IN MEMORIAM FLOR1AN CAJORI // SECONDARY ALGEBRA BY GEORGE EGBERT FISHER, M.A., PH.D. 1 1 AND ISAAC J. SCHWATT, Pn.D. ASSISTANT PROFESSORS OF MATHEMATICS IN THE UNIVERSITY OF PENNSYLVANIA PHILADELPHIA FISHER AND SCHWATT 1900 COPYRIGHT, 1900, BY FISHER AND SCHWATT. Nortoooto J. S. Gushing & Co. - Berwick & Smith Norwood Mass. U.S.A. PREFACE. IN the preparation of this book, the aim of the authors has been to give the student a working knowledge of the ele- mentary processes of algebra, with a conviction of the truth of principles through illustrations and particular examples. Each principle, or method, is therefore first clearly illustrated by numerous and simple exercises worked in the text. But the student is not left to assume that the principles are thereby proved. Even a beginner should not be encouraged, by text- book or teacher, to accept an illustrative example as a proof, or he will lose much of the educational value of the study. Particular attention has been paid to the grading of the exercises. The introductory chapter extends the familiar processes of arithmetic to the corresponding processes of algebra. The pupil is led by simple exercises, similar to those in arithmetic, to understand the use of letters to represent general and unknown numbers. Negative numbers are naturally intro- duced in connection with the extension of subtraction of arithmetical numbers. The meaning and use of positive and negative numbers, in the fundamental operations, are properly emphasized. Equations and problems are distributed throughout the book. The importance of equivalent equations is not over- looked, but is very briefly and simply considered in Chapter IV. Until that chapter is reached, the solutions of equations should be checked. iii iv PREFACE. All the matter in the book is printed in large type, and much pains has been taken to make the pages open and attractive. Any suggestions from teachers and others will be greatly appreciated. The authors have much pleasure in expressing their satisfac- tion with the excellence of the mechanical execution of the work, due to the ability and painstaking care of Messrs. J. S. Gushing & Co. and Messrs. Berwick & Smith, of the Norwood Press. G. E. F. I. J. S. UNIVERSITY OF PENNSYLVANIA, PHILADELPHIA. CONTENTS. CHAPTER I. PAGE INTRODUCTION ........... 1 Genera] Number 1 Equations without Transposition 9 Problems 10 Positive and Negative Numbers .18 CHAPTER II. FUNDAMENTAL OPERATIONS WITH ALGEBRAIC NUMBERS ... 24 Addition 24 Subtraction 25 Multiplication .......... 27 Division . . ... . . . . . .30 Parentheses .......... 34 Positive Integral Powers ........ 39 CHAPTER III. FUNDAMENTAL OPERATIONS WITH INTEGRAL ALGEBRAIC EXPRES- SIONS 42 Addition and Subtraction ........ 44 Parentheses .......... 51 Equations with Transposition, and Problems .... 53 Multiplication t 57 Equations and Problems 66 Division 69 CHAPTER IV. INTEGRAL ALGEBRAIC EQUATIONS . . . . . . .79 Equivalent Equations 81 Problems 84 v vi CONTENTS. i CHAPTER V. PAGE TYPE-FORMS IN MULTIPLICATION AND DIVISION .... 89 CHAPTER VI. FACTORS AND MULTIPLES OF INTEGRAL ALGEBRAIC EXPRESSIONS . 99 Integral Algebraic Factors ....... 99 Highest Common Factors 116 Lowest Common Multiples 118 Solutions of Equations by Factoring 127 CHAPTER VII. FRACTIONS 129 Reduction to Lowest Terms . . . . . . .131 Reduction to Lowest Common Denominator .... 133 Equations and Problems 136 Addition and Subtraction ........ 141 Multiplication .......... 149 Division 153 Complex Fractions ......... 155 CHAPTER VIII. FRACTIONAL EQUATIONS . 159 Problems . . . 162 CHAPTER IX. LITERAL EQUATIONS 167 General Problems 169 Interpretation of Solutions 172 CHAPTER X. SIMULTANEOUS LINEAR EQUATIONS ....... 177 Elimination by Addition and Subtraction ..... 178 Elimination by Comparison 180 Elimination by Substitution . . . . . . . 182 Linear Equations in Three Unknown Numbers . . . .183 Problems 192 CONTENTS. vii CHAPTER XI. PAGE INEQUALITIES . . 204 CHAPTER XII. INDETERMINATE EQUATIONS . . . . , . . . 210 CHAPTER XIII. INVOLUTION 213 Powers of Binomials 215 Powers of Multinomials . . . . . . . ..217 CHAPTER XIV. EVOLUTION 218 Hoots of Monomials 221 Square Roots of Multinomials . . . . . . . 222 Cube Roots of Multinomials . 226 Roots of Arithmetical Numbers 228 CHAPTER XV. SURDS 235 Reduction of Surds 236 Addition and Subtraction of Surds 239 Reduction of Surds of Different Orders to Equivalent Surds of the Same Order 240 Multiplication of Surds . 241 Division of Surds . 246 Rationalization 247 Surd Factors 248 Evolution of Surd Expressions 250 Properties of Quadratic Surds ....... 251 Square Roots of Simple Binomial Surds ..... 252 Approximate Values of Surd Numbers 253 Irrational Equations ...... . 255 viii CONTENTS. CHAPTER XVI. PAGB IMAGINARY AND COMPLEX NUMBERS 257 Imaginary Numbers 257 Complex Numbers 2(il Complex Factors . . 263 CHAPTER XVII. DOCTRINE OF EXPONENTS ........ 265 Zeroth Powers 266 Negative Integral Powers ........ 267 Fractional (Positive or Negative) Powers ..... 269 CHAPTER XVIII. QUADRATIC EQUATIONS ......... 277 Pure Quadratic Equations . .. . . . . .277 Solution by Factoring 280 Solution by Completing the Square 282 Relation between the Roots and Coefficients .... 287 Nature of the Roots 288 Irrational Equations 290 Higher Equations 293 Problems 295 CHAPTER XIX. SIMULTANEOUS QUADRATIC EQUATIONS 300 Problems 307 CHAPTER XX. RATIO, PROPORTION, AND VARIATION 311 Ratio 311 Proportion 313 Variation 320 CHAPTER XXI. PROGRESSIONS ........... 324 Arithmetical Progression ........ 324 Geometrical Progression ........ 331 Harmonical Progression ........ 339 CONTENTS. ix CHAPTER XXII. PAGE BINOMIAL THEOREM FOR POSITIVE INTEGRAL EXPONENT . . 342 CHAPTER XXIII. VARIABLES AND LIMITS , 348 Variables 348 Limits . . . . . . . - . . . 348 Indeterminate Fractions 352 Interpretation of Solutions ....... 353 CHAPTER XXIV. UNDETERMINED COEFFICIENTS . . . . . . . . 358 Convergent and Divergent Series 358 Expansion of Rational Fractions 361 Expansion of Surds ......... 364 Partial Fractions ......... 364 CHAPTER XXV. THE BINOMIAL THEOREM FOR ANY RATIONAL EXPONENT . . 370 CHAPTER XXVI. LOGARITHMS ........... 373 Principles of Logarithms ........ 374 Systems of Logarithms ........ 377 Properties of Common Logarithms ...... 378 To find the Logarithm of a Given Number . . . .381 To find a Number from its Logarithm ..... 385 C logarithms 388 Applications 389 Exponential Equations 393 Compound Interest and Annuities 395 TABLE OF LOGARITHMS . . . . . . . - i-xviii CHAPTER I. INTRODUCTION. GENERAL NUMBER. 1. Algebra, like Arithmetic, treats of number. 2. The examples 2 3 23 77 7 7 1111 11 11 are particular cases of the following principle : The sum of two fractions which have a common denominator is a fraction whose denominator is the common denominator, and whose numerator is the sum of the numerators; or, more briefly stated, 1st num. 2d num. _ 1st num. + %d num. com. den. com. den. com. den. This principle can be stated still more concisely by letting letters stand for the two numerators and the common de- nominator. Let a stand for 1st num., b for 2d num., and c for com. den. We then have a b _ a 4- b c c c This relation states by means of letters, or symbols, all that is contained in the verbal statement. The letters a, 6, and c stand for the terms of any two fractions, and therefore denote any numbers whatever. In the first example above, a = 2, b = 3, c = 7; in the second, a = 5, b = 4, c = 11. 1 2 ALGEBRA. [Cii. I 3. In ordinary Arithmetic all numbers are represented by the Arabic numerals, 1, 2, 3, etc. Each of these symbols stands for a definite number. The symbol 7, for instance, stands for a group of seven units, the symbol 5 for a group of Jive units. But in Algebra, such symbols as a, 6, x, y, are used to repre- sent numbers which may have any values whatever, as in Art. 2. For the sake of brevity we shall say the number a, or simply a, meaning thereby the number denoted by the symbol a. 4. The numbers represented by letters are, for the sake of distinction, called Literal or General Numbers. EXERCISES I. If p is the product obtained by multiplying a by 6, express in symbols the following principles of multiplication : 1. The multiplicand is equal to the product divided by the multiplier. Let p = 35, a 7, b = 5 -, p 24, a = 3, b 8. 2. The multiplier is equal. to the product divided by the multiplicand. Let p = 63, a = 9, b = 7 , p = 40, a 5, b = 8. If q is the quotient obtained by dividing m by n, express in symbols the following principles of division : 3. The dividend is equal to the divisor multiplied by the quotient. Let q = 9, m = 99, n = 11 ; q = 6, m = 42, n = 7. 4. The divisor is equal to the dividend divided by the quo- tient. Let q = 5, m = 45, n = 9 ; q = 6, m = 72, n = 12. State in verbal language the principles which are expressed in symbols in the following : 6 a c _ a x c b d b x d a _ a d d x n d d -*- n 11-16. In Exs. 5-10, let a = 8, 6 = 7, c = 5, d = 6, ?i = 2 ; a = 15, 6 = 8, c = 11, d 5, n = 5. 3-8] GENERAL NUMBER. S 5. As was assumed in Art. 2, the operations of Addition, Subtraction, Multiplication, and Division are denoted by the same symbols in Algebra as in Arithmetic. Just as 5 + 3, read five plus three, means that 3 is to be added to 5 ; so a + b, read a plus b, means that b is to be added to a. Just as 5 3, read Jive minus three, means that 3 is to be subtracted from 5 ; so a b, read a minus b, means that b is to be subtracted from a. Just as 5 x 3, read Jive multiplied by three, means that 5 is to be multiplied by 3 ; so a x b, read a multiplied by b, means a is to be multiplied by b. Just as 10 -T- 5, read ten divided by Jive, means that 10 is to be divided by 5; so a-t- b, read a divided by b, means that a is to be divided by b. 6. Since 5x3 = 3x5, either number may be taken as the multiplier, the other as the multiplicand. If the number on the left be taken as the multiplier, the symbol of multiplication is read times or into. As, 5x3, read five times three, if 5 be regarded as the multiplier. A dot () is frequently used, instead of the symbol x, to denote multiplication ; as a b for a x b. 7. The symbol of multiplication between two literal num- bers, or one literal number and an Arabic numeral, is fre- quently omitted. E.g., the product x x y X z, or x y z, is usually written xyz, and is read x-y-z. But the symbol of multiplication between two numerals cannot be omitted without changing the meaning. E.g., if in the indicated multiplication, 3x6, or 3-6, the symbol, x , or , were omitted, we should have 36, not 18. 8. In a chain of additions and subtractions the operations are to be performed successively from left to right. E.g., 7 + 4-3 + 2 = 11 -3 + 2 = 8 + 2 = 10. 4 ALGEBRA. [Cn. I In a chain of multiplications and divisions the operations are to be performed successively from left to right. E.g., 12 x 2 -^ 3 x 4 = 24 -=-3x4 = 8x4 = 32. In a chain of additions, subtractions, multiplications, and divisions, the multiplications and divisions are first to be per- formed, and then the additions and subtractions. Kg., 2x3 + 8-v-4 = 6 + 2 = 8. 9. An Algebraic ^Expression is a number expressed by means of the signs and symbols of Algebra ; as x, mn, ab cd, etc. 10. The Symbol of Equality, =, read is equal to, is placed between two numbers to indicate that they have the same or equal values ; as 3 -f- 2 = 5. 11. The Symbol of Inequality, >, read is greater than, is used to indicate that the number on its left is greater than that on its right ; as 7 > 5. 12. The Symbol of Inequality, <, read is less than, is used to indicate that the number on its left is less than that on its right ; as 3 < 4 + 2. 13. The use of letters to represent general numbers may be illustrated by a few simple examples. Ex. 1. If a boy has 3 books and is given 2 more, he will have 3 + 2 books. If he has a books and is given 5 more, he will have a -+- 5 books. If he has m books and is given n more, he will have m + n books. Ex. 2. If a man buys 5 city lots at 120 dollars each, he pays 120 x 5 dollars for the lots. If he buys a lots at 150 dollars each, he pays 150 a dollars for the lots. If he buys u lots at v dollars each, he pays vu dollars for the lots. Ex. 3. If a train runs 60 miles in two hours, it runs 60 -=- 2 miles in 1 hour. If it runs a miles in 5 hours, it runs a -4- 5 miles in 1 hour. If it runs p miles in q hours, it runs p -*- q miles in 1 hour. 8-13] GENERAL NUMBER. 5 Ex. 4. If, in a number of two digits, the digit in the units? place is 3 and the digit in the tens' place is 5, the number is 10 x 5 + 3. If the digit in the units' place is a and the digit in the tens' place is 6, the number is 10 b + a. Ex. 5. Just as 2 = 1 + 1, and 3 = 1+1 + 1, so 2 a = a + a, and 3a = a + a + a. Therefore, just as 3 + 2 5, so 3 a + 2 a = 5 a. In like manner, 5 x 3 x = 2 x^ and $x + %x = %x. EXERCISES II. Read the following expressions : 1. a + b. 2. ra n. 3. a x b. 4. m -t- n. 5. x + 2y. 6. 3m-Sn. 7. 4 a x 5 6. 8. 7 aj -J- 3 y. 9. a + 6 + c. 10. a? y + 2. 11. m n p. 12. 4 a c + 3 d. 13. a& + be ac. 14. 3 xy 5 6cd. 15. A father is n years older than his son. How old is the father, if the son is 10 years old ? If the son is x years old ? 16. A boy rides his bicycle x miles and then walks y miles. How many miles does he go altogether ? 17. A man has $ d. If he spends $ 10, how many dollars has he left ? If he spends $ z, how many dollars has he left ? 18. A man is now n years old. How old was he 8 years ago ? m years ago? How long must he live to be 75 years old? How long to be y years old ? 19. If one pencil costs 3 cents, how much do 5 pencils cost? x pencils ? 20. If one pencil costs c cents, how much do z pencils cost ? 21. How many square feet are there in a floor 15 feet long and 20 feet wide ? In a floor a feet long and b f eet wide ? 6 ALGEBRA. [Cn. 1 22. A train runs m miles in 1 hour. How many miles will it run in 4 hours ? In b hours ? 23. A train runs m miles in 8 hours. How many miles will it run in 1 hour? If it runs m miles in h hours, how many miles will it run in 1 hour ? 24. A boy paid c cents for 5 note-books. How much did he pay for each ? If he paid c cents for n note-books, how much did he pay for each ? 25. In 3 dimes there are 10 x 3 cents. How many cents in d dimes ? In x dimes ? 26. How many cents in a dollars and b dimes ? In x dollars, y dimes, and z cents ? 27. 10 x 2, 10 x 3, 10 x 4, etc., are particular multiples of 10. Write any multiple of 10. 28. Write a number containing 8 units and 5 tens. Contain- ing u units and t tens. 29. Write a number containing h hundreds, t tens, and u units. Containing a hundreds, b tens, and c units. W T hat are the values of the following expressions ? 30. a + a. 31. a + 2 a. 32. x + 3x. 33. a a. 34. 2 a a 35. 3z z. 36. 3 c + 5 c. 37. 5 d 3 d. 38. 8 x + 5 x. 39. Sx 5x. 40. x + %x. 41. x ^x. 42. f a 4- i a. 43. | a i a. 44. 5 m f m. 45. a 4 2 a + 3 a. 46. a + 2 a 3 a. 47. 5z + 8z + 4z. 48. 8z 52-f43. 49. 9x4- 3 x 8 x. 50. 9# 4?/ :$.". 51. A man has $10#. If he receives $8x, how many dol- lars will he have? If he spends $6x, how many dollars will he have left ? 52. A boy paid Sx cents for pencils and Sx cents for note- books. How much did he pay for both ? How much more for note-books than for pencils? 13-15] GENERAL NUMBER. 7 53. A girl has x dimes and 3 x cents. How many cents has she? 54. A girl has a dollars. If she spends 7 a dimes, how many dimes will she have left? If she spends 85 a cents, how many cents will she have left ? 55. A man has $ 45 x. If he spends $ 7 x for a lot, and $ 32 x for a house, how many dollars will he have left? 56. A boy rides a wheel x miles and then walks 160 x rods. How many rods did he go altogether ? How many rods more did he ride than walk ? 57. The width of a room is x yards, and the length is 2 x feet greater than the width. How many feet are there in the length of the room? Axioms. 14. An Axiom is a truth so simple that it cannot be made to depend upon a truth still simpler. Algebra makes use of the following mathematical axioms : (i.) Every number is equal to itself. E.g., 1 = 1, a = a. (ii.) The whole is equal to the sum of all its parts. E.g., 7 = 3 + 4, 5 = 1 + 1+1 + 1+1. (iii.) If two numbers be equal, either can replace the other in any algebraic expression in ivhich it occurs. E.g., If a-\-b = c, and b = 2, then a+2 = c, replacing b by 2. (iv.) Two numbers which are each equal to a third number are equal to each other. E.g., If a = b, and c = b, then a = c. (v.) The ivhole is greater than any of its parts; and, con- versely, any part is less than the whole. E.g., 3 + 2 > 2 and 2 < 3 + 2. 15. Literal numbers, as has been stated, are numbers which may have any values whatever. But it is frequently necessary to assign particular values to such numbers. 8 ALGEBRA. [Cn. I 16. Substitution is the process of replacing a literal number in an algebraic expression by a particular value. See axiom (iii.). Simple examples in substitution have already been given in Art. 2. Ex. 1. If, in a + b, we let a = 3 and b = 5, then a + 6 = 3 + 5 = S, or a + 6 = S. Ex. 2. If, in a + b 2 a + 3 b c, we let a = 6, 6 = 11, c = 1, we have a + 6-2a + 36-c = 6 + ll-2x6 + 3xll-l = 6 + 11-12 + 33-1-37. Ex. 3. If, in the last example, a 3, 6 = 1, and c = 1, we have a + 6-2a + 36-c = 3 + l-6 + 3-l=4-6 + 3-l. We cannot further reduce 46+3 1, since we are unable, as yet, to subtract 6 from 4. EXERCISES III. When a = 10, 6 = 5, c = 3, find the values of the following expressions : 1. a + 6. 2. a 6. 3. a6. 4. a -r- 6. 5. a + 6 c. 6. a 6 + c. 7. a _ & _ c . 8. c + 3 a. 9. 5 6 3 c. 10. 2a + 36-5c. 11. 5a-26-6c. 12. 3a-56 + 8c. 13. lab. 14. 2a6c. 15. 3a66. 16. 2 a6 + 5 ac. 17. 3 ac 5 6c. 18. 5 aa 3 66. 19. 2 a6 3 ac + 5 6c. 20. 5 aa 3 66 + 6 cc. Fundamental Principles. 17. The following principles are obtained directly from the axioms : (i.) If the same number, or equal numbers, be added to equaf numbers, the sums will be equal. 16-19] GENERAL NUMBER. 9 (ii.) If the same number, or equal numbers, be subtracted from equal numbers, the remainders will be equal. (iii.) If equal numbers be multiplied by the same number, or by equal numbers, the products will be equal. (iv.) If equal numbers be divided by the same number (except 0), or by equal numbers, the quotients will be equal. E.g., if 3 x = 6, then 3^ 2 = 6 2 3x-5 = 6-5, Equations. 18. An Equation is a statement that two expressions are equal ; as 7 x 9 = 63, 4 x 7 + 3 = 31. The First Member of an equation is the expression on the left of the symbol = ; the Second Member is the expression on the right of the symbol =. 19, Ex. 1. What is the value of x in the equation 3 x + 8 x = 22 ? Since 3 x + 8 x = 11 x, we have 11 a? = 22. Dividing both members by 11 [Art. 17, (iv.)], x = 2. To check this result we substitute 2 for x in the equation 3a + 8z = 3x2+8.x2 = 6-fl6 = 22. Ex. 2. If 8 x 3 x has the value 20, what is the value of x ? We have 8 x - 3 x = 20. Or, since Sx 3x = 5x, 5 a? = 20. Dividing both members by 5, x 4. Check: 8x4-3x4-32-12 = 20. 10 ALGEBRA. [Cii. I 20. An Unknown Number of an equation is a number whose value is to be found from the equation. The Known Numbers of an equation are the numbers whose values are given. In the equation x -f 1 = 3, the unknown number is a?, and the known numbers are 1 and 3. Unknown numbers are usually represented by the final letters of the alphabet, x, y, z, etc., as in the above examples. EXERCISES IV. Find the value of x in each of the following equations : 1. 3x = 9. 2. 6 x = 18. 3. 5x = 0. 4. ix = 4. 5. ix = 5. 6. ia = 0. 7. fo; = 6. 8. |a? = 15. 9. a = 21. 10. x + x = &. 11. a? + 5 a; = 24. 12. 5 # + 4 ;E = 45. 13. 5a-4# = 3. 14. 6x 3x = 3. 15. 7a;-5z=12. 16. x 4- 3 x + 5 a; = 18. 17. 2oj + 5a; + 3a; = 20. 18. 7a + 3oj + 5a; = 90. 19. 5 .T + 4 a? - 6 x = 15. 20. 8 a; 5 x + x = 12. 21. 11 a; + 7 a; 5 a; = 26. 22. x + x = 6. 23. a? i a = 10. 24. 24^ + | = 149. 25. 3a; + f = 30. 26. 5x-%x = 33. 27. 2%x-x = U. 28. x + ^x + f^ = 28. 29. 2.T-^x + |^ = 34. 30. Jaj + f a;-ia; = 54. 31. 5 x - \x - ^x = 62. Problems solved by Equations. 21. A Problem is a question proposed for solution. Another use of literal numbers is shown by the following problems : Pr. l. The older of two brothers has twice as many marbles as the younger, and together they have 33 marbles. How many has the younger? 20-21] GENERAL NUMBER. 11 The number of marbles the younger brother has is, as yet, an unknoivn number. Let us represent this unknown number by some letter, say x. Then, since the older brother has twice as many, he has 2 x marbles. The problem states, in verbal language : the number of marbles the younger has plus the number the older has is equal to 33 ; in algebraic language, x + 2 x = 33, or, 3 x = 33. Dividing both members of the last equation by 3, we have x = n, the number of marbles the younger has. The older has, 2 x, = 2 x 11, = 22 marbles. To check this result, we substitute 11 for x in the equation of the problem : x + 2x=ll+22 = 33. Notice that the letter x stands for an abstract number. The beginner must never put x for marbles, distance, time, etc., but for the number of marbles, of miles, of hours, etc. Pr. 2. Divide 52 into three parts, so that the second shall be one-half of the first, and the third one-fourth of the second. Let x stand for the first part. Then | x stands for the second part, and J x \ x, = J x, stands for the third part. The problem states, in verbal language : the first part, plus the second part, plus the third part, is equal to 52 ; in algebraic language, x + \ x -f- \ x 52, or, J x = 52. 12 ALGEBRA. [Cn I Dividing both members of the last equation by 13, *-* Multiplying both members of this equation by 8, x = 32, the first part. Then the second part is 1 x , = i x 32, = 16, and the third part is -L x , = i X 32, =4. Check: x + x + x = 32 + 16 + 4: = 52. 22. In stating problems in algebraic language, the beginner should observe the following directions : (i.) Read the problem carefully, and note what are the numbers whose values are required. (ii.) Let some letter, say x, stand for one of the required numbers. (iii.) The problem will contain statements about the values' of other numbers. Use these statements to express their values in terms of x. (iv.) Express concisely in verbal language a statement in the problem which furnishes an equation. (v.) Express this statement in algebraic language. EXERCISES V. 1. What number is five times x? Twelve times x? 2. Five times a number is 80. What is the number ? 3. Twelve times a number is 132. What is the number ? 4. The greater of two numbers is four times the less. If the less is x, what is the greater ? What is their sum ? Their difference ? 5. The greater of two numbers is four times the less. If their sum is 75, what are the numbers ? 21-22] GENERAL NUMBER. 13 6. The greater of two numbers is seven times the less. If their difference is 72, "what are the numbers ? 7. A father is three times as old as his son. If the son is x years old, how old is the father ? What is the sum of their ages ? How much older is the father than the son ? 8. A father is three times as old as his son, and the sum of their ages is 48 years. How old is each ? 9. A father is five times as old as his son. If the father is 32 years older than his son, what are their ages ? 10. At an election A received twice as many votes as B, and his majority was J.38. How many votes did each receive ? 11. In a company are 32 persons. The number of children is three times the number of adults. How many are there of each? 12. Two trains leave Philadelphia in opposite directions. After one hour they are 60 miles apart. If one has gone three times as far as the other, how many miles is each from Phila- delphia ? 13. Two trains leave Chicago in the same direction. After one hour they are 20 miles apart. If one has gone twice as far as the other, how far is each from Chicago ? 14. A man pays $ 55 in one-dollar bills and ten-dollar bills. If he pays the same number of one-dollar bills as of ten-dollar bills, how many of each does he pay ? 15. In a number of two digits, the tens' digit is three times the units' digit, and their sum is 8. What are the digits ? W^hat is the number ? 16. In a number of two digits, the units' digit is twice the tens' digit, and their difference is 3. What is the number ? 17. What is the sum of twice x and six times x ? The difference ? 18. If twice a number is added to six times the same num- ber, the sum will be 96. What is the number ? 14 ALGEBRA. [Cn. I 19. If four times a number is subtracted from seven times the same number, the remainder will be 72. What is the number ? 20. A traveller first rides his bicycle 9 miles an hour. He then rides the same number of hours in a car 35 miles an hour. If he travels 132 miles, how many hours did he ride his bicycle ? 21. Two trains run out of New York in opposite directions. One runs 42 miles an hour, the other 34 miles an hour. After how many hours will they be 228 miles apart ? 22. Two trains run out of New York in the same direction. One runs 38 miles an hour, the other 34 miles an hour. After how many hours will they be 32 miles apart ? 23. A boy has 75 cents in dimes and five-cent pieces. He has the same number of dimes as of five-cent pieces. How many coins of each kind has he ? 24. A owes B $ 40. He pays his debt in ten-dollar bills, and receives in change the same number of two-dollar bills. How many ten-dollar bills did A pay B ? 25. A cistern has two pipes. One lets in 8 gallons a minute, and the other 15 gallons a minute. If the cistern holds 207 gallons, how many minutes will it take the pipes to fill it ? 26. A cistern has two pipes. One lets in 11 gallons a minute, and the other lets out 6 gallons a minute. How many minutes will it take the one pipe to let in 85 gallons more than the other lets out ? 27. What is the sum of x, four times x, and seven times x? Of oj, twice x, and five times x ? 28. The sum of a certain number, four times the number, and seven times the number is 96. What is the number ? 29. Three boys, A, B, and C, together have 21 pencils. B has twice as many as A, and C four times as many as A. How many has A ? How many has each ? 30. Divide 147 into three parts, so that the second part shall be four times the first, and the third part twice the first. 22] GENERAL NUMBER. 15 31. A merchant receives $ 64 in ten-dollar bills, five-dollar bills, and one-dollar bills. He receives the same number of each kind. How many of each does he receive ? 32. At an election 726 votes were cast. A, B, and C were candidates. B received three times as many votes as C, and A twice as many as C. How many votes did each receive ? 33. A cistern has three pipes. The first lets in 6 gallons a minute, the second 9 gallons a minute, and the third 12 gallons a minute. If the cistern holds 243 gallons, how long will it take the pipes to fill it ? 34. A cistern has three pipes. The first lets in 5 gallons a minute, the second 14 gallons a minute, and the third lets out 10 gallons a minute. How many minutes will it take the two pipes to let in 108 gallons more than the third pipe lets out ? 35. An estate of $ 9600 is divided among 2 sons and 2 daughters. The sons receive equal amounts, and a daughter receives three times as much as a son. How many dollars does each receive ? 36. What is twice 3 x ? Seven times 5 x ? Four times 9 x ? 37. A receives x dollars, B receives three times as much as A, and C receives twice as much as B. How many dollars does C receive ? How many dollars do all receive ? 38. Three boys, A, B, and C, together receive $ 70. B re- ceives three times as much as A, and C twice as much as B. How many dollars does each receive ? 39. A merchant's profits doubled each year for three years. If his profits for the three years were $ 8750, what were his profits the first year ? 40. In a company are 50 persons. The number of women is three times the number of men, and the number of children is twice the number of women. How many of each are in the company ? 41. What number is \ of a? ? f of x ? 16 ALGEBRA. [Cn. I 42. If J of a number is 16, what is the number ? 43. The less of two numbers is f of the greater. If the greater is x, what is the less ? What is their sum ? Their difference ? 44. The less of two numbers is f of the greater. If their sum is 91, what are the numbers ? 45. A and B together have $ 1133. If B has -^ as much as A, how many dollars has each ? 46. A has $ 31 more than B. If B has J as much as A, how many dollars has each ? 47. Two boys, A and B, catch 36 fish. If A catches f as many as B, how many fish does each catch ? 48. A workman pays f of his wages for board. If he has left $ 8 each week, what are his wages ? 49. Two boys together solve 65 problems. If the first solves -f as many as the second, how many problems does each solve ? 50. A solves 21 more problems than B. If B solves J- as many as A, how many problems does each solve ? 51. A tree 126 feet high is broken by the wind. If the part left standing is T 3 T of the part broken off, how long is each part ? 52. What is the sum of -J- of x and f of x ? The difference ? 53. If i of a number is added to J of the same number, the sum will be 39. What is the number ? 54. If -| of a number is subtracted from J of the same num- ber, the remainder will be 3. What is the number ? 55. If to a number is added -J- of itself and f of itself, the sum will be 50. What is the number ? 56. Three boys, A, B, and C, together have 29 pencils. B has | as many as A, and C has } as many as A. How many pencils has each ? 22-23] GENERAL NUMBER. 17 57. Divide 104 into three parts, so that the first shall be three times the second, and the third ^ of the second. 58. A man makes a journey of 69 miles. He goes f as far by boat as by train, and as far by stage as by train. How many miles does he go by each conveyance ? 59. What is of three times x? Twicefofar? f off of a? 60. The second of three numbers is three times the first, and the third is -i- of the second. If the first number is x, what is the second ? The third ? What is the sum of the three numbers ? * 61. The sum of three numbers is 99. The second is four times the first, and the third is f of the second. What are the numbers ? 