UC-NRLF 1 III $B 531 fib2 1 if'lil; I ^>'' ^4si2i?i-)^^^l ^J IN MEMORIAM FLORIAN CAJORI 'ft a^ (tY- WELLS' MATHEMATICAL SERIES. Academic Algebra. Higher Algebra. University Algebra. College Algebra. Plane Geometry. Solid Geometry. Plane and Solid Geometry. Plane and Spherical Trigonometry. Plane Trigonometry. Essentials of Trigonometry. Logarithms (flexible covers). Elementary Treatise on Logarithms. Special Catalogue and Terms ok application. SHORT COURSE IN HIGHEE ALGEBKA ACADEMIES, HIGH SCHOOLS, AND COLLEGES. BT WEBSTER WELLS, S.B., ASSOCIATE PROFESSOR OF MATHEMATICS IN THE UA88ACHUSKTTS INSTITUTE OF TECHNOLOGY. LEACH, SHEWELL, & SANBORN, BOSTON AND NEW YORK. COPTRIGHT, 1889, Bt WEBSTER WELLS. J. S. CiTSHiNQ & Co., Printers, Boston. 2.^ PREFACE. THIS volume has been prepared in response to a demand from numerous teachers for a work intermediate in its scope between the author's Academic and University Algebras. The first 24f pages of the Higher Algebra are the same as the corresponding pages of the Academic Al- gebra. The work on pages 249 to 305 of the latter has been rewritten with reference to the new matter, and 71 pages have been added to the book ; and it is now put forth as a complete preparatory text, containing all the topics required for admission to any of the Colleges, Universities, or Scien- tific Schools of the country. The new matter is contained principally in the following chapters : \ XXVI. Inequalities. XXVII. The Theory of Limits; Interpretation of the - a a -,0 forms -, — , and — ^ XXIX. Variation. XXXII. Harmonical Progression. XXXIV. The Theorem of Undetermined Coefficients. XXXV. The Binomial Theorem ; Fractional and Nega- tive Exponents. XXXVII. Compound Interest and Annuities. XXXVIII. Permutations and Combinations. XXXIX. Continued Fractions. There is also given in connection with the chapter on Logarithms, "a discussion of Lo^garithmic and Exponential Series. !Oi:?OR036 iv PREFACE. Attention is respectfully invited to the following among the many new features of the book : 1. The method of factoring quadratic expressions, Art. 283 ; and the examples in the same article illustrating the factoring of expressions of six terms. 2. The method of interpreting the forms - and ~ ; Arts. 300 and 301. ^ 3. The fundamental ideas with regard to convergency and divergency of series ; Arts. 371 to 373. 4. The method given in Ex. 2, Art. 381, for finding the coefficients when separating a fraction into its partial frac- tions. 5. The proof of the Binomial Theorem for any form of exponent, Arts. 389 to 391 ; especially the general proof, in Art. 390, of the law of formation of the successive coeffi- cients. 6. The proof of the formula for the number of permuta- tions of n quantities taken r at a time ; Art. 447. WEBSTER WELLS. COl^TEI^TS. TAQM I. Definitions and Notation 1 Symbols of Quantity 1 Symbols of Operation 2 Symbols of Relation 4 Symbols of Abbreviation 4 Algebraic Expressions 5 Axioms 10 Solution of Problems by Algebraic Methods ... 10 Negative Quantities 13 n. Addition 15 Addition of Similar Terms 16 Addition of Polynomials 18 III. Subtraction 20 Subtraction of Polynomials 21 IV. Use of Parentheses 24 V. Multiplication 27 Multiplication of Monomials 29 Multiplication of Polynomials by Monomials . . 30 Multiplication of Polynomials by Polynomials . . 31 VI. Division 36 Division of Monomials 37 Division of Polynomials by Monomials 38 Division of Polynomials by Polynomials .... 39 VII. FORMUL.E 44 VIIL Factoring 51 VI conte:n^ts. PAGE IX. Highest Common Factor 65 X. Lowest Common Multiple 75 XL Fractions 80 General Principles 80 To Reduce a Fraction to its Lowest Terms ... 82 To Reduce a Fraction to an Entire or Mixed Quan- tity 86 To Reduce a Mixed Quantity to a Fractional Form 88 To Reduce Fractions to their Lowest Common Denominator 89 Addition and Subtraction of Fractions 91 Multiplication of Fractions 97 Division of Fractions 99 Complex Fractions 101 XII. Simple Equations 106 Transposition 107 Solution of Simple Equations 108 Solution of Equations containing Fractions . . . Ill Solution of Literal Equations 115 Solution of Equations involving Decimals .... 117 XIII. Problems leading to Simple Equations con- taining One Unknown Quantity .... 119 XIV. Simple Equations containing Two Unknown Quantities 132 Elimination by Addition or Subtraction . . . . 133 Elimination by Substitution 135 Elimination by Comparison 136 XV. Simple Equations containing more than Two Unknown Quantities 144 XVI. Problems leading to Simple Equations con- taining more than One Unknown Quan- tity 148 CONTENTS. vii PAGE XVII. Involution 158 Involution of Monomials 158 Square of a Polynomial 159 Cube of a Binomial 161 Any Power of a Binomial 162 XVIII. Evolution 165 Evolution of Monomials 165 Square Root of Polynomials 167 Square Root of Numbers . 170 Cube Root of Polynomials 173 Cube Root of Numbers 176 XIX. The Theory of Exponents 180 XX. Radicals 190 To Reduce a Radical to its Simplest Form . . 190 Addition and Subtraction of Radicals .... 193 To Reduce Radicals of Different Degrees to Equivalent Radicals of the Same Degree . . 195 Multiplication of Radicals 196 Division of Radicals 198 Involution and Evolution of Radicals .... 199 To Reduce a Fraction having an Irrational De- nominator to an Equivalent Fraction whose Denominator is Rational 200 Imaginary Quantities 203 Multiplication of Imaginary Quantities . . . 204 Properties of Quadratic Surds 206 Square Root of a Binomial Surd 207 Solution of Equations containing Radicals . . 208 XXI. Quadratic Equations 211 Pure Quadratic Equations 211 Affected Quadratic Equations 212 First Method of Completing the Square . . . 213 Second Method of Completing the Square . . 216 Solution of Quadratic Equations by a Formula . 222 XXII. Problems involving Quadratic Equations. 223 viii COI^TENTS. PAGlS XXIII. Equations in the Quadratic Form .... 228 XXIV* Simultaneous Equations involving Quad- ratics 233 XXV. Theory of Quadratic Equations .... 246 Factoring 250 Discussion of the General Equation .... 254 XXVI. Inequalities . « 256 XXVII. The Theory op Limits 261 Interpretation of - 262 Interpretation of — 263 The Problem of the Couriers 263 XXVIII. Ratio and Proportion 267 Properties of Proportions 268 XXIX. Variation « 276 XXX. Arithmetical Progression 280 XXXI. Geometrical Progression 289 XXXII. Harmonical Progression 298 XXXIII. The Binomial Theorem; Positive Integral Exponent 301 XXXIV. The Theorem of Undetermined Coefficients, 306 The Theorem of Undetermined Coefficients . . 308 Application to the Expansion of Fractions into Series 310 Application to the Expansion of Radicals into Series 313 Application to the Decomposition of Rational Fractions 314 Application to the Reversion of Series .... 320 CONTENTS. IX PAGB XXXV. The Binomial Theorem; Fractional and Negative Exponents 323 XXXVI. Logarithms 330 Properties of Logarithms 332 Use of the Table 340 Applications 343 Arithmetical Complement 345 Exponential Equations 348 Exponential and Logarithmic Series .... 350 XXXVII. Compound Interest and Annuities .... 354 Annuities 357 XXXVIII. Permutations and Combinations .... 360 XXXIX. Continued Fractions 366 Properties of Convergents 370 Answers. ALGEBRA. I. DEFINITIONS AND NOTATION. 1. Algebra is that branch of mathematics in which the relations of numbers are investigated, and the reasoning abridged and generalized by means of symbols. Note. Writers on Algebra employ the word " quantity " as synony- mous with " number " ; this definition of the word will be understood throughout the present work. 2. The Symbols of Algebra are of four kinds : 1. Symbols of Quantity. 2. Symbols of Operation. 3. Symbols of Relation. 4. Symbols of Abbreviation. SYMBOLS OF QUANTITY. 3. The symbols of quantity generally used are the figures df Arithmetic, and the lettei's of the alphabet. Figures are used to represent known quantities and deter- mined values ; while letters may represent any quantities whatever, known or unknown. 4. Known Quantities, or those whose values are given, when not expressed b}" figures, are usually represented by the first letters of the alphabet, as a, 6, c. 5. Unknown Quantities, or those whose values are to be determined, are usually represented by the last letters of the alphabet, as x, y^ z. 2- ALGEBRA. 6. Quantities occupying similar relations in the same problem, are often represented by the same letter, distin- guished b}" different accents; as a', a", a'", read " a prime," " a second," " a third," etc. They may also be distinguished by different subscript figures; as «!, as, ag, read "a one," "a two," "a three," etc. 7. Zero, or the absence of quantity, is represented by the symbol 0. SYMBOLS OF OPERATION. 8. The Sign of Addition, +, is called ^^ plus.'' Thus, a + 6, read "a plus 6," indicates that the quantity h is to be added to the quantity a. 9. The Sign of Subtraction, — , is called " minus.'' Thus, a — b, read "a minus b," indicates that the quan- tity b is to be subtracted from the quantity a. Note. The sign '^ indicates the difference of two quantities ; thus, a^^ b denotes that the difference of the quantities a and b is to be found. 10. The Sign of Multiplication, x, is read '^ times," " into,'' or " multiplied by." Thus, a X b indicates that the quantity a is to be multi- plied by the quantity b. The sign of multiplication is usually omitted in Algebra, except between arithmetical figures ; the multiplication of quantities is therefore indicated by the absence of any sign between them. Thus, 2ab indicates the same as 2 x a X &. A point is sometimes used in place of the sign x between two or more figures ; thus, 2 • 3 • 4 denotes 2x3x4. 11. Quantities multiplied together are called factors, and the result of the multiplication is called the product. Thus, 2, a, and b are the factors of the product 2ab, DEFINITIONS AND NOTATION. 3 12. A Coefficient is a number prefixed to a quantity to indicate how man}^ times the quantity is to be taken. Thus, in Aax, 4 is the coefficient of ax, and indicates that ax is to be taken 4 times; that is, 4: ax is equivalent to ax -\- ax -j- ax -\- ax. When no coefficient is expressed, 1 is understood to be the coefficient. Thus, a is the same as 1 a. When any number of factors are multiplied together, the product of any of them ma}' be regarded as the coefficient of the product of the others. Thus, in abed, ah is the coefficient of cc? ; b of acd ; abd of c ; etc. 13. An Exponent is a figure or letter written at the right of, and above a quantity', to indicate the number of times the quantity is taken as a factor. Thus, in a.-^, the ^ indicates that x is taken three times as a factor ; that is, y? is equivalent to xxx. 14. The product obtained by taking a factor two or more times is called a power. A single letter is also often called i]iQ first power of that letter. Thus, a^ is read " a to the second power," or " a square,*^ and indicates aa ; a^ is read " a to the third power," or " a cube,'" and indi- cates aaa; a* is read " a to the fourth power," or '•'• a fourth" and indicates aaaxi ; etc. When no exponent is written, the first power is under- stood ; thus, a is the same as a^. 15. The Sign of Division, -^, is read " divided by." Thus, a-i-b denotes that the quantity a is to be divided by the quantity b. Division is also indicated by writing the dividend above, and the divisor below, a horizontal line. Thus, - indicates a ^ the same as a -5- 6. When thus written, - is often read " a over b." ALGEBRA. SYMBOLS OF RELATION. 16. The symbols of relation are signs used to indicate the relative magnitudes of quantities. 17. The Sign of Equality, =, read ^^ equals,'' or "is equal to,'' indicates that the quantities between which it is placed are equal. Thus, x = y indicates that the quantities x and y are equal. A statement that two quantities are equal is called an equatio7i. Thus, a;+4=2a;— lisan equation, and is read " x plus 4 equals 2x minus 1." 18. The Sign of Inequality, > or < , read ' ' is greater than " and " is less tha7i " respectively, when placed between two quantities, indicates that the quantity toward which the opening of the sign turns is the greater. Thus, x>y is read " ic is greater than y" ; x — 6 = -, c = -, and a; = 2 : 19. (2a + 35 + 5c) (8a + 36 -5c) (2a -36 + 15c). 21. a:*-(2a + 36)a.-3 + (3a-26)ar^-ca; + 5c. 22. 6 2 a; 8 6c — a a + 6 + c 41. Put the following into the form of algebraic expres- sions : 1. Five times a, added to twice 6. 2. Two times a;, minus y to the second power. 3. The product of a, 6, c square, and d cube. 4. Three times the cube of a, minus twice the product of a square and 6, plus the cube of c. 6. The product of x + y and a. 6. The product ot x -^ y and a — 6. 7. a square, divided by the product of 6 and c. 8. a square, divided by 6 — c. 10 ALGEBRA. 9. X divided by 3, plus 2, equals three times y minus 11. 10. The product of m and a -f 6 is less than the recipro- cal of X cube. AXIOMS. 42. An Axiom is a truth assumed as self-evident. Algebraic operations are based upon definitions, and the following axioms : 1. If equal quantities be added to equal quantities, the sums will be equal. 2. If equal quantities be subtracted from equal quanti- ties, the remainders will be equal. 3. If equal quantities be multiplied by equal quantities, the products will be equal. 4. If equal quantities be divided by equal quantities, the quotients will be equal. 5. If the same quantity be both added to and subtracted from another, the value of the latter will not be changed. 6. If a quantity be both multiplied and divided by another, the value of the former will not be changed. 7. Quantities which are equal to the same quantity are equal to each other. SOLUTION OF PROBLEMS BY ALGEBRAIC METHODS. 43. The following examples will illustrate the application of the notation of Algebra in tlie solution of problems. 1. The sum of two numbers is 30, and the greater is 4 times the less. What are the numbers? AVe will first solve the problem by the method of Arith- metic, and afterwards by Algebra. The marginal letters refer to the corresponding steps of the two methods ; that is, the operation (a) in the algebraic solution is equivalent to the operation (a) in the arithmetical ; and so on. In this way the student can compare the two processes step by step. DEFINITIONS AND NOTATION. H Solution by Akithmetic. The less number, plus the greater numher, equals 30. (a) Hence the less number, plus 4 times the less number, equals 30 (b) Therefore 5 times the less number equals 30. (c) Hence the less number is one-fifth of 30, or 6. (rf) Then the greater number is 4 times 6, or 24. Solution by Algebra. Let X = the less number. Then 4 or = the greater number. (a) By the conditions, x+4x = 30. (b) Or, 6x= 30. (c) Dividing by 5, x = 6, the less number. {d) Whence, 4a: = 24, the greater number. 2. A, B, and C together have $GG. A has one-half as much as B, and C has as much as A and B together. How much has each ? Let X = the number of dollars A has. Then 2 x = the number of dollars B has, and x + 2x, or 3 a: = the number of dollars C has. By the conditions, x+2x-{-Sx = 66. Or, 6x = 66. Whence, x = 11; the number of dollars A has. Therefore, 2 a: = 22, the number of dollars B has, and 3a: = 33, the number of dollars C has. 3. The sum of the ages of A and B is 109 years, and A is 13 years younger than B. What are their ages? Let X = the number of years in A's age. Then a: + 13 = the number of years in B*s age. By the conditions, a: + a: + 13 = 109. Or, 2x-|-13=109. Whence, 2 a: =96. And, a: = 48, the number of years in A's age Therefore, a: + 13 = 61, the number of years in B'a agfr 12 ALGEBRA. PROBLEMS. 4. The greater of two numbers is 5 times the less, and their sum is 42. What are the numbers ? 5. The sum of the ages of A and B is 68 years, and B is 6 years older than A. What are their ages ? 6. Divide $1200 between A and B, so that A may receive $128 less than B. 7. A man had $3.72 ; after spendiug a certain sum, he found that he had left 3 times as much as he had spent. How much had he spent? 8. Divide $260 between A, B, and C, so that B may i*eceive 3 times as much as A, and C 3 times as much as B. 9. Divide the number 125 into two parts, one of which is 21 less than the other. 10. The sum of three numbers is 98 j the second is 3 times the first, and the third exceeds the second by 7. What are the numbers? 11. A, B, and C together have $127 ; C has twice as much as A, and $13 more than B. How much has each? 12. My horse, carriage, and harness together are worth $400. The horse is worth 11 times as much as the harness, and the carriage is worth $175 less than the horse. What is the value of each? 13. The sum of three numbers is 108. The first is one- third of the second, and 33 less than the third. What are the numbers? 14. Divide the number 210 into three parts, such that the first is one-half of the second, and one-third of the third. 15. A man bought a cow, a sheep, and a hog, for $75 ; the price of the sheep was $27 less than the price of the cow, and $6 more than the price of the hog. What was the price of each? DEFINITIONS AND NOTATION. 13 NEGATIVE QUANTITIES. 44. The signs -f and — , besides indicating the opera- tions of addition and subtraction, are also used, in Algebra, to distinguish between quantities which are the exact reverse of each other in quality or condition. Thus, in the thermometer, we may speak of a temperature above zero as -f, and of one below as — . For example, + 25° means 25° above zero, and — 10° means 10° below zero. In navigation, north latitude is considered -f , and south latitude — ; west longitude is considered -{-, and east lon- gitude — . For example, a place in latitude —30°, longi- tude + 95°, would be in latitude 30° south of the equator, and in longitude 95° west of Greenwich. Again, in financial transactions, we may consider assets as +? £^nd debts or liabilities as — . For example, the statement that a man's property is — $100, means that he owes or is in debt $100. And in general, when we have to consider quantities the exact reverse of each other in quality or condition, we may regard quantities of either quality or condition as positive, and those of the opposite quality or condition as negative. 45. The thermometer affords an excellent illustration of the relation between positive and negative quan- i Q titles. 4- 5 Let OA represent the scale for temperatures [j^g above zero, and OB the scale for temperatures + 2 below zero ; and let us consider the foUowinoj B "^ Q problem : — 1 At 7 A.M. the temperature is — 6° ; at noon it is ~g 11° warmer, and at 6 p.m. it is 9° colder than at — 4 noon. Required the temperatures at noon and at 2q 6 P.M. — 7 Beginning at the scale-mark —6, and counting 14 ALGEBRA. 11 degree-spaces upwards, we reach the scale-mark +5; and counting from the latter 9 degree-spaces downwards, we reach the scale-mark — 4. Hence, the temperature at noon is -f 5°, and at 6 p.m. — 4°. EXERCISES. 46. 1. At 7 A.M. the temperature is — 8° ; at noon it is 7° warmer, and at 6 p.m. it is 3° colder than at noon. Required the temperatures at noon and at 6 p.m. 2. A certain city was founded in the year 151 B.C., and was destroyed 203 years later. In what year was it de- stroyed ? 3. At 7 A.M. the temperature is + 4° ; at noon it is lO'' colder, and at 6 p.m. it is 6° warmer than at noon. Re- quired the temperatures at noon and at 6 p.m. 4. What is the difference in latitude between two places whose latitudes are -|-56° and —31°? 5. A man has bills receivable to the amount of $2000, and bills payable to the amount of $3000. How much is he worth ? 6. At 7 a.m. the temperature is — 3°, and at noon it is -f 11°. How many degrees warmer is it at noon than at 7 A.M.? 7. What is the difference in longitude between two places whose longitudes are + 25° and — 90° ? 8. The temperature at 6 a.m. is — 7°, and during the morning it grows warmer at the rate of 3° an hour. Re- quired the temperatures at 8 a.m., at 9 a.m., and at noon. 47. The absolute value of a quantity is the number repre- sented by the quantity, taken independently of the sign affecting it. Thus, the absolute value of — 5 is 5. ADDITION. 15 . II. ADDITION. 48. Addition, in Algebra, is the process of collecting two or more quantities into one equivalent expression, called the sum. Thus, the sum of a and b is a + b (Art. 8). 49. If either quantit}^ is negative, or a polynomial, it should be enclosed in a parenthesis (Art. 20) ; thus. The sum of a and — 6 is indicated by a -|- (— 6). The sum of a — b and c — d is indicated by (a — 6) 4- (c — d) . 50. Required the sum of a and — b. Using the interpretation of negative quantities as ex- plained in Art. 44, if a man incurs a debt of $100, we may regard the transaction either as adding —$100 to his prop- erty, or as subtracting $100 from it That is. Adding a negative quantity is equivalent to subtracting a positive quantity of the same absolute value (Art. 47). Thus, the sum of a and — & is obtained by subtracting b from a ; or, a + ( — Z>) = a — 6. 51. It follows from Arts. 48 and 50 that the addition of monomials is effected by uniting the quantities with their respective signs. Thus, the sum of a, — 6, c, — d, and — e, is a — b-]-c — d — e. It is immaterial in what order the terms are united, pro- vided each has its proper sign. Thus, the above result may also be expressed ^ c -{- a—e — d — b, — d — 6 + c — e+a, etc. 16 ALGEBHA. ADDITION OF SIMILAR TERMS. 52. 1. Required the sum of 5 a and Sa. 6 a signifies a taken 5 times (Art. 12), and 3 a signifies q taken 3 times. We have, therefore, a taken in all 8 times, or 8 a. That is, 5a + 3a = 8a. 2. Required the sum of — 5 a and — 3 a. — 5 a signifies a taken 5 times subtractively (Art. 25) , and — 3a signifies a taken 3 times subtractively. We have, therefore, a taken in all 8 times subtractively, or —8 a. That is, — 5a — 3a = — 8a. Therefore, To add two similar (Art. 31) terms of like sign, add the coefficients, affix to the result the common symbols, and pi'ejix the common sign. 53. 1. Required the sum of 8 a and — 5 a. Since 8 a is the sum of 3 a and 5 a (Art. 52, 1), the sum of 8a and —5a is equal to the sum of 3a, 5a, and —5a, which is 3a + 5a — 5a. (Art. 51.) But, by Art. 26, 5a and —5a cancel each other, leaving the result 3 a. Hence, 8a+(— 5a) = 3a. 2. Required the sum of —Sa and 5a. Since — 8a is the sum of — 3a and — 5a (Art. 52, 2) , the sum of —8a and 5a is equal to the sum of —3a, —5a, and 5 a, which is — 3a — 5a4-5a, or —3a. Hence, ( — 8a) + 5a = — 3a. ADDITION. 17 Therefore, To add two similar terms of unlike sign, subtract the less coefficient from the greater, affix to the result the common sym- bols, and prefix the sign of the greater coeffixiient. Note. A clear understanding of the nature of the processes in Arts. 52 and 53 may be obtained by comparing them with the following, the negative quantities being interpreted as explained in Art. 44. 1. If a man owes $5, and incurs a debt of §3, he will be in debt to the amount of $8. That is, the sum of — $ 5 and - $3 is — 6 8. 2. If a man's assets amount to $8, and his liabilities to $6, he is worth $3. That is, the sum of §8 and - $5 is $3. 3. If a man's liabilities amount to $8, and his assets to $5, he is in debt to the amount of $3. That is, the sum of — $8 and $ 5 is —.$3. EXAMPLES. 54. Add the following : 1. 11 and -5. 2. - 13 and 3. 3. 12 and -1. 4. — 4 and — 7. 5. —2a and la. 6. b and -36. 13. Required the sum of 2 a, Since the order of the terms is immaterial (Art. 51), yre may add the positive terms first, and then the negative, and finally combine these results by the rule of Art. 53. The sum of 2 a, 3 a, and 6 a is 11a. The sum of —a and — 12a is — l3a. Hence, the required sum is 11a — 13 a, or — 2a. Ans. Add the following : 14. 7a, — a, and —3a. 15. — 6m, m, — 11m, and 5m. 16. 13 a6, — 7a6, — 8a6, and —%ab. 7. — 11m and —Sm. 8. be and 166c. 9. — 2 ax and 7 ax. 10. -Sa'b^ and— a'b\ 11. 1 2 mn2 and -19mn\ 12. — 13 abc and 22 a6c. — a, ,3a,— 12a, and 6a. 18 ALGEBRA. 17. ln\ -n\ -3ii2, \\n\ and -\0n\ 18. 13 ax^, — aa^, — 20aa^, 6aa^, and —5aaf. If the terms are not all similar, we may combine the simi- lar terms, and unite the others with their respective signs. 19. Required the sum of 12a, —5x, —Sy, —5a, 8a:, and —3x. The sum of 12 a and — 5a is 7a. The sum of — 5 a:, 8 x, and — 3 a? is 0. Hence, the required sum is 7a — 3y. Ans. Add the following : 20. 6 ax, —115, —ax, and 65. 21. 2a, 55, -3c, -85, and 9c. 22. 5 m, — 2n^, n, —2 m, —5^, and 37i^ 23. 3a;, —y, ^x, 6, —8y, —2x, 4^, and —5. ADDITION OF POLYNOMIALS. 55. A polynomial may be regarded as the sum of its monomial terms (Art. 51). Thus, 2a — 35 + 4c is the sum of the terms 2 a, —3 5, and 4 c. Hence, the addition of two or more, polynomials is effected by uniting their terms with their respective signs. Thus, the sum of a — 5 and c — d \^ a — h -\- c — d. 56. Required the sum of 6a — 7a;, 3a; — 2a4-3y, and 2 a; — a — mn. It is convenient in practice to set the expressions down one underneath the other, similar terms being in the same vertical column. Thus, 6a— 7a; — 2a + 3a; + 3i/ — a-\-2x —mn 3a — 2x + 3?/ — mn, Ans. ADDITION. 19 From the above principles we derive the following rule : To add tico or more expressions, set them down one under- neath the other, similar terms being in the same vertical colunni. Find the sum of the terms in each column, and unite the results with their respective signs. EXAMPLES. 57. Add the following : 1. 2. 3. 2a-lx -Sab-\-2cd — 11a— 5mp^ — a + 4tx — 7ab-{-Scd 8a-{-nmp^ a+ X 4a6-6cd — da— 7 mp^ 4. 2a — 36 4- 5c and 6 — 5c + 2d. 6. 9 mn^ + x^y, — mn^ -f- 3 x^y, and — 6 mn^ — 7 x^y. 6. aP-2ab-\-b\ a- -{- 2 ab -\- b\ and2a2-262. 7. 3a2 + 2a6H-462, oa^-Sab-\-b\ and -6a'-^Dab-5b\ 8. 6a^-7x-4:,x^-x-2,SindSx-dxr-x^. 9. 4 mn -f 3 a6 — 4 c, 3 a; — 4 a6 + 2 mn, and 3 m^ — 4 a:. 10. 3x — 2y — z, Gy — 5x — Iz, 82 —y — x, and 4a;— 92/. 11. Qx — ^y-\-lm, 2n — x + y, 2y — 4:X — 5m — 9n, and m — 2 x. 12. 2if^-6x^-x-\-7, 3x^ — 2-6ar + 8x, a; + 3x'-4, and 1 + 2 a,'^ — 5 x. 13. 2a-36 + 4d, 26-3d + 4c, 2a —11a; Ay -ISmn Ua-b 12. From 9 ab take - 2 ab. 16. From - ar^/ take 5x^y\ ' 13. From xy take — cd. 17. From — 70 abc take — b2abc. 14. Froml7mHake41??i^ 18. From - 7 ?w- take - 8 n^. 15. From -5a; take 3. 19. From -33a;»2/^ take 19ar^2/*. 20. From bab take the sum of 9a6 and —2ab. 21. From the sum of — Ua:^ and Sa? take the sum of - 10 ar' and 4ar'. SUBTRACTION OF POLYNOMIALS. 63. When the subtrahend is a polynomial, each of its terms is to be subtracted from the minuend. Hence, To subtract one polynomial from another^ change the sign of each term of the subtrahend, and add the result to the minuend. It will be found convenient to place the subtrahend under the minuend, similar terms being in the same vertical column. 64. 1. Subtract ba?y — 3ab -{- m^ from 3 x^y — 2 a6 -f- 4 ?i. Changing the sign of each term of the subtrahend, and adding the result to the minuend, we have 3ar>-2a6 + 4?i — ba?y-\-3ab — m^ —2a^y-\- ab-\-4:n — m^ 22 ALGEBRA. Note. The student should endeavor to perform mentally the opera- tion of changing the sign of each term in the subtrahend, as shown in the following example : 2. From 5a^-7b^~2a^b subtract Sa'b^ 4:ab'-2b'-^aK 6a^-2a'b -^7b^ a^-\-3a^b-Aab'-2b^ EXAMPLES. Subtract the following : 3. 4. ab -\- cd— ax 7x-\-by — Sa Aab — Scd — 4:ax x — 7y-i-6a — 4: 5. From a — 6 + c take a + b — c. 6. From a^-^2ab-\-b^ take a'-2ab-{- b\ 7. From 7abc—nx+oy—4:8 take lla6c4-3a;-f-72/+100. 8. Subtract 3m-\-y-—5a — 7 from 5m — 3/+ 7a — 6. 9. Subtract 17a^ + 5^/^- 4a5 -f 7 from 31a^-3/-f a6. 10. From 6a + 36— 5c-fl subtract 6a— 36 — 5c. 11 . From Sm — 5n + r — 2s take 2 /• + 3 n — m — 5 «. 12. Take 4a — 5 + 2c — 5c? from d — Sb + a — c. 13. From m^ + 3 w^ subtract — 4 m^ — 6 w^ + 71 a;. 14. From46 — 3 6 — 5d + 2ic take 3a -\- 8d—b — 6c. 16. From a — b — c take the sum of — 2a + 6+c and a^b -^c. 16. Froma;* + 2a^-3a; + 4 take 3a^ + 30^+ 5a;- 7. 17. From 4 a^ - 3 afe^ _ 5 6^ subtract 6 a^b - aft^ + 4 b\ 18. Froma^ — 8H-2a*-3aHake 6a- U — Sa^ — 2aS SUBTRACTION. 23 19. Take 2x^ — y'^ from the sum of x^ — 2xy -^^y^ and xy — i y^. — 20. From the sum of a;-}- 22/ — 32; and 3y — 4ic + a take z — bx-\-by. 21. From 7a3 + 3-5(X^ + a-5a2 subtract 2a—^o?—2a?-\-^ — \la\ 22. From —72/^ + 3 a^2/ — 2it'* + 6 a.-^/^ subtract Sa^y — 2 xy^ -{- a:^ — 9 y^. 23. From the sum of 2a:^ — x-\- 5 and ar^ + 8 a; — 11 take the sum of ay^ — 9x^ —llx and — 42/-^ + 3 a.*^ — 6. -^ 24. From the sum of a- -f- a6 + 6^ and a^ — 4 a6 -f 5 6^ take the sum of 4a2 + 76"- 2a6 and Sab-a--2b^. "i 26. From3ar^-7?/-2+a:2/-5/ subtract — 5xy -\-(jx — 2x^ — S-\- 2y^. 26. From 3ar' — 8a.'* + 3a.'^ — 5a^ — 2aj subtract - 3a^ + 4a^+ 6a^- 6a; + 2. 27. From the sum of 2 3i? — 3?y — 6 xif and 3a^y — 5xy^ — 4y^ take the sum of —2a^—7ofy—6y^ and — 6a:y^+5?/*. -^ 28. From the sum of a* — 1 and 2a^ — lOa^— 7a subtract the sum of —3a*-\-2a^ — 5a and — 5 a^— 12 a^ 4-3. Note. In Arithmetic, addition always implies augmentation, and subtraction diminution. In Algebra this is not always the case; for example, in adding — 2 to 5, the sum is 3, which is less than 5. Again, in subtracting — 2 from 5, the remainder is 7, which is greater than 5. Thus the terms Addition, Subtraction, Sum, and Remainder have a much more general signification in Algebra than in Arithmetic. ,24 ALGEBRA. IV. USE OF PARENTHESES. 65. The use of parentheses (Art. 20) is very frequent in Algebra, and it is necessary to have rules for their removal or introduction. 66. The expression 2a-3b+{bb-G-\-2d) indicates that the quantity 6b — G-\-2d is to be added to 2a — Sb. If the addition be performed, we obtain (Art. 55) 2a-3b-\-5b-G + 2d. Again, the expression 2a_35-(56-c + 2d) indicates that the quantity 5b — c-\-2d is to be subtracted from 2 a — 3 b. If the subtraction be performed, we obtain (Art. 63) 2a-36-56 + c-2d. 67. It will be observed that in the first case the signs of the terms within the parenthesis are unchayiged when the parenthesis is removed ; while in the second case the sign of each term within is changed, from + to — , or from — to +. We have then the following rule for removing a paren- thesis : A parenthesis preceded by a -\- sign may be removed without altering the signs of the enclosed terms. A parenthesis preceded by a — sign may be removed, if the sign of each enclosed term be changed, from ■\- to —, or from — to +. 68. Since the brackets, the braces, and the vinculum (Art. 20) have the same signification as the parenthesis, the rule for their removal is the same. PARENTHESES. 25 It should be observed in the case of the vinculum that the sign apparently prefixed to the first term unde rneath, is in reality the sign of the vinculum. Thus, -j- a — 6 and —a — b are equivalent to + (a — 6) and — (a — 6) , respectively. EXAMPLES. 69. 1. Remove the parentheses from 2a-36-(5a-46)-f(4a-6). By the rule of Art. 67, the expression becomes 2a — 36 — 5a + 46-|-4« — 6 = a, Ans. Parentheses are often found enclosing others. In this case they may be removed in succession by the rule of Art. 67, and it is better to remove first the innermost pair. 2. Simplify the expression We remove the vinculum first, and the others in succession. Thus, 4a;— 53a; + (— 2if — a; — a)5 = 4.x-\Zx-\-{-2x-x-^a)\ = A:X — \^x — 2 X — X + a\ = 4a; — 3a: + 2fl;-f-.T — a = 4a; — a, Ans. Reduce the following expressions to their simplest forms by removing the parentheses, etc., and uniting similar terms : 3. a-(&-c) + (-cZ + e). 4. 5a;-^2a;-32/^-[-2a; + 4?/]. 5. a-~h-{-c — a-\-h — c — c — h — a. m^. 6. m- — 2 ?i + J rt — ?i + 3 m^J — 5a + 3?i 7. ar-V- {or - 2ab + 5") - [a- + 2a6 + 6-]. 8. 3a-(2a- Ja + 2;). 26 ) ALGEBRA. 9. a- ~(b + \ -e + d}- -e). 10. a - -[(-b + c)-(d- -«)]• 11. Sx-l2y-{-x-y']-\-[3y-2x-\~yl^. 12. Ux-{ox-9)-\4.-Sx-(2x-3) 13. 2m-[n-j3m-(2n-m)J]. 14. Sx — (5 X -{-[_ — 4:X — y — x']) — (—x — 3y). 16. 3c-f-(2a-[5c-j3a + c-4a5]), 16. 5a-(4a-J-3a-[2a-a-l]J). 17. 8x-l5x-(Sx-A)-\7x-i-{-9x-{-2)l'] i 18. 2m - [3m - Jm - (2m - 3m + 4) ; - (5m ~2)]. 19. c - [2 c - (6 a - 6) - J c - (5 a + 2 &) - (a - 3 6) J ] . / 20. Sa-\b-lb-(a-j-b)-\-b-{b-a-b)l']\. 70. To enclose any number of terms in a parenthesis, we take the converse of the rule of Art. 67 : Any number of terms may be enclosed in a parenthesis pre- ceded by a -\- sign, without altering their signs. Any number of terms may be enclosed in a parenthesis pre- ceded by a — sign, if the sign of each term be changed, from -{• to —, or from — to -\-. 71. 1 . Enclose the last three terras of a — b + c — d-\-e in a parenthesis preceded by a — sign. Result, a — b— (— c -j- d — e) , In each of the following expressions, enclose the last three terms in a parenthesis preceded by a — sign : 2. a-\-b-{-c-{-d. 5. oc^y — nc^y'^ — x^ + y^. 3. 3a-2& + 5c-4d. 6. x' -Zo? -\-2x' -bx-8. 4. m^ + 5m2-6m + 3. 7. a^ -b^ - c^ -{-2a'b -\-2ac. 8. In each of the above results, enclose the last two terms in an inner parenthesis preceded by a — sign. MULTIPLICATION. 27 V. MULTIPLICATION. 72. Multiplication, in Algebra, is the process of taking one quantity as many times as there are units in another. Thus, the multiplication of a b}' 6, which is expressed ab (Art. 10), signifies that the quantity a is to be taken b times. 73. The Multiplicand is the quantity to be multiplied or taken. The Multiplier is the quantity which shows how many times it is to be taken. The Product is the result of the operation. The multiplicand and multiplier are called /acfors. 74. In Arithmetic, the product of two numbers is the same in whatever order they are taken ; thus, we have 3x4 or 4 X 3, each equal to 12. Similarly, in Algebra, we have a x 6 or 6 x a, each equal to ab. That is, the product of the factors is the same in whatever order they are taken. 75. Required the product of c and a — b. In Arithmetic, if we wish to multiply 87 by 98, we may express the multiplier in the form 100 — 2 ; we should then multiply 87 by 100, and afterwards by 2, and subtract the second result from the first. Similarly, in Algebra, to multiply c by « — 5, we should multiply c by a, and afterwards by 6, and subtract the sec- ond result from the first. Thus, the required product is ac — be. 76. Required the product of a — 6 and c — d. As in Art. 75, we should first multiply a — b by c, and 28 • ALGEBRA. afterwards by d, and subtract the second result from the first. The product of a — & and c is ac — he (Art. 75) . The product of a — h and d is ad — hd. Subtracting the second result from the first, the required product is „<.-6c-«d + M. 77. We observe, in the preceding article, that the product is formed by multiplying each term of the multiplicand by each term of the multiplier, with the following results in regard to signs : The product of the terms + a and + c gives the term -f- ac. The product of the terms — h and + c gives the term — he. The product of the terms + a and — d gives the term — ad. The product of the terms — h and — d gives the term + hd. From these considerations we may state what is called the Rule of Signs in Multiplication, as follows : + multiplied hy + •> c^'^^d — multiplied hy — , produee + ; + multiplied hy — , arid — multiplied by -f , produce — . Or, as it is usually expressed with regard to the product of two terms. Like signs produce + , and unlike signs produce — . 78. Required the product of 7a and 2h. Since the factors may be written in any order (Art. 74) , we have 7 a X 2h = 7 X 2 X a X h = lAah. That is, the coefficient of the product is the product of the coefficients of the factors. 79. Required the product of a^ and a^. By Art. 13, a^ = a x a x a, and a^==axa. Hence, a^xa^ = axaxaxaxa = a^. MULTIPLICATIOK. 29 That is, the exponent of a letter in the product is the sum of its exponents in the factors. Thus, a'xa^Xa = a'+^^ = aK MULTIPLICATION OF MONOMIALS. 80. We derive from Arts. 77, 78, and 79 the following rule for the product of two monomials : To the product of the coefficients annex the literal quantities^ giving to each letter an exponent equal to the sum of its expo- nents in the factors. Make the product -\- when the factors have the same sign, and — when they have different signs. EXAMPLES. 1. Multiply 2 a'' by 7a\ By the rule, 2a^ x 7a* = 14a^+^'' = 14a», Ans. 2. Multiply a''^6-c by -5a'bd. a^b^c X ( — 5 a-bd) = — 5 a^b^cd, Ans. 3. Multiply -7. T"* by 5 ar\ - 7 a;'" X 6x^ = - 35a;"'+^ Ans. 4. Multiply —11a;'" by —Sx'". -nx'^x{-Sx^) = 88x-'^, Ans. Multiply the following : 6. 13 by -19. 11. -ll^i^yby -57i<^. 6. - 18 by 12. 12. - 6a^bc by a^bm. 7. -22 by -51. 13. - 12a2a; by -20;^. 8. 15m«n« by 3mn. 14. - 2a'"6" by 5a^6". 9. 17a6c by —8 a5c. 15. 3 aV/ by 1 1 aa^y . 10. -17aVby3aV. 16. 3a'"6" by -5a"5^ 80 ALGEBRA. It is evident from the Rule of Signs (Art. 77) that the product of three negative terms is negative ; of four negative terms, positive ; and so on. Hence the product of three or more monomials will be pos- itive or negative, according as the number of negative factors is odd or even. 17. Required the product of — 20%^ Qthd^^ and — Ic^d. - 2 a^d^ X 6 6c^ X - 7 c^d = 84 a^ftVd, Ans. In this case the product is positive, as there are two nega- tive factors. Multiply the following : 18. 5a, —66, and 7c. 19. -2a2, -Ua^, and -9 a. 20. - 3 aV, - 2 6r , and 7 c^. 21. ^x'^y^, — ic"2/V, and \hyh\ 22. -2a, -3a2, -4a^ and -5a^ 23. - a^hc, 2 V^cd, — 5 d'cd, and - 3 ahH\ 24. — 7m'*a^, m^aj^, 2a^, and —%my^\ 25. 6072/^, —o?z^ 32/V, — 2fl7r^, and —^yz, MULTIPLICATION OF POLYNOMIALS BY MONOMIALS. 81. In Art. 75 we showed that the product of a — 6 and c was ac — bc. We have then the following rule for the prod- uct of a polynomial by a monomial : Multiply each term of the multiplicand by the multiplie'i\ and connect the results with their proper signs. EXAMPLES. 1. Multiply 2ic2 _ 5a; _ 7 by 8aj3. By the rule, the product is 1 6 a^ — 40a/''' — 560.-^, Ans, MULTIPLICATION. ^ 31 2. Multiply - 5 ab^ by 3 a^ft - 4 aW. 3a26_4a63 Multiply the following : 3. 3a; — 5 by 4a;. 8. m^ + mn-\-n^hy m^n^. 4. a^h + a&2 by - ah. 9. - 2m by Sm^-5mn-n\ 5. Sa^bc -d by 5a(P. 10. - a;.^- lOa^^+S by - 2a;3^ 6. a.-2-2a;-3 by -4ar. 11. a^ + l^ab - 6b^ by 4:ab\ 7. -2a^by3a^-^ex-7. 12. -Ga^c by 5 - 6ac-8a^ 13. 5a;3_4a^_33,_^2 by -Gar'. 14. a^b^ by a^ - 3 a^b + 3 ab^ - 6^. MULTIPLICATION OF POLYNOMIALS BY POLYNOMIALS. 82. In Art. 7G it was shown that the product of a — b and c — d was oo — 6c — afZ -f bd. We have then the following rule for the product of two polynomials : Multiply each term of tJie multiplicand by each term of the multiplier^ and add the partial products. EXAMPLES. 1. Multiply 3a — 26 by 2a — 56. In accordance with the rule, we multiply 3 a — 26 by 2a, and then by —56, and add the partial products. A convenient aiTangement of the work is shown below, similar terms being in the same vertical column. 3a — 26 2a - 56 Qa^-~ Aab -15a6 + lQ6'^ 6a2-19a6 + 1062, Ans. 32 ALGEBRA. I 2. Multiply x^ + l — x^ — xhyx-\-l. It is convenient to have both multiplicand and multiplier arranged in the same order of powers (Art. 36) , and to write the partial products in the same order. Arranging the expressions according to the ascending powers of x, we have l—x + x^ — a^ l-\-x 1 —x-i-x^ — a^ x — x^-\-x^ — x* 1 —x'^, Ans. 3. Multiply 6 a6- 8 52 + 4 a^ by -4b' -^2a^ -Sab. Arranging according to the descending powers of a, we have 4a2 + 6a&-862 20" -Sab -W Sa^ + 12a%-16a'b' -12a^b- 18a^b'-{-24:ab^ -Ua^b'-24:ab^-{-S2b* 8a* -60aW +32&S Ans. Note. The correctness of the answers may be tested by working the examples with the muhiplicand and multipHer interchanged. Multiply the following : 4. 3a; + 2 and 5a; -7. 6. 3a-26 and -2a + 4&. 5. 6a;— 5 and 3 ~2«. 7. 3 — Sa;^/ and — 6 — lOa^y. 8. a^-\-ab-{- b' and b — a. 9. 2a^b-S ab^ and 5a'b + 6 ab\ 10. 1 + a; 4- a^ + a^ and ax — a. 11. 3a;2-2a;2/-/ and 2a;-42/. 12. m^ — mn — Sn^ and 2 m^ — 6 mn. 13. a^-h2a;+l anda;2-2a;-h3. MULTIPLICATION. 33 14. ba' + ib'-Sab and 6a- 5b. 15. 4a;^4-6ic-7 and 2x2-3. 16. a-hb —c and a — b-^c. 17. 2x^-3x-{-5 sindx^-j-x-1. 18. 3x'-7x-\-4:B.nd2x'-i-9x~5. 19. 2x3-3x2-5x-l and3a;-5. 20. 6 m — 2 7n,2 _ 5 _ ^^^3 j^j^^i ^^^2 _f_ 2 — 2 m. 21. 2a;3 4-5x2_g^_7 ^^j^^^ ^_ .^_^^^ 22. a'b - a'b' - 4 ab^ and 2 a^ft - ab^ 23. a;'"+V __ 3 ^y--^ and 4x'»+y - 4xy. 24. a;2 -^.y^ — xy and xy + / -}- ^2^ 25. 2a6 + 62_^4a2and4a2_2a6 + 62. 26. 6.T*-3a^-x2+6x-2and2.x'2 + a; + 2. 27. m* - ^i^/i + 7n-7r - wiji^ _^ ^^4 ^^^^ ^^^2 _ 2 ^^^^^ _ 3 ^2^ 28. 27x^H-9x22/ + 3V + 2/-^and9a.-2_6x?/4-2/^ 29. a^-Sa'b-\-3ab^-b^andd'-2ab-^bK 30. a;2 + i/2 + 2;2^a^_2^2_;2a;anda; + ?/-|-«. 31. 2a:3-3x'2-f 5a;-l and 3ar^-x2__2a;-5. 32. a& + ccZ -\-aG-{-bd and a6 + cd — ac — bd. *.33. 2a3-5a2_6a + 4 and 4^^^ lOa^- 12a-8. Find the product of the following : 34. X — 3, a; + 4, andic — 7. 35. a-{-b, a^-ab + b% and a^ — b^. 36. 2m- 1, 3m + 4, and 6m -5. 37. a7 + l, 3aj--2, and3a;2__^_2. 38. aj'4-a5 + l,a^-a; + l,anda;^-ar^+l. 34 ALGEBRA. 39. a + b, a — b, a^ + b\ Siud a'^ + b*. 40. mH- 1, m— 1, m + 2, and m — 2. 41. 2i»— 1, 3a;+2, 4x — 3, and 5a; + 4. 42. a-{-b, a-b, a-[-2b, and a? -2a''b-ab^ -\-2W. 83. The product of two or more polynomials may be indi- cated by enclosing each of them in a parenthesis, and writing them one after the other. Thus, the product of a; + 2, » — 3, and 2 a? — 7 is indicated by * (a; + 2)(a;-3)(2a;-7). Similarly, the expression (a+6 + c)^ indicates that a -f 6-+- c is to be multiplied by itself (Art. 13). When the operations indicated are performed, the expres- sion is said to be expanded or simplified. EXAMPLES. 1. Simplify the expression {a — 2xy—2{x-\-^a){a — x). To simplify the expression, we should expand {a—2xY and 2 (a; + 3 a) (a — a?) , and subtract the second result from the first. {a— 2xY= o? — A: ax -\- 4:0? 2{x + ^a){a-x) =6a^ - 4:ax-2x^ Subtracting the second result from the first, we have a^—4:ax-\-4:X^ — 6a^-i-4:ax-\-2o(?=6x^—6a^y Arts. Simplify the following : /2: (a-hb + c-^dy, \Z. {a-b)(c-d) -\-(a-c)(b-d). 4. (2a; - 3)2 +(1- a;) (3a; -9). 5. (a-\-b-hcy-(a-b-\-cy. 6. (2a-56)2-4(a-26)(a-36). MULTIPLICATION. 86 7. (a - by (a + by. 8. {l+x){l-j-x*){l-x-^x'-o^). 9. (l+ay-{l-a){l -\-a'). 10. lx-{2y + ^z)]lx-{2y-3z)2. ' ' 11. (x-^y){a^-f)[x^-y{x-y)}. 12. {a-^b){b-\-c) - {c-\-d){d-{-a) - {a -\- c) {b - d) . 13. (a 4- 6 4- c)2 + (a - 6 - c)2+ (6 - c-a)^-}- (c-a-6)^ 14. (a-6)(6-c) + (6-c)(c-a)-f(c-a)(a-6). 15. if(ic-22/)+y(2/-22;) +z(;2 - 2 a;) - (x- 2/-2;)2. 16. a; (a; + 1) (a; + 2) (a; 4-3) 4- 1 - (ar 4- 3x4- 1)'. 17. (a 4- 6 4- c)2 - (a - 6 - cy+ {b-c-ay- (c-a-by. 18. [(m 4- 2ny - {27n - n)^] [(2m 4- ny - (m - 2n)2]. 19. (x-\-y + zy-(x^-\-f-j-z^)-3(y-j-z)(z-hx){x-^y), 84. Since ( 4- a) ( 4- 5) = a6, and ( — a) ( — 6) = a6, it fol- lows that in the indicated product of two factors all the signs of both factors may be changed without altering the value of the expression. Thus, (a — 6) (c — d) is equal to {b — a){d — c) , Similarly, we may show that in the indicated product of any number of factors, the signs of any even number of factors may be changed without altenng the value of the expression. Thus, (ci — &) (c — d) (e — /) , by changing the signs of the second and third factors, may be written in the equivalent form (iyide8-{-18x^-66x'hy -ex^ + 4:-\-8x. Arranging according to the ascending powers of x, 8_56a^+18aj^ 4: + 8x-6a^ 8+16a; -12a^ 2 — 4a; — 3a^, Ans. -16x -Ux' + 18x'^ -16a;-32a^ + 24fl^ -12a^-24a;3_^lg^4 -12a^-24g;^4-18a;^ 3. Divide 9ab^ + a^-9b^- bo?h by Sfe^ + a^- 2db, Arranging according to the descending powers of a, a^-2a'b + Sab^ Divide the following : 4. 6a;2_ ^_ 35 |3y 3^^ Y^ 5. 2 - 3aa; - 2aV by 1 - 2ax. 6. a2-4a6 + 4?)2by a-26. 7. 59a;-56-15ar^by 3a;-7. 8. Sb^ + Sab^-Aa'b-^a^hyb + a, 9. 2 a^x — 2 aa:^ by ax — a^. 10. 18a^-5a; + l by 6a:2_|_2a;_i. 11. 8m3 + 35-36mby 5-|-2m. 12. 27a^ + 2/^ by 3x4-2/. 13. 16m^-l by 2m-l. 14. a^-b'-i-c^-2aGhy a-i-b-c. 15. 8 a^ -\- 36 a-b-\-54:ab'-+- 27 b^hy 2a -{-Sb. 16. x' + 2/ -{-xhf by x' -^y' + xy. 42 ALGEBRA. 17. 2x*-19a^ + 9 by 2ar^ + 6a;2_a;-3. __ 18. 8 m^ -\- S n^ — 4: mhi — 6 mn^ by 2m — n. 19. 4a;^-8a^-6a?2 + 24 by 2a;-4. 20. 23a^ -48 4- 6rc*- 2a;- 31 a^ by 6 4-3.^2-50?. '21. 4a^ + 27-a«by 9-3a3 4-4a2 + 2a4_6a. 22. x^ — dsc^ — exy-y^hy a^-^Sx + y. 23. a8-816n)y a2 + 36. 24. a^ — 2/" + 2^/2; — 2;- by a; 4- 2/ — 2!. 25. 3a;4-14a?2 4-8by a;-2. 26. y^ + x'^y hy x + y. 27. 