORY OF EQUATIONS 
 
 as a factor, whence — 1 is a root of the equation and the equation 
 is of the third kind; but if the corresponding terms have unlike 
 signs, by § 111, 1, a; — 1 is a factor of each group, 1 is a root of 
 the equation, and the equation is of the second kind. 
 
 If n is even, f(x) has an odd number of terms and there is a 
 middle term provided the equation is complete. If corresponding 
 terms have like signs, the middle term satisfies the condition that 
 terms equally distant from the ends are equal, and the equation 
 is of the first kind ; but if corresponding terms have unlike signs, 
 the middle term cannot satisfy the condition that terms equally 
 distant from the ends have unlike signs except when it is zero. 
 In this case, then, the middle term is missing and f(x) takes the 
 form 
 
 f{x) = i(a;" - 1) + sa;(aj"-2 — 1) + ra^(x''-* - 1) H . 
 
 Since n, n — 2, n — 4, ••• are even, by § 111, 1 and 2, f(x) is 
 divisible by both x — 1 and x + 1. Hence 1 and — 1 are roots and 
 the equation is of the fourth kind. 
 
 It should be observed that only standard reciprocal equations 
 have an odd number of terms in the first member. 
 
 702. Let pffl?"^ +2)iar''"-i _| f-p^x™ ^ \-p^x +Po = (1) 
 
 represent a standard reciprocal equation. 
 
 Grouping corresponding terms and dividing by «"•, 
 
 Po(a^ + a;-™) + p,{x^-' + a^-C"-^)) + • • • + i>„_,(a^ -f x-"^ 
 
 +P^-i{x+x-')+p„ = Q. (2) 
 
 By the binomial formula, 
 
 {x + x-'^f = a^ + 2 +x-'- = (a^ + a;-^) + 2, 
 
 (a: -f x-y = ar' + 3 a; + 3 x-i + --c"' = (x" + x'^) + 3 (a; + a;"^), 
 
 {x + x-y = «* + 4 a,-2 + G + 4 a;-2 + x'* = (a;* + x'') + 4(a^ + x'") + 6, 
 
 and so on. Transposing and substituting y for x + x~'^, 
 
 x" -f- .T-2 = (;c + x-y -2 = f-1, 
 
 x" + x'^ = {x-\- x-y - 3(a; + x"^) = y^-3y, 
 
 a^ + X-* = (x + x-y - W + x-^) - 6 
 
 = 2^-%2- 2) -6 = 2/^-4/4-2, 
 
THEORY OF EQUATIONS 601 
 
 and so on. Continuing this process until the value of a?"* + a?"*" is 
 found in terms of t/™ and lower powers of y, and substituting all 
 these values in (2), the transformed equation after reduction takes 
 the form 
 
 M™ + qiy""^ ^ h qm-\y + ^m = 0. Hence, 
 
 Principle 9. — Every standard reciprocal equation may he trans- 
 formed into an equation of half its degree. 
 
 Other reciprocal equations may be depressed to the standard form by 
 dividing by x + 1, x — 1, or both (§ 699), and then transformed to equations 
 of half the degree of the depressed equations. 
 
 Examples 
 
 1. Solve 10 a;*- 77 a^ + 150 a;2- 77 a; + 10 = 0. 
 Solution. — Grouping the terms with like coefficients, 
 
 10(x* + 1) _ 77 x(x^ + 1) + 150x2 = 0. 
 Dividing both members by x^, 
 
 Putting in the quadratic form by adding and subtracting 20, 
 10(x2 + 2 + l^-77^x + i]+ 130 = 0. 
 
 Let X + - = y ; then, 10 j/2 - 77 y + 130 = 0. 
 
 X 
 
 Factoring, (2 y - 5) (5 y - 26) = 0. 
 
 ,.,^, + U5^,26. 
 X 2 5 
 Solving these two equations as quadratics, x = 2, ^, 5, ^. 
 
 The given equation is a standard reciprocal equation. 
 
 2. Solve 10a^-67a;*-|-73ar'^ + 73a^-67a; + 10 = 0. 
 
 Solution. — Grouping terms with like coefficients, it is evident that the 
 first member is divisible by x + 1. 
 Depressing to the standard form, 
 
 10-67+ 73+ 73-67 + 101-1 
 
 - 10 + 77 - 150 + 77-10 
 10 - 77 + 150 - 77 + 10 
 The rest of the solution is as given for Ex, 1. 
 .•.x = -l, 2, i, 6,^. 
 
602 THEORY OF EQUATIONS 
 
 3. Solve a:" + 2 ic* - 4 »* + 8 a;'' - 4 aj2-f 2a; + 1=0. 
 
 Solution. — The equation Is a standard, reciprocal equation. 
 Grouping terms with like coefficients and dividing by x^, 
 
 (^+i) + 2(..+ i)-4(. + l)+8=0. 
 