62. The width of a field is \ of its length, and the distance around it is 88 rods. What is the width and the length of the field ? 63. The sum of $ 420 is divided among A, B, and C. B receives f as much as A, and C as much as A and B together. How many dollars does each receive ? 64. A sells a number of apples at 2 cents apiece, and B sells \ as many at 3 cents apiece. If they receive together 87 cents, how many apples does each sell ? Parentheses. 23. Parentheses, (), and Brackets, [], are used to indicate that whatever is placed within them is to be treated as a whole. E.g., 10 (2 + 5) means that the result of adding 5 to 2, or 7, is to be subtracted from 10 ; that is, 10 - (2 + 5) = 10 - 7 = 3. But 10 2 + 5 means that 2 is to be subtracted from 10 and 5 is then to be added to that result ; that is, 10 _ 2 + 5 = 8 + 5 = 13. 18 ALGEBRA. [Cn. I In like manner, [27 (3 -f 2) x 5] -H 2 means that the result of multiplying the sum 3 -f 2 by 5 is first to be subtracted from 27, and the remainder is then to be divided by 2 ; that is, [27 - (3 + 2) x 5] -* 2 = [27 - 25] + 2 = 2 -*- 2 = 1. Likewise, (a+fr)c is the result of multiplying a+b by c, etc. EXERCISES VI. Find the value of each of the following expressions : 1. 10 + (3 + 2). 2. 10 -(3 + 2). 3. 10 + (3 -2). 4. 10 -(3 -2). 5. 27 -(18 -11). 6. 53 + (40 + 7). 7. 97 -f (11 -8). 8. 58 -(15 -7). 9. 99 + (18- 17). 10. 5(8 + 2). 11. 6(11-6). 12. (10 + 15)-*- 5. 13. 10 + (15-s-5). 14. (12 -4) -4- 2. 15. 12 -(4-5-2). 16. (15 - 3) + (18 - 6). 17. (16 - 2) - (20 - 8). 18. (4 + 5) (8 -3). 19. (8 + 12) - (7 - 2). 20. 20 + [11 - (5 + 2)]. 21. 28 -[16 -(5 + 3)]. 22. [26 - (14 + 6)] x 5. 23. [27 - (18 - 12)] -j- 7. When a = 12, b = 6, c = 3, find the values of : 24. a + (b c). 25. a (6 + c). 26. a (b c). 27. c + 5(a-6). 28. 4a-2(6 + c). 29. 6[c + (a-6)]. 30. a [a-~K&+c)]. 31. [ a -(6-c)]-s-c. 32. [& + ( a -c)]-s-c. POSITIVE AND NEGATIVE NUMBERS, OR ALGEBRAIC NUMBERS. 24. In ordinary Arithmetic we subtract a number from an equal or a greater number. We are familiar with such opera- tions as 543 minuend 333 subtrahend (i.) 210 remainder 23-26] POSITIVE AND NEGATIVE NUMBERS. 19 But such operations as 210 minuend 333 subtrahend (ii.) ? ? ? remainder have not occurred in ordinary Arithmetic. In Arithmetic we cannot subtract from a number more units than are contained in the number. 25. Now, in the operations (i.) above, the remainder 2 indi- cates that the subtrahend is two units less than the minuend ; the remainder 1 that the subtrahend is one unit less than the minuend ; and the remainder that the subtrahend is equal to the minuend. In the operations (ii.), we must indicate by the remainders that the subtrahend is one, tivo, three, etc., units greater than the corresponding minuend. We do this by placing the sign ~ before the symbols for one, two, three, etc. ; as ~1, ~2, ~3, etc. The remainders in these cases are called Negative Numbers ; as ~1, ~2, ~3, etc., read negative one, negative two, negative three, etc. For the sake of distinction, the remainders in the operations (i.) are called Positive Numbers. They are indicated by the sign + ; as +1, + 2, + 3, etc., read positive one, positive two, positive three, etc. Positive and negative numbers are called Algebraic or Rela- tive Numbers. 26. We pos. 5 pos. 3 can now pos. 4 pos. 3 write (i pos. 3 pos. 3 ) and pos. pos. (ii.) as follows 2 pos. 1 3 pos. 3 pos. 3 inin. sub. rem. min. sub. rem. pos. 2 +5 +3 +2 pos. 1 +4 +3 +1 +3 +3 neg.l +2 +3 neg.2 +1 +3 -2 neg. -f 3 3 (iii.) 20 ALGEBRA. [Cn. I 27. We thus have in Algebra the series of numbers, . +* +4 +3 +<> +1 o -1 -9 -3 -A -*> > | *J J -'J > U J - 1 -? "J ) ^7 O, ", wherein the signs, , indicate that the succession of numbers continues without end in both directions. This series is usu- ally written with the positive numbers on the right, as ..., -5, -4, -3, -2, -1, 0, +1, +2, +3, +4, +5, .... 28. In this series the numbers increase by one from left to right, and decrease by one from right to left. Or, a number is greater than any number on its left, and less than any number on its right. Thus, + 2 is one unit greater than + 1, two units greater than 0, three units greater than ~1, etc. Again, ~3 is three units greater than ~6, two units less than ~1, three units less than 0, etc. 29. The signs + and ~ are called signs of quality; the signs -f and , signs of operation. The two sets of signs must, as yet, be carefully distinguished. 30. The Absolute Value of a number is the number of units contained in it without regard to their quality. E.g., the absolute value of + 4 is 4, of ~5 is 5. 31. From the results of the preceding articles, we obtain the following general relations : (i.) Of two positive numbers, that number is the greater which has the greater absolute value ; and that number is the less which has the less absolute value. (ii.) Of two negative numbers, that number is the greater which has the less absolute value; and that number is the less which has the greater absolute value. For example, ~3 > ~5, or ~5 < ~3, since ~5 is five units less than 0, and ~3 is only three units less than 0. 27-34] POSITIVE AND NEGATIVE NUMBERS. 21 32. It is important to notice that a negative remainder does not mean that more units have been taken from the minuend than were contained in it ; such a remainder indicates that the subtrahend is greater than the minuend by as many units as are contained in the remainder. Thus, in +15 +25 = ~10, the remainder, "10, indicates that the subtrahend is 10 units greater than the minuend. 33. It is evidently necessary thus to enlarge the meaning of subtraction in such an expression as a b. For, if a and b are to have any values whatever, the case in which b is greater than a, that is, in which the subtrahend ft greater than the minu- end, must be included in the operation of subtraction. 34. Negative numbers have been introduced by extending the operation of subtraction. But it is necessary to treat them as numbers apart from this particular operation. As in Arithmetic, so in Algebra, any integer is an aggregate of like units. Just as 4 = 1 + 1 + 1 + 1, so +4 =+1 ++1 ++1 ++1, and ~4 =~1 +~1 +~1 +~1. Just as | = i + 1, so + (|) = + (J) + + Q), and "(I) =-(i) +-(i). EXERCISES VII. Simplify the following expressions : 1. +17- +4. 2. +17 -+17. 3. +17 -+27. 4. +25 -+18. 5. +25- + 25. 6. +25 -+35. 7. +88 -+95. 8. +56- +27. 9. +101 - +105. What value of x will make the first member of each of the following equations the same as the second ? 10. #-+5 =+7. 11. x -+5 = 0. 12. a; +5 =-2. 13. a -+11 =+9. 14. a -+15 =-13. 15. #-+12 =-10 2'2 ALGEBRA. [Cn. I How many units is each of the following numbers greater or less than ? 16. +5. 17. -3. 18. -11. 19. +14. 20. -20. Which of each of the following pairs of numbers is the greater, and by how many units ? 21. +7, +4. 22. ~7, ~4. 23. +7, -4. 24. ~7, +4. 25. +19, +9. 26. -29, +1. 27. +32, ~3. 28. -14, -3. Positive and Negative Numbers are Opposite Numbers. 35. In Arithmetic we have : the remainder added to the sub- trahend is equal to the minuend. This principle, like all princi- ples of Arithmetic, is retained in Algebra. We therefore have from (iii.) Art. 26 : +3 +3 +3 +3 +3 +3 subtr. +2 +J. _0 2! I? I? rem - +5 +4 +3 +2 +1 min. 36. The equation +3 -f ~3 = gives us the following impor- tant principle : TJie sum of a positive number and a negative number having the same absolute value is equal to zero; i.e., two such numbers cancel each other ivhen united by addition. E.g., +!+-! = 0, +3 + ~3 = 0, ~17J + +171 = 0. In general, + n -f ~n = 0. For this reason, positive and negative numbers in their rela- tion to each other are called opposite numbers. When their absolute values are equal, they are called equal and opposite numbers. 37. Any quantities which in their relation to each other are opposite, may be represented in Algebra by positive and negative numbers ; as credits and debits, gain and loss. 34-37] POSITIVE AND NEGATIVE NUMBERS. 23 Ex. 1. 100 dollars credit and 100 dollars debit cancel each other. That is, 100 dollars credit united with 100 dollars debit is equal to neither credit nor debit; or, 100 dollars credit + 100 dollars debit = neither credit nor debit. If credits be taken positively and debits negatively, then 100 dollars credit may be represented by +100, and 100 dollars debit by -100. Their united effect, as stated above, may then be represented algebraically thus : +100 +-100 = 0. The result, 0, means neither credit nor debit. Similarly for opposite temperatures. Ex. 2. If a body is first heated 10 and then cooled down 8, its final temperature is 2 above its original temperature ; or, stated algebraically, +10 +-8- +2. The result, +2, means a rise of 2 in temperature. EXERCISES VIII. Express algebraically each one of the following statements : 1. $ 45 gain and $ 45 loss is equivalent to neither gain nor loss. 2. $ 95 gain and $ 50 loss is equivalent to $ 45 gain. 3. $ 37 gain and $ 57 loss is equivalent to $ 20 loss. 4. If a man travels 220 miles due west and then 220 miles due east, he is at his starting place. 5. If a man ascends 2250 feet in a balloon and then descends 200 feet, he is 2050 feet above the earth. 6. If a man walks 90 feet to the right and then 110 feet to the left, he is 20 feet to the left of his starting point. 7. A rise of 20 in temperature, fol-lowed by a fall of 27, is equivalent to a fall of 7. 8. A rise of 15 in temperature, followed by a fall of 12, is equivalent to a rise of 3. CHAPTER II. THE FOUR FUNDAMENTAL OPERATIONS WITH ALGEBRAIC NUMBER. ADDITION OF ALGEBRAIC NUMBERS. 1. TJie Addition of two numbers is the process of uniting them into one aggregate. The numbers to be added are called Summands. Addition of Numbers with Like Signs. 2. Ex. 1. Add +3 to +4. The three positive units, + 3, when added to the four positive units, + 4, give an aggregate of four plus three, or seven, positive units. That is, +4 + +3 =+(4 + 3) =+7. In like manner, Ex.2. -4 +-3 =-(4 + 3) =-7. These examples illustrate the following method of adding two numbers with like signs : Add arithmetically their absolute values, and prefix to the sum their common sign of quality. Addition of Numbers with Unlike Signs. 3. Ex. 1. Add -2 to +5. The two negative units, ~2, when added to the five positive units, + 5, cancel two of the Jive positive units. There remain then Jive minus two, or three, positive units. That is, +5+-2=+(o-2)=+3. 24 1-5] SUBTRACTION OF ALGEBRAIC NUMBERS. 25 Ex. 2. Add +2 to -5. The two positive units, + 2, when added to the five negative units, "5, cancel two of the Jive negative units. There remain then Jive minus two, or three, negative units. That is, -5 ++2 =-(5 -2) =-3. Observe that in both examples the sum is of the same quality as the number which has the greater absolute value. Also, that the absolute value of the sum is obtained by subtracting the less absolute value, 2, from the greater, 5. These examples illustrate the following method of adding two numbers with unlike signs : Subtract arithmetically the less absolute value from the greater. To that remainder prefix the sign of quality of the number which has the greater absolute value. The examples given in Ch. I, Art. 37, are concrete illustra- tions of the preceding principles. 4. Observe that a positive number increases a number to which it is added, while a negative number decreases it. EXERCISES I. Add: 1. 2. 3. 4. 5. 6. +2 -4 +9 -8 +13 -21 +6 -5 +3 -7 +19 -15 7. 8. 9. 10. 11. 12. +8 -8 -7 +13 -21 +37 -3 +3 +4 -17 +32 -22 SUBTRACTION OF ALGEBRAIC NUMBERS. 5. Subtraction is the inverse of addition. In addition two numbers are given, and it is required to find their sum, as in +9 ++2 =+11. 26 ALGEBRA. [Cn. II In subtraction the sum of two numbers and one of them are given, and it is required to find the other number, as in +11 - + 2 = ( + 9 ++2) -+2 =+9. That is, if from the sum of two numbers either of the numbers be subtracted, the remainder is the other number. In general, (a + 6) a b. 6. Ex. l. A man's net profits last year were 1200 dollars. This year his income is 150 dollars less, and his expenditures are the same. What are his net profits this year ? To take aivay 150 dollars income is equivalent to adding 150 dollars expenditures. If net profits and income be taken positively, and expendi- tures negatively, the last statement, expressed algebraically, is +1200 -+150 =+1200 +-150. Ex. 2. A man's net profits last year were 1200 dollars. This year his income is the same and his expenditures are 150 dol- lars less. What are his net profits this year ? To take away 150 dollars expenditures is equivalent to adding 150 dollars profits. The algebraic statement of this relation is +1200 - -150 =+1200 ++150. These examples illustrate the following principle : To subtract one number from another number, reverse the sign of quality of the subtrahend, and add. E.g. 9 +2 -+3 =+2 +-3, =-1. -2 -+3 =~2 +~3 =~5. +2 -~3 =+2 ++3, =+5. -2 -~3 =~2 + + 3 = +1. 7. It is important to notice that the preceding examples do not prove this principle. The following examples illustrate a method of proof which may be used. 5-9] MULTIPLICATION OF ALGEBRAIC NUMBERS. 27 Ex. 1. Subtract +5 from +7. In +7 + o, the minuend, + 7, is to be expressed as the sum of two numbers, one of which is + 5. Since ~5 H- + 5 = 0, we may write +7 =+7 +-5 + + 5 = (+7 +-5) ++5. That is, + 7 may be regarded as the sum of two numbers, one of which is + 7 + ~5, and the other is + 5. Therefore, by defini- tion of subtraction, +7 -+5 = [(+7 +-5) ++5] - + 5 = +7 +-5 =+2, That is, to subtract + 5 is equivalent to adding ~5. Ex. 2. Subtract ~5 from +7. We have +7 -~5 = [( + 7 ++5) +~5] -~5 = +7 ++5 =+12, That is, to subtract ~5 is equivalent to adding + 5. 8. We thus see that every operation of subtraction is equiva- lent to an operation of addition. On this account it is conven- ient to speak of a chain of additions and subtractions as an Algebraic Sum. EXERCISES II. Subtract : 1. 2. 3. 4. 5. 6. +9 +2 +8 +3 -9 -4 +2 +9 +3 +8 -4 -9 7. 8. 9. 10. 11. 12. -8 -7 +5 -6 + 6 -(> MULTIPLICATION OF ALGEBRAIC NUMBERS. 9. In multiplication, the multiplicand and multiplier are called Factors of the product. 28 ALGEBRA. [Cn. n 10. In ordinary Arithmetic, multiplication by an integer is denned as an abbreviated addition. Thus, 4 x 3 = 4 + 4 + 4; that is, the number 4 is taken three times as a summand. But 3 = 1 + 1 + 1. We thus see that the product 4 x 3 is obtained from 4 just as 3 is obtained from the positive unit, 1. We are thus naturally led to the following definition of multiplication : The product is obtained from the multiplicand just as the mul- tiplier is obtained from the positive unit. 11. The above definition is an extension of the meaning of arithmetical multiplication when the multiplier is an integer, and gives an intelligible meaning to arithmetical multiplication when the multiplier is a fraction. Thus, |- is obtained from the unit, 1, by taking one-third of the latter twice as a summand ; or In like manner, to multiply 5 by -f, we take one-third of 5 twice as a summand ; or 12. There are two cases to be considered in the multiplica- tion of algebraic numbers. (i.) The Multiplier Positive. Ex. 1. Multiply +4 by +3. By the definition of multiplication, the product, + 4 x +3, is obtained from +4 just as + 3 is obtained from the positive unit. But +3=+l+ + l++l. Consequently the required product is obtained by taking +4 three times as a summand, or +4 x +3 =+4 ++4 ++4 =+(4 + 4 + 4) =+(4 x 3) =+12. 10-13] MULTIPLICATION OF ALGEBRAIC NUMBERS. 29 Ex. 2. Multiply -4 by +3. By the definition of multiplication, we have -4 x + 3 =~4 +-4 +~4 =-(4 + 4 + 4) =~(4 x 3) =-12. (ii.) The Multiplier Negative. Ex. 3. Multiply +4 by ~3. By the definition of multiplication, the product, + 4x-3, is obtained from + 4 just as ~3 is obtained from the positive unit. But -3 =~1 +-1 +-1 = -+1 -+1 -+1 ; that is, ~3 is obtained by subtracting the positive unit, + 1, three times in succession from 0. Consequently, the required product is obtained by subtracting the multiplicand, + 4, three times in succession from ; or, +4 x-3 = - + 4-+4-+4 = +-4+-4+-4=-(4 x 3). Ex. 3. Multiply ~4 by ~3. By the definition of multiplication, we have -4 x -3 = --4 --4 --4 = ++4 ++4 ++4 =+(4 x 3). 13. These examples illustrate the following Rule of Signs for Multiplication: The product of two numbers having like signs is positive ; and the product of two numbers having unlike signs is negative. Or, stated symbolically, + a x + b = + (ab\ ~a x+6 =-(a6), -a x ~b = + (ab\ + a x ~b =-(a6). EXERCISES III. Multiply : 1. 2. 3. 4. 5. 6. +3 -3 +3 -3 +8 -7 +4 +4 Ii ^4 5 -6 7. 8. 9. 10. 11. 12. -9 +8 -12 -15 +2Q +16 +2 -6 -5 +4 +7 ~5 30 ALGEBRA. [Cn. II DIVISION OF ALGEBRAIC NUMBERS. 14. Division is the inverse of multiplication. In multiplica- tion two factors are given, and it is required to find their product. In division the product of two factors and one of them are given, and it is required to find the other factor. E.g., Since -28=-4x + 7, therefore, ~28 -h + 7 = ~4, and "28 -j-~4 = +7. 15. From the definition of division we infer the following principle : If the product of two numbers be divided by either of them, the quotient is the other number. 16. Since +ax + b=+(ab), therefore +(ab) -=- + a = + b; since ~a x + 6=~(a6), therefore ~(ab) ^-~a = + 6; since ~a x~b= + (ab), therefore + (a6) -5-~a =~b; since + a x~b=~(ab), therefore ~(ab) -r- + a =~6. From these equations we derive the following Rule of Signs for Division : Like signs of dividend and divisor give a positive quotient; unlike signs of dividend and divisor give a negative quotient. E.g., +8 -=-+2 =+4; -8-=--2=+4; +8-s--2=-4; -8 -s- +2 =-4. EXERCISES IV. Divide : 1. 2. 3. 4. +4)^8 +4)^8 -4) + 8 -4)2 6. 7. 8. 9. 10. ~7) + 28 +6)-30 +8) + 24 14-19] SIGNS OF QUALITY AND OPERATION. 31 ONE SET OF SIGNS FOR QUALITY AND OPERATION. 17. Most text-books of Algebra use the one set of signs, + and , to denote both quality and operation. We shall in subsequent work follow this custom. For the sake of brevity, the sign + is usually omitted when it denotes quality ; the sign is never omitted. Thus, instead of + 2, we shall write + 2, or 2 ; instead of ~2, we shall write 2. 18. We have used the double set of signs hitherto in order to emphasize the difference between quality and operation. It should be kept clearly in mind that the same distinction still exists. We now have +3 _|_+2 = + 3 + (+ 2) = 3 + 2, omitting the signs of quality, + ; +3 _l_-2:= -}-3 +( 2), wherein + denotes operation, and de- notes quality. +3 _+2 = + 3 (+ 2)= 3 2, omitting the signs of quality, + ; +3 _~2 = + 3 ( 2), wherein the first sign, , denotes opera- tion, the second sign, , denotes quality. 19. In the chain of operations (+2) + (-5)-(+2)-(-ll) the signs within the parentheses denote quality, those without denote operation. That expression reduces to (+ 2) -(+5) -(+2) + (+H), or 2-5-2 + 11, dropping the sign of quality, +. In the latter expression all the signs denote operation, and the numbers are all positive. 32 ALGEBRA. [Cn. II 20. The following examples illustrate the double use of the signs + and . Ex.1. +4 ++3 = + 4 + (+3) = 4 + 3 = 7. Ex.2. -5++2 = -5 + (+2) = -5 + 2 = -3. Ex. 3. +7 --5 = + 7 - (- 5) = 7 - (- 5) = 7 + 5 = 12. Ex.4. ~4x + 3 = -4x (+3)=-4x3 = -12. Ex.5. -4x-3 = -4x(-3) = 12. Continued Products. 21. The results of Article 13 may be applied to determine the value of a chain of indicated multiplications, i.e., of a continued product. E.g. (+ a) (+ 6) (+ C ) = (+ ab) (+ e) = + abc, (+ a) (+ 6) (- G) = ( + ab) (-c) = - abc, (+ a) (- 6) (- c) = (- a6) (- c) - + abc, These equations illustrate a more general rule of signs : A continued product which contains no negative number, or an even number of negative numbers, is positive; one that contains an odd number of negative numbers is negative. In practice the sign of a required product may first be deter- mined by inspection, and that sign prefixed to the product of the absolute values of the numbers in the continued product. E.g., the sign of the product 2x(-3)x(-7)x(+4)x(-5) is negative, since it contains three negative numbers ; the product of the absolute values is 840. Consequently, 2x (-3) x (-7)x (+4) x (-5) = -840. 20-21] SIGNS OF QUALITY AND OPERATION. 33 EXERCISES V. In the expressions in Exx. 1-4, which signs denote quality and which operation ? 1. +5 -}- (-3) -(+8). 2. .-7 + (+5) -(-9). 3. _3 + (-5)x(+4). 4. (+12)H-(-4)x(-3). 5-8. Find the value of the expressions in Exx. 1-4. Find the values of the expressions in Exx. 9-20, first chang- ing them into equivalent expressions in which there is only one set of signs -f and : 9. +8 + +2. 10. + 7- + 3. 11. +3- +7. 12. ~5 + ~7. 13. -8 - +3. 14. -9 - -5. 15. +4 x +5. 16. +5 x ~2. 17. ~5 x ~2. 18. +12 -*- +3. 19. +12 -f- -3. 20. -12 -r- -3. Simplify the following expressions : 21. 10-4. 22. 4-10. 23. -8-7. 24. 9-2. 25. 2-9. 26. -10 + 10. 27. 8x5. 28. - 8 x 5. 29. 8 x (- 5). 30. (_8)x( 5). 31. 20 -=-4. 32. 20-*- 4. 33. 20 -s- (4). 34. (-20)-f-(-4). 35. -45-T-9. 36. 3 x 5 4- 4 x 2. 37. 3 x (5 + 4 x 2). 38. 8 x 6 - 10 -3- 5. 39. (8 x 6 - 10) -*- 5. 40. 12 -s- 4 10 -5- 2. 41. 12 -j- (4 - 10 -5- 2). When a = 16, b = - 8, c = - 2, d = - 4, find the values of : 42. a + b -f- c. 43. a + b c. 44. a 6 -f- c. 45. a b c. 46. a (b c). 47. c (6 a). 48. a&c. 49. ab -s- c. 50. a -*- (6c). 51. a -=- b x c. 52. abed. 53. (aft) -f- (cd). 54. abc -r- d. 55. ab + cd. 56. a -s- b d -=- c. 57. A's assets are $ 2600 and B's are $ 2200. How much do A's assets exceed B's, taking assets positively ? 34 ALGEBRA. [Cn. II 58. A owes $200, and B's assets are $1800. How much do A's assets exceed B's, taking assets positively ? 59. The temperature in a room is 72 above zero, and out of doors it is 8 above zero. How much higher is the tem- perature in the room than out of doors, taking degrees above zero positively? 60. The temperature in a room is 70 above zero, and out of doors it is 4 below zero. How much higher is the tem- perature in the room than out of doors, taking degrees above zero positively ? PARENTHESES. 22. The Terms of an algebraic sum are the additive and sub- tractive parts of tne sum. E.g., the terms of 2 5 2 + 11 are + 2, 5, 2, + 11 The Sign of a Term is its sign + or . A Positive Term is one whose sign is + ; as +2. A Negative Term is one whose sign is ; as 5. Removal of Parentheses. 23. We have 9 + (5 + 6) = 9 + 5 + 6, since to add the sum 5 + 6 is equivalent to adding successively the single numbers of that sum. Again, 9 + (5 - 6) = 9 + [5 + (- 6)], since to add 6 is equivalent to subtracting 6. Therefore, removing brackets, 9 + (5 - 6) = 9 + 5 + (- 6), =9 + 5-6. The above example illustrates the following principle : (i.) When the sign of addition, -f, precedes parentheses, they may be removed, and the signs, + an d > within them be left unchanged; that is, yy+(+fl + 6) = /Y + a + 6, N+(+a-b)=N + a-b, etc. 21-25] PARENTHESES. 35 It is important to notice that if the first term within the parentheses has no sign, the sign + is understood. 24. We also have 9-(5 + 6) = 9-5-6, since to subtract the sum 5 + 6 is equivalent to subtracting successively the single numbers of that sum. Again, 9 - (5 - 6) = 9 - [5 + (- 6)], since to add 6 is equivalent to subtracting 6. Therefore, removing brackets, 9 _ (5 _ 6) = 9 - 5 - (- 6), =9-5 + 6. This example illustrates the following pri^-iple: (ii.) When the sign of subtraction, , precedes parentheses, they may be removed, if the signs within them be reversed from + to _ ? and from to + ; that is, yy-(+ fl + 6) = yv-a-6, N-(+a-b) = N-a + b, etc. Observe that the sign before the parentheses affects each term within them. Insertion of Parentheses. 25. The insertion of parentheses is the converse of the process of removing them. (i.) An expression may be inclosed within parentheses preceded by the sign + , if the signs of the terms inclosed remain unchanged. E.g., 7-5 + 3 -4 = 7 + (-5 + 3- 4), = 7-5 (ii.) An expression may be inclosed within parentheses pre- ceded by the sign , if the signs of the terms inclosed be reversed, from + to and from to +. E.g., 7-5 + 3-4 = 7- (5 -3 + 4), = 7-5-(-3 36 ALGEBRA. [Cn. II EXERCISES VI. Find the value of each of the following expressions, first removing parentheses : 1. 9 + (4+3). 2. 9 + (4-3). 3. 10-(3+4). 4. io_(3-4). 5. 12+ (6+8). 6. 12-(6+8). 7. i2_(6-8). 8. 12 + (-6+8). 9. 12-(-6+8). 10 . 15 + (9-6+2). 11. 15-(9-6+2). 12. 20-(7-9-l). 13. is _45-8. 14. l8--45-8. Insert parentheses in 10 7 + 4 6 and 7 + 8 9 4, 15. To inclose the last two terms, preceded by the sign + ; preceded by the sign . 16. To inclose the last three terms, preceded by the sign + ; preceded by the sign . The Associative Law. 26. The principle for inserting parentheses enables us to group successive terms in algebraic addition. Kg., 8 + (4 + l) = (8 + 4) + l, or 8 + 5 = 12 + 1. In general, a + (b + c) = (a + 6) + c. That is, the algebraic sum of three or more numbers is the same in whatever way successive numbers are grouped or asso- ciated in the process of adding. . This principle is called the Associative Law for addition and subtraction. 27. In finding the value of a continued product in Art. 21 9 the indicated operations were performed successively from left to right. E.g., 4 x 3 x (- 2) = 12 x (- 2) = - 24. But the result will be the same if 3 be first multiplied by 2, and then 4 be multiplied by this product. 25-29] PARENTHESES. 37 E.g., 4 x [3 x (-2)] = 4 x (- 6) = -24. In general, (ab)c = a (be). That is, the product of three or more numbers is the same in whatever way two or more successive numbers are grouped or associated in the process of multiplying. This principle is called the Associative Law for multiplication. The Commutative Law. 28. In an indicated addition, the number on the right of the sign + is to be added to the number on the left. E.g., in 5 + 3, =8, 3 is added to 5 ; while in 3 + 5, =8, 5 is added to 3. But the results are the same. That is, 5+3=3+5 In like manner, 8 5 = 5 + 8. In general, a + b = b + a ; a + 6 c = a c + b = etc. That is, the algebraic sum of two or more numbers is the same in whatever order they may be added. This principle is called the Commutative Law for addition and subtraction. 29. In an indicated multiplication, the number which follows the symbol of multiplication is the multiplier. E.g., in 4x3 = 4+4 + 4 = 12, the multiplier is 3 ; while in 3x4 = 3 + 3 + 3 + 3 = 12, the multiplier is 4. But the results are the same. That is, 4x3 = 3x4. In general, a x b = b x a ; a x b x c = a x c x b = etc. 38 ALGEBRA. [Cn. II That is, the product of two or more numbers is the same in whatever order they may be multiplied. This principle is called the Commutative Law for multiplica- tion. 30. By the preceding articles we have : 8-3 + 2-5 = 8 + 2-3-5= 10- 8 = 2 (i.) 25 x 27 x 4 = 25 x 4 x 27 = 100 x 27 = 2700, (ii.) 75 x 29 -r- 25 = 75 -r- 25 x 29 = 3 x 29 = 87. (iii.) In changing the order of the operations, it is important to carry the symbol of operation with the number. 31. Thus, by the methods of the preceding article, we secure the following advantages: In a succession of additions and subtractions, add the positive terms separately, then the negative terms, and unite the results, as in (i.). In a succession of multiplications and divisions, we may, by changing the order of the operations, frequently simplify the work, as in (ii.) and (iii.). EXERCISES VII. Find the value of each of the following expressions : 1. 8-3 + 2-5 + 9. 2. -6 + 4-14 + 12-7. 3. i9_7 + 3_5_io. 4. 16-7 + 4-9 + 3. 5. 17 + 2-3 + 9 -18. 6. 15-19 + 6-7 + 5. Find, in the most convenient way, the value of each of the following expressions : 7. 89-115 + 11. 8. 45f-85 + 54f. 9. 996 + 1008 + 4-8. 10. 98 + 96 + 92 + 2 + 4 + 8. 11. 25 x 32 x (- 4). 12. 12| x (- 29) x 8. 13. - 39 x 16| x 6. 14. 45 x 28 -h 9. 15. - 121 -j- 20 x 8. 16. 10 -- 42 x 21. 29-37] POSITIVE INTEGRAL POWERS. 39 POSITIVE INTEGRAL POWERS. 32. The Sign of Continuation, , is read, and so on, or and so on to; as 1, 2, 3, , read, one, two, three, and so on; or 1, 2, 3, , 10, read, one, two, three, and so on to 10. 33. A continued product of equal factors is called a Power of that factor. Thus, 2 x 2 is called the second power of 2, or 2 raised to the second power ; aaa is called the third power of a, or a raised to the third power. In general aaa to n factors is called the nth power of a, or a raised to the nth power. The second power of a is often called the square of a, or a squared; and the third power of a the cube of a, or a cubed. 34. The notation for powers is abbreviated as follows : a 2 is written instead of aa ; a 3 instead of aaa ; a n instead of aaa - to w factors. 35. The Base of a power is the number which is repeated as a factor. E.g., a is the base of a 2 , a 3 , , a". 36. The Exponent of a power is the number which indicates how many times the base is used as a factor, and is written to the right and a little above the base. E.g., the exponent of a 2 is 2, of a 3 is 3, of a n is n. The exponent 1 is usually omitted. Thus, a 1 = a. 37. The base of a power must be inclosed within parentheses to prevent ambiguity : (i.) When the base is a negative number. Thus, (_5) 2 =(-5)(-5) = 25; while - 5 2 = - (5 x 5) = - 25. 40 ALGEBRA. [Cn. II (ii.) When the base is a product or a quotient. Thus, (2 x 5) 3 = (2 x 5)(2 x 5)(2 x 5) = 1000; while 2 x 5 3 = 2 (5 x 5 x 5) = 250. :.*,. (f)'=| x f4 <*>.f-4--g. (iii.) When the base is a sum. Thus, while 2 + 3 2 = 2 + 3x3 = 2 + 9 = 11. (iv.) When the base is itself a power. Thus, (2 3 ) 2 = 2 3 x 2 3 = (2 x 2 x 2) (2 x 2 x 2) = 64. while 2 s2 = 2 3X3 =2 9 = 2x2x2x2x2x2x2x2x2 = 512. EXERCISES VIII. Express each of the following powers in the abbreviated notation : 1. a x a. 2. 4 x 4. 3. 2 x 2 x 2. 4. (- a) (- a). 5. -ax a. 6. (-3) (-3) (-3) (-3). 7. -nnnnn. 8. 2 x 2 x 2 ... to 8 factors. 9. ( a) ( a) ( a) .. to 9 factors. 10. (a + 6) (a + 6) (a + 6). 