15m^4-50m2 4-15-32m-32m3by 3m2 4-5-4m. 28. l-h4:X^ + Sx^hy (x + iy. 29. 21a''-216«by la-lh. 30. 64a;^4-l by 8a;2-4a;4-l. 31. 50a5 4-9a;*4-24-67a;2i^y ^^^2_g_ 32. a;^ 4- 2/* ~ ^^if ~ 4a;^2/ + Ga^^^/^ by x^ -\-y'^ — ^Xfy. 33. aj^-4a^4-2.'«-4-4a;4-l by (a;-l)2-2. 34. 9a!^4-4?/^-37a;y by 3a^-2/4-5a;2/. 35. a* 4- a'^' 4- 25 6^ by (a- 6) (a- 56) + 3a&. _36. 3a^4-4a;4-6ar'-lla;^-4by 3a;2-4. 37. 6ar5 4-15ar^4-51a;-18 by 2a;3-4a^4-7a?-2. 38. 2a;*-lla;-4a;2_i2_3a;3i3y 4_^2a.-2 4-a;. 39. m* — 48 — 17m3 4- 52m 4- 12m2 by m — 2 4- w^ . 40. a;"+^ 4- a;"?/ — xy'^ — y'^-^^ by x^ — y^. 41. x'y — x]f hy y? ■\- if" -\- xy'^ 4- a^^?/* 42. x^ — ^^-x-^\>y x?-\-'lx-\-^. 43. 2a'' 4- SSa^fts _ 49 6^ - la^y" -^a% by 2a^-5ab -751 DIVISION. 43 45. 2a^-6y^-V2z^-{-xy-2xz-i-17yzhy 2x + 4:Z-Sy. 46. a^n _ 52« _j_ 2 ^-c'" — c^'" by a'^ + &"* - c". 47. a;«_ 1 _ 6»^-3a^ by -20.-2 -aj + a^-l. 48. 12 a^-14a*& + ^Oa'V - a-^^ _ 8a6^+ 46^ by 6a^- 4 cr6- 3 aft^H- 2 6^. The operation of division may be abridged in certain cases by the use of parentheses. 49. Divide (a^ + ab)x^ + (2ac + 6c + ad)x + c{c + d) by ax -{- c. (a2+ ab)os^ + (2ac + be -\- ad)x + c{c + d) (a^-\-ab)a^ +( ac + bc )x ax-\-c ( ac ( ac -\-ad)x-\-c{c-{-d) + ad)x-\-c(c + d) {a+b)x+{c+d), Ans, Divide the following : 50. a^ + (a-{-b-i- c)x- + («6 + bc + ca)x + abc by fl^+(&+c)ic + 6c. 51. (6 + c)a2H-(6-4-36c + c2)a + 6c(6 + c) by a + &+c. 52. (x+yy-^(x + y) + 6hy {x + y)-2. 53. (a + 6)3_j-i by (a + 5) + l. 54. ar^ + (a 4- 6 — c)x^ + (a& — 6c — ca)a; — a6c by x'^-{-{b — c)x — be. 65. (m-7i)*-2(m-n)2 + l by (m — n)2 — 2 (m — n) + 1 . 56. if^-f- (a — 6 + c)ar + (ac — ab — bc)x — abc by a; + c. 57. a-^ + (3 - 6).^•^ + (c - 3 6 - 2) a^ + (2 6 + 3 c)a; - 2 c by or + 3 x — 2. 58. a-(6 + c) + a(62 + 6o + c') - 6c(6 + <■) by a + 6+ c. 44 ALGEBRA. VII. FORMULAE. 94. A Formula is the algebraic expression of a general rule. 95. The following results are of great importance in abridging algebraic operations : a +b a-b a +6 a -\-b a-b a-b a'-^-ab 0?— ab a^-\-ab ab +62 -ab +62 a2_2a6 + 62 -ab- -b' a:'+2ab + b^ a' -b' In the first case, we have (a-\- by = a'^ -\- 2 ab -j- b^. (1) That is, the square of the sum of two quantities is equal to the square of the first, plus twice the product of the two, plus the square of the second. In the second case, we have (a — by = a^—2ab-\- 6^. (2) That is, the square of the difference of two quantities is equal to the square of the first, minus twice the product of the two, plus the square of the second. In the third case, we have (a + 5) (a — 6) =a^ — b^. (3) That is, the product of the sum and differeyice of two quan- tities is equal to the difference of their squares. EXAMPLES. 96. 1. Square 3o+26c. The square of the first term is 9a^, twice the product of the terms is 12 abc, and the square of the second term is 4 6^c^. Hence, b}^ formula (1), (3 a + 26c)2 =9a2+ 12 a6c + 46V. FORMULA. 45 Note. The following rule for the square of a monomial is evident from the above : Square the coefficient, and multiply the exponent of each letter by 2. Thus, the square of ba% is 250*62. 2. Square 4fl7 — 5. By formula (2), (4a; -5)^ = 163^- 40a; + 25, Ans, 3. Multiply 6 a^ + & by 6 a2 - 6. By formula (3) , {a- + b) (6a^ - 6) = 36 a* - b^ Ans. Write by inspection the values of the following : 4. (a; -4)2. 16. (3^3 .^13)2^ 5. (S + ay. 17. {ea'-b'cy. Q. (a; 4- 3) (a; -3). 18. {oa+7b'){oa^7ly'). 7. (3a + 5)^ 19. (13a& + 5ac)2. 8. (2a; + l)(2a;~l). 20. (a^ -^5x){a^ - 5x), 9. (7-2a;)2. 21. (l-Uxyzy, 10. {2m + Sny. 22. {4.a^ -\-3f)(4:x'-3f). 11. {4.ab-xy. 23. {10a^-\-9a^y. 12. (5 + 7a;) (5 -7a;). 24. (4a^-56«)^. 13. (x'-fy. 25. («"* + «") (a"* -a"). 14. (3a;+ll)(3a;-ll). 26. (la^ + Uxy. 15. (a;V + 4)2. 27. (5a'«-a'*)^ 28. Multiply a + 6 + cbya + &— c. (a + 6 + c)(a + 6-c) = [(a + 6) + c][(a + 6)-c] = (a 4- 6)' - cS by formula (3) = a2 + 2a6 + 62_c2, ^^s. 46 ALGEBRA. 29. Multiply a + b — chya — b + c. {a-hb-c){a-b-\-c) = la + {b-c)'][a-(b-c)^ = a'-(b-cy = a'-(b'-2bc + c') = a'-b^-i-2bc-c', Ans. Expand the following : 30. {x-\-y-{-z)(x-y + z). 32. (1 + a- 5) (1- a + &). 31. {x-\-y + z){x-y-z). 33. (x^ -^x+1) {x^ -x-1). 34. (a-j-b — c)(a-b-c). 35. {a^ + 2a + l)(a'-2a + l). 36. {x^ + 2x-S){xP~2x-{-S), 37. (m^ + ^^ + ^^) (^^^ — "^^ + ^^) • 97. We find by multiplication : X +5 .^• -5 X +3 X —n si^ + 6x a^ — 5x + 3ar+15 -3i» + 15 a^ + 8aj+15 a^-8a;+15 a; +5 X —5 X -3 X +3 ar^ + 5a; af — 5x -3a;-15 + 3a;-15 a^^2x-15 ' a^-2a;-15 We observe in these products the following laws : I. The coefficient of x is the algebraic sum of the numbei^ in the factors. II. The last term is the product of the numbers. By aid of the above laws the product of two binomials of the form x + «, x-\-b may be written by inspection. FORMULAE. 47 1. Required the value of (x — 8) (x + 5) . The coefficient of « is — 3 ; and the last term is —40. Hence, (^x-8){x + 5) = af -3x-4:0, Ans. EXAMPLES. Write by inspection the values of the following : 2. (x-{-7){x + o), 10. (x + 9){x-o). 3. (a;-3)(a;-4). 11. {x-8)(x-d). 4. {x + 8){x-2). 12. (x-\-4:m){x + Q7n), 5. {x-S){x-\-l). 13. (x-5a)(it' + a). 6. {x-5){x + 6). 14. (a4-6)(a-46). 7. (.'» + l)(a;-|-12). 15. {a + i)b)(a + Sb). 8. (a;-7)(a;-f-2). 16. (a.-^ - 3) (a^ - 7) . 9. {x-8)(x-6). 17. (aj3 + 2a)(ar'^-6a). 98. The following results may be verified by division : (1) ^!LZL^ = a-b. (3) ^L±^ = a'-ab + bK (2) tzL^ = a + b. (4) ^^^' = a*+a64-6^ a— & a— 6 Formulae (3) and (4) may be stated in words as follows : If the sum of the cubes of two quantities be divided by the sum of the quantities, the quotient is equal to the square of the first quantity^ minus the product of the two, plus the square of the second. If the difference of the cubes of two quantities be divided by the difference of the quantities, the quotient is equal to the square of the first quantity, plus the product of the two, plus the square of the second. 48 ALGEBRA. EXAMPLES. 1. Divide 36?/V - 9 by ^yz" + 3. By formula (1) , — ^ „ ~ = 62/2^ — 3, Ans. ^ ^ 62/22-1-3 2. Divide 1 + Sa^ by 1 + 2a. By formula (3) , LLM = 1 - 2 a + 4 aS Ans. ^ ^ ^ l+2a 3. Divide 27a^-63by 3a-6. By formula (4) , ^^^'"^ = 9 a^ + 3 a6 + 5^, ^715. EXAMPLES. Write by inspection the values of the following : 4. a^-81 a;-9 9. 27 + 0^ 3 + a; 14. x'-f x^-f 5. 25-16a2 5 + 4a 10. a^-16aj2 a^ + 4a; 15. ^■\-xf 6. 0^+1 x+1 11. ajs_64 a;-4 16. 49a2-121&^ 7. 1 — m 12. l-2m 17. 64 m^ + n^ 8. a«-8 a-2 13. a3+343 a + 7 18. x-^5y Divide the following : 19. 27aj»2/^-642^by 3a;2/-4«. 20. 25 a^ - 81 b^c"^ by 5 a^ _ 9 6c». 21. 343 + 125a;^y'lt)y 7 + 5a;2/. FORMULAE. 49 22. Um^-216n^hy Am-6n\ 23. 729xy+5122«by 9aT^2/ + 822. ). By actual division we obtain : a-{-b a'-b* = a?- d'b + ab^ - b\ = a?-^a'b + ab''-{-W. a — b = a*- a% 4- a'b^ - aW + b\ «* + &'_.4 a + 6 a — b = a^ + o?b + a?l^-\-dt^-\-b^\ etc. In these results we observe the following laws : I. The number of terms is the same as the exponent of a in the dividend. II. The exponent of a in the first term is less by 1 than the exponent of a in the dividend, and decreases by 1 in each succeeding term. III. The exponent of b in the second term is 1, and increases by 1 in each succeeding term. IV. The terms are all positive when the divisor is a — 6, and are alternately positive and negative when the divisor is a + 6. 100. In connection with Art. 99, the following principles are of great importance : If n is any whole number, (1) a** + 6" is divisible by a-\-b if n is odd, and by neither a-\-b nor a — b if n is even. (2) a" — 6" is divisible by a — b if n is odd, and by both a-{-b and a — b if n is even. 50 ALGEBKA. EXAMPLES. 101. 1. Divide cH - H' hy a-b. Applying the laws of Art. 99, we have ^Sll. = a« + a'b + a'b' + a'b^ + a'b' + ab' + b', Ans, a — b 2. Divide a;^ - 81 by a; + 3. Since 81 = 3^, we have ^^~^^ = a^-3a;^ + 3'a;-3^ = a^-3a^ + 9a;-27, Ans, fl; + 3 Write by inspection the values of the following : 3. «°-»''. 8. a — b 4. ^:^. 9. x + y 5. "^' + < 10. 6. ^'-< 11. 7. i-Hi^. 12. l-x Divide the following : 18. m*-16n«by m-27i2. 20. ^2d -^b^ hy 2a + b. 19. Q^^i^T^hy x — yz. 21. m*- 243n'' by m — 3». 22. 256x^-2/«by 4a; + y^ x' -16 X -2 1- -a^ 1 — a a^ + 1 a + 1 1- -n« 1 — n 03^ -81 13. aj«-32 0^-2 14 a«-64 a + 2 16. ai« + &^' a2 + 62 16. aj^-128 x-2 17. x^ + 243 a; 4-3 FACTORING. 61 VIII. FACTORING. 102. Factoring is the process of resolving a quantity into its factors. (Art. 11.) 103. The factoring of monomials may be perfoi-med by inspection ; thus, 12a^6-c=2.2.3aaa56c. A polynomial is not always factorable ; but there are certain forms which can always be factored, the more important of which will be considered in the succeeding articles. Case I. 104. When the terms of the polynomial have a common monomial factor. 1. Factor a^ 4-3 a. Each term contains the monomial factor a. Dividing the expression by a, we have a^-f-S. Hence, a^-f3a = a(a^4-3), Ans. 2. Factor Uxy* — 35 a^/. Uxy^-35a^y^=7xy\2y'--5ay'), Ans. EXAMPLES. Factor the following : 3. ar + 6x. 8. 5a;=^-f 10a.-2 + 15a;. 4. Sm^-12m\ 9. a' -2a^ + Sa^ -a\ 5. 16a^-12a. 10. SQx^y -GOx'y* -84:xy. 6. 27c*cC^-\-d(^d. 11. 21m^n-\-35mn^-Umn. 7. 60mV-12m3. 12. SAx'f -UOx^y^-j-lOx^- 62 ALGEBRA. 13. Factor the sum of 54a^6^ - 72 aV, and - 90 a^d. 14. Factor the sum of 96c*c?^ 1200^^^ and - U4.(fd\ Case II. 105. When the polynomial consists of four terms, of which the first two and the last two have a common binomial factor, 1. Factor am — hm + an — hn. Factoring the first two and last two terms as in Case I, we have m{a — h) -\- n{a — h) . Each term now contains the binomial factor a~b. Divid- ing the expression hj a — b, we obtain 7n + n. Hence, am — bm-\-an — bn = (^a — b){m-\- n) , Ans. 2. Factor am — bm — an + bn. am — bm — an-\-bn = am — bm — {an — bn) = m(a — b)—n(a — b) = (a — b) (m — n) , Ans. Note. If the third term is negative, as in Ex, 2, it is convenient, before factoring, to enclose the last two terms in a parenthesis preceded by a — sign. EXAMPLES. Factor the following : 3. ab -\- bx -\- ay -{- xy. 8. a^ — a^b — ab'^ + W. 4. ac — cm -\-ad — dm. 9. o^ + ax — bx — ab. 5. x^-\-2x — xy — 2y. 10. mx^ — my^ + ?ix^ — ny'^. 6. Q(? — ax — bx-\-ab. 11. x^ -\- x^ -\- x -\- \ . 7. o?-a%-\-ab''-b\ 12. ^x^ ^\x? -9x-^. 13. ^cx—\2cy^2dx — Zdy. 14. 6n-21m2n-8m + 28m^- FACTORmG. 53 106. If a quantity can be resolved into two equal factors, it is said to be imperfect square^ and one of the equal factors is called its square root. Thus, since 9 a^6^ equals 3a-6 X 3a^6, it is a perfect square, and 3 a^h is its square root. Note. 9 a^b- also equals — 3 a^i x — 3 a%, so that its square root is either 3 d-b or — 3 a-b. In the examples in this chapter we shall con- sider the positwe square root only. 107. The following rale for extracting the square root of a monomial is evident from Art. 106 : Extract the square root of the coefficient^ and divide the exponent of each letter hy 2. For example, the square root of ^ba^^fz^ is hs^y^z. 108. It follows from Art. 05 that a trinomial is a perfect square when its first and last terms are perfect squares and positive, and the second term is twice the product of their square roots. Thus, 4a^ — 12£c?/ + 9?/^ is a perfect square. 109. To find the square root of a perfect trinomial square, We take the converse of the rules of Art. 95 : Extract the square roots of the first and last terms ^ and wnnect the results by the sign of the second term. Thus, let it be required to find the square root of ^x^-l2xy-\-^y-. The square root of the first term is 2 a;, and of the last term 3 2/; and the sign of the second term is — , Hence the required square root is 2a; — 3y, 54 ALGEBRA. Case III. 110. Wlien a trinomial is a perfect square (Art. 108). 1. Factor a^ + 2 ab' + b\ By Art. 109, the square root of the expression is a + b^. Hence, a2 4- 2 ab' -\-b'=(a-i- b') (a + b') , or (a + b'Y, Ans. 2. Factor 4tx^ - 12xy + 9y\ 4:X^—12xy-{-dy^ = {2x-3y){2x-3y) ^{2x-3yy, Ans. Note. The given expression may be written di/^ — 12xi/-{-4x'^; whence, 9j/2_ 12x^4- 4^-2= {Sy-2x){3r/-2x) = {Sy-2x)^; which is another form of the answer. EXAMPLES. Factor the following : 3. x^-\-2xy-}-y\ 16. 36 m^ - 36 mn + d nK 4. 4 + 4m + m2. 17. 4:a^ -{-Uab -h "i 21 b\ 5. a^- 14a; + 49. 18. x^ -\-8x^ -\-16x\ 6. a^- 10 a 4- 25. 19. a'b* -\- 18 ab'c -\- 81 c^.^ 1, 2/^ + 22/ + l. 20. 2bx--10xyz-{-4.^y-z^. 8. m2-2m+l. 21. 'do^ -66x^ -^121x\ 9. a;*+12a;2 + 36. 22. da^ -{-60a''bG'd+100b^(^d\ 10. n«-20n3 + 100. 23. 64»«- 1600;^ + 100a;«. 11. icy+ 160^2/ + 64. 24. la^l)' -{-h2a^b^ ^U^a^bK 12. l-l0ab + 2ba'b\ 25. 16a;*-120mna^+225mV. 13. 16m'^-8am + a\ 26. {a-by -\-2{a-b)+l. 14. a^ + 2a^-{-a\ 27. {x-^yy-16{x-\-y)^64.. 15. x"^ - 4.x' + 4:x\ 28. {x'-xy + 6(x'-x) + ^. FACTORIKG. 55 Case IY. 111. When an expression is the difference of two perfect squares. Comparing with the third case of Art. 95, we see that such an expression is the product of the sum and difference of two quantities. Therefore, to obtain the factors, we take the converse of the rule of Art. 95 : Extract the square root of the first term and of the last term; add the results for one factor, and subtract the second result from the first for the other. 1. Factor 36 a.-2- 49/. The square root of the first term is 6 a;, and of the last term ly. Hence, by the rule, 36ar-49/ = (6x-f 72/)(6ic-72/), Ans, 2. Factor (2 a; -3 2/) 2 -(a;-?/) 2. {2x-^yY-{x-yy = [i2x-Zy)-\-{x-y)-][{2x--^y)-{x-y)-] = {2x-3y + x-y){2x-3y-x-\-y) = (Bx — 4:y){x — 2y), Ans. EXAMPLES. Factor the following : 3. x'-y^ 7. 9a.'2-16/. 11. Adm^-100n\ 4. ar'-l. 8. 2oa'-b'. 12. 36x^-81y\ 5. 4:-a\ 9. l-49a^/. 13. 64a2- 1216V. 6. 9m=^-4. 10. a'b^-c'd\ 14. lUxY-225z''. 56 ALGEBRA. 15. (a-\-by~{c-\-dy. 19. (x-cy-(y-dy. 16. (a-c)2-62. 20. (a-3y-{h + 2y. 17. m^-{x-yy. 21. (2x + my -(x-7ny. IS. m'-{m-iy. 22. {3a-\-5y -{2a-3y. It is sometimes possible to express a polynomial in the form of the difference of two perfect squares, when it may be factored by the rule of Case IV. 23. Factor 2 m7i + m^ - 1 -f 7i\ The expression ma}' be written m^ + 2 mn + n^ — 1, which, by Case III., is equivalent to (m-|-?i)^— 1. Hence, by the rule, (m + ?i)^ — 1 = (?7i + ?i + 1) (m + w — 1) , A71S. 24. Factor 2xy -{- 1 -x^ -f. 2xy + 1 - x" - y'- = 1 - x" + 2xy - y'- = 1- (ay^- 2xy+y^) , By Case III., this may be written l—(x — yy. Hence the factors are ll+{x-~y)^[l-(x-y)-]^(l-}-x-y)(l-x + y), Ans. 25. Factor 2xy + W - x^ - 2ab — y"^ -\- a\ 2xy-\-b^-x^-2ab-y^-\-a' =^ a" -2ah -{-b'' - x" + 2xy - y^ = a'-2ab + b'-(i^-2xy-hy^) = (a - 6)2 _ (a; _ 2/)2, by Case III. = [(a-6)+(a.-2/)][(a-6)-(a;-y)] = (a — 6 + a; — 2/)(a — 6 — a; + 2/), Ans. Factor the following : 26. x'-^2xy + y^-A. 28. a2-62 4-26c- c^. 27. a2-2a6 + &'-c2. 29. a^-ft^- 26c-c2. FACTORING. 57 30. c^ _ 1 + fZ2 -f- 2cd. 32. 4& - 1 - 46- + 4m^ - 31. 9_a^_2/2_^2a:2/. 33. 4a' -{-b' -9d' - 4.ab. 34. • a' - 2am -]-m' ~b' -2bn- nK 35. x'-y'-{-c'-d''-2cx + 2dy. 36. a2 -b'-j- m- - ii' + 2 am + 2 6n. 37. a'-W + c'-d'-ir2(M-2bd. Case V. 112. When an expression is a trinomial of the form a^-\- ax-{-b. In Art. 97 we derived a rule for the product of two bino- mials of the form x-{-a^ a; -}- 6, by considering the following cases in multiplication : 1. (ic-f 5)(.^'4-3) = a^ + 8a;-^15. 2. (x-r:>){x-B) = x^-8x-\'l5. 3. (a; + 5)(x-3) = a^ + 2a;-15. 4. {x-6)(x-i-S) = x^-2x-U. In certain cases it is possible to reverse the operation, and resolve a trinomial of the form af -\-ax-{-b into the product of two binomial factors. The first term of each factor will obviously be x ; and to obtain the second terms, we take the converse of the rule of Art. 97 : Pi7id two numbers whose algebraic sum is the coefficient of x, and whose product is the last term. Thus, let it be required to factor a^ — 5 a; — 24. The coeflScient of a; is — 5, and the last term is — 24 ; we are then to find two numbers whose algebraic sum is — 5, and product — 24. By inspection we determine that the numbers are —8 and 3. Hence, a^-5x-24 = (x-8)(x-i-3). 58 ALGEBRA. 113. The work of finding the numbers may be abridged by the following considerations : 1. When the last term of the product is +, as in Exs. 1 and 2, the coefficient of x is the sum of the numbers ; both numbers being + when the second term is +, and — when the second term is — . 2. When the last term of the product is — , as in Exs. 3 and 4, the coefficient of x is the difference of the numbers (disregarding signs); the greater number having the same sign as the second term, and the smaller number the opposite sign. We may embody these observations in two rules, which will be found more convenient than the rule of Art. 112 in the solution of examples : 1. If the last term is -f, Jlnd two numbers whose sum is the coefficient of x, and whose product is the last term ; and give to loth numbers the sign of the second term. II. If the last term is —^find two numbers whose difference is the coefficient of ^^ and whose product is the last term; give to the greater number the sign of the second term^ and to the smaller number the opposite sign. ITote. By the expressions " coefficient of x " and " last term," in the above rules, we understand their absolute values, without regard to sign. EXAMPLES. . 114. 1. Factor a^+14a; + 45. According to Rule I., we find two numbers whose sum is 14, and product 45. The numbers are 9 and 5 ; and as the second term is +, both numbers are -|-. Hence, a^+14i» + 45 = (a;-h9)(a; + 5), Ans. 2. Factor ic^ — 6 a; + 5. By Rule I., we find two numbers whose sum is 6, and FACTORlKa. 59 product 5. The numbers are 5 and 1 ; and as the second term is — , both numbers are — . Hence, x^—6x-^6={x — o){x — l), Ans, 3. FsLctorx^-j-5x-U. By Rule II., we find two numbers whose difference is 5, and product 14. The numbers are 7 and 2 ; and as the second term is -f , the greater number is -h , and the smaller number — . Hence, x'^6x-U = (x+7){x-2), Ans. 4. Factor a^ — 5x — 24. By Rule II., we find two numbers whose difference is 5, and product 24. The numbers are 8 and 3 ; and as the second term is — , the greater number is — , and the smaller number +. Hence, a^-5x-24 = (a;-8)(a; + 3), Ans, Factor the following : 5. x' + bx + G. 17. x'-ex-U. 6. x'-3x+2. 18. m^ + Um + eS, 7. 2/2 + 22/ -8. ' 19. a^-loa + U. 8. m2-7m-30. 20. / + 72/-60. 9. a^-lla+lS. 21. a;--lla;+10. 10. x^ + x-6. 22. m^ + 27n-80, 11. c2 + 9c + 8. 23. w2-f-23n+102. 12. 2/' -22/ -35. 24. a^-9a;-90.' 13. a2+13a-48. 25. a- -11 a -26. 14. xr-10x + 2\. ' 26. x^-\-x-4:2. 16. ar^+13a7 + 36. 27. c2-18c4-32. 16. n2-n-90. 28. m2-8m-33. 60 ALGEBRA. / 29. a^ + 20x-{-75. 37. o^^- 19.^'2- 120. 30. 3^ + 405-96. 38. c« + 12c3 + ll. 31. /-17?/-110. 39. a^y^ + 2xf-120. 32. a^-19x+78. 40. a254_ 7^52_ ^44^ 33. x^'+lx-dS. '^■^. 41. nV + 25 7ia; + 100. 34. a2 + 22a + 105. 42. 2/^-20/ + 91. 35. a^-23i» + 130. 43. o.:^b^ -2a'b^-4.8. 36. a^ + 10a2-144. 44. m* + 26m2-87. 45. Factor ir^ + 5 abaP - 84 a^^^. "We find two numbers whose difference is 5, and product 84. The numbers are 12 and 7 ; and, by the rule, the greater is +, and the smaller — c Hence, a-* + 5 abx" - 84 a'b^ = (x^ + 1 2 ab) {p^-1 ab) , Ans, 46. Factor 1 — 6 a — 27 o^. The numbers whose difference is 6, and product 27, are 9 and 3. Hence, l-6a-27a2= (i_9a)(l4-3a), Ans. Factor the following : 47. a'-^ax + 2x'. 66. (a + 6)^ + 5 (a + 6)'+ 4. 48. ix?-\-bxy-e>Q,y\ 67. l-9a + 8a2. 49. l + 13a + 42a2. 68. b^ + ^ aW - b2 a\ \ 60. m^— 15m?i + 56w^ .69. (m — 71)^+ (m — 71) — 2. 61. o?-ab-b(Sb\ 60. x^ -bx^ -b()x^. 62. a^b^ + 4:abG-4:5c\ 61. a^ + 8ab + 12b\ 63. l-3a;-10a^. 62. 1 -ISxy + 4:0a^yK 64. a^ + 15a3 + 44a2. 63. (a- 6)2- 3(a- 6) -4. 65. ;s2_iOcc2/''2-39ajy. 64. a^y + 80:22,22; _ 43 ;22^ J'ACTORING. 61 115. If a quantity can be resolved into three equal fac- tors, it is said to be a perfect cube, and one of the equal factors is called its cube root. Thus, since 27 aW equals Sa^b x Sa-b x 3a^6, it is a per- fect cube, and 3a^6 is its cube root. 116. It is evident from the above that the cube root of a monomial may be found by extracting the cube root of the coefficient and dividing the exponent of each letter by 3. Thus, the cube root of 126 scf^y^x^ is bx^y^z. Case VI. 117. }Vhen an expression is the sum or difference of two perfect cubes. By Art. 98, the sum or difference of two perfect cubes is divisible by the sum or difference of their cube roots ; and in either case, the quotient may be written by inspection by aid of the rules of Art. 98. EXAMPLES. 1. Factor a^ + l. The cube root of a^ is a. and of 1 is 1 ; hence, one factor is a + 1 . Dividing the expression by a + 1 , we have the quotient a^ - a + 1 (Art. 98) . Hence, a3+l=(a + l)(a'-a + l), Ans. 2. Factor 21:x?-Q4.f. The cube root of 21 x^ is 3ic, and of 64^/^ is 4?/; hence, one factor is 3a; — 41/. By Art. 98, the other factor is ^a?+l2xy + Uy''. Hence, 21o^-Q>^f = {^x-Ay){^x^-\-l2xy'{'lQy'-), Ans, 62 ALGEBRA. Factor the following : 3. a^ + a?. 8. a^ + h\ 13. m^-Un\ 4. m^-n^. 9. x^ + \. 14. 64ar^-125. 6. a^-1. 10. 27a^-l. 15. 125a'^ + 27m». 6. a^y' + i?. 11. 8c«-d^ 16. 64c3(f + 27. 7. l-Sic^. 12. 27 + 8a3. 17. Uh-B,a^b\ Case VII. 118. TFi^ew an expression is the sum or difference of two equal odd powers of two quantities. By Art, 100, the sum or difference of two equal odd powers is divisible by the sum or difference of the quantities ; and in either case, the quotient may be written by inspection by aid of the laws of Art. 99. EXAMPLES. 1. Factor a** 4- &^ By Art. 100, one factor is a-\-h. Dividing the expression by a + 6, the quotient is a^ - a^b-\-a"b'^ - ab^ + b^ (Art. 99). Hence, a'' + 6^ = (a + b) {a^ - a% + a'W - ab^ + b') , Ans, Factor the following : 2. a^ — b'. 5. mJ + n\ 8. c^-m^nK 3. ar' + l. 6. x^ -y\ ^ 9. l+32n^ 4. l-a\ 7. a'-l. 10. 243^;'^-/. 11. a;^4-128. 12. 32 -243 a''. 119. By applying one or more of the rules already given, an expression may often be separated into more than two factors. FACTORING. 68 1 . Factor 2 ao^y^ — 8 axy'^. By Case I. , 2 aarV — 8 ax]^ = 2 axy^'i^ — 4y^) . Factoring the quantity in the parenthesis by Case IV., %a7?f — %ax^ = 2axy^{x^'ly){x-1y), Ans, 2. Factor m® — n^. By Case IV. , m® — w* = (m" + n^) (m^ — n') . By Case VI. , m* + n^ = (m + n) (m^ — mn + n^) , and m^ — w^ = (m — n) (m^ + mw + ?i^) . Hence, m^— n®= (m+n) (m— n) (m*— ?nn4-n^) (m^-hwin+w*), -47i«. 3. Factor oi? — ]/^. By Case IV., ^^f^{x'^J^y'){x'^-^-) = (a;* + 2/0 (^ + 2/0 (^' + y) (^ - 2/) » ^w«- MISCELLANEOUS EXAMPLES. 120. In factoring the following expressions, the common monomial factors should be first removed, as shown in Example 1 of the preceding article. 1. ^a^^-^o!^x. 8. 5a3-5. 2. i_4a; + 4ar^. 9. cv^U-^dK 3. x«-l. 10. x'-\^. 5. a;2 _|_ ti.^ _|_ 2>^. _^ ci^,. 12. 3a^ + 27a' + 42. 6. m--7m-8. 13. ar^- (2^/ - S;^)^. 7. 2ar^ + a;. 14. r'= -i- eOa6 + lOOd^. 64 ALGEBRA. 15. 5 a'bc- 10 ab^'c- Id abc\ 21. l + 12a5 + 27fi;^ 16. 3a^-21a^-\-30a\ 22. 18a^''y-2xy\ 17. a^ + 8f^. 23. x'-x". 18. 2a^-2a. 24. 4ar22/*4- 28a;,v-+ 49. 19. l-a2-62 + 2«6. 25. a^ + 6a-^-40. 20. a^-8a;+7. 26. a^- 18«Z;-4062. 27. 2 ar''?/ + 2 xf - 2 a;^/^^ + 4 icy . 28. 12m3n - 18mV + 24 ^mi^ 29. 32a^5 + 4a6^ 40. (x^ + y^ -z^ -4:x'y'. 30. a;^ — 81. _ 41. a"bc - ac^d - ab^d -\- bcd\ 31. 2/-2/^ 42. a2-14a5 + 3362. 32. 0^3 + 2^2 -a; -2. 43. 3a;V + 3a;/. 33. x^-{-7x^-S0x\ 44. 477i^- 20m2/i + 25n2. 34. (Sx-^yy-(x-2yy. 45. 3 a-^6 + 3 a^fe^ _ ^ ^^3^ 35. mV-8maj3-65. 46. a'x^ -a-y- -b^x^ +by. 36. 1350^ -5ar^ 47. (a - 26)2-2(a-26)-8. 37. 2a^y-2xY-60xy\ 48. lOOa^?/*- 81^'. 38. 80a^f-5x^y. 49. a«-64. 39. 3a^b-{-18a'b-\-27ab. 50. i«^-(a;-6)^ 51. (a2 + 3a)2-14(a2 4-3a)+40. 52. (4w + w)2-(2m-3n)2. 53. (a2-62_c2)2_462c2. 57. a''-\-b^-c'-d^~2ab-2cd. 54. 1000 + 27m«. 58. (ar^ + 4)2- 16a^. 55. x^ — X" — x-{-l. 59. a^ — y^ — Sxy(x — y). 56. 3 (a2- 6^) -(a -6)2. 60. (a^ + a-^/-^. HIGHEST COMMON FACTOR. 66 IX. HIGHEST COMMON FACTOR. 121. A Common Factor of two or more quantities is a quantity which will divide each of them without a remainder. Thus, 2an/^ is a common factor of 12ary and 20x^y^. 122. A prime quantity is one which cannot be divided, without a remainder, by any integral quantity except itself or unity. For example, a, 6, and a-\-c are prime quantities. 123. Two quantities are said to be prime to each other when they have no common factor except unity. Thus, 2a and 36* are prime to each other. 124. The Highest Common Factor of two or more quanti- ties is the product of all the prime factors common to those quantities. It is evident from this definition that the highest common factor of two or more quantities is the expression of highest degree (Art. 33) which will divide each of them without a remainder. Thus, the highest common factor of a^if and x^y*^ is a?y^, 125. In determining the highest common factor of alge- braic quantities, it is convenient to distinguish three cases. Case I. 126. When the quantities are monomials. 1. Find the H.C.F. of ^2a^b\ 70 a'bc, and 98a*b^cP. 4.2a'b' = 2'3-7'a^b'' 70a-bc=2-6'7'a'bc 98a*b^d' = 2'7'7'a'b^d^ Hence, the H.C.F. = 2.7- a'b{Axt. 124) = Ua'b, Ans, ee ALGEBRA. RULE. To the highest common factor of the coefficients^ annex the common letters^ giving to each the lowest exponent with which it occurs in any of the given quantities, EXAMPLES. Find the highest common factors of the following : 2. a^ix^, 7a*x, 5. 18 mn\ 46mhi, 72 m^n\ 3. 16cd\ dcH. 6. lUxfz^ Uia^y:^. 4. bAa^b, 90ac2. 7. ISa^o;, 45ay, 60aV. 8. lOSa^yV, lUxfz*, 120a^2/V. 9. d6a'b\ 120 a^b', 168 a%^ 10. SlaWii, 85 aSn^x, llda*mY. Case II. 127. WTien the quantities are polynomials which can be readily factored by inspection, EXAMPLES. 1. Find the H.C.F. of 5a^y-16x^ysiiid 10xPy + 40oi^y — 210xy. By the methods of Chapter VIII. , 5a^2/ — 16afy=:6x^y{x — S) 10ix^y + 40x^y — 210xy=10xy{x^-^4:X~21) = 10xy(x + 7)(x-S). In this case the common factors are 5, a?, 2/, and a? — 3. Hence, the H.C.F. = 5xy(x — S), Ans. 2. Find the H.C.F. of 4a52_4a; + l, 4a^-l, and 2ax-a-2bx + b. HIGHEST COMMON FACTOR. 67 4ic2-l=(2a; + l)(2a;-l) 2ax-a-2bx + b= (a - 6) (2 a; - 1) Hence, the H.C.F. = 2a; — 1, Ans, Find the highest common factors of the following : 3. 3 aoi^ — 2 a^x and aV — 3 abx, 4. x^ — y^ and a^ + y*. 5. 9a^-462and(3a2-26)«. 6. 2a^-2ar' and 6a^- 6a;. 7. 3ca; + 21c-3da;-21cf andar^ — 3a7-70. 8. m^w + 2mV + mn'' and m*n + m7i*. 9. 3ar'' + Oa.-^- 120a; and Saar'-Oaa;- 30a. 10. 3a;y-4?/ + 3a!2;-42; and Qa.-^— 16. 11. a;2_a;-42, ar'- 4a; -60, and ar^+ 12a; 4- 36. 12. a2-l, a«+l, anda2 + 2a4.1. 13. 4ar^-12a; + 9, 4ar^-9, and4m2wa;-6m2w. 14. a;^ — a;, a.'^ + ^a;^ — 10 a;, and a."^ — a;. 16. a3-86», a2-a5-262, Siud a^ - 4 ab + UK 16. 2ar»+2a;2_4^^ 3a;^+6a;»-9a;2^ and 4ar^-20a;*-f-16a;*. 17. 8m»-125, 4m2-25, and 4m«- 20m4-25. 18. a;*-16, a.*2-a;-6, and (a;2_4^2^ 19. 3aa;«-3aar', aa;^ - 9 aa;^ ^ g ^^.^ and 2 aar^ — 2 aa;. 20. a^ -h\ab-b^ + (m- be, and a^ - a^b + ab"" - b\ 21. 12aa!-3a + 8ca;-2c, 16ar^-l, and 160;^ -8a;-f 1, 68 ALGEBRA, Case III. 128. WJien the quantities are polynomials which cannot be readily factored by inspection. The rule in Arithmetic for the H.C.F. of two numbers, is Divide the greater number by the less; if there is a re- mainder, divide the divisor by it; and so on; continuing the operation until there is no remainder. Then the last divisor is the highest common factor required. For example, required the H.C.F. of 169 and 546. 169)546(3 507 39)169(4 156 13)39(3 39 Therefore 13 is the H.C.F. required. 129. We will now prove that a similar rule holds for the H.C.F. of two algebraic quantities. Let A and B be two expressions, the degree of A being not lower than that of B. Suppose that B is contained in A p times with a remainder C ; that G is contained m B q times with a remainder D ; and that D is contained in C r times with no remainder. To prove that D is the H.C.F. of A and B. The operation of division is shown as follows : B) A {p pB ~C) B(q q_C D)C(r rD We will first prove that D is a common factor of A and B. HIGHEST COMMON^ FACTOR. 69 From the nature of subtraction, the minuend is equal to the sum of the subtrahend and remainder (Art. 59). Hence, A=pB + C (1) B = qC+D (2) Substituting the value of G in (2) , we have B = qrD-\-D = D{qr-^l) (3) Substituting the values of B and (7 in (1), we have A ==pD {qr-\-l) + rD = D (pqr -i-p + r) (4) From (3) and (4) we see tliat D is a counnon factor of A and B. We will next prove that every common factor of A and B is a factor of D. Let K be any common factor of A and jB, such that A = mA", and B = nK. From the operation of division, we see that C = A-pB (5) D = B-qG (6) Substituting the values of A and B in (5) , we have C = mK—pnK, Substituting the values of B and C in (6), we have D = 7iK— q {mK—piiK) = K{n — qm +pqn) . Hence K is a factor of D. Therefore, since every common factor of A and J5 is a factor of Z), and since D is itself a common factor of A and B^ it follows that D is the highest common factor of A and B. 130. Hence, to find the H.C.F. of two algebraic expres- sions, A and B, of which the degree of A is not lower than that of B, 70 ALGEBRA. Divide Ahy B; if there is a remainder, divide the divisor by it; and continue thus to make the remainder the divisor, and the preceding divisor the dividend, until there is no re- mainder. Then the last divisor is the highest common factor required. Note 1. Each division should be continued until the remainder is of a lower degree than the divisor. Note 2. It is important to keep the work in the same order of powers of some common letter, as in ordinary division. 1. Find the H.C.F. of 18ic3_51a!24-13rK + 5 and 6i»2_i3aj-5. 18a^-39aj2-15a7 -12a^+28a;+ 5 ~12ar+26a;+10 2x- 5)6a^- -IZx- -5(3. ^+1 6a^- -Ibx 2x- -5 2x- -5 Hence, 2fc — 5 is the H.C.F. required. Note 3. Either of the given expressions may be divided by any quantity which is not a factor of the other, as such a quantity can evidently form no part of the highest common factor. Similarly, any remainder may be divided by a quantity which is not a common factor of the given expressions. 2. Find the H.C.F. of 6a^-26a:^ + 14:XSind 6ax^ +nax—10a. Dividing the first expression by x, and the second by a, we have ex'-25x-{-U)6af+nx-10{l 6a^-26x-j-U 36a?- 24 HIGHEST COMMON FACTOR. 71 Dividing the remainder by 12, 3x - 2)60^ - 26x + U(2x - 7 6a^- Ax -21x-\-U -21X+U Hence, 3a; — 2 is the H.C.E. required. Note 4. If the first term of a remainder is negative, the sign of each term may be changed. 3. Find the H.C.F. of 2o^ --Sx-2 and 2x^ ~ 5x -S, 2ar'_3a;-2)2ar^-5a;-3(l 2a;^-3a?-2 -2aj-l Changing the sign of each term of this remainder, 2a;+l)2ic2-3a;-2(a;-2 2aP+ X -4a;-2 -4x-2 Hence, 2a; + 1 is the H.C.F. required. Note 5. If the first terra of the dividend or of any remainder is not divisible by the first term of the divisor, it may be made so by multiplying the dividend or remainder by any quantity which is not a factor of the divisor. 4. Find the H.C.F. of 2a^-7a^ + 6x-6 and Sx^ - 7x^ -7x-{-3. Since 3 a;^ is not divisible by 2 a^, we multiply the second quantity by 2. 2a^-7x^+6x-Q)6x^-Ux^-Ux-^ 6(3 6a;^-21a;^ + 15a;-18 7ay'-2Qx+2i 72 ALGEBRA. Since 2a^ is not divisible by 7a^, we multiply each term of the new dividend by 7. 7a^-29a; + 24)14a^-49a^+35a;-42(2a; 14a;^-58a^ + 48a; dx^-13x-4:2 Multiplying this by 7 to make its first term divisible by 7a^-29a;+24)63a^- 91a;-294(9 63x^-261 a? +216 170a;-510 Dividing by 170, x-S)7x^-2dx-\-24:(7x-^S lx^-2lx - 8a; + 24 - 8a; + 24 Hence, a;— 3 is the H.C.F. required. Note 6. If the given quantities have a common factor which can be seen by inspection, remove it, and find the H.C.F. of the resulting expressions. This result, multiplied by the common factor, will give the H.C.F. of the given quantities. 6. Find the H.C.F. of 6 a;^ — aa;^ — 5 a^a; and 21 x^ — 26 aa;^ + 5 a?x. Removing the common factor a;, we find the H.C.F. of 6 a;^ — aa; — 5 a^ and 21 a;^ — 26 ax + 5a^ Multiplying the lat- ter by 2, 6ar^-aa;-5a2)42a;2-52aa;+ 10^2(7 42a;^- 7aa;-35a^ — 45 aa; + 45 o? Dividing by — 45 a, x — a)Qo?— ax~bo?{Qx-\-ba ^0? — Q>ax 5 aa; ~ 5 a^ 5 aa; — 5 a^ HIGHEST COMMON FACTOR. 73 Multiplying x — a by a, the common factor, we have «{x — a) ox x^ — ax as the H.C.F. of the given expressions. EXAMPLES. 13L Find the highest common factors of the following : 1. a^ + a; — G and 2a;' — llit' + 14. 2. 6ar^-7a;-24and 12a;2_^g^,_15 3. 2a2-5a + 3 and4a3-2a2-9a + 7. 4. 'l^x'-\-\\ax- 28a2 and 403:^ _ 5^^ q^. _^ i4^j^ 5. 8a=^-22a2 + 5aand6a26-23a64-206. 6. a^ — 5 mx^ + 4 rri^x and 7^ — mit?' + 3 m-x^ — 3 m^x. 7. 5mV + 58m?i2 4.337i2and 10m3 4-31m2- 20m - 21. 8. 2a* 4- 3a3a; - 9 aV and Ga^ -IJa^a; -f Uaar^ - 3a^. 9. a.'^-S anda^-6a^4-lla;-6. 10. 2a.'»-3a^-a; + l and6a;3-a^ + 3a;-2. 11. 8m--22m7i+5n2 and 6 m*- 29771^71 +43 mV- 20 mn^ 12. aa:^ + 2aa^ 4- aaJ + 2a and 33.-^ -V2x^ - 3a^ - 6a;. 13. aa;*— aa;^— 2 aa:^+ 2 aa; and aa;'— 3 ax'* +2 aa^^+aa;^— aa;. 14. 2a;*-2a;» + 4ar^ + 2a; + 6 and 3a;* + 6a:3_33,_g^ 16. a*4-ci^-6a2 + a + 3 and a* + 2a3- Ga^- a + 2. 16. :x^ - x"^ - 6a^ + 20? -^-Q^x B.Tidi x^ -{- x^ - y? -23? -2x. 17. 15a-V-20a-ar^-65a-a;-30a2 and 126a:3 + 206a.-2_166a;-166. 18. a^-\-a?x-\-a'x^ + a3?-\x'^ and a^ -!- 2 a^x + 8 a V 4- 4 aa;^ _ j q aj*. 74 ALGEBRA. 19. x*-^x^-\-x^-l &ndx^-\-3x^-\-2x. 20. x*-a^y-3a^y^-]-5xf-6y'' , and Sx'^ — 5x^y — x^y^ — 7xy^-{- lOy*. 21. 2x'^—5a^-\-5x^-5x+3 and 2x^-7 x^-\-4.:f^ -{-5x-?j. 22. 3a^-2a^6 + 2a262_5a63-26* and 6a^- a^b + 2 a^S^ -2ab^- h\ 132. To find the H.C.F. of three or more quantities, find the H.C.F. of two of them ; then of this result and the third quantity, and so on. The last divisor will be the H.C.F. of the given quantities. EXAMPLES. Find the highest common factors of the following : 1. 2x^-bx — ^2, 4aj2 + 8ic-21, and 6a^4-23a.' + 7. 2. 12a?2_28a;-5, 14 a^- 39 a; +10, and 10a^-lla;-35. 3. 6m2+7mn + 2w2, ^m^ — Im^n- 12 mn'^ - 4. n"^, and 15m^-f 4mn — 4w^. 4. 6a2+13a-5, 6a3+19a2+ 8a- 5, and 3a3 + 2a2 + 2a-l. 5. a^ + 3a^-6a;-8, i»^ + 5a^ + 2a;-8, andar^-3a^-16a; + 48. 6. a^-7a; + 6, a;^ + 3a^-16a; + 12, anda^-5a^4-7a;-3. 7. 2a»-3a2-5a + 6, 2a^ + 3a2_8a-12, and2a3-a2-12a-9. LOWEST COMMON MULTIPLE. 75 X. LOWEST COMMON MULTIPLE. 133. A Common Multiple of two or more quantities is a quantity which can be divided by each of them without a remainder. Hence, a common multiple of two or more quantities must contain all the prime factors of each of the quantities. 134. The Lowest Common Multiple of two or more quantities is the product of their different prime factors, each being taken the greatest number of times which it occurs in any one of the quantities. It is evident from this definition that the lowest common multiple of two or more quantities is the expression of lowest degree which can be divided by each of them without a remainder. Thus, the lowest common multiple of oi^y'^ y^z, and a^2^ is When quantities are prime to each other, their product is their lowest common multiple. 135. In determining the lowest common multiple of alge- braic quantities, we may distinguish three cases. Case I. 136. When the quantities are monomials. 1. Find the L.C.M. of ma^x, 60ay, and 8^ca;». SGa^o; =2'2-^'^'a''x 60ay=2-2'3-6'ay 84ca^ =2.2.3.7.caj3 Hence, the L.C.M. = 2.2.3.3.5.7. a^ca^f (Art. 134) = 1260 a^cxhf, A71S. 76 ALGEBRA. RULE. To the lowest common multiple of the coefficie7its, annex all the letters which occur in the given quantities^ giving to each the highest exponent which it has in any of the quantities. EXAMPLES. Find the lowest common multiples of the following : 2. Qa^b, a^b\ 6. a'b% da^b\ 12a^b\ 3. 10x^y,12y^z. 7. 16x^y, 4:2y^z. 4. 30m\ 21.n\ 8. 8c^d^ lOac, ISa^d. 5. Q>ab, 106c, 14ca. 9. 24m3a^, SOw^y, ^2xyK 10. 2>^xyh\ 63arV, 28x^2^. 11. ^Ott^bd?, 90 ac^d^ bWcd^. Case II. 137. When the quamtities are p)olynoniials which can be readily factored by inspection. 1. Find the L.C.M. oi x^+x-Q>,x^ - ^x-\- 4., and a:^- 9a;. x^j^x-Q,=^ (a;+3) (a;-2) a;S_9a; = a;(fl;+3) (a;-3) Hence, the L.C.M. = x{x - 2)\x + 3) (a; - 3) , (Art. 134) = a;(a;-2)2(a,'2-9), Ans. EXAMPLES. Find the lowest common multiples of the following : 2. y? — y^ and xy — y^. 3. xr — 1 and ar — 7a; — 8. LOWEST COMMON MULTIPLE. 77 4. Sa'b + 8ah' and 6a -6b. 5. m^ — n^ and m^ — n^. 6. a — b and cr — iab + Sb^. 7. ay^ — 2xy-\- 1/ and a^y — xy^. 8. 2a2 + 2a6, 3ab-Sb\ and 4: crc- 4 b^c. 9. aj2 + 2aaj-35a2andiB2-2aa;-15a2. 10. mn -f- n^, mw — ?i^, and ??i^ — ??^. 11. ax-2a-^bx-2b Sinda^-2ab-3b\ 12. aic^ + a^x, ^ — a^, and x^ — a^. 13. 8(a--62), G(a4-6)2, and 12(a-6)2. 14. a^-10x2 + 21a:andaar^ + 5aa;-24a. 15. x'-l,x'-2x+l,iinda^ + 2x-^l. 16. 2-2a^, 4-4a?, 8 + 8a;, and 12 + 12ar'. 17. x2 + 5ic + 4, a^ + 2£c-8, andar*+7ic + 12. 18. a(a; — 6) (a;— c) , 5(a; — c) (a; — a) , and c(x — a){X'- b). 19. (2m-l)2, 4m2-l, and8m3-l. 20. a^ -^a,a^- a^, and a« + a^. 21. a^ — 4a-f 3, a2_|_«_i2, and a^ — a — 20. / 22. l-cc*, l + 2a;2^^4^ ^^jj(j l_2a;2_^^4^ / 23. (a + 6)2 _ c« and (a - c)^ - b\ 24. aa; — a?/ — 6a; + by, {x — y) ^, and 3 a^^ — 3 ab^. 25. 9a^ + 12a^ + 4a;, 18aaj*-12aaj3 + 8aa^, and 27ar^ + 8. 26. x^-y'^-z'-i-2yz&ndx'-y^ + z' + 2^. 78 ALGEBRA. Case III. 138. When the quantities are polynomials which cannot be readily factored by inspection. Let A and B be two expressions ; let F be their highest common factor, and M their lowest common multiple. Sup- pose that A = aF and B — bF; then, AxB = abF' (1) Since a and b can have no common factor, the L.C.M. of aF and bF is abF; that is, Jf= abF; whence, FxM=abF' (2) From (1) and (2) we have AxB = Fx M (Art. 42, 7). That is, the product of any two quaritities is equal to the product of their highest common factor and lowest common multiple. Hence, to find the L.C.M. of two quantities. Divide their product by their highest common factor ; or. Divide one of the quantities by their highest common factor, and multiply the quotient by the other quantity. 139. 1. Find the L.C.M. of 6a^-nx-^12 andl2x^-4:X--21. 6x^-nx-{-V2)12x^- 40^-21(2 12a^-34a; + 24 30x-4:5 2x- S)Gx^-nx + 12(Sx-4: - 8.'«+12 — 8aj-f-12 That is, the H.C.F. of the quantities is 2a; — 3. Dividing 6x^ — nx-i-12 by 2a; — 3, the quotient is 3a; — 4. Hence, the L.C.M. = (3a;-4)(12a;-- 4a;- 21) = 36a;3-60a;--47a; + 84, Ans. LOWEST COMMON MULTIPLE. 79 EXAMPLES. Find the lowest common multiples of the following : 2. 2x'^ + x-Q and4:X^ — 8x + S. 3. 6«2_f.i3^_28 and 12a;2-31a;-|-20. 4. 8ic2_^30a.^7 and 12a;2-29aj-8. 5. ea^-8a^-S0xsindQax^-{-19ax+15a. 6. a' -8 ah -\-lW and a^ -9a?h-\- 23 ab^ - 15 h\ ^ 2m^n — Smn — 2n and 2m* — Qm^ + 6m^ — 87n -j- 8. 8. Goic^-ci^a;- 12a3and lOaa^ — 17a2a; + 3a^ 9. a^ + a2-8a-6 and 2a3 — 5a2-2a + 2. 10. 2a^ + a^-a; + 3 and 2ar» + 5a^-a;-6. 11. «» -2a'b + 2ab^- b^ and a^ + «'& - aM_ 63. 12. x* + 2oi^-{-2a^-\-xsindax^—2ax — a. 13. 2ic^-lla^ + 3a;2 4-10a;and 3.t^- Ua:^ _ (3^^_5a,^ 14.-a;*-a;»-8a; + 8 and x' -8x^ -^9x -2. 140. To find the L.C.M. of three or more quantities, find the L.C.M. of two of them ; then of this result and the third quantity ; and so on. EXAMPLES. Find the lowest common multiples of the following : (l. a^— 1, 2ic2_9a;4-7, and 2ar' + 3a; — 5.* <2. 3a2_2a-l, 6a2- a- 1, and 9 a^- 3a -2. V3. 2a:2_5^^2, 4ar^4-4a;-3, and lOar^- 7a; + 1. 4. ix--6x-18, 4:X^-i-Aa^-3x, and 6x* -}- oa^ -6a^, 6. a^-Ga^H-lla-e, a«-cr-14a + 24, and a^ + a^-Ha +15. 80 ALGEBRA. XI. FRACTIONS. 141. The expression - signifies a-i-b; in other words, - b b denotes that a units are divided into b equal parts, and that one part is taken. Or, what is the same thing, - denotes that one unit is b divided into b equal parts, and that a parts are taken. 142. The expression - is called a Fraction ; a is called the b numerator, and b the denominator. By Art. 141, the denominator shows into how many parts the unit is divided, and the numerator shows how man}^ parts are taken. The numerator and denominator are called the terms of the fraction. 143. An Entire Quantity or Integer is one which has no fractional part ; as 2xy, ov a-\-b. Every integer may be considered as a fraction whose de- nominator is unity ; thus, a = — 144. A Mixed Quantity is one having both entire and fractional parts ; as a + -, or a; H % 2 y + z GENERAL PRINCIPLES. 145. If the nnmerator of a fraction be multiplied, or the denominator divided, by any quantity, the fraction is multi- plied by that quantity. FRACTIONS. 81 I. Let - be any fraction. Multiplying its numerator by etc CLG a c, we have — To prove that — is c times — h b b In each of these fractions the unit is divided into b equal parts ; in the first case ac parts are taken, and in the second case a parts. Since c times as many parts are taken in — as in -, it follows that II. Let — be any fraction. Dividing its denominator by oc c, we have — To prove that - is c times — • b b be In each of these fractions a parts are taken ; but since in the first case the unit is divided into b equal parts, and in the second case into be equal parts, the parts in - will be c times as great as in ». Hence, ' be 146. If the numerator of a fraction be divided, or the de- nominator multiplied, by any quantity, the fraction is divided by that quantity. I. Let — be any fraction. Dividing its numerator by c, we have -• To prove that ^ is — divided by c. b b By Art. 145, (1), cx^ = ^. b Whence it follows that ^ = ^ -^. c. b b 82 ALGEBRA. II. Let - be any fraction. Multiplying its denominator h by c, we have — To prove that — is - diyided by c. be be b By Art. 145, (2), cx^ = ^- be b Whence it follows that — = - -^ c. be b 147. If the numerator and denominator of a fraetion be both multiplied, or both divided, by the same quantity, the value of the fraction is not altered. For, by Arts. 145 and 146, multiplying the numerator mul- tiplies the fraction, and multiplying the denominator divides it. Hence, the fraction is both multiplied and divided by the same quantity, and its value is not altered. Similarly we may show that if both terms are divided by the same quantity, the value of the fraction is not altered. TO REDUCE A FRACTION TO ITS LOWEST TERMS. 148. A fraction is in its lowest terms when its numerator and denominator are prime to each other. Case I. 149. When the numerator and denominator can be readily factored by inspection. Since dividing both numerator and denominator by the same quantity, or canceling equal factors in each, does not alter the value of the fraction (Art. 147), we have the fol- lowing rule : Resolve both numerator and denominator into their prime factors, and cancel all which are common to both. FRACTIONS. 83 1. Reduce — r- to its lowest terms. 45 (rltx 18 a^6^c ^ 2 ♦ 3 • 3 » a%-c Aba'b^x S'S'd-a'b^x Dividing both terms by 3 • 3 • a^b^, we have — -, Ans. X rj^ 27 2. Reduce — to its lowest terms. ic2-2a;-3 x' — 'll _ (a; - 3) (a^ + 3a; + 9) _ar^ 4- 3a; + 9 ^^^^ ar2_2a;-3"~ (a;-3)(a; + l) ^+1 ' Note. If all the factors of the numerator be removed by cancella- tion, unity (which is a factor of all algebraic expressions) remains to form a numerator. If all the factors of the denominator be removed, the result is an entire quantity; this being a case of exact division. EXAMPLES. g a^y^J g 32?7i7t g 15mxy^ xif^ ' 56m*w^ Ibmx^y^ ^ 2a'b'c - ebx'f:^ -^ 115c^a;^y ba%(^' ' 2Qx'fz' ' 2Z&^' 5 12 V 3 54 aV ^^ 154mV ' 32ar^* * 12(h^bc ' 8Sm^xf' ^2 2a^cd-^2abccl ^q 6a^b-j-3a^b\ Ga'xy + Qabxy ' 3a-b^-^Qab^' ^g 3ar^-6a;^y ^^ 4c^-20c + 25 _ 6x'y'-12xf' ' 4cS-25c 14 ^a^y-Q^y , 13 m^- 10m + 16 • aj2_g3,_|.i5* • m2 + m-72 * j^g g^ — 2 a — 15 , ^^g 9an^ — 4 a a2+i0a + 21 96712-126^1 + 46 84 ALGEBRA. 20. .^^-/^' „ . 25. 21. ^i£_iir .. 26. a2 4-a6-662 Sa^-j-f 4:a:^-2x'7j + xy' ac — ad — bc-^bd a^-W aa? — Aa a?-^:x?J^Ux 27?/^- 125 x^- -a^-\-2x- -2 2x' + x^ + Ax + 2 x'- -4:X-{-16 ax' ^ + 64 ace a?- -(b-^cY 23 ga^ — 4 g oo (a.-^— 4) (a;^— 3a;+2) *"""■* ' (a^-4aj+4)(a.'2+x-2) 24 27y^-125 ^9 (a-by-jc-dy 92/' -30?/ + 25 * (a-c)2-(6-(Z)2 Case II. 150. When the numerator and denominator cannot be readily factored by inspection. Since the H.C.F. of two quantities is the product of their common prime factors, we have the following rule : Divide both numerator and denominator by their highest common factor. EXAMPLES. 1. Reduce — -^ ^^^ to its lowest terms. 6a2_a-12 By the rule of Art. 130, the H.C.F. of 2a^-5a + 3 and 6a' — a — 12 is 2a — 3. Dividing the numerator by 2 a — 3, the quotient is a — 1 ; and dividing the denominator, the quotient is 3 a + 4. Hence, 2g^-5a + 3 ^ a-1 ^^^ 6a'-a-12 3a + 4' FRACTIONS. 85 Reduce the following to their lowest terms : x^^Gx-\-5 .- a^ + af-Sx-2 8. 2a^-\-5x^-2x-{-3 g 6y^-19/ + 7y + 12 _ 6f-26y'-\-ny + 20 2a^-3a'-a-2 4. 