 If ic + - = 2/, as in § 702, a;^ + — = w2 - 2, and x^ + j_ - yS _ 3 „ 
 X x:^ 'x? 
 
 Substituting these values and reducing, the given equation is transformed 
 into the cubic equation 
 
 y8 + 2 j/2 _ 7 2, + 4 = 0, 
 
 whose roots, by trial, are found to be 1, 1, — 4. 
 Solving the quadratics y = x + - = l,l, or— 4, 
 
 X 
 
 a;=Kl+ v^3), ^(l_V33), ^(i + V^), 1(1 -V^), -2+V3, -2-^3. 
 The student should show that these roots are reciprocals in pairs. 
 
 4. Solve 2a;^-15a;'» + 37x3_37a^^l5a,_2==o. 
 
 5. Solve 12a;^ + 23x^-135ar''-135a;2^23a; + 12 = 0. 
 
 6. Solve a;«-2aT'-7a;* + 7ar^ + 2a;-l = 0. 
 
 7. Solve a;« + a;^-14a;* + 17a;3-14ar' + a; + l=0. 
 
 8. Transform a;* + a;^4-5a;^ + 4a;54-9a;* + 4af' + 5a:2 + a; + l = 
 into a quadratic equation. 
 
 BINOMIAL EQUATIONS 
 
 703. An equation of the form a" = a, or a;" — a = 0, is called a 
 Binomial Equation. 
 
 Thus, x^ — 1 = and x^ + 1 = are binomial equations. 
 
 704. The n nth. roots of a real number. 
 
 1. The n nth roots of unity are the n roots of the equation 
 iB» - 1 = 0. 
 
 By Descartes' rule, if n is odd, this equaition has one real root, 
 + 1, and n—1 imaginary roots ; if n is even, it has two real roots, 
 -f- 1 and — 1, and n — 2 imaginary roots. 
 
THEORY OF EQUATIONS 603 
 
 Since af — 1 and its first derived function nx^~^ have no com- 
 mon factor, af — 1 has no multiple roots. 
 
 Hence, the n nth roots of unity are all imaginary except one when 
 n is odd or two when n is even, and all are different. 
 
 2. The n nth. roots of any real number a are the n roots of the 
 binomial equation af = a, or ic" — a = 0. 
 
 Let any nth root of unity be represented by a (the Greek letter 
 alpha) and let Va be the principal nth root of a. 
 
 Then, (a Va)» = a" ( Va)" = 1 . a = a ; 
 
 that is, the n nth roots of a are found by multiplying the corre^ondr 
 ing nth roots of unity by the principal nth root of a. 
 
 705. Relation between the n nth roots of unity. 
 Let a be any imaginary nth root of unity. 
 Then, a" = 1, 
 
 and if m is a positive integer, 
 
 (a")"* = 1, or (tt")" = 1 ; 
 that is, a" is an nth root of unity. 
 
 Hence, any power of an imaginary nth root of unity is equal to 
 some one of the nth roots of unity. 
 
 706. The three cube roots of a number. 
 
 On page 299 it is shown that the three cube roots of 1 are 1, 
 \{— 1 +V— 3), and |-(— 1— V— 3), the roots of tiie equation 
 x^ - 1 = 0. 
 
 The student may verify the following statements : 
 
 The sum of the cube roots of 1 is zero. 
 
 The two imaginary cube roots of 1 are conjugates, squares, and 
 reciprocals of each other. 
 
 The cube roots of 1 are equal to any three consecutive powers of 
 
 i(-i+V^). . __ 
 
 In the following pages |(— l+V— 3) will be represented by w 
 (the Greek letter omega), and [i( — 1 + V^)]^ or i(-l — V^^), 
 by o)''. 
 
 Then, the cube roots of 1 are 1, <o, and w^; or w, wVand w^; or 
 ii?, <o', and to* ; etc. For w^ = 1, w* = w^ • w = w, etc. 
 
604 THEORY OF EQUATIONS 
 
 Let a be any of the cube roots of I. Since the equation 
 a? — a? = () gives the three cube roots of a^, or of I, and since this 
 equation is derived from a;^ — 1 = by multiplying the roots of 
 the latter by a, the cube roots of I are a, aw, and aur. 
 
 Hence, the three cube roots of any number are obtained from any 
 one of them by multiplying by 1, w, and w^, respectively. 
 Thus, the three cube roots of 64 are 4, 4 w, and 4 w^. 
 
 ALGEBRAIC SOLUTIONS 
 
 707. So far the theory of equations has been applied to the 
 solution of numerical equations and has been found sufficient to 
 determine, exactly or approximately, the real roots of numerical 
 equations of any degree. In the following pages the theory of 
 equations will be applied to the solution of certain literal equations. 
 