11. (x-yy)(x-yy)(x-yy). 12. (a + 6) (a + 6) (a + 6) to 12 factors. Express each of the following powers as a continued product : 13. 3 6 . 14. 6 3 . 15. -4 3 . 16. ( 4) 8 . 17. xf. 18. (xy) s . 19. (-a) 4 . 20. -a 4 . Write : 21. Four times #. 22. x to the fourth power. 23. The sum of the cubes of a and b. 24. The cube of the sum of a and b. 25. The length of a side of a square floor is a feet. How many square feet in the floor ? 37-38] POSITIVE INTEGRAL POWERS. 41 26. A field is 3 a rods long and 2 a rods wide. How many square rods in its area ? 27. A box is 4 # feet long, 3 x feet wide, and 2 x feet high. How many cubic feet does it contain ? Properties of Positive Integral Powers. 38. (i.) All (even and odd) powers of positive bases are positive. E.g., 2 3 = 2x2x2 = 8. 3 4 = 3 x 3 x 3 x 3 = 81. (ii.) Even poivers of negative bases are positive; odd powers of negative bases are negative. E.g., (- 2)4 m (- 2) (- 2) (- 2) (- 2) = 16 ; (_ 5)3 =(-5) (-5) (-5) = -125. In general, ( + a) m = + a m ; ( ay 2 " = a 1 " ; ( a) 2w+1 = a 2w+1 . EXERCISES IX. Find the value of each of the following powers : 1. 2' 5 . 2. 5 2 . 3. (-2) 6 . 4. -2 6 . 5. (-3) 5 . 6. (-2) 8 . 7. -3 3 . 8. (-3) 3 . 9. (-a) 6 . 10. (-a) 9 . Express as powers of 2 : 11. 8. " 12. 32. 13. 128. 14. 1024. 15. 4096. Express as powers of 3 : 16. 9. 17. -27. 18. -243. 19. 729. 20. -2187. Find the value of each of the following expressions : 21. 2 2 + 3 2 . 22. (2 + 3) 2 . 23. 3 3 -2 3 . 24. (3-2) 3 . 25. (4x3) 2 . 26. 6x4 2 . 27. 2(-3) 3 . 28. [2(-3)] 3 . When a = 5, b = 4, c = 2, find the value of each of the following expressions : 29. a c . 30. b a . 31. (ab) c . 32. bc a . 33. (abc) e . 34. a __ c 2. 35. a 2_ 6 2_ c 2_ 36 a .2 CHAPTER III. THE FUNDAMENTAL OPERATIONS WITH INTEGRAL ALGEBRAIC EXPRESSIONS. DEFINITIONS. 1. An Integral Algebraic Expression is an expression in which the literal numbers are connected only by one or more of the symbols of operation, +, , x, but not by the symbol -;-. E.g., 1 -f x + x-j o a 2 b + f cd 2 , etc. 2. The word integral refers only to the literal parts of the expression. E.g., a -f- b is algebraically integral ; but when a = ^, 6 = f , we have 3. Coefficients. In a product, any factor, or product of fac- tors, is called the Coefficient of the product of the* remaining factors. E.g., in 3 abc, 3 is the coefficient of abc, 3 b of ac, etc. A Numerical Coefficient is a coefficient expressed in figures. E.g., in Sab, 3 is the numerical coefficient of ab. A Literal Coefficient is a coefficient expressed in letters, or in letters and figures. E.g., in 3 ab, a is the literal coefficient of 3 b, and 3 a of 6, The coefficients + 1 and 1 are usually omitted. 4. A coefficient must not be confused with an exponent. E.g., 4a = aH-a + a-fa; while a 4 = a x a x a x a. 42 1_7] DEFINITIONS. 43 5. The sign +, or the sign , preceding a product, is to be regarded as the sign of its numerical coefficient. Thus + 3 a means the product of positive 3 by a ; 5 x means the product of negative 5 by x. In particular, -f- a means the product of positive 1 by a, and a means the product of negative 1 by a, unless the contrary is stated. EXERCISES I. What is the coefficient of x in l. 2x? 2. -3x? 3. 5 ax? 4. -7bx? 5. If the sum, a -f a -f a -f- a, be represented as a product, what is the coefficient of a ? 6. If the algebraic sum, b b b b b, be represented as a product, what is the coefficient of b ? Of b? 7. If the sum ax -f ax + ax -f- to 10 terms be represented as a product, what is the coefficient of ax ? Of x? 6. Like or Similar Terms are terms which do not differ, or which differ only in their numerical coefficients. E.g., in the expression + 3 a + 6 a& 5a + 7 ab, -h 3 a and 5 a are like terms ; so are + 6 ab and -f- 7 ab. Unlike or Dissimilar Terms are terms which are not like. .g. 9 -f 3 a and 7 ab in the above expression. 7. A Monomial is an expression of one term ; as a, 7 be. A Binomial is an expression of two terms ; as 2 a 2 + 3 be. A Trinomial is an expression of three terms. E.g., a + b-c, - 3 a 2 + 7 b* - 5 c 4 . A Multinomial * is an expression of two or more terms, in- cluding, therefore, binomials and trinomials as particular cases. E.g., a + b 2 , a 2 + b- c 3 , ab + bc-cd- ef. * The word Polynomial is frequently used instead of Multinomial. 44 ALGEBRA. [On. Ill ADDITION AND SUBTRACTION. Addition of Like Terms. 8. Like Terms can be united by addition into a single like term. Just as 2 = 1 + 1, so 2 xy = xy + xy ; just as 3 = 1 + 1 + 1, so 3 xy = xy + ## + xy. Therefore, just as 2 + 3 = 5, so 2 any + 3xy = (2 + 3)ojy = 6ajy. That is, to add fo'fce terms, add their numerical coefficients and annex to the sum their common literal part. x Ex. l. Add - 7 ab to 4 a&. We have 4 a& + (- 7 06) = [4 + ( 7)] a& = - 3 a&. Ex. 2. Find the sum of 3 a, 5 a, 8 a, 4 a. Uniting the positive terms by themselves, and the negative terms by themselves, we have Ex. 3. Add ax to to. Since the sum of the coefficients of x is a + 6, we have aa; -f- bx = (a + &) % EXERCISES II. Add: 1. 2. 3. 4. 5. 6. a -36 2x - 3m 7a 5# 2a -56 7cc 15m 12 a -4ai 3a -26 5 a? 11 m -5a 30 7. 8. 9. 10. 11. 12. -9a 2 11 a 2 -5a 2 4o?i/ 15 #2/ 12 xy 7 a? 2 ?/ aa? 6 x*y ex ay -by cy mx 2 nx 2 px 2 8-9] ADDITION AND SUBTRACTION. 45 Find the sum of : 13. 4a, 5 a, 7 a, 9a. 14. 5 a, 3x, -9a;, -13a?. 15. 6a 2 , -3a 2 , lla 2 , -2a 2 . 16. 11 xy, 17 xy, 5xy, 4 icy. 17. 8a 2 &, -3a 2 6, 27a 2 6, -Ila 2 &, -21a 2 6. 18. m + n, 5 (m + n), 9(m + n), 4 (m + n). 19. 3(a 2 + ^), -8(a 2 + &), -14(a 2 + 6), a 2 + 6. i Simplify the following expressions : 20. 5 x 13 x + 9 cc. 21. 7 a 9 a 4 a. 22. 5m + 13m-8m. 23. a 2 - 7 a 2 -f- 5 a 2 . 24. - a?b + 15 a 2 6 - 8 a?b 25. 2a 3 - 26. -7 27. 12 a 3 6 - 15 a 3 6 - 8 o 8 6 + 20 a 3 6 - 28. a + & - 3 (a + &) 4- 8 (a + 6) + 5 (a + 6) - 10 (a + &). 29. 5(x 2 + y 2 ) + 8(x 2 4-2/ 2 )-ll(^ 30. aj + ^aj-|a-Jaj. 31. 32. ia ^a + fa fa + fa + fa J gi-a. Simplify the following expressions, first removing paren- theses : 33. 2a [ 4a ( 6a)]. 34. m +[2m - (3m - 4m)]. 35. 6y-[5y-4y-(-3y + 2y)]-y. 36. x x 2xx 3x x Subtraction of Like Terms. 9. Like Terms can be united by subtraction into a single like term. Just as 5-2 = 3, so 5 a 2 a = (5 2) a = 3 a. That is, to subtract like terms, subtract their numerical coeffi- cients, and annex to the remainder their common literal part. 46 ALGEBRA. [Cn. Ill Ex.1. Subtract -5x 2 y from -7 tfy. We have = (- 7 EXERCISES III. Subtract : 1. 2. 3. 4. 5. 6. a 3x 5m 2m -Sy 3a 11 x -ox 7. 8. 9. 10. 11. 12. -la -llm -12m 3 a 2 5a 2 7m 3 8m 3 -3ai - 8 JJ 13. 13 a?b from 15 a?b. 14. -Itff from 15. f ^?/ 2 from | ay 2 . 16. | ab 5 from | a6 5 . 17. 2 (a + 6) from - 3 (a + 6). 18. or 2 + / from - 2 (a 2 + 2/ 2 ). Addition of Multinomials. 10. Unlike Terms are added by writing them in succession each preceded by the sign + . Ex.1. Add36to2a. Wehave2a + 36. Ex. 2. Add - 3 x 2 to 2 y\ We have Such steps as changing + ( Sx 2 ) into Sx 2 , should be performed mentally. 11. A multinomial consisting of two or more sets of like terms can be simplified by uniting like terms. Ex. 1. 2a-36-5a 46 = 2a-5a 12. If two or more multinomials have common like terms, these terms can be united. 9-12] ADDITION AND SUBTRACTION. 47 Ex. l. Add -2a + 36to3a-56. We have (3a-56) + (-2a +'3 6) =3a-5b -2 a + 36, = a-26. In adding multinomials, it is often convenient to write one underneath the other, placing like terms in the same column. Ex. 2. Find the sum of - 4 or 2 + 3 / - 8 z 2 , 2x 2 3z 2 , and We have -x 2 + 3y 2 -Sz 2 -3z 2 It is evidently immaterial whether the addition is performed from left to right, or from right to left, since there is no carry- ing as in arithmetical addition. EXERCISES IV. Add: 1. 2. 3. 4. 5. 6. a 2 3 -4 a m 1 b x c b n 7. a to a 2 . 8. # to x 2 . 9. 2m to n. 10. x 2 to -2^. 11. /y*yt "\"O 7/2 12. a 2 b to ab 2 . Simplify the following expressions by uniting like terms : 13. a + 2 + a- 2. 14. 56-3-46 + 4. 15. Wx-$ + 5-7x. 16. 9m-3n-8m-f 17. _ a s _ 5 a 2 + 4 a 2 + 2 a 2 + 2 a 3 . 18. ab - 3 a 2 6 2 + 5 ab - 8 a 2 6 2 + 4 06 + a 2 6 2 . 19. - 3(a 2 + 6)+ 5 (a + 6 2 )+ 4(a 2 + 6)- 4 (a + b 2 ). Simplify the following expressions, first removing paren- theses : 20. a + 1 (2 3 a). 21. 5x ( 2y + 3 a?). 22. 22 + 3a; 3a 4. 48 ALGEBRA. [Cn. Ill 23. 2w + 3w (5m 4n) ( 24. 2a# + 5y -(2ajy - 3yz) + 2ajy ~(3ajy - 2 ys) + 5 2/2. 25. a [3a- (2a + 6)] (36 5). 26. 3a [> + 3y (y 2a)]. 27. 8m [m (3m ri) + (2m 3w)]. Find the values of the expressions in Exx. 20-24, 28. When a = 1, a; = 3, y = 5, z = 10, m = 4, n = 7. 29. When a = - 3, a; = 6, y = 7, z = 8, w = 1, w = 2. Find the sum of the following expressions : 30. 5 a + 2 6, 3a 46, 7 a + 3 6, 9 a 6. 31. 7x 3y, 5a + 42/, -10ic + 4?/, 3x-7y. 32. a + 26-3c, 2a-36 + c, 2a + 56 + 2c. 33. a 2 + 2 a 4- 1, a 2 3 a 2, a 2 + 4 a + 2. 34. ^-5a + 6, 3ar 2 + 2z-7, Ga^ + Saj + l. 35. 2 ab + 3 oc 4, 3 a6 5 ac + 2, 5 ab + 3 ac + 8. 36. a 2 - 3 ab + 6 2 , 2 a' 2 + 2 a6 - 6 2 , db - 2 a 2 . 37. a 3 - 5 a 2 6, 7 a 2 6 - 6 3 , a 3 + 6 3 . 38. a 3 3^ + 58 1, 7^ + 2^-6^ + 4, 39. x 3 + 5 x*y 7 xy 2 2 f/ 3 , 2 a? 3 + 6 of?/ + 11 ay 2 15 y^, 40. ct 4 + 2a 2 -5a-3, - 3 a 4 + 2 a 3 + 6 a - 4, 2 a 4 - 7 a 3 + 3 a 2 + 9, 5 a 4 - 7 a 3 - 5 a 2 + a. 42. a? 2 (a + 6) 2 + 6 2 , a 2 + 3 (a + 6) 2 , a 2 2 6 2 . 44. |a 2 6-ia6 2 , -ia 2 6 + i6 2 , -|a 2 6- 45. l-x 2 -^ 12-14] ADDITION AND SUBTRACTION. 49 Subtraction of Multinomials. 13. Unlike Terms are subtracted by writing them in succes- sion, each preceded by the sign . Ex. Subtract llm from 2 n. We have 2 n ( 11 m) = 2n + llm. 14. If two multinomials have common like terms, these terms can be united. Ex. 1. Subtract 2 a + 3 b from 3 a 5 b. We have (3a-5&)-(-2a Ex.2. Subtract 2x 2 -6x-3 from 3x 2 -5x + l. Changing mentally the signs of the terms of the subtrahend and adding, we have 3x 2 -5x + l 2x 2 -6x-3 Ex. 3. Subtract 2x 2 -3z 2 from -4x 2 + 3y 2 , and from the result subtract 2y 2 + 5z 2 . When several multinomials are to be subtracted in succes- sion, the work is simplified by writing them with the signs of the terms already changed. We then have 2X 2 +3z 2 -2y 2 -5z 2 y 2 -2z 2 EXERCISES V. Subtract : 1. 2. 3. 4. 5. 6. 1 3 x x 2 mn a-b a b y x m ab 2 50 ALGEBRA. [Cn. Ill 7. 3 a 2 b from 4 a 3b. 8. 5 # -}- 4 ?/ from 4 -f- 5 ?/. 9. 7 m + 2 ?i from 3 m -f- 3 n. 10. 2^ 3 a; from 3 0^ 2 a;. 11. 5a- 12. 2^ 13. 2 xy + 5 cez 7 ?/z from 5 a;?/ + 3 xz 6 yz. 14. 2a 2 -3a6-126 2 from3a 2 -a6-115 2 . 15. 16. a*- 17. 2 a 3 - a 2 6 - b 3 from 3 a 3 + 2 a 2 6 + 3a& 2 . 18. o*-x-l from a 3 + 2^. 19. 2 (a? + ?/) 5 2 from 3 (a; + y) 4 2. 20. 6(a-fe)-3a + 6 2 from 5(a 6) - 2a + a 2 . 21. From the sum of 5ic 5^ + 32; and 4#-f-4;?/ 22 sub- tract 8#-2?/-2z. 22. From a 2 ab + 6 2 subtract the sum of 2 a 2 - 3 a& + 5 b 2 and a 2 + ab 4 6 2 . 23. How much does m 2 -f- n 2 exceed m 2 n 2 ? 24. How much does 1 # 2 exceed 2 3 a? 2 ? 25. What expression must be added to2a 36-f-4cto give 26. What expression must be added to xy + xz -\- yz to give a-2 4. 2/2 + 3,2 ? 27. What expression must be subtracted from a 2 + ab + Z> 2 to give a 2 - 2 a& + Ir ? 28. What expression must be subtracted from x 2 2xy + y* to give^ + 2^ + 2/ 2 ? 29. What expression must be added to x 2 -f- x + 1 to give ? Ifo? = 2a-36 + 4c, y = 3a + 2b -7 c, z = 9a-7b + 6c, find the values of 30. x+y+z. 31. a; y-\-z. 32. x+yz. 33. xyz. 14-17] PARENTHESES. f>1 If A=$x-ly+$z, B=-$ find the values of 34. A + B + C. 35. A - B + C. 36. A + B - C. 37. A - B - C. PARENTHESES. 15. The use of parentheses has been briefly discussed in Ch. II., Arts. 23-25. It is frequently necessary to employ more than two sets of parentheses, and to distinguish them the following forms are used : Parentheses, ( ) ; Brackets, [ ] ; Braces, { j . A Vinculum is a line drawn over an expression, and is equivalent to parentheses inclosing it. E.g., (a + b) (c d) = a + b - c d. 16. The principles given in Ch. II., Arts. 23-24, are to be applied successively when several sets of parentheses are to be removed from a given expression. 17. In removing parentheses we may begin either with the inmost or with the outmost. The following example will illustrate the method of remov- ing parentheses, beginning with the inmost : Ex. 4 a - 1 3 a + [2 a - (a - 1)] } = 4a 4a 1= 1. When, in such examples, we come to one of a pair of paren- theses, (, or [, or J, we must look for the other of like form. We then treat all that is contained in each pair as a whole. EXERCISES VI. Simplify each of the following expressions : 1. 2x-3y-[5x-(2y-3x-y)]. 2. a + 2&-[6a-{3&-(6a-6&){]. 52 ALGEBRA. [Cn. Ill 4. 6a-[7a-J8tt-(9a- 5. a - {5 6 - [a - (3 c - 3 6) + 2 c - (a - 2 6 - c)] {. 6. (7a-6)-{4ar-[2a-l-(3-4a-5)]}. 7. a - [a? - 1 2 a: - 3 - [4 x - 5 - (6 x - 1 x - 8)] \ ] . 8. a -3a-a-b + 5a-b-7a-6 _8a 6 , 9. Find the values of the expressions in Exx. 1-5, when = - 3, 6 = 4, c = 5, x = 8, y = - 9. 18. Ex. l. Express &(x y)+y x as a product, of which one factor is x y. We have 4 (x y) + ?/ x = 4 (a? y) (x y) = 3 (x y). The sign -f- r before a pair of parentheses can evidently be reversed from + to , or from to +, if the signs of the terms within the parentheses be reversed. Ex. 2. 7aj-l-3l-oj=7a?- EXERCISES VII. Write each of the following expressions as a product, of which the expression within the parentheses is one of the factors : 1. 3(a &) a + &. 2. 3. 3ra-5n-4(5n-3ra). 4. 1 - a n + 3(a n - 1). 5. 5(x 2 x+1) x* + x 1. 6. a t/ 2 6(y + z x). Write each of the following expressions as a single product, of which the expression within the first parentheses is a factor . 7. 2oj-l-3l-2oj. 8. 22m- 9. 5 (x 2 y 2 ) + 2 (y 2 x 2 ). 10. 7 (xy z) (z xy). Simplify the following expressions without removing the parentheses : 11. (a &) c + (6 a) c. 12. 5 (x y) z -f 5 (y x) z. 17-21] EQUATIONS AND PROBLEMS. 53 EQUATIONS AND PROBLEMS. 19. Ex. Find the value of x from 2x 5 = T + x. Adding 5 to both members of the equation, we obtain 2x 5 + 5 = or, since 5 + 5 = 0, 2# = Subtracting x from both members of the last equation, we have 2x x = 7 + 5+iP a; or, since x x = 0, 2 x x = 1 + 5. (1) Uniting terms, x = 12. Check: 2x12-5 = 7 + 12, or 24-5 = 7 + 12, or 19 = 19. 20. Observe that equation (1), Art. 19, could have been obtained directly from the given equation by transferring the term 5, with sign changed, to the second member, and the term + aj, with sign changed, to the first member. That is, any term may be transferred from one member of an equation to the other, if its sign be reversed from + to , or from to +. 21. Ex. Find the value of x from the equation x 3=8 3. Adding 3 to both members of the equation, we obtain : x _ 3 + 3 = 8 - 3 + 3 ; or, since 3 + 3 = 0, x = S. Check: 8 - 3 = 8 - 3, or 5 = 5. Observe that this step is equivalent to dropping the common term 3 from both members. That is, the same term, or equal terms, may be dropped from both members of an equation. This step is called cancellation of equal terms. T>4 ALGEBRA. [Cn. Ill 22. These examples illustrate the following method : Transfer all the terms containing the unknown number to one member of the equation, usually to the first member, and all the terms containing known numbers to the other member. Unite like terms. Divide both members by the coefficient of the unknown number. Check by substituting the value thus obtained in the given equation. 23. Pr. A boy being asked his age, replied, " If 10 is added to twice the number of years in my age the sum will be 40." How old was the boy ? Let x stand for the number of years in his age. Then 2 x stands for twice that number of years. The problem states, in verbal language : twice the number of years in the boy's age plus 10 is equal to 40 ; in algebraic language : 2 x + 10 = 40. Transferring 10, 2 x = 30. Dividing by 2, x = 15. The boy was 15 years old. Check: 2 x 15 + 10 = 40, or 30 + 10 = 40, or 40 = 40. EXERCISES VIII. Solve each of the following equations : 1. # + 4 = 9. 2. 3 + .T = 10. 3. x 5 = 6. 4. 15-a = 10. 5. ll-x = 13. 6. 3^ + 2 = 11. 7. 5x-3 = 17. 8. 7 + 12 a; = 31. 9. 41 -17 a; =7. 10. 15 = 3 + 4#. 11. 19 = 13-6x. 12. 14 = 8-3o;. 13. 9 + 5 x = 13 + 4 x. 14. 8 x - 5 = 10 x - 11. 15. 18-5x = 33-8x. 16. 14 a?- 13 = 7 a? + 29. 17. 3z-4 + 5o; = 7a; + 9. 18. 4a;-9 = 8a;-3-2aj. 22-23] EQUATIONS AND PROBLEMS. 55 19. 2 20. 11 # 15 4 x = 2x + 5 5 x. 21. 13^-25 + 7 x = 87 + 9 x + 9. 22. 5 + 14-8x = 3x-16-4ic. 23. 6 x + 7 - 15 x + 23 = 36 x + 15. 24. 6 a; - 25 + 3 x - 14 x = 25 - 3 a. 25. 4 a + (2 a 3) = 15. 26. 2 x- (5 t 7. 8. 9. 11. 12. -6 5 J -* -^ 7 a 2 : 5 2 5 2 abc 2 5 a;-?/^ 3 3xy~z 2 13. 2(a + Z>) by 3(a + ^) 2 . 14. - 5 (a - yf by Simplify the following continued products 15. 3 ab x 5 be x 6 ac. 16. - 7 afy X (- 2 ?/ 2 ^) x 3 xz 2 . 17. xy 2 x7 bx-z x 2 6^2. 18. x 2 ?/^ 1 x 5 x m y*" x ( Multiply : 19. 2a 3 6 2 c, -3& 2 c 3 , a*& 4 c, 5a6c 4 . 20. a m+2 , a 2 "*, a 3 ~ TO , a TO ~ n , a 2 "" 81 ". 21. aj" +1 , 5 a?"- 1 , 2 x-~ m , x m+2 . 60 ALGEBRA. [Cn. Ill Multiplication of a Multinomial by a Monomial. 30. If the indicated operation within the parentheses in the- product, 4 (2 -f- 3 1), be first performed, we have 4(2+3-l) = 4x4 = 16. But if each term within the parentheses be multiplied by 4 and the resulting products be then added, we have 4x2 + 4x3-4x1 = 8 + 12 -4 = 16, as above. Therefore 4(2 + 3-1) = 4x2 + 4x3-4x1. This example illustrates the following method of multiplying a multinomial by a monomial : Multiply each term of the multinomial by the monomial, and add algebraically the resulting products. That is, a (b + c (/) = ab + ac ad. This principle is called the Distributive Law for multiplica- tion. 31. Ex. 1. Multiply (x - y) by 3. We have 3(x y) = 3x 3y. Ex. 2. Multiply 3x 2y 7z by 4 #. We have = - 12 x 2 + 8 xy + 28 xz. Such steps as changing (4 a;) (3 a;) into 12 # 2 , ( into +8xy, and (4 x) (7 z) into +28x2!, should be per- formed mentally. The work may be arranged as in arithmetic, by placing the multiplier under the multiplicand : 3x-2y-7z - 12 x' 2 + 8 xy + 28 xz It is customary to multiply from left to right, instead of from right to left as in arithmetic. 30-31] MULTIPLICATION. 61 EXERCISES XII. Multiply : 1. x + 1 by 3. 2. a -3 by 5. 3. 2m + 5 by 3. 4. 3x-7by-8. 5. 2a + 36by3a, 6. 5 x 3y by 2 x. 7. 6a 2 -56 by 56. 8. 3x-5y 2 by -6xy. 9. Stf + Sy 2 by 2xy. 10. a + 6 - c by 5. 11. a-^-z by -3. 12. 3a + 26-5c by 4. 13. 5ra 3w -4_p by 3. 14. 2a 76 + 3cby 5 a. 15. 5 x- + 3 ?/ 2 2 z 2 by 2 cc^. Multiply a 2 - 3 a + 1 by 16. 2 a. 17. -36. 18. 5a6. 19. 6a 2 6 3 . Multiply x*y + 3 xy 5 ?/ 2 by 20. -3s* 21. -5i/ 2 . 22. ~6x*y. 23. Sx 2 ?/ 2 . Simplify the result of substituting a + 6 c for x, and a 6 + c for y, in the following expressions : 24. bbx-lay. 25. 3arbx-Uab 2 y. 26. 7abx + 2bcy. Find the values of the results of Exx. 24-26 27. When a = - 2, 6 = 3, c = - 4. 28. When a = 5, 6 = 7, c = 5J. Multiply 5 x n - 3 x n ~ 3 y 2 -f 4 x n ~y + 2/ n ~ 4 by 29. x 3 . 30. -5x 2 y. 31. 3a?Y- 32 - - Simplify the following expressions : 33. 4o?-2}[a?-3(2-aj)]aj-4}. 34. 13a~13{10[7(4a-3)-6]-9a}. 35. -206-2J-5[3-2x-6(4ic-7)]-3(5-2a;)|. 36. x + *x-2-lx-x 2 '2 ALGEBRA. [Cn. Til Multiplication of Multinomials by Multinomial? 32. Ex. Multiply 7-5 by 2 + 3. If we let a stand for 7 5, we have Now replacing a by 7 5, we obtain (2 + 3) (7 - 5) = 2 (7 - 5) + 3 (7 - 5) = 2x7-2x5 + 3x7-3x5. This example illustrates the following method of multiply- ing a multinomial by a multinomial : Multiply each term of the multiplicand by each term of the mul- tiplier, and add algebraically the resulting products. In general, (a + b) (c + d - e) = a (c + d - e) + b (c + d - e) = ac + ad ae + be + bd be. 33. 1. Multiply -3a + 26by 2a-3b. We have -3a + 26 2a - 36 4 aft 9a6-66 8 - 6 a 2 + 13 6 - 6 6 2 The work is arranged as follows : Write the multiplier under the multiplicand; the first partial product, i.e., the product of the multiplicand by the first term of the multiplier, under the multi- plier ; the second partial product under the first; and so on, placing like terms of the partial products in the same column. Ex. 2. Multiply x + a by x + 6. We have x + a x + b x~ + ax ab 4. ( a 4. ft) a- 4. -33] MULTIPLICATION. Ex. 3. Multiply 4a 2 + l-2a-8a 3 by 1+2 a. Arranging to ascending powers of a, we have 2a-4a 2 +8a 3 -16a* 1 - 16 a 4 Ex. 4. Multiply tf + yZ + l xy x y by x Arranging to descending powers of x, we have x 2 xy x + / y -f 1 a? + y + 1 a? 3 x 2 ?/ 3? -\- My 2 xy -\-x x 2 y xy*- xy _ + ^ xy x + / y + 1 a? -Sajjy +^ +1 Ex. 5. Multiply 2 a m+1 5 of 1 + 7 x 7 "- 1 by x 2m x* m ~\ We have 5 x m -f- 7 x"- 1 2 ^'" +1 - 7 or 1 * + 12 or 3 1 - 7 EXERCISES XIII. Multiply : l. a + 1 by a + 2. 2. a + 1 by a - 2. 3. m 5 by m + 3. 4. i/ 6 by T/ 5. 5. m - 12 by m - 3. 6. a - 12 by a - 15. 7. 2 a; + 1 by x + 3. 8. 3 a + 5 by 2 a - 3. 9. llm 6 by 2m 5. 10. 15 05 8 by 10 a 3. 11. x + y by x y. 12. 2 a + b by 3 a b. 64 ALGEBRA. [Cn. Ill 13. 3m 2 n by 5 m + 3 n. 14. 5x 6y by 3 x 2 y. 15. 2# 2 + 7y by 5x 2 -3y. 16. 11 m 2 + On by 5m 2 - 7 n. 17. 2a 2 + 36 8 by 4a 2 -56 2 . 18. 3ar + xy by 2x 2 + 3xy. 19. 7a 2 + 2a& by 3a 2 -5ab. 20. 6a 2 -5a?/ by 3x 2 -2xy. 21. x 2 + x + l by x 1. 22. 0^ 0; + 1 by av+ 1. 23. a 2 + 5a-6 by a - 3. 24. x 2 -llx + l2 by a? -8. 25. 2a 2 + 3a-o by 3a-2. 26. 6aj 2 -7aj + 2 by 6aj-7. 27. 5 x 2 - 2 a? + 1 by 5 x + 2. 28. 3 x 2 + 4 aj - 5 by 3 x - 4. 29. x 2 + 2xy + y 2 by a- + ?/. 30. a 2 2 ab + 6 2 by a &. 31. X s x 2 oj + 1 by aj + 1. 32. x 3 + x 2 H-ic+l by a; 1. 33. 8a? 4a^ + 2o; 1 by 2a? + l. 34. cc 3 + a? 2 ?/ + ^2/ 2 + y 3 by x y. 35. 27a 3 + 18a 2 6 + 12a& 2 + 8& 3 by 3a-26. 36. 2a + 36 + ocby2aH-36 5c. 37. Ga^ + Sa + l by Ga^- 38. l + ^ + ^y by 1-ajy- 39. 2a 2 -3a6 + 56 2 by 2a 2 40. 2 2 + 3 xy + 4 / by 3 x 2 - 4 xy + ?/ 2 . 41. 0^-2^ + 3^-1 by x 2 -3x + 2. 42. a5 4 5i 2 + 6.T 3 by x 2 + 5x 4. 43. x 4 -6c 3 + 2x + 5 by 3^-2^ + 5. 44. x 3 4 or?/ 4- 2 a# 2 t/ 3 by x 2 3 xy + y 2 . 45. x 4 + 2^ + a? 2 -4ar-li by ^-2^4-3. 46. x 2 xy 4- y 2 -f # + ?/ + 1 by # 4- y 1. 47. x n 2 a**- 1 3 x n ~ 2 5 cc n ~ 3 by x + 1. 48. 5 a;" + 3 a;"- 1 8 a; n ~ 2 3 by aj 2. 49. a n+1 -5a n -f 7a n ~ 1 -3 by a 2 + a + l. 50. aj 3 " o; 2n + x n 1 by a; n + 1. 51. a 2n - 2 a n l n + & 2 " by a 2 " + 2 a"6 n + ft 2 ". 33-34] MULTIPLICATION. 65 52. (x + m)(x + n). 53. (x m) (x ri). 54. (x + m)(x ri). 55. (x m) (x + n). 56. [> 2 - (a 4- 6) a; + 06] (a - c). 57. x 2 + a-&a Simplify each of the following expressions : 58. 59. 60. X - 61. 62. 63. 64. ^(y - 2) + f(z - x) + Z 2 - y) + (a? - y) (y -)(- a?). 65. (x-^+^-^ + ^-^ 66. (2m 2 + 3m-2)(m-l)(2m 67. (a? 68. (a? 69. (a 2 -a + l)(a 2 + a + l)(a 4 -a 2 + l). Simplify each of the following expressions : 70. (Ja 2 + i& 2 )(ia 2 + i& 2 ). 71. Qa 2 72. (fa-f& + fc)(f-f& + fc). 73. (Ja;_} y -|-52!)(4*-3y-i2!). 74. (21 -3-^ + 31 x 2 ) (i x 2 + 21 x + It). 75. (3a 2 -f-i6 2 -ic 2 )(fa 2 -i6 2 --|c 2 ). 76. (ax + i bx 2 + icx 3 ) (t aaj + t &* - Zero in Multiplication. 34. Since N- = N(b b), by definition of 0, = Nb - Nb = 0, we have N - = and - M = 0. That is, a product is if one of its factors be zero. 66 ALGEBRA. [CH. Ill EXERCISES XIV. 1. What is the value of 2 (a &), when b = a? 2. What is the value of (a + 6) (c d), when c = d? 3. What is the value of (b -\- c) (a -+- b c), when c = a -\-b? 4. What is the value of (a? 2 9) (x 4 7 a? + 2 x 9), when x = 3? For what values of x does each of the following expressions reduce to : 6. (x-)(x + 7)? 7. ( x -l)(x-d)? 25)? 9. x(x-a)(x-b)(x-c)? Equations and Problems. 35. Ex. Find the value of x from the equation 3(a;-4) + 5 = 4(> -3). Removing parentheses, 3x 12 -f- 5 = 4 a? 12. Cancelling -12, 3z + 5 = 4#. Transferring terms, 3# 4# = 5, or x = 5. Dividing by 1, # = 5. Check: 3(5 -4) +5 = 4(5 - 3), or 3 + 5 = 4x2, or 8 = 8. To solve such equations : Remove parentheses, and proceed as in Art. 22. Pr. A number of persons were to raise a fund by paying $ 5 each. Had there been 4 persons more, each would have had to contribute only $ 3. How many persons were there ? Let x stand for the number of persons. Then the number of dollars contributed was 5 x. Had there been 4 persons more, there would have been x + 4 persons. Then the number of dollars contributed would have been 3 (x + 4). 34-35] MULTIPLICATION. 67 The problem implies, in verbal language : the number of dollars contributed in the one case is equal to the number of dollars contributed in the other; in algebraic language : 5 x = 3 (x -f 4). Kemoving parentheses, 5 x = 3 x + 12. Transferring 3 a?, 2x = 12. Dividing by 2, x = 6. Check : 6 persons contributed 6 x 5, = 30 dollars ; 6 + 4, or 10, persons would have contributed 10 x 3, = 30 dollars. EXERCISES XV. Solve the following equations : 1. 6( + l) = 6. 2. 4 (2.3-1) = 5. 3. 3(z + 5) + 17 = 26. 4. 14 + 3(7-2z) = 29. 5. 15 + 4(8-2z) = 7. 6. 25-3(5-4a) = 22. 7. 27 + 4(2z -8) = 12. 8. 11(2- 5 a?) =47 -30 a. 9. 12(4 5) = 13 98a?. 10. 7x- 6(10 0?) = 33. 11. 4(2a5 + 3) _3(2a? + 4) = 10. 12. 5(3x + 4)-2(4z-3) = 54. 13. 7(2-3)-ll(5aj-4) = 64. 14. ( X -3)(x-4:)=x z + 5. 15. (a; - 4) (a; 6)= a; (a; 9). 16. ( + !)( + 2) = (a; -3) (a; -4). 17. 18. The sum of two numbers is 50. If five times the less exceeds three times the greater by 10, what are the numbers ? 19. Two boys, A and B, had the same number of apples. A said to B : " Give me 5 apples and I will have twice as many as you will have left." How many apples had each ? 68 ALGEBRA. [C'n. Ill 20. Add 10 to a certain number, and multiply the sum by 2, or subtract 8 from the same number, and multiply the difference by 5. The results will be equal. What is the number ? 21. A is 30 years old, and B is 12 years old. After how many years will A be twice as old as B? 22. A father is 30 years older than his son ; 5 years ago he was four times as old. What are the ages of father and son ? 23. A and B invested equal amounts. A gained $ 200, and B gained $2600. If B then had three times as much as A, how much did each invest ? 24. Three boys, A, B, and C, catch 128 fish. If B catches 10 more fish than A, and C catches three times as many as A and B together, how many fish does each boy catch. 25. In one room there are twice as many persons as in a second room. If 10 persons pass from the first room into the second, there will be three times as many persons in the second as in the first. How many persons are there in each room ? 26. A woman has enough money to buy 11 yards of cloth of one kind, or 8 yards of another kind. If the latter costs 30 cents more a yard than the former, how much does a yard of each kind cost ? 27. In a stairway there are 45 steps of a certain height. If the steps had been made 1 inch higher, there would have been only 40. How high are the steps ? 28. The capacity of a certain vessel is 90 gallons. One pipe lets in 2 gallons a minute and a second pipe 1 gallon. If the first pipe is opened 15 minutes before the second, how long after the first pipe is opened will the vessel be filled ? 29. A farmer has two fields containing together 5 acres. A offers to pay $ 62 an acre for the first field and $ 72 an acre for the second. B offers to pay $ 60 an acre for the first field and $ 75 an acre for the second. If both offers amount to the same, how many acres are there in each field ? 35-37] DIVISION. 69 30. The capacity of a certain cistern is 2200 gallons. One pipe lets in 80 gallons in a minute, and a second pipe 50 gallons. How many minutes must the first pipe be opened before the second in order that the cistern may be filled 4 minutes after the second pipe is opened? 31. One cask contains 70 gallons, and another 50 gallons. If three times as many gallons are drawn from the larger as from the smaller, the contents of the smaller will be equal to three times the contents of the larger. How many gallons are drawn from each cask ? 32. A man has $ 115 in two-dollar bills and five-dollar bills. If he has 35 bills altogether, how many of each kind has he ? 33. A rides his bicycle 12 miles an hour, and B his 10 miles an hour. A rides a certain number of hours, and B rides 2 hours longer. If they ride the same distance, liow many hours does each ride ? 34. Twenty-five men were to raise a certain fund by con- tributing equal amounts. But 5 men failed to contribute, and in consequence each of the remaining men had to contribute $ 2 more. What was to be the original contribution of each ? What was the amount of the fund ? DIVISION. 36. One power is said to be higher or lower than another according as its exponent is greater or less than the exponent of the other. E.g., a 4 is a higher power than a 3 or 6 2 , but is a lower power than a 6 or b 7 . Quotient of Powers of One and the Same Base. 37. Ex. a 7 -r- a 3 = (aaaaaad) -r- (aaa). = (aaaa) x (aota) -f- (aaa) = aaaa = a 4 = a 7 " 3 . 70 ALGEBRA. [Cn. Ill This example illustrates the following method of dividing a higher power by a lower power of the same base : The exponent of the quotient is the exponent of the dividend minus the exponent of the divisor; or, stated symbolically, We also have a m -r- a" = 1, when mn. E.g., a? -r- a 2 = 1. EXERCISES XVI. Express each of the following quotients as a single power : 1. 2 5 + 2. 2. 3 5 ^-3 2 . 3. x? + x 2 . 4. a 6 --a 4 . 5. x 7 -s-x\ 6. a 6 -=- a 5 . 7. (a) 6 -7- a 5 . 8. (3 x) 5 -5- (3 x). 9. (ab) 7 +(-aby. 10. 5 n --5 3 . 11. a n+1 -f- a. 1 o n>n+7 . n,n -i o />a+3 . /yia+1 -i 4. /v2n . /-w 1 J.^. X -r" X . AO. X -r- - 5 = 25. 43. The number 20, obtained by the first step of the division, is called the Partial Quotient at that stage. It is the greatest number whose product by the divisor is equal to or less than the dividend. In general, if D be the given dividend, d the given divisor, and q the partial quotient, the principle used above, stated symbolically, is : 0^d=q + (D-qd)^d. 44. The following example illustrates the application of this principle in dividing one multinomial by another. Ex. Divide x* + 3x + 2 by We have (1) (2) =a;+(2a;+2)-Ka;+l) (3) =a?+2+0-s-(a;+l) =x+2, since 0-s-(a;+l)=0. 74 ALGEBRA. [Cn. Ill We take the quotient of the term containing the highest power of x in the dividend by the term containing the highest power of x in the divisor as the partial quotient at each step. The work may be arranged more conveniently thus : a? + l x + 2, quotient. to be subtracted from x 2 + 3 x + 2 ; see (1) and (2) above. Remainder to be divided by x + 1 ; see (3) above. 