3x' + 4.x-7 lOa^-a-21 2a'-7a + 6 2m2-5m-f-3 12wi2-28m + 15 x'-2x-S af-2x'-2x-S 12m2 + 16mn — 3n* jj a^-4ar^y + 4a^^-y^ ^ / 10m2 + m?i-2l7i^ ""' a^-2oi^y-{-4:X7f-3f" 151. Since a fraction represents the quotient of its numerator divided by its denominator, it is positive when its terms have the same sign, and negative when they have different signs. Thus, if?=^a;, then = X, and — — = — - = — x. — 6 0—0 152. It follows from Art. 151 that the fraction - can h written in any one of the forms a — a a 6' ~h~' ^^ ^' That is, if the signs of both numerator and denominator are changed^ the value of the fraction is not altered. But if the sign of either one is changed ^ the sign before the fraction is changed. 86 ALGEBRA. 153. If either numerator or denominator is a polynomial, care must be taken, on changing its sign, to change the sign of eoc^ of its terms. Thus, the fraction ~ , by changing the signs of both c — ct numerator and denominator, can be written in the form ^ ' or (Art. 67). — {c—d) d—c rom Art. 151 that the fraction cd 154. It follows from Art. 151 that the fraction — can be written in any one of the forms (-^)?> (-^)(-b) (-a){-b) c{-dy cd ' (_c)(-d)' (-a)6 ab (-a)(-5) or, — , — — -, — — , eic. cd (—c)d c(—d) From which it appears that If the terms of a fraction are composed of factors^ the signs of any even number of factors may be changed without altering the value of the fraction. But if the signs of any odd number of factors a7'e changed, the sign before the frac- tion is changed. Thus, the fraction — can be written in any {x — y){x — z) one of the forms a — b b — a b — a . ' 7 ^7 ^' —- ^^ ^, etc. (y — x){z — x) {y — x){x — z) {y — x){z-x) TO REDUCE A FRACTION TO AN ENTIRE OR MIXED QUANTITY. 155. Since a fraction is an expression of division, we have the following rule : Divide the numerator by the denominator. FRACTIONS. 87 1. Reduce , ^ "" ^^— ^q g^ mixed quantity. Dividing each term of the numerator by the denominator, 3x 3x Sx Sx 3x 8a^-12a^-9a; + 10 , . , 2. Reduce ' to a mixed quantity. 4ar — 3 - 4a;2-3)8a;^- 12a^-9a;+10(2ic-3 8af -6x -12ar^-3ic -12a^ 4-9 -3x-\-l A remainder whose first term will not contain the first term of the divisor, may be written over the divisor in the form of a fraction, and added to the quotient. Thus, the result is 3a; -fl 2a;-3 + 4a;2-3 Or, since the sign of each term of the numerator may be changed, if at the same time the sign before the fraction is changed (Art; 152), we have 8a^-12x^-9x^l0 ^^^_^_3x-l ^^^^ 4a;2-3 4:0^-3 EXAMPLES. Reduce the following to mixed quantities : 3. 5a^-10a; + 4 6. 2^-^l 5x a;-3 A 6a? — 3a^+dx- 3x -2, Y a3-a2-a-2 a^ 4- a - 1 5 a^ + 2f Q Uo?-8x + l x+y 4a;-l 88 ALGEBRA. 9 ^L±A\ 12 aT^ + 2a^ + 3a;4-4 ' a-\-b' ' x'-^x-^l 2 ??i — 3 72 a; + y 11 2a^-a^-9a^+14 ^^ 6ar^-13a;^+ 6a;- 6 TO REDUCE A MIXED QUANTITY TO A FRACTIONAL FORM. 156. The operation being the converse of that of Art. 155, we have the following rule : Multiply the integral part by the denominator ; odd the numerator to the product when the sign before the fraction is + , a7id subtract it when the sign is — ; and write the result over the denominator. 1. Reduce — ^^^ f- ^ — 2 to a fractional form. 2a;-3 By the rule, a^-5 .^ ^^ a;-5+(a^-2)(2a;-3) 2a;-3 2a;-3 ^ a;-5 + 2a^-7a; + 6 2a;-3 ^2^-6^+1 ^^^^^ 2a;-3 (j,2 7^2 K 2. Reduce a-\-b to a fractional form. a — b a^-b'- 5 _ (a + b)(a- b )-(a'~b'-5) a-\- — — — » —^ a—b a—b a—b a—b Note. If the numerator is a polynomial, it will be found convenient to enclose it in a parenthesis, Avhen the sign before the fraction is — . FRACTIONS. 89 EXAMPLES. Reduce the following to fractional forms : 3. 0^+1 +?±i. 11. ^±1'-1. 4. a; + l — . 12. m-?i + — ^ — 5. 1- m — w. 13. a^ — ab-\-b^ — 6. 7x_3-5i^^ll2. 14. ^_3x-Ml=^. 8 a; — 2 7. l_!!Lz:^. As. m» + »^ , -(m-«). m + 71 f m'* + mn + 71'' 8. a + 6-^-±^. 16. l-f-2.T4-4ar^ + -^^tl. a + 6 2a;-l 9. 2 H-3a;-2. 17. x^2y- ^ "^-^f 2« + l iB2_4iC2/4-42/' 10. ^2_^2_^a&(ct + ?>). 18. a^-2^'+3-'^+^^^-^ a — 6 ic^4-3.^'— 2 TO REDUCE FRACTIONS TO THEIR LOWEST COMMOl^ DENOMINATOR. 157. 1. Reduce —V? -^^ and — ^ to equivalent frac- tions having the lowest common denominator. The lowest common denominator is the lowest common multiple of 3a^6, 2a6^, and 4a^6, which is \2a^lr. By Art. 147, both terms of a fraction may be multiplied by the same quantity without altering its value. Hence, ^0 ALGEBRA; Multiplying both terms of -^ by 4a6, we have ?^£^ Multiplying both terms of — - by 6 a^, we have — -• ^^ ^ 2ab^ ^ 12a^b^ Multiplying both terms of — j- by 36, we have f-^* Therefore the required fractions are 20 abed 18 a^mx -. dbny a inr-, :ri and ^, Aj[is. \2a^b^' 12 a'b'' 12 a'b'' It will be observed that. the terms of each fraction are multiplied by a quantity which is obtained by dividing the lowest common denominator by its own denominator. Hence the following rule : Find the lowest common multiple of the given denominators. Divide this by each denominator separately, multiply the cor- responding numerators by the quotients, and write the results over the common denominator. Note. Before applying the rule, each fraction should be in its lowest terms. EXAMPLES. Reduce the following to equivalent fractions having the lowest common denominator ; 2 Sab 2ac ^ 6bc . , , and 14' 21' 6 3. — — and — • a^a^ aa?^ a?x . 4c-l , 36-2 4. — and — 8a62 12 ah 6a^y 8fz ^^ lOxz' FRACTIONS. 91 2^ and ^^ a^ + a - 6 a' - 4 7. -A- and ^ iB2-l a;«-l 8. m, -, and — r — 9. -^, -5-, and ^ r 10 ^y ^^ and ^^ a6 ^^^ m-n am — 6m + aw — 6/1 2 a^ — 2 a6 ^ + ^ , ^^+L_, and ^^ + ^ . ADDITION AND SUBTRACTION OF FRACTIONS. 158. It follows from the definition of Art. 141 that a h a-\-h , a h a — b _-!--=: — I — , and = . c c c c c c Hence the following RULE. To add fractions, reduce them, if necessary, to equivalent fractions having the lowest common denominator. Add the numerators of the resulting fractions, and write the sum over the common denominator. To subtract one fraction from another, reduce them to equivalent fractions having the lowest common denominator. Subtract the numerator of the subtrahend from that of the minuend, and write the result over the common denominator. Note. The final result should be reduced to its simplest form. 92 ALGiifeM. 1. Required the sum of and — The lowest common denominator is 12 ab'^c. Multiplying the terms of the first fraction by 3 6^, and of the second by 2 a, we have 4a-l S-5b^ ^ l2ab^-Sb^ 6a-10ah^ 4ac 66^0 Uab'c 12ab'c _ 12ab^-Sb^-^6a- lOab^ ~ 12 ab^c ^2ab^-Sb^ + ea^^^^ 12 ab'c 2. Subtract ^^-1 from ^^-2. 2x 3a The L.C.D. is 6 ax. Hence, 6a — 2 4a;— 1 12ax — 4:X 12ax-3a 3a 2x 6ax 6 ax 12ax-4:X- - {12ax-3a) 6 ax 12ax — 4:X — ■12ax-[-3a 6 ax _3a-4:X . IS. Qax Note. If a fraction, whose numerator is a polynomial, is preceded by a — sign, care must be taken to change the sign of each term in the numerator before combining it with the others. It is convenient in Buch a case to enclose the numerator in a parenthesis, as shown in Ex.2. EXAMPLES. Simplify the following : 3. 2x-5 J 3a; + ll 4, ^ ■• 12 18 5a62 2a^h f^RACtiONS. 93 g 2a + 3 3a + 5 y 6 — 4a a + 56 6 8 * * 24a 30& ' 6 ^ — 2 2 — 3mn^ g a— 6 2a+& 3a— 6 2mn 3mV 4 6 8 9 q' + l 6a' + l &-2 So" 12a3 66 * 10 2a;-l 2rc + 3 6a;+l 12 "^15 20 ' -| m + 2 mH-2 m + 3 7 14 21 12 2 2a;-l 3x^ + 1 * 3 Qx day" ' 13 a; — 2 3a;+l 6a; — 5 3 14 3a + l 26-1 4c-l 6d + l , 12a 86 16c 24d 15. Simplify -J— + ^ a; + ^*^ a? — ar^ The L.C.M. of a; + a^ anda; — a;^ is a;(l -|-a;) (1 — a;), or a;(l— a;^). Multiplying the terms of the first fraction by 1 — a;, and of the second by 1 + a?, we have 1 1 ^ 1-a; 1+x x-\-QC^ x—af a;(l— a.*^) a;(l — a^) 1 — x + l+aJ a;(l —X') a;(l — ar) 94 ALGEBRA. 16. Simplify « + * «-* *«^ a — h a-\-h a^ — h^ The L.C.D. is o? - b\ Hence, a-^b a — h Aab a — b a-\-b a^ — b^ ^ {a-hby (a -by Aab a2_62 ^2_J2 ^2_52 ^ {a-\-by-(a-by-4:ab a'-b' ^ g^ + 2 a6 + &^ - (a^ -2ab + b^)- 4:ab o?-b' Simplify the following : 17. l+x l-x 24. {m — ny ' m^ — r? 18. 26. 1 1 a2_4a + 4 a^ + a-G 19. 1 1 26. X Zx 2xy x — y x-\-y a? — y^ 20. a b a — b a-f-6 27. a b 2ab a-\-b a-b a^-b^ 21. a+6 a—b a— b a-\-b 28. m 11 mn — n^ m — n n 22. x+y 2xy + x'^ 29. x+y x-y ^ x—y x+y y y{^ + y) 23. 1-^x 1 — x 1—x 1 +» 30. 2 3 2a;-3 X 2x-\ 4a^-l FRACTIONS. 96 Si -1- I 1 2a go 1 X 3?-^x 'a-^l'^a-h a' + W ' l-x {l-xY {l-xf I 1 2cd 33. ah — cd ab-{- cd -a^b^ — (?d? 34 fl;-3 x-\-\ a; + 13 x-'l x-\-h ^-\-'6x-\^ «g x — a X— h {a — hy x—h x — a {x — a){x — b) 1 1 . ^ x{x + l) x{x-l) ar^-1 37 «4-^ . ^-f-c . c + g (6-c) (c-a) (c-a) (a-b) {a-b) (6-c) 38. -i ^-f-^. og 2x-6 x+2 a? + l 40. ^-y I y-^ g-a; (a;+2;) (2/+0) (aj+y) (a;+2;) (a;+y) (.7+2;) In certain cases, the principles of Arts. 152 and 154 en- able us to change the form of a fraction to one which is more convenient for the purposes of addition and subtraction. 41. Simplify ^+^i:_. a — b b^ — or Changing the signs of the terms in the denominator of the second fraction, and at the same time changing the sign before the fraction (Art. 152), we have _3 25 + a a-b a'-b^' 96 ALGEBRA, The L.C.D. is now a^ — 6^. Hence 3 2b-j-a _ 3{a + b) 2b + a a-b o?-b''~ d'-b^ a^-W 3(a + &)-(26 + o^) a?-b^ 3a + 36-2&-a _2a + 5 . 42. Simplify 1 {x-y){x-z) (y-x){y-z) {z-x){z-y) By Art. 154, we may change the sign of the factor y — x in the second denominator, at the same time changing the sign before the fraction ; and we may change the signs of both factors of the third denominator. The expression then becomes I + 1 ?_. (x-y){x-z) {x-y){y-z) {x-z){y-z) The L.C.D. is now {x — y){x — z)(y — z). Hence the result = (y-g)4-(a?-g) — (a;-2/) ^ y -z-{-x-z-x-{-y {x-y){x-z){y-z) {x-y){x-z){y -z) 2y-2z ^ 2(y-g) (x-y){x-z)(y-z) (x-y)(x-z){y -z) 2 (x-y){X'-z) , Ans. Simplify the following : 43 _JL_+_2 45 _l_-f_L_. *"• a'-ab^b'-ab ' Sx-x'^x'-Q 44 5a + l 3a-l ^ 1 1__ ' 3a — 3 2 — 2a ' m^ — mn n^ — m? FRACTIONS. 47. (a-2){x-\-2) {2-a){x-^a) 48. a a 2oL- a + 5 h-a a?-W 49. X X x^ l+x l-x x^-1 50. 2-x x-^ «2_5a._,.g 51. ^ I ^ 1 (a-6)(5-c) {h-a){a-c) (c- -a)(c- -6) 52. 2 3 1 97 (a;-2)(a;-3) (3-a;)(4-a;) (a;-4)(2-a;) MULTIPLICATION OF FRACTIONS. ft /• 159. Required the product of - and — b d 2 5 5 In Arithmetic, - times - signifies two-thirds of -• Similarly, in Algebra, ^ X - signifies a bths of -• That cl d is, we divide -by b, and multiply the result by a, d By Art. 146,11., |-& = ^- a oa By Art. 145,1., ^xa = ^. bd bd TT a ^c ac Hence, I X 3 = 13* b d bd 98"" ALGEBRA. "We have therefore the following rule for the multiplication of fractions : Multiply the numerators together for the numerator of the product, and the denominators for its denominator. Mixed quantities should be reduced to a fractional form before applying the rule. Common factors in the numerators and denominators should be canceled before performing the multiplication. EXAMPLES. 36V L. Multiply 1^ by i^ ^^ 96aj2 ^ 4.a lOa^ 3 6V ^ 10 . 3 . a^6V y 96a^ 4 ay d-A-a^xy'' 5b^x 6y Ans. 2. Multiply together 3^-^^, ^, and ^^ + ^ ■ ar — 2a; — 3 ar — x x^-{-x — 6 X^ —2X sy ^ —^ sy X^+ X X ~~z X ^ a;(a;-2) (a; + 3)(a;-3) a;^a;+l) (a;-3)(a;H-l) a;(a;-l) (a; + 3)(a;-2) -, Ans. x-1 Multiply the following : o ba^hc -,0 t2a6c -,56 3. -andSmn. 6. — , — , and — Umn" 36 5a d>c 4. ?^and^. 6. «4l^%ndi4. 5a/ ^ahy? 9y^ le;^^ 10 a;^ FRACTIONS. 99 7. 3«^, 3:^, and §^. 12. ^^2a&+^ ^^^ 6 4cc?' 26(i' 96c ' a + b ax—hx 9. ^A^ and -11- 14. i^!^^!±f and 4^- 10. ^^16 and ^=25 . 15. 1 , 4 _ 5 „^ 3aj a^4-5x a^ — 4a; x o? x^-j-x — 2 11. _«^iL and 4r^^ 16. i^^ i^, and -J- (i^ + 2 a6 a^ — ah \ —y x + or \—x jg^ a;^y-4y ^^^^ ^-^-^ . {x-yf-z^ xyr\-2y 19. ^-y" ^ ^ + 2/" ^andl | ^ 20. ^^-(^-^)^ and ^^-(^ + < ^a + cY-h' {a-cy-b' DIVISION OF FRACTIONS. 160. Required the quotient of - divided by -- b d By Art. 85, we are to find a quantity which, when multi- plied by -, will produce -• d b That quantity is evidently ^ ; hence, be a . c __ad b d ^hc TOO ALGEBRA. We observe that the quotient is obtained by multiplying the dividend, -, by -, which is the divisor inverted. We h c have then the following rule for the division of fractions : Invert the divisor^ and proceed as in multiplication. Mixed quantities should be reduced to a fractional form 6efore applying the rule. If the divisor is an integer, it may be written in a frac- tional form, as explained in Art. 143. EXAMPLES. ^' ^"^^^5^^^' 10^- By the rule, 6a^b . 9a^b^ ^ ea^b lOar^y^ ^ Ay ^^^^ 5a^2/' * lOitY 5a.y 9a^b^ Wx 2. Divide — ;r- by — ^^-— lo 5 a^-9 . fl^ + 2a;-3 ^ (a; + 3)(a;-3) 5 15 * 5 15 (a; + 3)(a;-l) ^-^ .,Ans 3(aj-l) Divide the following : 6. .rrJ-T^by 1 o?j^a-12 -" a2-f-3a-18 4: x" ^ 12^3 FRACTIONS. 101 w ar^ — 25 a; , x^ — hx 8. "t-^:„ by ^ o^ + ^ab + y ' G?-IP- m^ — 2m7i + 7i^ m — n 10. 4+f by !^=^. 11. 9+-^by3 + -A2L. j2 a^-8a&^ ^ a^ + 2a^6 + 4a6» a2-2a6-362 ^ a-36 18. A_A_f_Aby-? ^. 32/' xy 23? ^ 3f It? COMPLEX FRACTIONS. 161. A Complex Fraction is one having a fraction in its numerator or denominator, or both. It may be regarded as a case in division ; its numerator answering to the dividend, and its denominator to the divisor. EXAMPLES. 1 Reduce ^^ its simplest form. ,_c bd — c bd — c bd — c "d d 102 ALGEBRA. It is often advantageous to simplify a complex fraction by multiplying both numerator and denominator by the lowest common multiple of their denominators. 2. Reduce t — ^ — 2Jl^ to its simplest form. b . a a—b a+b The L.C.M. of a + 5 and a — 6 is (a + b)(a — b). Multi- plying each of the component fractions by (a -\-b)(a — b)t we have a(a-\-b) — a(a — b)_a^-\-ab — a^ + ab_ 2ab j b{a-\-b)-ha{a-b) ~ ab -j- b' + a' - ab ~ a' + b'' 3. Reduce to its simplest form. X 1 =^+1 ^+\Ans. X Reduce the following to their simplest forms : or—^ 2,1 a^ + 4v^ . , ^ x^ -\-- — ! — - — 4 a; 4. ^. * 6. -. 8. l. l+i 1+- X « 1 1 1 T-~ a-2+- K a m a 0. -. 7. . a__6 1 _i 6 a a 1 y 2 ' X 9. x-l^ 12 ■ a; a; + 3- 18 X FRACTIONS. 17. m m — n n^ mn 1+ '^ 1-x' 1 2 I ^^ 1 —X^-\- 1-ar^ . 12 a4- 6 ■ g — 6 aj + 3 c — d c H-d 12. 2 L^. 19. ^ ^ 3 + -^ aj4 ^ ^-2 1 I ^ + 1 2 3-a; a_6^ a;4-2y x 13. -i-^. 20 ^+^ ^ a J 6 * a;4-22/ a; b a y x-\-y 14. t—-^. 21. -4±-' y X a-^x 1 1 a?^h^ a?-h^ 103 j^ l-x \+x a'-b' a^-j-b^ 1 . 1 * a+^ g— 6 1 —X 1 +« a— b a + b ^ 26 — 2c m — ?i m^ + y^^ g + 6 — c m + 71 m^ — 71^ 16. 4^ 23. j^ ■ 2 c m^ m^n + yi^ g--6 — c m — n (m — ny 104 ALGEBRA. MISCELLANEOUS EXAMPLES. 162. Eeduce the following to their simplest forms : J 26 a + bx-j-ca^ g 10 a^ -{- SO ab + 20 b^ 3m«-75m V ^A V g a^(l + xY-a^(l + xy g l-ax + a(x+a) {l+xy ' * (l-axy+(x + ay A a^ — b^ , a^ — ab ^ {m-\-ny ' bm -\-bn ' (i-^D-e-;) f> 1 +2a^ 2 + a; ^q b(b — ax) + a(a + &a;) 2 4-2a^ 2 + 2a;* * {b - axy + {a + bxy J J aa; b . ax(Sb — ax) ax + b ax~b (^y? — b^ _^ j2 _Jrx-^^_^ 10a; + 10a^ 13. 2 + 4a;+2a^ 9-18a; + 9a;* 6n^-487i^ 971^ + 1871^^ + 36923* 14. a:2_2^2_6^1ii^Vf^. ^ x-Sy ^5. ^ + & <^ 462 a — b a-\-b a? — ''■ (^+^+^^)(^-3- -- 2a; — 1 2a; + l 2a;2_2a;4.i 2a;2^2a; + l' FRACTIONS. 105 1 18. + 14-^ 2\a; — a x-\-aJ a' X 1Q ^ — y ^ + y 20 ^ — 9^ + 26 a; — 24 ""^^ ~' ' ar^- 123.-2 + 47a; -60 ai 2 o J. oj" 62(7a + 6&) 21. a^ — Sab — 2b- ^^ —■ — ^• a — 3b 22. M±l_l y a^' a; + 2/ 2/ — a^ a^ — i/* no (a? + y + ;g)^ + (a;-y)2+(y-g)' + (g-a;)^ 24 1 4 8(1 -aO ' a;-2 (.T-2)2 (a;-2)» «- a* — a*b — ab^ + 6^ a* — a^& — a-62 + ab^ 26 (4a: + .v)2-(a:-2i/)^ • (^Sx-4yy-(2x + Syy 27. _1_ + J_ + _1 (a + 6 + c)^ a + 6 6 + c c + a (a + 6) (6 + c) (c + a) 28 ^(1-^^-) 1-2^ I 2 (14-aj)(l + 9a;) (l.+ a;) (1 +4a;) l+4a; gg 1 1 . 3a;2 3a^ x-1 x+1 a^+1 a:3_i OQ f ct-\-b a^ + 6^ . fa — b a^ — b b a'-by \a-\-b a^-\- 106 ALGEBRA. XII. SIMPLE EQUATIONS. 163. An Equation is a statement of the equality of two expressions. Tlie First Member of an equation is the expression on the left of the sign of equality, and the Second Member is the expression on the right of that sign. Thus, in the equation 2a; — 3 = 3a; — 5, the first member is 2a; — 3, and the second is 3a; — 5. The sides of an equation are its two members. 164. A Numerical Equation is one in which all the known quantities are represented by numbers ; as 2a; — 3 = 3a; — 5. 165. A Literal Equation is one in which some or all the known quantities are represented by letters ; as 2a; + 3a = 6a;-4. 166. An Identical Equation is one whose two members are equal, whatever values are given to the letters involved ; as a? — a^ — {x-\- a) {x — d). 167. The Degree of an equation, in which there is but one unknown quantity, is denoted by the highest power of the unknown quantity in the equation. Thus, 2x — 3 = 3a; 5 ) „ \ are equations of the ^rs^ degree. and orx —be —d) 3a;^ — 2a; = 65 is an equation of the second degree^ etc. 168. A Simple Equation is an equation of the first degree. SIMPLE EQUATIONS. 107 169. The Root of an equation containing but one unknown quantity, is the value of the unknown quantity ; or, it is the value which, when put in place of the unknown quantity, makes the equation identical. Thus, the equation 5a;— 7 = 3a;-|-l, when 4 is put in place of a;, becomes 20 — 7 = 12 + 1, which is identical. Hence the root of the equation, or the value of a:, is 4. Note. A simple equation has but one root ; but it will be seen here- after that an equation may have two or more roots. 170. The solution of an equation is the process of finding its roots. A root is verified, or the equation satisfied, when, on sub- stituting the value of the root in place of its symbol, the equation becomes identical. 171. The operations required in the solution of an equa- tion are based upon the following general principle, which is derived from the axioms of Art. 42 : IftJie same operations he performed upon equal quantities, the results will he equal. Hence, Both members of an equation may he increased, diminished, multiplied, or divided hy the same quantity, without destroying the equality. TRANSPOSITION. 172. Any term may he transposed from one side of an equation to the other hy changing its sigii. For, consider the equation x-\-a=zh. Subtracting a from both members (Art. 171), we have x-\-a — a=zh — a ; or, by Art. 26, x = h — a. 108 ALGEBRA. where -\- a has been transposed to the second member by changing its sign. Again, consider the equation x — a — b. Adding a to both members (Art. 171), we have x — a-{-a = b-\-a; or, xz=b-^a. where — a has been transposed to the second member by changing its sign. Note. If the same term appear in both members of an equation affected with the same sign, it may be suppressed. 173. The signs of all the terms of an equation may be changed without destroy trig the equality. For, consider the equation a — x=b — c. Transposing each term (Art. 172), we have c — b = x — a; or, x — a=c — b, which is the same as the original equation with every sign changed. SOLUTION OF SIMPLE EQUATIONS. 174. 1 . Solve the equation 5x — 7 = 3x-{-l. Transposing the unknown quantities to the first member, and the known quantities to the second, we have 5x — Sx = 7 + U Uniting the similar terms, 2x = 8. Dividing both members by 2 (Art. 171), aj= 4, Ans, SIMPLE EQUATIONS. 109 Note. The result may be verified by substituting the value of x in the given equation, as shown in Art. 169. We have then the following rule for the solution of a simple equation containing but one unknown quantity : Transpose the unknown terms to the first member^ and the knoicn terms to the second. Unite the similar terms, and divide both members by the coefficient of the unknown quantity. EXAMPLES. 2. Solve the equation 14 — 5 .t = 19 + 3a;. , Transposing, — 5a; — 3a;=19 — 14. Uniting terms, —8x= 5. 5 Dividing b}^ —8, « = — -, Ans. 8 Note. To verify this result, put x = — - in the given equation. Then, 8 H-5(-|)=19H.3(-|) Or, " + f = l«-f Or, 1^ = 131; which is identical. 8 8 Solve the following equations : 3. 8a; = 5a; + 42. 9. 5a; + 14= 17 -3a;. 4. 7a;4-5 = -30. 10. 3a; - 31 = 11a;- 16. 5. 7a; + 5 = a; + 23. 11. 18 - 7a; = 18a;- 7. 6. 9a; + 7 = 3a;-ll. 12. 27 -f 10 a; = 13 a; + 23. 7. 3a;-8 = 5a; + 8. 13. 19a;- 11 = 15 + 6a;. 8. 5-6a;=l-4a;. 14. 32a;- 15 = 7 + 65a;. 110 ALGEBRA. 16. 13a;-81 = 5a;-31a;-159. 16. 12a;-20a;4-13 = 9a;-259. 17. Solve the equation (2x-3y-x{x + l) = S{x-2)(x-\- 7) -5. Performing the operations indicated, we have 4 ar* - 1 2 a; + 9 - ic2 - a; = 3 0^ + 1 5 a; - 4 2 - 5 . Transposing, • 4iB2__i2a;-a^-a;-3a:2-15a; = -42-5-9. Uniting terms , — 28 a; = — 5 6 . Dividing by — 28, a; = 2, Ans» Solve the following equations : 18. 3 + 2(2a;4-3)=2a;-3(2a; + l). 19. 2a;-(4a;-l) = 5a;-(a;-l). 20. 7(aj-2)-5(a; + 3) = 3(2aj-5)-6(4a;-l). 21. 3(3aj4-5)-2(5a;-3) = 13-(5a;-16). 22. (2a;-l)(3a;4-2) = (3a;-5)(2a;+20). 23. (5-6a;)(2a;-l) = (3a; + 3)(13-4a;). 24. (a;-3)2-(5-a;)2 = -4a;. 26. (2a; - 1)2- 3 (a; - 2) + 5 (3 a; - 2) - (5 - 2xy=i^ 0. 26. 2(a;-2)2-3(a;-l)2+a^ = l. 27. (a; - 1) (a; - 2) (a; + 4) = (a; + 2) (a; + 3) (a; - 4) . . 5(7+3a;)-(2a;-3)(l-2a;)-(2a;-3)2-(5-fa;) = 0. . (5a;-l)2-(3a; + 2)2-(4a;-3)2 + 4 = 0. 30. (2a; + l)» + (2a;-l)3=16a;(a^-4)-228. SIMPLE EQUATIONS. Ill SOLUTION OF EQUATIONS CONTAINING FRACTIONS. 175. 1 . Solve the equation = ^ 3 4 6 8 The L.C.M. of 3, 4, 6, and 8 is 24. Multiplying each term of the equation by 24, we have 16a;- -30 = 20a;- -27 16aj- -20a; = 30 - -27 -4a; = 3 X 3 Ans. We have then the following rule for clearing an equation of fractions : Multiply each term by the lowest common multiple of the denominators. EXAMPLES. Solve the following equations : 2. a; + ^ + ^ = -ll. 7. ^-^-1^^^ 2 3 '^ "" ^'^ 3. ^_:^ + i- = o. 4 6 18 4. 2a;-^ = i^-^. 4 14 7 5. 7^_7 = ^^-i^. 10. a;-^+20 = ^ + ^ + 26. 4 3 4 7 2 4 * 6 2a; 4 12a;* ' a; 2a; 12 3a;* 5 20 10 4 8. X 2x 2x 9. X . 11 a;_a; 3a; 2 ~6 3~6~T' 112 ALGEBRA. 12. Solve the equation ^^^^ - i^^ = 4 + 1^+^. ^ 4 5 ^ 10 Multipljing through by 20, the L.C.M. of 4, 5, and 10, 15a; - 5 - (16 a; - 20) = 80 + 14a; -f- 10 15 a; - 5 - 1 6 a; + 20 = 80 + 14 a; + 10 15a; - 16a; - 14a; = 80 + 10 + 5 - 20 — 15 a; = 75 a;= — 5, Ans. Note. K a fraction whose numerator is a polynomial is preceded by a — sign, care must be taken to change the sign of each term of the numerator when the denominator is removed. It is convenient, in such a case^ to enclose the numerator in a parenthesis, as shown in the above example. 13. 3^ + ^^^ = ^^- 14. x-^-^±l = 5x-L 7 2 5 3 15. 7a;-li^^ = 3a;+7. 16. 4a; -^^^+- (a; -9) = 5a;. 3 2^ ^ 17. a;-(3a;-4)-5-=^=2. 18 ^^ =x 7 1 ^-^^ . 19 a; + l a; + 4 _a;-4 '21 15 * '2 5 7 ' 20. 2-I^:zi = 3x-i^^+^. 6 4 OI 5a; — 2 3a; + 4 7a; + 2 a;— 10' ^^- 3 4 6 2 22. i(^ + l)- 2a;-5_lla; + 5 a;-13 5 10 3 SIMPLE EQUATIONS. 113 3 9 2^ ^ 6 i+^^l/3^_2)_li^±2_l(2_9aj). 7 2^ ^ 14 3^ ^ OK 2a;H-l 4a; + 5 8 + 3? , 2a;4-5 26 ^^~^ _ 7 — 3a; _ 10a; — 3 _ 3 — 5ag 2 3a; ~ 4 2a; * 27 3 ^' + 7 _ 4(ar^-2) _ a;«+16 ^ 7 2 3a; Bx" 2 28. Solve the equation — -^ — = 0. ^ a;-l a;+l ar^-1 Multiplying each term by ar^ — 1, the L.C.M. of the denom- inators, 2(a;+l) -3(a;-l) -1=0 2a; + 2-3a; + 3-l = 2a;-3a; = -2-3 + l — a; = — 4 X = 4, Ans. oa o 1 i.u i.- 6a;H-l 2a; — 4 2a;— 1 29. Solve the equation — — = ^ 15 7a;-16 5 Multiplying each term by 15, ^ 30a;-60^g^_3^ 7a; -16 Transposing and uniting terms, 4 = — ^-^^ Multiplying by 7a; -16, 28a;- 64 = 30a; -60 -2a; = 4 a;= — 2, Ans. 114 ALGEBRA. Note. If the denominators are partly monomial and partly polyno- mial, it is often advantageous to clear of fractions at first partially ; multiplying by a quantity which will remove the monomial denomi- nators. Solve the following equations : 30. — ^ = 0. 34. 3a;-7 3a;H-7 31. 2^nl = 2^±I. 36. 3a; + 4 3a;-f2 32. l^zi7^±A_3. 36. 2ar2 + 5a;-13 33. ^-i2_^5fl^ + 7. 37^ «(« — 1) a^— 1 2 1 38. X x^ — 3 ^x- ■6x -7 .2 '3' {x-¥by ^ 5a ' + 1 x-^ 5 1 1 2 a; 3 a. + l' aj + 2 + 3 3a; + 2 2aj- -1 a; 6 3ic- -7 2 1 a; — 2 a;~3 a;^ — 5a;H-6 39 6a; + 7 2 (a;- 1) ^ 2a; + 1 . 15 7a;-6 5 40. -^ ? ^ = 0. 1 —X 1 -fa; 1 — ar ?a^4-3a; 1 2a; + l 3a; 41. 2^+^4-^ = «; + l. 42. 2f^-±iU3f^ = 5. \x + 2)^ \x+\) 43. ^ 1 3a;4-l x + l 2x 44 ^ = ^ + 1 _ 7 — 2a;^ _ 9~ 3 l-9x' 45 (a; + i)^^a;-4 (a; 4- 2)2 a; -2 SIMPLE EQUATIONS. 115 3a^4-a;— 1 3a; + l' 47 a;-l a; + 1 ^ 2(ar^ + 4a; + l) ^ a;_2"^a;4-2 (a; + 2)2 ^ 4a;4-3 12a; — 5 2a;— 1 _q 10 5a; -29 5 ~ ' a;— 1 a; — 2 a; — 3 a; — 4 49. x — 4: a; — 5 SOLUTION OF LITERAL EQUATIONS. 3.76. 1 . Solve the equation 2aa; — 36 = a; + c — 3aa;. Transposing and uniting terms, 5aa; — a; = 36 + c. Factoring the first member, a; (5a— 1) =36 + c. Dividing by 5 a — 1 , x = — i-^ , Ans. 2. Solve the equation (b — cxy — (a — cxy = 6(6 — a). Performing the operations indicated, 62 _ 26ca; + c^ar^ - (a" -2acx + c2^)= 6^ - ab b'-2bcx-^(^x^-a^ +2acx-c^x^ =b^-ab 2 oca; — 2 6ca; = a^ — ab Factoring both members, 2 ca;(a — 6) = a(a — 6) Dividing by 2c{a - 6), x= ^(^-^) 2c(a — 6) = ^, Ans. 2c 116 ALGEBRA. EXAMPLES. Solve the following equations ; 3. 2ax-{-d = 3c — bx. 4. 6 bmx — 6 an =15 am — 2 bnx. 5. x + l = 2ax—a^{x—l), o d? , h W . a e. — !-- = Ht* X 2 X 4: 7. {d'-2xY = (J.x-3a'){x + a^), 8. (2m + 3a;)(2m-3a7) = n2-(3a;-n)^ 9. ^II^_^±^+2 = 0. h a 10. (a;-a-&)2-(x — a)(a;-6) + a5 = 0. a; x + 2h a^ + ft^ 11 a; — a' ic+a x^ — a^ j2 (6-3a;)(c+2a;) ^ ^ 2(a;-c)(6-3c-3a;) . 13. (ic+a)2-(£c-a)»-a(3a;-a)(2ic+a) =a;(a-|-l)+3. x+ 2n 15. (a-a5)(6-a;) -a(& + l) = - + a;^ 16. ^_3+-^ = A-2a(2-3a). 2a 4a« 3a2 ^ ^ 17 a? I 1 — 2aa; i 2a;— 1 _q ^^^ 2"^ 2a a^ ~ * f"^ \ SIMPLE EQUATIONS. 117 18. JL-=l±m = JL-^m-l). mn Sn 3n 19. a;4-2a x-3a ^^ X — a ic + a 20. i^ZL^-^-±-^=l. 2x — a x — a 21 E _ <^ — ^^^ _ ^ _ 0^ — 4 5a; 2 26c ~ 6c 3bc ax-i-b 36 aV + 62 22. ax — b ax-\-b a V — 6^ ax — b bx — a a — b f\ ax-\-b bx-^a {ax -\- b) {bx -\- a) QA x — n x^ — mx — '^ _ ., , m mx — w mx — inf SOLUTION OF EQUATIONS INVOLVING DECIMALS. 177. 1. Solve the equation .2ic-. 01-. 03a; = .113a;+. 161. Changing the decimals into common fractions, 2x 1 3x 113a; , 161 10 100 100 1000 1000 Clearing of fractions, 200a; - 10 - 30ic = 113a; + 161 57a; = 171 a; = 3, Am. 118 ALGEBRA. Or, we may solve the equation as follows : Transposing, .2a; — .03x — .113a; = .01 + .161. Uniting terms, .057a;=.171. Dividing by .057, a; = 3, Ans. EXAMPLES. Solve the following equations : 2. .23fl;- 2.05 = . 02a;— 1.882. 3. .001a;-.32=.09a; — .2a;-.653. 4. .3a;-.02-.003a; = .7-.06a;-.006o 5. .3(1.2a;-5)=U+.05a;. >/. 6. .7(a;+.13) = .03(4a;-.l)+.5. 8. 4.25 -I? = 11-1-=:^. Q .6a; + .044 .5a; — .178 oo W. =.d8. .4 .6 f 10 2 — 3a; 5a; 2a; — 3 _ a; — 2 25^ ^ * 1.0 1.25 9 ~ 1.8 9° PROBLEIVIS. 119 XIII. PROBLEMS. LEADING TO SIMPLE EQUATIONS CONTAINING ONE UNKNOWN QUANTITY. 178. For the solution of a problem by Algebra no gen- eral rule can be given, as much must depend on the skill and ingenuity of the student. A few suggestions, however, may be found of service : 1. Express the unknown quantity^ or one of the unknown quantities^ by one of the final letters of the alphabet. 2. From the given conditions, Jind expressions for the other unknown quantities, if any, in the problem. 3. Form an equation in accordance with the conditions of the problem. 4. Solve the equation thus formed. PROBLEMS. 179. 1. What number is that to which if four-sevenths of itself be added, the sum will equal twice the number diminished by 27? Let X = the number. 4x Then, -—= four-sevenths of it, 7 and 2x = twice it. By the conditions, x -\ = 2a: — 27 7x + ix=Ux-189 -8a:=-189. Whence, x = 63, the number required. 2. A is three times as old as B, and eight years ago he was seven times as old as B. Required their ages at present. 120 ) ALGEBRA. Let X = B's age. Then, Sx= A's age. Also, j: — 8 := B's age 8 years ago, and Zx — S = A's age 8 years ago. By the conditions, 3;r — 8=7(x — 8) Sx-S = 1x-b6 -4t=_48. Whence, x = 12, B's age, and 3ar==36, A's age. Note. In the ahove solution we say " Let x = B's age," meaning " Let X = the number of years in B's age." Abbreviations of this nature are often used in Algebra ; but it should be remembered that they are in fact abbreviations, and that x can only represent an ab- stract number. 3. A had twice as much money as B ; but, after giving B $35, he had only one-third as much as B. How much had each at first? Let X = what B had at first. Then, 2x = what A had at first. After giving B $35, A had left 2 0; — 35 dollars, while B had ar + 35 dollars. Then, by the conditions, a: + 35 = 3(2x-35) ar + 35 = 6 a; - 105 -5a: =-140. Whence, x = 28, B's money at first, and 2 a: = 56, A's money at first. 4. What number is that whose double exceeds its half by 45? 5. Divide 34 into two parts such that four-sevenths of one part may be equal to two-fifths of the other. 6. What number exceeds the sum of its third, tenth, and twelfth parts by 58 ? 7. Divide 59 into two parts such that the sum of one- seventh the greater and one-third the less shall be equal to 18. PROBLEMS. 121 8. A is four times as old as B, and in 30 years he will be only twice as old as B. What are their ages? 9. A is 62 years of age, and B is 36. How many years is it since A was three times as old as B? 10. A had one-half as much monej' as B ; but after B had given him $42, he had four times as much as B. How much had each at first? 11. Divide 207 into two parts such that one-fourth the greater shall exceed two-sevenths the less by 3. 12. What two numbers are those whose difference is 3, and the difference of whose squares is 51 ? 13. A drover paid $1428 for a lot of oxen and cows. For the oxen he paid $55 each, and for the cows $32 each ; and he has twice as many cows as oxen. How many has he of each? 14. Divide 80 into two parts such that if the greater is taken from 62, and the less from 48, the remainders are equal. 15. A gentleman left an estate of $1872 to be divided betweeii his wife, three sons, and two daughters. The wife was to receive three times as much as either of the daugh- ters, and each son one-half as much as each of the daugh- ters. How much did each receive? 16. Divide $70 between A, B, and C, so that A*s share may be three-eighths of B's, and C's share two-ninths of A's. 17. In a garrison of 2744 men, there are 12|- times as many infantry as cavalry, and twice as many cavalry as artillery. How many are there of each kind ? 18. A is 34 years older than B ; and he is as much above 50 as B is below 40. Required their ages. 122 ALGEBRA. 19. A man travelled 3036 miles. He went four-sevenths as many miles on foot as by water, and two-fifths as many miles on horseback as by water. How many miles did he travel in each manner? 20. Divide a into two parts such that m times the first part shall be equal to n times the second. Let X = the first part. Then, a — x = the second part. By the conditions, mx = n{a — x). Or, mx + nx = an. "Whence, x = — ^^, the first part, w + n Therefore, a-x = a ^^ = -^HL.^ the second part. m + w m -{■ n 21. Divide a into two parts such that m times the first shall be equal to the second divided by n. 22. Find four consecutive numbers whose sum is 94. 23. Divide 43 into two parts such that one of them shall be three times as much above 20 as the other lacks of 17. 24. Divide $47 between A, B, C, and D, so that A and B together may have S27, A and C $25, and A and D $23. 25. If a certain number is increased by 15, one-half the result is as much below 80 as the number itself is above 100. Required the number. 26. Divide 205 into four parts such that the second is one-half of the first, the third one-third of the second, and the fourth one-fourth of the third. 27. Eleven years ao^o, A was 4 times as old as B, and in 13 years he will be only twice as old. Required their ages at present. 28. Find two consecutive numbers such that the differ- ence of their squares added to three times the greater num- ber exceeds the less number bv 92. PROBLEMS. 123 29. What number is that, five-sixths of which as much exceeds 25 as one-ninth of it is below 9 ? 30. A is m times as old as B, and in a years he will be n times as old. Required their ages at present. 31. Divide a into three parts such that the first may be n times the second, and the second n times the third. 32. A can do a piece of work in 8 days which B can perform in 10 days. In how many days can it be done by both working together ? Let X = the number of days required. Then, - = what both can do in one day. X Also, - = what A can do in one day, 8 and — = what B can do in one day. By the conditions, - H — = — 8 10 ar 6x-H4a- = 40 ' 9x = 40. Whence, ar = 4f, the number of days required. 33. A can do a piece of work in 15 days, and B can do the same in 18 days. In how many days can it be done by both working together ? 34. A can do a piece of work in 3J hours which B can do in 2 J hours, and Q \\i2\ hours. In how many hours can it be done by all working together? 35. The stones which pave a square court would just cover a rectangular area whose length is 6 yards longer, and breadth 4 yards shorter, than the side of the square. Required the area of the court. 36. A, B, and C found a sum of money. It was agreed that A should receive $15 less than one-half, B $13 more than one-fourth, and C the remainder, which was $27. How much did A and B receive ? 124 ALGEBRA. 37. A can do a piece of work in a hours which B can do in b hours. In how many hours can it be don© by both working together? 38. A vessel can be filled by three taps ; by the first alone it can be filled in a minutes, by the second in b minutes, and by the third in c minutes. In what time will it / be filled if all the taps are opened ? 39. A sum of money, amounting to $4.32, consists entirely of dimes and cents, there being in all 108 coins. How many are there of each kind ? Let X = the number of dimes. Then, 108 —x = the number of cents. Also, 10 a: = the value of the dimes in cents. By the conditions, 10 a: + 108 -x = 432 9 a: = 324. Whence, x = 36, the number of dimes, and 108 — a: = 72, the number of cents. 40. A man has $4.04 in dollars, dimes, and cents. He has one-fifth as many cents as dimes, and twice as many cents as dollars. How many has he of each kind? 41. A man has 3 shillings 7 pence in two-penny pieces and farthings ; and he has 19 more farthings than two- penny pieces. How many has he of each kind? 42. I bought a picture for a certain sum, and paid the same price for a frame. If the frame had cost $ 1 less, and the picture 75 cents more, the price of the frame would have been only half that of the picture. Required the cost of the picture. 43. A laborer agreed to serve for 36 days on condition that for every day he worked he should receive $1.25, and ♦ for every day he was absent he should forfeit 50 cents. At the end of the time he received $17. How many days did he work, and how many was he absent? Ca PROBLEMS. 125 44. A has $105, and B $83. After giving B a certain sum, A has only one-third as much money as B. How much was given to B ? 45. A has a dollars, and B h dollars. After giving B a certain sum, A has c times as much money as B. How much was given to B ? 46. A vessel can be emptied by three taps ; by the first alone it can be emptied in 80 minutes, by the second in 200 minutes, and by the third in 5 hours. In what time will it be emptied if all the taps are opened ? 47. The second digit of a number exceeds the first by 2 ; and if the number, increased by 6, be divided by the sum of its digits, the quotient is 5. Required the number. Let X = the first digit. Then, x + 2 = the second, and 2 X + 2 = the sum of the digits. The number itself is equal to 10 times the first digit, plus the second, which is lOx + X + 2, or 11 X + 2. Hence, by the conditions, llx + 2 + 6_5 2x+2 llx + 8 = 10x + 10. Whence, x = 2. Therefore, 11 a: + 2 = 24, the number required. 48. The first digit of a number exceeds the second by 4 ; and if the number be divided by the sum of its digits, the quotient is 7. Required the number. 49. The first digit of a number is three times the second ; and if the number, increased by 3, be divided by the differ- ence of its digits, the quotient is 16. Required the number. 50. A merchant has grain worth 9 shillings per bushel, and other grain worth 13 shillings per bushel. In what pro- portion must he mix 40 bushels, so that the mixture may be worth 10 shillings per bushel? 126 - ALGEBRA. 61. Gold is 19J times as heav}' as water, and silver lOl. times. A mixed mass weighs 4160 ounces, and displaces 250 ounces of water. How many ounces of each metal does it contain ? 62. The second digit of a number exceeds the first by 3 ; and if the number, diminished by 9, be divided by the sum of its digits, the quotient is 3. Required the number. 63. Two persons, A and B, 63 miles apart, start at the same time and travel towards each other. A travels 4 miles an hour, and B 3 miles an hour. How far will each have travelled when they meet? Let X = the distance A travels. Then, 63 — a: = the distance B travels. Also, - = the time A takes to travel x miles, 4 and — ^— ^ = the time B takes to travel x miles. By the conditions. 3 X 63 4 3 3a: = 252-4a: 7x = 252. ^ - "Whence, x = 36, the distance A travels, and 63 — a: = 27, the distance B travels. 54. A person has 4^ hours at his disposal. How far can he ride in a coach which travels 5 miles an hour, so as to return home in time, walking back at the rate of 3 J miles an hour? 55. A courier who travels a miles daily is followed after n days by another, who travels 6 miles daily. In how many days will the second overtake the first ? 56. Two men, A and B, 26 miles apart, set out, B 30 minutes after A, and travel towards each other. A travels 3 miles an hour, and B 4 miles an hour. How far will each have travelled when they meet ? ^, 'Ai probli:ms. 1^7 57. A capitalist invests | of a certain sum in 5 per cent bonds, and the remainder in 6 per cent bonds ; and finds that his annual income is $180. Required the amount in each kind of bond. 58. What principal at r per cent interest will amount to a dollars in t years ? 59. In how many years will p dollars amount to a dollars, at r per cent interest ? 60. Separate 41 into two parts such that one divided by the other may give 1 as a quotient and 5 as a remainder. Let X — the divisor. Then, 41 — x = the dividend. By the conditions, ^^~^ = 1 + - X X 41— x = T+6 - 2 X = - 36. Whence, x = 18, the divisor, /C/ and 41 - X = 23, the dividend. — 61- Separate 37 into two parts such that one divided by the other may give 3 as a quotient and 1 as a remainder. 62. Separate 113 into two parts such that one divided by the other may give 2 as a quotient and 20 as a remainder. 63. A general, arranging his men in a solid square, finds he has 21 men over. But attempting to add 1 man to each side of the square, he finds he wants 200 men to fill up the square. Required the number of men on a side at first, and the whole number of troops. 64. Separate a into two parts such that one divided by the other ma}^ give 6 as a quotient and c as a remainder. 65. The denominator of a fraction exceeds the numerator by 6 ; and if 8 is added to the denominator, the value of the fraction is \. Required the frnction. 128 ALGEBRA. 66. The sum of the digits of a number is 6, and the num- ber exceeds its first digit bj' 46. What is the number? 67. At what rate of interest will p dollars amount to a dollars in t years ? 68. A man bought a picture for a certain price, and paid three-fourths the same amount for a frame. If the frame had cost $2 less, and the picture 60 cents more, the price of the frame would have been one-third that of the picture. How much did each cost? 69. The denominator of a fraction exceeds the numerator by 1. If the denominator be increased by 2, the resulting fraction is less by unity than twice the original fraction. Required the fraction. 70. At what time between 3 and 4 o'clock are the hands of a watch opposite to each other? Let OM and OH represent the positions of the minute and hour- M hands at 3 o'clock, and OM' and OH' their positions when opposite to each other. Let X = the arc MHH'M' over which the minute-hand has passed since 3 o'clock. Then, — = the arc HW over which the houB ' 12 hand has passed since -3 o'clock. Also, the arc MH — 15 minute-spaces, and the arc WM' = 30 minute-spaces. arc MHH'M' = arc MH+ arc H'M' + arc HH'. ^=15+30 -f:^ 12 ar = 540 -f ar llx = 540. a: = 49^. Hence the required time is 49^ minutes after 3 o'clock. 71. At what time between 7 and 8. are the hands of a watch opposite to each other? Now, That is, "Whence, PROBLEMS. 129 72. At what time between 2 and 3 are the hands of a watch opposite to each other ? 73. At what time between 5 and 6 are the hands of a watch together? 74. At what time between 1 and 2 are the hands of a watch together ? 76. A woman sells half an egg more than half her eggs. Again she sells half an egg more than half her remaining eggs. A third time she does the same ; and now she has sold all her eggs. How many had she at first? 76. A man has two kinds of money, dimes and half- dimes. If he is offered $1.35 for 20 coins, how many of each kind must he give ? 77. A man has a hours at his disposal. How far can he ride in a coach which travels b miles an hour, so as to return home in time, walking back at the rate of c miles an hour? 78. At what time between 6 and 6.30 o'clock are the hands of a watch at right angles to each other ? 79. At what times between 10 and 11 o'clock are the hands of a watch at right angles ? 80. A banker has two kinds of money. It takes a pieces of the first kind to make a dollar, and b pieces of the sec- ond kind. If he is ofiered a dollar for c pieces, how many of each kind must he give ? 81. A alone can perform a piece of work in 12 hours ; A and C together can do it in 5 hours ; and C's work is two- thirds of B's. The work must be completed at noon. A commences work at 5 a.m. ; at what hour can he be relieved by B and C, and the work be just finished in time? 82. At what time between 4 and 5 is the minute-hand of a watch exactly 5 minutes in advance of the hour-hand ? 130 ALGEBRA. 83. A man buys a certain number of eggs at the rate of 3 for 10 cents. He sells one-third of them at the rate of 2 for 7 cents, and the remainder at the rate of 4 for 15 cents ; and makes 16 cents by the transaction. How many eggs did he buy ? 84. A merchant increases his capital annually by one- thu'd of it, and at the end of each year sets aside $2700 for expenses. At the end of four years, after deducting the amount for expenses, he finds that his capital is reduced to $2980. What was his capital at first? 86. A man owns a harness valued at $25, a horse, and a carriage. The harness and carriage are together worth two- thirds the value of the horse, and the horse and harness are together worth $15 more than twice the value of the carriage. Required the value of the horse, and of the carriage. 86. Two men, A and B, 107 miles apart, set out at the same time and travel towards each other. A travels at the rate of 13 miles in 5 hours, and B at the rate of 11 miles in 4 hours. How far will each have travelled when they meet? 87. A mixture is made of a pounds of coffee at m cents a pound, h pounds at n cents, and c pounds at p cents. Required the cost per pound of the mixture. 88. A, B, and C together can do a piece of work in 6 days ; B's work is one-half of A's, and C's work is two- thirds of B's. How many days will it take each working alone ? 89. A and B start in business, A putting in f as much capital as B. The first year, A gains $150, and B loses \ of his money. The next j^ear, A loses \ of his money, and B gains $300 ; and they have now equal amounts. How much had each at first? 90. At what time between 9 and 10 is the hour-hand of a watch exactly one minute in advance of the minute-hand ? PROBLEMS. 131 91. A and B together can do a piece of work in 1^ days, A and C in IJ days, and B and C in 2f days. How many days will it take each working alone? 92. A man buys two pieces of cloth, one of which con- tains 3 yards more than the other. For the larger piece he pays at the rate of $5 for 6 yards, and for the other at the rate of $ 7 for 5 yards. He sells the whole at the rate of 3 yards for $4, and makes $8 by the transaction. How many yards were there in each piece ? 93. A gentleman distributing some money among beggars, found that in order to give them a cents each, he should need b cents more. He therefore gave them c cents each, and had d cents left. Required the number of beggars. 94. A man let a certain sum for 3 years at 5 per cent compound interest ; that is, at the end of each year there was added -^ to the sum due. At the end of the third year there was due him $2315.25. Required the sum let. 95. A man starts in business with $4000, and adds to his capital annually one-fourth of it. At the end of each year he sets aside a fixed sum for expenses. At the end of three years, after deducting the fixed sum for expenses, his capital is reduced to $2475. What are his annual expenses r 96. A man invests one-third of his money in 3^ per cent bonds, two-fifths in 4 per cent bonds, and the balance in 4J per cent bonds. His income from the investments is $595. What is the amount of his property ? 