 708. The general, or algebraic, solution of a literal equation con- 
 sists in finding such an expression involving the general coeffi- 
 cients as will represent all of the roots at the same time. 
 
 Thus, the algebraic solution of ax^ + fta; + c = consists in finding the 
 
 expression x = — which has two values, inasmuch as 6^ _ 4 ac 
 
 2 a 
 has two square roots. 
 
 It is only a matter of common agreement that these two values shall be 
 denoted by ± before the radical, which is then taken to mean the positive 
 square root only. 
 
 In the following pages algebraic solutions of cubic and biquadf 
 ratic equations will be given. In 1825 the mathematician Abel 
 demonstrated the impossibility of obtaining algebraic solutions of 
 equations of higher degree than the fourth. 
 
 709. Cardan's solution of the cubic equation. 
 
 By §§ 673, 676, the general cubic equation may be transformed 
 into an equation of the form 
 
 x^-\-ax-\-b = 0. (1) 
 
 Since any number can be separated into two parts having a 
 given product, it is permissible to suppose that each root of (1) 
 may be divided into two parts whose product is — \a. 
 
 Let the parts be y and z. 
 
THEORY OF EQUATIONS 605 
 
 Then, x = y+z, (2) 
 
 and 3 2/2: = — a. (3) 
 Substituting the values of x and a in (1), 
 
 2/3 + 28 4- & = 0. (4) 
 Irom (3), z = — ^. Substituting in (4), 
 
 32/ 
 
 ^-2^ + ^ = ^' 
 
 a' 
 
 or y6^6^_^=0. (5) 
 
 Solving (5) as a quadratic in 2/^, 
 
 
 whence, by (4), ^ = " I ^ V^+f • (7) 
 
 '27 4 
 Taking the cube roots in (6) and (7), by (2), 
 
 The lower signs are omitteopbecause they give the same value 
 as the upper signs, y and z being interchangeable. 
 
 This is commonly called Cardan^ s formula for the roots of the 
 cubic equation a? + ax -\-h = Q. 
 
 Since every number has three cube roots, if the indicated cube 
 
 roots in (8) are taken in all possible combinations, nine values of 
 
 X result. But by (3) the parts of x must be so taken that their 
 
 product is — \a, a rational number. Tliis relation between the 
 
 parts of each root limits the number of roots to three. 
 
 b fc^ &^ 
 For representing the three cube roots of — 7;+\/;7^ + -7 by 
 
 h, ho}, and Ato^, and the three cube roots of the conjugate expression 
 by k, k(i>, and kui^, it is seen that h • k, hw • kw% and ho)^ • kto are 
 rational products while each of the six remaining products, Ji • kw, 
 h • kw^, hu) • k, ho) ' km, h(o^ • k, and hm^ - kw^, is irrational. 
 
606 THEORY OF EQUATIONS 
 
 Hence, Cardan's formula is to be applied so as to give these 
 roots 
 
 X = h-\- Jc, hu) + ku)^, hu)^ + kw. (9) 
 
 Examples 
 
 1. Solve a^-2ar' + 48a;-96 = 0by Cardan's method. 
 
 Solution. — Multiplying the roots by |, Prin, 6, 
 
 2/3 - 3 2/2 + 108 2/ - 324 = 0, 
 
 in which y = ^x. Decreasing the roots of this equation by 1 (§ 676), the 
 transformed equation lacking the second term is 
 
 03 + 105 z - 218 = 0, 
 
 in which 2 = 2/ — l=|a; — 1. 
 
 Substituting 105 for a and — 218 for b in Cardan's formula, 
 
 « = ^109 + V54766 + Vl09 - a/54756 = ^^^iS + V- 126 
 = 7 — 5or7w — 6w2or7co2_5w 
 
 = 2 or - 1+ 6V^ or - 1 - 6V^^. 
 
 But since 2! = fx-l, x = |(s + 1). 
 
 .-. x = 2, 4 V^ 3, - 4>/^. 
 
 Solve by Cardan's method : 
 
 2. ar* 4- 24 x + 56 = 0. 4. 27 ar'-117cc2+105aj+49=0. 
 
 3. 8x'-20x^ + Ux-'S = 0. 5. 7 a^ + Wx^ + 12 x + 4: = 0. 
 
 710. Discussion of the roots of a cubic equation. 
 
 the three roots of a^ + ax -\-b = are, § 709, (9), h + k, hw + ku,^ 
 and hoi^ + k<a. 
 
 52 ... 
 
 1. Since — is always positive, if a is positive or if a is negative 
 
 a^ W 
 
 and — is numerically less than — , h and k are real and unequal. 
 
 27 -^4 
 
 Hence, one root is real and the other two are imaginary. 
 
 Each of the equations x3 + 6x + 2 = and x3_2x-4 = has one real 
 and two imaginary roots. 
 