2 (x + 1) to be subtracted from 2 x + 2 ; see (4). 45. The method of applying the principle of Art. 43 to the division of multinomials, as illustrated by this example, may be stated as follows : Arrange the dividend and divisor to ascending or descending powers of some common letter, the letter of arrangement. Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. Multiply the divisor by this first term of the quotient, and sub- tract the resulting product from the dividend. Divide the first term of the remainder by the first term of the divisor, and write the result as the second term of the quotient. Multiply the divisor by this second term of the quotient, and subtract the product from the remainder previously obtained. Proceed with the second remainder and all subsequent remainders, in like manner, until a remainder zero is obtained, or until the highest power of the letter of arrangement in the remainder is less than the highest power of that letter in the divisor. In the first case the division is exact ; in the second case the quotient at this stage of the work is called the quotient of the division, and the remainder the remainder of the division. 46. Ex. 1. Divide x 2 - 4 x - 5 by x - 5. We have x 2 4 x 5 !B 5 x 2 - 5 x x-5 x-5 44-46] DIVISION. Ex. 2. Divide a*b - 15 6 4 -f 19 ab 3 + a 4 - 8 a 2 b 2 by a 2 - 5 6 2 + Sab. Arranging to descending powers of a, we have 75 ^ 4. a 3 6 - 8 a L 7> 2 + 19 aW - 15 6 4 _ 5 a .2 6 a a 2 + 3 afr - 5 6 2 - 2 a 3 6 - 3 a 2 6* + 19 a& 3 a 2 - 2 a& + 3 6 2 3a 2 6 2 + 3a 2 6 2 + 156* Ex. 3. Divide 8 x 3 - y 3 by 2 ^ + 4 a 2 + ^/ 2 . Arranging the divisor to descending powers of x, we have : + 2 a;? 4 a? 2 ?/ 2 xy' 2 y 3 2x - y Observe that the remainder after the first partial division is arranged to descending powers of x. Ex.4. Divide 12a" +1 + Sa n - 45 a- 1 + 25a w ~ 2 by 6a-5. We have 6a-5 12 a n+1 - 10 a n 18 a* 46 a 1 18 a n - 15 a"- 1 30 a n ~ l -f 25 a n ~ 2 _ 3Q fl -i + 25 a"" 2 Ex. 5. Divide a,- 3 + (a + 6 + c) cc 2 4- (a& + ac + 6c) CP + abc by a? 2 + (a + 6) x + 06. We have x 3 + (a -f- b + c) x 2 + (ab + ac + be) x + a&c I x 2 + (a - or + (a + 6 ) x 2 + a&a I a? 4- c ex 2 + (ac + be) x + a&c ex 2 H- (ac + 6c) x -f- # &c 76 ALGEBRA. [Cn. Ill EXERCISES XIX. Find the values of the following indicated divisions : 1. (a- 3 + 2 a + 1) -*- ( + 1). 2. (x 2 + llaj+30)-5-(aj+5). 3. (a-a + x _ 90) _=_(_ 9). 4 . .(a-2 _ 5 x + 6) -s- (x - 3). 5. (a 2 + 7.T-44)-f-(x + ll). 6. (a*-3aj-40)-s-(aj-8). 7. (3^-13^-10) -=-(3x4-2). 8. (2 a 2 + a-6)-f-(2 a-3). 9. (15x 2 -7a-2)-:-(5a+l). 10. (6 z 2 - 23 a +20)-=- (2 #-5). 11. (oj- 4 as 2 -20 a + 3) -s- (a: + 3). 12. (a; 8 - 7 x 2 + 13 x - 15) -- (x - 5). 13. (4 x 3 - 3 a; 2 - 24 a; - 9) -s- (a? - 3). 14. (3a?-13aj s + 23aj-21)-fr-(3o;-7). 15. (18 a? + 7 x + 10) -s- (3 x + 2). 16. (50 x 3 - 23 x -j- 6) -s- (5 x - 2). 17. (a 2 + 2 aZ> + 6 2 ) -*- (a + 6). 18. 2x 2 + 6a 2 + 7aa?_=-2a; + 3a.. 19. 20. 21. (8 x 2 / - 65 z?/z 2 - 63 z 4 ) -5- 22. 23. 24. 25. (a c - 6 a 4 + 9 a 2 - 4) -s- (a 2 -!). 26. (21 a 6 6 + 20 6 4 - 22 tfb 3 - 29 a 4 6 2 ) - (3 a 2 6 - 5 6 2 ). 27. (a? + 8 x 2 + 9 x - 18) -s- (x 2 + 5 x - 6). 28. (a; 4 + ar 5 -4x 2 + 5a-3)--(a,' 2 + 2.T-3). 29. (6 .T 4 - x 5 - 11 x 2 - 10 x - 2) -- (2 a 2 - 3 x - 1). 30. (a? - 1) -*- (a 2 + + 1). 31. (a 3 + 8) -*- (a 2 - 2a + 4). 32. (^ 6 -642/ 3 )-(x 2 -4y). 33. (aV + /) -5- (ax + y). 46] DIVISION. 77 34. (04 _|_ x 2 + 1) - (x 2 - X + 1). 35. (aV 4- 64 a?) -r- (4 ax + aV + 8). 36 . (4 a 4 _ 25 c 4 - 30 6-c- - 9 b 4 ) - (2 a 2 + 5 c 2 + 3 6 2 ). 37. (27x 4 -6cV44c 4 )-f-(c 2 -6cx4-9z 2 ). 38. (8 aV + 32 a 6 + 1 ?i 6 ) -5- (4 an + ?i 2 + 4 a 2 ). 39. (16 a 4 Z> 2 +9 a*b*-l2 a s 6 s -8 a 5 6+3 a 6 ) -(a 4 +3 a 2 6 2 -2 a 3 6). 40. (28 a 5 c - 26 a 3 c 3 - 13 a 4 c 2 4- 15 a 2 c 4 ) -s- (2 a 2 c 2 + 7 a 8 c - 5 ac 3 ). 41. (81 z* - 90 6 4 ^ 4 + 81 6 6 ^ 2 - 20 Z> 8 ) -- (9 2 4 + 9 b 2 z 2 - 5 6 4 ). 42. 43. 44. (a 2 + 2 a& + b 2 - x* + 4a?y - 4 ?/ 2 )-f-(a + 6 - + 2 y). 45. (a 2 + 2 ac - 6 2 - 2 6d + c 2 - d 2 )-s-(a + c - b - d). Find the values of the following indicated divisions : 46. [a? 2 + (a + 1) # + a] -*- (* + a)- 47. [aj 2 -(a + 6)aj + a*]-s-(a;-6). 48. [cor (abc + 1) x + a6] -?- (x aft). 49. [(6 + c) x- box + aj 3 6c (6 4- c)] -=- ( 2 6c). 50. [or 3 +(a + b + c)^ 2 +(a& + ac + &c)# 4- a6c]-i- (aj + 6). 51. [a? (c/ 4- b 4- c) x 2 4- (a& 4- ac 4- 6c) # a&c] -s- (a? c). 52. (6 & - 25 x 2 " 4- 27 aJ" - 5) - (2 x w - 5). 53. (6 ^ - 11 x 4w 4- 23 ar 3w 4- 13 x 2w - 3 i w + 2) -(3 x w 4- 2). 54. (6 x* n+l - 29 tf n 4- 43 x 2 "- 1 - 20 x 2M ~ 2 ) -%- (2 x n - 5 x' 1 - 1 ). 55. (1 4- a fa - 2 a 3 *) -=- (3 a 2a5 + 2 a 31 + 2 a* 4- a 4 * 4- 1). 56. 57. 58. (- T 9 e ^ 6 4- tt 2 ^ 4 - f a 4 ^ 2 + J a)-(f ^ - | a 2 x + 59- (t 4 + tV & 4 + tfV - if c 4 )-(| a 2 4- i & 2 - | c 2 ). 78 ALGEBRA. [Cn. Ill 47. In the equation D + d = q + (D qd) + d, I) qd is the remainder at any stage of the work, and q is the corresponding partial quotient. If, for brevity, we let R stand for the remainder at any stage, we have D + d=q + R + d. (1) That is, the result of dividing one number by another is equal to the partial quotient at any stage, plus the remainder at this stage divided by the given divisor. E.g., 29 -=-6 = 4 + 5-=-6 = 4 + ; (a* _ x + 2 ) - (x + 1) = (aj - 2) + 4 - (x + 1). 48. If both members of the equation D -H d = q + R -r- d be multiplied by d, we have D-r-d X d = (q = qd + R -j- d x d = qd 4- R, since * d x d = -T- 1. Therefore, D = qd + R. That is, the dividend is equal to the product of the quotient at any stage and the divisor, plus the remainder at this stage. E.g., 29 = 4 x 6 + 5, and tf - x + 2 = (x - 2) (x + 1) -f 4. EXERCISES XX. Find the remainder of each of the following indicated divi- sions, and verify the work by applying the principle of Art. 48 : 1. (or 2 - 7 a + 11) -:- (a - 2). 2. (3aj 8 + 5*-9)-5-(oj-4). 3. (a 8 -17 x 2 + 15x-13)H-(2o;-5). 4. CHAPTER IV. INTEGRAL ALGEBRAIC EQUATIONS. We will now distinguish between two kinds of equations. Identical Equations. 1. An example of the one kind is : (a + 6) (a - 6) = a 2 - 6 2 . The first member is reduced to the second member by per- forming the indicated multiplication. 2. Such an equation is called an Identical Equation, or more simply, an Identity. 3. Notice that identical equations are true for all values that may be substituted for the literal numbers involved. E.g., if a = 5 and 6 = 3, the above equation becomes 8 x 2 = 25 - 9, or 16 = 16. Conditional Equations. 4. An example of the second kind is : x + 1 = 3. The first member reduces to the second member, when x = 2. It seems evident, and it is proved in School Algebra, Ch. IV., that x -|- 1 reduces to 3 only when x = 2. 5. Such equations impose conditions upon the values of the literal numbers involved. Thus, the equation in Art. 4 imposes the condition that if 1 be added to the value of x, the sum will be 3. 79 80 ALGEBRA. [Cii. IV A Conditional Equation is an equation one of whose members can be reduced to the other only for certain definite values of one or more letters contained in it. Whenever the word equation is used in subsequent work we shall understand by it a conditional equation, unless the con- trary is expressly stated. 6. An Integral Algebraic Equation is an equation whose mem- bers are integral algebraic expressions in an unknown number or unknown numbers. E.g., 3x? 4 = 2x, and f # + 5y | are integral equations. 7. The Degree of an integral equation is the degree of its term of highest degree in the unknown number or numbers. 8. A Linear or Simple Equation is an equation of the first degree. E.g., x -f- 1 = 6 is a linear equation in one unknown number. 9. A Solution of an equation is a value of the unknown num- ber, or a set of values of the unknown numbers, which, if substituted in the equation, converts it into an identity. E.g., 2 is a solution of the equation x + 1 = 3, since, when substituted for x in the equation, it converts the equation into the identity 2 + 1=3. The set of values 1 and 2, of x and y, respectively, is a solu- tion of the equation x -f- y = 3, since 1 + 2 = 3 is an identity. 10. To Solve an equation is to find its solution. An equation is said to be satisfied by its solution, or the solu- tion is said to satisfy the equation, since it converts the equation into an identity. 11. When the equation contains only one unknown number, a solution is frequently called a Root of the equation. E.g., 2 is a root of the equation x + 1 = 3. 5-13] INTEGRAL ALGEBRAIC EQUATIONS. 81 Equivalent Equations. 12. Consider the solution of the equation fa?-5 = l. (1) Adding 5 to both members, }* = 6. (2) Dividing by 3, %x = 2. (3) Multiplying by 4, x = 8. (4). It is evident that 8 is a root of equations (1), (2), (3), and (4). In thus applying the principles of Ch. L, Art. 17, we re- place the given equation by a simpler one, which has the same root, this equation by a still simpler one, which again has the same root, and so on. Such equations as (1), (2), (3), and (4) are called Equivalent Equations. In general, two equations are equivalent when every solution of the first is a solution of the second, and every solution of the second is a solution of the first. 13. It is important to notice that the use of the principles given in Ch. L, Art. 17, may lead to incorrect results. Thus, by (iii.), we should be permitted to multiply both members of an equation by an expression which contains the unknown number. E.g., the equation x 3 = has the root 3. Multiplying both members by x 2, we obtain 0-3)0-2) = 0. This equation has the root 3, since (3 - 3) (3 - 2) = 1 = ; and also the root 2, since (2 -3)(2-2) = -1 -0 = 0. 82 ALGEBRA. [Cn. IV But 2 is not a root of the given equation, since 2 3 does not equal 0. That is, in multiplying both members by x 2, we gained a root 2. Observe that this root is the root of x 2 0. The derived equation is therefore not equivalent to the given one. Again, by (iii.), we should be permitted to multiply both members of an equation by 0. Multiplying both members of x 3 = 0, by 0, we have 0(a;-3) = 0. Any number is a root of this equation, since 0(1 - 3) =0, 0(2 - 3) = 0, 0(3-3) = 0, 0(4- 3) =0, etc. Finally, by (iv.), we should be permitted to divide both members of any equation by an expression which contains the unknown number. E.g., the equation (x 1) (x -f- 1) = 3 (a; 1), has the root 1, since (1 - 1) (1 + 1) = 3 (1 - 1), or x 1 = 3 x 0, or = j and the root 2, since (2 - 1) (2 + 1) = 3(2 - 1), or 1x3 = 3x1. Dividing both members by x 1, we obtain x + 1 = 3. This equation has the root 2 only, and not the root 1 of the given equation. That is, in dividing both members by x 1, we lost the root 1. Observe that this root is a root of x 1 = 0. The derived equation is therefore not equivalent to the given one. 14. The correct statements of the principles which are applied in solving equations are, therefore, as follows: (i.) Addition and Subtraction. TJie equation obtained by adding to, or subtracting from, both members of an equation the same number or expression is equivalent to the given one. 13-14] INTEGRAL ALGEBRAIC EQUATIONS. 83 (ii.) Multiplication and Division. The equation obtained by multiplying or dividing both members of an equation by the same number, not 0, or by an expression which does not contain the unknown number or numbers, is equivalent to the given one. These principles are proved in School Algebra, Ch. IV. In the solutions of equations in the preceding chapters, we multiplied or divided only by Arabic numerals. Nevertheless, we required each result to be checked. EXERCISES I. Solve each of the following equations : 1. x(x + 3) = x(x-5). 2. 3x(x-5) = 3x(x + 2). 3. 2(x + 1) - 3(x + 1) + 9(oj + 1) + 18 = 7(x + 1). 4. 5 (x - 7) - 4 (oj - 7) + 11 (oj - 7) = 10 + 2 (x - 7). 5. _8(3aj-5) + 5(3aj-5)-17-2(3a;-5) = 3. 6. ( + l) + a>(a + 2) = (aj + 3)(2a;-l). 7. (5 a -2) (3 a -4) = (3 a + 5) (5 a -6). 8. 2z + 2( + 3 = 2( + 2a;-5. 9. 10. (16 a + 5) (9 # -f- 31) = (4 o,' + 14^36 a; + 10). 11. ar J _.Tl-x-23-^ = ic + l. 12. 13. 14. 4 4{4 4[4 4(4 a?)]} = 15. 16. 4{4[4(4aj-3)-3]-3j-3 = l. 17. 3[5{5(a?-3)-3j-7] = 2(aj + 2 84 ALGEBRA. [Cn. IV Problems. Pr. 1. A man has $ 4.50 in dimes and dollars, and he has five times as many dimes as dollars. How many coins of each kind has he ? Let x stand for the number of dollars. Then 5 x stands for the number of dimes. We must tirst express the dimes as fractional parts of dollars, or the dollars as multiples of dimes. The latter method is the simpler. Since one dollar is 10 dimes, x dol- lars are 10 # dimes. The man evidently has 45 dimes. The problem states, . in verbal language : ten times the number of dollars plus the number of dimes is equal to 45 ; in algebraic language : 10 x -f 5 x = 45, 15^ = 45; whence x = 3, the number of dollars. Then 5 x, = 15, the number of dimes. Evidently the value of the coins is 3 + i-J dollars, or $ 4.50. As in this problem, the magnitudes of all concrete quantities of the same kind must be referred to the same unit ; if x stand for a certain number of yards, then all other distances must likewise stand for numbers of yards, not of miles or of feet. Pr. 2. I have in mind a number of six digits, the last one on the left being 1. If I bring this digit to the first place on the right, I shall obtain a number which is three times the number I have in mind. What is the number ? Let x stand for the number which is composed of the five digits on the right of 1. Then the original number is 100,000 + x. When 1 is moved to the first place on the right, each digit in x is moved one place to the left. Therefore, the resulting number is 10 x -f 1. 14] INTEGRAL ALGEBRAIC EQUATIONS. 85 The problem states, in verbal language : the resulting number is equal to three times the original number ; in algebraic language : 10 x + 1 = 3 (100,000 + x), whence 7 x = 299,999, and x = 42,857. Therefore the required number is 142,857. Pr. 3. A man asked another what time it was, and received the answer : " It is between 5 and 6 o'clock, and the minute- hand is directly over the hour-hand." What time was it? At 5 o'clock, the minute-hand points to 12 and the hour- hand to 5. The hour-hand is therefore 25 minute-divisions in advance of the minute-hand. Let x stand for the number of minute-divisions passed over by the minute-hand from 5 o'clock until it is directly over the hour-hand between 5 and 6 o'clock. Since the minute-hand must pass over 25 more minute- divisions than the hour-hand in order to overtake the latter, the number of minute-divisions passed over by the hour-hand is x 25. The problem states, or implies, in verbal language : the number of minute-divisions passed over by the minute-hand is 12 times the number of minute-divi- sions passed over by the hour-hand; in algebraic language : x = 12 (x 25). From this equation we obtain x = 27^. Consequently, the two hands coincide at 27 T 8 T minutes past 5 o'clock. EXERCISES II. 1. The sum of three consecutive numbers exceeds the second by 42. What are the numbers ? 2. A and B divide a sum of money. A receives $ 3 as often as B receives $5. If A receives $ 3 x, how many dollars does B receive ? 86 ALGEBRA. [Cn. IV 3. A and B divide $ 1200. A receives $ 3 as often as B receives $ 5. How many dollars does each receive ? 4. The length of a room is four times its width. If it were 12 feet shorter and 12 feet wider, it would be square. What are the dimensions of the room ? 5. A man travels 144 miles by train, boat, and stage. He travels 20 miles farther by boat than by stage, and three times as far by train as by boat and stage together. How many miles does he travel by each conveyance ? 6. A man paid a debt in four monthly payments. He paid $45 more each month than the preceding. If his debt was three times his last payment, how much was his first payment ? How much was his debt ? 7. In a number of two digits, the tens' digit is three times the units' digit. The number itself exceeds four times the units' digit by 54. What is the number ? 8. In a number of two digits, the tens' digit is twice the units' digit. If the digits are interchanged, twice the resulting number exceeds the original number by 9. What is the number ? 9. Three boys, A, B, and C, have a number of marbles. A and B have 55, B and C have 62, and A and C have 57. How many marbles has each boy ? 10. A man, wishing to give alms to several beggars, lacks 15 cents of enough to give 22 cents to each one. If he were to give 20 cents to each one, he would have 1 cent left over. How many beggars are there ? 11. A, travelling 25 miles a day, has 3 days' start of B, who travels 30 miles a day in the same direction. After how many days will B overtake A ? 12. The sum of two numbers is 47, and their difference increased by 7 is equal to the less. What are the numbers ? 15] INTEGRAL ALGEBRAIC EQUATIONS. 87 13. The sum of three consecutive even numbers exceeds the least by 42. What are the numbers ? 14. Atmospheric air is a mixture of four parts of nitrogen with one of oxygen. How many cubic feet of oxygen are there in a room 12 yards long, 5 yards wide, and 17 feet high ? 15. A merchant paid $ 7.50 in an equal number of dimes and five-cent pieces. How many coins of each kind did he pay? 16. A man has $5.70 in dimes and quarters, and he-has 6 more quarters than dimes. How many coins of each kind has he ? 17. In my right pocket I have as many dollars as I have cents in my left pocket. If I transfer $6.93 from my right pocket to my left, I shall have as many dollars in my left pocket as I shall have cents in my right. How much money have I in my left pocket ? 18. One barrel contained 36 gallons, and another 60 quarts, of wine. From the first three times as much wine was drawn as from the second; the first then contained twice as much wine as the second. How much wine was drawn from each ? 19. A regiment moves from A to B, marching 18 miles a day. Two days later a second regiment leaves B for A, and marches 26 miles a day. At what distance from A do the regiments meet, A being 212 miles from B ? 20. A man travels 3 miles in one hour. During the first half-hour, he goes 10 yards farther every minute than during the second half-hour. How many yards a minute does he go the first half-hour ? 21. The greatest of three vessels holds 28 gallons more than the second, and 45 gallons more than the third. If the con- tents of the second and third, when full, are poured into the first, when empty, the latter will lack 8 gallons of being filled. What is the capacity of each vessel ? 88 ALGEBRA [Cn. IV 22. A father leaves $ 25,800 to his four sons. The first receives twice as much as the second, less $ 300 ; the second three times as much as the third, less $ 600 ; and the third four times as much as the fourth, less $ 900. How many dollars does each son receive ? 23. Two bodies move from the same point in the same direc- tion, one at the rate of 24 feet a minute, the other at the rate of 30 feet a minute. If the second starts 35 minutes after the iirst, where will it overtake the first ? When will the distance between them be 270 feet before they meet ? When 270 feet after they meet ? 24. A child was born in November. On the 10th of Decem- ber the number of days in its age was equal to the number of days from the 1st of November to the day of its birth, inclu- sive. What was the date of its birth ? 25. A person attempts to arrange a number of coins in the form of a square. On the first attempt, lie has 31 pieces left over. When he adds 2 to each side of his square, he lacks 25 coins of enough to complete this square. How many coins has he ? 26. In a certain family each son has as many brothers as sisters, but each daughter has twice as many brothers as sisters. How many children are in the family ? 27. A merchant's investment yields him yearly 33 J% profit. At the end of each year, after deducting $ 1000 for personal expenses, he adds the balance of his profits to his invested capital. At the end of three years his capital is twice his original investment. How much did he invest ? 28. I have in mind a number of four digits, the first one on the right being 2. If I bring this digit to the last place on the left, I shall obtain a number which is less than the number I have in mind by 2106. What is the number? 29. At what time between 3 and 4 o'clock will the minute- hand of a watch be directly over the hour-hand ? At what time between 9 and 10 o'clock ? CHAPTER V. TYPE-FORMS. 1. We shall in this chapter consider a number of products and quotients which are of frequent occurrence. They enable us to shorten work by writing similar products and quotients without performing the actual multiplications and divisions. They are called Type-Forms. TYPE-FORMS IN MULTIPLICATION. The Square of a Binomial. 2. By actual multiplication, we have (a + b) 2 =(a + b)(a + b) = a 2 + ab + 6a + 6 2 = o 2 + 2 a6 + 6 2 . That is, the square of the sum of two numbers is equal to the square of the first number, plus twice the product of the two num- bers, plus the square of the second number. E.g., (2x + 5y) 2 = (2 xf + 2 (2 oj)(5 y} + (5 yf 3. By actual multiplication, we have (a - 6) 2 = (a - b) (a - 6) = a 2 - ab - ba + 6 2 = a 2 - 2 ab + b\ That is, the square of the difference of two numbers is equal to the square of the first number, minus twice the product of the two numbers, plus the square of the second number. E.g., (3x-7y) 2 = (3 *) 2 - 2 (3 x)(7 y) + (7 y) 2 = 9 x 2 - 42 xy + 49 y 2 . 90 ALGEBRA. [Cn. V 4. Observe that this type-form is equivalent to that of Art. 2, since a b = a + (&). E.g., (3x-7y) 2 = (3x) 2 + 2 (3 a?) (- 1 y) + (- 7*,) 2 = 9 x 2 42 x?/ -f 49 ?/ 2 , as above. The signs of all the terms of an expression which is to be squared may be changed without changing the result. For, (a - b) 2 = [ - (b - a)] 2 = (b - of 5. In applying the type-forms in this Chapter, it will be necessary to raise a monomial to any required power. We have (5 a 3 & 4 ) 2 = 5 - 5 a s a s b 4 b*= 5 2 a 3+3 & 4+4 = 5 2 a 2x3 6 2x4 = 25 a 6 6 8 . That is, to square a monomial : Square the numerical coefficient, and multiply the exponent of each literal factor by 2. In general, to raise a given monomial to any required power : Raise the numerical coefficient to the required power, and mul- tiply the exponent of each literal factor by the exponent of the required poiver. E.g., (3 ab 2 ) s = 3W x8 = 27 o 8 6 8 . EXERCISES I. Write, without performing the actual multiplications, the values of : 1. (a + 1) 2 . 2. (x-3) 2 . 3. (a + 5) 2 . 4. (a_4) 2 . 5. (3x + 2) 2 . 6. (4-oz) 2 . 7. (mn + 6) 2 . 8. (ab-S) 2 . 9. (xy + z) 2 . 10. (4z 2 -3) 2 . 11. (3xy + 5z) 2 . 12. (2ab-6bc)*. 13. (xy' 2 -3x 2 y) 2 . 14. (2a~b y - 9c 2 ) 2 . 15. (4 a 2 6 3 - 8 c 4 ) 2 . 16. (z n + l) 2 . 17. (x m -y n ) 2 . 18. (a n+1 + a' 1 - 1 ) 2 . Simplify the following expressions : 19. a * + V-(a-V)*. 20. (x-y) 2 -(x + y) 2 . 21. x 2 + y 2 - 4:x + 6y + 3, when x = a -f 1, y = a-2. 22. (a + b - c)(a + 6) + (a - 6 + c)(a + c) + (6 + c - a) (6 + c). 4-6] TYPE-FORMS IN MULTIPLICATION. 91 Verify the following identities : 23. (a 2 + b 2 ) (or + y 2 ) - (ax + by) 2 = (ay - bx) 2 . 24. a 2 + b 2 + 4c 2 + 2ab + 86c = 4 (a + c) 2 , when b = a. Product of the Sum and Difference of Two Numbers. 6. By actual multiplication, we have (a + 6) (a - b) = a 2 - ab + ba - b 2 = a 2 - 6 2 . That is, the product of the sum of two numbers and the difference of the same numbers, taken in the same order, is equal to the square of the first, minus the square of the second. Ex. 1. (2x + 3y)(2x-3y) = (2x) 2 -(3yy = 4:X 2 -$y 2 . The product of two multinomials can frequently be brought under this type-form by properly grouping terms. Ex. 2. (x* + x + I)(x 2 -x + 1) = [> 2 + 1)+ x- Ex.3. x- EXERCISES II. Write, without performing the actual multiplications, the values of: 1. ( + 2)(a-2> 2 . (a -6) (a + 6). 3. (m + 9)(m-9). 4. (2 a + 1) (2 a - 1). 5. (5 a -7) (5* + 7). 6. (9-6)(9 + 5a;). 7. (2a + 36)(2a-36). 8. (5x 6y)(5x + 6y). 9. (-8w+5n)(8m+5?i). 10. (a& + 1) (a& - 1). 11. (3 ax 4) (3 ax + 4). 12. ( ay + z) (ajy + z). 13. (-2a&4-c)(2a6 + c). 14. (5 a?y - 3 z) (5 xy + 3 2). 92 ALGEBRA. [Cn. V 15. 2 + 1)(> 2 -1). 16. (3 a 3 + 4) (3 a 3 - 4). 17. (5a 4 -26)(5a 4 4- 26). 18. 19. (3 a" 4- 5) (3 a" -5). 20. (-5 # n+1 4- 9 a?"- 1 ) (5 aj" +1 + 9 a;' 1 - 1 ). 21. [a 2 4- 6 (a + 6)] [a 2 - 6 (a + 6)]. 22. (aj + y + 5)(aj + y-5). 23. (4 a - 3 b - 7) (4 a - 3 b + 7). 24. (^ + / + z 2 )(-* 2 + 2/ 2 + Z 2 )' 25. (a 2 - ab 4- 6 2 ) 2 + 2a;-l)(> 2 -2z-l). 27. (x 4 - x 2 4- 1) (V 4- a^ - 1). 28. (-c^- Simplify the following expressions : 29. (l 30. (2x 31. (a; - 3) (a -1) (a? + !)( + 3). 32. (a x)(a + x) (a 2 + x~) (a 4 33. (aj'- 34. (aj z - 35. (a + & - c) (a + c - 6) (6 + c - a) (a + & -f c). The Product (x + a) (x + 6) . 7. By actual multiplication, we have (JT + a)(jf + 6) = * 2 + a* + 6jr + a6 = jr 2 + (a + 6)jr + a^ ; (JT + a)(jr 6) = jr 2 + ajr 6jr ab = x 1 + (a b)x ab ; (JT a)(jr 6) = x 2 ax bx + ab = x 2 (a + 6)jf -f aA. We thus derive the following method for multiplying two binomials which have a common first term : The first term of the product is the square of the common first terms of the binomials. 6-7] TYPE-FORMS IN MULTIPLICATION 93 The coefficient of the second term of the product is the algebraic sum of the second terms of the binomials. The last term of the product is the product of the last terms of the binomials. Ex. 1. Write the product (x + 3) (a; + 7). The first term is x* ; The second term is (3 -f- 7) a;, = 10 x ; The third term is 3 x 7 = 21. Therefore (x + 3) (x + 7) = x* + 10 x + 21. Ex. 2. Write the product (x 8) (x + 2). First term : x 2 ; second term : ( 8 + 2) x, = 6 a? ; third term : - 8 x 2 = - 16. Therefore (a; - 8) (x + 2) = x 2 - 6 x - 16. Ex. 3. Write the product (a 2 + 9) (a 2 - 3). First term : (a 2 ) 2 , = a 4 ; second term : (9 3) a 2 , = 6 a 2 ; third term: 9 x (-3), =-27. Therefore (a 2 + 9) (a 2 - 3) = a 4 + 6 a 2 - 27. Ex. 4. Write the product (x 5y)(x7 y). First term: x 2 ; second term : (5y 7y)x, = 'L2xy J third term : 5yx(7y), = 35 y 2 . Therefore (x - 5 y) (x - 7 y) = x 2 - 12 xy -f 35 y 2 . EXERCISES III. Write, without performing the actual multiplications, the values of : 1. (aj + 2)(aj + 3). 2. (x + 2) (x - 3). 3. (x - 2) (x -f 3). 4. (aj-2)(aj-3). 5. (a? + 5) (a? + 8). 6. (a? + 6) (a? - 8). 7. 0-5)( + 8). 8. 0-5)0-8). 9. (8 + m)(w 9). 10. (5 + a) (a -6). 11. (7 + 3 a?) (7 -a?). 12. (- 3 + 5 a) (6 + 5 a). 13. a + aj + 2. 14. x + yx 94 ALGEBRA. [Cn. V 15. ( x -y)(x-2y). 16. (ab + 1) (ab - 3). 17. (#y + 7) (xy - 8). 18. (ab + 3 c) (a& - 5 c). 19. (or' + 8) (or 2 9). 20. (afy - 5) (xhj + 11). 21. (a#* + 9 a) O/ - 6 a). 22. (a 8 + 3 aft) (a 2 - 2 a&). 23. (a n + 2) (a n - 5). 24. (x TO+l - 3)(^ w+1 + 8). 25. ( a + t + 3)(a + & 7). 26. (a - y + 3z) (x - y - 5z). The Product (ax + 6)(r + - lOx - 24. 17. a? + 3x - 40. 18. x 2 - 18 x - 40. 19. x 2 + 6 #-40. 20. x 2 --39x-40. 21. x 2 - 4x-60. 22. # 2 + 7x-30. 23. x 2 + 12x + 32. 2 4. a 2 -3x-40. 25. x 2 - 12 a; + 35. 26. x 3 -17 x 2 +72 x. 27. x 2 + 13 x - 30. 28. ex-x^x 3 . 29. 35 + 2x-x 2 . 30. a; 4 + 4 a 2 21. 31. x 4 + 8#2 + 15. 32. x 4 -24x 2 + 63. 33. 3x 6 +39x 3 +66. 34. x 6 - x 3 - 56. 35. a*+6 x"-112. 36. x 2 " - 16 x n + 55. 37. a? 2 + (a -f &) a? + a&. 38. cc 2 (m + w) x + wii. 39. ic 2 + (_p q)x pq. 40. x 2 + (3 r 2 s) a? - 6 rs. 41. a^ + 7a 2 + 6a 3 . 42. x 2 + 2xy-15y 2 . 43. # 2 -4 ax -12 a 2 . 44. ic 2 - 7 ax + 12 a 2 . 45. 2xy-26xy + 84x2/ 4 . 46. or - 11 xm + 30 m 2 . 47. x 2 z 2 + 12xz- 13. 48. a 2 b 2 -7ab + 10. 49. m 2 n 2 - 20 mn + 99. 50. 1 - 25 ay + 126 or /. 51. !-23a 2 6 + 132a 4 6 2 . 52. oV - 23 a 2 x + 120. 53. ajy -7x 2 y 2 - 78. 54. a 4 6 6 + 3 a s 6 - 108. 55. a 4 6 8 + 5 a 2 6 4 ^ - 84 x 4 . 56. a 2 "6 2n - 2 a n &"c? - 15 c 4 . 13. From Ch. V., Art. 8, we have (fljr + 6) (ex + (/) = act 2 + (at/ 4- 6c) JT + M. A trinomial which can be factored by this type-form must satisfy the following conditions : (i.) One term of the trinomial is the product of the first terms of its binomial factors. (ii.) A second term of the trinomial is the product of the second terms of its binomial factors. (iii.) The remaining term of the trinomial is the sum of the cross-products. Ex. 1. Factor 6 x 2 + 19 x + 10. The first terms of the required binomial factors are factors of 6 x 2 , the second terms are factors of 10, and the sum of the cross-products is 19 x. 12-13] INTEGRAL ALGEBRAIC FACTORS. 107 The factors of 6 x 2 are : x and 6 x, 2 x and 3 x ; and the factors of 10 are : 1 and 10, 2 and 5. The following arrangements represent possible pairs of factors : a- + 1 a? + 10 x + 2 6x + h. x 16 a? 61a; 17 x 2# + 10 2x + 2 XXX & + 1 3.+ 5 3z + 2 32x 16x 19x Since the sum of the cross-products in the last arrangement is equal to the middle term of the given trinomial, we have 6 or 2 + 1 9 a + 1 = (2 a + 5) (3 a + 2) . Ex. 2. Factor 5 x 2 - 6 xy - 8 y 2 . The factors of 5 x 2 are x and 5 x ; and the factors of 8 y z are : y and 8 y, y and 8y,2y and 4 y, 2y and 4 y. x 2y Since the sum of the cross-products in the arrange- ment on the left is equal to the middle term of the given trinomial, we have Ex. 3. Factor 10 a 4 + a?b - 21 6 2 . The factors of 10 a 4 are: a 2 and 10 a 2 , 2 a 2 and 5 a 2 ; and the factors of - 21 W are : b and - 21 6, - b and 21 6, 3 b and - 7 6, - 3 b and 7 b. 2 a 2 + 3 b since the sum of the cross-products in the arrange- Xment on the left is equal to the middle term of the - 2 _ , given trinomial, we have ~tfb~ 10a 4 + a 2 6 - 21 b 2 = (2 a 2 + 3 b) (5 a 2 - 76). 108 ALGEBRA. [C H . VI 14. The following directions may be observed in factoring trinomials which come under this type-form : (i.) When all the terms of the trinomial are positive, only posi- tive factors of the last term are to be tried. (ii.) When the middle term is negative and the last term is positive, the factors of the last term must be both negative. (iii.) When the middle term and the last term are both negative, one factor of the last term must be positive, the other negative. (iv.) /Select those pairs of factors of the jirst and last terms which, by cross-multiplication, give the middle term of the trinomial. EXERCISES V. Factor the following expressions : 1. 