97. At what time between 8 and 9 o'clock is the minute- hand of a watch exactly 35 minutes in advance of the hour- hand? 98. A fox is pursued by a greyhound, and has a start of 60 of her own leaps. The fox makes 3 leaps while the greyhound makes but 2 ; but the latter in 3 leaps goes as far as the former hi 7. How many leaps does each make before the greyhound catches the fox ? 132 ALGEBRA. XIV. SIMPLE EQUATIONS. CONTAINING TWO UNKNOWN QUANTITIES. 180. If we have a simple equation containing two un- known quantities, as x-\-y — 12^ it is impossible to deter- mine the values of x and y definitely ; because, if any value be assumed for one of the quantities, we can find a corre- sponding value for the other. Thus, if a; =9, then 9 + 2/ =12, or 2/= 3; if flj = 8, then 8 -f 2/ = 12, or 2/ = 4 ; etc. Hence, any of the pau's of values, a; = 9,2/ = 3; a; = 8,2/ = 4; etc., will satisfy the given equation. Similarly, the equation cc — 2/ = 4 is satisfied by any of the following pairs of values : a; = 9,2/ = 5; a; = 8, 2/=4; etc. Equations of this kind are called indeterminate. But suppose we are required to find a pair of values which will satisfy both x+y=^12 and a; — 2/ = 4 at the same time. It is evident by inspection that the values a; = 8, 2/ = 4 satisfy both equations ; and no other pair of values can be found which will satisfy both simultaneously. 181. Simultaneous Equations are such as are satisfied by the same values of their unknown quantities. Independent Equations are such as cannot be made to assume the same form. SIMPLE EQUATIONS. 133 Thus, x + y = 9 and x — y=l are independent equations. But x-\-y=9 and 2a; -f 22/ = 18 are not independent, since the first equation may be obtained from the second by dividing each term by 2. 182. It is evident from Art. 180 that two unknown quan- tities require for their determination two independent, simul- taneous equations. Two such equations may be solved by combining them so as to form a single equation containing but one unknown quantity. This operation is called Elimination. 183. There are three principal methods of elimination : 1. By Addition or Subtraction. 2. By Substitution. 3. By Comparison. ELIMINATION BY ADDITION OR SUBTRACTION. 184. 1. Solve the equations | 5^' - ^2/ = 19 (1) I 7a; + 42/= 2 (3) Multiplying (1) by 4, 20 a;- 122/= 76 Multiplying (2) by 3, 21x + l2y=: 6 Adding these equations, 41 a; = 82 Whence, a; = 2 Substituting the value of a; in (1), 10— 32/ = 19 -32/= 9 Whence, y = — 3 Ans. a;= 2, y = — 3. This solution is an example of elimination by addition. i4 ALGEBRA. 2. Solve the equations 1 (1) .24 (2) Multiplying (1) by 2, 30a; + 16?/ = 2 (3) Multiplying (2) by 3, 30a;-21^ = - -72 (4) Subtracting (4) from (3), 37y = 74 2/ = 2 Substituting this value in (2) , 10a;-14 = - •24 10fl; = - •10 X — — • 1 Ans. £c = — ■ l,2/ = 2. This solution is an example of elimination by subtraction. RULE. Multiply tJie given equations by such numbers as will make the coefficients of one of the unknown quantities equal. Add or subtract the resulting equations according as the equal coefficients have unlike or like signs. Note. If the coefficients which are to be made equal are prime to each other, each may be used as the multiplier for the other equation. If they are not prime, such multipliers should be used as will produce their lowest common multiple. Thus, in Ex. 1, to make the coefficients of y equal, we multiply (1) by 4, and (2) by 3. But in Ex. 2, to make the coefficients of x equal, since the L.C.M. of 15 and 10 is 30, we multiply (1) by 2, and (2) by 8. EXAMPLES. Solve the following by the method of addition or subtrac- tion: 3 <7x + 2y==31. ^ (2x-3y= 4. \3x-4:y = 2S. ' {ex- 3/=28. ^ (3x-i-7y = 33. ^ (7y-5x = -n. 1 x + 2y=10. ' \l5x-Uy = 82. SIMPLE EQUATI0:N^S. 135 -ll?/ = -58. 2. 8 \Sx-{-2y= 3. * 1 15a;-f 8y== I 907-13?/= 76. jg (ll2/-18a;= 2. U5a;+ 42/ = 101. * l24a;- 52/ = -22. g^ (24a;-M3y = -27. ^^ | 24a;- 18?/ = -43. l36a;+ll2/ = -15. * U2ic + 30^= 17. ^Q (152/- 8a;=12. ^g |lla;-122/ = -32. l25t/ + 12a;= 1. ' lll2/-12ic= 14. 11. f5^-72/=15. ^Q r dx-ny= 24. l3a;-52/=13. ' llOa;+ 9y = -37. ^^ I 12a; + 21 2/ = -23. 1 15a; + 282/ = — 30. ELIMINATION BY SUBSTITUTION. 185. 1. Solve the equations j 7^ - ^2/ = - 62 (1) l2y-ox= 44 (2) Transposing 5a; in (2), 2y = 5a; + 44 Or, 2/ = ^^ (3) Substituting this in (1) , 7x-sf^^^±^\ = - 62 Or, , ^^_15a;+132^_ 2 Clearing of fractions, 14a;— 15a;— 132 = — 124 — x= 8 Whence, a; = — 8 Substituting this value in (3) , t/ = -^^+^^ ^ 2 Ans. a; = — 8, 2/ = 2. 136 ALGEBRA.. RULE. Find the value of one of the unknown quantities in terms of the other from one of the given equations, and substitute this value for that quantity in the other equation. EXAMPLES. Solve the following by the method of substitution : Q J ojo-\-ly = — 19. - 7y= i 4:y — 15x = — 7. Q jlOa;^ 7y= 9 1 4v — 15x = 10 I ex-5y = ^7, llOa; + 32/= 11. g (7x — 2y^ 8. jj (9x + 2y=:15. ' l82/~5i« = -9. ' Uaj-fV2^= 8. _ ^g r 9a;-42/ = -4. ^g ( 8a;+ 72/ = -23. \ * ll5a;-f 82/ = -3. ' 1 5y- 12a; = - 12. - (2iC-72/== 8. j jg (72/ --3a; =139. Xiy-dx^ld. I ' l2a; + 5y= 91. ELIMINATION BY COMPARISON. 186. 1. Solve the equations j2aJ-52/ = -16 \Bx + 7y= 5 Transposing — 5yin (1), 2x = 5y — 16 Or, « = 5jiZLL6 (3) Transposing 72/ in (2), 3x=z6 — 7y 0) (2) SIMPLE EQUATIONS. 187 ^ . , , ., 5V-16 5-7y Equating these values of a?, -^— — = — r-^ Clearing of fractions, ISy - 48 = 10 - 14y 29 2/ = 58 Substituting this value In (3) , x— r — = — 8 ^715. a; = — 3, y = 2. RULE. i^nc? ^fee va?ue o/ ^^e same unknown quantity in terms of the other from each of the given equations^ and place these values equal to each other. EXAMPLES. Solve the following by the method of comparison : g (5x-\-Gy= 24. (92/-8a; = -26. g (7x- 8y = -n. 1 a;-122/= 12. jQ C 5x-12y=7. (lOo;- 92/ = 4. ^j (72/-12a;=17. 1 8a; 4- 11?/ = 20. j2 I 7x + Sy = e. lnx + 9y = 8. jg J ii>x-\-6y = — 7. ^ ~ 21a;=« 18. ri55 (8y 138 ALGEBRA. MISCELLANEOUS EXAMPLES. 187. Before applying either method of elimination, each of the given equations should be reduced to its simplest form. 1. Solve the equations C 7 3 = 3a;-9 = 0, or ly-Zx = 13 19 From (1), 72/+ 28 From (2), xy — ^x — xy -\- b y = ^ 13, or by — 2x — — \^ Multiplying (3) by 2, Uy-Qx = - 38 Multiplying (4) by 3, 151/ - 6 a; = - 39 Subtracting (5) from (6), 2/ = — 1 Substituting in (4) , Solve the following ; 'll?/ + 6a;=115. 2x _ ii_^ _ _ 5. 3 6 2* -5-2a;=-13 -2a; = - 8 a; = 4. Ans. a; = 4, 2/ = — 1. 3. + 3y = -46. ^+3a;=66. 2 3 ^ + ^ = -4. L8^6 [5 6 90* (1) (2) (3) (4) (5) (6) .2a; -.05 2/ =.25. .03 a; + .32/ =.96. .5a;+ 22/= .01. .lla;+.32/ = -.009. SIMPLE EQUATIONS. 189 8. 2 3 3 4 9. Q 2x-^Sy ^y 5 3 4:y — 3x __3x . . 6 ""4: a;(22/-3) = 22/(a;+l), 10. ^ 3 ^_5_,^o. x-l 2/ + 2 11. a^(2/-3)-y(« + 4) = 22. (y + l)(a:-2)-(y + 3)(a;-4) = 6. 12. ' x + y 5 flj-y 3 16. 2a; + 3y ^1, a: + y + 13 2* 5a; 7y-2 _j^^ 18. -12 ^ + 8. 4 x + y _2y-x__^^ 5 4 16. — St. 2— ^^ 6 10 2y — 3 5 — 3a? _ y- 14. y- 3a; + 2 ^y + 2 5 3 2y + l_a; — 6 17. ^ a; + 3y ^ 3 2a; — 2/ 8* 7y — x 24-a; + 2y = -17. 18. x — 5 2a; — y — l _ 2y — 2 4 3 5 ' 2y + a; — l _ a; + y 9 4 140 ALGEBRA. 19. 3x _y X 2y T~3 2 7" 13 4 41/ — 3a;= 11. ' 2^x-by 3^ 2a; + y 2 "^ 5 ' ^__ x-2y ^x y 4 2 8 21. ^ 2a; + 2/ ^17 2y + a; 3 12 4 5 2a; — y _ 2y — a? 4 4 "^ 3 * 22. 2^ __ 3^ __ a; + 2i/ _ q __ 5a? — 6y ^ 3 5 4 ■" 4 * 2^^ 5 ^15 a?-2y 3aj 2a; — 42/-1 6a;-l 3_5w 4a;-13 0» il — . a; + 2 '4ar'4-4a;y + 272= (a; + 2/)(4a5+17). y(a;-i/) + 54 ^ 5y + 27 ^ x — y 5 25. 0^ _ 42,2 _17 = (a; 4- 22^ - 2) (a; - 22/ +1) . xy — b , 1 — 2 a; 2^-2 y-1 SIMPLE EQUATIONS. 141 Note. In solving literal simultaneous equations, the method of elimination by addition or subtraction is usually to be preferred. 26. Solve the equations ( ax-\-by = c \a'x + b'y=:e' (1) (2) Multiplying (1) by 6', ab x + bb'y=b'c Multiplying (2) by b, a'bx + bb'y = be' Subtracting, ab X — a'bx = b'c — be' Whence, b'c-bd ^ ab' - a'b Multiplying (1) by a', aa'x + a'by = a'c (3) Multiplying (2) by a, aa'x -f ab'y = ac' (4) Subtracting (3) from (4) , ab 'y — a'by= ac'—a'c Whence, ac' — a'c ^ ab'-a'b Solve the following equations : 2^ (2x-3y = a. XSx + Ayz^b. 33. ^ a b 2g ( ax + by = m. 1+5 = "- ,c a 29. 1 ^^ — ^2/ = c. 1 a;— y = d. -f ' x-\-ay=ia{a + 26). 30. 1 aa;-6y = 0. 1 mx-\-ny=p. 31. |«^ + ^2/ = '^- \ ex — dy = n. 32. \'^^-f^y=P' \ m'x — n'y =: p' . ^- 86. ' ax -{-by = 2. .ab{ay — bx): a + 6) -6^. 142 37. mx-^ny nx-\-my ALGEBRA. mn 36. (a + h)x — (a — h)y = 4a&. (a-6)a?-(a + 6)2/ = 0. ha ah x — h _ y — a _ (J? — h^ a h ah ■\ X y _a^-{-m^—n^ 40. ^ ot m-|-w~ a(mH-n) (m -\-ny{m — n)x = a^y. 41. -^4--l-=— i— . f X y a^h^a-h a'-h' hri:7. + -^=^^- 42. J ^ + ^ a — ^_+_l___J_ ^a — 6 a + h a^ — fl? — y = 4a6. Note. Certain fractional equations, in which the unknown quanti- ties occur in the denominators, are readily solved without previously clearing of fractions. 43. Solve the equations i2-£= 8 X y X y (1) (2) Multiplying (1) by 5, ^-15= 40 Multiplying (2) by 3, ^i + l^^-S X y Adding, Z*= 37 SIMPLE EQUATIONS. 148 37 x= 74 x= 2 Substituting in (1), 5-2 = y y 2/ = -3 Ans. a; = 2, y = — 3. Solve the following equations : 44. X y 2_3 La; y 48. \ X V "~ a; y 45. ?-? = X y 15__8 X y 17 3* I ^ y c d - -f - =3 n. X y 46. 11 X 2/ 2' X y 50. r_2__ 9a;' 2y = -3. 5- + i. = II. L3a; 42/ 6* 47. 2-^ = 16. a; 22/ 2a; y 61. -^ + - = ^^(^+^)' - + — = wi* -H n*. X y 144 ALGEBRA. XV. SIMPLE EQUATIONS. CONTAINING MORE THAN TWO UNKNOWN QUANTITIES. 188. If there are thi-ee simple equations containiDg three unknown quantities, we may combine two of them by the methods of elimination explained in the last chapter, so as to obtain an equation containing onl}" two unknown quanti- ties. We may then combine the third equation with either of the others, and obtain another equation containing the same two unknown quantities. By solving the equations thus obtained, we derive the values of two of the unknown quantities. These values being substituted in either of the given equations, the value of the third unknown quantity may be determined. A similar method may be used when the number of equa- tions and of unknown quantities is greater than three. The method of elimination by addition or subtraction is usually the most convenient. 189. 1. Solve the equations Qx- 4?/- 7z= 17 (1) 2x- ly-Uz^ 29 (2) 10a;- 5y- 3z== 23 (8) Multiplying (1) by 3, 18a;- 122/- 21 2= 51 Multiplying (2) by 2, 18a; - Uy - 322; = 58 Subtracting, 2y + nz = - 7 (4) Multiplying (1) by 5, 30a; - 20y -3oz= 85 (5) Multiplying (3) by 3, S0x-16y- 9z= 69 (6) Subtracting (5) from (6), 5y-\-26z = -U (7) SIMPLE EQUATIONS. 145 Multiplying (4) by 5, 10?/ + 552! = -35 Multiplying (7) by 2, 10?/ 4- 522; = -32 Subtracting, 2>z = - 3 z = — 1 Substituting in (4) , 22/-ll = - 7 .-.2/= 2 Substituting the values of y and 2; in (1), 6a;-8 + 7= 17 .•.a;= 3 Ans. x = 3,y = 2, z 'u+x+y= 6 (1) x + y+z^' 7 (2) y+2+«= 8 (3) L z + u + x= 9 (4) -1. In certain cases the solution may be abridged by aid of the artifice which is employed in the following example. 2. Solve the equations Adding the given equations, 3w + 3a; + 32/ + 32; = 30 Whence, u + x-\-y-\-z = 10 (5) Subtracting (2) from (5) , u= S Subtracting (3) from (5), x= 2 Subtracting (4) from (5), 2/= 1 Subtracting (1) from (5), 2=4 EXAMPLES. . Solve the following equations : x-^y= 2. (2x — 5y= — 19. 3. ^2/ + 2 = -l- 4. ■l3y-\-4:Z= 13. z4-x= 3. (22:-5aJ= 12. 146 10 ALGEBRA. 3a;-2?/ = -l. r7ir + 4?/— 2; = — 50 5^ + 42 = — 6. 13. )^x — by — ^z= 20 x — y — 3z = ll. ( a; — 3?/ — 42!= 30 2x- y = 5. ZX'\-2y— 2; = 6. x — ^y-\-2z=l. 14. ( a;-62/ + 42= 3. 3 4a; + 4y-32!=10. L2X+ 2/ + 62! = 46. ■■{' Kx x-\- y+ z— 53. -{-2y-\-Sz=107. a; + 3y+42=137. Sx— y — 2z = ~23. 8. Hx-{-2y + 3z= 15. 4:X+3y— z = — 6, aj-fy — 2; = 8. y + 2; — a;= 1. 2; +ic — 2/ = — 11. ra;— 22/H-32;= . jy — 2z-hSx = — 26 Lz—2x-\-3y= 9 r6a;-32/+22! = 41. 11. }2x+ y- 2 = 17. (5aj + 42/-22 = 86. 2a;+ 2/+ 2J = -2. 12. ^ a: + 2^+ z= 0. a;_|. y4.22! = -4. 16 r 8x-9y-7z = . -J 12ir- y-3z = L &x—2v— 2 = 16. 17. 18. 19. 8x—9y—7z = — S6. 36. 2y- 2= 10. 4aj-32/ + 22 = 40. 5a;4-92/-72! = 47. 937 + 82^-32 = 97. 2 3 4 3 4 4 2 3 2u-Sx Sx — 4:y 4:y — 5z bz -Qu -43. 34. -50. f2y + 2-f-2w = -28. 2/ + 82 =-2. 4»+z = 13. | + 3t^ = -20. 3 SIMPLE EQUATIONS. 147 20. 21. 23. 24. ■l*\ = 1. \*\ 3 2* hi = 2. r3__2 X y = -13. y ^ = 14. 3 2 ^Z X = 18. ■ax + DtV = 2. a^2/ + a«2;=2. .a^z + i x^a; = a«+l 3w — 2;=22 — » — 2t/. 4ic-2/ = 35-32;. 4^ — 2?/= 19 — 3a;. 2 = 39-2^-42/. 12 3 - + - + - = -7. X y z ?-? + f = 9. X y z 26. 27. 28. x + z x—y x—z 5 6 = 1. = 0. L±^_^dLy = _4 4 2 ay -f 5a; = c. ex -\-az = h. bz -{-cy = a. 4 3 25-12(a; + z) = -2/. ' 2x-\-y y-^z ^i 4 3 a; + 3y a; — z _ p 3 4 ~ * 2! + 2/ 2; + a;_ 3 [3 4 ~ 2 »•) 'aa;+2/— 2; = a2+a— 1. ay-{'Z—x = a^—a-{-l. ^az-\-x—y=ia. 30. a; — ay + a'^z = a*. x — hy-^-hh^ 53. ^ — cy + c^z = c*. * Add the equations together. 148 ALGEBKA. XVI. PROBLEMS. LEADING TO SIMPLE EQUATIONS CONTAINING MORE THAN ONE UNKNOWN QUANTITY. 190. In solving problems where more than one letter is used to represent the unknown quantities, we must obtain from the conditions of the problem as many independent equations as there are unknown quantities to be determined. 1. Divide 81 into two parts such that f the greater shall exceed f the less by 7. Let X = the greater part, and y = the less. By the conditions. / x-t-y = ox ( 5 9 Solving these equations, a; = 45, y = 36. 2. If 3 be added to both numerator and denominator of a fraction, its value is -| ; and if 2 be subtracted from both numerator and denominator, its value is ^. Required the fraction. Let X = the numerator, id y = the denominator. ' x + 3 2 By the conditions, - y + 3 3 x-2 1 [y-i~2 Solving these equations, x=7,y = 12. Therefore the f ract ion is 1. PROBLEMS. 149 PROBLEMS. 3. Divide 50 into two parts such that three-eighths of the greater shall be equal to two- thirds of the less. 4. Find two numbers such that 7 times the greater ex- ceeds 4" the less by 97, and 7 times the less exceeds -J- the greater by 47. 5. If one-fifth of A's age were added to two-thirds of B's, the sum would be 19^ years ; and if two-fifths of B's age were subtracted from seven-eighths of A's, the remain- der would be 18^ years. Required their ages. 6. If 1 be added to the numerator of a certain fraction, its value is ^; and if 1 be added to its denominator, its value is \. Required the fraction. 7. A gentleman at the time of his marriage, found that his wife's age was j of his own ; but after they had been married 12 years, her age was f of his. Required their ages at the time of their marriage. 8. A and B engaged in trade, A with $240 and B with $96. A lost twice as much as B ; and on settling their accounts, it appeared that A had three times as much remaining as B. How much did each lose? 9. Eight years ago, A was 4 times as old as B ; but in 12 years he will be only twice as old. Required their ages at present. 10. If 5 be added to both terms of a fraction, its value is ^ ; and if 3 be subtracted from both, its value is \. Required the fraction. 11. A and B agreed to dig a well in 10 days ; but having labored together 4 days, B agreed to finish the job, which he did in 16 days. In how many days could each of them alone dig the well? 150 ALGEBRA. 12. If the greater of two numbers be divided by the less, the quotient is 2 and the remainder 12 ; but if 4 times the less be divided by the greater, the quotient is 1 and the remainder 14. Required the numbers. 13. If the numerator of a fraction be doubled, and the denominator increased by 7, its value is |; and if the denominator be doubled, and the numerator increased by 2, the value is f . Required the fraction. 14. If a — 1 be subtracted from the numerator of a cer- tain fraction, its value is a + 1 ; and if a be added to its denominator, its value is a. Required the fraction. 15. A gentleman's two horses, with their harness, cost $300. The value of the poorer horse, with the harness, was $ 20 less than the value of the better horse ; and the value of the better horse, with the harness, was twice that of the poorer horse. What was the value of each? 16. A merchant has three kinds of sugar. He sells 3 lbs. of the first quality, 4 lbs. of the second, and 2 lbs. of the third, for 60 cents ; or, 4 lbs. of the first quality, 1 lb. of the second, and 5 lbs. of the third, for 59 cents ; or, 1 lb. of the first quality, 10 lbs. of the second, and 3 lbs. of the third, for 90 cents. Required the price per pound of each quality. 17. A sum of money was divided equally between a cer- tain number of persons. Had there been 3 more, each would have received $1 less; had there been 6 less, each would have received $5 more. How many persons were there, and how much did each receive ? Let X = the number of persons, and y = what each received. Then, xy = the sum dirided. By the conditions, Ux^S){y-l) = xy I {x~6)iy-^6) = xy. Solving these equations, a: = 12, y = 5. PROBLEMS. 151 18. A boy spent his money for oranges. If he had got five more for his money, they would have cost a half -cent each less ; if three less, they would have cost a half -cent each more. How much money did he spend, and how many oranges did he get? 19. A merchant has two kinds of grain, worth 60 and 90 cents per bushel respectively. How many bushels of each kind must he take to make a mixture of 40 bushels, worth 80 cents per bushel ? 20. My income and assessed taxes together amount to $50. If the income tax were increased 50 per cent, and the assessed tax diminished 25 per cent, they would together amount to $52.50. Required the amount of each tax. 21. A man purchased a certain number of eggs. If he had bought 20 more for the same money, they would have cost a cent apiece less; if 15 less, a cent apiece more. How many eggs did he buy, and at what price ? 22. If a certain lot of laud were 8 feet longer and 2 feet wider, it would contain 656 square feet more ; and if it were 2 feet longer and 8 feet wider, it would contain 776 square feet more. Required its length and width. 23. If B gives A $5, they will have equal amounts ; but if A gives B $ 15, B will have ^ as much as A. How much money has each ? 24. Find three numbers such that the first with half the other two, the second with one-third the other two, and the third with one-fourth the other two, may each be equal to 34. 26. There are four numbers whose sum is 136. Twice the first exceeds the second by 46, twice the second ex- ceeds the third by 44, and twice the third exceeds the fourth by 40. Required the numbers. 152 ALGEBRA. 26. The sum of the digits of a number of three figures is 13. If the number, decreased by 8, be divided by the sum of its second and third digits, the quotient is 25 ; and if 99 be added to the number, the digits will be inverted. Re- quired the number. Let X — the first digit, y = the second, and z = the third. Then, 100x-{-10y + z = the number, and 100 z-\- lOy -\- x= the number with its digits inverted. By the conditions, x + y-\-z=13, lOOx+lO.y + g-S ^og and 100 a: + 10^ + ^+99= 100;? + 10y + a:. Solving these equations, x = 2, y z=S, z = S. Therefore the number is 283. 27. The sum of the digits of a number of two figures is 11 ; and if 27 be subtracted from the number, the digits will be inverted. Required the number. 28. The sum of the digits of a number of three figures is 1 1 , and the units' figure is twice the figure in the hundreds' place. If 297 be added to the number, the digits will be inverted. Required the number. 29. A and B can perform a piece of work in 6 days, A and C in 8 days, and B and C in 12 da^^s. In how many days can each of them alone perform it? 30. If I were to make my field 5 rods longer and 4 rods wider, its area would be increased by 240 square rods ; but if I were to make its length 4 rods less, and its width 5 rods less, its area would be diminished by 210 square rods. Required its length, width, and area. 31. Find three numbers such that the sum of the first and second is c, of the second and third is a, and of the third and first is b. PROBLEMS. 153 32. There is a number of three figures, whose digits have equal dififerences in their order. If the number be divided by half the sum of its digits, the quotient is 41 ; and if 396 be added to the number, the digits will be inverted. Re- quired the number. 33. A sum of money is divided equally between a certain number of persons. Had there been m more, each would have received a dollars less ; if n less, each would have received 6 dollars more. How many persons were there, and how much did each receive ? 34. A gentleman left a sum of money to be divided between his four sons, so that the share of the eldest should be \ the sum of the shares of the other three, of the second \ the sum of the other three, and of the third \ the sum of the other three. It was found that the share of the eldest exceeded that of the youngest by S140. What was the whole sum, and how much did each receive? 35. A grocer bought a certain number of eggs, part at 2 for 5 cents and the rest at 3 for 8 cents, and paid for the whole $1.71. He sold them at 36 cents a dozen, and made 27 cents by the transaction. How many of each kind did he buy ? 36. If a number of two figures be divided by the sum of its digits, the quotient is 7 ; and if the digits be inverted, the quotient of the resulting number, increased by 6, divided by the sum of the digits, is 5. Required the number. 37. If 45 be added to a certain number of two figures, the digits will be inverted ; and if the resulting number be divided by the sum of its digits, the quotient is 7 and the remainder 6. Required the number. 38. A and B can do a piece of work in m days, B and C in n days, and C and A in p days. In what time can each alone perform the work? 154 ALGiyBRA. 39. A crew can row 10 miles in 50 minutes down stream, and 12 miles in an hour and a half against the stream. Find the rate in miles per hour of the current, and of the crew in still water. Let X — the rate of the crew in still water, and y = the rate of the current. Then, x -\- y = the rate rowing clown stream, and X — y = the rate rowing up stream. Since the distance divided by the rate gives the time, we have, by the conditions, ^ 10 ^5 x-^y 6 12 ^ 3 x-y 2* Solving these equations, a: = 10, 3/ = 2. 40. A crew can row a miles in h hours down stream, and c miles in d hours against the stream. Find the rate in miles per hour of the current, and of the crew in still water. 41. A boatman can row down stream a distance of 20 miles, and back again, in 10 hours ; and he finds that he can row 2 miles against the current in the same time that he rows 3 miles with it. Required his time in going and in returning. 42. A number consists of three digits whose sum is 21. The sum of the first digit and twice the second exceeds the third by 8 ; and if 198 be added to the number, the digits will be inverted. Required the number. 43. A merchant has two casks of wine. He pours from the first cask into the second as much as the second con- tained at first ; he then pours from the second into the first as much as was left in the first ; and again from the first into the second as much as was left in the second. There are now 16 gallons in each cask. How many gallons did each contain at first? PROBLEMS. 155 44. A number consists of two figures. If the digits be inverted, the sum of the resulting number and the original number is 121 ; and if the number be divided by the sum of its digits, the quotient is 5 and the remainder 10. Required the number. 45. A man has $30,000 invested at a certain rate of interest, and owes $20,000, on which he pays interest at another rate ; and the interest which he receives exceeds that which he pays by $800. Another man has $35,000 invested at the second rate of interest, and owes $24,000, on which he pays interest at the first rate ; and the interest which he receives exceeds that which he pays by $310. What are the two rates of interest ? 46. A certain sum of money, at simple interest, amounted in 2 years to $132, and in 5 years to $150. Required the sum, and the rate of interest. 47. A certain sum of money, at simple interest, amounted in m years to a dollars, and in n years to h dollars. Re- quired the sum, and the rate of interest. 48. A train running from A to B meets with an acddent which causes its speed to be reduced to one-third of what it was before, and it is in consequence 5 hours late. If the accident had happened 60 miles nearer B, the train would have been only 1 hour late. What was the rate of the train before the accident? Let ^x = the rate of the train before the accident. Then, x = its rate after the accident. Let y = the distance to B from the point of detention. By the conditions, ^ = JL 4- 5 X 3x ■V-60 ^ .V-60 J ^ X Sx Solving these equations, x = 10. Hence the rate of the train before the accident was 30 miles an hour. 156 ALGEBRA. 49. A man rows down a stream, whose rate is S^ miles per hour, for a certain distance in 1 hour and 40 minutes. In returning, it takes him 6 hours and 30 minutes to arrive at a point 2 miles short of his starting-place. Find the distance which he rowed down stream, and his rate of pulling. 50. If a certain number be divided by the sum of its two digits, the quotient is 6 and the remainder 1. If the digits be inverted, the quotient of the resulting number increased by 8, divided by the sum of the digits, is 6. Required the number. 51. A train running from A to B meets with an accident which delays it 30 minutes ; after which it proceeds at three- fifths its former rate and arrives at B 2 hours and 30 minutes late. If the accident had occurred 30 miles nearer A, the train would have been 3 hours late. What was the rate of the train before the accident? 52. A, B, and C together have $24. A gives to B and C as much as each of them has ; B gives to A and C as much as each of them then has ; and C gives to A and B as much as each of them then has. The}^ have now equal amounts. How much did each have at first? 53. A and B are building a fence 126 feet long. After 3 hours, A leaves off, and B finishes the work in 14 hours. If 7 hours had occurred before A left off , B would have finished the work in 4|- hours. How many feet does each build in one hour? 54. Divide 115 into three parts such that the first part increased by 30, twice the second part, increased by 2, and 6 times the third part, increased by 4, may all be equal to each other. 55. Four men. A, B, C, and D, play at cards, B having $1 more than C. After A has won half of B's money, B one-third of C's, and C one-fourth of D's, A, B, and C have each $18. How much had each at first? PROBLEMS. 157 56. A gives to B and C as much as each of them has ; B gives to A and C as much as each of them then has ; and C gives to A and B as much as each of them then has. Each has now $48. How much did each have at first? 57. A, B, and C, were engaged to mow a field. The first day, A worked 2 hours, B 3 hours, and C 5 hours, and together they mowed 1 acre ; the second day, A worked 4 hours, B 9 hours, and C 6 hours, and all together mowed 2 acres; the third day, A worked 10 hours, B 12 hours, and C 5 hours, and all together mowed 3 acres. In what time could each alone mow an acre ? 58. A man invests $3600, partly in 3^ per cent bonds, and partly in 4 per cent bonds. The income from the 3^ per cent bonds exceeds the income from the 4 per cent bonds by $6. How much has he in each kind of bond? 59. A and B run a race of 480 feet. The first heat, A gives B a start of 48 feet, and beats him by 6 seconds ; the second heat, A gives B a start of 144 feet, and is beaten by 2 seconds. How many feet can each run in a second ? 60. The fore-wheel of a carriage makes 4 revolutions more than the hind-wheel in going 96 feet ; but if the cir- cumference of the fore-wheel were f as great, and of the hind-wheel ^ as great, the fore-wheel would make only 2 revolutions more than the hind-wheel in going the same dis- tance. Find the circumference of each wheel. 61. A and B together can do a piece of work in 4|^ days ; but if A had worked one-half as fast, and B twice as fast, they would have finished it in 4-^^ days. In how many days could each alone perform the work? 62. A and B run a race of 300 yards. The first heat, A gives B a start of 40 yards, and beats him by 2 seconds ; the second heat, A gives B a start of 16 seconds, and is beaten by 36 yards. How many yards can each run in a second? 158 ALGEBRA. XVII. INVOLUTION. 191. Involution is the process of raising a quantity to any required power. This is effected, as is evident from Art. 13, by taking the quantity as a factor a number of times equal to the exponent of the required power. 192. If the quantity to be involved is positive, all its powers will evidently be positive ; but if it is negative, all its even powers will be positive, and all its odd powers nega- tive. Thus, (-a)2 = (-a)x(-a) = + a2 (_a)3 = (-a)x(-a)x(-a) ^-a? (_a)^ = (— a) X (—a) X (— «) X (— a) = + a'*; etc. Hence, the even powers of any quantity are positive; and the ODD powers of a quantity have the same sign as the quan- tity itself. INVOLUTION OF MONOMIALS. 193. 1. Find the value of {ba^y, (5 d'cY = 5 a^c X 5 a^c X 5 a'c x 5 a^c = 625 aV, Ans, 2. Find the value of (-3m^)3. (-3m*)3 = (-3m*) X (-3m4) x (-3m^) = -27m^S Ans, From the above examples we derive the following rule : Raise the numerical coefficient to the required power ^ and multiply the exponent of each letter by the exponent of the required power. Give to every even power the positive sign, and to every odd power the sign of the quantity itself. mvoLUTioisr. 159 EXAMPLES. Write by inspection the values of the following : 3. (-a6V)*. 7. (-6V)^ 11. (3a'6*c)«. 4. (-ba^by. 8. (a^ftV)". 12. (-Gx^'Y. 6. (x^y)'^. 9. (-Sm^n)*. 13. (4a'"62n)5. 6. (2mnV)^ 10. (4(1263^4)3. 14, (-ja^y^i^)*. A fraction is raised to any power by raising both numera- tor and denominator to the required power, ^ , / 2a;'"V 16a;*^ For example, ( — -— • ) = — — -• Write by inspection the values of the following : "• ©•• "■ (-^" .,.(- Ixyy 3nJ '»■ (If)" IS. (!«)•. »(- bcx'^Y 4aV* SQUARE OF A POLYNOMIAL. 194. We find by multiplication : a + b + c a + b + G a^-\- ab+ ac + ab +52^ 5c + ac + 6c + c2 a^ + 2ab-{-2ac-{-b^-\-2bc + c^ This result, for convenience of enunciation, may be written as follows : (a + 6 + c)2 = a« + 6* + c^ H- 2a5 + 2ac + 26c. 160 ALGEBRA. In a similar manner, we find *. ■i-2ab-h2ac + 2ad-\-2bc-\-2bd-\-2cd', and so on. We have then the following rule for the square of any polynomial : Write the square of each term, together with twice its product by each of the following terms. EXAMPLES. 1. Square 2 0^ — 3 « — 5. The squares of the terms are 4 a;'', 9a^, and 25. Twice the first term into each of the following terms gives the results, —12ix^ and — 20a^ ; and twice the second term into the following term gives the result, SOx. Hence, {2a^-3x-5y=:ix^-\-di^ + 26-12x^-20a^ + S0x = 4:X^-12x^-nx' + 30x-]-25, Ans, Square the following expressions : 2. a-b-\-c. 11. x^-2x-\-5. 3. a + b-c. ^ ^^' 2a^ + 3a^ + l. 4. 2ic2-fa; + l. 13. 3a2-2a6-562. 6. a^-3a;4-l. 14. 4m^ ■i-mn^ -Sn\ 6. a^+4a;-2. 15. a — b-c-^d. 7. 2x^-x-S. le, a-b + c-d. 8. 3a2-5a + 4. 17. l+aj + a^ + a^. 9. 2a^ + 5a;-7. 18. 3x^-2x^-x-{-4:. 10. x + 2y-Sz. 19. x^-Ax'-2x-3. INVOLUTION. 161 CUBE OF A BINOMIAL. 19S We find by multiplication : {a + by=a'-\-2ab +b' a -i-b a^-{-2a'b+ ab"- a'b-{-2ab^-j-b^ (a -hby = a^ -^-Sa'b -\-3ab^ ■+■ {a-by = o?-2ab +&' a —b - a'b + 2ab''-b^ (a _ 5)3 = a^ -^a'b + 3ab^- W That is, The cube of the sum of two quantities is equal to the cube of the first, plus three times the square of the first times the second, plus three times the first times the square of the sec- ond, plus the cube of the second. The cube of the difference of two quantities is equal to the cube of the first, minus three times the square of the first times the second, plus three times the first times the square of the second, minus the cube of the second. EXAMPLES. 1. Find the cube of a-\-2b. (a + 26)« = a3-f 3a2(26) + 3a(26)2 + (26)3 = a» + 6a26 + 12a62 + 86^ Ans. 2. Find the cube of 2x — 3tf. {2x-3yy=^{2xy-3{2xy{3f) + ^2x){3fY-{Zyy = 8a^-36afy' + 54.xy*-27f, Ans. 162 ALGEBRA. Find the cubes of the following : 3. x-hS. 7. 3m2-l. 11. 2a^-3x. 4. 2a;- 1. 8. a^ + 4. 12. Qx^-j-xij. 5. ab-cd. 9. a + 56. 13. 3m + 5n. _ 6. a + 46. 10. 2x-6y. 14. 3a;2/-4a2. The cube of a trinomial may be found by the above method, if two of its terms be enclosed in a parenthesis and regarded as a single term. 16. Find the cube ot x^ — 2x — l, (x'-2x-iy = l(a^-2x)-l2^ = (x^-2xy-3{i>f-2xy + S(x'-2x)-l = a^-6af + 12a^-8x^-S(x^-4:i^-^4:X^) + 3(a^-2a?)-l = a^ — 6a^+9i»^ + 4a^-9i^_(3^_l^ j^^g^ Find the cubes of the following : 16. a^-x—1. 18. a + b — c. 20. x^ + s^;.^!. 17. a-b + 1. 19. x'-2x-\-2. 21. 2ar^-3aj-l. ANY POWER OF A BINOMIAL. 196. By actual multiplication, we obtain : (a-hby^a^+2ab + 6^ (a-hby^a^ + Sa'b + S ab^ + b^ (a-\-by=a' + 4:a^b + 6 a'b^ + Aab^ + b*] etc. (a-'by=a^-2ab +b^ (a-by = a^-Sa^b + 3ab^-b^ (a - by = a'-4.a'b + 6 a'b' ^iab' + b'] etc. INVOLUTION. 163 In these results we observe the following laws : I. The number of terms is one more than the exponent of the binomial. II. The exponent of a in the first term is the same as the exponent of the binomial, and decreases by 1 in each suc- ceeding term. III. The exponent of b in the second term is 1, and increases by 1 in each succeeding term. IV. The coeflficient of the first term is 1 ; and of the sec- ond term, is the exponent of the binomial. V. If the coeflBcient of any term, be multiplied by the exponent of a in that term, and the result divided by the exponent of b increased by 1 , the quotient will be the coeflS- cient of the next term. VI. If the second term of the binomial is negative, the terms in the result are alternately positive and negative. By aid of the above laws, any power of a binomial may be written by inspection. EXAMPLES. 1. Expand (a + xy. The exponent of a in the first term is 5, and decreases by 1 in each succeeding term. The exponent of x in the second tenn is 1 , and increases by 1 in each succeeding term. The coefficient of the first term is 1 ; of the second term, 5 ; multiplying the coefficient of the second term by 4, the exponent of a in that term, and dividing the result by the exponent of x increased by 1, or 2, we have 10 for the coefficient of the third term ; and so on. Hence, (a + xy = a^-\-5a*x-\- 10 aV + lOaV -\-5ax*-\- x', Ans, 164 ALGEBRA. Note. The coefficients of terms equally distant from the beginning and end of the expansion are equal. Thus the coefficients of the lat- ter half of an expansion may be written out from the first half. 2. Expand (1 - xy. (1 - a;)« = 1^ - 6 . 1^ . a; -f 15 . 1* . aj2 - 20 . P . a^ = 1 -6x-{-15x'-20af + 16x'^-6x^-\-x^, A71S. Note. If the first term of the binomial is numerical, it is con- venient to write the exponents at first without reduction. The result should afterwards be reduced to its simplest form. Expand the following : 3. (a -by. 7. (1-0^)^ 11. {x-^y. 4. (a + by. 8. (x-^yy. 12. (a-Sy. 5. (a -by. 9. {m-ny. 13. (a + 2)^ 6. {x-iy. 10. (2-\-xy. 14. (x-2y. 16. Expand {Sm-ny. = (3m)*-4(3m)3(n2) + 6(3m)2(n2)2 -A{3m)(n'y-\-(7i'y = 81 m* — 108 m^n^ + 54 mV — 12 mn^ + n^ Ans. Note. If either term of the binomial has a coefficient or exponent other than unity, it should be enclosed in a parenthesis before apply- ing the laws- Expand the following : 16. (a-3xy. 18. (a' + bcy. 20. (2a'-{-by. 17. (3 + 26)^ 19. (a;3-4)^ 21. {2m^-Sny. EVOLUTION. 165 XVIII. EVOLUTION. 197. If a quantity be resolved into any number of equal factors, one of these factors is called a Boot of the quantity. 198. Evolution is the process of finding any required root of a quantity. This is effected, as is evident from the preceding article, by finding a quantity which, when raised )i to the proposed power, will produce the given quantit3\ ' / 199. The Radical Sign, ^, when prefixed to a quantity, indicates that some root of the quantity is to be found. Thus, -y/a indicates the second or square root of a ; ■^a indicates the third or cube root of a ; ■^a indicates ihe fourth root of a ; and so on. The ijidex of the root is the figure written over the radical sign. When no index is written, the square root is under- stood. EVOLUTION OF MONOMIALS. 200. Required thfe cube root of a^b^c^. By Art. 198, we are to find a quantity which, when raised to the third power, will produce a^6V. That quantity is evidently a6V. Hence, ■V^iFW^ = ab'(^. That is, any root of a monomial is obtained by dividing the exponent of each factor by the index of the required root. 201. from the relation of a root to its corresponding power, it follows from Art. 192 that : 1. The odd roots of a quantity have the same sign as ihe quantity itself. Thus, Va* = a, and ■{/ — a^ = — a. 166 ALGEBRA. 2. The even roots of a positive quantity are either positive or negative. For the even powers of either a positive or a negative quantity are positive. Thus, -y/a"^ = a or — a ; that is, -^a^ =±a. Note. The sign ±, called the double sign, is prefixed to a quantity when we wish to indicate that it is either + or — . 3. The even roots of a negative quantity are impossible. For no quantity when raised to an even power can pro- duce a negative result. Such roots are called imaginary quantities. 202. From Arts. 200 and 201 we derive the following rule : Extract the required root of the numerical coefficient, and divide the exponent of each letter by the index of the 7'oot. Give to every even root of a positive quantity the sign ± , and to every odd root of any quantity the sign of the quantity itself. Note. Any root of a fraction may be found by taking the required root of each of its terms. EXAMPLES. 1. Find the square root of 9a^6V. By the rule, V9 a*6V = ± 3 a^^c^ Ans. 2. Find the fifth root of — 32 aV'". V-32a^V'" = -2a2a;'", Ans. Find the values of the following : 3. ■^-125aTy. 7. 8. 9. 10. ■roduct from the remainder. If other periods remain, proceed as before, doubling the part of the root already found for the next trial-divisor. Note 1. It should be observed that decimals require to be pointed to the right. Note 2. As the trial-divisor is an incomplete divisor, it is sometimes found that after completion it gives a product greater than the remain- der. In such a case, the last root-figure is too large, and one less must be substituted for it. Note 3. If any root-figure is 0, annex to the trial-divisor, and bring down to the remainder the next period. EXAMPLES. 207. 1. Find the square root of 49.449024. 49.449024 7.032, Ans, 49 1403 4490 4209 14062 28124 28124 Since the second root-figure is 0, we annex to the trial- divisor 14, and bring down to the remainder the next period, 90. 172 ALGEBRA. Extract the square roots of the following : 2. 45796. 6. .247009. 10. 446.0544. 3. 273529. 7. .081796. 11. .0022448644. 4. 654481. 8. .521284. 12. 811440.64. 5. 33.1776. 9. 1.170724. 13. .68112009. If there is a final remainder, the given number has no exact square root ; but we may continue the operation by annexing periods of ciphers, and thus obtain an approximate value of the square root, correct to any desired number of decimal places. 14. Extract the square root of 12 to five figures. 3.4641..., A71S. 12.06060606 9 64 300 256 eSG 4400 4116 6924 I: 28400 27696 69281 70400 69281 1119 Extract the square roots of the following to five figures : 15. 2. 18. 11. 21. .7. 24. .001. 16. 3. 19. 31. 22. .08. 25. .00625. 17. 5. 20. 17.3. 23. .144. 26. 2.08627. The square root of a fraction may be obtained by taking the square roots of its terms. EVOLUTio:Nr. 173 If the denominator is not a perfect square, it is better to reduce the fraction to an equivalent fraction whose denomi- nator is a perfect square. 3 Thus, to obtain the square root of -, we should proceed as follows : J5 = JA=V6 = 2,M?M^ = . 61237,.. \8 \16 4 4 Extract the square roots of the following to five figures : 27. 1 4 "f- -1 33. 11 8* 35. ^. 18 ^■a- »i. -M 34. 9 10* 72 CUBE ROOT OF POLYNOMIALS. 208. Since (a 4- &) ^ = a' + 3 a^^ + 3 aW + 6■^ we know that the cube root of a^ -f 3 a^h + 3 cib'^ -f- h^ is a + 6. It is required to find a process by which, when the quan- tity a^ -f 3 a^6 + 3 ab^ -f- h^ is given, its cube root, a + 6, may be determined. a-^b 3a2 + 3a6-f-62 ^a'h + Zah^ + h^ The cube root of the first term is a, which- is the first term of the root. Subtracting its cube from the given expression, the remainder is ^a}h + 3a&2 + h^, or (Sa^ + 3a6 + 6^) h. Dividing the first term of this remainder by 3 a^, or three times the square of the first term of the root, we obtain h, the second term. Adding to the trial-divisor 3 aft, that is, three times the product of the first term of the root by the second, and 6^, that is, the square of the last term of the root, completes the divisor, Sa^ + 3a6 + h"^. This being multiplied by h, and the product, Za% -{-Zab'^ ■\- h^, subtracted from the remainder, completes the operation. 174 ALGEBRA. From the above process we derive the following rule : Arrange the terms according to the powers of some letter. Find the cube root of the first' term ^ write it as the first term of the root, and subtract its cube from the given expression. Divide the first term of the remainder by three times the square of the first term of the root, and write the quotient as the next term of the root. Add to the trial-divisor three times the product of the first term of the root by the second, and the square of the second term. Multiply the complete divisor by the term of the root last obtained, and subtract the product from the remainder. If other terms remain, proceed as before, taking three times the square of the root already found for the next trial- divisor. EXAMPLES. 209. 1. Find the cube root of S:(^ -2>Qd^y + 64.x^y^ — 272/«. 8a^ 12a^-18a^2/+92/' 2 a^ — 3 2/, Ans. - 36 x*y + 54:x'y^- 27 f -S6x'y + 54taPy^-27f The cube root of the first term is 2 x"^, which is the first term of the root. Subtracting Sx^ from the given expression, we have —S6x*i/ as the first term of the remainder. Dividing this by three times the square of the first term of the root, 12a:*, we obtain — 3y as the second term of the root. Adding to the trial-divisor three times the product of the first term of the root by the second, — 18 x'^y, and the square of the second term, 9y^, completes the divisor, 12 x* — lSx^y-{-9y^. Multiplying this by —Sy, and subtracting the product from the re- mainder, there is no remainder. Hence, 2^2 — 3y is the required cube root. 2. Find the cube root of 40a^-6««- 96a; + aJ«- 64. EVOLUTION. 175 An-anging according to the descending powers of ic, o^^2x~4:. Ans. 3a^_6a^4-4a^ 3aj^-12a;3+12a^ 3x*-12a^ + 24a;+16 -12a;^+48a^-96a;-64 -12x*+ASa^-dQx-64: The second complete divisor is formed as follows : The trial-divisor is 3 times the square of the root already found ; that is, 3 {x^ -2x)\ or 3a:* - 12^3 + 12x2. Three times the product of the root already found by the last term of the root is 3(— 4)(x' — 2a:), or — 12a:2 + 24 a:,* and the square of the last root-term is 16. Adding these, we have for the complete divisor 3x* — 12 a:' -f- 24x + 16. Find the cube roots of the following : 3. i-Qy-^i2y'-8f. 4. 27iB«H-27a;*-f 9ar+l. 6. 54:xf-^27f+S6x'y-\-8a^. 6. 64a»- lUa'xy-{- 108 ax'y^ - 27 a? f, 7. a36-f-6a^-40a^-f 96a;-64. 8. f-1^5f-Sy'-Sy. 9. 15x*-6x-6a^-\-15x^-{-l-\-a^-20a?. 10. 9a^-21ic2-36ar'-|-8a^-9a;+42a^-l. 11. 8a«-12ci«-54a* + 59a3 4-135a2-75a-125. 12. 30ic2-12a^-12a;-f8-25a^+8a;« + 30aj*. 13. xf^ + Sxi'y-3xY-na^f-{-6a^y*-\-12xf-8f. 14. 27 a« - 54 a'^ft 4- 9 a%^ -f 28 a^b^ - 3 a%^ - 6 aft* - 6«. 176 ALGEBRA. CUBE ROOT OF NUMBERS. 