THEORY OP EQUATIONS 607 
 
 2. If a is negative and— ::r is numerically equal to — , the nuni- 
 
 27 i 
 
 bers h and A; are equal and consequently hu) + kuy^ — hm^ + ka) ; 
 also, /i and A; are real, and since w + w^ = — 1, h + k, h<a + fcw^, 
 and /io)^ + fcw are real. 
 
 Hence, the roots are real and two of them are equal. 
 
 The roots of x^ — 12 x + 16 = are real and two of them are equal. 
 
 3. If 1 — is negative, h? and T^ are complex numbers. 
 
 Since there is no algebraic method of finding the cube root of a 
 complex number, this is called the Irreducible Case. 
 
 In this case, however, the roots may be found by trigonometri- 
 cal formulae, and it may be proved that the roots are real and 
 unequal. In the irreducible case, then, it is easy to find the roots 
 by trial or by Horner's method. 
 
 By Cardan's formula the roots ofx* — 7x + 6 = are given by 
 
 X = V- 3 + V- -\^ + 9 + v - 3 - V- -W- + 9 
 
 = ^- 3 + ^ y/^^i + a/ _ 3 - ify/^^, 
 
 which we cannot evaluate by algebra. But by trial the roots are found to be 
 1, 2, and — 3. 
 
 711. Descartes' solution of the biquadratic equation. 
 
 By §§ 673, 676, the general biquadratic equation may be trans- 
 formed into an equation of the form 
 
 X* -f qa? + ra; -f s = 0. (1) 
 
 Assume a^ -{- q:x? -\- rx -\- s = {a? + Ax + B)(a^ — Ax -^ C), (2) 
 
 in which A, B, and O are to be determined. Expanding (2), 
 
 x!' + qa^ + rx + s = x* + (- A"" + B + C)x' + {AC - AB)x + BC, 
 
 from which, § 590, -A^ + B+C=^q,AC-AB = r,BC==s. (3) 
 
 Eliminating B and C from the equations in (3), 
 
 ^« -I- 2 qA^ -f- (^2 - 4 s)A^ -7-2 = 0. (4) 
 
 Equation (4) is a cubic in A?. Hence, § 685, Cor. 1, A^ always 
 has at least one positive real value, and when this has been found, 
 the values of B and C may be found by means of (3). Finally, 
 
608 THEORY OF EQUATIONS . 
 
 substituting the values of A, B, and G in (2), the roots of the 
 given biquadratic are obtained by solving the quadratics 
 
 x' + Ax + B = 02in&x'-Ax+C = 0. 
 
 Equation (4) is called the reducing cubic of (1). 
 
 Examples 
 
 1. Solve the equation a;''-6a^+3a^ + 22a;-6 = 0. 
 
 Solution. — Multiplying the roots by 2 so that the coefficient of the third 
 power of the unknown number may be a multiple of 4, 
 
 yi _ 12 2/3 + 12 2/2 + 176 2/ - 96 - 0, in which ?/ = 2 x. 
 
 Decreasing the roots of this equation by J^^, or 3, 
 
 0* - 42 22 ^. 32 2 + 297 = 0, in which 2 = 2 x - 3, 
 
 Since this equation is in the form of (1), g = — 42, r = 32, and s = 297. 
 Substituting these values in the reducing cubic (4), 
 
 ^6 - 84 ^* + 576 ^2 _ 1024 = 0. 
 Simplifying by dividing the roots by 4, 
 
 Ai^ - 21 Ai^ + 36 ^1 - 16 = 0, in which Ai = \ A^. 
 By trial, vli = 1 is found to be one root of this equation. 
 
 .•. \ J.2 = 1, whence A= ^/^•\ = 2. 
 Substituting 2 for A, - 42 for g, 32 for r, and 297 for s in (3), 
 
 B+ C = -38, C-5=16, andBC=297. 
 Solving, B = - 27 and C = - 11. 
 
 2* - 42 ^2 4- 32 z + 297 = {z^ + Az + B){z^ - Az + C) 
 
 = (22 + 2«-27)(22_2«-ll) 
 
 Since 2 = 2 x - 3, = 16(x2 - 2 x - 6)(x2 - 4 x + 1). 
 
 That is, X* - 6 x3 + 3 x2 + 22 X - 6 = 16Cx2 - 2 x - 6)(x2 - 4 x + 1). 
 Equating the quadratic factors to zero and solving, 
 
 X = 1 ± V7, 2 ± V3. 
 
 Solve the following equations : 
 
 2. x*-2x'-^x-'6 = 0. 
 
 3. a^-4ar'-8a; + 35 = 0. 
 
 4. x"" - 12 :(? -^ ^5 a? - 102 a; + 124 = 0. 
 
 5. x^- 6x3 + 11 3^^-100; + 2 = 0. 
 
 6. a;*-2x3-15ar^ + 16a; + 14 = 0. 
 
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