60? + a; 12. 2. 6x*-x-l2. 3. 35x 2 + 32x-12. 4. 35 a; 8 + a? -12. 5. 35 a; 2 + 16 a; 12. 6. 35^ 7. 2a? 2 + 5a; + 2. 8. 10 9 . 6 + 13^-63^. 10. 3^ + 13^ + 12. 11. 40 + 2^-2^. 12. 25or 5 + 25.T 2 -6#. 13. 36 x 4 - 18 a 2 -10. 14. 12 a; 6 a; 2 90 a?. 15. 10x 2 +Jx-33. 16. 8^-19^-15. 17. 40 + 6Z-270 2 . 18. 49 2 -35x + 6. 19. 64^-92^ + 30. 20. 6- 21. 6^-41^-56. 22. 30^-89^ 23. 18x 2 -3xy-5y\ 24. 3a 2 -5ab-2b 2 . 25. abx 2 -(a 2 + b 2 )x + ab. 26. abx 2 + (a 2 - b' 2 ) x - ab. 27. 5aV-4a 2 xz-96z 2 . 28. - 10 a 4 + 7 a 2 b 2 + 12 b 4 . 29. 4x* xy-3y 2 . 30. 10a 2 + 11 ab - 6b 2 . 31. 9ar 2n -4af-5. 32. 2 x* r+2 - 3 af +1 - 2. 33. 6x 2m + x m y n -Wy 2n . 34. 10(a + &) 2 +7c(a+6)-6c 2 . 35. 7(x-y)*-37z(x-y) + Wz>. 36. 6 (x 2 + y*f - 9 (x 2 + f) z 2 - 15 2 4 . 37. 2(a 2 -c 2 ) 2 -46(a 2 -c 2 )-66 2 . 14-16] INTEGRAL ALGEBRAIC FACTORS. 109 Binomial Type-Forms. 15. From Ch. V., Art. 6, we have a 2 b 2 = (a + b) (a b). That is, the difference of the squares of two numbers can be written as the product of the sum and the difference of the numbers. Ex. 1. a?x- - 1 ft 2 = (ax) 2 - (J ft) 2 = (ax -f- -J- ft) (ax -J- ft). Ex. 2. 32 m*n -2n 3 = 2 n (16 m 4 - n 2 ) = 2n[(4m 2 ) 2 -n 2 ] = 2 n (4 m 2 -f w) (4 m 2 - n). 16. The difference of any even powers of two numbers can be written as the difference of the squares of two numbers, and should therefore first be factored by applying this type-form. Ex. a 4 - ft 4 = (a 2 ) 2 - (ft 2 ) 2 = (a 2 + ft 2 ) (a 2 - ft 2 ) EXERCISES VI. Factor the following expressions : 1. X 2 -1. 2. 4 -a 2 . 3. 16 - y 2 . 4. 25 x 2 y 2 9. 5. 36 a 2 - 49 ft 2 . 6. &X 2 y |4 7. 86 2 - 14 2 . 8. 57 2 - 43 2 . 9. 372 _ 2' 10. 81 a 4 - -16. 11. 4 a 2 6 2 _ |5 C 2 d : \ 12. 16 a 6 - 25 ft 4 c 6 . 13. o?l AJo _ ~ i 14. i a V- T i__x ( 5 . 15. a 2 " - 1. 16. a 2 ' 17. x 2n+2 4. 18. 9 a 2n ft 2 - -4C 2 "*. 19. 7- -11 2 x 4 20. 16 a 4 - f. 21. a 8 - ft 8 . 22. 1 - 256 afy 8 23. x - 2/ 16 . 24. a 16 -!. 25. 5a 2 - 180 ft 4 . 26. 2 a6 2 _ 1 a( , 2> 27. 5 XV* -5.CCZ 6 . 28. 75a 2 ft 4 -108c 2 d 4 . 29. 243 ft 5 c 6 - 75 ft 7 . 30. a 4 * ft 4 *. 31. 144 a" a; w+2 . 32. 110 ALGEBRA. [Cn. VI 33. a 2 - b 2 -f (a + b)c. 34. a 2 - x 2 + a - x. 35. a 4 a 3 -f a 1. 36. x 2 xz yz if. 37. a 2 - a?n + an 2 - n 2 . 38. a 4 - 2 a& 3 - 6 4 + 2 a 3 6. 39. a?y x f + x i y + xy 2 . 40. a 2 + 3 or 5 - a 4 - 3 a. 41. (a -f- n) (a 2 - x 2 ) - (a - x) (a 2 - n 2 ). 42. (n - x) (5n 2 - 4x- 2 ) - (3 x 2 - 4 ti 2 ) (a - n). 17. This type-form may frequently be applied to multi- nomials. Ex. 1. ^- Ex. 2. = (2ac + a 2 - b 2 + c 2 ) (2ac - a 2 + 6 2 - c 2 ) = (a + c + 6) (a + c 6) (6 + a c) (6 a + c). EXERCISES VII. Factor the following expressions : 1. (a + 6) 2 -c 2 . 2. (a-6) 2 -c 2 . 3. (n + I) 2 - n 2 . 4. n 2 -(w-l) 8 . 5. 9-(3-z) 2 . 6. 49-4(a + 5) 2 . 7. (2 a -f- 6) 2 - 9 c 2 . 8. (4^-3) 2 -16a^. 9. 2oa 2 -4(6 + c) 2 . 10. 36 'X 2 - 81 (x - 2) 2 . 11. (a + 6) 2 -(c + cZ) 2 . 12. (a-&) 2 -(c-d) 2 . 13. (a + Z>) 2 -(a-&) 2 . 14. ( + 2) 2 -(a?-l) 2 . 15. (5x-2) 2 -(4z-3) 2 . 16. (3 ^-4) 2 - (2^-6) 2 . 17. ( a + 6-c) 2 -(a-6 + c) 2 . 18. (x+y-3) 2 -(x-y+5) 2 . 19. (x 2 + x-fl) 2 -(^ 20. (a + &) 2 -l-2(a+6 21. (a-26) 2 -9-3(a- 22. ^-2x2/4-?/ 2 -z 2 . 23. a ~-2 24. z 2 x 2 2xy y 2 . 25. 9 or 2 -f- 2 xy y 2 . 16-18] INTEGRAL ALGEBRAIC FACTORS. Ill 26. a 2 -n 2 + 2np-p 2 . 27. a 2 + 2 be - b 2 - c 2 . 28. 25 + 12a2/-9^-42/ 2 . 29. 25Z 2 - 49?/ 2 - 10 a? + 1. 30. a 2 - 2 a& + b 2 - x 2 - 2 xy - y 2 . 31. x 2 - 32. a a + 33. a 2 + 34. 25x 2 35. a 4 - 36. 4a 4 37. a 2 + 38. a 2 + 6 2 -c 2 -d 2 -2a6-cd). 39. 2 (06 + cd) - (a 2 + b 2 - c 2 - d 2 ). 40. a 2 - b 2 + 2 fo - 2 aa + aj 2 - z 2 . 41. 4a 2 6 2 -(a 2 + 6 2 -c 2 ) 2 . 42. a 2r - a 47 " - 2 a 7r - a 10r . 43. a 4 + 4a 2 c-46 2 + 45d 2 + 4c 2 - 44. 4 (ad + 6c) 2 - (a 2 - b 2 - c 2 + d 2 ) 2 . 18. From Ch. V., Art. 10, we derive Ex.1. Ex. 2. 512 x*-y s = = = (8 x 2 - y) (64 cc 4 + 8 x 2 y + 2/ 2 ) Ex. 3. a 6 - 729 6 6 = (a 3 ) 2 - (27 6 3 ) 2 = (a 3 + 276 3 )(a 3 -276 3 ) = (a+3 6)(a 2 -3 a6+9 fe 2 ) (a-3 6)(a 2 112 ALGEBRA. [Cn. VI Ex. 4. (1 - a;) 8 - 8 a 8 = (1 - x)* - (2 xf 19. The sum of the like even powers of two numbers, whose exponents are divisible by an odd number, except 1, can be factored by applying the type-forms of Art. 18. Ex. aP " = i* EXERCISES VIII. Factor the following expressions : l. 3 + l. 2. a? -8. 3. a 3 + 27. 4. 64 ar 3 -!. 5. 8 or 5 -/. 6. 8^-27. 7. 125 ofy 6 + 8. 8. 3 a 2 -24 a 5 . 9. 27a-a 4 6 fi . 10. 27 a 5 -/. 11. 125 ar 5 -?/ 12 * 12 . 12. 2arV + 432/. 13. 27aW+l. 14. 64zyz 9 -125. 15. 8wV-343p 9 . 16. a 6 -64. 17. x 6 + 2/ 6 . 18. X 9 + /. 19. .T 9 -l. 20. a 12 -!. 21. a 12 + 6 12 . 22. 1-2 18 . 23. a 18 + ^ 18 . 24. a 3 " -ft 3 ". 25. 8 ar*ty m - 729 tT+V. 26. (a; + y) 8 -l. 27. l_(aj-^) 8 . 28. 27 (3 + 2 a?) 8 . 29. (a + by 4- (a - 6) 8 . 30. (2 x - I) 3 - (a: - 2) 3 . 31. (2a4-^) 3 + (a-2) 8 . 32. (a 4- ^) 3 - (c + d) 3 . 33. x* y* 2xry + 2xy 2 . 34. 4 ^ + 4^ x 5 . 35. ar 5 a? 1 - a; 2 -f 1. 36. a,- 8 - 8 - Qx 2 + 12 a?. 37. a 3 - 4 a"c 4 ac 2 H- c 8 . 38. ?i 6 + 5 71 V + 5 ?i V 4- a 20. From Ch. V., Art. 11, we derive : (i.) 7%e swm o/ the like odd jwwers of two numbers contains the sum of the numbers as a factor. 18^22] INTEGRAL ALGEBRAIC FACTORS. (ii.) The difference of the like odd powers of two numbers con- tains the difference of the numbers as a factor. Ex. 1. x> + y 5 = (x + y) (x 4 - tfy + X 2 y 2 - xf + y*). Ex. 2. x 7 - y 7 = (x - EXERCISES IX. Factor the following expressions : 1. a 5 + & 5 . 2. x> 1. 3. 4. a 7 -!. 5. 32a- 5 -6 10 . 6. 243 a 10 -?/ 5 . 7. a 10 + fe 10 . 8. x 10 -!. 9. # 15 + 1. 10. 128x 7 + l. 11. a 5 6 5 32. 12. ar' 10 - 1024 1 Special Devices for Factoring. 21. A factorable expression can frequently be brought to some known type-form by adding to or subtracting from it one or more terms. Ex. l. Factor x* + x 2 y 2 + y\ This expression would be the square of a^ + y 2 , if the co- efficient of x~y 2 were 2. We therefore add x 2 y 2 ; and, in order that the value of the expression may remain the same, we- subtract x 2 y 2 . We then have x* + 2 x 2 y 2 + if- x 2 y 2 =(x 2 + y 2 ) 2 - x 2 y 2 22. Another device consists in separating a term into two or more terms, and grouping these component terms with others of the given expression. Ex. Factor x 3 - 3 x 2 + 4. Separating 3 x 2 into 2 x 2 and ic 2 , we obtain = (x 2)(x 2 -x-2) 114 ALGEBRA. [On. VI EXERCISES X. Factor the following expressions : 1. l + 4a 4 . 2. 1 + 64 a 4 . 3. a 4 " + 4 1/ 4 ". 4. l + 3a 2 + 4a 4 . 5. l-7a 2 + a 4 . 6. 7. aj'-ay+lG y\ 8. a 4 +2/ 4 -llxy. 9. 16 x*- 10. x 4 + 4 2T 4 - 12 ajy. 11. x 4 + 2/ 8 + tf 2 ?/ 4 . 12. x 8 + y 8 - 142 afy* . 13. or 5 -6 ^ + 16. 14. ^-15^ + 250. 15. or ? + 6.T 2 + 10.T + 4. 16. 0^-9 ^4- 32 a? -42. 17. jc 8 - 15 a; 8 + 72 a? - 110. 18. 8 or 3 -36 or 2 + 48 aj - 18. EXERCISES XI. Factor the following expressions by the methods given in this chapter : 1. a 4 + 2a 3 6-2a& 3 -& 4 . 2. ax 2 + (a + b + c)x + & + c. 3. 10c 4n+1 -5c 7 " +1 -5c n+1 . 4. 2 ?/ 2 + 17 xy + 16. 5. x 6 + 64. 6. a 6 6 6 + l. 7. 2 336+s 64. 8. 9. 2a 4 -16a&. 10. aj* + 2a; 2 + 9. 11. 24a? B -(3&-8a)a;-a*. 12. 6 2 - c 2 + a (a - 26). 13. x- 2m - 2 + 2 a m+n + a; 2n+2 . 14. x 4 - 2 a?- 1 + 2x. 15. or + 11 a + 24. 16. a 2 -a6-66 2 . 17. tff 4 xy 5. 18. ic 2 + x + y y 2 . 19. aft (x 2 + if) + spy (a 2 + & 2 ). 20. 28 (x + 3) 2 - 23 (x 2 - 9) - 15 (a? - 3) 2 . 21. aa? 5 + 6a^ + c,y? ax 2 bx c. 22. (a + 6) y? + (a - 2 6) x - 3 b. 23. a 2 - 6 2 - c 2 - 2 a + 2 6c + 1. 24. 49 x\f + 42 z 7 ?/ 9 + 9 ^y 2 . 25. x 2 -13^/ + 40/. 26. a 2 -5a& + 6& 2 . 27. m 2 n 2 + 6 ?/i7i 55. 28. & 2 + ac c 2 + a&. 22] INTEGRAL ALGEBRAIC FACTORS. 115 29. xy-xz-\-2yz-y 2 -z 2 . 30. x 2 - 2x + 1 - ?/ 2 . 31. 15 a 2 + # 40. 32. x 3 x 2 z -f- #z 2 z 3 . 33. a 3 _ 1 _j_ c _ ac . 34. a 2 _ a 1 a?c + ac + c. 35. 2a 2 + a-4 : ax-x + 2x 2 . 36. 20^-123^ + 180. 37. x 3 -5x 2 -x + 5. 38. 2 (a + 1)- 6 2 (6 + 1). 39. 25 a 4 6 4 + 70 aW + 49 c 4 . 40. x 4 y + zx* - xy - z. 41. ^-9^ 2 -4?/(2/-f 3z). 42. a? 8 - 2 ajy + / - 4 a?y ( 2 - ?/ 2 ) 2 . 43. a 3 + a 2 c + a&c + 5 2 c - 6 3 . 44. 5 a 4 - 10 a 3 - 75 a 2 . 45. 3(a-l) 3 -(l-a). 46. 47. 2 -ax -bx + db. 48. 49. a; 2 + 9-2x(3 + 2^ 2 ). 50. 51. 3x Q + 8^-8^-3. 52. 53. (tf+xy+y^-^-xy+y*) 2 . 54. 55. (a^ + I) 3 - (y 2 + I) 3 . 56. aftaj 8 + a; + ab + 1. 57. 36a 4 -21a 2 + l. 58. 10 x 4 - 47 x 2 + 42. 59. (x 2 + ^-2/ 2 ) 2 -(^-^- 2 / 2 ) 2 . 60. x 2 + c (a + 6) ic + a6 (a + c) (c 6). 61. 5 a 2 - 180 b 2 . 62. ^ ab + ?/) + 77. 73. (a 2 - 6 2 ) or 2 - (a 2 + 6 2 ) a + ab. 74. 300a&c 2 -432a&d 2 . 75. 75 a 2 6 2 - 108 c 2 ^ 2 . 76. ^abx 2 y 2 -^ 9 -abz 2 . 77. 18 aV - 98 &y. 78. 18 (x + t/) 2 + 23 (x 2 - y 2 ) -6(x- y) 2 . 79. Express (a 2 b 2 ) (c 2 d 2 ) as the difference of two squares. 116 ALGEBRA. [Cn. VI HIGHEST COMMON FACTORS. 23. If two or more integral algebraic expressions have no common factor except 1, they are said to be prime to one another. E.g., ab and cd ; 5 tfy and 8 z 8 ; a 2 + b 2 and a 2 - b' 2 . 24. The Highest Common Factor (H. C. F.) of two or more integral algebraic expressions is the expression of highest degree which exactly divides each of them. E.g., the H. C. F. of ax 2 , ba?, and ex 4 is evidently a?. 25. Monomial Expressions. The H. C. F. of monomials can be found by inspection. Ex. 1. Find the H. C. F. of x 2 y 5 z, x 4 y*z 2 , and afyV. In the expression of highest degree which exactly divides each of the given expressions, the highest power of x is evi- dently x 2 , of y is y*, and of z is z. Therefore the required H. C. F. is x 2 y s z. Observe that the power of each letter in the H. C. F. is the loivest power to which it occurs in any of the given expressions. If the monomials contain numerical factors, the Greatest Common Measure (G. C. M.) of these factors should be found as in Arithmetic. Ex. 2. Find the H. C. F. of 18 aWd, 42 a?bc 4 , and 30 aW. The G. C. M. of the numerical coefficients is 6. The lowest power of a in any of the given expressions is a 2 ; the lowest power of b is b ; the lowest power of c is c 2 ; and d is not a common factor. Therefore the required H. C. F. is 6 arbc 2 . 26. In general, to obtain the H. C. F. of two or more monomials : Multiply the G. C. M. of their numerical coefficients by the product of their common literal factors, each to the lowest power to which it occurs in any of the given monomials. 27. Multinomial Expressions. The method of finding the H. C. F. of multinomials by factoring is similar to that of find- ing the H. C. F. of monomials. 23-27] HIGHEST COMMON FACTORS. 117 Ex. l. The expressions and x 2 + x - 2 = (x - 1) (x + 2), have only the common factor x 1. This is their H. C. F. In general, the H. C. F. of two or more multinomial expres- sions is the product of their common factors, each to the lowest power to which it occurs in any of them. Ex. 2. Find the H. C. F. of a 2 x 2 a 2 , 2 ax 2 + 2 ax 4 a, and 4 ax 2 12 ax 4- 8 a. We have a?x* a 2 a 2 (x + 1) (a; 1), 2 ax 2 + 2 ax - 4 a == 2 a (a; + 2) (a? - 1), 4 ax 2 - 12 ax + 8 a = 4 a (a; - 2) (x - 1). Therefore the required H. C. F. is a (x 1). EXERCISES XII. Find the H. C. F. of each of the following expressions r 1. 36 a 2 , 27 a 4 . 2. 20 ab 2 , 35a 2 6. 3. 45ary, 12afys. 4. a 2 bx\ aWx 2 , ab s x\ 5. 56a?y, 70 x 2 /, . 98 afy 2 . 6. 24a 2 6a?*, 42 az 3 , 18a 8 afy. 7. 15 mV/, 40mVa;, 35m 3 nx 2 . 8. 9(a? + y), 6(a-h2/) 2 . 9. 12y 2 (a-b), 30 y (a b) 2 . 10. 2 -9, x 2 + 3x. 11. 3^-3^, 5aj-5ajy*. 12. (a + 6) 2 , a 2 -6 2 . 13. a^ 2 - a, ax 2 + 2 ax -f a. 14. x 2 - 25 /, x 2 + a^ - 30. 15. (a 2 6 - a& 2 ) 2 , a6 (a 2 - 6 2 ). 16. 27a^ + /, 9 or 2 -/. 17. a 3 - 4 ab 2 , a*-8b s . 18. x 2 -2i-15, 2 + 10 a? + 21. 19. oj 2 -2a?-24, aj 2 + 9a; + 20. 20. 3x 3 -32/ 3 , x 2 -by + bx-xy. 21. 3 -/, x 4 + 3^ L V 2 -42/ 4 . 22. or + z?/-30/, ^ 2 -2^-15/. 23. x 2 y 2 -xy s -A2y 4 , 6 orfy + 18 a; 2 / - 108 a#* 24. 25. 118 ALGEBRA. [Cn. VI 26. a 3 -f-2a 2 + 2a + l, a 3 -f 1. 27. x 2 4- a& ax ~bx, x 2 ab ax + 60;. 28. a 2 - (6 - c) 2 , (a - c) 2 - b 2 . 29. a? 3 y 3 , # 4 + a? 2 ?/ 2 -f- y 4 . 30. # 2 -3^ # 2 -9, x- 2 -6z + 9. 31. a 8 -8, o' + Taj-lS, ^-8x + 12. 32. a 2 -3x-40, z 2 + 3x-10, ^-oj-SO. 33. x 2 -\- 2 xy -\- y 2 z 2 , ax + ay -f- ^z .34. (y - z) 2 -x 2 , (x + y) 2 - z 2 , y 2 - (z - *) 2 . LOWEST COMMON MULTIPLES. 28. A Multiple of an integral algebraic expression is an ex- pression which is exactly divisible by the given one. E.g., multiples of a + b are 2 (a + 5), (a? y) (a + 6), etc. 29. The Lowest Common Multiple (L. C. M.) of two or more integral algebraic expressions is the integral expression of lowest degree which is exactly divisible by each of them. E.g., the L. C. M. of aa? 2 , bx 3 , and ex 4 is evidently abcx*. 30. Ex. 1. Find the L. C. M. of a*b, a?bc?, and 6 2 c 4 . In the expression of lowest degree which is exactly divisible by each of the given expressions, the lowest power of a is evi- dently a 3 , of b is b 2 , and of c is c 4 . Therefore their L. C. M. is + 2az-3a 2 . 20. m 2 + 2 win 15 w 2 , m 2 + 3 m?i 10 n 2 . 21. a 3 aj 3 , a 2 x 2 , a. 22. x 2 y 2 , (x i/) 2 , or 3 y 3 . 23. a? a, a 2 x 2 , x 4 a 4 . 24. 1 2 x, 4 or 2 1, 1 + 4 x 2 . 25. aj - 11 a; + 24, aj"-6aj-16, x 2 - a; - 6. 26. a^ 4a? 45, a? 2 -7^-18, o^ + 7^ + 10. 27. 3^ + 24^ + 45, 6^ + 18^-60, 8^-24^ + 16. 28. 4.^ + 4^-224, 6^ + 24^-462, 8^ + 64a;-264. 29. x 2 - 4 ax + 3 2 , x 2 + 4 az - 5 a 2 , a? + 2 ax - 15 a 2 . 30. x 2 + 2 mx - 3 m 2 , or 2 + 7 mx - 8 m 2 , x 2 - 6 ma? - 27 m 2 . 31. a?-4a 2 , a? + 2a 2 + 4a 2 a; + 8a 3 , ^ - 2 aar + 4 a?x - 8 a* 32. a - - (6 + c) 2 , 6 2 - (a + c) 2 , c 2 - (a + 6) 2 . 120 ALGEBRA. [Cn. VI H. C. F. AND L. C. M. BY DIVISION. 32. If the given expressions cannot be readily factored, their H. C. F. can be obtained by a method analogous to that used in Arithmetic to find the G-. C. M. of numbers. 33. The expressions whose H. C. F. is required should be arranged to powers of a common letter of arrangement. If one of two expressions be divisible without a remainder by the other, which must be of the same or lower degree in the letter 'Of arrangement, then the latter (the divisor) is the required H. C. F. For it is a factor of the other expression. But if the one expression be not divisible without a remain- der by the other, their H. C. F. is found as follows : (i.) Divide the expression of higher degree in a common letter of arrangement by the one of lower degree ; if the expressions be of the same degree, either may be taken as the first divisor. (ii.) Continue the division until the remainder is of loicer degree than the divisor in the letter of arrangement. (iii.) Divide the first divisor by the first remainder, the first remainder (second divisor) by the second remainder, and so on, until a remainder is obtained. The last divisor ivill be the required H. C. F. 34. Ex. Find the H. C. F. of 2x 3 -5x 2 -5x + S and 3.2 _ 4 x _j_ 3 We have 6x 3x 2 -llx 3x 2 -12x-\-9 x 2 - x -3x -3x By Art. 33 (iii.), the H. C. F. is x - 1. 32-36] H.C.F. AND L.C.M. BY DIVISION. 121 35. The validity of the preceding method is based upon the following principle : If an integral algebraic expression be divided by another (of the same or loiver degree in a common letter of arrangement) 'and if there be a remainder, then the H. C. F. of this remainder and the divisor is the H. C. F. of the given expressions. E.g., the H. C. F. of ^_ 10 ^ + 35 X 2_ 50 x + 24? = (<,. _ i) (3 _ 2) (a? - 3)0 - 4), (1) and x 3 -7x 2 + llx-5, = (a? - 1) (a? - 1) (a? - 5) (2) is evidently x 1. The remainder obtained by dividing (1) by (2) is 3o 2 -12a + 9, =3(a;-l)(a?-3). (3) The H. C. F. of this remainder and the divisor (2) is evi- dently also x - 1, the H. C. F. of (1) and (2). Notice that the H. C. F. of the remainder and the dividend (1) is (a; 1) (a; - 3), and is not the H. C. F. of (1) and (2). Since this principle can be applied at any stage of the work, the H. C. F. of any remainder and the corresponding divisor is the required H. C. F. When the last remainder is 0, the last divisor is the H. C. F. of itself and the corresponding divisor, that is, of the preceding remainder and divisor, and is, therefore, the required H. C. F. If a remainder which does not contain the letter of arrange- ment, and which is not 0, is obtained, the given expressions do not have a H. C. F. in this letter of arrangement. The proof of the principle enunciated is given in School Algebra, Ch. VIII. 36. The following principle will frequently simplify the work of finding the H. C. F. of two expressions : Either of the expressions may be multiplied or divided by any number which is not already a factor of the other expression. For a factor introduced by multiplication into one expression will not be common to both of them, and therefore will not be introduced into their H. C. F. 122 ALGEBRA. [Cn. VI In like manner, the factor removed by division from one expression was not common to both of them, and therefore would not have been a factor of their H. C. F. Ex. Find the H. C. F. of 2^ + 5^-3 and 2^ + ^ We have 2x 2 + 5x-3)2x 3 + x 2 - 3x _4a 2 - 2x The next step would introduce fractional coefficients. To avoid these, we divide 8 x 4 by 4, since 4 is not a factor of 2 x 2 + 5 x 3) and take 2 x 1 as the divisor of the second stage : 2X 2 - 6x-3 6a;.-3 The required H. C. F. is 2 x - 1. 37. Before proceeding with the division, remove from the given expressions any monomial factors and set aside their H. C. F. as a factor of the required H. C. F. Ex. Find the H. C. F. of * - 6 or 3 + 6 x 2 - 3 x + 2), - 12 x 2 y We set aside xy, the H. C. F. of 2 xy 2 and 3 x 2 y ) as a factor of the required H. C. F., and find the H. C. F. of the remaining factors by division. The first of these expressions cannot be divided by the second without introducing fractional coefficients. To avoid 36-38] H.C.F. AND L.C.M. BY DIVISION. 123 these we multiply the first by 2, since 2 is not a factor of the other expression. 7 x 2 - + 4.x- 8 - 35 x 2 2d divisor, 5 x 2 - - 9 x + 4 To avoid fractional coefficients in the next stage of the work, we multiply the last divisor by 5 : - 20(2x-7 8a? X 5)_ 7 X 2 + 27 x- 20 1350; -100 - 28 - 72 3d divisor, x l)5x* 9a; + 4(5a;- 4 -4a? To avoid fractional coefficients, we multiplied the partial remainder of the first division by 2, divided the remainder of the first division by 5, multiplied the partial remainder of the second division by 5, and divided the remainder of the second division by 72. The required H. C. F. is xy(x 1). 38. If the divisor and dividend at any stage of the work can be factored readily, it is better to find their H. C. F. by factoring than by continuing the method of division. Ex. Find the H. C. F. of a- _ 10 a 3 + 35or> - 50 x + 24, (1) and tf-Tx 2 + llx-5. (2) 124 ALGEBRA. [Cn. VI We have : x 4 Tar'+ll^ 2 5x 9 3 The remainder x 2 4 x + 3, = (a; 1) (a; 3), is readily fac- tored. Dividing X s 1 X s + 11 x 5 by x 1, we have ^3 _ 7 tf + n x ._ 5 = (g. _ !) (^ _ 6 ^ + 5 ) = ^ _ ^2 ^ _ 5 ^ The H. C. F. of the first remainder and (2), and therefore the required H. C. F., is x 1. Lowest Common Multiple by Means of H. C. F. 39. If the given expressions cannot be readily factored, their L. C. M. can be obtained by first finding their H. C. F. Ex. Find the L. C. M. of x*-2x 2 -2x?y + 4:xy + x-2y and x* - 2x 2 y + xy 2 - 2f. The H. C. F. of these expressions is found to be x 2 y. Consequently the other factors of the given expressions can be found by dividing each of them by their H. C. F. We have From the definition of the H. C. F., as also by inspection, we know that these second factors, x 2 2 x + 1 and x 2 + ?/ 2 , have no common factor, and therefore that the L. C. M. of the given expressions must contain both of them as factors. Consequently the required L. C. M. is This example illustrates the following principle : The L. C. M. of two integral algebraic expressions is the product of their H. C. F. by the remaining factors of the expressions. 38-40] H.'C.F. AND L.C.M. BY DIVISION. 125 Relation between H. C. F. and L. C. M. 40. The following example illustrates an important relation between the H. C. F. and the L. C. M. of two integral algebraic expressions. Ex. TheH.C.F. of and a? 8 - ! = (- !)( + !) is (x - 1). The L. C. M. of the same expressions is The product of the two given expressions is In general, The product of two integral algebraic expressions is equal to the product of their H. C. F. and their L. C. M. It follows from this principle that the L. C. M. of two integral algebraic expressions can be found by dividing their product by their H. C. F. EXERCISES XIV. Find the H. C. F. and L. C. M. of the following expressions : 1. a* + 4 a - 5, x*-2x 2 + 6x-5. 2. 2 X 3 ^ r 3x 2 -x- 12, 6 x s -17x 2 + 2 x + 15. 3. X s - 3 x + 2, y? + 2 x 2 - x - 2. 4. 2 aj 8 - 17 x 2 + 19 x - 4, 3 X s - 20 x 2 - 10 x + 27. 5. X 3 _ 5 tf + 9 x _ 9? X 4 _ 4 ,# + 12 x - 9. 6. x 3 - x 2 - 9 cc + 9, cc 4 - 4 x 2 + 12 a? - 9. 7. ^-3^ + 4, aj 8 -2aj 2 -4oj + 8. 8. ^_3x + 2, x 4 -6a 2 + 8a-3. 9. 2^ + 3^-2, 4^ + 16 aj 2 - 19 a? + 5. 10. ^-3^ + 4, 3 z 3 - 18 ^ + 36 a -24. 11. x 3 (a + 6 c) ar> + (06 ac be) x + abc, X s (a b + c) ^ + (ac a6 6c) a? + 126 ALGEBRA. [Cn. VI 12. or 5 + ar 2 -5# + 3, 13. 3 oj 8 - 8 or 2 - 36 a? + 5, 9 x 3 - 50 a? + 27 a - 10. 14. 4y-3ajy-4ajy + 3, 5 or 5 ?/ 3 + 8 afy 8 + a?y - 14. 15. a*-3xy*-2y*, 2 y* - 5 a^ - xy 2 + 6^. 16. a 3 - a 2 - 5 a + 2, 3 a 3 - a 2 - 8 a + 12. 17. x s + 2 x 2 + 2 x + 1, a 8 - 4 x 2 - 4 a? - 5. 18. 30.0? 25oor ! + 8a 2 a; a 3 , 18 or 5 - 24 a 2 + 15 a 2 a - 3 a 3 . 19. 2 a 4 -3^ + 4 a 2 -5z- 4, 2aJ 4 -a 8 + a-12. 20. 40 3 So^ + Sa; 3, 2# 4 -3a 3 + 6a 2 -3a; + 2. 21. 4 a 4 - 8 x 3 - 3 x 2 + 7 a - 2, 3 x 8 - 11 or 2 + 2 a? + 16. 22. 36 a 6 + 9 a 3 - 27 a 4 - 18 a 5 , 27 a 5 6 2 - 9 a s 6 2 - 18 a 4 6 2 . 23. 3^-10^ + 15^ + 8, ^_2x 4 -6^ + 4^ + 13a 24. 2 x ut> a* - aar - 6* + ab' 2 ' 10-12] FRACTIONS. 133 ' ' 33^-14 x 2n + 3 x n + 2 , ** + * 30 a*-(b-c) 2 ' 22 ' 22 (g + fr) 2 -4c 2 a 2 + a + b - ' ' 33. Z if * + ?=*-- 66 10 a 6 x 2 if - nc -*- ^ t og *" " if "y \ y (1 + ax) 2 (ct + x) 2 x 4 x*y x 2 y 2 + xy 3 37. 5^_5a; + 7 39. ^~* + 2 . *-] I t AO T 3 ^ I 7* I 1 A Jj ds ~~T~ dJ "T"^ -LTt Reduction of Two or More Fractions to a Lowest Common Denominator. 11. Two or more fractions are said to have a common de- nominator when their denominators are the same. E.g., 2 and 5; - and *" . 2 2 The Lowest Common Denominator (L. C. D.) of two or more fractions is the L. C. M. of their denominators. E.g., the L. C. D. of and is 6 2 c 2 . b 2 c .be 2 12. The value of a fraction is not changed if both numerator and denominator be multiplied by the same number, not 0. -Q a xa xxa + x) a 2 x 2 a -h x (a + a;) x (a + x) (a + x) 134 ALGEBRA. [Cn. VII Let the value of the fraction - be denoted by v, or Multiplying by b, vb = - x b = a. Multiplying by n, vbn = an. Dividing by bn, v = an -+- bn = But v = 2. b Therefore - = ^L. b bn 13. Ex. 1. Reduce and to equivalent fractions having b 2 c be a lowest common denominator. Their required L. C. D. is 6 2 c 2 . Multiplying both terms of -2- by b 2 c 2 -t-tfc. =c, we have ; b 2 c b'-cr and both terms of by frV-f-te 2 , =&, we have Ex. 2. Reduce x, = -, and ^ to equivalent fractions having a lowest common denominator. The required L. C. D. is x y. Multiplying both terms of - by x y, we have 11^1 . 1 x y and both terms of ^ by 1, we have, * x y x y Ex. 3. Reduce 1 1 _ 3 x + 2 ' (a? - 1) (x - 2)' and ^_ = 2 to equivalent fractions having a lowest common denominator. The required L. C. D. is (a; - 1) (a; - 2) (x + 1). 12-14] FRACTIONS. 135 Multiplying both terms of the first fraction by we have ; (x 1) (x 2) (x + 1) and both terms of the second fraction by (aj - 1) (3 - 2) (aj + l)-Ka?- !)( + !), =-2, we have 14. These examples illustrate the following method : Take the L. C. M. of the denominators as tlie required denomi- nator Divide this denominator by the denominator of each fraction; and multiply both numerator and denominator of the fraction by the quotient. EXERCISES II. Reduce the following fractions to equivalent fractions having a lowest common denominator : 1 x 4m 5n 5a*6 2a&* 7 4 ~3~> T 14 ' ~2T a 2 l-i-4m 1 K 9 * 4. 1 a ? - 5. m, : 6. a -f- 1 m 4 q >7 9^/y^J_9/r 7. -4-, 1, _-L_. 8. ^^^, 5if. 9. 1 1 5 1 3 2 3 " ' ' 14. 1 ^L *4^ 15. 16 17 ' ' __ ? 1 - x 2 136 ALGEBRA. [Cn. VII m y 1 + m ig ax b a bx 1 y(x y)' m(y x) J my ax + atf bx + b 2 ' a~b' 2 3 5 20. r~f -^ , =^h=- 21. 22. Z7 a 2 -a a 2 -! " 20-2' a; 2 -2x-+l' 1-x 2 1 3 nm m n M rm W ffflV 23. -i -, - 24. 25. (a-c)(a-b)' (b-a)(b-c)' (c-a)(c-b) Equations. 15. Ex. 1. Solve the equation 2 a; + - = 9. Multiplying by 4, 4 x 2x + 4 x | = 4 x 9; (1) or. since 4 x 7 = ic, 8 a; + x = 36. 4 Uniting terms, 9 x = 36. Dividing by 9, x = 4. The step represented by (1) is called clearing the equation of fractions, and should be performed mentally. To clear of fractions, we multiplied by 4, the denominator of the fractional term. If the equation contains more than one fraction, we multiply by their L. C. D. Ex. 2. Solve the equation - t = 3 x. The L. C. D. is 15. Multiplying by 15, 3oj-5(2aj-l) = 15 (3 - x). (1) Removing parentheses, 3 x 10 x + 5 = 45 15 x. Transferring terms, 3x 10ic + 15ic = 45 5. Uniting terms, 8 x = 40. Dividing by 8, x = 5. 14-17] FRACTIONS. 137 16. Observe that the sign of a fraction affects each term of the numerator, or the dividing line between the numerator and the denominator has the same effect as parentheses. Kg., _a-6 + c == _ (a _ 6 + c) _ f _ d = ( a + 6 c)-r-d a + 6 c d 2 x _ \ Thus, in Ex. 2, Art. 15, the sign before the fraction o changes the signs of both terms in its numerator, and not simply the sign of the first term, when the denominator is removed. This caution should be kept in mind, and step (1) omitted in clearing of fractions. 17. Ex. Sol ve the equation The L. C. D. is 24. Multiplying by 24, 4 Transferring terms, 4 x 3 x 2 x = 3. Uniting terms, x = 3. Dividing by 1, x = 3. EXERCISES III. Solve each of the following equations : 1. X x ^2 = 18. , .-% = 5. 3. 10 + 6 = x. 4. X 2 .x 4 = 15. 2x x 3 2 = 5. 6. 3x 5 " H- X -2 3-x o a 4 5- - X ~ 3x-2 3x+2 3 2 5 4 4 5 T n X + 3 \ X A x-l aj o 2 4 5 1 o 3 /y 2 x .. TO 6-5 5x 3 5 4 4 3 14. 3 x a; + 4 1 *5 5 a a - 10 4 6 8 6 138 ALGEBRA. [Cn. VII 16 5 ft 1 _ g 1? 5 a a 4 _ # 3 32 12 15 ~2 "2~ ~3~ ~6~ ~3~ ~5~^ 10" 5-3a; 5a;-4 25-19 a? x-2 2x 5 4-3^ 14> ~T~ ~10~ T5~ ' ~3^~" ^^ ~5~ "HT" x y z 5y 3z x 47 14 18. X ~ 3a-46 2a-6-c 15a-4c a ^~ ~~ 12 21 20. - 21. -^T + 2~^-T' 22 - ^Ill_^^|. 23. ll_-2 24. W^_!^. 25 . _J JL_ ic+2 mn m-\-n a w 4-l a w . 27 | ' a 2 1 a + 1 a 2 4 ?/ 2 ac + 2 cy 3a-l l-3a 3a-16 a 2 -9 a+3 144 ALGEBRA. [Cn. VII 3 a a 2 ax x 1 \x x5 \ a + x a-x a 2 -x 2 6#+24 or'-lG 3z-12 2m-3 ,3,2 2 , x-6 1 32. ~T~ ~~r~ " OO. I 1 4 m 2 1 2 m ra & 34. 9 a 2 - 25 6 2 6 ad + 10 6d 6 ad - 10 6d 2 3 | 4 4^ 2/K 36. 35. ^-5 ax 2 a 2 37. -^-r- 3 , 38. L - + a I a 3 1 x + 1 a 2n a n 1 a 3 + ^ 3 2 n art 2 + n 3 an -f- w 2 1 3 nra m n n m n 3 m 3 m 2 -f- mn -\- n 2 . -4= -^ -+- oj_8 x-7 ' x + S 21. Frequently the denominators are multinomials in the same letter of arrangement, but not arranged to the same order of powers. TT, c<- TJ? x 2x 2x Ex. 1. Simplify -. 1 - It is better first to change the second fraction so that the denominators shall be arranged in the same order. We then have, by Art. 7 (ii.), x 2x 2x t x - l tf - I x + 1 The L. C. D. is x 2 - 1. 20-21] FRACTIONS. 145 Therefore, x 2x 2x = a?(o; + l) 2a; 2a?(a? 1) ~ or* x 2 1 or 1 05 + 1 As in this example, the result of the addition should be reduced to its lowest terms. Ex. 2. Simplify (a b)(a c) (ba)(bc) (cd)(cb) Changing the fractions into equivalent fractions, whose denominators, taken in pairs, have one common factor, we have (a b) (a c) (a b)(b c) (a c) (b c) b c a c (a - b) (a - c) (b-c)~ (a- b) (b - c) (a - c) a b _ _ b c a -}- c -\- a &_ (a c) (b c) (a b) ~ (a b) (a c) (b c) ~~ EXERCISES VI. Simplify the following expressions : 36 a 3. 5. 7. X 2 - 4 2-oj 3 a a 2 3 a b a 4 a_l_4 ^_2 a 2 - ab b 2 ab 5 o; 10 6 - 3 o; 3 7 4-20 x m 2mn 2m 2xl ' 2#+l l-4ar> m ?i w 2 m 2 m +-w a I/ . -i / J -4- X 1 1 a 2 + 1\ 1 1 a + 1 1^1 - a ' a 2 lj iC : B -3oj-f2 1- it- 2 1 1 , 1 + a; 2 + 1 a - 1 - a 2 'a? + 1 + or 2 10 - ; i r^ + ALGEBRA. [Cn. VU 1 1 11 f l+ + * i c + <* (b - c) (c - a) (c - a) (a - 6) (a - b) (b - 12. ab + * + ca (6-c)(c-a) (c-a)(a-b) ( a -b)(b- 13. -^ ,+ 1 ' 1 a(a-6)(a c) b(b-a)(b-c) c(c-a)(c 6) 14. ! \ + ^ -f H (a 6) (a c) (b a) (b c) (c a) (c 15. ^- -+ * ' * (a b)(a c) (b a) (b c) (c a) (c 16 be ac a(a 2 - b 2 ) (a 2 - c 2 ) b (b 2 - a 2 ) (6 2 - c 2 ) c( 17> a 2 -6c , 6 2 + ac , c 2 + ab Reduction of Mixed Expressions to Improper Fractions. 22. A Proper Fraction is one whose numerator is of lower degree than its denominator in a common letter of arrange- ment. E.g., oj + + ^- An Improper Fraction is one whose numerator is of the same or of a higher degree than its denominator in a common letter of arrangement. Wn If both integral and fractional terms occur in an expression, it is sometimes called a Mixed Expression. 23. Ex. 1. Eeduce a-f-- to an improper fraction. First c reducing a to the form of a fraction with denominator c, we have .b_acb_ac-)-b c c c c 21-24] FRACTIONS. 147 This example illustrates the following method : Multiply the integral part by tJie denominator of the fractional part. To this product add algebraically the numerator of the fractional part) and write the sum as the required numerator. Ex. 2. Simplify 1 x + x 2 Wehave 1 - x x 1 + a; 1 + a; EXERCISES VII. Simplify the following expressions : 1. 2a-. 2. 7+i- 3. m +i. 3 a m 4 a X 1 1 * 1 I 6 3 <7 1 JO t)u J_ 3 7 m 3m ~ 5 ^- B a ^ oil 8a 4 a + 6 a - ft) 2 O X 12. a + 4 9ic + 20 13 5a: 42 a; 14 ^y. -U 7) IB 7 a-7 4 a/j i j . ab ab 16. I -fa- \ 17. a 2 + a.T 1 VW ! v , , + ay a x Reduction of Improper Fractions to Mixed Expressions. 