210. The method of Art. 209 may be used to extract the cube roots of arithmetical numbers. The cube root of 1000 is 10; of 1,000,000 is 100; etc. Hence, the cube root of a number less than 1000 is less than 10 ; the cube root of a number between 1,000,000 and 1000 is between 100 and 10 ; and so on. That is, the integral part of the cube root of a number of one, two, or three figures, contains one figure ; of a number of four, five, or six figures, contains two figures ; and so on. Hence, If a point he placed over every third figure in any integral number^ beginning with the units' place ^ the number of points shows the number of figures in the integral part of its cube root. 211. Let it be required to find the cube root of 157464. 50 -f- 4 Pointing the number according there are two figures in the integral part of the cube root. Let a denote the value of the number in the tens' place in the root, and h the number in the units' place. Then a must be the greatest multiple of 10 whose cube is less than 157464 ; this -we find to be 50. Subtracting a^, that is, the cube of 50 or 125000, from the given num- ber, the remainder is 32464. Dividing this remainder by Sa^, that is, 3 times the square of 50 or 7500, we obtain 4 as the value of h. Adding to the trial-divisor 3 ah, that is, 3 times the product of 50 and 4, or 600, and b^, or 16, we have the complete divisor 8116. Multiplying this by 4, and subtracting the product, 32464, there is no remainder. Hence, 50 + 4 or 54 is the required cube root. The ciphers being omitted for the sake of brevity, the work will stand as follows : L57464 1 a3= 125000 3a2 =7500 32464 3a6= 600 &^= 16 8116 32464 EVOLUTION. 157464 54 125 7500 32464 600 16 8116 32464 177 From the above process, we derive the following rule : Separate the number into periods by pointing every third figure^ beginning with the units' place. Find the greatest cube in the left-hand period^ and write its cube root as the first figure of the root; subtract its cube from the number^ and to the result bring down the next period. Divide this remainder by three times the square of the root already founds with two ciphers annexed^ and write the quotient as the next figure of the root. Add to the trial-divisor three times the product of the last root figure and the part of the root previously found, with one cipher annexed^ and the square of the last root figure. Multiply the complete divisor by the figure of the root last obtained, and subtract the product from the remainder. If other periods remain, proceed as before, taking three times the square of the root already found for the next trial-divisor. The notes ta Art. 206 apply with equal force to examples in cube root, except that in Note 3 ttco ciphers should be annexed to the trial-divisor. 212. In the illustration of Art. 208, if there had been more terms in the given quantity, the next trial-divisor would have been three times the square of a + 6 ; that is, 3a^ -{-Qab-\-3b^. We observe that this is obtained from the preceding complete divisor, 3 a^ + 3 a6 + 6", by adding to it its second term, 3 ab, and twice its third term, 25^ We may 178 ALGEBRA. then use the following rule for forming the successive trial- divisors in the cube root of numbers : To the preceding complete divisor^ add its second term and twice its third term; and annex two ciphers to the result. EXAMPLES. 213. 1. Find the cube root of 8.144865728. 8.144865728 2.012, Ans. 8 120000 144865 600 1 120601 120601 600 24264728 2 12120300 12060 4 24264728 1213236 4 Since the second root-figure is 0, we annex two ciphers to the trial- dirisor 1200, and bring down to the remainder the next period, 865. The second trial-divisor is formed by the rule of Art, 212. The pre- ceding complete divisor is 120601 ; adding its second term, 600, and twice its third term, 2, we have 121203; annexing two ciphers to this, we obtain the result 12120300. Extract the cube roots of the following : 2. 29791. 3. 97.336. 4. .681472. 5. 1860867. 6. 1.481544, 7. .000941192. 8. 8.242408. 9. 51478848. 10. 10077.696. 11. .517781627. 12. 116.930169. 13. .031855013. 14. .724150792. 15. 1039509.197. 16. .000152273304. EVOLUTION. 179 Extract the cube roots of the following to four figures : 17. 2. ^Id. 7.2. 21. |. >23. ^. 18. 6. -{^20. .03. 22. |. -< 24. |. 214. When the index of the required root is the product of two or more numbers, we may obtain the result by suc- cessive extractions of the simpler roots. For, by Art. 198, ("^^/'a)""* = a. Taking the nth root of both members, ('»^a)-=-;ya. (1) Taking the mth root of both members of (1), -ya= V(-v/a)- That is, TJie mnth root of a quantity is equal to the mth root of the nth root of that quantity. For example, the fourth root is the square root of the square root ; the sixth root is the cube root of the square root; etc. EXAMPLES. Find the fourth roots of the following : 1. 16a;*-96a^?/H-216a^/-216a;2/3_f_gi2/4^ 3. a^-8a;^+16aj«+16ar'-56a;^-32a;s+64a;2^g4a,^lg^ Find the sixth roots of the following : 4. ai2-6ai« + 15a«-20a« + 15a^-6a2 4.i. 0. 64a^-|-192«« + 240a^+160a;» + 60ar'-f-12a; + l. 180 ALGEBRA. XIX. THE THEORY OP EXPONENTS. 215. In the preceding chapters we have considered an exponent only as a positive whole number. It is, however, found convenient to employ fractional and negative expo- nents ; and we proceed to define them, and to prove the rules for their use. 216. In Art. 13 we defined a positive integral exponent as indicating how many times a quantity was taken as a factor ; thus, a"* signifies axax«X ••• torn factors. "We have also found the following rules to hold when m and n are positive integers : I. a'^ X a^ = a''*+'*. (Art. 79.) II. ((X'«)" = a'"». (Art. 193.) 217. The definition of Art. 13 has no meaning unless the exponent is a positive integer, and we must therefore adopt new definitions for fractional and negative exponents. It is convenient to have all forms of exponents subject to the same laws in regard to multiplication, division, etc., and we shall therefore assume Rule I. to hold for all values of m and w, and find what meanings must be attached to fractional and negative exponents in consequencco 218. Required the meaning of a^. Since Rule I. is to hold universally, we must have a^ X a^ X a^ = a^ "^^"^^ = a^. That is, a^ is such a quantity that when raised to the third power the result is a^. Hence (Art. 198), a^ must be the cube root of a^ ; or, a^ = ^a^. THEORY OF EXPONENTS. 181 We will now consider the general case : p Required the meaning of a?, where p and q are positive integers. p p p By Rule I., a^ x a^ x a^ X ••• to q factors P That is, ai is such a quantity that when raised to the gth p power the result is a^. Therefore a" must be the gth root of 9^'i or, a^ = -{/a^. Hence, in a fractional exponent, the numerator denotes a power and the denominator a root. For example, a* = -y/a^ ; c^ = ^c^ ; x^ = -^x ; etc. EXAMPLES. 219. Express the following with radical signs : 1. aK 3. 2ci 5. x^y^. 7. 4a^5t. 9. 5yh^. 2. b^. 4. 3am*. 6. m^ni 8. 2c^di 10. ab^c^dK Express the following with fractional exponents : 11. ^x\ 13. V^- 15. S^m\ 17. ^a*^&. 12. -^2/'- 14. ^c. 16. 4-^a9. 18. V^-v^^/'- 19. SVwi'a/w*- 20. 2a^x^y"'. The value of a numerical quantity affected with a fractional exponent may be found by first extracting the root indi- cated by the denominator, and then raising the result to the power indicated by the numerator. Thus, (-8)* = (-\Ar8)2 = (-2)^=4. 182 ALGEBRA. Find the values of the following : 21. 9i 23. 36i 25. (-27)i 27. 64^ 22. 27i 24. lei 26. (-32)i 28. (-216)i 220. Required the meaning of a". Since Rule I. is to hold universally, we must hare Therefore aP must be equal to 1 . That is, any quantity whose exponent is is equal to 1. 221. We pass next to the case of negative exponents. Required the meaning of a~^. By Rule I., a~^ x a^ = a-^+s = a« = 1. (Art. 220.) Hence, a~^=—' a^ We will now consider the general case : Required the meaning of a~'^^ n being integral or fractional. By Rule I., a"" x a^ = a-^+" = a^= 1. Hence, a"^ = — a^ For example, 1 _*!_,_! _; a ^ = --; Sx~y 2 _. ,-2_A. ^-^_ - ..3a.-y?^ 3 . etc. »2 a^ xy^ 222. In connection with Art. 221, the following principle may be noticed : Any factor of the numerator of a fraction may he transferred to the denominator^ or any factor of the denominator to the numerator, if the sign of its exponent he changed. THEORY OF EXPONENTS. 183 Thus, the fraction ^ can be written in any of the forms cdr o' 0^^ _i etc EXAMPLES. 223. Write the following with positive exponent* i 1. a^y-\ 5. a-'b-\ 9. 6a-'b-^c. 2. x-'y^. 6. 3ah-^. 10. 2m-«w-*. 3. m^rrK 7. 2x-*y~K 11. ^x'^y'K 4. ^xy~K 8. a-^6-V. 12. a-26-^c"i Transfer the literal factors from the denominators to tha numerators in the following : 13. 1. X 16. A. Transfer the literal factors from the numerators to the denominators in the following : oo ^^ OK 2c"^ OQ -a 1 22. -—• 25. — - — 28. m ^w^. 3 5 23. ^*. 26. 3ai 29. ^. 24 £!! 27 ^^~^^ 30 i^^ll^. * 2 ' * A* ' * Sc' * 16. 1 2a;* 17. 3c 18. ad" 19. 5a2 2 6c3 20. 2x^y^ 21. ^x 184 ALGEBRA. 224. Since the definitions of fractional and negative expo- nents were obtained on the supposition that Rule I., Art. 216, was to hold universally, we have for any values of m and w, «»« X a** = «"*+**. For example, a^ X a a xa~^ = a^~2- =a,~^; etc. EXAMPLES. Find the values of the following : 1. a^ X a"^ 6. SaXoTK 11. 2c"^x3a^c8. 2. a? X a-\ 7. 5c-3x3c"i 12. 2a-364xa6~^ 3. x-^ X x-\ 8. a' X -^a^. 13. a^,-tx-/. 4. n^ X n~^. 9. x-^ X ^a;-^ 14. ^a;x 5^x-^ 5. 2x^ X x~^. 10. m^X ,^ . 15. 1 X ^ 16. Multiply a 4- 2 a^ - 3 a* by 2 - 4 a"^ - 6 a"^. a + 2a^ -3a* 2 — 4a~* — 6a"^ 2a + 4a^ -6a* -4a^ -8a* + 12 -6a*-12 + 18a~* 2a -20a* +18a *, ^ws. Note. In examples like the above, it should be borne in mind that any quantity whose exponent is is equal to 1. (Art, 220.) THEORY OF EXPONENTS. 185 Multiply the following : 17. a2-24-ci~'by a2+2 + a~^. 18. a^ + a^x^-^ x^ by a^ — a#. 19. x~^ — x~^-\-x~^-lhyx~^-\-l. 20. a;-2-2a3Fi + l-2a;by a;-3 + 2a;-2. 21. 3a-i - a-26-i + a-^h'^ by e>a?h^ + 2a^h + 2a. 22. 2a;^— 3a;* — 4 + a;~^by 3a;^ + a; — 2a;^. 23. x-^y^—x-^y-2x-^ by 2a:^2/-^ + 2a:»2^-2-4a?V~'. 24. a^«~^ + 2-{-a'^x^ by 2a~^a;^ — 4a"^a;^ -^-^a'^x^. 26. 3a^6-i + a*-2a-*6by 6a^&-i-2a"*-3a-^6. 225. The rule of Art. 89 for the division of exponents holds universally ; for, it follows from Art. 222 that — = a"* X a-** = a"*-**. (Art. 224.) -f For example, — = a~^~^ = cT^ ; a~* -3+1 -JU — r = a ^^ =a ^; etc. EXAMPLES. Divide the following : 1. a«bya-\ 4. a"Hy a~^. 7. a-Hy ~ 2. abya^ 5. 3c-i by ^cr'. 8. 15aby3a-i^6. 3. aHya*. 6. m^ by -^m-^. 9. Gaj-^^/^ by 3a;2/"^. 186 ALGEBRA. 10. Divide 2a^ — 20 + ISa"^ by a + 2a^-SaK 2a^ + 4a*-6 a + 2a'-3a* 2a * — 4a ^ — 6a-^, Ans. 4a*-14 + 18a"^ -4a* - 8 + 12a~^ -6-12a~* + 18a"^ -6-12a-* + 18a"^ Note. It is important to arrange the dividend and divisor in the same order of powers, and to keep this order throughout the work. Divide the following : 11. a-bhy a^-b^. 12. a'^ + l by a"^ + 1. 13. x-^-6x-^-4:6-4:0xhy x'^ + 4. 14. x-^—lhyx~^ — x~^-i-x~^ — l. 15. m — Bm^n^-j-Sm^n^—nhym^—n^, 16. x-^y~^ - 3 x-^y-^ + x-^y-^ by x'^y-^ + x'^y'"^ - x-*y-\ 17. a + ah^ + bhy a^ — a^b^-j-b^. 18. m~^ + m~^n~^ + n~* by m~^ + m~^n~'^ + rrT^n'^. 226. We will now prove that Rule II., Art. 216, holds for all values of m and ti. We will consider three cases, in each of which m may have any value, positive or negative, integral or fractional. Case I. Let n be a positive integer. Then, from the definition of a positive integral exponent, (a'^Y = a"* X a"* X a*" X ••• to n factors __ ^TO -j- n* + m +••• to w terms __ ^wn THEORY OF EXPONENTS. 187 Case II. Let n be a positive fraction, which we will de- note by -?. Then, (a'")?"= VJary, by the definition of Art. 218, = Vo^, by Case I., = aT, by Art. 218, = a«^f . Case III. Let n be a negative quantity, which we will denote by — «. Then, (a'»)- = ^!— , by the definition of Art. 221, = — , by Cases I. or II., We have therefore for all values of m and n, (a"*)'» = a*"". For example, (a-s)"^ = a~'''~* =a; etc. EXAMPLES. 227. Find the values of the following : 1. {o?YK 5. {x-^y\ 9. (-^m3)l 13. ("— ¥• 2. (a-2)2. e. (a-^)i 10. (^r')~'. 14. (—\^. 3. (a3)r 7. (^i)l. 11. (^i.^t. 15^ ^[(x-^)^]. 4. (c-^)"^. 8. (Va^)"*. 12. (xiyl 16. (a^-^);^. 188 ALGEBRA. 228. To prove that {aby = a^b"", for any value of n. In Art. 193 we showed the truth of the theorem for a positive integral value of n. Case I. Let n be a positive fraction, which we will denote by 5. By Art. 226, [(a2>)*]* = («&)". By Art. 193, [a* 6^]* =a^6^ = (aZ))^ Therefore, [(a6)?"]«= [afft^J^. Taking the gth root of both members, p p p {ab) 9 =aib9. Case II. Let n be a negative quantity, which we will denote by — s. Then, (ah) - = — ^ = -L, by Art. 193, or Case I. ' ^ ^ (aby a'b' ^ zzza-'b-'. MISCELLANEOUS EXAMPLES. ^ 229. Square the following (Art. 95) : 1. a^-6i 2. a-^ + 2a. 3. x'^f-Sx'y-K Extract the square roots of the following : 4. a-^xK 5. 9mni 6. ^-^. 7. ^'. ^^r cd*e^ 10. a^b~^-4:ah~^-\-6-4:a~h^ + a-^b^. THEORY OF EXPONENTS. 189 Extract the cube roots of the following : 11. ah\ 12. -%x-'y^. 13. %lmH~^, 14. -^. 64 a;^ 15. 82/-2-122/"^' + 62/"^-2/"^. 16. a* - 9a;^ + 33a;^ - 63a; + 66a;^ - 36a;^ + 8a;i Reduce the following to their simplest forms : 1 17. a*-«'+2'a2-+i'-3*a'. 20. [a;-'-"' «^'— *]«-*. ^ + 2m-r * ' \ «" / V^'-V 19. (aj-)-^-j-(a;-'')-*. 22. [(aja-tf -Ja+k. 23 ^i.^ — ^^) - ^^ (^"^ - a?"^) . 2(a* + a;*)(a^-a;^) 24. ^-^ - 25. (l-^^^-)'+(^^+^^)' 26. (4a;3_3^)(a:2^1)-i^3a;(a^ + l)f. 27 (l-3a; + a?')--a;(a;-3)(l-3a; + a^)"^ l-3a; + a:2 28 ^ra?"^+(m + a;)"^1 ^ m + 2a; 2[a;^ + (m + x)^] 2a;^ (m + a;)* 29. ^+ri+(l+a^)^P 190 ALGEBRA. XX. RADICALS. 230. A Radical is a root of a quantity indicated by a rad- ical sign ; as, Va, or -vx + 1. If the indicated root can be exactly obtained, it is called a rational quantity ; if it cannot be exactly obtained, it is called an irrational or surd quantity. 231. The degree of a radical is denoted by the index of the radical sign ; thus, Vx-f- 1 is of the third degree. 232. Similar Radicals are those of the same degree, and 5 with the same quantity under the radical sign ; as, 2-\/ax and SVo^. 233. Most problems in radicals depend for their solution on the following important principle (Art. 228) ; 11 1 For any value of n, {ah) « = a« x h-»-. That is, ^ab = Va x Vh. TO REDUCE A RADICAL TO ITS SIMPLEST FORM. 234. A radical is in its simplest form when the quantity under the radical sign is not a perfect power of the degree denoted by any factor of the index of the radical, and has no factor which is a perfect power of the same degree as the radical. Case I. 235. When the quantity under the radical sign is a perfect power of the degree denoted by a factor of the index, 1. Reduce VS to its simplest form. -^8 = a/23 = 2^ = 2^ = V2,* Ans. RADICALS. 191 EXAMPLES. Reduce the following to their simplest forms : 2. \/25. 6. ^27. 8. ^64. 11. ■v/49mV. 3. ^9. 6. ^/lOO. 9. \/'25^. 12. -^125 a'b\ 4. ^8. 7. a/81. 10. ^v/32^. 13. "Vo^. Case II. 236. WTien the quantity under the radical sign has a factor which is a perfect power of the same degree as the radical. 1. Reduce V54 to its simplest form. a/54 = a/27 X 2 = a/27 X a/2 (Art. 233) = 3 ^2, Ans, 2. Reduce A/l8a26^ — 27a^6* to its simplest form. Vl8a26^-27a364 = Vo a25^(2 6 - 3 a) A/9a26*X A/26-3a = 3a6^A/26 — 3 a, ^7i8. RULE. Resolve the quantity under t1ie radical sign into two factors^ one of which is the highest perfect power of the same degree as the radical. Extract the required root of this factor, and pre- fix the result to the indicated root of the other. Note. If the highest perfect power in the numerical portion of thp quantity cannot be determined by inspection, it may be found by re- solving the number into its prime factors. Thus, >/1944 = V¥~xJ^ = y/Wx^ X \^"3<3 = 2 X 32 X V6 =r 18 V6. 192 ALGEBRA. EXAMPLES. Reduce the following to their simp 3. V50. 6. -5/320. 4. 3V24. 5. V72. 7. 8. 2 a/so. ■VdSa^l)'. 12. V25a^2/^- -50a;y. 13. ■i/54ta'b'-\-lS5a^b\ 9. a/8T^. 10. 7V63a^6V. 11. -v/250^/?. 14. V(a^-2/')(aj + 2/): 16. -Vax^ — 6 ao; -f 9 a. 16. V20ic2+60a; + 45. 17. V3 m^ - 54 m^n + 243 mnK If the quantity under the radical sign is a fraction, multiply both terms by such a quantity as will make the denominator a perfect power of the same degree as the radical. Then pro- ceed as before. 18. Reduce \/ — -to its simplest form. .|X = ^|Ii^ = J_l_x2a = -?-V2^, Ans, \8a^ \16a^ \16a^ 4ta^ Reduce the following to their simplest forms : 19. 20. 21. 28. ^||■ - «-Alf '^U a^c — 2 a^bc -\- ab^c RADICALS. 193 237. Conversely^ the coefficient of a radical may be intro- duced under the radical sign by raising it to the power denoted by the index. I . Introduce the coefficient of 2 aV^o^ under the radical sign. \ 2aV^= A/8a^x"3^= \/24aV, Ans. Note. A rational quantity may be expressed in the form of a radical by raising it to the power denoted by the index, and writing the result under the corresponding radical sign. EXAMPLES. Introduce the coefficients of the following under the radical signs : 2. 3V5. 4. 3a/2. 6. 4V5a6. 8. baV2^. 3. 2^/7. 5. 4a/5. 7. a'hV'^K 9. 3mnH*l^'- \ 27 10. (a,_i)J^±I. 12. Hl^./IE«. II. {l^x)S^^^^^\, 13. ^^~\ l ^- --1. ADDITION AND SUBTRACTION OF RADICALS. 238. The sum or difference of two similar radicals (Art. 232) may be found by prefixing the sum or difference of their coefficients to their common radical part. 1. Find the sum of V20 and V45. By Art 236, V20 = Vi^TB = 2 V5, V45 = V9l<^ = 3V5. Hence, V20 + V45 = 2V5 + 3V5 = 5V5, Ans. 194 ALGEBRA. 2. Simplify ^i+^|-^|. O 4: RULE. Reduce each radical to its simplest form. Unite the similar radicals, and indicate the addition or subtraction of the dissimilar, EXAMPLES. Simplify the following : 3. y27 + V12. 9. V4^+V96^. 4. V96 + V54. 10. V75 + V48 - V245. 5. V180-V45. 11. ^16 + -^54+ ^128. 6. ^162-^48. l2.-^|_^i+^^. 7. V128 + V98 + V50. 13. ^/5 - ^1 + ^A. 15. 7V27-V75-24^i-27^±. 16. V27aP + V75a3+(a-36)V3a. 17. V9a^ + 18«^&- V4:a6« + 8&'' RADICALS. 195 18. -^24 + 5^54 --^250 -^192. 19. V2Sa^x-28ax-\-7x-^7a^x-{-^2ax + 6^x. 20. x — y, Ix-^y Sv' — a^ 1-3, -^ TO REDUCE RADICALS OF DIFFERENT DEGREES TO EQUIVALENT RADICALS OF THE SAME DEGREE. 239. 1. Reduce V^? \/3, and ^5 to equivalent radicals of the same degree. By Art 218, V2 = 2^ = 2^ = ^2« = ^64. ^3 = 3^ = Z^ = ^3* = ^81. ^5 = 5* = 5T^ = ^53 = 2J/125. RULE. Write the radicals with frax^tional exponents^ and reduce these exponents to a common denominator. Note. The relative magnitude of radicals may be compared by re- ducing them to the same degree. Thus, in Ex. 1, 1^125 is greater than ^^81, and ?^81 than 2^64. Hence, ^5 is greater than ^3, and ^3 than v^. EXAMPLES. Reduce the following to equivalent radicals of the same degree : 2. -^2 and V^. 5. a^, v'36, and VTc. 3. V^' -v^^' ^^^ \/^- ^- ^^2/j '^y^i and -Vzx. 4. -^5, ^6, and -^7. 7. Va^h and v'o^. 196 ALGEBRA. 8. Which is the greater, ^2 or ^3? 9. Which is the greater, -^3 or ^5 ? 10. Arrange in order of magnitude -y/3, ^4, and -^7. MULTIPLICATION OF RADICALS. 240. 1. Multiply V6 by Vl5. By Art. 233, V6 X Vl5 = V6"xT5 = V90 = 3 VlO, Ans. 2. Multiply V2a by VSaK Reducing to equivalent radicals of the same degree, V2a =(2a)* =(2a)* = a/ (2^^ = -^/g^ -?/3^ = (3a2) 3 = (3a2)^ = -v/(3^ = ^9^ Hence, V2a X -V3a^= ^ + 2 :=V6 + V2, Ans. 208 ALGEBRA. 2. Extract the square root of 22 — 3 V32. V22 - 3 ^732 = V22 - v'288 = V22 - 2 V72. We then separate 22 into two parts whose product is 72c The parts are 18 and 4 ; hence, V22 - 3 V32 = Vl8-2V72 + 4 = Vl8-V4 = 3V2~2, Ans, EXAMPLES. 256. Extract the square roots of the following : 1. 12 + 2V35. 6. 8-V60. 11. 23+V360. 2. 7-2V12. 7. 15 + 4V14. 12. 24-2V63. 3. 9 + 2VS- 8. 12~-V108. 13. 33+20V2. 4. 9-4V5. 9. 20-5V12. 14. 47-6V10. 6. 16 + 6V7. 10. 14 + 3V20. 15. 67-7V72. 16. 2m~2Vm2-7i2. 17. 2 a + x -{- 2 VoTfax* SOLUTION OF EQUATIONS CONTAINING RADICALS. 257. 1. Solve the equation Var^ — 5 — a; = — 1 , Transposing, V flr* — 5 = a; — 1 . Squaring both members, a^ — 5 = a^— 2a;-f-l. Transposing and uniting terms, 2a; = 6. x — 3, Ans. RADICALS. 209 2. Solve the equation V2a;— 1 -f V2a; -f 6 = 7. Transposing y/2x — 1 , V2a;-f6 = 7-V2a;-l. Squaring, 2a; + 6 = 49 - 14 V2a;-1 + 2a; - 1. Transposing and uniting, 14V2a;-l = 42. Or, V2^^ = 3. Squaring, 2 a; — 1 = 9. 2a;=10. a; =5, -4715. RULE. Transpose the terms of the equation so that a radical term may stand alone in one member; then raise both members to a power of the same degree as the radical. If there are still radical terms remaining , repeat the opera- tion. Note. The equation should be simplified as much as possible before performing the involution. EXAMPLES. 3. V5^^-2 = l. 8. i/a^-6x'-x-{-2 = 0. 4. 6=-\/2x-\-S. 9. Va; + Va; + 5 = 5. 6. A/4a; 4-3 = 3. 10. Va;- 32 + Va;= 16. 6. V4a;2_i9_2a;= -1. 11. Va;-3 - Va;-t-12 = -3. 7. Va;2-3a; + 6 = 2-a;. 12. V2a;-7+ V2a; + 9 = 8. 210 ALGEBRA. 13. V3a; + 10~V3a; + 25 = -3. 14. ^{x-ay + 2ab-{-b"- = x-a + b, 15. Va^-3a; + 5-Vaj2-5a;-2 = l. 16. Vx-V^':r3 = _2_. 17. Vx— l+Vi» + 4 = V4a; + 5. 18. V«2 + 4a; + 12-fVar^- 12a;- 20 = 8. 19. Va; — 3 _ Va; — 4 20. V3^+V3a;H-13 = -^ 91 V3X+13 21. Va; + l+Va;-2-V4x-3=0. 22. Va;4-V^T^= ^^ . Vic + a 23. Vl9 + icV^^^=^J = a;-3. 24. a a; = V5 — ic. Va — a; V6 — a; 25. Va; + a+Va; + 6 = V4a; + a + 3&. 26. Vll+ajVa^+lBH^^ + l- 27. V{a^-2aa;4-a;^V3a-a;i = «-^' QUADRATIC EQUATIONS. 211 XXL QUADRATIC EQUATIONS.' 258. A Q,uadratic Equation, or an equation of the second degree (Art. 167), is one in which the square is the highest power of the unknown quantity. A Pure Quadratic Equation is one which contains only the square of the unknown quantit}^ ; as, aaP = b. An Affected Quadratic Equation is one which contains both the square and the fust power of the unknown quan- tity ; as, aic^ -j- 6ic -H c = 0. PURE QUADRATIC EQUATIONS. 259. A pure quadratic equation is solved by reducing it to the form sc^ = a, and then extracting the square roots of both members. 1. Solve the equation Sx^-}-7 =— -\-35. Clearing of fractions, 12ic2 + 28 = 5a^+ 140. Transposing and uniting, 7a^=112. Or, x'^W. Taking the square root of both members, a; = ±4, Ans. Note 1. The double sign is placed before the result because the square root of a number is either positive or negative (Art. 201). 2. Solve the equation Tar' — 5 = 5 x^ — 13. Transposing and uniting, 2a^ = — 8. Or, ' a^ = -4. Whence, a;=±V— 4 Note 2. Since the square root of a negative quantity is imaginary (Art. 246), the values of x can only be indicated. 212 ALGEBRA. EXAMPLES. Solve the following equations : 3. 4:3^-7=29. 6. 4-V3a^ + 16^=6. 4. bx' + b^Sx'-^-bb, 7. ax^ + b = c. g_5 L = _M. 8 ^ ^ ^ 6ic2 4a^ 16 4-x 3 4 + a; 9 2(x + 3) (aj - 3) = (ic + 1)^ - 2ic. 10. (3 a; - 2) (2 a; + 5) + (5 a; + 1) (4a; - 3) - 91 = 0. 11. ^_34.i^ = JL_a^ + §M. 2 12 24 24 12 2a7^-5 3a;^4-2 a^-lQ ^Q 3 7 6 * 13 <^ ^ 6 14 4a;^-3 ^ 2(9a^ + 2) ar^-6 a^-a* ' 2a:2_i 3(3a^+2) 16. (2 a; - a) (a; - 6) + (2 a; + a) (a; + &) = a^ + b\ jg 5a^-l 3a^H-l 89 _« a^-3 a;2_|_2 (a^_ 3)(a;2^ 2) 17. a; + V^T3 = 18. L= 1 =:^. l_Vl-a;^ 1 + Vl-a^ a^ AFFECTED QUADRATIC EQUATIONS. 260. An affected quadratic equation is solved by adding to both members such a quantity as will make the first mem- ber a perfect square ; an operation which is termed complet- ing the square. QUADRATIC EQUATIONS. 213 FIRST METHOD OF COMPLETING THE SQUARE. 261. Every affected quadratic equation can be reduced to the form ^ +px = q ; where p and q represent any quantities whatever, positive or negative, integral or fractional. Let it be required to solve the equation a^ -\-3x = 4:. In any trinomial square (Art. 108), the middle term is twice the product of the square roots of the first and third terms ; hence the square root of the third term is equal to the second term divided by twice the square root of the first. Therefore the square root of the quantity which must be added to or + 3 a; to make it a perfect square, is — , or — . Zx A 3 9 Adding to both members the square of - , or -, we have iB2 + 3aj + - = 4-|-- = — • 4 4 4 Extracting the square root of both members, a; + - = ± — 2 2 Transposing |, a; = _ | + 1, or - 1 - 1. Whence, x~\ or —4, Ans. 2Q2l. From the above operation we derive the following rule : Reduce the equation to the form oc^ +px — q. Complete the square by adding to both members the square of half the coefficient of x. Extract the square root of both members^ ciud solve the sim- ple equation thus formecl,, 214 ALGEBRA. 1. Solve the equation da^ — 8x= — 4:. Dividing by 3, a^-^ = -i, 3 3 which is in the form a^ + pa? = g. Adding to both members the square of -, or — , 3 9' 3 "^ 9 3"^ 9 9' Extracting the square root, 3 3 Whence, a-^^l-t^gor?, Ans, 3 3 3 Note. These values may be verified as follows : Putting a: = 2 in the given equation, 12 — 16 = — 4. Puttinga:=?, 4_16^_^ 3 -^ 3 3 * If the coefficient of a^ is negative, it is necessary to change the sign of each term. 2. Solve the equation — 3a^— 7ic = — • Dividing by - 3, x^-\-~ = -~ Adding to both members the square of -, or — , 6 36 3 36 9 36 36* Extracting the square root, 6 6 Whence, (c=-^ ± ? = -? or -5, ^^5. 6 6 3 3 QUADRATIC EQUATIONS. 215 EXAMPLES. Solve the following equations : 3. a^4-4a;=5. 8. 2iK2 + 5a; = -2. 4. ic2-5a; = -4. 9. 4:aF-8x + S = 0, 6. a^_7a; = -12. 10. 4a^-3 = llx. 6. a^-\-x=6. 11. S-x-2x^ = 0. 7. 3a^_4aj=:4. 12. 14 4-15a?-9«2=:0. 263. If the coefficient of a^ is a perfect square, it is con- venient to complete the square directly by the principle of Art. 261 ; that is, by adding to both members the square oj the quotient obtained by dividing the second term by twice the square root of the first. 1. Solve the equation 9 a;^ — 5 a; = 4. The quotient of the second term divided by twice the 5 5 square root of the first, is -. Adding the square of - to both members, 9a--5a.+?^=4 + ?5 = i5?. 36 36 36 Extracting the square root, Q 5 . 13 3a; == ± — 6 6 6 6 3 Whence, a; = l or — -, Ans* «/ Note. K the coefficient of x^ is not a perfect square, it may be made so by multiplication. Thus, in the equation 18 x'^ + 5 a: = 2, the coefficient of x^ may be made a perfect square by multiplying each term by 2. If the coefficient of x"^ is negative, the sign of each term must be changed. 216 ALGEBRA. EXAMPLES. Solve the following equations : 2. 4a^ + 3ic=10. 7. 8a^ + a;-34 = 0. 3. 9ic2+2ic=ll. 8. 11x^12-360^ = 0, 4. 25a^-15a; = -2. 9. 6x'-bx = -l. 5. 4a^-7a;=-3. 10. 32a;2 + 20£C- 7 = 0. 6. 2a;2 + 15a; = -13. 11. 48a^-32aj = 3. SECOND METHOD OF COMPLETING THE SQUARE. 264. Every affected quadratic can be reduced to the form aaP -\-bx = c. Multiplying each term by 4 a, we have 4 a^ic^ H- 4 a6a; = 4 ac. Completing the square by adding to both members the square of b (Art. 263), 4aV + 4a6ic + &2 = 62 4- 4aa; + ca; = (a + c) (a - 6) . 53. a6x^ + ^:= ^^^ + ^^-2^^ -^. c c^ c 54. (3a2 + 62)(a^_a; + i) = (362 + a2)(a^ + a;4-l). 222 ALGEBRA. SOLUTION OF QUADRATIC EQUATIONS BY A FORMULA. 267. It was shown in Art. 264 that if aoi^ + bx = c, then 5±VF-+4ac ^ 2a ^ ^ This result may be used as a formula for the solution of quadratic equations, as follows : 1 . Solve the equation 3a:^ + 5x=12. In this case, a = 3, 6 = 5, c = 12; substituting these values in (1), _ - 5 ± V25 4- 144 _ - 5 ± ^169 X — — — — — 6 6 -5 ±13 o 4 . 3 or -, Alls. 6 3 2. Solve the equation 110cc2-21a;=-l. In this case, a=110, & = — 21, c = — 1; therefore, ^ 21±V441-440 21 ±1 1 1 J X = = = — or — , A71S. 220 220 10 11 Note. Particular attention must be paid to the signs of the coeffi- cients in substituting. EXAMPLES. Solve the following equations : 3. 2iB2 + 5»=18. 8. 5a^-lla; = -2. 4. Sa^^2x = 5. 9. 4:a^-Sx-5 = 0. 6. ar^-7a; = -10. 10. 6a^ + 25a; + 14 = 0. 6. bx' + x^lS, 11. 30x-16 = dx', 7. 6ic2 + 7a; = -l. 12. 27 + 39a;- lOa;^^ 0. QUADRATIC EQUATIONS. 223 XXII. PROBLEMS. INVOLVING QUADRATIC EQUATIONS. 268. 1. A man sold a watch for $21, and lost as much per cent as the watch cost him. Required the cost of the watch. Let X = the cost in dollars. Then, x = the loss per cent, and X X — = -^ = the loss in dollars. 100 100 By the conditions, -— - = a: — 21. 100 Solving this equation, x = 70 or 30. That is, the cost of the watch was either $ 70 or 1 30 ; for each of these values satisfies the given conditions. 2. A farmer bought some sheep for $72. If he had bought 6 more for the same money, they would have cost $ 1 apiece less. How many did he buy ? Let X = the number bought. 72 Then, — = the price paid for one, X 72 and = the price if there had been 6 more. x + 6 ^ By the conditions, — = — ^^ + 1. ^ X x + 6 Solvmg, a: = 18 or - 24. Only the positive value of x is admissible, as the negative value does not answer to the conditions of the problem. The number of sheep, therefore, was 18. Note 1. In solving problems which involve quadratics, there will always be two values of the unknown quantity ; but only those values should be retained as answers which satisfy the conditions of the prob- lem. 224 ALGEBRA. Note 2. If, in the given problem, the words " 6 more *' had been changed to " 6 fewer" and " $ 1 apiece less " to " $ 1 apiece more" we should have found the answer 24. In many cases where the solution of a problem gives a negative result, the wording may be changed so as to form an analogous prob- lem to which the absolute value of the negative result is an answer. PROBLEMS. 3. I bought a lot of flour for $175 ; and the number ot dollars per barrel was \ of the number of barrels. How many barrels were purchased, and at what price ? 4. Separate the number 15 into two parts the sum of whose squares shall be 117. 6. Find two numbers whose product is 126, and quotient 6. I have a rectangular field of corn containing 6250 hills. The number of hills in the length exceeds the num- ber in the breadth by 75. How many hills are there in the length, and in the breadth? 7. Find two numbers whose difference is 9, and whose sum multiplied by the greater is 266. 8. The sum of the squares of two consecutive numbers is 113. What are the numbers? 9. A man cut two piles of wood, whose united contents were 26 cords, for $35.60. The labor on each cost as many dimes per cord as there were cords in the pile. Required the number of cords in each pile. 10. Find two numbers whose sum is 8, and the sum of whose cubes is 152. -• 11. Find three consecutive numbers su-ch that twice the product of the first and third is equal to the square of the second increased by 62. PROBLEMS. 225 12. A grazier bought a certain number of oxen for $240. Having lost 3, he sold the remainder at $8 a head more than they cost him, and gained $59. How many did he buy? 13. A merchant bought a quantity of flour for $96. If he had bought 8 barrels more for the same money, he would have paid $2 less per barrel. How many barrels did he buy, and at what price ? 14. Find two numbers, whose product is 78, such that if one be divided by the other the quotient is 2, and the remainder 1. 15. The plate of a rectangular looking-glass is 18 inches by 12. It is to be framed with a frame all parts of which are of the same width, and whose area is equal to that of the glass. Required the width of the frame. 16. A merchant sold a quantity of flour for $39, and gained as much per cent as the flour cost him. What was the cost of the flour? 17. A certain company agreed to build a vessel for $6300; but, two of their number having died, the rest had each to advance $ 200 more than they otherwise would have done. Of how many persons did the company consist at first? 18. Divide the number 24 into two parts, such that the sum of the fractions obtained by dividing 24 by them shall be||. 19. A detachment from an army was marching in regular column, with 6 men more in depth than in front. When the enemy came in sight, the front was increased by 870 men, and the whole was thus drawn up in 4 lines. Required the number of men. 20. A merchant sold goods for $16, and lost as much per cent as the goods cost him. What was the cost of the goods ? 226 ALGEBRA. 21. A certain farm is a rectangle, whose length is twice its breadth. If it should be enlarged 20 rods in length, and 24 rods in breadth, its area would be doubled. Of how many acres does the farm consist? 22. A square court-yard has a gravel-walk around it. The side of the court lacks one yard of being 6 times the breadth of the walk, and the number of square yards in the walk exceeds the number of yards in the perimeter of the court by 340. Find the area of the court and the width of the walk. 23. A merchant bought 54 bushels of wheat, and a cer- tain quantity of barley. For the former he gave half as many dimes per bushel as there were bushels of barley, and for the latter 40 cents a bushel less. He sold the mixture at $1 per bushel, and lost $57.60 b}^ the operation. Re- quired the quantity of barley, and its price per bushel. 24. A certain number consists of two digits, the left- hand digit being twice the right-hand. If the digits are inverted, the product of the number thus formed, increased by 11, and the original number, is 4956. Find the number. 25. A cistern can be filled by two pipes running together in 2 hours 55 minutes. The larger pipe by itself will fill it sooner than the smaller by 2 hours. What time will each pipe separately take to fill it? 26. A and B gained by trade $1800. A's money was in the firm 12 months, and he received for his principal and gain $2600. B's money, which was $3000, was in the firm 16 months. How much money did A put into the firm? 27. My gross income is $1000. After deducting a per- centage for income tax, and then a percentage, less by one than that of the income tax, from the remainder, the income is reduced to $912. Find the rate per cent of the income tax. PROBLEMS. 227 28. A man travelled 102 miles. If he had gone 3 miles more an hour, he would have performed the journey in 5| hours less time. How many miles an hour did he go? 29. The number of square inches in the surface of a cubical block exceeds the number of inches in the sum of its edges by 210. What is its volume? 30. A man has two square lots of unequal size, contain- ing together 15,025 square feet. If the lots were contigu- ous, it would require 530 feet of fence to embrace them in a single enclosure of six sides. Required the area of each lot. 31. A set out from C towards D at the rate of 3 miles an hour. After he had gone 28 miles, B set out from D towards C, and went every hour -^^ of the entire distance ; and after he had travelled as many hours as he went miles in an hour, he met A. Required the distance from C to D. 32. A courier proceeds from P to Q in 14 hours. A sec- ond courier starts at the same time from a place 10 miles behind P, and arrives at Q at the same time as the first courier. The second courier finds that he takes half an hour less than the first to accomplish 20 miles. Find the dis- tance from P to Q. 33. A person bought a number of $20 mining-shares when they were at a certain rate per cent discount for $ 1500 ; and afterwards, when they were at the same rate per cent premium, sold them all but 60 for $1000. How many did he buy, and what did he give for each of them? 228 ALGEBRA. XXIII. EQUATIONS IN THE QUADRATIC FORM. 269. An equation is in the quadratic form when it is expressed in three terms, two of which contain the unknown quantity ; and of these two, one has an exponent twice as great as the other; as, a^ _j_ ipl = 72 ; (a;2-i)2_}-3(a^_l) = 18; etc. 270. The rules for the solution of quadratics are applica- ble to equations having the same form. 1 . Solve the equation ic^ — 6 a^ = 1 6 . Completing the square, fl;6_6a;3 4-9 = 16 + 9 = 25. Extracting the square root, a^-3 = ±5. Whence, a^ = 3±5 = 8or — 2. Extracting the cube root, ic= 2 or — -^2, Ans. Note. There are also four imaginary roots, which may be obtained by the method explained in Art. 282. 2. Solve the equation 2x-h3-y/x=27. Since ^x is the same as x^, this is in the quadratic form. Multiplying by 8, and adding 3^ or 9 to both members, 16a; + 24Va' + 9 = 216 + 9 = 225. Extracting the square root, .4Va^ + 3 = ±15. Or, 4Va; = -3±15 = 12or -18, QUADRATIC EQUATIONS. 229 9 Whence, V^ = 3 or — - • 81 Squaring, a; = 9 or — , Ans, 3. Solve the equation 16 a;" ^ — 22 ic" ^ = 3. Multiplying by 16, and adding 11^ to both members, le^a;"^ - 16 X 22a;~* + 121 = 48 + 121 = 169. Extracting the square root, 16aj"*-ll = ±13. Or, 16a;"^ = 11 ± 13 = - 2 or 24. Whence, a;~* = — - or -• ' 8 2 Extracting the cube root, Raising to the fourth power. 1 X /'3\f -1= JL or ' ^^ 16 Inverting both members, a;= 16 or ( - j , Ans, V Note. In solving equations of the form xi = a, first extract the root corresponding to the numerator, and afterwards raise to the power corresponding to the denominator. Particular attention should be paid to the algebraic signs ; see Arts 192 and 201. EXAMPLES. Solve the following equations : 4. a;*-25ic2 = -144. 7. aj--*- 9a;-2 = - 20. 5. »«+20iB8-^69 = 0. 8. 81a^4-- = 82. 6. a;i« + 31ar''-32 = 0, ' 9. 8x^-216 = 373:^. 230 ALGEBKA. 10. (3a^-2)2-ll(3a;2_2) + l0 = 0. 11. (a^-5)2 = 241-29a^. 12. a^--a;^ = 56. 17. 2x~'-j-61x~^ -96=^0, 13. x^-{-x^=766. 18. 4x-15 = 17VaJ. 14. 2/.+ 3.-«-56 = 0. 19. ^g±2=4_i. ^ 4+V^ V^ 15. 3 aj^ + 0^^ = 3104. 20. 3a5^-^'=- 592. 16. 3(c^4-26a;^ = -16. 21. 8aj~^- 15a;~^- 2 = 0. 271. An equation may be solved with reference to an expression, by regarding it as a single quantity. 1. Solve the equation (x — 5y — 3{x — 5)^ =z 40. Regarding x — 5 as a single quantity, we complete the square in the usual way. Multiplying by 4, and adding 9 to both members, 4(a; - 5)3 - 12(a; - 5)* + 9 = 160 + 9 = 169. Extracting the square root, 2(a;-5)^-3 = ±13. Or, 2(a;-5)^ = 3± 13 = 16 or -10. Whence, (a; — 5)^ = 8 or -5. Extracting the cube root, (a;-5)^ = 2or --^5. Squaring, a? — 5 = 4 or ^25. Transposing, a;= 9 or 5 +-^25, Ans. QUADRATIC EQUATIONS. 231 An equation of the fourth degree may sometimes be solved by expressing it in the quadratic form. 2. Solve the equation a;* -|- 12 ar^ 4- 343^- 12 a;- 35 = 0. We may write the equation as follows : (a;^+12aT^ + 36a^)-2aj2-12a; = 35. Or, ^ (a^ + 6a;)^-2(a^+6a;) = 35. Completing the square, Extracting the square root, (a:^ + 6a;) — 1 = ± 6. Whence, a?-\'ex=l±6= 7 or— 5. Completing the square, »^ + 6 a; + 9 = 16 or 4. Extracting the square root, x + 3= ±4or ±2. Whence, a;= -3 ± 4 or -3 ± 2 = 1, —7, —1, or —5, Ans, Note. In solving equations like the above, the first step is to form a perfect square with the .r* and x^ terms, and a portion of the x^ term. By Art. 261, the third term of the square is the square of the quotient obtained by dividing the x^ term by twice the square root of the x* term. 3. Solve the equation 2a;2^. V2ar' + 1 = H. Adding 1 to both members, (2ar^ + 1) + V2^Tl = 12. Completing the square, 4(2a;2 + l)4.4V2ar^-f 1 + 1 = 48 + 1=49. Extracting the square root, 2 V2FT1 + 1 = ± 7. Or, 2V2a:2 4.i = _i±7 = 6or-8. Whence, \^2x^-{-l = 3 or - 4, 232 ALGEBRA. Squaring, 2ic^ 4- 1 = 9 or 16. 2a^ = 8 or 15. a^ = 4or— . 2 Extracting the square root, a; = ± 2 or ± - V^O, Ans, Note. In solving equations of this form, add such quantities to both members that the expression without the radical in the first mem- ber may be the same as that within, or some multiple of it. EXAMPLES. Solve the following equations : 5. aJ*+10a^+17i»2-- 40a; -84 = 0. 6. ic2 - 10a; -2VV-^10^+18+ 15 = 0. 7. a;2_j_5+Va;2_j.5^i2. 8. 2a;2+3a;-5V2^N-3¥+9 = -3. 9. fl;4 + 2iB5-25a;2_26a; + 120 = 0. 10. iB^-6a;3-29a;2_,_ii4a,^30. 11. a2_6a._|.5Va;2_g^^20 = 46. 12. Va;+10-){x-\-^) = 0. Therefore, cc = ; flj — 8 = 0, ora; = 8; and a;H-3 = 0, or a; = — 3. Ans. ic = 0, 8, or —3. 3. Solve the equation a:^4-4aj2 — a; — 4 = 0. Factoring the first member (Art. 105), (a; + 4)(a.'2-l) = 0. Therefore, a; + 4 = 0, ora; = — 4; and a^ — 1 = 0, oraj=±l. Ans, aj = — 4 or ± 1 , THEORY OF QUADRATIC EQUATIONS. 249 4. Solve the equation x^ — 1 =0. Factoring the first member, (.T-l)(ar + a;-j-l) = 0. Therefore, a? — 1 = 0, or a; = 1 ; and a^ + .T+l = 0. (1) 2 Solving (1) by the rules for quadratics, x = Ans. a; = 1 or EXAMPLES. Solve the following equations : 5. /'a;_?Ya; + I'j = 0. 10. 2ic3-18a; = 0. 6. ^x'-x^O. 11. (3a; + l)(4a^-25) = 0. 7. {ax + h){hx-a) = 0, 12. 3a^ + 12a.-2 = 8. (a^-4)(aj2-9) = 0. 13. {x'-a'')(x'-ax-h)=0, 9. (a;-2)(x24-9a;+20)=0. 14. "14.^ -^x" ^I2x = 0. 15. ic(2a; + 5)(3a;-7)(4a;+l) = 0. 16. (a^ - 5a; + 6) (a^ + 7a; + 12) {x'-^x - 4) = 0. 17. a?-\-\=:Q, 19. a.-3-a;2-9a; + 9 = 0. 18. a;«-l=0. 20. 2a;3 + 3a;2_2a;_3 = 0. Note. The above examples are illustrations of the important prin- ciple that, the degree of an equation indicates the number of its roots ; thus, an equation of the third degree has three roots ; of the fourth degree, four roots ; etc. It should be observed that the roots are not necessarily unequal; thus, the equation x^ — 2x + l = may be written (x — l)(a: — 1) = 0, and therefore the two roots are 1 and 1. 250 ALGEBRA. FACTORING. 283. A quadratic expression is a trinomial expression of the form ay?-\-hx-\- c. Any such expression may be resolved into two simple fac- tors by the artifice of completing the square (Art. 260), in connection with Art. 111. EXAMPLES. 1. Factor 6a;2 + 7a;-3. 6ar^+7a;-3=6(a^ + 7^_1\ 6 2/ By Art. 262, the expression in the parenthesis will become a perfect square if the third term is ( — j ; hence, "'^— [-+¥-(fj-(fj-i] = (2a;-}-3)(3ic-l), Ans, 2. Factor 4 a^ — 20a; + 19. 4 ar^ - 20a; + 19 = 4a;2 _ 20a; + 25 - 25 + 19 = (2a; -5)2- 6 = (2a;-'5+V6)(2^-5-V6)i ^ns. THEORY OF QUADRATIC EQUATIONS. 251 Note. If the x^ term is negativey the entire expression should be en- closed in a parenthesis preceded by a — sign. 3. Factor 102 + 11a; - a.-^. 102) :(-^)"-f] 102 = - X i'-'i^m 11 2S\ 2 2 J = (x-{-6)-{-l){x-17) = (6+ a;) (17 — a;), Ans. 4. Factor ar^ — a;2/ — 21/^ — 5 a; -I-2/-I- 6. a^ — xy — 5x — 2y^-\-y-\-Q i'-'-^) y^+lOyH-25 8y^ 4 4?/ -24 _/^ .V + 5V 9y^ + 6y + l (-f') = a;- .V + 5 +^X'"-4^'-^) (^^*-"^' = (a; + 2/ — 2)(a; — 2?/ — 3), ^ws. Factor the following : 5. ft^_4a;_60. 6. a^ + 13a; + 40. 7. x'-Ux + lS, 8. 2a.-2-7a;-15. 9. 4a^-15a; + 9. 10. 5x'-}-S6x-j-7. 11. 39-10a;-a:2^ 12. 2 + X-6X', 13. a;2 + 4a; + l. 14. 9a;2-6a;-4. 252 ALGEBRA. 15. Sx^-lSx + d, 22. 1-Sx-af. 16. 6-x-2ay'. 23. 16 -\- 26x-24.a^. 17. 5-\-4:X-12x^. 24. 25a;2-20ic-2. 18. 9a^-12a;+l. 25. 6a;2- 13aa;- ISa^. 19. 5-18ir-8a^. 26. 20 a^ -i- 41 mx + 20 m\ 20. 10a;2-23aj + 6. 27. 12 x^ -\- 7 xy - 10 y\ 21. 16a^ + 34a; + 15. 28. 21a:2_ 53^^^^ 21 7^^. 29. x'^ + xy -Qy^-\-x-{-lSy-6. 30. a^ + 3rc2/ + 2?/- + 3.T + 4?/4-2. 31. Q-5y-j-x-6y^-\-6xy-x^. 32. aj^ — 5 x?/4- 62/^ — 372; + 142/2 + 40^. 33. 2a^-i»2/-7/ + 3aj+3?/-2. 34. 3a2 + 4a6 + 62 + 5a-6-12. Certain expressions of the fourth degree may be resolved into two quadratic factors by the artifice of completing the square. 35. Factor a* + a^^s 4- 54. By Art. 108, the expression will become a perfect square if the middle term is 2a^b^. Hence, a* + a'b^ + 6^ = (a* + 2 a'b^ + b') - a'b^ = (a' -\- by - a'b' = (a^ -j-b^-j- ah) {a" + 6^ _ ^j) (^^.^^ m) = (^2 4. a6 + 62) (^2 _ab + b') , ^ns. 36. Factor 9a;^- 39 a^ + 25. 9 ic^ - 39 a^ + 25 = (9 o;^ - 30a^ + 25) - 9 0.-2 = {Sx^-5y-9x' = (Sx^ - 6 + Sx) {Sx'' - 5 - Sx) = (3a;2 + 3a;-5)(3a;2-3a;-5), Ans, THEORY OF QUADRATIC EQUATIONS. 25S 37. Factor a;* +1. = (x^ + iy-(x^2y Factor the following : 38. x^ + x'-hl. 46. a*-5aV + a;\ 39. a;^-7a^ + l. 47. ic^ + 81. 40. 4a*-8a262 + 6^ 48. 4.a* -\- 15 a'b^-^Ub'. 41. m^-14?MV + n^ 49. 16a:^- 49mV4-9m*. 42. l-1362 + 46\ 50. 9a;*-6a;2-f-4. 43. X*- 12x^7/ -\-4y\ 61. 9a*4- 14aW + 25m*. 44. 4a* + Sa^ + 9. 52. 4-32n2 + 49n*. 45. 4m^-24?/i2+25. 53. 16 x* - 49 x^f + 25 y\ 284. The equation a;*-f- 1 =0 may be solved, as in Art. 282, by placing the factors of the first member (Ex. 37, Art. 283) separately equal to zero ; thus, -v/2 + a/ 2 a^ + a;V2 + l = 0; whence a; = — YJl±^ f; and 7? — X -y/2 -|- 1 == ; whence x = - ~ Therefore, 2 ±V2±V^^ EXAMPLES. Solve the following equations : 1. x*-\-16 = 0. 4. x'-{-a* = 0. 2. »^-6a;2 4-1 = 0. 5. 0^^-8.^2 + 4 = 0. 3. x'-x' + l = 0. 6. a;*- — + 1 = 0. 254 ALGEBRA. DISCUSSION OF THE GENERAL EQUATION. 285. The roots of the equation a;- -\-px= q are '2 '2 We will now discuss these values for different values of p and q. I. Suppose q positive. Since p^ is essentially positive (Art. 192), the quantity under the radical sign is positive and greater than p^. Therefore the value of the radical is greater than p. Hence Vi is positive and rg is negative. If p is positive, rg is numerically greater than r^ ; that is, the negative root is numerically the greater. If p is 2;ero, the roots are numerically equal. If p is negative, r^ is numerically greater than r.2 ; that is, the positive root is numerically the greater. II. Suppose q=0. The quantity under the radical sign is now equal to p^, so that the value of the radical is p. If p is positive, ?'i= 0, and r^ is negative. If p is negative, ri is positive, and r.^ = 0. III. Suppose q negative, and 4g numerically p^. The quantity under the radical sign is now negative ; hence, by Art. 201, both roots are imaginary. The roots are both rational or both irrational according as p^ + 4g is or is not sl perfect square. EXAMPLES. 1. Determine by inspection the nature of the roots of the equation 2 a;^ — 5a; — 18 = 0. 5x The equation may be written x^ = 9. Since q is positive and p negative, the roots are one posi- tive and the other negative ; and the positive root is numeri- cally the greater. 25 169 In this case, p^ + 4^= }- 36 = ; a perfect square. 4 4 Hence the roots are both rational. Determine by inspection the nature of the roots of the fol- lowing : 2. a;2_,_2a;_i5 = o. 6. 6a^-7ar-5 = 0. 3. ic2 + 5a;-f 6 = 0. 7. 9ar + 30a; = - 25. 4. ic2-10a; = -25. 8. 9ar^-f 8 = 18a;. 6. 3x'2-5a;-f-4 = 0. 9. 10 - 3a; - 18a;2^ 0^ 2b6 ALGEBRA. XXVI. INEQUALITIES. 2586. An Inequality is a statement that one of two quanti- ties is greater or less than the other ; as, a > &, or m < n. The terms greater and less are here taken in the algebraic sense ; that is, of any two quantities a and 6, a is the greater when a — 6 is positive, and the less when a — & is negative. 287. The expression on the left of the sign of inequality is called the First Member, and that on the right the /Second Member, of the inequality. 288. Two inequalities are said to subsist in the same sense when the first member is the greater or the less in each. Thus, a > 6, and c>d; or m < n, and p 6, and c < d are inequalities which subsist in a contrary sense. 290. An inequality will continue in the same sense after the same quantity has been added to, or subtracted from, both members. For consider the inequality a^b. Then by Art. 286, a — 6 is positive. Therefore each of the quantities (a-^c) — (b + c), and (a — c) — (b — c) is positive, since each is equal to a — b. INEQUALITIES. 257 Whence by Art. 286, a + c > 6 -h c, and a — c>h — c. It follows from the above that a term may he transposed from one member of an inequality to the other by changing its sign. 291. If the signs of all the terms of an inequality are changed, the sign of inequality must be reversed. For consider the inequality a — b>c — d. Transposing each term (Art. 290) , we have d — c>b — a. That is, b —ab. By Art. 286, a — 6 is positive. Hence, if m is positive, each of the quantities ,x , a — b m or, ma — mb, and 7 m m is positive. ma > m6, and — > — ' mm Therefore, 293. If both members of an inequality are multiplied or divided by the same negative quantity, the sign of inequality must be reversed. For multiplying or dividing by a negative quantity changes the signs of all the terms, and hence the sign of inequahty must be reversed (Art. 291). 258 ALGEBRA. 294. If any number of inequalities^ subsisting in the same sense, are added member to member, the resulting inequality will also subsist in the same sense. For consider the inequalities a>b, a'>b', a">b", .... Then each of the quantities a — b, a' — b', a" — b", ..., is positive. Therefore their sum a-b + a'-b'+a"-b"-\-'", or, a + a'+ a"+ {b + 6'+ &"+ •••) is positive. Whence, a + a'+ c^"+ •••> 5 + &'+&" + •••. Note. If two inequaUties, subsisting in the same sense, are sub- tracted member from member, the resulting inequality will not neces. sarily subsist in the same sense. Thus, if a > 6 and a' > 6', then a — - 6 and a' — b' are positive. But a — b — (a' — b'), or its equal, a — a' — (b — b'), may be eithei positive, negative, or zero ; and hence it does not necessarily follow that a — a'>6 — 6'. EXAMPLES. 295. 1. Find the limit of x in the inequality „ 23 ^2a; , . 7x < \-5, 3 3 Clearing of fractions (Art. 292), we have 21a;— 23 < 2a; +15. Transposing (Art. 290) , and uniting terms, 19a;<38. Whence by Art. 292, a;< 2, Ans. INEQUALITIES. 259 2. Find the limits of x and y in the following : |3a; + 2y>37 (I) (2a; + 3y = 33 (2) Multiplying (1) by 3, and (2) by 2, 9a; + 62/>lll Subtracting, 5 a; > 45 Whence, a;>9. Multiplying (1) by 2, and (2) by 3, 6aj + 42/>74 6a; + 9y=99 Subtracting, — 5 ?/ > — 25 Whence by Art. 293, y 9, and 2/ < 5, Ans, Find the limits of x in the following : 3. (6a; + l)2-105<(4a;-3)(9a; + 4). 4. (2a;4-3)(3a;-l)>(2a; + 7)(3x-2)+l. 5. (x + l)(a; + 2)(a;-3)>(a;-l)(a;-4)(a; + 5). 6. 3aa; + 14a6> Ga^H- 76a;, if 3 a — 76 is negative. 7. ^~^ < ^~ , if a and 6 are positive and a > 6. h a Find the limits of x and y in the following : g (5a;4-72/>38. g |2a; + 32/<57. 1 a;- y = -2. * l3a; + 72/ = 93. 10. Find the limits of x when 2a;— 9> 21 — 4a;, and 3 a; — 11 > 5a; — 41. 11. A certain positive whole number, plus 23, is less than 6 times the number, minus 12 ; and 9 times the number, minus 54, is less than twice the number, plus 9. What is the number? 260 ALGEBRA. 12. A teacher being asked the number of his pupils, re- plied that 29 was less than twice their number, diminished by 7; and that 5 times their number, diminished by 5, was less than twice their number, increased by 55. Required the number of his pupils. 13. A shepherd has a number of sheep such that twice the number, diminished b}* 45, exceeds 79, diminished by twice the number; and 5 times the number, increased by 1, is less than 3 times the number, increased by 69. How many sheep has he ? 14. Prove that if a and b are positive, a Since the square of any quantity is positive, (a-by>0. That is, a2_2a6+62>o, or, a'-{-b^>2ab. Dividing each term of the inequality by ab (Art. 292) , we have b a 15. Prove that for any value of a;, .'^ — 3 a? + 4 > If. 16. Prove that for any values of a and 6, (2a + 6)(2a-&)>26(6a-5 6). 17. Prove that for any values of a, 6, and c, a^ + 62 + c^ > 2 {ab -\-bc- ca) . 18. Prove that (a^ - b^) (c^ - d') < (ac - bdy. J9. Prove that if a^ + 6^ ^ 1 and c^ + d'^=l, then THE THEORY OF LIMITS. 261 XXVII. THE THEORY OF LIMITS. INTERPRETATION OF THE FORMS -» -. AND -• <» Note. The symbol oo is called Infinity. 296. A variable quantity^ or simply a variable^ is a quan- tity which may assume, under the conditions imposed upon it, an indefinitely great number of different values. A constant is a quantity which remains unchanged through- out the same discussion. 297. A limit of a varinble is a constant quantity, the dif- ference between whicii and the variable may be made less than any assigned quantity however small, without ever becoming zero. In other words, a limit of a variable is a fixed quantit}' to which the variable approaches indefinitely near, but never actually reaches. The variable is said to approach indefinitely to its limit. 298. Suppose, for example, that a point moves from A towards B under the condition that it shall move, during successive equal , , i i i intervals of time, first from A to O, half-way between A and B ; then to D, half-way between C and B ; then to E, half-way between D and B ; and so on indefinitely. In this case the distance between the moving point and B can be made less than any assigned quantity however small, but cannot be made equal to zero. Hence the distance from A to the moving point is a vari- able which approaches indefinitely the constant value AB as a limit, without ever reaching it. Again, the distance from the moving point to 5 is a vari- able which approaches the limit 0. 262 ALGEBRA. 299. The Theorem of Limits. If two variables are always equal, and each approaches a limit, the tiuo limits are equal. A 1 M t C 1 B 1 A' 1 M' \ B> .,,1 Let AM and AM' be two equal variables which approach the limits AB and A'B', respectively. If possible, suppose AB > A'B\ and lay off AC= A'B'. Then the variable AM may assume values between AC and AB, while the variable A'M' is restricted to values less than AC', which is contrary to the hypothesis that the vari- ables should always be equal. Hence AB cannot be > A'B', and in like manner it may be proved that AB cannot be 7i. In this case the values of x and y are positive. Hence the couriers are together at some time after 12 o'clock, and at some point to the right of P. This corresponds with the supposition made ; for if m is greater than n, A is travelling faster than J5, and it is evi- dent that he will eventually overtake him at some point in advance of their positions at 12 o'clock. THE PROBLEM OF THE COURIERS. -265 2. m w or m<. n. In this case we have x = and y = 0. Hence the couriers are together at 12 o'clock, at the point P. This corresponds with the hypothesis ; for if a = 0, and m and n are unequal, the couriers are together at 12 o'clock, and are travelling at unequal rates ; hence they never could have been together before that time, and they never will be together afterwards. 266 ALGEBRA. 5. a = 0, and m = n. " ' In this case both x and y take the form — According to the supposition made, the couriers are together at 12 o'clock, and are travelling at the same rate. Therefore they always must have been, and always will be together. There is in this case no single answer nor finite number of answers to the problem ; for any value of x whatever, to- gether with the corresponding value of y^ will satisfy the given conditions. Hence, a result - indicates that the problem is indetermi- nate; that is, the number of solutions is indefinitely great. RATIO AND PROPORTION. 267 XXVIII. RATIO AND PROPORTION. 304. Ratio is the relation with respect to magnitude which one quantity bears to another of the same kind, and is expressed by writing the first quantity as the numerator and the second as the denominator of a fraction. Thus the ratio of a to 6 is - ; and it is also expressed a:b. b 305. A Proportion is an equality of ratios. Thus, if the ratio of a to 6 is equal to the ratio of c to d, they form a proportion, which may be written in either of the forms: ^ ^ a:b = c:d, - = -, or a:b::c:d. b d 306. The first term of a ratio is called the antecedent^ and the second term the consequent. Thus in the ratio a : 6, a is the antecedent, and b is the consequent. The first and fourth terms of a proportion are called the extremes^ and the second and third terms the means. Thus in the proportion a:b = c:d, a and d are the ex- tremes, and b and c the means. 307. In a proportion in which the means are equal, either mean is called a Mean Proportional between the first and last terms, and the last term is called a Third Proportional to the first and second terms. A Fourth Proportional to three quantities is the fourth term of a proportion whose first three terms are the three quantities taken in their order. Thus in the proportion a:b = b:c^ b is a mean propor- tional between a and c, and c is a third proportional to a and b. In the proportion a : 6 = c : d, d is a fourth proportional to c:, 6, and c. 268 ALGEBRA. 308. A Continued Proportion is a series of equal ratios, in which each consequent is the same as the following ante- cedent; as, a : b =^ b : c = c : d = d : e. PROPERTIES OF PROPORTIONS. 309. In any proportion the product of the extremes is equal to the product of the means. Let the proportion be a : 5 = c : d. Then by Art. 305, ^ = -- •^ b d Clearing of fractions, ad = be. 310. A mean proportional between two quantities is equal to the square root of their product. Let the proportion he a : b = b : c. Then by Art. 309, b^=ac. Whence, b = Vac. 311. From the equation ad = bc, we obtain be J , ad a = — , and b= — d c That is, in any proportion either extreme is equal to the product of the means divided by the other extreme ; and either mean is equal to the product of the extremes divided by the other mean. 312. (Converseof Art. 309.) If the product of two quantities is equal to the product of two others^ one pair may be made the extremes, and the other the means ^ of a proportion. Let ad = be. ^. .,. , , , ad be a c D.v.d.ngbyM, ^ = ^'"'1=1 Whence, a:b = c:d. RATIO AND PROPORTION. 269 In a similar manner we ma}^ prove that : a : c = 6 : d, b : d = a : c, c:d = a:b, etc. 313. In any proportion the terms are in proportion by Alternation ; that is, the first term is to the third, as the second term is to the fourth. Let a : b = c : d. Then by Art. 309, ad = be. Whence by Art. 312, a:c = b:d, 314. In any proportion the terms are in proportion by Inversion ; that is, the second term is to the first, as the fourth term is to the third. Let a : b = c : d. Then, ad = be. Whence, b : a = d : c. 315. In any proportion the terms are in proportion by Composition ; that is, the sum of the first two terms is to the first term, as the sum of the last two terms is to the third term. Let a : b = c: d. Then, ad=:bc. Adding both members to ac, ac-{-ad = ac + be, or, a(c + d) = c(a-\-b). Whence (Art. 312), a-f6:a = c + d:c. Similarly we may prove that a + b:b = c-\-d'.d. 270 ALGEBRA. 316. In any proportion the terms are in proportion by Divis- ion ; that is. the difference of the first two terms is to the first term, as the difference of the last two terms is to the third term. Let a : b — c : d. Then, ad =bc. Subtracting both members from ac, ac — ad = ac — be, or, a{c — d) = c{a — b). Whence, a — b:a = c — d:c. Similarly, a — b : b = c — d : d. 317. /n any proportion the terms are in proportion by Composition and Division ; that is, the sum of the first two terms is to their difference, as the sum of the last two terms is to their difference. (1) (2) Let a:b — c:d. Then by Art. 315, a + b c-\-d a c And by Art. 316, a— b c—d a c Dividing (1) by (2), a-{-b c-{-d a—b c—d Whence, a + b : a— b — c-\-d: c d. 318. In a series of equal ratios, any antecedent is to its consequent, as the sum of all the antecedents is to the sum of all the consequents. Let a:b — c:d=e:f. Then by Art. 309, ad = be, and a/= be. Also, ab = ba. Adding, a(b-\-d -{•f) = b (a+c + e). Whence (Art. 312), a:6 = a + c + e:6 + d +/. RATIO AND PROPORTION. 271 319. In any number of proportions^ the products of the corresponding terms are in proportion. Let a:h = C'.d^ and e : /= g-.h. h d f h Multiplying these equals, b f d h' bf dh Whence, ae : bf= eg : dh. 320. In any proportion ^ like powers or like roots of the terms are in proportion. Let a:b = c:d. rm, a c Therefore, t- = t • ' 6* d^ Whence, a" : ft** = c" : d"". In a similar manner we may prove that :^a'.yb = yc: ^d. 321. In any proportion., if the first two terms are multiplied by any quantity., as also the last tivo, the resulting quantities will be in proportion. Let a:b = c:d. Then, Therefore, Whence, ma :mb = nc: nd. a — = b c = — . d ma mb _nc nd 272 ALGEBRA. In a similar manner we may prove that a b __c d m' m~ n' n Note. Either m or n may be unity ; that is, either couplet may be multiplied or divided without multiplying or dividing the otlier. 322. In any proportion, if the first and third terms arh multiplied by any quantity, as also the second and fourth terms, the resulting quantities will be in proportion. Let a : b = c : d. Then, Therefore, Whence, ma : n6 = mc : nd. In a similar manner we may prove that a b _ c d m' n m' n Note. Either m or n may be unity. 323. If three quantities are in continued proportion, the first is to the third as the square of the first is to the square of the second. a _ _ c ~d ma nb _mc Let a:b = b: c. Then, a b b" c Therefore, b c~b b Or, a^a^ c~b' Whence, a:c = a^'.W, RATIO AND PROPORTION. 273 324. If four quantities are in continued proportion^ the first is to the fourth as the cube of the first is to the cube of the second. Let a:b = b:c = c:d. Then, ^ — 5 — ^. bed Therefore, 5x^x^ = 2x?x^. c d b b Whence, a:d = a^: b^. Note. The ratio a^ : b^ is called the duplicate ratio, and the ratio a* : 6^ the triplicate ratio, of a : 6. PROBLEMS. 325. 1. Solve the equation, a;+l : X — 1 = a -\- b : a — b. By Art. 317, 2a;: 2 = 2a: 26. Whence by Art. 321, .t : 1 = a : 6. Therefore, x= -■> Ans. b 2. If x:y = (x-{-zy:(y-{-zy, prove that 2; is a mean proportional between x and y. From the given proportion, by Art. 309, y{x-\-zy = x{y-\-zy. Or, 0!^y-\-2 xyz -f- yz"^ = xy^-{-2 xyz + xz^. Or, 3?y — xy^ = xz^ — yz^. Dividing by a? — y, ^ = 2;^. Therefore 2; is a mean proportional between x and y. 274 ALGEBRA. 3. Find the first term of the proportion whose last three terms are 18, 6, and 27. 4. Find the second term of the proportion whose first, third, and fourth terms are 4, 20, and 55. 5. Find a fourth proportional to |, |^, and ^. 6. Find a third proportional to f and |. 7. Find a mean proportional between 8 and 18. 8. Find a mean proportional between 14 and 42. 9. Find a mean proportional between 2^ and -f^. Solve the following equations : 10. 2ic-5:3aj + 2 = a;-l : 7a; + l. 11. ar^-4:ar^-9 = a^-5a; + 6:aj2 + 4a;4-3. 12. x-i-Vl-x':x-VV^^ = a+^/¥^^':a-VW' 13. \^''^--— 14. ix: x:y= 3:5. -- ( x + y: x — y=a -{-b : a—b. 4=15:1/. ■ Xx'-^-f^a'b'ia' + b'). 16. Find two numbers in the ratio of 2|^ to 2, such that when each is diminished by 5, they shall be in the ratio of H to 1. 16. Divide 50 into two parts such that the greater in- creased by 3 shall be to the less diminished by 3, as 3 to 2. 17. Divide 12 into two parts such that their product shall be to the sum of their squares as 3 to 10. 18. Find two numbers in the ratio of 4 to 9, such that 12 is a mean proportional between them. 19. The sum of two numbers is to their difference as 10 to 3, and their product is 364. What are the numbers? 20. Ifa — 6:& — c = 6:c, prove that 6 is a mean propor- tional between a and c. RATIO AND PROPORTION. 276 21. If 5a + 46: 9a4-26 = 5 6 + 4c:96 4-2c, prove that 6 is a mean proportional between a and c. 22. If (a + 6+c + d) (a-6-c + d) = (a-64-c-d) (a + 6 — c — d) , prove that a : h = c : d. 23. If ax — by:cx—dy = ay — bz:cy — dz, prove that y is a mean proportional between x and z. 24. Find two numbers such that if 3 is added to each, they will be in the ratio of 4 to 3 ; and if 8 is subtracted from each, they will be in the ratio of 9 to 4. 25. There are two numbers whose product is 96, and the difference of their cubes is to the cube of their difference as 19 to 1. What are the numbers? 26. Divide $564 between A, B, and C, so that A's share may be to B*s in the ratio of 5 to 9, and B's share to C's in the ratio of 7 to 10. 27. A railway passenger observes that a train passes him, moving in the opposite direction, in 2 seconds ; whereas, if it had been moving in the same direction with him, it would have passed him in 30 seconds. Compare the rates of the two trains. 28. Each of two vessels contains a mixture of wine and water. A mixture, consisting of equal measures from the two vessels, contains as much wine as water ; and another mixture, consisting of four measures from the first vessel and one from the second, is composed of wine and water in the ratio of 2 to 3. Find the ratio of wine to water in each vessel. 29. Divide a into two parts such that the first increased by b shall be to the second diminished by 6, as a + 3 Z) is to a -3b, 276 ALGEBRA. XXIX. VARIATION. 326. Ooe quantity is said to vary directly as another when the ratio of any two values of the first is equal to the ratio of the corresponding values of the second. Note. It is customary to omit the word *' directly," and say simply that one quantity varies as another. 327. Suppose, for example, that a workman receives a fixed sum per day. The amount which he receives for m days will be to the amount which he receives for n days as m is to n ; that is, the ratio of any two amounts received is equal to the ratio of the corresponding numbers of days worked. Hence the amount which the workman receives varies as the number of days during which he works. 328. One quantity is said to vary inversely as another when the first varies directly as the reciprocal of the second. Thus, the time in which a railway train will traverse a fixed route varies inversely as the speed ; that is, if the speed is doubled^ the train will traverse its route in one-half the time. 329. One quantity is said to vary as two others jointly when it varies directly as their product. Thus, the wages of a workman varies jointly as the amount which he receives per day, and the number of days during which he works. 330. One quantity is said to vary directly as a second and inversely as a third, when it varies jointly as the second and the reciprocal of the third. Thus, in physics, the attraction of a bod}^ varies directly as the quantity of matter, and inversely as the square of the distance. VARIATION. , 277 331. The symbol oc is used to express variation ; thus, a ccb is read " a varies as 6." 332. If xocy, then x is equal to y multiplied by a constant quantity. Let a;' and ?/' denote a fixed pair of corresponding values of X and y, and x and y any other pair. Then from the definition of Art. 326, X y x' - = -,' or x = -y. x' y' y' x' Denoting the constant ratio — by m, we have x = my. 333. It follows from Arts. 328, 329, 330, and 332 that: 1. If X varies inversely as y, x= — 2. If X varies jointly as y and z, x = myz. Z. If X varies directly as y and inversely as z, x = — ^« z 334. Problems in variation are readily solved by convert- ing the variation into an equation by aid of Arts. 332 or 333. EXAMPLES. 335. 1. If a; varies inversely as y, and is equal to 9 when 2/ = 8, what is the value of x when y—lS? If X varies inversely as y, we have by Art. 333, m x = — y Putting a; = 9 and y = 8, we obtain 9 = -, orm=72. 8 72 Whence, x = — y 72 Hence, if ?/ = 18, we have x = — = 4, Ans. 18 278 ALGEBRA. 2. Given that the area of a triangle varies jointly as its base and altitude, what will be the base of a triangle whose altitude is 12, equivalent to the sum of two triangles whose bases are 10 and 6, and altitudes 3 and 9, respectively? Let B, H, and A denote the base, altitude, and area, respectively, of any triangle, and B' the base of the required triangle. Then since A varies jointly as B and H, we have A = mBH{Avt. 333). Therefore the area of the first triangle is m X 10 X 3, or 30 m, and the area of the second is m x 6 x 9, or 54 ?n. Hence the area of the required triangle is 30m + 54m, or 84m. But the area of the required triangle is also m X B'x 12. Therefore, 12mJB'= 84 m. Whence, B' = 7, Ans, 3. If yccx, and is equal to 36 when a? = 4, what is its value when x—7? 4. If yocz^^ and is equal to 15 when 2=3, what is the value of y in terms of z^? 5. If X varies inversely as y^ and is equal to 4 when 2/= 2, what is the value of y when x = ^? 6. If z varies jointly as x and 2/, and is equal to 90 when x = S and 2/ = 6, what is the value of z when x = 2 and y =7? 7. If X varies directly as y and inversely as z, and is equal to 4 when y=2 and 2; = 3, what is the value of x when y = Sd and 2; = 15? 8. If2a; — 3oc32/-f7, and x = 3 when y=l, what is the value of x when ?/ = — 1 ? 9. If a^oc2/^ and a; =6 when y = S, what is the value of y when a; = 2 ? VARIATION. 279 10. Two quantities vary directly and inversely as x, re- spectively. If their sum is equal to 7 when a; = 2, and to — 13 when a;= — 3, what are the quantities? 11 . Given that the volume of a pyramid varies jointly as its base and altitude, what will be the altitude of a pyramid whose base is 12, equivalent to the sum of two pyramids ' whose bases are 5 and 8, and altitudes 12 and 6, respectively? 12. If the illumination from a source of light varies in- versely as the square of the distance, how much farther from a candle must a book, which is now 3 inches off, be removed so as to receive just half as much light? 13. Two circular plates of gold, each an inch thick, the diameters of which are 6 and 8 inches, respectively, are melted and formed into a single circular plate one inch thick. Find its diameter, having given that the area of a circle varies as the square of its diameter. 14. Three spheres of lead whose diameters are 3,4, and 5 inches, respectively, are melted and formed into a single sphere. Find its diameter, having given that the volume of a sphere varies as the cube of its diameter. 15. If 5 men in 6 weeks earn $57, how many weeks will it take 4 men to earn $ 76 ; it being given that the amount earned varies jointly as the number of men, and the number of weeks during which they work. 16. If the volume of a cylinder of revolution varies jointly as its altitude and the square of its radius, what will be the radius of a cylinder, whose altitude is 18, equivalent to the sum of two cylinders whose altitudes are 5 and 12, and radii 6 and 9, respectively? 17. Given that y is equal to the sum of two quantities, of which one is constant and the other varies as xy. If y is equal to 1 when a; = — 2, and to — J when a? = 2, what is the expression for y in terms of x ? 280 ALGEBRA. XXX. ARITHMETICAL PROGRESSION. 336. An Arithmetical Progression is a series of terms, each of which is derived from the preceding by adding a constant quantity called the common difference. Thus, 1, 3, 5, 7, 9, 11, ... is an increasing arithmetical progression, in which the common difference is 2. Again, 12, 9, 6, 3, 0, —3, ... is a decreasing arithmetical progression, in which the common difference is — 3. 337. Given the first term, a, the common difference, d, and the number of terms, n, to find the last term, I. The progression is a, a-\-d, a-{-2d, a + 3d, ••• It will be observed that the coefficient of d in any term is one less than the number of the term. Hence, in the nth, or last term, the coefficient of d will he n—1. That is, l=a + {n-l)d. (I.) 338. Given the first term, a, the last term, I, and the num- ber of terms, n, to find the sum of the series, jS. S=a-j-(a-\-d) + {a-\-2d)-\ \-(l-d)-^L Writing the series in reverse order, jS=1 +(l -d)-^(l -2d)-\-'--+(a-\-d.) + a. Adding these equations, term by term, 2S = (a+l) + (a+l) + (a+O + ••• + (a+0 + («+0 = n(a-}- 1). Therefore, S = -(a + i). (II.) 339. Substituting in (II.) the value of I from (I.) , we have >S = |[2a-f-(r^-l)c^]. ARITHMETICAL PROGRESSION. 281^ EXAMPLES. 340. 1. In the series 8, 5, 2, — 1, —4, ... to 27 terms, find the last term and the sum. In this case, a = 8, d = — 3, 7i = — 27. Substituting in (I.) and (II.), Z = 8 + (27-1) (-3) = 8 -78 = - 70. ^ = ?I(8-70) = 27 X (-31) =-837. Note. The common difference may be found by subtracting the first term from the second. Thus, in the series 5 1 o 1 ; 1 5 11 -, , —2, •••, we have a = = 3 6 6 3 6 In each of the following, find the last term and the sum of the series : 2. 1, 6, 11, ... to 15 terms. 3. 7, 3, -1, ... to 20 terms. 4. -9, -6, -3, ... to 23 terms. 5. -5, -10, -15, ... to 29 terms. 6. -, -, , ... to 35 terms. 4 2 4 7. ?, — , ... to 19 terms. 5 15 8. |, 5, 5 ... to 16 terms, 3 4 6 9. i -^, ... to 22 terms. 10. — 3, — -, ... to 17 term 11. 2 1 , , ... to 14 terms. 5' 3' 282 ALGEBRA. 341. If any three of the five elements of an arithmetical progression are given, the other two may be found by sub- stituting the given values in the fundamental formulae (I.) and (II.)? a^^ solving the resulting equations. 1. Given a = ,n=20,S = ; find d and Z. Substituting the given values in (I.) and (II.) , we have / = -2 + 19d. (1) 3 \ 3 J 6 3 ' ^ From (2), Z = ^-l = l ^ ^ 3 6 2 Substituting in (1), 2 3 ' 6 Arts. d= -. 1 = — 6 2 2. Given d = -3, ? = -39, aS = - 264; find a and w. Substituting in (I.) and (II.), _39 = a + (^-l)(-3), or a = 3n-42. (1) -264 = -(a-39), or an - 39n = - 528. (2) z Substituting the value of a from (1) in (2), 3^2_ 4271 -39n = - 528, or, w2-27n = -176. Whence, n = ^Z.±^^M=I1 = ?7±^ = 16 or 11. 2 2 Substituting in (1), a =48 -42, or 33-42 = 6 or -9. Ans, a= 6, n = 16 ; or, a = — 9,n= 11. ARITHMETICAL PROGRESSION. 283 Note. The interpretation of the two answers is as follows : If a = 6, and n = 16, the series is 0, 3, 0, - 3, - 6, - 9, - 12, - 15, - 18, ^ 21, - 24, - 27, -30, _33, _36, -39. If a = — 9, and n = 11, the series is _ 9, _ 12, _ 15, - 18, - 21, - 24, - 27, - 30, - 33, - 36, - 39. In each of these the last term is — 39, and the sum is — 264. 113 3. Givena = -, d = -, S = — -; find Z and w. o 1 ^ ^ Substituting in (I.) and (II.) i 3 ^ 'V 12/ 12 -| = ^(^ + '), or « + 3;» = -9. (2) Substituting the value of I from (1) in (2), 5n-n^ ^_g ^^ n2.-.9?i=36. 4 Solving this equation, ?i = 12 or — 3. The second value is inapplicable, for the number of terms in a progression must be a positive integer. Substituting the value n= 12 in (1), 5-12 7 (1) 12 12 Ans. Z = -, 71 = 12. 1 ^ Note. A negative or fractional value of n is inapplicable, and should be rejected together with all other values dependent upon it. EXAMPLES. 4. Given c? = 4, ? = 75, ?i = 19 ; find a and S. 5. Given d = — l,n = 15, S = ; find a and I. 284 ALGEBRA. "^ 2 6. Given a = , w = 18, / = 5 ; find d and 8. o 7. Given a = , n=l^ JS = —7 ; find d and I. 4 8. Givena = |, Z = -^, .^ = -?|l; finddandTi. 9. Given Z = -31, n= 13, ^ = - 169 ; find a and c«. 10. Given d = -^, S = - 328, a = 2 ; find Z and n. 11. Given a = 3, Z = 42| , cZ = 2^ ; find n and S. 12. Given d = — 4, rj = 17, /S' = — 493 ; find a and /. 13. Given Z = ^, d = i, >S = 20 ; find a and n. 14. Given Z = ^, n = 21, /S^ — ; find a and d. 15. Given a = -l, Z = --, ;S = - — ; find d and w. o o 3 16. Given a = - ?, n = 15, /S = 120 ; find d and l. 4 17. Given Z = -47, d = -l, ^^ = -1118; find a and n. 18. Given a = 6, d = - -, .S = - — ; find n and I. 3 3 From (I.) and (II.) general formulae for the solution of cases like the above may be readily derived. 19. Given a, d, and S ; derive the formula for n. Substituting the value of I from (I.) in (II.), 2/S = w[2a-h(n-l)cZ], or dn" ■^{'^a- d)n = 2 S. This is a quadratic in ti, and may be solved by the method of Art. 265. ARITHMETICAL PROGRESSION. 285 Multiplying by 4d, and adding (2 a — dy to both members, 4 dV -{-4:d(2a-d)n + {2a-dy = 8dS-^{2a^ dy. Extracting the square root, 2d7i + 2a-d=±-V8dS + {2a-dy. Whence, n 20. Given a, I, and n 21. Given a, ?i, and S 22. Given d, «, and S 23. Given a, c?, and / 24. Given d, /, and n 25. Given Z, 7i, and >S' 26. Given a, d, and /S" 27. Given a, Z, and >iS' 28. Given d, Z, and >S' - d-2a±-V8dS -\- {2a-dy ^ 2d derive the formula for d. derive the formulae for d and I. derive the formulae for a and I. derive the formulae for n and S. derive the formulae for a and S. derive the formulae for a and d. derive the formula for I. derive the formulae for d and n. derive the formulae for a and n. 342. To insert any number of arithmetical means between two given terms. For example, let it be required to insert 5 arithmetical means between 3 and — 5. This signifies that we are to find an arithmetical progres- sion of 7 terms, whose first term is 3, and last term — 5. Substituting a = 3, 7 = — 5, and ?i = 7 in (I.), we have _5 = 3 + 6d, or (i = -^- Hence the required series is o 5 1 . '33 7 II 3' "T -5. 28(3 ALGEBRA. 343. Let X denote the arithmetical mean between a and 6. Then, by the nature of the progression, jc — a = 6— ic, or2ic=a-f-6. Whence, aj = :^L±i. 2 That is, tlie arithmetical mean between two quantities is equal to one-half their sum. EXAMPLES. 344. 1. Insert 5 arithmetical means between 2 and 4. 2. Insert 7 arithmetical means between 3 and — 1. 3. Insert 4 arithmetical means between — 1 and — 7. 4. Insert 6 arithmetical means between — 8 and — 4. 1 13 5. Insert 8 arithmetical means between - and 2 10 Find the arithmetical mean between : 6. 24 and - 14. ^ , , ^ . 7. (a + &)'and-(a-6)2. * a-b^^ a-^b PROBLEMS. 345. 1. The sixth term of an arithmetical progression is 5 16 -, and the fifteenth term is — Find the first term. 6 3 By Art. 337, the sixth term is a + 6 rf, and the fifteenth term is a+ 14 c?; hence, r«+ 6rf=|. (1) \a^Ud==^. (2) Subtracting (1) from (2) , 9d = -, oTd = -' 2 ^ 1 /» jf Substituting in (2), a + 7 = — ; whence, a = — -, Ans. 3 3 / ARITHMETICAL PROGRESSION. 287 2. Find four quantities in arithmetical progression such that the product of the extremes shall be 45, and the product of the means 77. Let the quantities he x — Sy, x — y, x + y, and x + Sy. Then, by the conditions, rx2-9y2 = 45. \x^~ y^=ll. Solving these equations, a: = ± 9 and y = ±2. Therefore the quantities are 3, 7, 11, and 15; or, — 3, — 7, — 11, and -15. Note. In problems like the above it is convenient to represent the unknown quantities by symmetrical expressions. Thus if five quanti- ties had been required, we should have represented thera byx — 2y, X — y, x, ar + y, and x + 2 y. 3. Find the sum of the odd numbers from 1 to 100. 4. The seventh term of an ai-ithmetical progression is 27, and the thirteenth term is •— 3. Find the twenty -first tenn. 5. Find four numbers in arithmetical progression such that the sum of the first and third shall be 22, and the sum of the second and fourth 36. 6. A person saves $270 the first year, $245 the second, and so on. In how many years will a person who saves every year $145 have saved as much as he? 7. In the progression m, 2m — Sn, 3m — 6 n, ... to 10 terms, find the last term and the sum of the series. 8. The seventh term of an arithmetical progression is 5 a + 4 6, and the nineteenth term is 9a— 2b. Find the fifteenth term. 9. Find the sum of the even numbers beginning with 2 and ending with 500. 10. The sum of the squares of the extremes of four num- bers in arithmetical progression is 200, and the sum of the squares of the means is 136. What are the numbers? 288 ALGEBRA. 11. The seventh term of an arithmetical progression is 1 3 9 , the thirteenth term is -, and the last term is — Find 2 2 2 the number of terms. 12. Find five quantities in arithmetical progression such that the sum of the first, third, and fourth is 3, and the product of the second and fifth is — 8. 13. Two persons start together. One travels 10 leagues a day ; the other 8 leagues the first day, which he augments daily by half a league. After how many days, and at what distance from the point of departure, will they come together? 14. A body falls IQj^ ^^^^ ^^^^ ^^^* second, and in each succeeding second 32|^ feet more than in the next preceding one. How far will it fall in 16 seconds? 15. Find three quantities in arithmetical progression such that the sum of the squares of the first and third exceeds the second b}- 123, and the second exceeds one-third the first by 6. 16. After A had travelled 2| hours at the rate of 4 miles an hour, B set out to overtake him, and went 4|- miles the first hour, 4| the second, 5 the third, and so on, increasing his speed a quarter of a mile every hour. In how many hours would he overtake A? 17. If a person should save $100 a year, and put this sum at simple interest at 5 per cent at the end of each year, to how much would his property amount at the end of 20 years ? 18. The digits of a number of three figures are in arith- metical progression ; the first digit exceeds the sum of the second and third by 1 ; and if 594 be subtracted from the number, the digits will be inverted. Find the number. GEOMETRICAL PROGRESSION. 289 XXXI. GEOMETRICAL PROGRESSION. 346. A Geometrical Progression is a series of terms, each of which is derived from the preceding by multiplying by a constant quantit}' called the ratio. Thus, 2, 6, 18, 54, 162, ... is an increasing geometrical progression in which the ratio is 3. Again, 9, 3,1,-, -, ... is a decreasing geometrical pro- gression in which the ratio is — Negative values of the ratio are also admissible ; thus, — 3, 6, —12, 24, —48, ••• is a geometrical progression in which the ratio is — 2. 347. Given the first term, a, the ratio, r, and the number of terms, n, to find the last term, I. The progression is a, ar, ai^, ar^, ... It will be observed that the exponent of r in any term is one less than the number of the term. Hence, in the ?ith or last term, the exponent of r will be n — 1 . That is, l = ar''~\ (I.) 348. Given the first term, a, the last term, I, and the ratio, r, to find the sum of the series, S. /S = a-\-ar +ar^-\ f- ar""'^ + ar""-^ -f- ar""'^. Multiplying each term by r, rS = ar -\- a')'^ -\- ar^ -\ -f- ar^~^ + air^'^ -f ar*". Subtracting the first equation from the second, rS-S = ar^^-a', or, S = ^''" ~ ^ . ?• — 1 But from (I.), Art. 347, rl^ar"". Hence, S = rLz^. (II.) r — 1 290 ALGEBRA. EXAMPLES. 349. 1. In the series 3, 1, -, ... to 7 terms, find the last term and the sum. In this case, a = 3, r = -, n = 7. Substituting in (1.) and (11.), ^ [sj 3' 24 243 ix— -3 —-3 21^6 y_ 3 243 _ 729 _ 729 ^ 1093_ 1 _ 1 _ 2 _ 2 2^^ ' 3 3 3 Note. The ratio may be found by dividing the second term by the first. 2. In the series — 2, 6, — 18, 54, ... to 8 terms, find the last term and the sum. In this case, a = — 2, r = = — 3, n = 8. Hence, — z l = -2{- Sy= - 2 X ( - 2187) = 4374. ^^ -3 X 4374 -(- 2) ^ -13122 + 2 ^3^,gQ^ ^ — 3 — 1 —4 In each of the following, find the last term and the sum of the series : 3. 1, 2, 4, ... to 9 terms. 4. 3, 2, -, ... to 7 terms. o 6. -2, 8, -32, ... to 6 terms. 6. 2, -1, i, ... to 10 terms. 7. ^^ Ti Q» ••• to 11 terms. >& 4 o GEOMETRICAL PROGRESSION. 291 8. ? -1, -, ••• to 8 terms. 3 2 9. 8, 4, 2, ••• to 9 terms. 10. -, , — , ••• to 6 terms. 4 4 12 11. 3, -6, 12, ... to 7 terms. 12. , , , ••. to 10 terms. 3 3 6 350. If any three of the five elements of a geometrical progression are given, the other two may be found by sub- stituting the given vahies in the fundamental formulae (I.) and (II.)? and solving the resulting equations. But in certain cases the operation involves the solution of an equation of a degree higher than the second ; and in others the unknown quantity appears as an exponent, the solution of which form of equation can usually only be effected by aid of logarithms (Art. 427). In all such examples in the present chapter, the equations may be solved by inspection. 1. Given a = — 2, ?i = 5, / = - 32 ; find r and S. Substituting the given values in (I.), we have — 32 = — 2r* ; whence, r* = 16, or r = ± 2. Substituting in (II.), Ifr= 2, ^ = ^r-^2)-^(-^ =-64 + 2 = -62. Ifr = -2, ^= (-^)^-^^>-(-^) = 64 + 2^_ -2-1 -3 Ans. r=2, S = -62; or, r = -2, >S' = -22. Note. The interpretation of the two answers is as follows : If r = 2, the series is — 2, — 4, — 8, — 16, - 32, in which the sum is — (32. If r=- 2, the series is -2, 4,-8, 16, - 32, in which the sum is - 22, 292 ALGEBRA. 2. Given a = 3, r = , S = ; find w and L 3 729 Substituting in (II.)? 1640 3 Z4-9 -1/-3 729 _i_i 4 Whence, Z-f.9=^^; or, Z = —- 729 729 Substituting in (I.)> = 3-^ ; or, 729 V 37 V 37 2187 Whence, by inspection, n — 1 = 7, or n = 8. EXAMPLES. 3. Given r= 2, n= 10, Z= 256 ; find a and S. no 4. Given r = — 2, n = 6, /S'= — ; find a and l. 5. Given a = 2, n = 7, Z = 1458 ; find r and S. 6. Given a = 1, r = 3, Z = 81 ; find n and S, 7. Given»' = i,n = 8,^ = ^^; findaandL 3 6561 8. Given a = 3, ti = 6, I — ; find r and S. 1024 1 127 9. Given a=2,l =— , >S' = ^^^ ; find n and r. 32 32 10. Given a = -, r = - 3, aS = - 91 ; find n and /. id 11. Given Z=:--128, r=2, >S = -255; find a and ti. GEOMETRICAL PROGRESSION. 293 From (I.) and (II.) general formulae may be derived for the solution of cases like the above. 12. Given a, r, and S 13. Given a, I, and S 14. Given r, I, and S 15. Given ?•, n, and I 16. Given r, ?i, and /S 17. Given a, w, and I derive the formula for I. derive the formula for r. derive the formula for a. derive the formulae for a and S. derive the formulae for a and I. derive the formulae for r and S. Note. If the given elements are n, I, and S, equations for a and r may be found, but there are no definite formuloe. for their values. The same is the case when the given elements are a, n, and aS^. The general formulae for n involve logarithms; these cases are dis- cussed in Art. 427. 351. The limit (Art. 297) to which the sum of the terms of a decreasing geometrical progression approaches, as the number of terms increases indefinitely, is called the sum of the series to infinity. The value of S in formula (II.), Art. 348, may be written a — rl /S = 1 In a decreasing geometrical progression, the greater the number of terms taken, the smaller will be the value of the last term. Hence as the number of terms increases indefinitely, the term rl approaches the limit 0. \ ct — — rl n Therefore the fraction approaches the limit 1 — r 1 —r That is, the sum of a decreasing geometrical progression to infinity is given by the formula a ^=137- ■ (I"-) 294 ALGEBRA. EXAMPLES. 1. Find the sum of the series 4, —-,—,... to infiuit3^ Q In this case, a = 4, r = — -• 3 4 12 Substituting in (III.)? ^ = r = — ' ^^'• 1+- ^ 3 Find the sum of the following to infinity : 2. 2, 1,-, ... ' ' 2 6. 3 1 1 4'2'3'- 3. 4, -2, 1, .., 7. o 3 3 ^' 10' 100' - 4. -1 1 -^ ''3' 9' ... 8. 5 oO 5. o 3 -3, -^, - 3 25'-* 9. ' ^' ^^' - 352. T(9 Jind the value of a repeating decimal. This is a case of finding the sum of a geometrical pro- gression to infinity, and may be solved by the formula of Art. 351. 1. Find the value of .85151 ... .85151 ... = .8+.051 + .0005l4---- The terms after the first constitute a decreasing geometrical progression in which a = .051, and r = .01. Substituting in (III.), ^ ^ _:051_ ^^051 ^ _5]^ ^ J7 l-.Ol .99 990 330* Hence the value of the given decimal is ^ J^_281 . 10 "^330" 330' GEOMETRICAL PROGRESSION. 295 EXAMPLES. Find the values of the following : 2. .7272... 4. .7333... 6. .110303-.. 3. .407407... 5. .52121... 7. .215454... 353. To insert any number of geometrical means between two given terms. Example. Insert 4 geometrical means between 2 and This signifies that we are to find a geometrical progression of 6 terms, whose first term is 2, and last term 243 OA Substituting a = 2, ri = 6, and I = ^ — in (I.) , we have 64 „ , , ,32 , 2 — — = 2r \ whence, r = ■» and r = -. 243 '243 3 Hence the required series is « 4 8 16 32 64 z, -1 -? — 1 — t . 3 9 27 81 243 354. Let X denote the geometrical mean between a and b. Then, by the nature of the progression, X b 2 7- - = -, oYxr — ab. a X Whence, x = Va6. That is, the geometrical mean bettveen two quantities is equal to the square root of their product. EXAMPLES. 1 28 355. 1. Insert 6 geometrical means between 3 and — -• 729 2. Insert 5 geometrical means between - and 364|-. 296 ALGEBRA. 3. Insert 6 geometrical means between — 2 and — 4374. 3 3 4. Insert 7 geometrical means between - and 2 512 5. Insert 5 geometrical means between — 2 and — 128. 799 6. Insert 4 geometrical means between 3 and — ^ 1024 Find the geometrical mean between : 7. 11| and 2f 8. 4a.-2+12a;2/ + 92/^ and4a.'2-12a;?/ + 9/. Q a^ — ah -, 9. and ah + 6^ ah- h^ PROBLEMS. 356. 1. Find three numbers in geometrical progression, such that their sum shall be 14, and the sum of their squares 84. Let the quantities be a, ar, and ar^ ; then, by the conditions, f a-^ar-\- ar^ = 14. (1) '^ \ a2 + a2r2 + aV = 84. (2) Dividing (2) by (1), a-~ar-\- ar^ = 6. (3) Subtracting (3) from (1), 2ar = S, or r = -- (4) a Substituting in (1), a + 4 + — rz: 14. a Or, a2 _ 10 a = - 16. Solving this equation, a = 8 or 2. Substituting in (4), r = - or - = ^ or 2. 8 2 2 Therefore, the numbers are 2, 4, and 8. 2. The fifth term of a geometrical progression is 48, and the eighth term is — 384. Find the first term. 3. The sum of the first and second of four quantities in geometrical progression is 15, and the sum of the third and fourth is 60. What are the quantities? GEOMETRICAL PROGRESSION. 297 4. Find three quantities in geometrical progression, such that the sum of the first and second is 20, and the third exceeds the second by 30. 5. The fourth term of a geometrical progression Is — 108, and the eighth term is — 8748. Find the first term. 6. A person who saved ever}' year half as much again as he saved the previous year, had in seven years saved $2059. How much did he save the first year? 7. The elastic power of a ball, which falls from a height of a hundred feet, causes it to rise to 0.9375 of the height from which it fell, and to continue in this way diminishing the height to which it will rise, in geometrical progression, until it comes to rest. How far will it have moved? 8. The sum of four quantities in geometrical progression is 30, and the quotient of the fourth quantity divided by the 4 sum of the second and third is — Find the quantities. 9. The third term of a geometrical progression is — , and the sixth term is Find the eighth term. 512 10. Divide the number 39 into three parts in geometrical progression y such that the thii'd part shall exceed the first by 24. 11. The product of three numbers in geometrical progres- sion is 64, and the sum of the squares of the first and third is 68. What are the numbers? 