24. Ex. l. Reduce - ' x ' to a mixed expression. x + 1 We have 2 a? 2 + x + 5 I a; + 1 - aj-1 148 ALGEBRA. [Cn. VII But by Ch. III., Art. 47, we have (2x* -f- x + 5) -r- (x + 1) = 2x - 1 + 6 -s- (a; + 1), or = 2 # 1 H - This example illustrates the following method : Divide the numerator by the denominator, until the remainder is of lower degree than the divisor. Write the remainder -as the numerator of a fraction whose denominator is the divisor. Add this fraction to the integral part of the quotient. Ex. 2. Eeduce "*" ^ ~ x ^~ to a mixed expression. x 2 + 2 x 1 We have x -I -x + 2 Therefore, ^ + ^~ ** + 3 = x- 1 + EXERCISES VIII. Reduce each of the following fractions to equivalent frac- tional expressions, containing only proper fractions : _ 1 10a 2 -3a 6a 8 3 a x y a b _ ^ a;-5 . 21 x z + 20 x - a-1 oj + 1 3x + 2 m s -n s -l 1;i a; 3 -3 o^ + 2 a; -3 24-26] FRACTIONS. 149 m 3 mn 2 m 2 n+?i 3 +l _ 5x z 3 # 14 _. .. m-n x?-2 1a? + 9 ( _ ar 3 + a; 2 - 2 Multiplication of Fractions. 25. Multiply - by -. Let the value of - be denoted by v, and that of - by w ; or a (I i C v = -, and w = b d Multiplying the first equation by b, and the second by d, we have vb = ^ and wd = ^ Multiplying together corresponding members of these equa- tions, we have vb x wd = ac, or vw x bd = ac. Dividing by bd, vw = ac-t-bd= bd But vw = -x-- b d Therefore ^x = ^. b d bd This proves the following method of multiplying fractions : The numerator of the product is the product of the numerators; and the denominator of the product is the product of the denomi- nators. 26. Ex. 1. Find the product 22 xY 25 a' 2 b The factor 5 is common to the numerator of the first fraction and the denominator of the second. Since to cancel a com- mon factor before multiplication is equivalent to cancelling it after the multiplication, we should first cancel 5. For a similar 150 ALGEBRA. [Cn. VII reason we should cancel the factors, 2, a 1 , b, x, and y- before the multiplication. We then have 3 ab 1 21 ab X T~ 11 xy 3 5 55 xy* In general, if the numerator of one fraction and the denomi- nator of another have common factors, such factors should be cancelled before the multiplications are performed. 8a 2 Ex. 2. Find the product a 2 b 2 4 a Cancelling the common factors, 4 a and a -\-b, we have 2q a + 6 2a(a + b} ' /\ a b 27. Ex. Find the product X 9 ~ y 9 x (x + ar + r x We have Observe that a fraction is multiplied by an integer, by multi- plying its numerator by the integer. 28. If one of the factors is a mixed expression, it should first be reduced to an improper fraction. Ex. Find the product fj Y - Y x 1 1 W T e have \ xj\x* iy x 1 i "1 EXERCISES IX. Multiply : 1. - X 3. 2. X 5. 3. ay 26-28] 7. FRACTIONS. _ 15 oW 14 xy 2 _ I NX ' , Q ' ' " t7 y 25 2 6 21a 2 6 151 4 2 156 26c_ ' 565 x -fj x 3^ u. * x sa x 2 6 2 c 2 ax 3 12. 14. 16. 18. x(3a-26). 13. -J^Ux a6 2 ^> 3 a 3 aft 2 a 2 4- 26 2 a 2 -6 2 2 a 15. _ ^ ^, X ?- 5 a& (a 17. 19. ^Ix X 20. (+&) x b(a-b) 6 ax 15 5ic 8 ax + 4a 2 -256 2 22. 23. 24. 25. x y x + x 2 (a + c) x + ac x 2 b 2 a 2 -(b-c} 2 (x ( a _ 6)2 _ 26. -^ X x 2 -y 2 27. 28. 29. - 4 - 9 x 2 -8x + 15 ^-8;c + 12 a? y -f ic 2 + 2 #?/ -f- ?/ 2 z 12 a 2 4 a; 2 9 y 2 , 22 a 2 - 10 ab * 6 ax -9 ay ^10 60; + 15 6y" 152 ALGEBRA. [Cn. VII ' , 5-L. a 6 x 2 + x - 12 ff 2 - 3 x - 10 X x - 20 x - 6 x 2 - 4 (m -f- w) 3 12 m n x -\- y 32. (o?_ OJ4-1M-.4--4-1 ) 33. (- 34. a 36. 35 - -{- 37 . 38. a 2 - -2/ 2 m 2 (6 a) 2 (m a) 2 6 2 am afr + a 2 ' 40. 1 + x y-zj (x 2 -4c 2 Reciprocal Fractions. 29. The Reciprocal of a fraction is a fraction whose numera- tor is the denominator, and whose denominator is the numera- tor, of the given fraction. E.g.) the reciprocal of - is b a 30. The product of a fraction and its reciprocal is 1. For b a ba 28-32] FRACTIONS. 153 Division of Fractions. 31. Divide - by - Let the value of - be denoted by i>, b d b and that of - by w ; or, a -i c v = -, and w = b d Multiplying the first fraction by b, and the second by d } we have v b = a, and wd = c. Dividing the members of the first equation by the corre- sponding members of the second, we have vb a vb -r- wd = a -j- c, or = wd c Multiplying by f, 5xf-fx, v ad or - = w be But - = - + r w & d Therefore + =^ = x i o a be b c This proves the following method of dividing one fraction by another : Multiply the dividend by the reciprocal of the divisor. 32 Ex 1 4(a 2 -a6) . 6a = 4a(o-6) (a-6)(a + 6) (a + &) 2 ' a 2 -6 2 (a + &) 2 -6a = 2 (a - b) 2 3 (a + b) ' a? 4- 2/ 2 - y ^ 2 + 2/ 2 Observe that a fraction is divided by an integer by dividing its numerator by the integer. 154 ALGEBRA. [Cn. VII EXERCISES X. Simplify the following expressions : 2 ^ 4 . 3 6 a 6 . 2 2* 5 by ^ 15 / ' 7 ' ' 21 12 a ' 2 a 3 - a - a + 26 2a + 4& 7(or 3 -l) " (1 - a?) a 2 - (& - c) 2 . a - b + c a 8 -! . ' : 1 + n - n 3 - n 4 . n 2 - 1 : ' 19. 20. 21. 22. 23. a; a? 2 H- ?/ 2 or 3 s/ 3 ' x? y (2x y) 25 (q? + m) 2 - (.y + 7i) 2 . (a; - y) 2 - (n - m (x + y) 2 - (m + n) 2 ' (a? - m) 2 - (n - 2/) 32-34] FRACTIONS. 155 Complex Fractions. 33. A Complex Fraction is a fraction whose numerator and denominator, either or both, are fractions. 2 a+x 1 1 3 a x x E.G.* ~TJ - ; - ? - T* 4 a + y ^ __1 5 a y x Observe that the line which separates the terms of the com- plex fraction is drawn heavier than the lines which separate the terms of the fractions in its numerator and denominator. 34. Ex. 1. Simplify -^ 1 x Multiplying both numerator and denominator by a?, we obtain x. x)x. x) x To reduce a complex fraction to a simple fraction : Multiply both its terms by the L. C. D. of the fractions in the numerator and denominator. Ex. 2. x + L_ *+ * .*1=S 1+?LJ: _J_ 4 ^3-x 3-x 3 4 3a? + 3 x + 1 4 Observe that in this reduction the work proceeds from below upward. 156 ALGEBRA. [Cn. VII EXERCISES XI. Simplify the following expressions : ! ax x x + l 2. ?Ltf . 3> _1 a x 4. - . 5. a + -.- 6. a 1+a; a _a_ + 1 3. 1_ ^_&_ 2 1 1 _1_ /* 7>3 ,,3 8. , 1-5. 9. 1 a aj + 1 a 1 a + x 2x la x _ 1+ . a? a: + 2 a 2 + 4 a; 1 2 a; x a a -\-x x y 1 1 13. "... . ,. g 14. 1 + ^T-T 15. 1- i 1- 1 a; 1 # a + x a x a + 1 a 1 ,a a; a + x a I a-f-1 16. - 17. - - - 18. - - - - n n 4 ax a + 1 a 1 n x n -\-x a 2 x 2 a 1 a-f-1 x x x* 1 x-2 +-+! OB ' ar 3 a n x , ax f 1\Y IV ^ a n*- nx V gy V g. a n-a; ' ' / 1V/ I - x a V PJ V JP. 34] FRACTIONS. 157 EXERCISES XII. MISCELLANEOUS EXAMPLES. Simplify the following expressions: 3 a 4 . ^+r^W-+2=i\ \c + d c dj \c d c + dj 11 a & 2 5. a + & r T- 6. + 7. m - u-\-- -L -r- AT- -TT 6 a 6 a a 6 1 1 ra+w 2 1 + m , ^ + a )^_.)_(^ + ( ,)(___). 10. i 11. 1_2 J^a + z I o 03 1 ' x + 1 a 2 ax x 2 n-l n + lY A w 1 T O X a + n a 2 - n 2 a + i &+ i a+ T v ~ V H 9 ^ 9 . o 6 c a x x 2 + a 2 a; x 2 + a 2 15. -r X ^X T- 16. 3 h iJ ^ HI" 1 a + s" 1 "! a-a? i ^~TT ^*i 9 9 " > i -j a 6 c aa - + araa" + or 158 ALGEBRA. [Cn. VII 17. _ + A 18. 19. x 1 + 2 be In each of the following expressions make the indicated substitution, and simplify the result : m tt 22. In . ^m b 23. In 1 + -, let a + & + c = 2 s. 24. In ^l-^- 1 - let = a m a Verify each of the following identities = 37. -z), when CHAPTER VIII. FRACTIONAL EQUATIONS IN ONE UNKNOWN NUMBER. 1. A Fractional Equation is an equation whose members, either or both, are fractional expressions in the unknown number or numbers. E.g, x + 2 a + 1 a? + 1 3 2 2. Ex. 1. Solve the equation - - = -- - Multiplying by (a: + !)( + 2), 3 (a? + 1) = 2 (a? + 2). Transferring terms, 3 # 2 # = 4 3. Uniting terms, a? = 1. In clearing this equation of fractions, we multiplied by an expression, (x -f- 1) (x + 2), which contains the unknown num- ber. In such a case a root may be introduced. But it is proved in School Algebra, Ch. X., that, if a root is introduced in clearing of fractions, it must be a root of one of the factors of the L. C. D. equated to 0. Since 1 is not a root of x + 1 = 0, or of x + 2 = 0, it is a root of the given equation. 2 x -\- 19 17 3 Ex. 2. Solve the equation ^ = The L.C.D. is 5 (x 2 - 1), = 5 (x - 1) (x + 1). 159 160 ALGEBRA. [Cn. VIII Multiplying by 5 (a; 2 - 1), 2 x + 19 - 85 = - 15 x - 15. Transferring terms, 2 x + 15 x = 15 19 + 85. Uniting terms, 17 x = 51. Dividing by 17, a; = 3. Since 3 is not a root of x 1 = 0, or of x + 1 = 0, it is a root of the given equation. Ex. 3. Solve the equation 6x + l - 2a7 " 1 = 3x ~ l . 4 3x-2 2 When the denominators of some of the fractions do not con- tain the unknown number, it is usually better first to unite these fractions. Transferring 3 ^*, L1. Uniting first two fractions, Multiplying by 4 (3 x - 2), 9 x - 6 - 8 x + 4 = 0. Transferring and uniting terms, x = 2. Since 2 is not a root of 3 x 2 = 0, it is a root of the gfren equation. Ex. 4. If both members of the equation be multiplied by x 2 1, we obtain the integral equation - 2 x 2 - x(x + 1) = - x(x - 1) - 3 (x 2 - 1), 3r (aj + l)(-3) = 0. (2) Now observe that it was not necessary to multiply by x 2 1, = (x + !)(# 1), to clear the given equation of fractions. For, if the terms in the second member be transferred to the first member, we have _ _ or, uniting terms, - = 0, XT 1 2] FRACTIONAL EQUATIONS. 161 y. g or, cancelling x + 1, = 0. Clearing the last equation of fractions, we have 3-3 = 0; (3) whence x = 3. The root 3 of the derived equation (3) is found, by substi- tution, to be a root of the given equation. Had we solved equation (2), we should have obtained the additional root 1, which is not a root of the given equation. This root was introduced by multiplying both members of the given equation by the unnecessary factor x -f- 1, and is a root of the equation obtained by equating this factor to 0. EXERCISES I. Solve each of the following equations : -2 3 JM 4. 6. 8. 10. 19 x-3~ 5 7 oj + 2 5 5. 7 9 11 1.^ x-12 7 24- a; - 3 x + 17 x + 7 x-l 4 2o? + 3 3x + 5 2 01 rr / 6 6x + 2 x 1 14 = 5. 3. .34 7 8 x < 11 9 3^-2 5 6^ + 3 15 ;-l l-2x 6 l + 2a? 13 . -^fL. _9_ =6. 3 ( a?-5 = 2> 15 a?-9 ( ^- 16 -*- + -^. ' x + 2 x-2 a,* 2 - 4 " + 3. . ' ~ . - 3 x 2 - R , a J.1 a* 1 ^ 7 19 __ O ^/^-J. (/ _ rt J. JLA x +. 162 ALGEBRA. [Cii. VIII x x-l2x 2 -5x-2l ' 1 . x-4:2x*-7x-29 23. 24. 25. 26. 28. Problems. 3. Pr. 1. A number of men received $120, to be divided equally. If their number had been 4 less, each one would have received three times as much. How many men were there ? Let x stand for the number of men. Then each man re- 120 ceived dollars. If their number had been 4 less, each one x 19() would have received - dollars, a 4 The problem states, in verbal language: the number of dollars each would have received, if there had been four less, is equal to three times the number of dollars each received. in algebraic language : = 3 x x 4 x Whence, x = 6. Therefore there were six men. 2-3] FRACTIONAL EQUATIONS. 163 Pr. 2. A can do a piece of work in 9 days, B in 6 days ; and A, B, and C together in 3 days. In how many days can do the work ? Let x stand for the number of days it takes C to do the work. Then, in one day, A does - of the work; B does -; and C does - 9 6 x In 3 days, 3 33 A does - of the work; B does -; and C does 9 6 # Therefore, in 3 days, A, B, and C together do ? + ? + of the work. 9 6 # The problem states, in verbal language : the work done by A, B, and C together in 3 days is equal to all the work, or 1 ; 000 in algebraic language : - + - + - = 1. y o x Whence, x = 18. Therefore C can do the work in 18 days. Pr. 3. A cistern has 3 taps. By the first it can be emptied in 80 minutes, by the second in 200 minutes, and by the third in 5 hours. After how many hours will the cistern be emptied, if all the taps are opened ? Let x stand for the number of minutes it takes the three taps together to empty the cistern. Then, in 1 minute, the three together will empty - of the cistern. But, in 1 minute, the first will empty -fa of the cistern ; the second -pfa, and the third -^ ; and together they will empty iro + irio- + iroir of tne cistern. Therefore + + = - 80 200 300 x Whence x 48. It will take the three taps together 48 minutes, or of an hour, to empty the cistern. 164 ALGEBRA. [Cn. VIII EXERCISES II. 1. What number added to the numerator and denominator of f will give a fraction equal to f ? 2. The sum of two numbers is 18, and the quotient of the less divided by the greater is equal to ^. What are the numbers ? 3. The denominator of a fraction exceeds its numerator by 2, and if 1 be added to both numerator and denominator, the resulting fraction will be equal to f . What is the fraction ? 4. The sum of a number and 7 times its reciprocal is 8. What is the number ? 5. The value of a fraction, when reduced to its lowest terms, is f. If its numerator be increased by 7 and its denominator be decreased by 7, the resulting fraction will be equal to f . What is the fraction ? 6. What number must be added to the numerator and subtracted from the denominator of the fraction T 7 5 , to give its reciprocal ? 7. If \ be divided by a certain number increased by }, and i be subtracted from the quotient, the remainder will be J. What is the number ? 8. A train runs 200 miles in a certain time. If it were to run 5 miles an hour faster, it would run 40 miles farther in the same time. What is the rate of the train ? 9. A number has three digits, which increase by 1 from left to right. The quotient of the number divided by the sum of the digits is 26. What is the number ? 10. A number of men have $ 72 to divide. If $ 144 were divided among 3 more men, each one would receive $4 more. How many men are there ? 11. It was intended to divide J by a certain number, but by mistake \ was added to the number. The result was, never- theless, the same. What is the number ? 3] FRACTIONAL EQUATIONS. 165 12. A steamer can run 20 miles an hour in still water. If it can run 72 miles with the current in the same time that it can run 48 miles against the current, what is the speed of the current ? 13. A man buys two kinds of wine, 14 bottles in all, paying $ 9 for one kind and $ 12 for the other. If the price of each kind is the same, how many bottles of each does he buy ? 14. A farmer intended to feed 80 bushels of corn to a certain number of sheep. When 6 of the sheep died, he could have sold 24 bushels of corn and have had enough left to give each remaining sheep the same amount as before. How many sheep had he ? 15. It takes a pedestrian 5 hours to go from A to B. It takes a bicycle rider, who goes 6 miles farther every hour, 2 hours to go the same distance. How far is A from B ? 16. A can do a piece of work in 10 days, B in 6 days and A, B, and C together in 3 days. In how many days can C do the work ? 17. A and B together can do a piece of work in 2 days, B and C together in 3 days, and A and C together in 21 days. In how many days can A, B, and C together do the work ? 18. The circumference of the hind wheel of a carriage exceeds the circumference of the front wheel by 4 feet, and the front wheel makes the same number of revolutions in run- ning 400 yards that the hind wheel makes in running 500 yards. What is the circumference of each wheel ? 19. A cistern has 3 taps. By the first it can be filled in 6 hours, by the second in 8 hours, and by the third it can -be emptied in 12 hours. In what time will it be filled if all the taps are opened ? 20. An inlet pipe can fill a cistern in 3 hours, and an outlet pipe can empty it in 9 hours. After how many hours will the cistern be filled if both pipes are open one-half of the time, and the outlet pipe is closed during the second half of the time ? 166 ALGEBRA. [Cn. VIII 21. In a number of two digits, the digit in the tens' place exceeds the digit in the units' place by 2. If the digits be interchanged and the resulting number be divided by the original number, the quotient will be equal to |~|. What is the number ? 22. In a number of three digits, the digit in the hundreds' place is 2 ; if this digit be transferred to the units' place, and the resulting number be divided by the original number, the quotient will be equal to J^. What is the number ? 23. In one hour a train runs 10 miles farther than a man rides on a bicycle in the same time. If it takes the train hours longer to run 255 miles than it takes the man to ride 3 miles, what is the rate of the train ? 24. A cistern has three pipes. To fill it, the first pipe takes one-half of the time required by the second, and the second takes two-thirds of the time required by the third. If the three pipes be open together, the cistern will be filled in <5 hours. In what time will each pipe fill the cistern ? 25. A and B ride 100 miles from P to Q. They ride together at a uniform rate until they are within 30 miles of Q, when A increases his rate by ^ of his previous rate. When B is within 20 miles of Q, he increases his rate by ^ of his previous rate, and arrives at Q 10 minutes earlier than A. At what rate did A and B first ride ? 26. A circular road has three stations, A, jB, and C, so placed that A is 15 miles from B, B is 13 miles from C in the same direction, and C is 14 miles from A in the same direction. Two messengers leaving A at the same time, and travelling in opposite directions, meet at B. The faster messenger then reaches A 7 hours before the slower one. What is the rate of each messenger ? CHAPTER IX. LITERAL EQUATIONS IN ONE UNKNOWN NUMBER. 1. The unknown numbers of an equation are frequently to be determined in terms of general numbers, i.e., in terms of numbers represented by letters. The latter are commonly represented by the leading letters of the alphabet, a, b, c, etc. Such numbers as a, b, c, etc., are to be regarded as known. E.g., in the equation x -f- a = b, a and b are the known num- bers, and x is the unknown number. From this equation we obtain x = b a. 2. A Numerical Equation is one in which all the known num- bers are numerals ; as 2 a? -f- 3 = 7 ; 4 a? 3y = 7. A Literal Equation is one in which some or all of the known numbers are literal ; as 2 ax -f- 3 b = 5 ; ax + by = c. 3. Ex. 1. Solve the equation b a 2ab Clearing of fractions, 2 ax - 2 a 2 + 2 bx - 2 b 2 = - a 2 + 2 ab - b 2 . Transferring and uniting terms, Dividing by 2 (a + b), x = ^ Notice that the above equation, although algebraically frac- tional, is integral in the unknown number x. The equation which follows is fractional in the unknown number. 167 168 ALGEBRA. [Cn. IX Ex. 2. Solve the equation ^t_^ = ^LJL b + x 6 + 1 Multiplying by (&+)(& + !), (a+)(& + l) = (&+aj) ( + !) Simplifying, a& + 60; + a + x = ab + ax + 6 + a;. Cancelling terms, &e + a = ax -f- 6. Transferring and uniting terms, (b a)x = b a. Dividing by b a, x = 1. EXERCISES I. Solve the following equations : 1. a x = c. 2. mx + a, = b. 3. rax = wic + 2. 4. 3 ax 5 a& + 6 ax 7 ac = 2 ax + 2 ab. 5. 4 a 2 - 2 afrc + 6 2 + 3 a?x = 5a?-b 2 x + 2 a?x. 6. a (a H- a) - b(x - b) = Sax + (a - 6) 2 . 7. 05 (x + a) + a; (x + 6) 2 (a? + a) (x + 6) = 0. ,6 8. a -f- - = c. x b x a 2 10 ^JL^ = 5. 11 ^l_i_5_^ = ^. a a 4 ax a b x x + a 2 _ x a 12 = t 13 b + x 6 + 1* 2 + a 6x-f ^ a; ~~^_o 15 - + ^ _ x 4:X+b 2x a ' b + a ba ^ ' ' x-ab a 2 -ab + b 2 x - b (2x-b) 2 18 ^ 2 + a " x _1. ' 4JB 2 a 2 2x + a 4* ' 3-4] LITERAL EQUATIONS. 169 x-a x-b a?-c = l .1.1. 2bc 2ac 2ab a be l-2ax 2 1+200* 4 aba? 22. - 23. 2bx 2 g a? -\- x a 2 + a x + a a; a + 1 24 a 2 + a; _ a 2 -a; = 4 abx + 2 a 2 - 2 6^ 25 <* 2 + a + a~a? + aar + or a 4 + 2 a^ar + a; 4 a + a? 2 / 2 a + a? a 4 a 2 a? o?x + 2 aa; 2 a 3 a General Problems. 4. A General Problem is one in which the known numbers are literal. Pr. 1. The greater of two numbers is m times the less, and their sum is s. What are the numbers ? Let x stand for the less required number. Then mx stands for the greater. By the condition of the problem, we have x -f mx = s ; whence, x - , the less number, and mx= ms , the greater. 1+m 1+m If m = 3 and s = 84, we have x = J*i- = 21, and mx = 3 x 21 = 63. 1 + 3 When the numbers are equal, m = 1, and we obtain aj = , and ma?=, 170 ALGEBRA. [Cn. TX for all values of s ; that is, either of the two numbers is half their sum. Thus the solution of this general problem includes the solu- tions of all like problems. A solution for any like problem is obtained by substituting particular values for m and s, as above. Pr. 2. A cistern has two taps. By the first it can be filled in a minutes, and by the second in b minutes. How many minutes will it take the two taps together to fill the cistern ? Let x stand for the number of minutes it takes the two taps to fill the cistern. Then, in 1 minute, the two together will fill - of the cistern. x But, in 1 minute, the first will fill - of the cistern, the sec- ond -; and together they will fill - + - of the cistern. b a b Therefore 1 + 1 = 1. a b x Whence ab a + b This solution gives a general rule for solving problems of like character. In a particular example, a may be the number of minutes it takes a tap to fill a cistern, the number of hours it takes a man to build a wall, to dig a ditch, to plough a fieldj etc. Pr. 3. If one man can dig a ditch in 6 days, and a second man in 3 days, in how many days can they dig the ditch, working together ? Substituting a = 6, b 3, in the result of Pr. 2, we have 6x3 Therefore they can together dig the ditch in 2 days. 4] LITERAL EQUATIONS. 171 EXERCISES II. Find the general solution of each of the following problems, and from this solution obtain the particular solution for the numerical values assigned to the literal numbers in the problem. 1. Find a number, such that the result of adding it to n shall be equal to n times the number. Let n = 2 ; 5. 2. Divide a into two parts, such that of the first, plus - of m u- the second, shall be equal to b. Let a = 100, b = 30, m = 3, n = 5. 3. A sum of d dollars is divided between A and B. B receives b dollars as often as A receives a dollars. How much does each receive ? Let d 7000, a = 3, b = 2. 4. A father's age exceeds his son's age by m years, and the sum of their ages is n times the son's age. What are their ages ? Let m = 20, n = 4 ; m = 25, n = 7. 5. A farmer can plough a field in a days, and his son in b days ; in how many days can they plough the field, working together ? Let a = 10, b = 15. 6. What time is it, if the number of hours which have elapsed since noon is m times the number of hours to mid- night ? Let m = \. 7. One pipe can fill a cistern in a hours, a second in b hours, and a third in c hours. In how many hours can the three pipes fill the cistern, working together ? Let a = 2, b = 3, c = 6. 8. One pipe can fill a cistern in m hours, a second in n hours, and a third can empty it in p hours. After how many hours will the cistern be filled, if all pipes are open ? Let m 4, n = 6, p = 3. 9. Two couriers start at the same time and move in the same direction, the first from a place d miles ahead of the second. The first courier travels at the rate of m^ miles an hour, and the second at the rate of w a miles an hour. After 172 ALGEBRA. [Cn. IX how many hours will the second courier overtake the first? Let d = 15, mi = 17, m 2 = 20. From the result of the preceding example find the results of Exx. 10-12. 10. At what rate must the second courier travel in order to overtake the first after h hours ? Let d = 18, ?% = 15, h = 3. 11. At what rate must the first courier travel in order that the second courier may overtake him after h hours ? Let d = 12, m 2 = 22,h = 3. 12. How many miles behind the first courier must the second start in order to overtake the first after h hours ? Let m 1 = 18, m 2 = 21, h = 4. 13. In a company are a men and 6 women; and to every m unmarried men there are n unmarried women. H<3w many married couples are in the company ? Let a = 13, b = 17, m = 3, n = 5. INTERPRETATION OF THE SOLUTIONS OF PROBLEMS. 5. In solving equations we do not concern ourselves with the meaning of the results. When, however, an equation has arisen in connection with a problem, the interpretation of the result becomes important. In this chapter we shall interpret the solutions of some linear equations in connection with the problems from which they arise. Positive Solutions. 6. Pr. A company of 20 people, men and women, proposed to arrange a fair for the benefit of a poor family. Each man contributed $ 3, and each woman $ 1. If $ 55 were contributed, how many men and how many women were in the company ? Let x stand for the number of men; then the number of women was 20 x. The amount contributed by the men was 3 x dollars, that by the women 20 x dollars. By the condition of the problem, we have 3 x + (20 x) = 55 ; whence x = 17|. 4-8] INTERPRETATION" OF SOLUTIONS. 173 The result, 17^, satisfies the equation, but not the problem. For the number of men must be an integer. This implied con- dition could not be introduced into the equation. The conditions stated in the problem are impossible, since they are inconsistent with the implied condition. Negative Solutions. 7. Pr. A father is 40 years old, and his son 10 years old. After how many years will the father be seven times as old as his son ? Let x stand for the required number of years. Then after x years the father will be 40 + x years old, and the son 10 + x years old. By the condition of the problem, we have 40 + x = 7 (10 + a?), whence x = - 5. (1) This result satisfies the equation, but not the condition of the problem. For since the question of the problem is " after how many years ? " the result, if added to the number of years in the ages of father and son, should increase them, and there- fore be positive. Consequently, at no time in the future will the father be seven times as old as his son. But since to add 5 is equivalent to subtracting 5, we conclude that the question of the problem should have been, " How many years ago?" The equation of the problem, with this modified question, is : 40 x = 7 (10 x) ; whence x = 5. (2) Notice that equation (2) could have been obtained from equa- tion (1) by changing x into x. 8. The interpretation of a negative result in a given problem is often facilitated by the following principle : If x be substituted for x in an equation which has a negative root, the resulting equation will have a positive root of the same absolute value ; and vice versa. E.g., the equation x + 1= x 3 has the root 2 ; while the equation x + \ = x 3 has the root 2. 174 ALGEBRA. [Cn. IX 9. Pr. Two pocket-books contain together $100. If one- half of the contents of one pocket-book and one-third of the contents of the other be removed, the amount of money left in both will be $70. How many dollars does each pocket-book contain ? Let x stand for the number of dollars contained in the first pocket-book ; then the number of dollars contained in the second is 100 x. When one-half of the contents of the first and one-third of the contents of the second are removed, the number of dollars remaining in the first is -x, and in the second 2 |(100 x). By the conditions of the problem, we have $x + |(100 - a;) = 70, whence x = - 20. Substituting x for x in the given equation, we obtain - \x + |(100 + x) = 70, or (100 + x) - x = 70. This equation corresponds to the following conditions : If x stand for the number of dollars in one pocket-book, then 100 4- x stands for the number of dollars in the other ; that is, one pocket-book contains $ 100 more than the other. The second condition of the problem, obtained from the equation, is : two-thirds of the contents of one pocket-book exceeds one-half of the contents of the other by $ 70. Therefore the modified problem reads as follows : Two pocket-books contain a certain amount of money, and one contains $100 more than the other. If one-third of the contents be removed from the first pocket-book, and one-half of the contents from the second, the first will then contain $ 70 more than the second. How much money is contained in each pocket-book ? 10. These problems show that the required modification of an assumption, question, or condition of a problem which has led to a negative result, consists in making the assumption, question, or condition the opposite of what it originally was. Thus, if a positive result signify a distance toward the right from a certain point, a negative result will signify a distance toward the left from the same point ; and vice versa; etc. 9-11] INTERPRETATION OF SOLUTIONS. 175 Zero Solutions. 11. A zero result gives in some cases the answer to the ques- tion j in other cases it proves its impossibility. Pr. A merchant has two kinds of wine, one worth $7.25 a gallon, and the other $5.50 a gallon. How many gallons of each kind must be taken to make a mixture of 16 gallons worth $88? Let x stand for the number of gallons of the first kind ; then 16 x will stand for the number of gallons of the second kind. Therefore, by the condition of the problem, we have 7.25x + 5.5(16 - x) = 88 ; whence x = 0. That is, no mixture which contains the first kind of wine can be made to satisfy the condition. In fact, 16 gallons of the second kind are worth $ 88. EXERCISE III. Solve the following problems, and interpret the results. Modify those problems which have negative solutions so that they will be satisfied by positive solutions. 1. A and B together have $ 100. If A spend one-third of his share, and B spend one-fourth of his share, they will then have $ 80 left. What are their respective shares ? 2. A father is 40 years old, and his son is 13 years old ; after how many years will the father be four times as old as his son ? 3. The sum of the first and third of three consecutive numbers is equal to three times the second. What are the numbers ? 4. In a number of two digits, the tens' digit is two-thirds of units' digit. If the digits be interchanged, the resulting number will exceed the original number by 36. What is the number ? 5. A teacher proposes 30 problems to a pupil. The latter is to receive 8 marks in his favor for each problem solved, and 12 marks against him for each problem not solved. If the number of marks against him exceed those in his favor by 420, how many problems will he have solved ? 176 ALGEBRA. [Cn. IX 6. In a number of two digits the tens' digit is twice the units' digit. If the digits be interchanged, the resulting number will exceed the original number by 18. What is the number ? 