12. The product of three quantities in geometrical pro- gression is 8, and the sum of their cubes is 73. What are the quantities? 298 ALGEBRA. XXXII. HARMONICAL PROGRESSION. 357. Quantities are said to be in Harmonical Progression when their reciprocals form an arithmetical progression. Thus, 1, -, -, -, -, ... are in harmonical proojression, ''3579 ^ "^ because their reciprocals 1, 3, 5, 7, 9, ... form an arithmetical progression. 358. Any problem in harmonical progression, which is susceptible of solution, may be solved by taking the recipro- cals of the terms and applying the formulae of the arithmet- ical progression. There will be found, however, no general formula for the sum of the terms of a harmonical series. 359. Let X denote the harmonical mean between a and b. Then, by the nature of the progression, - is the arithmet- 11 ^ ical mean between - and — a b Whence, 1 =. ^_^ (Art. 343) = ^+^. X 2 ^ ^ 2ab Therefore, x = a + b 360. If any three consecutive terms of a harmonical series are taken, the first is to the third as the first minus the second is to the second minus the third. Let the terms be a, 6, and c. Then since -, -, and - are in arithmetical progression, a b c c b b a b — c a — b or, -— - = — — . be ab HARMONICAL PROGRESSION. 299 Multiplying both members by , we have a _ a — b c b — c EXAMPLES. 2 2 361. 1. In the series 2, -, -, ».. to 36 terms, find the last term. Taking the reciprocals of the terms, we have the arith- metical progression 1 3 5 2' 2' 2' '" In this case a = -, cZ= 1, n = 36. Substituting in (I.)» -^.rt. 337, we have ; = | + (36-l)xl = ^. 2 Taking the reciprocal of this, we obtain -^ as the last term of the given harmonical series. 2. Insert 5 harmonical means between 2 and — 3. Taking the reciprocals of the terms, we have to insert 5 arithmetical means between - and 2 3 Substituting a = -, Z = , and n = 7, in (I.) , Art. 337, 2 o we have — _ = -4-6a; or, a = 3 2 36 Then the arithmetical series is 1 15 ? Jl _i_ _-I _i. 2' 36' 9' 12' 18' 36' 3* Therefore the required harmonical series is 2, ^A ?, 12, -18, -55, -3. 13 2 ' 7 300 ALGEBRA. Find the last terms of the following : 3. -,,^ — , -, ... to 11 terms. 4 11 6 4. -, --, --,... to 17 terms. 5- -5 -, — ? •.. to 23 terms. 2 3 8 6. , , , ... to 26 terms. 3 2 7 7. —-,——-, , ... to 31 terms. 7 23 16 2 3 8. Insert 7 harmonical means between - and — 5 10 9. Insert 4 harmonical means between — 2 and — 8. 10. Insert 6 harmonical means between 3 and — 1. Find the harmonical mean between : 11. 3 and - 5. 12. ?^+^ and ^!^. a — b a + b 13. Find the last term of the harmonical series a, 6, ... to n terms. 14. If m harmonical means are inserted between a and by what is the second mean ? 15. The fourth and ninth terms of a harmonical progres- 3 1 sion are and , respectively. What is the seventh term ? 16. Prove that the geometrical mean between two quanti- ties is a mean proportional between their arithmetical and harmonical means. THE BINOMIAL THEOREM. 301 XXXIII. THE BINOMIAL THEOREM. POSITIVE INTEGRAL EXPONENT. 362. The Binomial Theorem is a formula by means of which any power of a binomial may be expanded into a series. Examples of its application have been given in Art. 196. Proof of the Theorem for a Positive Integral Exponent. 363. If we assume the laws of Art. 196 to hold for the expansion of (a + x)""-, where n is any positive integer : The exponent of a in the first term is n, and decreases by 1 in each succeeding term. The exponent of x in the second term is 1 , and increases by 1 in each succeeding term. The coefficient of the first term is 1 ; of the second term, n ; multiplying n, the coefficient of the second term, by n — 1, the exponent of a in that term, and dividing the result by the exponent of x increased by 1 , or 2, we have ^^^ ~ — - as the coefficient of the third term ; and so on. Thus, (a-\-xy = a^-{- na"-^a; _^ n(7t - 1) ^„_2^ » V I / ^ ^ 1.2 n(n-l)(n-2) ^._3^^ (1) 1.2.3 ^ ^ This result is called the Binomial Theorem. Multiplying both members by a + «, we have ^ ^ ^ ^ 1.2 , w(n — l)(n — 2) __2 Q , + ^ 1.2.3 -''"^+- + a"x + na"-'ar= + 'ii^Jziia'-^a;' + ... X * if 302 ALGEBRA. Collecting the terms which contain like powers of a and x, (a + g;)"+^ = a"+^ + {n + l)a*'x + r ^(^~ ^) +n a''-^^^ rn(n-l)(yi-2) n(n-l) ~| 2 , L 1.2.3 1.2 J = a"+i + (n + l)a»aj + nf ^^^^ + l1 a'*" V ^'— ^ + 1 |a--=^ar^ + ... 1-2 = a'^+i + (n H- l)a"a' + nf^^^lla'^-ia^ + n(n— 1) fn + 1' 1.2 L ^ = a'^+i + (n + l)a"cc + (^ + l)^ ^n-ia^ 1 . z (7i + l)n(n-l)^,_,^ 1.2.3 It will be observed that this result is in accordance with the laws of Art. 196. Hence, if the laws of Art. 196 hold for any power of a + a; whose exponent is a positive integer, they also hold for an exponent greater by 1. But in Art. 196, the laws were shown to hold for (a -\-xy, and hence they also hold for {a-\-x)^; and since they hold for (a + xy, they also hold for (a-\-xy; and so on. Therefore the laws hold when the exponent is any positive integer, and equation (1) is proved for any positive integral value of n. Note 1. The above method of proof is known as the Method of Induction. Note 2. In place of the denominators 1-2, 1 • 2 • 3, etc., it is cus- tomary to write [2, [3, etc. The symbol \n, read "factorial n," signifies the product of the natural numbers from 1 to n inclusive. THE BINOMIAL THEOREM. 303 I 364. Putting a= 1 in equation (1), Art. 363, we obtain EXAMPLES. 365. Note. The Notes on page 164 apply with equal force to the examples in the present chapter. If the second term of the binomial is negative, it is convenient to enclose it, negative sign and all, in a parenthesis, before applying the laws of Art, 196. In reducing after- wards, care must be taken to apply the principles of Art. 192. 1. Expand (m"^ — V^)*- (m~^ - -^ny = [(m"^) + ( - w^)]*. The exponent of (m~^) in the first term is 5, and decreases by 1 in each succeeding term. The exponent of (— n^) in the second term is 1, and increases by 1 in each succeeding term. The coefficient of the first term is 1 ; of the second term, 5 ; multiplying 5, the coefficient of the second term, by 4, the exponent of {m~^) in that term, and dividing the result by the exponent of { — n^) increased by 1, or 2, we have 10 as the coefficient of the third term ; and so on. Hence, ■i-10(m~^y(-7i^y+5{m~^){-n^y-{-(-n^y — m~^— 5m~^n- + 10m~^7i — lOm'^n^ -{-6m~^n^—n^, Ans. Expand the following : 2. (ct + c^^)^ 4. f^-^'. 3. (m~^-ny. 5. {x"' + 2y^y. 304 ALGEBRA. 6. (a« + 3V^)'. 12. (^ + -1- 7. ,^^-V^Y. 13. -M 8. ^i^ + il-Y- ^^- («^* + 32/"*)^ .w^^y ?r 9. m^-^ . 15. a Vft 2 Va;\^ 2/ V2a;^ a^h^ 10. (aU-*-a-^6^)''. 16. (3 a-'\/^ - ^"^ ^«)*- 11. (Va^-3>y. 17. (yll+^yl'ij' Note. A trinomial may be raised to any power by the Binomial Theorem if two of its terms are enclosed in a parenthesis and regarded as a single term. (Compare Art. 195.) Expand the following : 18. (i-x-ay^y. 20. (l + 2aj-a^)*. 19. (a^ + aj_2)4. 21. (1-x + a^y, 366. To find the rth or general term in the expansion of (a-\-xy. The following laws will be observed to hold for any term in the expansion of {a-\- x)'\ in equation (1), Art. 363 : 1. The exponent of x is less b}^ 1 than the number of the term. 2. The exponent of a is 7i minus the exponent of x. 3. The last factor of the numerator is greater by 1 than the exponent of a. 4. The last factor of the denominator is the same as the exponent of x. Therefore, in the rth term, the exponent of x will be r — 1. The exponent of a, will be n — (r — 1) , or w — r + !• The last factor of the numerator will be 7i — r + 2. The last factor of the denominator will be r — 1. THE BINOMIAL THEOREM. 306 Hence, the rth term 1.2.3...(r-l) EXAMPLES. 3G7. 1. Find the eighth term of (3a^ - b'^y^. In this case, r = 8, and ti = 11 ; hence the eighth term ^11.10.9.8.7.6.5 K.,^,y = 330(81 a-){-b-'') = - 26730a26-% Ans. Note. If the second term of the binomial is negative, it should be enclosed, sign and all, in a parenthesis, before applying the formula. Find the 2. Seventh term of (a + a^)"; 3. Sixth term of (l + m)^«. 4. Eighth term of (c - dy\ 6. Fifth term of {l-ay\ 6. Seventh term of ( - + - ) • \b a) 7. Fifth term of {x--yjxy^. 8. Sixth term of fa" "^ ab]. 9. Eighth term of («-^+ 22/^) 1^ 10. Fourth term of (a' -3a;-^)". / 2 \^ 11. Ninth term of ( V^i + -: — ) • 306 ^ ALGEBRA. XXXIV. THE THEOREM OF UNDETER- MINED COEFFICIENTS. 368. A Series is a succession of terms so related that each may be derived from one or more of the others in accord- ance with some fixed law. The simpler forms of series have already been exhibited in the progressions. 369. A Finite Series is one having a finite number of terms. An Infinite Series is one the number of whose terms is unlimited. The progressions in general are examples of finite series ; but in Art. 351 we considered infinite geometrical series. 370. Infinite series may be developed by the process of Division, when the divisor is not exactly contained in the dividend. Let it be required, for example, to divide 1 by 1 — x. 1-x) 1 (l4-a; + aj2+ar^-f ... 1-x X X- -x' x' x'-a^ Therefore, = 1-f a; + a^ + ic^H 1 — x Infinite series may also be obtained by the process of Evo- lution (see Examples 20 to 23, page 169), and by other methods, one of the most important of which will be consid- ered in Art. 376. UNDETERMINED COEFFICIENTS. 307 371. A series is said to be convergent either when the sum of the first n terms approaches a certain fixed quantity as a limit (Art. 297), when n is indefinitely increased, or when the sura of all the terms is equal to a finite quantity. A series is said to be divergent when the sum of the first n terms can be made to numerically exceed any assigned quan- tity, however great, by taking n sufficiently great. 372. Consider, for example, the infinite series I. Suppose X = iCi, where Xi is positive and < 1. The sum of the first n terms is now 1 -f- a.'i + a:,2 + 0^1^ H- . . . + xr' = V^^ (^'•*- ^^) 1 — Xi As n increases indefinitely, a^i* decreases indefinitely, and approaches the limit 0. 1 /- n J^ Therefore the fraction ^ approaches the limit I —Xi 1 — JCl That is, the sum of the first n terms approaches a certain fixed quantity as a limit, when n is indefinitely increased. Hence the series is convergent when x is positive and < 1. II. Suppose a; =1. In this case, each term of the series is equal to 1, and the sum of the first n terms is equal to n ; and this sum can be made to numerically exceed any assigned quantity however great, by taking n sufficiently great. Hence the series is divergent when x=l. III. Suppose ic > 1 . In this case, each term of the series after the first is >1, and the sum of the first w terms is > n ; and this sum can be made to numericall}' exceed any assigned quantity however great, by taking n sufficiently great. Hence the series is divergent when a; is > 1. 308 ALGEBRA. 373. If an infinite series is convergent, the greater the number of terms taken, the more nearly does their sum ap- proach to the value of the expression which produced the series ; but if it is diveYgent, the sum diverges more and more from the value of the expression. Consider, for example, the equation (Art. 370), — J_=l ^x + X^ + Otf-i-". 1 —X Putting a;=.l, in which case the series is convergent (Art. 372), the equation becomes y=l + .l + .01 + .001 + - In this case, however great the number of terms taken, the sum can never be made exactly equal to — , but it approaches y this value as a limit. (See Art. 352.) Again, putting x= 10, in which case the series is diver- gent, the equation becomes - i = 1 + 10 -f 100 -}- 1000 -f .- y „^ In this case it is evident that the sum of the terms diverges more and more from the value 9 It follows from the above that an infinite series cannot be regarded as representing the value of the expression which produced it, unless it is convergent. 374. The infinite series a -\- bx -}- cx^ -\- da^ -\ is convergent when a; = ; for the sum of all the terms is equal to a when x = 0. THE THEOREM OF UNDETERMINED COEFFICIENTS. 375. An important method for expanding expressions into series is based on the following theorem, known as the Theorem of Undetermined Coefficients. UNDETERMINED COEFFICIENTS. 309 376. If the series A-\- Bx-[- Coi? -\- Dd^ -\ is always equal to the series A' + B'x + C'x^ -f D'x^ -\ , when x has any value which makes both series convergent, the coefficients of like powers of x in the two series will be equal; that is, A = A', B = B\ C=C\ etc. For since the equation A-it Bx + Cx" ^ Da? -i- '- = A^ -{- B'x + Ox" + D'^ff' + '" is satisfied when x has any vaUie which makes both series convergent, and since both members are convergent when x = (Art. 374), it follows that the equation is satisfied when ic= 0. Putting a; = 0, we have A = A'. Subtracting A from the first member of the equation, and its equal A' from the second member, we obtain jBa; + Ca^ 4- ^ar^ + ••• = B'x + C'a^ + i)'ar^ +••• Dividing through by x, B-\-Cx-^Dx^-\-'-' = B'-\- C'x + D'x'+'" This equation also is satisfied when x has any value which makes both members convergent ; and i^utting a; = 0, we have B = B', In like manner we may prove C = C', D = D', etc. Note. The reason for limiting the theorem to values of x which make both series convergent, is that a convergent series evidently cannot be equal to a divergent series; and two divergent series cannot be equal, because two expressions neither of which is finite cannot be said to be equal. 377. Since a finite series is always convergent, it follows from the preceding article that if two finite series A + Bx-\-Cx^-}- ... 4- -ffic" and A'-\- B'x-^C'x^-{ \-K'x\ are equal for every value of x, the coeflEiciepts of like powers of X in the two series are equal, 310 ALGEBRA. APPLICATION TO THE EXPANSION OF FRACTIONS INTO SERIES. 378. 1. Expand Sx" X' l-2x-hSx' in ascending powers of x. We have seen in Art. 370 that a fraction of the above form can be expanded into a series b^^ dividing the numerator by the denominator ; we therefore know that the proposed ex- pansion is possible. Assume then 2-3i»2 ^A + Bx+Cx'-^-Dx^-^- Ex' + (1) l-2a;-f-3a;2 where A, B, C, D^ E, ..., are quantities independent of x. Clearing of fractions, and collecting the terms in the second member involving like powers of x, we have 2-3x'-a^ = A-\- B -2A x+ C x^-h D x'^- E -2B -20 -2D + 3^ + 35 + 30 a;^+... (2) The second member of (1) must express the value of the fraction for every value of x which makes the series con- vergent (Art. 373). Hence equation (2) is satisfied when x has any value which makes both members convergent, and by the Theorem of Undetermined Coefficients, the coefficients of like powers of X in the two series must be equal ; that is, A= 2. B-2A= 0; whence, 5= 2 J. =4. 0-2J5+3^=-3; whence, 0= 25-3^- 3 = - 1. D- 2 + 35=-!; whence, D=20-3 5- 1 =- 15. jS;-2Z)+30= 0; whence, -E^= 2 i>-3 =-27; etc. UNDETERMmED COEFFICIENTS. 311 Substituting these values in (1), we have ^-^x'-x^ ^2 + 4.x-:x^-lbx^-21x' , Ans. \-2x+2>a? The result may be verified by division. Note. A vertical line, called a bar, is often used instead of a paren- thesis; thus, + 5 I X is equivalent to (^ — 2 A) x. -2a\ If the numerator and denominator contain only even pow- ers of a;, the expansion will involve only even powers of x ; in this case the operation may be abridged by assuming a series containing only the even powers of x. Thus, if the fraction were — — — — -, we should assume l-^x'+bx^ it equal io A-^Ba?-^ Cx* -\- Dx^ -\- Ex^ + • - In like manner, if the numerator contains only odd powers of ic, and the denominator only even powers, we should assume a series containing only the odd powers of x. If every tenn of the numerator contains x, we may assume a series commencing with the lowest power of x in the nu- merator. EXAMPLES. Expand each of the following to five terms, in ascending powers of x 2. l-x 1-^x 3. 2 + 5a: 1-Sx A S-4.a^ l+5»2 n 2x X — 9t? l+X + X^ 10. M X — o Xi ~~' Xj ^ ^ 1 - «2 12. 9 I ~ ^^ 2^3 3-2ar^ * \-^2x-Z;ff ' ^-Zx-^x^ 2-2>x^-\^ \-\-2x-hx? :^^-2:^ 2-x-x^ 3^_a;-2»» 3_a;2-}-a53 l-3ar^ 312 ALGEBRA. If the lowest power of x in the denominator is higher than the lowest power in the numerator, we may determine by actual division what power of x will occur in the first term of the expansion ; we should then assume the fraction equal to a series commencing with this power of a;, the exponents of X in the succeeding terms increasing by unity as before. 1 14. Expand ^x'-o? in ascending powers of x. X Dividing 1 by 3 ar, the quotient is '- — ; we then assume 1 Clearing of fractions, l = 3vl + 3J5 = Ax~'' + Bx-^^ C+Dx + Ex^-\- (1) a; 4-30 - B ar^ + 3Z> - C - D Equating the coefficients of like powers of », 3^=1 ^B-A=0 SC-B=0 3D-G=0 SE~-D=0: etc. Whence, A 4- =!■- 1 = 27' --k Substituting in (1), we have 1 =f- .- 1 9 27 ^i + 243 + 3a^- a^ 243 Ans. etc. Expand each of the following to five terms, in ascending powers of a;: 2 ,« l-2a^-ir» 16. 16. 3ar'-4a^ x-2af + 3i^' 17. 18. x^ -\-a^ — x^ 3-2a;4-a;^ 2a^ 2a;« UNDETERMINED COEFFICIENTS. 313 APPLICATION TO THE EXPANSION OF RADICALS INTO SERIES. 379. 1. Expand Vl —x in ascending powers of x. We have seen in Art. 204 that the square root of an imper- fect square can be expanded into a series by the process of Evolution ; we therefore know that the proposed expansion is possible. Assume then (1) VT^^ = A -^ Bx -\- Ca^ -{• Dx^ -\- Ex* -^ '" Squaring both members, we have by Art. 194, l-x = A' + 2AB x^ ^ x" X' + C + ^AC + 2 AD + 2AE + 2BC + 2BD X* ■■{-'" Equating the coefficients of like powers of a;, A^= 1; whence, ^=1. whence, B= — whence, C= — J_ 2A 2A 1 n J5C 1 whence, i>= = A 16 whence, E= C^±2BD 2A 128 2AB^-\ B^-j-2AC= 2AD-j-2BC= C'-h2AE-\-2BD= etc. Substituting these values in (1), we have rz ^ X X^ X^ 5x* . ~ 2 8 16 128 ' The result may be verified by the method of Art. 204. Note. The equation A^=\ gives A= ±\; and taking the negative value of A, we should find B=~f C= -> D = — , etc. 2 8 16 Thus another answer to the example is 814 . ALGEBRA. EXAMPLES. Expand each of the following to five terms, in ascending powers of x : 2. Vl+2a;. 4. Vl-2ajH-3a;^ 6. -Vl-x. 3. Vl -Sx. 5. Vl + x-af. 7. s/l+a;-har^. APPLICATION TO THE DECOMPOSITION OF RATIONAL FRACTIONS. 380. If the denominator of a fraction can be resolved into factors, each of the first degree in ic, and the numerator is of a lower degree than the denominator, the Theorem of Unde- termined Coeflflcients enables us to express the given fraction as the sum of two or more j)ortial fractions^ whose denomi- nators are factors of the given denominator, and whose numerators are independent of x. Case I. 381. When no two factors of the denominator are equal. 19ic -4- 1 1. Separate — into partial fractions. ^ (3a;-l)(5a; + 2) ^ Assume ' = , (1) (3a;- 1) (5a; + 2) 3a;-l 5a;4-2 ^^ where A and B are quantities independent of x. Clearing of fractions, we have 19a; -f 1 = ^ (5a; + 2) + -S (3a; - 1) = {5A + SB)x + 2A-B. (2) The second member of equation (1) must express the value of the given fraction for every value of x. Hence equation (2) is satisfied by every value of x, and by Art. 377 the coefficients of like powers of x in the two members are equal. UNDETERMINED COEFFICIENTS. 315 That is, 5^ + 3jB = 19, and 2A- B= 1. Solving these equations, we obtain A = 2 and B = S, Substituting in (1), we have 19- + 1 _ 2 , 3 ^„^_ (3.r-l)(5a;4-2) Sx-1 5x + 2 The result may be verified by adding the partial fractions. X -\- 4: 2. Separate -~ — — into partial fractions. ji X — io — 2/ The factors of 2 a; — a.*^ — af^ are a;, 1 — a;, and 2 -f- a; (Art. 283). Assume then 2x — a^ — a? X 1 — x 2-^x Clearing of fractions, we have X -\- 4: = A{l-x) {2 -^ x) + Bx (2 -I- a;) 4- Cx(\-x) . This equation, being satisfied b}^ every value of a;, is satis- fied when a; = 0. Putting a; = 0, we have 4 = 2^, or A = 2. Again, the equation is satisfied when a; = 1. Putting a: = 1, we have 5 = 3 .B, or B = -. o The equation is also satisfied when a; = — 2. Putting a; = — 2, we have 2 = — 6 (7, or C= o Substituting in (1), we obtain , 5 _1 a; + 4 2 3 3 ! = - _i I 2a; — a:^ — a:^ x 1 — x 2-fa; ^a"*"3(l-a;)~3(2+a;)' Note. The student should compare the above method of finding A and B with that used in Example 1. 816 . ALGEBRA. EXAMPLES. Separate the following into partial fractions : g 18 a; +10 g 2a;^- 17a;-24 7 ixo/ — .t-a^LO - ^ 2 iC^ — 20 3. Ux -25 Aa^ -25 A 4x + 15 Sx^ + bx H x"- -45 13 0^+10 6i r^_13ar-5a; aa.'-Ua2 a.-2 -3aic-4a2 7.T + 9 8. ''"-^'^ . 11. (a^-4)(a^-l) 4ic-14 2a^-18a; 9 + 9aj-4aj2 4a;2-20a; + 23 Case II. 382. Wlien all the factors of the deriominator are equal. x^ 11a; + 26 Example. Separate - — ^ — into partial fractions. (x — 3) If we attempt to solve the example by the method of Case I., we should assume a^-lla;+26_ A B C {x-Zy ic-3 aj-3 x-'6 That is, ^^-1^^ + ^^ = A±A±G, {x-2>y x-z But this is evidently impossible, for the given fraction cannot be reduced to an equivalent fraction having a; — 3 for a denominator, and a numerator independent of x. Let us now substitute in the given fraction 2/ + 3 in place of a; ; we then have (y + 3)^-ll(y + 3)+26 ^ y^-5y + 2 ^1 5 ^ 2 yi f y y^ f Replacing 2/ by a; — 3, the result takes the form I 5 2 x-2, {x-^y {x-^y UNDETERMINED COEFFICIENTS. 317 This shows that the given fraction can be expressed as the sum of three partial fractions, whose numerators are in- dependent of X, and whose denominators are the powers of x — 3 beginning with the first and ending with the third. A similar result will hold in any example under Case II. ; the number of partial fractions being equal to the number of equal factors in the denominator of the given fraction. EXAMPLES. 383. 1. Separate — — ■ into partial fractions. In accordance with the principle stated in Art. 382, we assume the given fraction equal to the sum of two partial fractions, whose denominators are the powers of 3 a; + 5 be- ginning with the first and ending with the second; that is, 6a; + 5 ^ A B {Sx-j-5y~3x-\-5 {3x-j-5y Clearing of fractions, we have 6a; + 5 = ^(3a;H-5) -\- B = SAx-\-5A+B. Equating the coeflficients of like powers of x, SA=6, and 5A-^B = 6. - Solving these equations, we have A = 2 and B = — 5. Whence, 6fl; + 5 _J 5 ^^^ (3a;+5)2 3^.4.5 (Saj + S)^' Separate the following into partial fractions : 2 2a;-13 ^ 3x^-4. ' g x(5x-4) a.-2-f-10a^+25* * {x-\-iy ' {5x-2y' 3 ^ 5 18a;^+12a;-3 ,- x(x-\-2y {x-2y ' {3x-\-2y ' ' (x-{.iy' g 2a^-10a.-^+17a;-10 g 4a^-18a^ {x-iy ' ' {2x-3y' 318 ALOEBRA. Case III. 384. When some of the factors of the denominator are equal. 1. Separate — — ^into partial fractions. X\,X ~\~ J. ) The method in Case III. is a combination of the methods of Cases I. and II. ; we assume x{x-\-iY X x + i {x + \y {x + iy Clearing of fractions, ^x -\-2= A{x + ly + Bx{x + iy + Cx{x -\-l) + Dx = ( J. + B)x^ + (3^ + 25 + C)x^ + {^A-\-B+C+D)x + A, Equating the coefficients of like powers of «, A-\-B = 0, 3^ + 2J5+(7=0, 3u4 + 5+(7 + Z) = 3, and A = 2. Solving these equations, we have ^= 2, JB = -2, (7=-2, andZ)=l. Substituting in (1), 3a; + 2 2 2 2,1 . x{x-^\y X x + i {x + \y {x-\-\y Note. It is impracticable to give an illustrative example for every possible case ; but the student should find no difficulty in assuming the proper partial fractions if attention is given to the following general rule : jr A fraction of the form should be put equalto (. + a)(. + .) ... (. + m)'... X •\-a X ■\-h ar+jw (ar+ w)^ (^ + wj)' Single factors like x •\- a and x -{-h having single partial fractions corresponding, arranged as in Case I. ; and repeated factors like (^x-\-m)r having r partial fractions corresponding, arranged as in Case II. UNDETERMINED COEFFICIENTS. 319 EXAMPLES. Separate the following into partial fractions : S-^x-x" g 3a^-llar^ + 13a;-4 * x{x-\-2y' ' x{x-l){x-2y 3 3a;-l g 15 - 7a; + 3a^- 3a^ x'ix-j-iy ' a;* + 5cc3 (2a;-3)(2a:2_7^^g)* * sc»(^x-hiy 385. If the degree of the numerator is equal to, or greater than, that of the denominator, the preceding methods are inapplicable. x^ — Sa^ — 1 Thus, let it be required to separate into partial ar — X fractions. If we proceed as in Case I., we should assume x^-Sx^-l ^A B 3^ — X X X—1 Clearing of fractions and uniting terms, :x?-^x'-lz={A-\-B)x-A. Equating the coefficients of q(?^ we have 1 = 0, a result which shows that the method of Case I. is inapplicable. But by actual division, we obtain — = a;-2+— -. (1) XT — X XT — X 2x — 1 We can now separate — into partial fractions by xr — x the method of Case I. ; the result is 1 3 X x — 1 Substituting in (1), we have — = x — 2-] •> Ans. or — X X X — I 820 ALGEBRA. EXAMPLES. Separate the following into entire quantities and partial fractions : 1. 8a^-36a;^-2 (2rc-5)(2a; + l) 3 5a^+5af — 2x^-^S x^' + af 5. 4 3ar^-2a^ + 22a^ + 9; {x'-iy 2x^-2a^-7x^-i-2a^ + x-l x^ — oc^ APPLICATION TO THE REVERSION OF SERIES. 386. Note. To revert a given series y — a + bx'^ + cx^ + ... is to express x in terms of y. Example. Revert the series y = 2x + x^-2a^-Sx* -{-'•» Assume x — Ay-\- By^ -\- Cy^ + Dy^ -\ (1) Substituting in this the given value of 2/, we have a; = ^ (2a; -f- a;2 - 2a;3 _ 3 aj4 4- ...) + 5 (4a;2 + a-^ + 4.^3 - 8 a;^ + ...) 4-0(8a.'3 + 12a;^4-...) + i)(16a;*+. ..)+••• x^-\ That is, a; = 2 ^ar 4- A + 45 a?-2A + 45 + 8(7 x^- 3 A - IB + 12 C + 16i) Equating the coefficients of like powe 2^=1 rs of a;, -2^ + 4 ~3^-75 + 12C ^ + 45 = 5 + 80 = ^+16i) = = = 05 = 0; etc. u:n^determined coefficients. 321 Solving these equations, ^ = 2' ^ = -8'^ = ^'^ = - ■S' - Substituting in (1), we have r 8^ ^16^ 128^^ • ♦, Ans. If the eyen powers of x are wanting in the given series, the operation may be abridged by assuming x equal to a series containing only the odd powers of y. Thus, to revert the series y = x — '3i?-\-x' — i^-\ , we should assume x=^Ay^Bf■\-C]t^-Dy'^-•" If the odd powers of x are wanting in the given series, the reversion of the series cannot be effected by the method previously given. But by substituting another letter, say t^ for x^, we ma}' revert the series and express t in terms of y ; and by taking the square root of the result, x itself may be expressed in terms of y. If the first teim of the given series is independent of a;, it is impossible, by the method previously given, to express x definitely in terms of y ; but it is possible to express it in the form of a series in which y is the only unknown quantity. Let it be required, for example, to revert the series y=2^-2x-^y?-2^-Zx''^"' The series may be written 1/ - 2 = 2a; 4- «2 - 2a^ - Sx* + •.. We then assume a; = ^(2/-2) + 5(2/-2)=^-hC(2/-2)3+i)(2/-2)^+... Proceeding as in Ex." 1, we find a,= 1(2,-2) -i(y-2)^ + A {y-%Y- ^ (2/-2)<+ ... 322 ALGEBRA. EXAMPLES. 387. Revert each of the following to four terms 1. y = x-\-3(^ + x^-\-x^-\ * ^ 2 4 6 8 4. 2/ = l4-a; + ^ + - + -+ — I? |3 li 6. y = x — af-\-x^ — x'^-\-"' 6. 2/ = ^-^ + ^-^' + .- ^ 2 3 4 5 7. 2/ = 3a;4-5a^+7a^+lla;^-t- — ^ 3^5 7 THE BINOMIAL THEOREM. 323 XXXV. THE BINOMIAL THEOREM. FRACTIONAL AND NEGATIVE EXPONENTS. 388. It was proved in Art. 364 that when n is a positive integer, (1+ xy = l-^nx-\--^-- — ^a^H — ^^ -^ ^af + ... (1) 11 '_ Proof of the Theorem for any Exponent. 389. I. When the exponent is a positive /inaction. Let the exponent be ^, p and q being positive integers. Then, (l + a;)?" = V(l +a;)^ (Art. 218) = ^l+pa;+-,by (1). It is evident that a process may be found, analogous to those of Arts. 203 and 208, for expanding -^1 -\-px-\ in ascending powers of x ; and the first term of the result will evidently be 1. Assume then. Vl-fpa;+--- = l+3fa;4-^ar^ + -- (2) Raising both members to the ^th power, we have 1 -^px-{- '" =[1 -{.{Mx -h Nx^ -\- "•)2' = l-\-q{Mx+Nx--i- ...) + -, by (1). This equation being satisfied by every value of x which makes both members convergent, by the Theorem of Unde- termined Coefficients (Art. 376) the coefficients of x in the two series are equal. That is, p = qM, or J/ = -• Substituting this value in (2) , we have {l-{-xy=l+^x+... (3) 324 ALGEBRA. II. When'the exponent is a negative quantity. Let the exponent be — s, s being a positive quantity. Then, (1 + x)-' = ^ (Art. 221) (1 -j-X)" ^ by (1) or (3). l+sx-\- Whence by actual division, we obtain {l+x)-'=l-sx+"' (4) From (1), (3), and (4), we observe that whether n is positive or negative, integral or fractional, the form of the expansion is (1 + «)" = 1 4- waj + ^o.-^ -j- JBa^ + ... (5) X Writing - in place of x, we obtain (-!)■ a a- a^ Multiplying both members by a*", we have (a + a;)" = a'' + 7ia''^^x 4- Aa^'-^x^ +Ba''-^x' + ... (6) This result is in accordance with the second, third, and fourth laws of Art. 196 ; hence these three laws hold for any value of the exponent. 390. We will now prove the Jifth law of Art. 196 for any value of the exponent. Let P and Q denote the coefficients of ic'' and x'"+^ in the second member of (5) ; then (5) and (6) may be written (1 4- aj)" = 1 + 7ix + ... + iV + Qx^-^' + ... , (7) and (a 4- xy = a"+ na'^-^x -\ \- Pa'"'' x' -^ Qa''-'-^ x'+'^ -] (8) In (8) put a=l -\-y, and x = z', then, (1 4-2/ + ^)^ = (l 4-2/)*^ 4- '" 4-P(l 4-2/)'*-'-=?'-4- ••' (») THE BINOMIAL THEOREM. 325 Again, in (7) put x — z-\-y; then, (1 +z + yY = 1 + - + P(z + yy + Q{z-{-yy-^'-^ •>' Expanding the powers of z + yhj aid of (8), we have {1 -hz -{-yy = 1 -h '" + P[z^ -hrz^~'y + '-'] + Q [2'-+'+ (r + l)2^'-y +•••]+••• (10) The first members of (9) and (10) being identical, their second members are equal for every value of z which makes both series convergent ; and by the Theorem of Undeter- mined Coefficients, the coefficients of z" in the two series are equal ; that is, ^(1 + y)""'' = P+ Q(r + 1)2/ + terms in /, f, etc. Expanding the first member by aid of (7), this becomes P[H-C/» -r) 2/ +•••] = ^+Q(r 4-1)2/ + - This equation being satisfied by every value of y which makes both members convergent, the coefficients of y in the two series are equal. Therefore, F{n-r)= Q(r + 1), or Q = ^-=^. That is, the coefficient Q is equal to the coefficient of the preceding term in (8), multiplied by the exponent of a in that term, and divided by the exponent of x increased by 1. Thus the fifth law of Art. 196 is proved to hold for any value of the exponent. 391. By aid of the law proved in Art. 390, the coefficients of the terms after the second in the second member of (8) , Art. 390, may be readily found as in (1), Art. 363. Thus, {a + xy = a^-^ na'^-' x + ^'^'^ ~ ^^ ct'^-^ar^ L? and the Binomial Theorem is proved in its most general form. 326 ALGEBRA. If n is a positive integer, the number of terms in the series is w + 1 ; for all coefficients after the (?i+l)st contain the factor n — n, or 0. (Compare Art. 196.) But if n is fractional or negative, the expansion never terminates, since no one of the quantities /i— 1, n — 2, ..., can become equal to zero. The development in this case furnishes an infinite series, which however expresses the value of (a ■j-x)'' only for such values of a and x as make the series convergent. (Compare Art. 373.) EXAMPLES. 392. In expanding expressions by the Binomial Theorem when the exponent is fractional or negative, it is convenient to obtain the exponents and coefficients of the terms by aid of the laws of Art. 196, which have been proved to hold uni- versally. If the second term is negative, it should be enclosed, sign and all, in a parenthesis, as in Arts. 365 and 367, before applying the laws. 1. Expand (a + x)^ to four terms. 2 The exponent of a in the first term is -, and decreases by o 1 in each succeeding term. The exponent of x in the second term is 1 , and increases by 1 in each succeeding term. The coefficient of the first term is 1 ; of the second term, 2 2 1 - ; multiplying -, the coefficient of the second term, by , 3 3 o 2 the exponent of a in that term, and dividing the product, — -, J by the exponent of x increased by 1, or 2, we have — - as the coefficient of the third term ; and so on. Hence, (a-tx)^ = a^-i--a~^x — -ar^i^-\ a~^ x^ . Ans. ^ ^ 3 9 81 THE BINOMIAL THEOREM. 327 2. Expand (1 — 2a;~2)-2 to ^^^ terms. - 4.1-^. (-2a;"2)3_|. 5.1-6. (_2x-^)4-.. . 1 Mi-i 3. Expand j^=z to four terms. yia'^-^3x^ -v/a-^ + 3x^ (a-^ + 3a;^)^ -M(ai)-¥(3.^)3+... = a^ — a^a;- 4-2a^a; — — a'^^x^H , ^?is. o Expand each of the following to five terms : \- 6. {1-x)-^. 11. {x--^-Sy)^. 16 4. {a-\-x)K 9. ?— — • 14. (a;< + 4a6)^. (a — aj)3 1 7. V^^=^. 12. (a-2a^)-i 17. (4a2+a;-^)i 393. The formula for the rth term of (a + ^y (Art. 366) holds for fractional and negative values of n, since it was derived from an expansion which has been proved to hold universally. 328 ALGEBRA. EXAMPLES. 1. Find the seventh term of (a — 3a;~^)"^. (a -Sx~^y^ = [a -\-(-3x~^)yK In this case r = 7, and n = ; hence the seventh term o 10 13 16 1.2.3.4.5.6 Find the 2. Eighth term of (a 4- x)K 3. Twelfth term of (1 + m)-*. 4. Fifth term of (l-a'yK 5. Seventh term of (a — x)K 6. Sixth term of (a* + b^yK 7. Seventh term of (ic^i — ^"^)^. 8. Sixth term of 9. Eleventh term of (a^ + 2x)K 10. Ninth term of 3 _19 a „ — a 5 (_3a;-2)6 1 1 . Sixth term of (a^ + 3 x'^) ~i 12. Eighth term of (^^-^r* THE BtNTOMIAL THEOREM. 329 394. To find any root of a number approximately by the Binomial Theorem. 1. Find the approximate value of -^2^ to five places of decimals. ^25 = 25* = (27 - 2)^ = (3^ - 2)*. Expanding b}' the Binomial Theorem, we have [(3^) + (-2)]' = (3^)^ + |(30-*(-2)-i(3r^(-2)« + ^(3r^(-2)«-- ^g__2 £_ 40 3.32 9.3^ 81-3« Expressing the value of each fraction approximately to five places of decimals, we have ^25 = 3 - .07407 - .00183 - .00008 = 2.92402, Ans. RULE. Separate the given number into two parts, the first of tvhich is the nearest perfect power of the same degree as the required root. Expand the result by the Binomial Theorem. Note. If the second term of the binomial is small compared with the first, the terms of the expansion diminish rapidly; but if the second term is large compared with the first, it requires a great many terms to ensure any degree of accuracy. EXAMPLES. Find the approximate values of the following to five places of decimals : 2. VIO- *• V^' 6. ^17. 3. V^^- 5- 1, we have a"^ = — = -= (Art. 300). ' ^00 GO ^ ^ Whence by Art. 395, log^O = — oo. Note. As stated in Art. 301, no literal meaning can be attached to the result loga = — oo ; it must be interpreted as indicated in Art. 300. That is, if in any system whose base is greater than unity, a number approaches zero as a limit, its logarithm is negative, and increase* without limit in absolute value. LOGARITHMS. 333 406. In any system, the logarithm of a product is equal to the sum of the logarithms of its factors. Assume the equations «^ =''']; whence, by Art. 395, | ^ = ^^^«^^' a^=n ) ( y = log„ri. Multiplying, we have a' X a^ = m7i, or a' '^^ = mn. Whence, logo77i?i = a^-f^/- Substituting the values of x and y^ we have log„mn = log„m -|- loga?i. In like manner, the theorem may be proved for the product of three or more factors. 407. By aid of the theorem of Art. 406, the logarithm of any composite number may be found -when the logarithms of its factors are known. 1. Given log 2 = .3010, and log3 = .4771 ; find log72. log72 = log(2x2x 2x3x3) = log2 +log2 + log2 +log3 +log3 = 3xlog2 + 2 xlog3 = .9030 + .9542 = 1.8572, Ans. EXAMPLES. Given log2 = .3010, log3 = .4771, log5 = .6990, and log 7 = .8451 ; find: 2. log 21. 7. log 98. 12. log 135. 17. log 1134. 3. log 63. 8. log 105. 13. log 168. 18. log 5145. 4. log 56. 9. log 112. 14. log 147. 19. log 7056. 5. log 84. 10. log 144. 15. log 375. 20. log 14406. 6. log 45. 11. log 216. 16. log 343. 21. log 15552. 834 ALGEBRA. 408. In any system^ the logarithm of a fraction is equal to the logarithm of the iiumerator minus the logarithm of the denominator. Assume the equations "" = ™|; whence, j'" = '<'§« »»' a^ — n ) ^y— log„n. Dividing, we have — = — , or a* ^ = — a^ n n Whence, log^ ~ = x — y. Substituting the values of x and y, m log,- = log«m-log„7i. 409. 1. Given log 2 =.3010; find log 5. log 10 — log 2 1-. 3010 =.6990, Ans. log 5 = log — = log 10 - log 2 EXAMPLES. Given log 2 = . 3010, log 3=. 4771, and log 7 = .8451 ; find : 2. log-. 5. log 35. 8. log—. 11. log7f 3. log 12. 6. log—. 9. log 175. 12. log—. ^7 ^16 * ° 6 4. log3f 7. log 125. 10. log Hi. 13. log5f 410. In any system^ the logarithm of any power of a quan- tity is equal to the logarithm of the quantity multiplied hy the exponent of the power. . Assume the equation a'' = m\ whence, x = log„ m. Raising both members to the pth power, we have a^* = m^ ; whence, log„ m^ =px=p log^ m. LOGARITHMS. 335 411. In any system^ the logarithm of any root of a quantity is equal to the logarithm of the quantity divided by the index of the root. I 1 For, loga -{/m = log«(m'") = - log^ m (Art. 410) . 412. 1. Given log 2 = .3010 ; find the logarithm of 2i log 2^ = - X log 2 = 5 X .3010 = .5017, Ans. o o Note. To multiply a logarithm by a fraction, multiply first by the numerator, and divide the result by the denominator. 2. Given log3 = .4771 ; find the logarithm of ^3. log ^3=1^= '^^ = .0596, Ans, EXAMPLES. Given log 2=. 3010, log3 = .4771, and log 7 = .8451 ; find: 3. log3i 7. logl2i 11. logl5i 15. log ^5. 4. log2^ 8. log2li 12. logV7. 16. log ^35. 6. log7^ 9. logl4\ 13. log the amount at the end of the last one, in accordance with the law observed n above, will be PR*. n That is, A = PR'. (1) By logarithms, log^ = logP-f- logi2. (2) Example. What will be the amount of $7326 for 3 years and 9 months at 7 per cent compound interest, the interest being compounded quarterly ? COMPOUND INTEREST AND ANNUITIES. 355 In this case, P=7326, n = 3|, ^ = ^, i2= 1.0175, and -= 15. logP= 3.8649 log i2 = 0.0075 ; multiply by 15 = 0.1125 log^ = 3.9774 .•.A= $9492, Ans, 2. Given n, t, jR, A; to find P. From (2), log P = log ^ - - log i2. c Example. What sum of money will amount to $ 1763.50 in 3 years at 5 per cent compound interest, the interest being compounded semi-annually ? In this case, n = 3, i = |, P= 1.025, vl= 1763.5, and- = 6. log^ = 3.2464 logP = 0.0107 ; multiply by 6 = 0.0642 logP= 3.1822 .•.P= 8 1521.40, Ans. 3. Given P, ^ i2. A; to find n. From (2) , - log P = log ^ - log P. c Whence, ,,^ ^0o^-^ -log^) . \ogR Example. In how many years will 8 300 amount to $ 396.90 at 6 per cent compound interest, the interest being com- pounded quarterly? Here, P=300, t = \, P= 1.015, and ^=396.9. . ^ _ log 396.9 - log 300 ^ 2.5987 - 2.4771 ^ .1216 ' 4 log 1.015 4 X. 0064 ~ ,0256 = 4.75 years, Ans, 356 ALGEBRA. 4. Given P, w, t^ A; to find R. From (2), log 7? = l^ii^liSSf. n 7 Example. At what rate per cent per annum will $500 amount to $688.83 in 6 years and 6 months, the interest being compounded semi-annually ? Here, P= 500, n = ^, t=^,A = 688.83, and - = 13. log^= 2.8381 logP= 2.6990 13 )0.1391 logR= 0.0107 .'.B= 1.025. That is, the interest on one dollar for 6 months is $.025, and the rate is 5 per cent per annum. EXAMPLES. 439. 1. What will be the amount of $1000 for 18 years at 6 per cent compound interest, the interest being com- pounded annually? 2. What sum of money will amount to $870.50 in 7 years and 3 months at 3 per cent compound interest, the interest being compounded quarterly ? 3. In how many years will $968 amount to $1269.40 at 5 per cent compound interest, the interest being compounded semi-annually ? 4. At what rate per cent per annum will $2600 gain $416.40 in 3 years and 9 months, the interest being com- pounded quarterly? 5. In how many years will a sum of money double itself at 5 per cent compound iijterest, the interest being con}' pounded annually? COMPOUND INTEREST AND ANNUITIES. 357 6. In how many years will a sum of money treble itself at 7 per cent compound interest, the interest being compounded semi-annually? 7. What sum of money will amount to $1000 in 11 years and 8 months at 3J per cent compound interest, the interest being compounded every four months ? ANNUITIES. 440. The present value of a sum of money, due at the end of a given period, is the sum which when put at interest for the period in question will amount to the given sum. In finding the present value of an annuity, it is customary to allow compound interest. 441. To find the present value of an annuity to continue for n successive years, allowing compound interest. Let A = the annuity in dollars ; li = the amount of one dollar for one year ; P„, = the present value of the payment due at the end of m years ; P= the present value of the annuity. By Art. 440, the sum P^ will amount to A when put at compound interest for m years, the interest being compounded annually. In this case, w = m, and t=l ', whence by (1), Art. 438, A=P^Iir, or P^ = — • Br By aid of the above formula, the present value of the 1st payment = — ; 2nd payment = ^ ; nth payment = — 358 ALGEBRA. Hence the sum of the present values of the separate pay- ments, or the present value of the annuity, is A A iA_j-^ Thatis, P=^[± + ^^ + ... + i + y. The expression in brackets is the sum of the terms of a Geometrical Progression, in which a = -— , r = i2, and ? = 7, ; whence by (II.), Art. 348, Example. What is the present value of an annuity of $ 150 to continue for 20 years, allowing 4 per cent compound interest ? Here, ^=150, n = 20, i2=1.04, and i?-l = .04. mi .04 L (1. Whence, P=l^fi_^^]. log L_^ = 20 colog 1.04. ^(1.04)20 ^ colog 1.04 = 9.9830 20 9.6600 Number corresponding = .4571. Therefore, p= 3^ (1 _ .4571) = 3750 x .5429. log 3750 = 3.5740 log .5429 = 9.7347 log P= 3.3087 .-. P= $2035.70, Ans. COMPOUND INTEREST AND ANNUITIES. 359 442. If in (1), Art. 441, n is indefinitely increased, the limiting value of the second member is A E-1 (Art. 301). That is, the present value of a perpetual annuity is equal to the amount of the annuity divided by the interest on one dollar for one year. EXAMPLES. 443. 1. What is the present value of an annuity of $ 200 to continue 15 years, allowing 5 per cent compound interest? 2. What is the present value of a perpetual annuity of $600, allowing 3^ per cent compound interest? 3. What is the present value of an annuity of $1127 to continue 3 years, allowing 7 per cent compound interest? 4. What annuity to continue 10 years can be purchased for $2038, allowing 6 per cent compound interest? 5. A person borrows $5254; how much must he pay in annual instalments in order that the whole debt may be dis- charged in 12 years, allowing 4^ per cent compound interest ? 360 ALGEBRA. XXXVIII. PERMUTATIONS AND COMBINA- TIONS. 444. The different orders iu which quantities can be arranged are called their Permutations. Thus the permutations of the quantities a, 6, c, taken two at a time, are t. r. i. i, ' ab, ac, ba, be, ca, cb ; and their permutations taken three at a time, are abc, acb, bac, bca, cab, cba. 445. The Combinations of quantities are the different col- lections which can be formed with them, without regard to the order in which they are placed. Thus the combinations of the quantities a, 6, c, taken two at a time, are , , ' ab, be, ca ; for though ab and ba are different permutations, they form the same combination. 446. To find the number of permutations of n quantities taken two at a time. Let the quantities be %, aa? <^39 <^45 •'•-) ^«- The permutations of the quantities taken two at a time, having cti as the first element, are ttitta, ciiOt.3, aia^^, ..., ditt^ j the number of which is n — 1. In like manner, there are n — 1 permutations of the quan- tities takew two at a time, having ag as the first element ; and similarly for each of the remaining quantities ag, a^, ..., a„. Therefore the whole number of permutations of the quan- tities taken two at a time is equal to n(w— 1). PERMUTATIONS AND COMBINATIONS. 361 447. We will now consider the general case. To find the 7iumher of permutations of n quantities taken r at a time. Let the quantities be tti, 0^2' <^3? •••> ^r? ^r+l1 <^r+2) •••? <^n* One of the permutations containing r quantities will be that consisting of the first r quantities in their order ; that is, aia.2a^...ar. Placing after this the other n — r quantities one at a time, as follows, there are formed n — r dififerent permutations, each contain- ing r+l quantities. We may proceed in a similar manner with the remaining permutations containing r quantities, and in each case we shall obtain n — r permutations containing ?- -f 1 quantities. That is, the number of permutations of the quantities taken r at a time, multiplied by 7i — ?