7. A has $ 100, and B has $ 30. A spends twice as much money as B, and then has left three times as much as B. How much does each one spend ? Discuss the solutions of the following general problems. State under what conditions each solution is positive, negative, or zero. Also, in each problem, assign a set of particular values to the general numbers which will give an admissible solution. 8. A father is a years old, and his son is b years old. After how many years will the father be n times as old as his son ? 9. Having two kinds of wine worth a and b dollars a gallon, respectively, how many gallons of each kind must be taken to make a mixture of n gallons worth c dollars a gallon ? 10. Two couriers, A and B, start at the same time from two stations, distant d miles from each other, and travel in the same direction. A travels n times as fast as B. Where will A overtake B ? CHAPTER X. SIMULTANEOUS LINEAR EQUATIONS. SYSTEMS OF EQUATIONS. 1. If the linear equation in two unknown numbers 5 (1) be solved for y, we obtain y = 5 x. We may substitute in this equation any particular numerical value for oj, and obtain a corresponding value for y. Thus, when x = 1, y = 4 ; when x = 2, / = 3 ; when aj = 3, y 2 ; etc. In like manner the equation could have been solved for x in terms of y, and corresponding sets of values obtained. Any set of corresponding values of x and y satisfies the given equation, and is therefore a solution. 2. Solving the equation y-* = l <2> for y, we have ?/ = 1 -f- a?. Then, when x = 1, 2/ = 2 ; when jr = 2, / = 3 ; when # = 3, 2/ = 4 ; etc. Now, observe that equations (1) and (2) have the common solution, a; = 2, y = 3. It seems evident, and it is proved in School Algebra, that these equations have only this solution in common. Equations (1) and (2) express different relations between the unknown numbers, and are called Independent Equations. Also, since they are satisfied by a common set of values of the unknown numbers, they are called Consistent Equations. 177 178 ALGEBRA. [Cn. X 3. A System of Simultaneous Equations is a group of equa- tions which are to be satisfied by the same set, or sets, of values of the unknown numbers. A Solution of a system of simultaneous equations is a set of values of the unknown numbers which satisfies all of the equations. 4. The examples of Arts. 1-2 are illustrations of the follow- ing general principles : A system of linear equations has a definite number of solutions. (i.) When the number of equations is the same as the number of unknown numbers. (ii.) When the equations are independent and consistent. 5. Two systems of equations are equivalent when every solution of either system is a solution of the other. E.g.) the systems (I.) and (II.) : 2x-2y = are equivalent. For they are both satisfied by the solution, x = 2, y = 1, and, as we shall see later, by no other solution. 6. If the equations x -f- y = 7, * - y = 1, be added, we obtain 2 x = 8, in which the uriknown number y does not appear. We say that y has been eliminated from the given equations. 7. Elimination is the process of deriving from tw"o or more equations an equation which has one less unknown number. Elimination by Addition and Subtraction. 8. Ex. 1. Solve the system 3 x + 4 y = 24, (1) 5x-6y = 2. (2) To eliminate y, we multiply the equations by such numbers as will make the coefficients of // numerically equal. 3-9] SIMULTANEOUS LINEAR EQUATIONS. 179 Multiplying (1) by 3, 9 x + 12 y = 72. (3) Multiplying (2) by 2, 10 x - 12 y = 4. (4) Adding (3) and (4), 19 x = 76. Whence x 4. Substituting 4 for x in (1), 12 + 4y = 24. Whence y = 3. It is proved in School Algebra, Ch. XIII., that the above method is based upon equivalent equations. Consequently the required solution is x = 4, y = 3. This solution may be written 4, 3, it being understood that the first number is the value of x, and the second the value of y. Ex. 2. Solve the system 12x + 15y = 8. (1) 16o; + 9y = 7. (2) We will first eliminate x. Multiplying (1) by 4, 48 a? + 60 y = 32. (3) Multiplying (2) by 3, 48 x + 27 y = 21. (4) Subtracting (4) from (3), 33 y = 11. Whence y = -J-. Substituting i for y in (1), 12 a; + 5 = 8. Whence x = i. Consequently the required solution is \, ^. 9. The examples of the preceding article illustrate the fol- lowing method of elimination by addition and subtraction : Multiply both members of the equations by such numbers as will make the coefficients of one of the unknown numbers numeri- cally equal. Subtract, or add, corresponding members of the resulting equations, and equate the results. /Solve this equation in one unknown number. Substitute the value of this unknown number in the simpler of the given equa- tions, and solve for the other unknown number. The multipliers are obtained by dividing the L. C. M. of the coefficients of the unknown number to be eliminated by the coefficients of this unknown number. It is better to eliminate that unknown number which requires the smallest multipliers. 180 ALGEBRA. [Cn. X EXERCISES I. Solve the following systems of equations by the method of addition and subtraction : ! , . - , ~ , + y = a> x-y = 7. ( x -y = b. U + 4?/ = 19. 3s + y = 31, f4*-7y = 19, 1 5a-22/ = l5. ' U + 9y = 37. nx ay = 0, *'!-., 8. TTX ay = an. 10. ' ' " ' - -^+^ = c > 12. 16 x + 9 y = 7. I mx + ny =p. f 3 aj -f 16 y = 5, f 21 a? + 8 y = 66, I - 5 x 4- 28 y = 19. 13 i 28 x - 23 y = 13. f 18 a? - 20 y = 1, " L4 * ll5o; + 162/ = 9. 15 a; - 14 y = 33, f 25 x + 24 w = 98, 16. \ 17. \ 20 a; + 21 y = 24. 15 a; - 16 i/ = - 2. f 40 a; - 63 y = 57, f 15 a? + 28 y = 58 a, 18y = -87. ' Il8aj-36y = a. 18 \ 19 -I ' I35ic-18v = 87. Elimination by Comparison. 10. Ex. Solve the system 7x + 2y = 20, (1) 13a-32/ = 17. (2) To eliminate jc, w proceed as follows : Solving (1) for y, y = 2 ~ 7a; . (3) Solving (2) for y, = 13^-17, (4) o Equating these values of y, 20 -7 x 13 a; - 17 ,~ o o * (Pi 9-11] SIMULTANEOUS LINEAR EQUATIONS. 181 Whence x = 2. Substituting 2 for x in (3), y = 2Q ~ 14 = 3. It is proved in School Algebra, Ch. XIII., that the above method is based upon equivalent equations. Consequently the required solution is 2, 3. 11. This example illustrates the following method of elimi- nation by comparison : Solve the given equations for the unknown number to be elimi- nated, and equate the expressions thus obtained. The derived equation will contain but one unknown number. Solve this derived equation, and substitute the value of the unknown number thus obtained in the simplest of the preceding 4 equations, and solve for the other unknown number. EXERCISES II. Solve the following systems of equations by the method of comparison : = 28, (21x-23y = 2,. ' ' ' r8 + 3i/ = 58 (3a; -8 = - ^ ' 182 ALGEBRA. [Cii. X 63 a -46 y = 29, .y + c(x + l) = 0. = 9 a b, 18. Elimination by Substitution. 12. Ex. Solve the system 5x-2u = l, (1) 4a; + 5y = 47. (2) If we wish to eliminate a;, we proceed as follows : Solving (1) for x, x = 1 + 2y . (3) o Substituting "T ^ for a? in (2), o c- (4) Whence y = 7. Substituting 7 for y in (3), a? = 3. It is proved in School Algebra, Ch. XIII., that the above method is based upon equivalent equations. Consequently the required solution is 3, 7. 13. This example illustrates the following method of elimi- nation by substitution : Solve the simpler equation for the unknoivn number to be elimi- nated in terms of the other. Substitute the value thus obtained in the other equation. The derived equation will contain but one unknown number. Solve the derived equation, and substitute the value of the unknown number thus obtained in the expression for the other unknown number, and solve for the other unknown number. 11-14] SIMULTANEOUS LINEAR EQUATIONS. 183 EXERCISES III. Solve the following systems of equations by the method of substitution: 1. * = y. (2x = y. x = 2y-3, (x = 3y-7, { 5 x = 4 y. { x ny = 0. 9. -I 10 (8x-5y = Q. 11. _ 12. 13 4aj-15y = 22, flOa- 21 y = 75, 6a? + 7 =-26. Linear Equations in Three Unknown Numbers. 14. The following examples will illustrate a method of solving systems of three linear equations in three unknown numbers : Ex. 1. Solve the system 2x 3y + 5z = ll, (1) 5a + 42/-6z=-5, (2) 4# + 7y 8s = 14. (3) To eliminate x, we proceed as follows : Multiplying (1) by 5, Wx-15y + 25z = 55. (4) Multiplying (2) by 2, 10aj + 8y-12s = - 10. (5) 184 ALGEBRA. [Cn. X Subtracting (4) from (5), 23 y - 37 z = - 65. (6) Multiplying (1) by 2, 4 x - 6 y -f 10 z = 22. (7) Adding (3) and (7), y + 2 z = 8. (8) Solving (6) and (8), y == 2. 2 = 3. Substituting 2 for ?/ and 3 for 3 in (1), x = 1. Consequently the required solution is 1, 2, 3. Ex. 2. Solve the system ay cz = 0, (1) z-x = -b, (2) ax + % = a 2 + 6 (a + c). (3) Notice that by eliminating 3 from (1) and (2) we obtain an equation in x and y, which with equation (3) gives a system of two equations in the same two unknown numbers. Solving (2) for z, z = x- b. (4) Substituting x b for z in (1), ay ex + cb = 0. (5) Multiplying (3) by a, a?x + afo/ = a 3 4- a 2 6 + abc. (6) Multiplying (5) by 6, 6cx + c% = We. (7) Subtracting (7) from (6), (a 2 + be) x = a 3 4- orb + a&c + & 2 c &); (8) whence a? = a + 6. Substituting a-\-b for a; in (4), z = a. Substituting a for z in (1), ?/ = c. 15. These examples illustrate the following method : Eliminate one of the unknown numbers from any two of the equations j next eliminate the same unknown number from the third equation and either of the other two. Two equations in the same two unknoivn numbers are thus derived. 14-15] SIMULTANEOUS LINEAR EQUATIONS. 185 Solve these equations for the two unknown numbers, and sub- stitute the values thus obtained in the simplest equation which contains the third unknown number. EXERCISES IV. Solve the following systems of equations : _}. 3 y _[_ 3 % 19 ? x 4- y 4- z = 21, 4 y = 3 a?. 2. 6 x = 5 2, 1. 3. 3 = 2*. a = 42 -17, 5. x -f- z = 30, [ + z=32. 32/ + 22 =0. 15 ?/ 4- 4 2 7 x = 44, 2 = 5 y 33, [ x = 5 y 22. x 4- z = (y = 2 a. cc 4- 2 = 9. 8. 3y-z = 5, (3z-x = Q. (3x + 2y-z = 15, 10. 5 x 3 y 4- 2 2 = 28, 12. 14. 16. x + 25y- 6 = - 9. 11. 5 x = 34. 2 = c, y = b, 2x- y+ 92 = 28, 13. \ 7x+ 3y- 52 = 3, 15. 2a?-7y+ ^2= 3, -45, -95. 9^+3^-20 z= [ _l3o?+4y-30z= 35, f 80; 21 y- 92 = 37, 17. J 12x-2S y + loz = 64. 186 ALGEBRA. [Cn. X { ax + by = ?> 2 , f x + y -f z = a + 6 + c, 18. ] fry + cz = b 2 + c 2 , 19. | aa = by, [cz-\- ax = c 2 . az = cy. ax 4- &y cz = a 2 -f- & 2 > ax abz + 6 2 , 21. + a?/ + arz + a 3 = 0, x + 6y H- 5 2 z + 6 3 = 0, x 4- c?/ + c 2 z + c 8 = 0. 16. It is frequently necessary to simplify the equations be- fore applying one of the preceding methods : Ex. 1. Solve the system Clearing (1) and (2) of fractions, 28 + 4a-10x + 5i/ = 602/- 100, (3) 4 x - 3 + 15 y - 21 == 108 - 30 x. (4) Transferring and uniting terms, 6 x + 55 y = 128, (5) 34 x + 15 y = 132. (6) Multiplying (5) by 3, 18 x + 165 y = 384. (7) Multiplying (6) by 11, 374 x + 165 y = 1452. (8) Subtracting (7) from (8), 356 x = 1068 ; whence, x = 3. Substituting (3) for x in (5), 18 + 55 y = 128; whence, y = 2. Consequently, the required solution is 3, 2. 17. Certain fractional equations are to be solved for the reciprocals of one or both of the unknown numbers. 15-17] SIMULTANEOUS LINEAR EQUATIONS. 187 Q * Ex. 2. Solve the system - 1 *- = 8, (1) 2x oy y 2 2x-3y ' y-2 Let 2x 3y = u, y 2 = v. Then (1) and (2) become ? + - = 8, (3) u v 7 . 1*2*10. (4) U V We will solve this system for - and u v Multiplying (3) by 3, - 4- = 24. (5) \b \J Multiplying (4) by 5, ? + ~ = 50. (6) 26 Subtracting (5) from (6), - = 26. Dividing by 26, - = 1, or u = 1. . Substituting 1 for u in (3), 3 + - = 8, or - = 5. v Dividing by 5, - = 1, or v = 1. We now have to solve the system, y-2 = l. (8) From (8), y = 3. Substituting 3 for y in (7), 2 x 9 = 1, or 2 a? = 10. Dividing by 2, x = 5. Therefore the required solution is 5, 3. 188 ALGEBRA. [Cn. X Ex. 3. Solve the system, x -f- y = xy, (1) j 2x + 2z = xz, (2) (I.) 3y + 3z = yz. (3) J Observe that the given equations are neither linear nor frac- tional. Yet they can be transformed so that they will contain only the reciprocals of , y, and z. Dividing (1) by xy, (2) by xz, (3) by yz, we have : We will solve this system for -, -, x y z Multiplying (4) by 2, - + - = 2. (7) y x o o Subtracting (5) from (7), - - = 1. (8) Solving (6) and (8) for 1 and -, - = , ! = --!. Substituting for - in (4), - = . \2i y x \.2i Consequently, a solution of the given system is ^, -^ 12. It is important to notice that we cannot assume that the system (II.) is equivalent to the system (I.), since the equa- tions of (II.) are derived from the equations of (I.) by dividing by expressions which contain the unknown numbers. But if any solution of (I.) be lost by this transformation, it is a solution of the expressions (equated to 0) by which the equations of (I.) were divided ; that is, of xy = 0, xz = 0, yz = 0. (III.) The system (III.) has the solution 0, 0, 0, and this solution evidently satisfies the system (I.). We therefore conclude that the given system has the two solutions - 1 /, Y, - 12, and 0, 0, 0. 17] SIMULTANEOUS LINEAR EQUATIONS. 189 EXERCISES V. Solve the following systems of equations : 1. 3. 7. te+SI- nil 2x - 22, 2. = 20. -5 -7 oj-1 3 2/-l~4 7 x + 3 10 lly 5 - x-7 y- 2/ + 3 13 2x + 7y x + 1 3 2 4 x-7 y-5 g A n J 4 6 2x + 72/ a? -4- 7 2 3 2x+l 3y+2 6 3 3aj _4 42/-1 5 7 6 " 3x-l 7y + 2 2 5 5^_9 y _2 A "' /- ^x y. 4 6 a; ! y 1 7 2 * 1 y 1 w + 1 - -2-T + - n 1 w - 9. 10. 11. -1 n-1 y i r J m a m o x \ y i f i 2 - 1 5a; + 7# + 2 3x- n a Ti 6 h42/ + 7 3 4 7a: + 3# + 4 6a + 5^ + 7 4 5 cc _3x + 5 f + 17 = 52/ + 4x 7 ; 17 3 22-6?( 5x-7 x + 1 S.v + 5 3 11 6 18 3 + 7.y + l 2a;-3?/ + 8 K W> 5 o 5a;-77/H-10 3x + 2y + 6 3 5 ' 13. 190 ALGEBRA. [Cn. X 14. 17. 19. 22. 7-- = 16, 3 4 -< a l b 15. x y \ x y J.O. 3a-? = 4. ----2 6 + a -i y ay x y r 3 4 o 3 # 5 y Q x 4 y 1 ' x + 2 w 3 ia. - 9 2 J. O I/ iC Q i 4(^+2 y) 5 c 1 1 11 1 2 2 x + y 5 ' - + ~ = a, -H h - = 16, z y x y z 1 + 1 = 6, 20. i + i = 6, 21. i + ? + ?=15, a? 2! 2 C y Z X 1 + 1 = 7. - 11 199 - + - = c. i + f + f^U. a? y (z x y 1,3.4 - H h - = o, r 3 + 4_8 = 15 x y z ? -f - * M-?=, 5 1 + 2 -i a y z x 2y 9 6 ~x~l^ = 21 ' 4 # w EXERCISES VI. MISCELLANEOUS EXAMPLES. Solve the following systems of equations by the methods given in this chapter : 1. x + y = z + 10, aj-y + l 2. az = 3 (a; + z), [ a# *= 4 (a? -f 2/). 7 11 2-3y 4. = 06. 10?/-7 10 SIMULTANEOUS LINEAR EQUATIONS. 191 5. 7. - (x -[- a b x-b 2ft 1 y-a-b b y - a e. a a? -f- ny n lOw ny 3 x a cw + by , y + b e a + 1 x + ny ft f _i_ -> ft?/ A 1 4- 2 axby + . a 8. ^ __& + !. a ^_, = 1 + .2 2a ' 2a a?x - b 2 y = 0, (a 2 + &> + ( 2 - b 2 )y = a 4 + b*. (a + 6) x H- (a 6) ?/ = a 2 -f & 2 , ( a _ 6 )a; + (a .T?/ i r. n X2; h 12 \ W/j 111 111 13 2/ +2; ' x + 2/ + 2 = 6, 03 + 2! + U = 8, . y + z + w = 9. . "J 2 1^ 1 -1 y a? ?/ -f- 2? + u = 9, i/ 4- 1 , 5 a; Q l 15 5 x 4 y H- 1 3 16. 2x y a x .by .cz \T) i c 3x y x+y y+z 17. < i c *) c a a + b a \ y x a b 1 ") C y + az = a? + c?. y + x a -{-b 7 192 ALGEBRA. [Cn. X 20. 21. 19. rt(a- U(a- r a? 10 - ^ 2 a; + 3 ?/ 29 ' 7 a 8 ?/ + 24 80 4 * o h & _ C )aj + |( a - 6 + c) y = a 2 + (b - c) 2 , - b + c) x 4- (a + 6 c) y = a 2 (b c) 2 . .V .2 ,2 y-6 10 y+6 ri 2 -! a 2 -! 22. 5 3 Ia 2 -fl ' n 2 +l ar3x 3yxy 10 Problems. 18. Pr. 1. The sum of the two digits of a number is 12. If the digits are interchanged, the resulting number will exceed the original one by three-fourths of the original number. What is the number? Let x stand for the units' digit, and y for the tens' digit. Then the original number is.Wy + x. When the digits are interchanged, the resulting number is (1) The first condition of the problem states, in verbal language : the sum of the digits is 12 ; in algebraic language : x + y = 12. The second condition states, in verbal language : the resulting number minus the original number is equal to J of the original number; in algebraic language : 10 x + y (Wy + x) = f (Wy + x). (2) Solving (1) and (2), x = 8, y = 4. Therefore the required number is 48. Pr. 2. A tank can be filled by two pipes. If the first is left open 6 minutes, and the second 7 minutes, the tank will be filled; or if the first is left open 3 minutes, and the second 12 minutes, the tank will be filled. In what time can each pipe fill the tank ? 17-18] SIMULTANEOUS LINEAR EQUATIONS. 193 Let x stand for the number of minutes it takes the first pipe to fill the tank, and y for the number of minutes it takes the second pipe. Let the capacity of the tank be represented by 1. Then in 1 minute the first pipe fills - of the tank, and in / X rr 6 minutes - of the tank ; the second pipe fills - of the tank x y in 7 minutes. Therefore, by the conditions of the problem, 6,7 , 3 12 1 - + -=-L; H = 1. x y x y Whence x = lOi, y = 17. Pr. 3. The sum of the three digits of a number is 9. The digit in the hundreds' place is equal to one-eighth of the number composed of the two other digits, and the digit in the units 7 place is equal to one-eighth of the number composed of the two other digits. What is the number ? Let x stand for the units' digit, y for the tens' digit, and z for the hundreds' digit. Then, by the first condition, x + y + * = 9. (1) The number composed of the tens' and units' digits is Wy+x. Therefore, by the second condition, z = $(10y + 'x). (2) The number composed of the hundreds' and tens' digits is 10 z + y. Therefore, # = 1 (10 z + ?/). (3) Solving equations (l)-(3), we obtain, a> = 4, y = 2, 2 = 3. Therefore, the required number is 324. Pr. 4. The report of a cannon travels 172.21 yards with the wind toward A in the same time that it travels 167.97 yards against the wind toward B. Three seconds after it is fired it is heard at A and B, which are 2041.08 yards apart. What is 194 ALGEBRA. [On. X the velocity of the report in still air, and what is the velocity of the wind ? Let x stand for the number of yards the report travels a second in still air, and y for the number of yards the wind travels a second. Then, in 1 second the report travels x + y yards with the wind toward A, and x y yards against the wind toward B. 172 21 Therefore, it takes - seconds to travel 172.21 yards x + y - r*r^ C\^ toward A. and - seconds to travel 167.97 yards x-y toward B. Consequently, by the first condition, 172.21 167.97 (1) x -\- y x y In 3 seconds the report travels 3(x + y) yards to A, and 3 (x y) yards to B. Therefore, by the second condition, 3(x + y)+3(x-y)= 2041.08. (2) Solving equations (1) and (2), we obtain x : = 340.18, y = 4.24.' Therefore, the velocity of the report in still air is 340.18 yards a second, and the velocity of the wind is 4.24 yards a second. Pr. 5. Two boys, A and B, run a race from P to Q and return. A, the faster runner, on his return meets B 90 feet from Q, and reaches P 3 minutes ahead of B. If he had run again to Q, he would have met B at a distance from P equal to one-sixth of the distance from P to Q. How far is Q from P, and at what rates do A and B run ? Let x stand for the number of feet from P to Q, y for the number of feet A runs in 1 minute, z for the number of feet B runs in 1 minute. 18] SIMULTANEOUS LINEAR EQUATIONS. 195 When they first meet 90 feet from Q, A has evidently run x + 90 feet in - minutes, and B has run x 90 feet in - minutes. Therefore, 90 = 90. (1) A runs 2 x feet, from P to Q and return, in minutes, and 2x y B the same distance in minutes. z Therefore, by the second condition, Zi X Zi X Q /o\ - = -- O. (4 ) y * 13 x If A had again met B, he would have run 2x + x f = , 6 feet in - minutes, and B would have run 2x x, = - ~ y 6 ' feet in minutes. oz Therefore, by the last condition, !!* O r^ = i!. (3) 6z 6z' y z Solving equations (l)-(3), we obtain Therefore the distance from P to Q is 1080 feet ; A runs 13Q1-Q- feet a minute, and B runs HOy^ feet a minute. EXERCISES VII. 1. Find two numbers whose sum is 19 and whose difference is 7. 2. If one number be multiplied by 3 and another by 7, the sum of the products will be 58 ; if the first be multiplied by 7 and the second by 3, the sum will be 42. What are the numbers ? 196 ALGEBRA. [Cn. X 3. In a meeting of 48 persons, a motion was carried by a majority of 18. How many persons voted for the motion and how many against it ? 4. If one of two numbers be divided by 6 and the other by 5, the sum of the quotients will be 52 ; if the first be divided by 8 and the second by 12, the sum of the quotients will be 31. What are the numbers ? 5. Find two numbers, such that if 1 be subtracted from the first and added to the second, the results will be equal ; while if 5 be subtracted from the first and the second be subtracted from 5, these results will also be equal. 6. If 45 be subtracted from a number, the remainder will be a certain multiple of 5 j but if the number be subtracted from 135, the remainder will be the same multiple of 10. What is the number, and what multiple of 5 is the first remainder ? 7. If 1 be added to the numerator of a fraction, the result- ing fraction will be equal to J; but if 1 be added to the denominator, the resulting fraction will be equal to J-. What is the fraction ? 8. A said to B : " Give me three-fourths of your marbles and I shall have 100 marbles." B said to A : " Give me one- half of your marbles and I shall have 100 marbles." How many marbles had A and B ? 9. A bag contains white and black balls. One-half of the number of white balls is equal to one-third of the number of black balls, and twice the number of white balls is 6 less than the total number of balls. How many balls of each color are there ? 10. The sum of two numbers is 47. If the greater be divided by the less, the quotient and the remainder will each be 5. What are the numbers ? 11. A father said to his son : " After 3 years I shall be three times as old as you will be, and 7 years ago I was seven times as old as you then were." What were the ages of father and son? 18] SIMULTANEOUS LINEAR EQUATIONS. 197 12. A merchant received from one customer $26 for 10 yards of silk and 4 yards of cloth; and from another customer $ 23 for 7 yards of silk and 6 yards of cloth at the same prices. What was the price of the silk and of the cloth ? 13. A merchant has two kinds of wine. If he mix 9 gallons of the poorer with 7 gallons of the better, the mixture will be worth $ 1.37^ a gallon ; but if he mix 3 gallons of the poorer with 5 gallons of the better, the mixture will be worth $1.45 a gallon. What is the price of each kind of wine ? 14. A man has a gold watch, a silver watch, and a chain. The gold watch and the chain cost seven times as much as the silver watch ; the cost of the chain and half the cost of the silver watch is equal to three-tenths of the cost of the gold watch ? If the chain cost $ 40, what was the cost of each watch ? 15. A and B make a purchase for $ 48. A gives all of his money, and B three-fourths of his. If A had given three- fourths of his money and B all of his, they would have paid $ 1.50 less. How much money had A and B ? 16. A mechanic and an apprentice together receive $ 40. The mechanic works 7 days and the apprentice 12 days ; and the mechanic earns in 3 days $ 7 more than the apprentice earns in 5 days. What wages does each receive ? 17. I have 7 silver balls equal in weight and 12 gold balls equal in weight. If I place 3 silver balls in one pan of a balance and 5 gold balls in the other, I must add to the gold balls 7 ounces to maintain equilibrium. If I place in one pan 4 silver balls and in the other 7 gold balls, the balance is in equilibrium. What is the weight of each gold and of each silver ball ? 18. A tank has two pumps. If the first be worked 2 hours and the second 3 hours, 1020 cubic feet of water will be dis- charged. But if the first be worked 1 hour and the second 2J hours, 690 cubic feet of water will be discharged. How many cubic feet of water can each pump discharge in one hour ? 198 ALGEBRA. [Cn. X 19. It was intended to distribute $ 25 among a certain . number of the poor, each adult to receive $ 2.50 and each child 75 cents. But it was found that there were 3 more adults and 5 more children than was at first supposed. Each adult was therefore given $ 1.75 and each child 50 cents. How many adults and how many children were there ? 20. A man ordered a wine merchant to fill two casks of different sizes with wine, one at $ 1.20 and the other at $ 1.50 a quart, paying $ 88.50 for both casks of wine. By mistake the casks were interchanged, so that the purchaser received more of the cheaper wine and less of the dearer. The mer- chant therefore returned to him $ 1.50. How many quarts did each cask hold? 21. A and B jointly contribute $ 10,000 to a business. A leaves his money in the business 1 year and 3 months, and B his money 2 years and 11 months. If their profits are equal, how much does each contribute ? 22. One boy said to another : " Give me 5 of your nuts, and I shall have three times as many as you will have left." " No," said the other, " give me 2 of your nuts, and I shall have five times as many as you will have left." How many nuts had each boy ? 23. A father has two sons, one 4 years older than the other. After 2 years the father's age will be twice the joint ages of his sons ; and 6 years ago his age was six times the joint ages of his sons. How old is the father and each of his sons ? 24. If a number of two digits be divided by the sum of the digits, the quotient will be 7. If the digits be interchanged, the resulting number will be less than the original number by 27. What is the number ? 25. A man walks 26 miles, first at the rate of 3 miles an hour, and later at the rate of 4 miles an hour. If he had walked 4 miles an hour when he walked 3, and 3 miles an hour when he walked 4, he would have gone 4 miies farther. How far would he have gone, if he had walked 4 miles an hour the whole time ? 18] SIMULTANEOUS LINEAR EQUATIONS. 199 26. Two trains leave different cities, which are 650 miles apart, and run toward each other. If they start at the same time, they will meet after 10 hours ; but if the first start 4J- hours earlier than the second, they will meet 8 hours after the second train starts. What is the speed of each train ? 27. If the base of a rectangle be increased by 2 feet, and the altitude be diminished by 3 feet, the area will be dimin- ished by 48 square feet. But if the base be increased by 3 feet, and the altitude be diminished by 2 feet, the area will be increased by 6 square feet. Find the base and the altitude of the rectangle. 28. A number of three digits is in value between 400 and 500, and the sum of its digits is 9. If the digits be reversed, the resulting number will be |^- of the original number. What is the number ? 29. The report of a cannon travels with the wind 344.42 yards a second, and against the wind 335.94 yards a second. What is the velocity of the report in still air, and what is the velocity of the wind ? 30. The sum of three digits of a number is 14 ; the sum of the first and the third digit is equal to the second; and if the digits in the units' and. in the tens' place be interchanged, the resulting number will be less than the original number by 18. What is the number ? 31. The sum of the ages of A, B, and C is 69 years. Two years ago B's age was equal to one-half of the sum of the ages of A and C, and 10 years hence the sum of the ages of B and C will exceed A's age by 31 years. What are the present ages of A, B, and C ? . 32. The total capacity of three casks is 1440 quarts. Two of them are full and one is empty. To fill the empty cask it takes all the contents of the first and one-fifth of the contents of the second, or the contents of the second and one-third of the contents of the first. What is the capacity of each cask ? 200 ALGEBRA. [Cn. X 33. Three brothers wished to buy a house worth $ 70,000, but none of them had enough money. If the oldest brother had given the second brother one-third of his money, or the youngest brother one-fourth of his money, each of the latter would then have had enough money to buy the house. But the oldest brother borrowed one-half of the money of the youngest and bought the house. How much money had each brother ? 34. A father's age is twenty-one times the difference between the ages of his two sons. Six years ago his age was six times the sum of his sons' ages, and two years hence it will be twice the sum of their ages. Find the ages of the father and his two sons. 35. Find the contents of three vessels from the following data : If the first be filled with water and the second be filled from it, the first will then contain two-thirds of its original contents ; if from the first, when full, the third be filled, the first will then contain five-ninths of its original contents ; finally, if from the first, when full, the second and third be filled, the first will then contain 8 gallons. 36. Two messengers, A and B, travel toward each other, starting from two cities which are 805 miles distant from each other. If A starts 5| hours earlier than B, they will meet 6J- hours after B starts. But if B starts 5| hours earlier than A, they will meet 5| hours after A starts. At what rates do A and B travel ? 37. Each of two servants was to receive $ 160, a dress, and a pair of shoes for one year's services. One servant left after 8 months, and received the dress and $ 106 ; the other servant left after 9 months, and received a pair of shoes and $ 142. What was the value of the dress, and of the pair of shoes ? 38. On the eve of a battle, one army had 5 men to every 6 men in the other. The first army lost 14,000 men, and the second lost 6000 men. The first army then had 2 men to every 3 men in the other. How many men were there originally in each army ? 18] SIMULTANEOUS LINEAR EQUATIONS. 201 39. If the sum of two numbers, each of three digits, be in- creased by 1, the result will be 1000. If the greater be placed on the left of the less, and a decimal point be placed between them, the resulting number will be six times the number ob- tained by placing the smaller number on. the left of the greater, with a decimal point between them. What are the numbers ? 40. Three cities A, B, and C, are situated at the vertices of a triangle. The distance from A to C by way of B is 82 miles, from B to A by way of C is 97 miles, and from C to B by way of A is 89 miles. How far are A, B, and C from one another ? 