*, is equal to the number of permutations of the quantities taken r + 1 at a time. But the number of permutations of the quantities taken two at a time is equal to n{n — l) (Art. 446). Hence the number of permutations of the quantities taken three at a time, is equal to the number taken two at a time, multiplied by n — 2, or w(n — 1) (n — 2). The number of permutations of the quantities taken four at a time, is equal to the number taken three at a time, multi- plied by w — 3, or n{n—l) {n — 2) (/i — 3) ; and so on. We observe that the last factor in the number of permuta- tions is 71, minus a number one less than the number of quan- tities taken at a time. 362 ALGEBRA. Hence the number of permutations of the quantities taken r at a time is given by the formula w(n-l)(n-2)...[n-(r-l)], or, ?i(n-l)(n-2)...(n-r + l). ^ (1) 448. If all the quantities are taken together, r = n, and formula (1) becomes n{n-l){n-2)'"l^\n. (2) That is, the number of permutations of n quantities taken n at a time is equal to the product of the natural numbers from 1 to n inclusive. (See Note 2, Art. 363.) 449. To find the number of combinations of n quantities taken r at a time. The number of permutations of n quantities taken r at a time is n(n-l)(n-2)...(7i-r+l) (Art. 447). But by Art. 448, each combination of r quantities may have [r permutations. Hence the number of combinations of n quantities taken r at a time is equal to the number of permutations, divided by \rj, that is, n(n-l)(n-2)...(7i-r + l) /g) ll 450. Multiplying both terms of (3) by the product of the natural numbers from 1 to n — r inclusive, we have yi(?i-l)...(?i-r+l)X(n-r)---3-2»l ^ [^j: [r X 1 • 2 . 3 • • • (n — ?') \r_ \n — r ' which is another form of the result. 451. By Art. 450, the number of combinations of n quantities taken ?i — r at a time, is In \n \n — r\n — {n — r) \n — r\r PERMUTATIOKS AND COMBINATIOJTS. 363 But this is the same as the number of combinations of n quantities taken r at a time (Art. 450) . Hence, the number of combinations of n quantities taken r at a time is equal to the number of combinations of n quanti- ties taken n — r at a time.- EXAMPLES. 452. 1. How many changes can be rung with ten bells, taking 7 at a time ? Here n = 10, r = 7, and ?i — ?• 4- 1 = 4. Then by (1), Art. 447, the required number = 10. 9. 8.7.6. 5.4 = 004800, Ans. 2. How many different combinations can be formed with 16 letters, taking 12 at a time? By Art. 451, the number of combinations of 16 quantities taken 12 at a time is equal to the number of combinations of 16 quantities taken 4 at a time. Putting n — 16 and ?' = 4, in (3), Art. 449, we have 16.15.14.13 1.2.3.4 1820, Ans, 3. How many permutations can be formed of the 26 letters of the alphabet, taken 5 at a time ? 4. How many permutations can be formed of the letters in the word forming^ taken all together ? 5. How many combinations can be formed with the letters in the word triayigles, taking four at a time? 6. How many different numbers, of five different figures each, can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9? 7. From a company of 40 soldiers, how many different pickets of 6 men can be taken? 364 ALGEBRA. 8. How many combinations can be formed with 18 quan- tities, taking 11 at a time? 9. How many different words of 4 letters each can be made with 6 letters? How many of 3 letters each? How many of 6 letters each ? How many in all possible ways ? 10. How many combinations can be formed with 24 letters, taking 18 at a time? 11. How many different committees, consisting of 8 per- sons each, can be formed out of a corporation of 20 persons? 12. How many different numbers, of 4 different figures each, can be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, 0? 13. How many different words, each consisting of 4 con- sonants and 2 vowels, can be formed from 8 consonants and 4 vowels? The number of combinations of the 8 consonants, taken 4 at a time, is o r^ n n 1.2.3.4 The number of combinations of the 4 vowels, taken 2 at a time, is .0 But any one of the 70 sets of consonants may be associated with any one of the 6 sets of vowels. Hence there are in all 70 x6, or 420 sets, each containing 4 consonants and 2 vowels. Now each of these sets of 6 letters may have |^, or 720, different permutations (Art. 448). Therefore the whole number of different words is 420 X 720 = 302400, Ans. 14. How many different words, each consisting of 3 con- sonants and 1 vowel, can be formed from 12 consonants and 3 vowels? PERMUTATIONS AND COMBINATIONS. 365 15. How many different committees, each consisting of 2 Republicans and 3 Democrats, can be formed from 14 Repub- licans and 21 Democrats? 16. Out of 9 red balls, 4 white balls, and 6 black balls, how many different combinations can be formed, each con- sisting of 5 red balls, 1 white ball, and 3 black balls? 17. How many different words, each consisting of 4 con. sonants and 3 vowels, can be formed from 10 consonants, and 5 vowels? 18. Out of 11 physicians, 13 teachers, and 8 lawyers, how many different committees can be formed, each consist ing of 3 physicians, 4 teachers, and 2 lawyers? 19. How many words of seven letters each can be formeu from the letters a, Z>, c, d, e, /, g^ each word being such that the letters a, 6, c are never separated ? 366 ALGEBRA. XXXIX. CONTINUED FRACTIONS. 453. A continued fraction is an expression of the form a-\ — -; d e + ... or, as it is usually written in practice, h d We shall limit ourselves in the present chapter to con- tinued fractions of the form , 1 1 5+C+... where each numerator is unity, a is or any positive integer, and each of the quantities 6, c, ..., is a positive integer. 454. A terminating continued fraction is one in which the number of denominators is finite ; as, ,111 a-\ -• h-\. c+ d It may be reduced to an ordinary fraction by the process of Art. 161. An infinite continued fraction is one in which the number of denominators is indefinitely great. 455. In the continued fraction 1 1 1 «! + ^2+ «3+ a4+- a, is called the first convergent ; ttiH — is called the second convergent; «2 tti + is called the third convergent; and so on. ^2+ «3 CONTINUED FRACTIONS. 367 Note. If ai = 0, as in the continued fraction 1 1 1 y «3 + «3 + tti +"• then is considered the first convergent. 456. Any ordinary fraction in its lowest terms may he con- verted into a terminating continued fraction. Let the given fraction be -• Divide a by 6, and let aj denote the quotient and bi the remainder ; then, a , bi ,1 _,, + ^ = a, + ^. Divide b by 5i, and let ag denote the quotient and 62 tl^e remainder ; then, ? = a. + -l- = a. + -!— . bi Oi Again, divide 61 by 62? and let as denote the quotient and 63 the remainder ; then, ? = a, + i— - = «,+ ' b , 1 ^ , 1 O2 ©2 The process is the same as that of finding the Highest Common Factor of a and b (Art. 129); and since a and b are prime to each other, we must eventually obtain a remain- der unity, at which point the operation terminates. Hence any ordinary fraction in its lowest terms can be con- verted into a, terminating continued fraction. 368 • ALGEBRA. 62 Example. Convert — into a continued fraction. Zo 23)62(2 = ai 46 16)23(1 = a2 16 T)l6(2 = a3 U 6 1 Therefore, — =2+— -^i^, Ans. ' 23 1+2+3+2 457. A quadratic surd (Art. 250) may he converted intc an injinite continued fraction. Example. Convert ■^y6 into a continued fraction. The greatest integer in ^6 is 2 ; we then write V6=:2+-(V6-2). Reducing -y/Q — 2 to an equivalent fraction with a rational numerator (Art. 244) , we have ^ ^ V6 + 2 V6 + 2 = 2 +--7^ (1) V6 + 2 2 The greatest integer in ^^ is 2 ; we then write V6+-2 ^^ I V6-2 2 2 ^g (V6-2)(V6 + 2) ^^ 1 2(V6+-2) V6 + 2 CONTINUED FRACTIONS. 369 Substituting in (1), 1_ 1 V6 = 24- —, (2) 2 + V6 + 2 The greatest integer in ^6 + 2 is 4 ; we then write V6 + 2 = 4 + ( V6 - 2) = 4 + i^«^lMV6 + 2i -y/6 4" 2 V6+2 V6+2 Substituting in (2), we have 1 V6 = 24- 2 + ' 44 ' V6 + 2 2 The steps now recur, and we have V6 = 2-f— — J^ —^ Ans. ^ 2+4+2+4+- Note. An infinite continued fraction in which the elements recur, is called a periodic continued fraction. 458. A periodic continued fraction may always he expressed as the root of a certain quadratic equation. Example. Express as the root of a certain ^ 1+3+1+3+" quadratic equation. Let X denote the value of the fraction ; then, ^ ^ J 1_ ^ 3+- a; ^ 3 + a? 1+3H-X 3 + a; + l 4 + a;* Clearing of fractions, ^x + x^=^^+x, or ar'-f 3rc = 3. 370 ALGEBRA. Solving the equation, 3+V9 + 12_-3+V21^ ^ns. 2 2 Note. The + sign is taken before the radical, since x is evidently a positive quantity. PROPERTIES OF CONVERGENTS. 459. Let the continued fraction be ,11 11 and let p^ denote the numerator, and q^ the denominator, of the rth convergent (Art. 455) when expressed in its simplest form. 460. To determine the law of formation of the successive conver gents. The first convergent is dp The second is a^-\-- = Mdti. The third is a^^—l- = a, + — 'h_ ^M^^±ai±^. «2+ % «2«3 + 1 <*2<^3 + 1 The third convergent may be written in the form (aia2 + l)a3 + <^ . a2«3 H- 1 in which we observe that 1. The numerator is equal to the numerator of the preced- ing convergent^ multiplied by the last denominator taken, plus the numerator of the convergent next but one preceding. 2. The denominator is equal to the deiwminator of the preceding convergent, multiplied by the last denomiriator taken, plus the denominator of the convergent next but one 2^TQceding. We will now prove by Induction (Note 1, Art. 363) that the above laws hold for all convergents after the second, when expressed in their simplest forms. CONTINUED FRACTIONS. 371 Assume that the laws hold for all convergente as far as the nth. The nth convergent is Then since the last denominator is a„, we have Pn = (InPn-l +Pn-2 » and g„ = tt^^^.i + qn-2' (1) Whence, P_,^ <^nP.-.+P.-. , (2) Qn anqn-l + qn-2 The (n 4-l)st convergent is J 1_ 1 1 ^ a2+a3+ a„+«Xn+i' which differs from the nth only in having a^^H , or ^^ , m place of a„. Substituting ^"^"+' "*" ^ for a, in (2), wehave ;; Pn-1 + i5n-2 ?n-l -r yn-2 _ On+1 {anPn-l-\-Pn-2) + Pn-1 It is evident that the second member of (3) is the simplest form of the (n +l)8t convergent, and therefore i>n+l = «n+lPn+i>n-l, aud Qn+l = a^^+lQu -^ Qn-l' These results are in accordance with the laws stated on the preceding page. Hence, if the laws hold for all convergents as far as the nth, they also hold as far as the (n -f l)st. 372 ALGEBRA. But we know that they hold as far as the third convergent, and hence the}^ also hold as far as the fourth ; and since they hold as far as the fourth, they also hold as far as the fifth ; and so on. Therefore the laws hold for all convergents after the second. Example. Find the first five convergents of 1+ 2+ 3+ 4+- The first convergent is 1, and the second is 1 + 1, or 2. Then by aid of the laws just proved, ^, ^i- -, . 2-2+1 5 the tlurd is ' — = - ; 1-2+1 3 ^v, * ^1.- 5-3 + 2 17 the fourth is ' — = — ; 3.3+1 10 the fifth is !— = — 10-443 43 461. The difference between two coyisecutive convergents — and -^^ is equal to qn Qn+l Qnqn+l The difference between the first and second convergents is / 1\ 1 ai H — — tti = — Thus the theorem holds for the first and second conver- gents. Assume that it holds for the nth and (n +l)st convergents ; that is, Then, Pn+l _^ Pn+2 _ Pw+l _^ f^n+2Pn+\ 4" j^n /^j.^ 460) 9'w+l 9'n+2 9'n+l C^^+2 5'n+l + Q'n CONTINUED FRACTIONS. 373 ^ ((^n+2Pn+lQn+l-{-Pn+iqn) ^ (cin+2Pn+iqn+l +i>»gn+l) Qn+l{a^+2gn+l+qn) ^ Pn^.qn-^Pnqn^, (Art. 460) =3— i—, by (1). Q'n+1 qn+2 q,i+l qn+2 Hence if the theorem holds for any pair of consecutive con- vergents, it also holds for the next pair. But we know that it holds for the first and second conver- gents, and hence it also holds for the second and third ; and since it holds for the second and third, it also holds for the third and fourth ; and so on. Therefore the theorem holds universall}-. 462. It follows from Art. 461 that p„ and q^ can have no common divisor except unity ; for if they had, it would be a divisor of p„.+2Pn4-i +Pn Qn+2 <^n+1 ^n+1 + ^n The fraction itself is obtained from its (n-j-2)nd con- vergent by putting a^+o -\ in place of a„^9. Hence, denoting the value of the fraction itself by x^ we have 1 Pn+l+Pr. mPn+i-hPn an4-2 4- «n+34- hi+2 + 1 where m stands for a„^2 + 1 'tnqn+i + Qr, «n+3 + - Now, ^^Pn^^i>n-H+P.^P, Wl(Pn4-ign~P»gn+l) qn{mqn+i m Also, a; qn+1 mq^+l + gn '^ ^n+l qn) (Art. 461). (1) (2) qn+i i^nq^+i + qn) qn+1 i'inqn+i + qn) Since 1 ; that is, m is > 1. And since g„+i = a^+^q^ + g„_i (Art. 460) , q^+i is > g„. Therefore the fraction (2) is less than the fraction (1), for it has a smaller numerator and a greater denominator. Hence the (n+l)st convergent is nearer than the nth to the value of the fraction itself. CONTINUED FRACTIONS. 375 465. By Art. 464, the difference between the fraction itself and its nth convergent is — ; ^, or — -. (1) Since m is > 1 (Art. 464), the denominator qn(Qn+i + The denominator is also > q^qn+i- Hence the fraction (1) is > — ; r^? and < Qn{qn+, + qn) qnqn+l That is, the eiror made in taking the nth convergent for the fraction itself lies between the limits 1 , 1 and ^■Mn+l-hqn) qnqn+l EXAMPLES. 466. Convert each of the following into a continued frac- tion, and find in each case the first five convergents : 1. ^. 3. 3.61. 5. ^. 7. if«. 39 326 345 2 72 4 112 6 — 8 ?^. 91* ' 153* ' 89 * * 6961* Convert each of the following into a continued fraction, find in each case the first four convergents, and determine limits to the error made in taking the fourth convergent for the fraction itself : 9. V5- 10. V3. 11- Vll- 12. V7. Express each of the following in the form of a surd : i« 1 1 1 1 15 24-- —' ^^- ^3T^3T:^* 1^- ^ + 1+1 + ... ^ 1+4+1+4+- ^1+8+ 1+8+- 376 ALGEBRA. 17. The ratio of the circumference of a circle to its diame- ter is approximately equal to 3.14159 ; express this decimal as a continued fraction, and find the first four convergents. 18. The modulus of the common system of logarithms is approximately equal to .43429 ; express this decimal as a continued fraction, find its seventh convergent, and determine limits to the error made in taking this convergent for the fraction itself. 19. The base of the Napierian system of logarithms is 2.7183 approximately; express this decimal as a continued fraction, find its eighth convergent, and determine limits to the error made in taking this convergent for the fraction itself. 20. Express the positive root of the equation a^-a;-ll = as a continued fraction, and find the first five convergents. ANSWERS. Note. In the following collection of answers, all those are omitted which, if given, would destroy the utility of the example. 1. 19. 2. 1. 3. 24. 4. 15f. 6. iZ. 12 144 Art. 40 ; pages 8 and 9. 6. — . 11. 48. 16 7. 6|. 12. 114. 8. 9. 13. 24. 9. 5^. 14. 204. 135 10. IM?. 15. 310. 20. 48. 16. 4. 17. 9^ 5' 18. 50 39' 19. 36. 21. 3. 22. ^' Art. 43 ; page 12. 4. 35 and 7. 8. A, $20; B, $60; C, $180. 5. A, 31 ; B, 37. 9. 52 and 73. 6. A, $536; B, $664. 10. 13, 39, and 46. 7. $0.93. 11. A, $28; B, $43 ; C, $56. 12. Horse, $275; carriage, $100; harness, $25. 13. 15, 45, and 48. 14. 35, 70, and 105. 15. Cow, $45; sheep, $18; hog, $12. Art. 57 ; page 19. 4. 2a-26-f2d. 6. 4a^ 7. 2a^-ab. 2 ALGEBRA. 8. 5a^-Sx^-e. 12. -ar^ + 3a;+2. 9. 6mn-ab-4:C—x+Sm\ 13. 3a-j-Sb + 3c + ^d. 10. x-6y, 14. 5a3-3(x62_453^ 11. _a; + 3m-7n. 15. a3-4a^. Art. 64 ; pages 22 and 23. 6. 4a5. 10. 66 + 1. 7. _4a6c-14a;-22/-148. 11. 4m-8w-r + 3s. 8. 2m-42/2 + 12a + l. 12. 6d-26-3a-3c. 9. 14a^-8^2^5a5-7. 13. 57n^ + 9n^ -71x. 14. _3a-26 + 10c-13d + 2a;. 15. 2a-6-3c. 20. 2a;-3^. 16. x^-a^-Say'-Sx+n. 21. ea^ + 9a^ -^a^ -a-6. 17. 4a3-6a26-2a62-96^ 22. -3a^-5i»V+8a;2/2+22/3. 18. 4a^-3a3+6a2-6a+3. 23. 5a^ + 7a^ + 18.T.. 19. ^x^-xy. 24. -a2-4a6 + 6^ 25. 5a^ + 6iC2/-7/-6a;— 72/ + 6. 26. SxP-6x^-a^-na^ + Ax-2. 27. 4ajs+ 90^2/ -4a;/- 32/3. 28. 4a^ + 7a3_2a-4. Art. 69 ; pages 25 and 26. 4. 5x-y. 10. a+b-c-{-d-e. 16. -3a-l. 5. a-b + c. 11. 2/. 17. 4a;-2. 6. om2-6w-4a. 12. 14icH-2. 18. 67/1 + 2. 7. _a2-362. 13. 6m-3ri. 19. 0. 8. 2a + 2. 14. 2a; + 42/. 20. a + 26. 9. a— 6+c— c?+e. 15. a — c. Art. 82 ; pages 32 to 34. 4. 16x^-Ux-U. 6. -6a2+i6a6-862. 5. -12a;2_f_28a;-15. 7. 50a;22^2_i8^ ANSWERS. 8. W-a\ 11. 6a:«-16a^2/+6a;/4-4/. 9. 10a*62_3^353_i8^2j4^ 12. 2m*- 8m^7i + 18m«^ 10. ax^ — a. 13. iK* + 4aj4-3. 14. 30a-^-43a26 4-39a52-206^ 15. 8a^-14a^-18a; + 21. 16. a^ -62^_26c-c2. 17. 2iB*-a^ + 8a;-5. 18. 6a^ + 13a^-70a^ + 71a;-20. 19. 6fl^-19a^ + 22a;H-5. 20. -m«-37m2+70m-50. 21. -6ar'-25a^4-7ar'^H-81a^4-3a;-28. 22. 2 a''^^ - 3 a*63 _ 7 ^354 ^ 4 ^255^ 23. 4a^'"+V-16a;'"+<5?/''+^ + 12ar'/»*-^ 24. a;* + a.V + 2/'. 26. 12a^+7a^+5cc'+10a;-4. 25. 16a*4-4a262 + 6^ 27. ?7i6 - 3 m^n + mn«- 3 n«. 28. 243ar^-81a^2/-3a?2/^ + 2/^- 29. a** -ba^b + lOa^V" - lOa-ft^ + 5a6* - 6^ 30. a^ + 2/3^^_3^y2^ 31. 6a^--lla^+14a;*-12a^ + 6a:2-23a; + 5. 32. a^ft^ + c^d^-a^c^-ft^c^. 36. 36m3-49m + 20. 33. 8a«-98a*+152a2-32. 37. 9fl;*-13a^ + 4. 34. ic3-6x2-19a;+84. 38. a^ + aj^ + l. 35. a«-6«. 39. a8-6«. 40. ??i*-5m2 + 4. 41. 120x* + 26a^-lllaj2-14a;H-24. 42. a«-6a*62 + 9a25*-46«. Art. 83 ; pages 34 and 35. 2. a^H- ^2 4. c2 + d^ + 2a6 + 2ac 4- 2ad + 26c + 2hd + 2cd. 3. d&H-ac — 2acZ — 26c + &d+cd. 4. a^. 5. 4a&+46c. 7. a^ - 2 a'^h'' + h\ 9. 4.a-h2a? + 2aK 6. 6^ 8. 1-a^. 10. a;2-4a;2/+42/*^-92;2. 4 ALGEBRA. 11. a^-y\ 13. 4a2+462+4c2. 15. -4 2/2;, 12. b'-dK 14. ab-{-bc-\-ca-a^-b^-c\ 16. 0. 17. Sac. 18. -9m* + 82mV-9n*. Art. 93 ; pages 41 to 43. 19. 0. 4. 2ic-5. 7. 8-5a;. 10. 3a; — 1. 6. 2 + aa;. 8. Sb^-Aa"". 11. 4m2-10m + 7. 6. a -2b. 9. -2a;2_2aa;. 12. 9a^-3a;y+2/^ 13. 8m3 + 4m2+277i + l. 15. 4a2 4- 12a6 + 962. 14. a — b~c. 16. x^ — xy + ff. 17. a;-3. 19. 2«3-3a;-6. 21. 2a + 3. 18. Am^-^n\ 20. 2a^-7a;-8. 22. a^-3a;-2/. 23. a6-3a^6 + 9a262_276^ 24. a;-2/ + 0. 25. 3a^+6a^-2a;-4. 26. yix'-a^y^x'y'-xy^ + y'). 27. 5m2-4m + 3. 28. ^x^-2x-{-l. 29. 3(a4 4-a^& + a262 + a?>3_f.j4)^ 30. 8a^4-4a;+l. 33. a^-2a;-l. 36. 2a^-a;+l. 31. 9a^-9a;-4. 34. 3a^-5a??/-2r. 37. 3a;2 + 6ic + 9. 32. a^-2a;2/+/- 35. a2_|_3a64-5 52. 38. a:2_2^_3. 39. m^-m^-Um + 24. 40. a; + 2/. 50. x-\-a. 41. o?y — xy'^. 51. (5+c)a + 6c. 42. ar^-2i»2 + a;-2. 52. (0^ + 2/) -3. 43. o?-2a%-bab^ + lW. 53. (a + &)2- (a + ?>) + 1. 44. a^-2i»2_^_|_i^ 54^ ^_^^^ 45. a; + 22/-3^. 55. (m-n)2+2(m-n) +1. 46. a'*-^>"* + c^ 56. a;2_|. (^ _5)a;- a6. 47. a^ + 2a^-a; + l. 57. a^-6a;4-c. 48. 2o?-ab + 2b\ 58. a(^>+c)-6c. ANSWERS. Art. 96 ; page 46. 30. aP-y^ + 2xz-i-z\ 34. a^ -b^ -2ac + (^, 31. x'-y^-2yz-z\ 35. a^-2a' + l. 32. l-a^ + 2ab-b\ 36. x^- 4:X^-\-12x-9. 33. x^-a^-2x-l. 37. m^ + m'n^ + n\ Art. Ill ; pages 56 and 57. 26. (x-\-y + 2){x-\-y-2). 29. (a + 64- c) (a-6 -c). 27. (a-6 + c)(a-6-c). 30. (c + cZ4- l)(c + d- 1). 28. (a + 6-c)(a-6 + c). 31. (S + x-y){S -x + y). 32. (2m2 + 26-l)(2m2-26 + l). 33. (2a-6 + 3(Z)(2a-6-3d). 34. {a-{-b — m + n)(a — b — m — n), 35. (a; + ?/ — c — d) (a;— 2/ — c + d). 36. {a + b-\-m — n){a — b-{-m-\-n), 37. (a + 6H-c-f d)(a — ?> + c — d). Art. 120 ; page 64. 27. 2xy(x + y-{-z){x-{-y-z). 40. (x-\-y-j-z){x-\-y-z){x-y-\-z)(x-y^z). 46. (a + 6)(a_&)(x-f2/)(aj-2/). 50. {x-2){x-\-S){x'-x-\-6). 51. (a-l)(a-2)(a + 4)(a + 5). 53. (a + 6 -f c) (a + 6 — c) (a — 6 + c) (a — 6 — c) . 55. {x-iy{x-{-l). 58. (x + 2)2(a;-2)2. 56. 2{a-b)(a + 2b). 59. {x-yy. 60. (a -l)(a + 2)(a-2)(a + 3). Art. 131 ; pages 73 and 74. 1. x-2. 4. 8a;-7a. 7. 5m + 3. 10. 2a;-l. 2. 2x-i-3. 5. 2a-5. 8. 2a-3a;. 11. 2m-5n. 3. a-1. 6. x^-mx. 9. a;-2. 12. a; + 2. e ALGEBRA. 13. ax^-ax. 15. a^-a-1. 17. 3a; + 2. 19. x + l. 14. x^-hx+1. 16. a^-2a;. 18. a-x. 20. a;-2y. 21. 2a;-3. 22. 3a + 6. Art. 132 ; page 74. 1. 2a;4-7. 2. 2a;-5. 3. 3m + 2ti. 4. 3a-l, 5. aj + 4. 6. a;-l. 7. 2a + 3. Art. 136 ; page 76. 4. 270mV. 7. 336a^2/'2?. 9. iSOm^n'a^y^ 6. 210 a6c. 8. 360 aVd^. 10. 252 a^2/'^^ 11. lOSOa^ftVd*. Art. 137 ; pages 76 and 77. 2. y(x^-y^)' 9. {x + 3a)(x-5a)(x-\-7a). 3. (aj2_i)(a;_8). 10. ^(m^-n^). 4. 24:ab{a'-b^). 11. (a + 6) (a -36) (a; -2). 5. (m -\- n) (m^ — n^) . 12. aa?(a;4- a) (o^ — a^). 6. a2-4a6 + 36^ 13. 2^a + by{a-by. 7. a^(a; + 2/)(x-?/)2. 14. ax(^x-S)(x-7){x-\-8). 8. 12a6c(a2-62). 15. x4_2a;2 4.i. 16. 24(1 -a;^). 17. (a; + l)(a;-2)(a;+3)(aj4-4). 19. (2m + l)(2m-l)2(477i2_^2m + l). 20. a\a-l){a'-^l). 21. (a_l)(a-3)(a + 4)(a-5). 22. (l+a;)2(l -0^)2(1+0^)2. 23. (a + & + c)(a + & -c)(a-6-c). 24. Sab(a-b)(x-yy. 25. 2aa;2(3a; + 2)2(9ar^-6a; + 4). 26. (ic + 2/ + ^)(a^4-2/-2;)(a^'-2/+^)- ANSWERS. . 7 Art. 139 ; page 79. 2. 4a^-13a; + 6. 4. 24ic3+26a^- 219cc- 56. 3. 24a;3+22aj2-177a;+140. 5. 2aa;(6a^+a^-42ic-45). 6. a*- 16a36 + 86a262_ 176^^3 _^1056^ 7. 27i(2m^-5??i^ + 3m3-5?7i2 + 4^_^4). 8. a(30a^-llaa^-59a2a;+12a3). 9. 2a^ + a3-17a2-4a+6. 10. 2aj=+3a;*-4ar5+5a;-6. 11. a^ - a362 _(- a^ft^ _ 2,5^ 12. aa;(ar^ + a;^-a^-3a^-3aj-l). 13. a;(6ar^-31a;^-4ic3 4-44a;2_,_7^_lQ)^ 14. ic«+2a;5-4«^-7a^-16a:2_|_32a._8. Art. 140 ; page 79. 1. 4a^-4a^-39ar^ + 4a; + 35. 2. 18a*-33a3+14a2 + 3a-2. 3. 20a.'*-24a^-51a^ + 41ic-6. 4. 2a:2^i2ic^-32a;»-29ar' + 57a;-18). 5. a^ + 3a*-23a3-27a2+166a-120. 12. 13. 14. 15. 16. 17. Art. 149 ; pages 83 and 84. cd 3xy 18. m — 2 m + 9 24. 9/4-152/+25 3?/ -5 7? 2/ 19. a(37i-f2) 6(3n-2) 25. a;-l 2cc+l 2^y 20. a + 2& a + 36 26. 1 x-h aa;(a; + 4) a-5 a + 7 21. 2a;H-2/. 27. a + 64-c a — ?> + c 2a2^a6 a6 + 262 22. a^ + a6 + 62 28. 1. 2c-5 c(2c + 5) 23. a(aj+2) a:(a:-7) 29. a — 6 -j-c — c? a -|-6 — c — d ALGEBRA. Art. 150 ; page 85. 2. ""-^ ' 3x + 7 ^ m-1 6m — 5 g 6 m — n {. 3cc — 2 5m — In x-\-3 3 5a + 7 a-2 5. ^ + 1 . 7 a; + 2 9 22/-3 a;-3 ' 2^-5 10- o"^. -a-1 Art. 155 ; pages 87 and 88. 5. s^-xy + y^ + ^—' 10. 2m^-5mn-\-7n^-- — x-{-y 2m — 3n 6. 2x + Q-^^. 11. a^ '^ 3a-5 a; -3 2^2 -a- 3 7. a-2+ ^^^-^ . 12. ^+1+ .^ + ^ » 8. 30^-5-^^. 14. 2a;-3 ^^±-5 4a;-l 3a^-2a;4-l Art. 156 ; page 89. 3 (^+^^ g 2ab ^^ a^-Sx" x a-\-b M gy^ -\-4x — l g 6xP — x ^c x-i-S ' ' 2aj+l* ^ 5m^— 2m7i— 4n^ -^ a^4-6^ -^ 3m+n a—b 6. §^zif. 11. _ly_, 17. 8 a;-)/ 7. -^^. 12. ^2!^. 18. m + w m-\-n a^ + 3a; — 2 13. ^^'. a + 6 x-2 2n^ w? + mn + T? 9a^ 2a;-l 6a;y 2y — x x'^-h^-l ANSWERS. 9 Art. 157 ; page 91. (a + 3)(a2-4)' (a + 3)(a2-4)* 7 «^ + a^ + 1 a; 4-1 g m?i (m^ — n^) m? (m + n) mn^ n (m^ — n^) n {m^ — n^) ' w (m^ — n g 2(a^+c^-& + ci6-+^') 3(a^-ft-^6+a6^-6-^) 4(a^- rt.4 _ 7)4 » „4 _ 7.4 ' ^4 _ ^ 11. 12. 2a^b m^ — n^ 2 a (a — b) (m -{- n) 2 a {a — b) {m -\- n) ar^-9 07^-1 (a; _ 1) (a; _ 2) (a; - 3) ' (a; - 1 ) (» - 2) (a; - 3) ' a^-4 (a;-l)(»-2)(a;-3)* Art. 158 ; pages 92 to 97. g 12a;+7 g 3mV — 4 q 4a6 — 6 — 4a^ 36 ' * 6m^n^ * • * 12a^b M 6a — 5b - 5 6- -f 4 g^ ^q J_ ^^^ m^ lOa'ft^ * ' 120a6 * * 15* * 42 24 * * 24 ' * 18a;2 • l« !_ ^- 4 bed + 6 acd — 3 abd — 2 abc 60 ' 4:8 abed 17. -?— 20. ^^±A'. 23. ^^. 18 5 2^ 4a6 g^ 3m^-{-n^ ' e + x-a^' ' a^-b""' ' (771+71) (m-w)' 19. !^ 22. ^ 25. 5 a^ + 15a; + 56 x + y (ci + 3)(a-2)2 26. ^^ 10 27. ^^+i. a — b 28. 0. 4a^ x^-y^ 30 2 34. Ax x{4:a^- 1) a^ + 3x-10 31. iab' c^-b' 35. 2. 32. {\-xy 36. 0^-2 a.(a;2-l) 33. 0. 37. 0. 38. (^^ + ^)^ 40 ^(^-^) . 39. 1^-1^^ 43. ^^-^^ (x + 1) (x + 2) (ic - 3) ab{a - b) 44. ^9a-l ^g 2m +n ^g 2 a 6(a— 1) m{m^ — n^) a-\-b 45. -A_. 47. L_ 49. ^ 9a;-a^ (aj+2)(a;+a) a;2_i 50. —-^ 51. 0. 52. - ^ a^ - 5a; + 6 (a;-2) (aj-3) (07-4) Art. 159 ; pages 98 and 99. 4. 1. 8.-^. 12. ^^-^' . 16. i±^. a 10 ax + &a; a; 5 1. 9 3^-1 iQ ^y 17. —^ — aa;— 2 a 18. 6. -^. 10. ^-^-^^. 14. »• -2 a^- -aj-20 a^ a - -26 4a;2/ a^ a + 1 x—y+z 19. a^ + a;?/. 7. a«. 11. ^nl-^ 15. ^^±1^ a^ a^+2a; 20. 1. Art. 160; pages 100 and 101. 3. —^—. 4. ^V , 5. « + 6 , 106%V ' oma; ' a4-4 ANSWERS. 11 » X' -i^x — 20 Q m? — mn + n^ ^- Sx — 2y a; — 2 m^ — 7^71 * x + y 12 ^^ '^^ . 13 2a; — 3y * aH-6 * 2x-\-^y Art. 161 ; pages 102 and 103. 4. x-1. 10. ^!i±i?. 16. °-^-''- 6. -i^. 11. ^ii5. 17 1 a + 6 a; + l l+ic^ 6. o?-x-\-l. 12. ^.^Ili?. 18. ^2jzM, 7. a-l. 13. ^^^. 19. ^ a 3a; + 3 8. a?-2xy, 14. ^~^, 20. 1. xy 9. ^. 15. X. 21. ^-« a;-f-6 x-\-2a 22 ^^ 23 2 (^ — ^) • a2 + 62* (m + w)^' Art. 162 ; pages 104 and 105. I hx — a o m — 1 g "^ a;2 3m -15 (1+a;)* 4 ah-\-h\ 5 __E!zi1__. am + an * 2 (1 + a;)(l +a^) g 2(a4-6) 9 a_^. -lo 5a;2 7. ar^ + 1+1 10. ^ 1 + a^ ax + 6 8. _!-, 11. * A2_36 + l. 18. a3H-3a-^-3a2c+3«62-6a6c+3ac2+63-362c+36c2-c3. 19. x«-6ar^+18a;^-32a^ + 36a^-24a; + 8. 20. ic^ + 9a^ + 30a;^ + 45.'c^H-30a^+9x+l. 21. 8x^-36x'-\-i2x*-\-9a^^21x'-9x-l, Art. 196 ; page 164. 10. 16-^S2x-\-2ia^-\-8x'-j-x\ 11. x'-lQx^-{-d6x^-2o(jx-\-2D6. 12. a^-15a^ + 90a3-270a2 + 405a-243. 13. a^4-10a^ + 40a3+80a- + 80a + 32. 14. x"^ - 12x^ -\- 60x* - IGOaf -\- 240x^ - ld2x -\- 64:. 16. a^ - Ua'x + QOa^a^ - 270aV + 405a«^ - 243ar*. 17. 81 + 216& + 21662 4-96&«+166^ 19. a^-16i»^+9Ga;«-256a:^ + 256. 20. 64a^- + 192a^«6 + 240a852+160a^6H60a*6^4-12a^5'+6*. 21. 16m'' - 96mV + 216mV - 216m%« + 8171^. Art. 204 ; page 169. 3. a2-2a + l. 6. 5-^3ff + xK 9. 2a'-5ab-\-8b'. 4. 2ar^-a;-l. 7. 3a^-4a;-5. 10. l-7«-2a^. 6. S-2x + x^. 8. 7^ + 1--- m 11. a — b — c. 22 ALGEBRA. 12. x-2y + Sz. 13. 3ar^+5ar^-7. 14. 4c3-5c-3. 15. a3_2a2 + 5a + 3. 16. 2oif — x^y — xy^ — 2y^. 17. 2a^-3a^+4a;-5. 18. a6 H 3 ^2 19. 3x»-2a^2/ + a;i/2-22/^ 20. 1+^-^4-^. 2 8 16 21. i-a-^-^ 2 2 22. a_26-^^-^^ 36^ 36' 2a a^ 23. 2a. + X_^ + ^L 2a; IGa;'^ 640.-^ Art. 207 ; pages 172 aftid 173. 2. 214. 10. 21.12. 19. 5.5678. 27. 1.3229, 3, 523. 11. .04738. 20. 4.1593. 28. .43301, 4. 809. 12. 900.8. 21. .83666. 29. 1.0541, 5. 5.76. 13. .8253. 22. .28284. 30. .44721 6. .497. 15. 1.4142. 23. .37947. 31. 1.1547. 7. .286. 16. 1.7321. 24. .031623. 32. .64550, 8. .722. 17. 2.2361. 25. .079057. 33. 1.1726 9. 1.082. 18. 3.3166. 26. 1.4444. 34. .94868, 35. .62361. 36. .42492. Art. 209 ; page 175. 7. a^ + 2x -4. 10. 2x^- -3a;- -1. 13. x^-i-xy -2y\ 8. y'-y- -1. 11. 2a'- - a — 5. 14. 3a'-2ab-b\ 9. x^-2x + 1. 12. 2 -a; + 2 ar^ Art. 213 ; pages 178 and 179, 2. 31. 8. 2.02. 14. .898. 20. .3107. 3. 4.6. 9. 372. 15. 101.3. 21. .7211. 4. .88. 10. 21.6. 16. .0534. 22. 1.077. 5. 123. 11. .803. 17. 1.260. 23. .6376. 6. 1.14. 12. 4.89. 18. 1.817. 24. .8736. 7. .098. 13. .317. 19. 1.931. ANSWERS. 23 Art. 214 ; page 179. 1. 2a;-32/. 2. a^-x-{-l. 3. a^-2a;-2. Art. 219 ; page 182. 21. 243. 23. 216. 25. -243. 27. 128. 22. 81. 24. 32. 26. 16. 28. 1296. Art. 224 ; pages 184 and 185. 10. 4mi 11. QaS. 14. 5aj~3. 15. Sahj. 17. a4_2_|.a-*. 18. a'-a^. 19. oj-^-l. 20. X ''^Sx-^-4:X \ 21. 18a26- + 10 + 2a-26-2. 22. 6a.'2-7a;3-19a;^ + 5a;H-9a;^-2a;i 23. 2x-^y-10xy~^-\-8x^y-\ 24. 2 -4a~^a;^ + 2a"^a!3. 25. 18ab~^-23+a''h-\-6a-^b\ Art. 225 ; pages 185 and 186. 5. 3c-l 6. m'A 7. a;ii 11. J-{.ah^ + b^. 12. a"^-a~^ + l. 13. ic-^-9-lOa;. 14. a;"^ + l. 15. m^-2m^n*+rii 17. a^ -\-ah^-^bK 16. x~^y'^ — x-^y-^ — x-^y-\ 18. m~^ — ri-^ + m^w-^ Art. 227; page 187. 10. f. 13. c"^. 14. w 1. 15. x~K 16. a-. Art. 229; pages 188 and 189. 8. Sx'^-2x ^-1. 16. x^-Sx^-\-2xK 21. a'^. 9. 2x^-{-x-4:XK 18. a^*-"*. 22. a;. 10. a^6~^-2 + a-*6i 19. x'^^'^K ^, 15. 2y-^^-y-K 20. a:« ^ 2a^ 24 ALGEBRA. 24. cfil^-c^l^, 26. -^. 28. ^^^. 25. 1+a;. 27. (1 - Sa^ + a.-^)"'. 29. (l+ar')i Art. 235; page 191. 9. "V^. 10. V2a. 11. Vlwh^. 12. V5^. . 13. V^. Art. 236; page 192. 12. bxy^xy''-2x'y. 15. (.t-3)V«. 13. ^ah^'2aJ)' + bh. 16- (2.^•^-3)V5. 14. (a? + 2/) Va; — 2/. 17. (m — 97i)V3m. 19. iv«- 21- ^V21- 23. |-?/6i. 25. 1V7. 77 20. Iv30. 22. ^V3- 24. i^l.5. 26. :^,Vl06cci. 27. ^Vl4^. 28. 2(^ + ^) • 29. ^j^^^- Art. 237; page 193. 10. V^f^l. 11. Vl^^. 12. \^r^- 13. V4-4a^. Art. 238; pages 194 and 195. 3. 5V3. 9. (2a+3&)V^. 14. 5^2 + ^-^18. 4. 7V6. 10. 9V3-7V5. 15. 9^3. ^ 5. 3V5. 11. 9^2. 16. 6aV3^. ^' f^' 12. ^V^- ^'^- (3ci2-263)V^rr26. 7. 20 V2. 5 jg^ 10^2-2-^3. 8- ^- ^^- ^V6. 19. (a-4)V7^. 15 36 20. 2Va^-/, ANSWERS. 25 Art. 239 ; pages 195 and 196. 2. ^^4, ^27. 4. ^625, ^216, ^49. 3. ^"125, ^16, 7. ±2, ±V11- 14. -2, -^15. b. 3, --, 15. 0, 2, -, -. 9. 2, -3, 4, -5. 16. 1, -1, -6, -8. 10. 1, 2, -5, 8. 17. 0, -5, -, -— . 3 3 18. a + 6^ a4-3-\/36"^ ANSWERS. 31 Art. 274; page 234. Note. In this, and the three following articles, the answers are arranged in the order in which they are to be taken ; thus, in Ex. 2, the value a: = 1 is to be taken with y = —2, and x — with y = —. 25 ^ 25 2. a^ = i, -^; 6. .T = 3, -4; 11. . = 3,-1^; 25 .> 46 2/ = 4, -3. 13' -, 62 2/ = ~^9 :;7* 2/ = 2, — 25 7. 05=2,3; ^ 13 3. x=l, -8; y = -3, -2. 12. . = 1,-1; 2/=-8, 7. 8. a; = 5, -2; 2' 7/ = 2, -1. 4. a; = 9, -6; 2/ = 6, -9. ^ = -2,5. 13. -».f^ 5. .= 2,-1; 9. a; = 4, 2/ = 6. '-.-f A 5 10. a;=6, -4; 14. a; = 4, -2; 2, = 4,-. 2/ = -4, 6. 2/ = 8, -1. Art. 275 ; page 237. 4. a; = 3, -2; 10. x = b^ — 3 ; 15. 05=7, -12; 2/ = -2,3. 2/=3, -5. 2/ = -12, 7. 5. ic=9, —3; 11. a;=2, 1; 16. 05=15, -3; 2/ = 3, -9. 2/=l,2. 2/ = -3, 15. 6. a; = -3, -7; . = 8, -y, 15 c 2/ = V ~^- 2/ = 7, 3. 12. 17. 05 = -l, -4 7. a;=2, -3; 2/ = 4, 1. 2/ = -3, 2. 2 8. ic= ±4, ±3; y= ±3, ±4. 13. 05= ±7, ±6; 2/= ±6, ±7. 18. 05=6, -11; 2/=ll, -6. 9. a; = 3, -7; 14. 05 = 3, -7; 19. 05 = 7,-9; 2/ = -7, 3. 2/=7, -3. 2/ = -9, 7. 32 ALGEBRA. Art. 276; page 239. 2. x=±7, ±1^2; 7. x = ±3, ±|; 2, = ±2, t|v2. 2/ = ±5, ±^- 3^ ^ = ±1,^19. 3 9. x=±2, ±|fV31 4. x=±2, ±|V5; 2, = ±3, T^V31. i/=t3, ±|V5- 5. a;=±2, ±ivi^; 10. a; = ±2, i^Vll 2 5 11. a5=±l, ±2; 6. x=±3, ±2V2; y=±-, ±-- 2/ = ±l, ±V2. ^' ^ Art. 277; pages 242 and 243. 6. ^ = -5, --; 10. x=2m, -m; ^ y — m^ —2m. y=l, --• 3 11. a; = 3, 2, -3±V3; 6. a;=l, 8; 2/ = 2,3, -S^V^- 7. . = 4,-4; 1^- -=±2, ±-V5; ^ = ^''=''- . = :F3,TiV5. " 8. a; = 2, 3; 5 2/ = 3, 2. 13. a; = 8, 4; 20 V = 4, 8. 9. a: = -2, ^; ^ 8 14. a; =±2, ±14; 2/ = -^'n* 2/ = T3, :f5. ANSWERS. 33 10. flj = -, -; 18. x = -.-: 2 9 4 7 = 1 1 -_1 _1 ^~9'2* ^~ 7' 4* 16. x = 2a, —a — b; 19. x=S, 4, — 4±V— 11 ; y = a + b, -2a. 2/ = 4, 3, -4:f V^Hl. 17. x = -, -3; 2^^ a; = ±3, ±2; 2/=±i, ±V3. 2^ = :f 2, 1:3. 21. 05=1, -3, 1 ±V-2; 2/ = -2 3, 1, lq:V-2. 22. .= l,2,^±^-^^,= ~3±V-55 ' ^' 2 23. a; = 2 a, — a ; 32. ,_2,i, 3±V-19 y = a, — 2 a. 2 ^ 24. ic = 4,9; 3TV-19. y = 9, 4. 2 25. x= a —b, b — a; 33. 05-2,--, y = a-\-b, 2a. 2/ = -l, 2. 26. 05=2,3; 34. ic = 4, i; y = 3, 2. 9' 27. a; = 6 ± a ; 22 59 2/ = — 6 ± a. 2=3, — 28. a; = 4, 16, -12±V58; 9 2/=5, -7, -IqiV^S. 35. ^1 ^59 29. 0^ = 9, 5^2_g^; 4 31 . 20 2/ = t3, ±^- 2/ = 4, 8 ^ ' 117 a5=±2, ±iv7; 36. 05 = 3, -7; 30. 2/ = -l, -21. 2/=±5, iFyV7- 37. . = 3,2,-^^/^^^ 31. a;=2a — 6, a-26; 2/ = a — 26, 2a — b. ,= 2,3,-^^V309 34 ALGEBRA. Art. 278; pages 243 to 245. 1. 9, 5. 3. 900 square rods, 400 square rods. 2. 13, 6. 4. 3, 4. 5. Length, 10 rods; breadth, 6 rods. 6. 7, 4. 7. Duck, $1.75; turkey, $2.25. q _ig 9. - or 8. 21 or 12. 5 24 10. Length, 150 feet; breadth, 100 feet. 11. Length, 16 rods ; breadth, 10 rods ; or, length, 13i rods ; breadth, 12 rods. 12. Rate of the boatman, 4 miles an hour; of the stream, 2 miles an hour. 13. A, 40 acres at $8 ; B, 64 acres at $5. 14. 7, 5. 15. 5, 2. 16. Hind-wheel, 4 yards ; fore-wheel, 3 yards. 17. First rate, 7 per cent ; second, 6 per cent. 18. Length, 16 yards; width, 2 yards. Art. 281 ; page 247. 9. x^-9x = -20. 15. 6^2 + 31 a; = -35. 10. x' + 2x = S. 16. 3ic2-f 17a; = 0. 11. 5a;2-12a; = 9. 17. ;^-2ax -bx = -a'-ab+2b^. 12. 3a?2_2a;=i33. 13. 12a;2-17a; = -6. 14. 21ic2-|-44a; = 32. Art. 282 ; page 249. 6. 0, i. 9. 2,-4,-5. 16. -1, 2, ±3, ±4. 17 1 1 ±^^-S 17. -1, ^-— 3 _3±3V-3 18. X- — 2 mx =m'^ — m^. 19. X'-4:X = -1. 20. 4 a^ — 4 mx = n — m?. 10. 0, ±3. 12. 0,-4. 13. 14. . ^ a±Va2 + 4& ±a, 2 ""U' 3 18. 19. ±1 2' 4 ±1 ±V^^ ANSWERS. 35 20. -1,±V^. 21. 1, ±3. 22. 23. a, ±aV-l. 5. 6. 7. 8. 9. 10. 11. 12. 25. ±1. 24. ^ ±1 Art. 283 ; pages 251 to 253. (x-{-6){x-10). (x + 8)(x-\-5). {x-2){x-d). (2x-{-S){x-6). {4:X-S){x-3). {x-^7){6x-^l). (13+ a;) (3 -a;). (1 + 2a;) (2 -3a;). 13. 14. 15. 16. 17. 18. (a; + 2+V3)(a^+2-V3). {Sx-l-{-^o)(Sx-l-^5). (4:X-S){'2x-3). {2-\-x){S-2x). {l-h2x){5-Gx). (3x-2 + V3)(3a;-2-V3) 19. {5-{-2x)(l-4x). 20. (10x-3)(a;-2). 21. (2a; + 3)(8a;-|-5). 22. (V17 + 4 + a;)(V17-4-a;). 23. (5 + 12a;)(3-2a;). 24. (5a;-2 + V6)(5a;-2-V6). (6a; 4- 5a) (.T- 3a). 27. {ix + 5y)(Sx -2y). (4a; + 5m) (5a; -h 4m). 28. (7a; — 3m7i)(3a;— 7m7i). 29. (x^Sy-2)(x-2y-\-3). 30. (a;-h22/-h2)(a; + 2/-hl). 31. {S-\-2y-x)(2-Sy + x). 32. (x-2y-4:z)(x-Sy-z). 33. (2x-\-y-l)(x-y-{-2). 34. (a + 6 + 3)(3a-f-6-4). 38. (x'-\.x + \)(xP-x+l). 39. 40. 41. 42. 43. (a^-\-3x + l)(a^-3x+l). {2a'-h2ab- b^) {2a^-2ah- b^), (m^ -|- 4 mn -j- n^) (m^ — 4 mn -\- n^) . (l_|.36_262)(i_35_252). (^x^ ^4a:y + 2y'){a^ -4.xy -\-2f). 36 ALGEBRA. 44. (2a2 4-2a + 3)(2a2-2a + 3). 45. (2m2 + 2m-5)(2m=^-2m-5). 46. {d'-^ax^S-a^){a'-ax-^S-x''). 47. (x^'^Sx^2 + d){x^-Sx-^2-^9). 48. (2a2 + a& + 4&2)(2a2-a6 + 462). 49. (4 aj2 4- 5 7,ia; -Sm-) {4a^ - 5mx -3 m^) . 50. (3ic2 + 3a;V2 + 2)(3.T2_3^^2 + 2). 51. (3a2 + 4am + 5m2)(3a2-.4«m + 5m"). 52. (2 + 2w-7n2)(2-27i-7n2). 53. (4:X^-hSxy-5y')(4x'^-Sxy-6y'^). Art. 284 ; page 253. 1. ±V2± 2. 4. ±V2±V-2 2. ±1±V2. 5. ±1±VS ;. 3. ± V3 ± V- 2 -1^ Art. 6. ±^1^± 4 295 ; pages 259, 260. V- 2^ 3. a;<4. 7. x2, y>4. 12. 19. 5. ^>lf 9. a;<24, y>S. 13. 32 or 33. 6. a;<2a. 10. Art. a;> 5 and < 15. 325; pages 274, 275. 3. '■ '-s- 7. 12. 9. 5^3. 11. . 2,-3, 4. ■•■ «■ p- 8. 14 V3. 10. 3,-1. 12. -!■ 13. a;=±6, 15. 25, 20. 18. 8, 18. 14. 2/=±10. x = ± a'b, 16. 23, 27. 19. 26, 14. y = ± ab^. 17. 9, 3. 24. 17, 12. ANSWERS. 37 ^5. 12, 8. 26. A, $105; B, $189 ; C, $270. 27. 8:7. 28. First, 1:2; second, 2 : 1. 29. ^^, ^^. Art. 335 ; pages 278, 279. 3. 63. 8. 21 lO' 13 . 10 inches. 4. .=i'- 9. ±|V3. 14 . 6 inches. 5. 6. 10. 5xand-l X 16 . 10. 6. 70. 11. 9. 16 . 8. 7. 14. 12. 3(V2— 1) inches. 17 14 4 — 5a; Art. 340; page 281. 2. 3. S = 540. Z = -69, 6. i = f, 9. >S = -620 . s = o. 10. 1 = 5, 4. Z=57, S=17. /S' = 552. '■ '=!■ 11. J 137 ^" 15' 5. ; = -i45 >S' = -217 5. -!•■ « 917 ^^=15- Art. 341 ; pages 283 to 285. 4. a = 3, S= 741. '• "-h 9. a= 5, d = -S. 5. a=li, Z = -12^ 1 -. l=-li. 10. 71=16, I =-4:3. 6. '=h 8. d = -2i, 11. n=18, ;S = 39. n=13. /S' = 411. 38 ALGEBRA. 12. a = 3, 15. d = -l, 18. n = 14, ^ = -^^- n=.U.^ ' = -151- 13. a = i -I; 16. d = 5. 20. d '-« 21. d = g('^-«») »=10, 12. ; = ^. 14. a = -l, ,, ^* , M"-l) 2' 17. a = 4, — 5; d=2. ?i = 52, 43. ^ = 22^ ^^ 2S-n(n-l)d 2S-an 2n 24. a = Z-(n-l)cZ, OQ l — a-^d OK 2S — nl 23. n = ! — , 25. a = , d n j^^ {l + a)(l-a + d) ^^ ^(nl-S) 2d ' n{n-l)' 26 ;^ -^^ v'^^^+(^^-^? - It 07 ^ ^'-a' OQ 2Z+d±V(2/+d)2-8d^ 2/S — a — Z 2d 2S ^^ dq: V(2Z + d)^-8d>S a + Z* 2 * Art. 344 ; page 286. 1. d = l. 3. d = -\\^ 5. d = -i. 7. 2a6. Q 2. d = -i. 4. d = i 6. -i. 8. ^!±^'. 2 7 15 a^-V" Art. 345 ; pages 287, 288. 3. 2500. 6. 11. g 23a^ 4. -43. 7. Z=10m-27n, ' 3 * 5. 4,11,18,25. >S'=55m-135n. 9. 62750. 10. 2, 6, 10, 14; or, -2, -6, -10, -14. 11. 22. ANSWERS. 39 10 A ^ o K Q 59 165 11 7 12. -4, -1, 2, 5, 8; or, —,—,—,-—, --. 14 7 14 7 2 13. After 9 days, at a distance of 90 leagues. 14. 4117^ feet. 15. 3, 7, 11 ; or, 4|, 7^, 10^. 16. 8. 17. $2950. 18. 852. Art. 349 ; pages 290, 291. 3. Z = 256, 7 ^^JL_ 10 Z=— L, >S'=511. 2048 324 4. ,^_6i, ^^2047. ^^IL. 243 2048 162 Z=192, ^^2059 g ?==-I?2, 11. Z=192, 243 64 6. Z = 2048, ^^ 1261 >S'=:1638. 192* 6. Z = --i-, 9. 1 = —, 12. Z = ?-, 256 32 768 ^^341 ^ = yi. S = -—' 256* 32 ' 256* Art. 350; pages 292, 293. 3. a = i, 7. a=-, 11. a=-l, 2 3 ?i= 8. 12. ^^ a+(r-l)5 r ^^1023 ? = -l_ 2 * 6561 4. a = -?, 8. r = -^, 2 4 _ ^-a ^=48. S = ^^ ' " S-l' 1024* 5. r = 3, -3; 9. r = i, H. a=rM--l)^. /S'= 2186, 1094. ^ ?i= 7 6. n = 5, 10. l = - 15. a = -?-, 243 r"-^ >S'=121. 2 ' ^^ l(r--l) n = 6. r" \r — 1) 40 ALGEBRA. 16. „=(!lrl)^, i^r'-Hr-i)S^ r" — I r"" — \ 17. r = -v-' ^=irT7^ — ;r-i7 — Art. 351 ; page 294. 2. 4. 4. -i 6. 5. 8. -1^. 4 4 19 3. 1 5. -1^. 7. ?5. 9. ^ 3 4 11 a2 + aj2 Art. 352 ; page 295. ^ ii. 5. i^. 6. 11 ' 27 15 165 825 1100 2. £. 3. il. 4. ii. 5. i^. 6. ^. 7. 2^^ Art. 355 ; pages 295, 296. 1 2 - - — — — ' 3' 9' 27' 81' 243* o 4-5 9^_ 27 81 243 * 2' 2' "^ 2 ' 2 ' 2 * 3. -6, -18, -54, -162, -486, -1458. '^'i' 8' "^16' 32' 64' 128' 256* 5. ±4, -8, ±16, -32, ±64. 4' 16' 64' 256 ^ 6 Art. 356; pages 296, 297. 2. 3. 3. 5, 10, 20, 40; or, -15, 30, -60, 120. 4. 5, 15, 45 ; or, 40, - 20, 10. 5. ±4. 6. $64. .y QiAA^ ^ Q o . o 1^ 810 540 360 240 7. 3100 feet. 8. 2, 4, 8, 16 ; or, — , — • , , ' ' ' ' ' 13 13 13 13 9 ._§L. ^^- 3,9,27. 12. 1, 2, 4. 8192* 11. 2,4,8; or, - 2, 4, -8. ANSWERS. Art. 361 ; page 300. 3. 3 A 1 K 5 ft — 4. 0. b. 74 78 4 3 4' 7. - 8. 48 24 16 12 48 8 48 125' 65' 45' 35' 145' 25' 155 11. 15. 9. 40 20 40 ~r7' "T' ~n' ''• 12. a' - b' d'-\-b' 10. 21 7 21 21^ 7, -21, --, -, ^, ^^ 13. a6 j4 ab(m-^l) 7ia — nb — a-i-2b * 6m + 2 a — 6 15. 3 19* 41 3 142* Art. 365 ; pages 303, 304. 2. c^ + 4c2(r^ + 6c*d~^H-4c'cr^ -f «^~'^. 3. m~^ — 5m V + 10?/r^n* — lOm-^71^ + 5m~^n' — n^\ 4. oi^y^ — dxy~^ -\-21 x~^y — 27x~^y^. 5. .-r"" + lOx^'^y'' -f 40ar^ V + SOar^-^" + 80a;V* + S2y^'\ 6. a'^-{-12a^x^ + 64:a^x-{-l08a'x'-^8\x^. 7. m*n-^ — 6m^n~^ + 10m%~2 — lOm^n"^ + 5 m^n — m^nK 8. xy-^-{-3y-'-hSx^-^x-Y 9. 7M* — 2 m^rr' + - m*?i^ 7n^n^ H n^^. 2 2 16 10. ah~^ -5ah-^ + 10a''6"L I0a~^6^ + 5a'^6 - a'^di 11. a«-12a'^'4-54a"'^'-108a^4-81ai 12. 16a;V^ + 16a;V^ + ^^ + aj"V + -^^"V* 16 13. a-^2 - 2 a-%^ + -a-«a; _ ^^a-^ajt + -^a-^ar' 3 27 27 81^ "" +729 ' 42 ALGEBRA. 14. a^4-15a;"'^V^+ dOx^y~^ -{-270 xhj~^ -\- 4:05 x^y~'^ + 2^3 y-\ 15 . I a^b^x~ ^ - - b^x-s + 6 a-^b~'^x^ - 8 a. %~^x. 8 2 16. 81 a-%^ - 108a-26 + 54a-i - 126-i + ab~\ 17. a^ft 3 _|_ 12 a25-2 + 60a6-i + 160 + 240a-i6 + 192a-252_,_ 34^-3^3^ 18. l-4.x + 2x'^^x'-bx^-So^ + 2:^-\r^x'' -^-a?. 19. a^ + 4a;7— 2a;«-20af + a;^+40aj3-8a^-32a;4-16. 20. 1 +8a;4-20ar'4-8a.-3-26ic^-8a^+20a;^-8a;^ + a^. 21. l-5a;+15.^2-30a^ + 45a;4-51a^ + 45a;^-30a;^ H-15a^-5aj9 + a.'^°. Art. 367 ; page 305. 2. 462 aV. 6. 84a-36'. 9. IbmOx-^y^. 3. 4. 5. 252 m^ - 792 c^cf. 1001 a«. 7. 715 a;". 8. -^a-'b\ 16 10. 11. -4455a'^V^ 126720. Art. 378 ; pages 311, 312. 2. l-2x + 2a;2_2aj3_^2a;^ 3. 2H-lla; + 33a;2 4-99x3 + 2970;*+. •. 4. 3 -19a;2_^ 95^.4 _ 475^6 _j_2375a^ 5. ^x + ^o^ + -x^ + -x' + —x' + "' 3 9 27 81 243 6. l-2a; + 2a;3_2a;4 4.2a;6 7. x-Q(?-2;K?-bx^-\2a^ 8. 2-a; + 3a;2_^^3^4 9. l-2a; + 5ar'-16ar^ + 47a^ 10. 2-7a;+28a;2_9i^^322a;4 ANSWERS. 43 ie 2 _2 , 8 _i . 32^128^ ,512 «, 3 9 27 81 243 16. x-^ + S-h2x-6x'-Ua? 17. x-^-x-^-2x+2x^-4a^-\-'" 18. ^x-'-lx-'--x''-\- — -\-—x + "' 2 4 8 16 32 Art. 379 ; page 314. 2. \^x--x'-\--a^-^x*+'-' 2 2 8 3. X-^-x-^-x^-'^x^-'-^x^-.. 2 8 16 128 4. \-x-{-x^ + a^-h-x^-{-'" 2 8 16 128 6. \-lx-^-x?-^x''-^x^->>> 3 9 81 243 7. \J^\xJr^-^-—^^—x'^-- 3 9 81 243 Art. 381 ; page 316. 3. 4. 6. ^ 1 ^ 2aj + 5 2a;-5 3 5 X 3x4-5 5 1 1 6. -? + ^-4-^- a; 3a; 4-1 2a; -5 - 3a 2a a;4- a a; — 4a 8. -^ + ^. 2a; a;4-3 a;-3 3 + 4a; 3-a; 44 ALGEBRA. 9. -i-+ 2 3 10. 1 1 3.3 x-{-2 x — 2 x-\-l x—1 11. 1+V2 I 1-V2 2a;-5-hV2 2a;-5-V2 Art. 383 ; page 317. 23 3. -i-+ ^ oj-f 5 (aJ + 5)2 07-2 (ic- 2)2 ■ (£c-2) 4 3 6 1 a; -1-1 {x-\-iy (0^4- 1)^ 2 4 3 3a; + 2 (3a; + 2)" (3x+2)3 1 4 5 (5a; -2) 5 (5 a; -2)3* 7, -^+ 1 1 1 ic + i {x-hiy {x + iy (x-^iy Q 2 4^3 1 9. a;- 1 (a;- 1)2 (a;-!)^ (x - ly 1 27 27 2 (2a; -3) 2(2a;-3)« (2a; -3)^* Art. 384 ; page 319. 2. ? § 5_. 3 5_2 4: ,i; + 2 (a; + 2)'' a; re« ai+l (a; + l)' 4. -J '—+ ' x-2 2 (2a; -3) 2 (2a; -3)' .5. i+^+-i,+ 1 a; a;-l a; - 2 (a;-2)^ 6. 1-1 + 4- * a; ar^ x^ a;-f-5 7. 5-1 + 1==. ^ X x^ s^ x-\-l (a; + l)' ANSWERS. 45 Art. 385 ; page 320. 1. 2aj-5 !^+ ^ 2a;— 5 2a; + 1 2 3_^ 1 5 18 a; + 2 (.^ + 2)2 (a; + 2)' 3. 5a;2 4.1_l+ 3 1 X x^ X' x-ir\ 4. 3a; -2 ?— _^ ? — -+ ^ 4. ^ a;-f-l (a;+l)- a; - 1 (a;-!)^ 6. 2a;2_7_2_^i 5 X x^ x—1 Art. 387 ; page 322. 2. x=^-y-{-^y^--l-f--^y^^... ^^ 27^ 243^ 2187^ 3. a; = 22/4-6/4-Y/ + 9S2/' + - 4. a; = (2/-l)- I (y-l)2 + i(2^_l)3_ 1(2,-1)44..., 5. a; = 2/ + 2/' + 2y' + 5/+... 6. ^ = 22/ + |2/^ + ^2/3^464^.^... 7^1 5 , , 29 , 199 . 2/* + 8. . = , + 1/+^^ + !^,. + ... Art. 392 ; page 327. 7. a^-\a-^-x--ar'^x'-~a-ix? La-^a;*-... 2 8 16 128 8. l-la; + ?a;^-iia;3 + l^^._.,, 3 9 81 243 46 ALGEBRA. 9. a-^ + 3a-*x + 6a-^x^ -j-lOa-^ar^ -i-loa'^ X* -\- '" 10. c"^ - c-^d + c~^d- - G-^d^ + c-~^'d' 11. x~^ — 2x^y — x^y^ x'^y^ x^y^ — ••• o o 2 8 16 128 13. a-s - a'h-' + -a-'^"b'' --a-"^ b-' + — a-'b ' 4 2 16 14. a;3 + dx-^ab - -x-'a'b' + -x^'a'b' - Maj-i^a^^^ + ... 2 2 8 16. l-10xy-' + 80x'y-'-^x'y-'-^^^x'y-'-'- o o 16. a*+12a^2/-2+90a^2/'' + 540a^?/ « + 2835a«2/~«4---- 17. 12Sa^ + lV2a'x-^ + S6a^x~^ -h^ax-'-{-■P^a-^x~^- 8 256 18. m + 3m*n^ + -m^n^ + 5^m^n'+^m'^V+- ^ Z o Art. 393 ; page 328. 2. JLa-^a;^ 6. -^^Ma-^^^t. 10. SOOSn^'c-'^. 2048 256 «^. II » 44 JL4 „ ., 308 -34 3. -364m". 7. x^ y-\ 11. —a ^ x-\ 6561 3 128 8192 ^ ^ 5. --A_a-^a^. 9. -ll-a,-'?'a;i«. 1024 256 Art. 394 ; page 329. 2. 3.16228. 4. 2.08008. 6. 2.03055 3. 9.94988. 6. 2.97182. 7. 1.94729 ANSWERS. 47 Art. 407 ; page 333. 2. 1.3222. 7. 1.9912. 12. 2.1303. 17. 3.0545. 3. 1.7993. 8. 2.0212. 13. 2.2252. 18. 3.7114. 4. 1.7481. 9. 2.0491. 14. 2.1673. 19. 3.8484. 5. 1.9242. 10. 2.1582. 15. 2.5741. 20. 4.1585. 6. 1.6532. 11. 2.3343. 16. 2.5353. 21. 4.1915. Art. 409 ; page 334. 2. .3680. 6. 1.5441. 8. .2252. 11. .8539. 3. .1549. 6. .1182. 9. 2.2431. 12. .7660. 4. .5229. 7. 2.0970. Art. 412 ; 10. 1.0458. page 335. 13. .7360. 3. .2863. 9. 4.5844. 15. .1165. 22. .2601. 4. 2.7090. 10. 3.2620. 16. .3860. 23. .6884. 5. 4.2255. 11. .9801. 17. .2212. 24. .1840. 6. .1398. 12. .4225. 18. .1750. 25. .2215. 7. .7194. 13. .1590. 20. 2.6145. 26. .2494. 8. .6611. 14. .0430. Art. 414 ; 21. .1678. page 337. 27. .1449. 2. .2552. 7. 7.7; 323-10. 12. 2.4804 3. .3522. 8. 6.4983-10. 13. 8.7905 -10. 4. 9.2922- -10. 9. 3.8663. 14. 6.3588 5. 8.6811- -10. 10. .60 74. 15. .1964. 6. 1.5841. 11. 9.6511-10. 16. .1688. Art. 420; page 341. 7. 9.8878- ■10. 11. 1.3028. 15. 0.7144. 8. 3.0237. 12. 4.9659. 16. 3.0155, 9. 0.5177. 13. 9.6055-10. 17. 8.9379 -10. 10. 8.7164- •10. 14. 7.8560-10. 18. 9.0610 -10. 48 ALGEBRA. Art. 421 ; page 343. 6. 1.646. 10. .003318. 13. .2079. 16. 63329. 7. 8886. 11. 10221. 14. 44.48. 17. .01301 9. .01461. 12. 9.492. 15. .001109. 18. 502.9. Art. 426 ; pages 346 to 348. 1. 8.454. 19. -1.184. 39. .6443. 2. 10.73. 20. .000007038. 40. .5010. 3. - 2202. 21. 2.924. 41. 1.062. 4. .2179. 22. .9146. 42. -.9102. 5. .01157. 23. 4.638. 43. 1.093. 6. -.7032. 24. .0000639. 44. .7035. 7. 7.672. 25. 1.414. 45. .5807. 8. .6688. 26. 1.495. 46. -.6313. 9. -3.908. 27. -1.246. 47. 24.62. 10. 1782. 28. .6553. 48. .2979. 11. .3500. 29. .2846. 49. 98.50. 12. - .4748. 30. 2.372. 50. 1.660. 13. .4127. 31. -.5142. 51. 3.076. 14. -4.671. 32. .1588. 52. .8678. 15. .2415. 35. 5.883. 53. 1.134. 16. - .0725. 36. .7885. 54. .5881. 17. 13587. 37. 1.195. ' 55. 1.805. 18. .006415. 38. .6803. 56. 57. .003229. .03344. Art. 427 ; page 349. 3. .4581. 4. .1853 5. - .4949. 6. -.2601. 7. 8. m log h -\-n log a log a .l0g( -• 9 11 . 3, _i. 10. -3. log /-log a ^ log n — log m logr ANSWERS. 49 12. ^^ log[0--l)^ + «]-loga , 13. n = los: I — losf a + 1. log(>S-a)-log(6'-0 logr 16. 3.4598. 18. -3.467. 20. .9395. 17. -.1386. 19. 11.193. 21. -1.8204. 3. 4.479. Art. 437 ; page 353. 4. 7.19. 5. -1.07. 6. -2.4576. 1. $2853.75. 6. 14.198. Art. 439 ; pages 356, 357. 2. $702.86. 6. 16.01. 3. 5^. 4. 4. 7. $647.14. Art. 443; page 359. I. $2076.40. 3. $2959.18. 4. $277. 6. $576.50. Art. 452 ; pages 363 to 365. 3. 7893600. 5. 126. 7. 3838380. 4. 5040. 6. 15120. 8. 31824. 9. 360; 120; 720; 1956. 10. 134596. 14. 15840. 17. 10584000. II. 125970. 15. 121030. 18. 3303300. 12. 4536. 16. 10080. 19. 720. Art. 466 ; pages 375, 376. 1. 1 -\ ; 5th convergent, — -• 2+ 1+3+1+2 ^ ' 14 o 1 1 1 1 1 1 r.i 4. 15 2. ; 5tn convergent, — -• 1+ 3+1+3+1+3 ^ 19 50 ALGEBRA. 3. 3 H : 5th convergent, — 1+ 1+ 1+ 1+ 3+2+2' ^5 -11111111.,, ^8 4. ; otn convergent, — 1+ 2+ 1+ 2+1+2+1+2 ^ ' 11 6. 2 H :;— ;; I 5th convergent, — 3+ 2+ 1+3+2+1+2 ^ ' 37 o. 1 H ; 5th convergent, -• 1+1+1+1+1+1+1+1+2 ^ '5 ,y 1 , 1 1 1 1 1 1 1 . ., .24 7. 1 + : 5th convergent, — 3+ 1+ 3+1+3+1+3 "" 19 « 1 1 1 1 1 1 .., .68 o. — - — - — ; 5th convergent, 2+ 3+ 4+ 5+ 6+7 ^ 157 Q O L 1 1 .I^T, * 161 1 1 y. 2 H ; 4th convergent, ; , 4+ 4 + - ^ '72 27144 21960 10. 1 H ; 4th convergent, - ; — , — 1+ 2 + - *= ' 4' 60 44 11. 3 H ; 4th convergent, ; , « 3+6 + - ^ ' 60 ' 26340 22740 12. 2 + — — J: — ; 4th convergent, - ; — , — • 1+ 1+ 1+4+-' ^ 3' 51' 42 -q -3 4-V15 IK 3+V5 13. 15. — ^— . 14. -2 + 2V2. 16. -H-2V6. 17. 3 +— — —; 4th convergent, — • 7+ 15+ 1+- "^ 113 -g 111111 1 .76. 1 1__ * 2+ 3+ 3+ 3+ 1+ 1+ 7+- ' 175 ' 262325' 231700* 19 24-i- J--i--i--i- J-_L__L_. 19?. 1+ 2+ 1+ 1+ 4+ 1+ 1+ 19 + -' 71' 1 1 103589' 98548* 20. 3 +- J- -J— ; 5th convergent, — • 1+ 5+"' 41 UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which borrowed. This book is DUE on the last date stamped below. D 21-100m-ll,'49(B7146sl6)476 QA »^3 THE UNIVERSITY OF CALIFORNIA LIBRARY