41. A regiment of 600 soldiers is quartered in a four-story building. On the first floor are twice as many men as are on the fourth ; on the second and third are as many men as are on the first and fourth ; and to every 7 men on the second there are 5 on the third. How many men are quartered on each floor ? 42. Four men are to do a piece of work. A and B can do the work in 10 days, A and C in 12 days, A and D in 20 days, and B, C, and D in 7-J- days. In how many days can each man do the work, and in how many days can they all together do the work ? 43. The year in which printing was invented is expressed by a figure of four digits, whose sum is 14. The tens' digit is one-half of the units' digit, and the hundreds' digit is equal to the sum of the thousands' and the tens' digit. If the digits be reversed, the resulting number will be equal to the original number increased by 4905. In what year was printing invented ? 44. A vessel sails 110 miles with the current and 70 miles against the current in 10 hours. On a second trip, it sails 88 miles with the current and 84 miles against the current in the same time. How many miles can the vessel sail in still water in one hour, and what is the speed of the current ? 202 ALGEBRA. [Cn. X 45. A and B run a race of 400 yards. In the first heat A gives B a start of 20 seconds, and wins by 50 yards. In the second heat A gives B a start of 125 yards, and wins by 5 seconds. What is the speed of each runner ? 46. A and B formed a partnership. A invested $ 20,000 of his own money and f 5000 which he borrowed ; B invested $ 22,000 of his own money and $ 8000 which he borrowed at the same rate of interest as was paid by A. At the end of a year, A's share in the profits amounted to $ 1750 more than the interest on his $ 5000, and B's share to $ 2000 more than the interest on his $ 8000. What rate per cent interest did they pay, and what rate per cent did they realize on their investments ? 47. Two bodies move along the circumference of a circle in the same direction from two different points, the shorter distance between which, measured along the circumference, is 160 feet. One body will overtake the other in 32 seconds, if they move in one direction ; or in 40 seconds, if they move in the opposite direction. While the one goes once around the circumference, the distance passed over by the other exceeds the circumference by 45 feet. What is the circum- ference of the circle, and at what rates do the bodies move ? 48. A number of workmen, who receive the same wages, earn together a certain sum. Had there been 7 more work- men, and had each one received 25 cents more, their joint earnings would have increased by $ 18.65. Had there been 4 fewer workmen, and had each one received 15 cents less, their joint earnings would have decreased by $ 9.20. How many workmen are there, and how much does each one receive ? 49. A farmer has enough feed for his oxen to last a certain number or days. If he were to sell 75 oxen, his feed would last 20 days longer. If, however, he were to buy 100 oxen, his feed would last 15 days less. How many oxen has he, and for how many days has he enough feed ? 18] SIMULTANEOUS LINEAR EQUATIONS. 203 50. An alloy of tin and lead, weighing 40 pounds, loses 4 pounds in weight when immersed in water. Find the amount of tin and lead in the alloy, if 10 pounds of tin lose If pounds when immersed in water, and 5 pounds of lead lose .375 of a pound. 51. Two men were to receive $ 96 for a certain piece of work, which they could do together in 30 days. After half of the work was done, one of them stopped for 8 days, and then the other stopped for 4 days. They finally completed the work in 35-J- days. How many dollars should each one receive, and in what time could each one have done the work alone ? 52. It took a certain number of workmen 6 hours to carry a pile of stones from one place to another. Had there been 2 more workmen, and had each one carried 4 pounds more at each trip, it would have taken them 1 hour less to complete the work. Had there been 3 fewer workmen, and had each one carried 5 pounds less at each trip, it would have taken them 2 hours longer to complete the work. How many work- men were there, and how many pounds did each one carry at every trip ? 53. Three carriages travel from A to B. The second carriage travels every 4 hours 1 mile less than the first, and is 4 hours longer in making the journey. The third carriage travels every 3 hours If miles more than the second, and is 7 hours less in making the journey. How far is B from A, and how many hours does it take each carriage to make the journey ? 54: A fox pursued by a dog is 60 of her own leaps ahead of the dog. The fox makes 9 leaps while the dog makes 6, but the dog goes as far in 3 leaps as the fox goes in 7. How many leaps does each make before the dog catches the fox ? CHAPTER XI. INEQUALITIES. 1. One number is greater than a second number when the remainder obtained by subtracting the second number from the first is positive. Thus, since 6 4, = 2, is positive, 6 > 4. One number is less than a second number when the remainder obtained by subtracting the second number from the first is negative. Thus, since 5 2, = 7, is negative, 5 < 2. In general, a > b, when a b is 2^ositive, and a < b, when a b is negative. 2. An Inequality is a statement that two numbers or expres- sions are unequal ; as ar + b 2 > a 2 . The members or sides of an inequality are the numbers or expressions which are connected by one of the signs of in- equality, > or <. 3. Two inequalities are of the Same or Opposite Species, or are said to subsist in the same or opposite sense, according as they have the same or opposite sign of inequality. E.g., 8 > 3 and 5 > 7 are inequalities of the same spe- cies ; > 1 and < 1 are inequalities of opposite species. Principles of Inequalities. 4. A relation of inequality between two numbers can be stated in two ways ; as 7 > 3, or 3 < 7. That is, if the members of an inequality be interchanged, the sign of inequality must be reversed. 204 1-8] INEQUALITIES. 205 5. If one number be greater than a second, and this second number be greater than a third, then the first number is greater than the third; that is, If a > b and b > c, then a > c. In like manner, if a < b and b < c, then a < c. E.g., 3>2, 2>1, and 3>1; -3<-2, -2<0, and -3<0. 6. An inequality will continue to be of the same species, (i.) When the same number is added to, or subtracted from, each member. (ii.) When each member is multiplied or divided by the same positive number. That is, if a > b, then a + /? > 6 + /i, a /? > 6 /? ; and an > bn, a -r- n > b -r- n ; wherein n is positive. E.g., 8 > 4, and 8 + 2 > 4 + 2, 8-2>4-2; and 8x2>4x2, 8 -=- 2 > 4 -=- 2. 7. An inequality will be reversed, (i.) When each member is subtracted from the same number. (ii.) When each member is multiplied or divided by the same negative number. That is, if a > b, then n a < n b, a (/?)< 6 (/?), < n n E.g., 8 > 4, and 5 - 8 < 5 - 4, or - 3 < 1 ; 8(-2)<4(-2),or-16<-8; and _-<-i-,or _4<-2. 8. There is often an advantage in using the same letter with some distinguishing marks to represent different numbers in the same discussion. Thus, with subscripts: a lt a z , a s , etc., read a sub-one, a sub- two, a sub-three, etc., or simply a one, a two, a three, etc. 206 ALGEBRA. [Cn. XI A subscript must not be confused with an exponent. Thus, a 3 stands for the product aaa; while a 3 is a notation for a single number. Two or More Inequalities. 9. If the corresponding members of two or more inequalities of the same species be added, the resulting inequality will be of the same species. That is, if aj > 6 1? a 2 > 6 2 , > then a x -f a 2 --- > b l + 2 ---. E.0., -5>-7, 3>2, and -o+3>-7 + 2; or, -2>-5. 10. If all the members of two or more inequalities of the same species be positive, and if the corresponding members be multiplied together, the resulting inequality will be of the same species. That is, if ^ > 6 1? a 2 > b^ a 3 > 6 3 , then a^a^ > MA? wherein a 1? b l} a 2) b 2 , a 3 , b B are all positive. E.g., 12 > 4, 3 > 2, and 12 x 3 > 4 x 2, or 36 > 8. 11. If the members of one inequality be subtracted from, or divided by, the corresponding members of another inequality of the same species, the resulting inequality will not necessarily be of the same species. That is, if ^ > ^ and a 2 > b.,, then Oi a 2 may or may not > b b 2 , and may or may not > -* a 2 2 E.g., 11 > 6, 4 > 3, and 11 - 4 > 6 - 3, * > f ; 5 > 4, 3 > 1, but 5 - 3 < 4 - 1, | < f ; 8 > 6, 4 > 2, while 8-4 = 6-2; 8 > 6, 4 > 3, while | = f . These examples show the truth of the principle enunciated. 12. Transformation of Inequalities. The preceding princi- ples enable us to make the following transformations of in- equalities : 8-13] INEQUALITIES. 207 (i.) Any term may be transferred from one member of an in- equality to the other, if its sign be reversed. E.g., if a b > c, then a > b + c. (ii.) If the signs of both members of an inequality be reversed from -\-to-, or from to +, the sign of inequality must be reversed. E.g., 3 < 5, and 3 > 5. 13. Ex. 1. Find one limit of the values of x, if x> 5 x 10. Transferring 5 a?, 4 x > 10. Dividing by 4, x < 2-J-. That is, the inequality is satisfied by all values of x less than 2. Ex. 2. Find the limits of the values of x, if x-5<4-2x, (1) and 5-2a7-4. (2) Transferring in (1), 3 x < 9, whence x < 3 ; Transferring in (2), 2 x > 2, whence x > 1. Therefore the values of x lie between 3 and 1. Ex. 3. What values of x and y satisfy the inequality 5x + 3y>tt, (1) and the equality 3 x -f- 5 / = 13 ? (2) Multiplying (1) by 3, 15 a? + 9 y > 33. (3) Multiplying (2) by 5, 15 x + 25 ?/ = 65. (4) Subtracting (4) from (3), - 16 y > - 32, or y < 2. Multiplying (1) by 5, 25 x + 15 y > 55. (5) Multiplying (2) by 3, 9 x + 15 ?/ = 39. (6) Subtracting (6) from (5), 16 x > 16, or x > 1. 208 ALGEBRA. [Cn. XI Pr. l. A man receives from an investment an integral num- ber of dollars a day. He calculates that if he were to receive $ 6 more a day his investment would yield over $ 270 a week ; but that, if he were to receive $ 14 less a day, his investment would not yield as much as $ 270 in two weeks. How much does he receive a day from his investment ? Let x stand for the number of dollars which he receives a day. Then, by the first condition, 7 (x _|_ 6) > 270 ; whence x > 32f And, by the second condition, 14 (a; - 14) < 270 ; whence x < 33f Therefore he receives $ 33 a day from his investment. EXERCISES I. Determine one limit of the value of x in each of the following inequalities : 1. x _ 8 > 4. 2. - 3 (x + 10) > - 20 3x 8 37 2 x Q 4 11 a x a x ~~ ~~~~ 5. 5 --_eL_0, ' ' Determine the limits of the values of x and y in each of the following systems: 9 (2x + 3y = -, 1Q J7a> + y = 15, [x-y> 2. \3x-2y>14:. 11. What integers have each the property that one-half of the integer, increased by 5, is greater than four-thirds of it, diminished by 3 ? 13-15] INEQUALITIES. 209 12. What integers have each the property that, if 9 be sub- tracted from three times the integer, the remainder will be less than twice the integer, increased by 12 ? 13. A has three times as much money as B. If B gives A $ 10, then A will have more than seven times as much as B will have left. What are the possible amounts of money which A and B have ? Identical Inequalities. 14. Many inequalities hold for all values of the literal num- bers involved ; as a 2 + b 2 > a 2 . Such inequalities are analogous to identical equations. 15. Prove that if a is not equal to 6, then a 2 + 6 2 > 2 ab. We have (a - 6) 2 > 0, (1) since the square of any positive or negative number is positive, and therefore greater than 0. From (1), a 2 -2ab + b 2 >0; whence a 2 + b 2 > 2 ab, by Art. 12 (i.). EXERCISES II. Prove the following inequalities, in which the literal num- bers are all positive and unequal : 1. a 2 -f b 2 + c 2 > ab + ac -f be. 2. a 2 b 2 + b 2 c 2 + a 2 c 2 > abc (a + b + c). 3. ab (a -f b) + be (b + c) -f ac (a + c) > 6 abc. 4. If I 2 + m 2 + n 2 = 1, and I? + m? + nf = 1, then Hi + wm! + nn < 1. 5. a 3 + b s > a 2 b + ab 2 . 6. a 4 + 6 4 > a 3 & + a& 3 . 7. (a + 6) (6 -f c) (c + a) > 8 a&c. 8. 3(a 2 -f b 2 + c 2 ) > (a + b + c) 2 . CHAPTER XII. INDETERMINATE LINEAR EQUATIONS. 1. It was shown in Ch. X., Art. 1, that the linear equation in two unknown numbers is satisfied by an indefinite number of sets of values of x and y. An Indeterminate Equation is an equation which, like the above, has an indefinite number of solutions. Evidently the number of solutions will be more limited if only positive integral values of the unknown numbers are admitted. In this chapter we shall consider a simple method of solving in positive integers linear indeterminate equations. 2. Ex. 1. Solve 4 x -\- 7 y = 94, in positive integers. Solving for x, which has the smaller coefficient, we obtain (i) < < or O V y Since x and y are to be integers, * must be an integer. That is, y must have such a value that 2 3 y shall be divisi- ble by 4. 2 3 v Let ^ = m, an integer. 4 Then y = ~~ m , an inconvenient form from which to de- termine integral values of y. But since the expression - is to be an integer, any multiple of it will be an integer. We therefore multiply its numerator by the least number which 210 1-4] INTERMEDIATE LINEAR EQUATIONS. 211 will make the coefficient of y one more than a multiple of the denominator, i.e., by 3. We then have *=y =l-2y+ 2^1, an integer. 2 v Therefore, - - = m. an integer. 4 Whence y = 2 - 4 m. (2) Then, from (1) and (2), x = 20 + 7 m. (3) Any integral value of m will give to x and y integral values. But since y is to be positive, m < 1 ; and, since x is to be positive, m > 3. Therefore the only admissible values of m are 0, 1, 2. Whenw = 0, a = 20, y= 2; m = 1, a? = 13, ?/ = 6 ; m = 2, 0;= 6, y = 10. 3. An Indeterminate System is a system of equations which has an indefinite number of solutions. Thus, if the system x -f y z = 9, be solved for x and y, we obtain x = U-2z,y = 3z-5. In these values of x and y we may assign any value to z and obtain corresponding values of x and y. 4. In solving a system of two linear equations in three unknown numbers, we first eliminate one of the unknown num- bers, and apply to the resulting equation the preceding method, Pr. A party of 20 people, consisting of men, womfcn, and children, pay a hotel bill of $ 67. Each man pays $ 5, each woman $ 4, and each child $ 1.50. How many of the company are men, how many women, and how many children ? Let x stand for the number of men, y for the number of women, z for the number of children. 212 ALGEBRA. [Cn. XII Then, by the conditions of the problem, aj + y + = 20, (1) 6a? + 4y + f3 = 67. . (2) Eliminating z, 7 # + 5 y = 74. Solving this equation, we obtain x = 2 5 m, y = 12 + 7 w, 2 = 6 2 m. When m = 0, x = 2, y = 12, z = 6; m = 1, x = 7, y= 5, 2 = 8. EXERCISES. Solve in positive integers : 1. 5x+8?/=29. 2. 3a+5y=10. 3. 12a+13y=175. 4. 25 0^+15 2/=215. 5. 5o,-+132/=229. 6. 34 a; + 89^=407. 7z = 68. 8> \lx-2y- z = 8. Solve in least positive integers : 9. 89a; 144y=l. 10. 14 a; 49^=133. 11. 67x-43y=5. 12. Divide 1000 into two parts so that one part shall be a multiple of 13, and the other a multiple of 53. 13. What positive integers when divided by 4 give a re- mainder 3, and when divided by 5 give a remainder 4 ? 14. A farmer received $ 16 for a number of turkeys and chickens. If he was paid $2 for each turkey and $.75 for each chicken, how many of each did he sell ? 15. A gardener has fewer than 1000 trees. If he plants them in rows of 37 each, he will have 8 left ; but if he plants them ift a different number of rows of 43 each, he will have 11 left. How many trees has he ? 16. A said to B : " If I had eight times as much money as I now have, and you had seven times as much money as you now have, and I were to give you $ 1, we should have equal amounts." How many dollars had each ? CHAPTER XIII. INVOLUTION 1. Involution is the process of raising a number to any required power. Powers of Powers. 2. Ex. 1. (a 4 ) 5 = a 4 tt 4 a 4 a 4 a 4 = a 4+4+4+4+4 = a 4x5 = a 20 . Ex. 2. (a 9 ) 10 = W - - to 10 factors __ ,09+9+9+ -to lOsummands _ ^,9x10 __ ^,90 These examples illustrate the following method of finding any required power of a given power : Multiply the exponent of the given power by the exponent of the required power ; or, stated symbolically, (fl m ) w = a mn . Eor, (a m ) M = a m a m a m - - to n factors n m-\-m+m+ to n summands fl mn U/ til a Powers of Products. 3. Ex. 1. (aby = (a&) (a6) (a6) (a6) = (aaaa) (bbbb) = a 4 b 4 . Ex. 2 (xy) w = (a^) (xy) (xy) - - to 10 factors = (xxx ... to 10 factors) (yyy to 10 factors) = x w y. These examples illustrate the following method of finding any required power of a product : Take the product of the factors, each raised to the required power ; or, stated symbolically, (ab)" = a n b n ; (abc)" = a"b"c" ; etc. 213 214 ALGEBRA. [Cn. XIII For, (db) = (ab) (db) (ab) - - to n factors = (aaa to n factors) (bbb -"ton factors) = a n b n . In like manner, (abc) n = a n b n c n ; and so on. 4. The converse of the principle of Art. 3 is evidently true. That is, a m b m = (ab) m ; a m b m c m = (abc) n ; etc. 5. The principles of Arts. 2-3 prove the method, already given in Ch. V., Art. 5, of raising a monomial to any required power. Raise the numerical coefficient to the required power, and multiply the exponent of each literal factor by the exponent of the required power. Ex. 1. (4 a s b) 2 = 4 2 a s * 2 b 2 = 16 6 6 2 . Ex. 2. ( - 3 aV) 3 = ( - 3) 3 a 4x V x3 = - 27 a% 6 . Powers of Fractions. F *\*_2x 2 2x> (2^) 2 4z 4 -~ x B ~ SS " Ex. 2. -= - x - x - .- to 9 factors \bj bbb _ aaa to 9 factors _ a 9 ~bbb to9factors~6 9 ' These examples illustrate the following method of raising any fraction to a required power : liaise each term of the fraction to the required power; or, stated symbolically, For, - = - x - x - to w factors \bj bbb _ aaa to n factors _ a n ~~ bbb to n factors b n 3-7] INVOLUTION. 215 EXERCISES I. Write the cubes and the fourth powers of : 1. x 2 . 2. x 4 . 3. 2x 7 . 4. Sab. 5. 5ab 2 . 6. 4ary. 7 - Zmfaf. 8. 5 aW. * -r --i --is- Write the squares, the cubes, and the ?ith powers of : 13. a m+ \ 14. x m ~ 2 . 15. 2x m+n y. 16. 3a m+n ~ l if'. Find the values of each of the following powers : 17. (-3ojV) 8 . 18. (5a a 6 8 c) 2 . 19. (-4xyV) 3 . 20. (2xf#)\ 21. (cPxf)*. 22. (-2m 2 n 3 ) 5 . /3a 2 6V 24 25 _ ' ' ' ' ' Powers of Binomials. 7. By actual multiplication, we obtain (a + 6) 3 = (a 2 + 2a6 + 6 2 ) (a + b) = a 3 + 3a 2 6 + 3a6 2 + A 3 , ( a _ 6) 3 = (a 2 - 2ab + 6 2 ) (a - 6) = a 3 - 3a 2 6 + 3a6 2 - b 3 , (a + 6) 4 = (a 2 + 2a6 + 6 2 ) (fl 2 + 2a6 + b 2 ) = a 4 + 4fl 3 6 + 6a 2 6 2 + 4a6 3 + 6 4 , (a - 6) 4 = a 4 - 4a 3 6 + 6a 2 6 2 - 4a6 3 + 6 4 . The result of performing the indicated operation in a power of a binomial is called the Expansion of that power of the binomial. In the preceding expansions the following laws are evident : (i.) The number of terms exceeds the binomial exponent by 1. (ii.) The exponent of a in the first term is equal to the binomial exponent, and decreases by 1 from term to term. (iii.) TJie exponent of b in the second term is 1 and increases by 1 from term to term, and in the last term is equal to the binomial exponent. 216 ALGEBRA. [Cn. XIII (iv.) The coefficient of the first term is 1, and that of the second term, except for sign, is equal to the binomial exponent. (v.) The coefficient of any term after the second is obtained, except for sign, by multiplying the coefficient of the preceding term by the exponent of a in that term, and dividing the product by a number greater by 1 than the exponent of b in that term. E.g., the coefficient of the fourth term in the expansion of (a + b) 4 is 6 x 2 -f- 3, = 4. (vi.) TJie signs of the terms are all positive when the terms of the binomial are both positive ; the signs of the terms alternate, -f and , ivhen one of the terms of the binomial is negative. Observe, as a check : (vii.) The sum of the exponents of a and b in any term is equal to the binomial exponent. (viii.) The coefficients of tivo terms equally distant from the beginning and the end of the expansion are equal. In a subsequent chapter the above laws will be proved to hold for any positive integral power of the binomial. 8. Ex. 1. (2 a - 3 b) 4 = (2 a) 4 -4(2 a) 3 (3 b) + 6(2 a) 2 (3 b) 2 _4(2a)(36) 3 + (36) 4 = 16 a 4 - 96 a*b + 216 a 2 6 2 - 216 ab 3 + 81 V. Ex. 2. EXERCISES II. Raise each of the following expressions to the required power : 1. (a + 1) 3 . 2. (a-3) 3 . 3. (2a + 3) 3 . 4. (5 -2 y) s . 5. (2ab + 3) s . 6. (5x-6y) 3 . 7. (x 2 -8) 3 . 8. (5^-3i/) 3 . 9. (6^ 10. (a -I) 4 . 11. (2^ + 3) 4 . 12. (3Z-2 13. (a + b) 5 . 14. (2 m 3 ?i)^. 15. (x yf. 7-9] INVOLUTION. 217 Powers of Multinomials. 9. We have (a + b + C ) 2 = [(a + 6) + c] 2 = (a + b) 2 + 2 (a + 6)c + c 2 = .a 2 + 2 a& + b 2 + 2 ac -f 2 6c + c 2 . Therefore (a -f 6 + c) 2 = a 2 + 6 2 + c 2 + 2 a6 + 2 ac + 2 6c. In like manner, (a + b - c) 2 = a 2 + 6 2 + c 2 + 2 a6 - 2 ac - 2 6c. ( a _ b - c) 2 = a 2 + 6 2 + c 2 - 2 a6 - 2 ac + 2 fc. By repeated application of this principle we can obtain the square of a multinomial of any number of terms. We have ( a + 6 + c + d) 2 = [(a + b + c) 2 + d] 2 = a 2 4. 52 + C 2 + 2 06 + 2 ac + 2 6c + 2 (a b 2 + c 2 + d* That is, the square of a multinomial is equal to the sum of the squares of the terms, plus the algebraic sum of twice the product of each term by each term which follows it. Ex.1. (3x+5y-7zy=(3x)*+(5yy+(-7z)*+2(3x)(oy) EXERCISES III. Raise each of the following expressions to the required power : 1. (a + 6 + 1) 2 . 2. (x-y-T)' 2 . 3. (2a + 3& + l) 2 . 4. (3a-4& + 5c) 2 . 5. (a 2 + a + l)*. 6. 2 -a + l) 2 . 7. (a? + xy + y*)*. 8. (a 2 - 3 a6 + & 2 ) 2 . 9. (a + 6 + c) 3 . 10. (a-6-c) 3 . 11. (a 2 _a + l)3. 12. (2a-& + 5) 3 . 13. (a + b + c + d) 2 . 14. (a-6-c + d) 2 . 15. (a 3 - a 2 + a - I) 2 . 16. (aj 8 + 2x 2 - 3^ 4- 4) 2 . CHAPTER XIV. EVOLUTION. 1. A Root of a number is one of the equal factors of the number. E.g., 2 is a root of 4, of 8, of 16, etc. 2. A Second, or Square Root of a number is one of two equal factors of the number. E.g., since 5 x 5 = 25 and ( 5) ( 5) = 25, therefore + 5 and 5 are square roots of 25. A Third, or Cube Root of a number is one of three equal factors of the number. E.g., since 3 x 3 x 3 = 27, therefore 3 is a cube root of 27 ; since ( 3) ( 3) ( 3) = 27, therefore 3 is a cube root of - 27. In general, the qth root of a number is one of q equal factors of the number. E.g., a qth root of x q is x. 3. The Radical Sign, -y/, is used to denote a root, and is placed before the number whose root is to be found. The Radicand is the number whose root is required. The Index of a root is the number which indicates what root is to be found, and is written over the radical sign. The index 2 is usually omitted. E.g., ^/9, or -y/9, denotes a second, or square root of 9 ; the radicand is 9, and the index is 2. 4. A vinculum is often used in connection with the radical sign to indicate what part of an expression following the sign is affected by it. 218 1-10] EVOLUTION. 219 E.g., -y/9 4- 16 means the sum of ^/9 and 16, while V9 + 16 means a square root of the sum 9 + 16. Likewise -f/a 3 x b c> means the product of ^/a 3 and 6 6 , while -\/a 3 x 6 6 means a cube root of a?b 6 . Parentheses may be used instead of the vinculum in connec- tion with the radical sign ; as y (9 4- 16) for V9 + 16. 5. It follows from the definition of a root that the square of a square root of a number is the number, the cube of a cube root of a number is the number, and so on. E.g., (V4) 2 = 4; (- ^X> Number of Roots. 8. Since ( 4) 2 = 16, therefore V 16 = 4 ; since ( a) 4 = a 4 , therefore -^/a 4 = a. These examples illustrate the principle : A positive number has at least two even roots, equal and oppo- site; i.e., one positive and one negative. 9. Since (- 3) 3 = - 27, therefore J/ 27 = - 3 ; since 2 5 = 32, therefore ^/32 = 2. These examples illustrate the principle : A positive or a negative number has at least one odd root of the same sign as the number itself. 10. Since (+ 4) 2 = -f 16 and ( 4) 2 = + 16, there is no num- ber, with which we are as yet familiar, whose square is 16. 220 ALGEBRA. [Cn. XIV Consequently -^16 cannot be expressed as a positive or as a negative number; that is, in terms of the numbers as yet used in this book. Such roots are called Imaginary Numbers, and will be con- sidered in Ch. XVI. Evolution. 11. Evolution is the process of finding a root of a given number. 12. In the following articles the radicands are limited to positive values, and the roots to positive roots. 13. (i.) Since (a 2 ) 3 = a 6 , therefore -fya 6 = a 2 = a*. This example illustrates the principle : The root of a power is obtained by dividing the exponent of the power by the index of the root. E.g., -*/a* = a; -^/a 15 = eft = a 3 . 5? In general, Qja nc i = a 1 = a". nq For, since (a n ) q = a nq , therefore J/a nq = a n = a' 1 . (ii.) Since (a&) 2 = a 2 b 2 , therefore V( a *^) = ab = V a ' X V &2 - This example illustrates the principle : The root of a product of two or more factors is equal to the product of the like roots of the factors, and conversely. E.g., V( 16 X 25 ) = V 16 X V 25 = 4 X 5 = 20 ; ,*/(8 a 3 6 6 ) = -#8 X s!/a 3 x J/b 6 = 2 x a x b 2 = 2 ab\ In general, fy(aib For, since | = |, therefore $ = - Roots of Monomials. 14. The positive root of a positive number can be found by applying the principles of Art. 13. The negative even root of a positive number is found by pre- fixing the negative sign to its positive root. Since ^/-8 = -2, and _^/8 = -2, therefore ^-8 = --^/8. That is, the negative odd root of a negative number is found by prefixing the negative sign to the positive root of the radi- cand taken positively. Ex. 1. V( 16 a2 b 4 ) = V 16 X V^ 2 X = 4 ab 2 , the positive square root. Therefore V(16 a 2 6 4 ) = 4 ab 2 . In the following examples we shall give only the positive even roots. Ex. 2. j/(- 27 ajtyk 8 ) = ^/- 27 x ^ x -tyjf x These examples illustrate the following method : Tafce f/ie required root of the numerical coefficient, and divide the exponent of each literal factor by the index of the required root. Ex 3 4 /16 a 8 fr 12 /16 a 8 6 12 /16 a^ J ^ 2 a 2 222 . ALGEBRA. [Cn. XIV 15. It is frequently of advantage to separate a number ex- pressed in figures into its prime factors before taking the root. Ex. 4. V( 15 X 40 x 216) = V(o 3 x 2 3 5 x 2 3 3 3 ) = V(o 2 - 3 4 2 6 ) = 5 3 2 2 3 = 360. EXERCISES I. Simplify the following expressions : 3. 6. 9. 12. \ c w d 2n ' \ a 8 6 16 Find the values of each of the following expressions : 25. V 643 - 26 - V 49 "- 27 - A/ 216 '- 28 - S/-27 4 . 29. V( 40 x 15 x 6). 30. V( 56 x 40 x 35). 31. v 1024 - 32 - V 2025 - 33 - V 12544 - 34. ^/(6 x 20 x 225). 35. ^/(84 x 18 x 49). 36. V( 45 x y X 35 o?2 x 63 2/2). 37. -3/(36 a 2 6c x 75 a6V x SQUARE ROOTS OF MULTINOMIALS 16. The square root of a trinomial which is the square of a binomial can be found by inspection (Ch. VI., Art. 9). 17. Since (a + b) 2 = a 2 + 2 ab + & 2 , we have V( a2 + 2 a6 + 6 2 ) = a + &. 15-18] SQUARE ROOTS OF MULTINOMIALS. 223 From this identity we infer : (i.) The first term of the root is the square root of the first term of the trinomial; i.e., a= A /a 2 . (ii.) If the square of the first term of the root be subtracted from the trinomial, the remainder will be Twice the first term of the root, 2 a, is called the Trial Divisor. (iii.) The second term of the root is obtained by dividing the 'first term of the remainder by the trial divisor ; i.e., b = ~ ^ < ' The trial divisor plus the second term of the root is called the Complete Divisor. (iv.) If the product of the complete divisor by the second term of the root be subtracted from the first remainder) the second remainder will be 0. The work may be arranged as follows : 2 ab + b 2 2ab 2ab a + b 2 a trial divisor - 2 a = 6, second term of root 2 a + b complete divisor = (2a + 6)6 18. Ex. 1. Find the square root of 4 a 4 - 12 x*y + 9 f. The work, arranged as above, writing only the trial and the complete divisor, is : 4 x 4 - 12 x 2 y - -12x*y - 12 o?y + 9 if 4 a/* 2 The square root of 4 a? 4 is 2 a? 2 , the first term of the root. The trial divisor is 2 (2 a? 2 ), = 4 x~. The second term, of the root is '- -, = 3y. The complete divisor is 4 x~ 3 y. ~r Ou 224 ALGEBRA. [Cn. XIV Ex. 2. Find the square root of 4 x 4 - 12 x 3 + 29 x 2 - 30 x + 25. The work follows : - 29 x 2 -30 a + 25 2x 2 -3x + 5 -12x* 9x 2 20 x 2 20 x 2 - 30x4- 25 Only the trial divisor and the complete divisor of each stage are written, the other steps being performed mentally. The square root of 4 a? 4 is 2 x 2 , the first term of the root. The trial divisor is 2 (2 x 2 ), = 4 x 2 . The second term of the root is 1 x 8 ^ , = 3 x. The complete divisor is 4 or 2 3 #, which is 4 X multiplied by the second term of the root, giving 12 X s + 9 x 2 . The first term of the second remainder is 20 x 2 . 20 x 2 The third term of the root is . = 5. 4 a 2 To form the complete divisor at this stage, we multiply the part of the root previously found, 2 x 2 3 x, by 2, and to the product add the term just found. We thus obtain 4 x 2 6 cc+5. This complete divisor we multiply by the last term of the root. In the preceding examples the terms were arranged to de- scending powers of x. They could equally well have been arranged to ascending powers. 19. The preceding method can be extended to find square roots which are multinomials of any number of terms. The work consists of repetitions of the following steps : After one or more terms of the root have been found, obtain each succeeding term, by dividing the first term of the remainder at that stage by twice the first term of the root. 18-19] SQUARE ROOTS OF MULTINOMIALS. 225 Find the next remainder by subtracting from the last remainder the expression (2 a + b) b, wherein a stands for the part of the root already found, and b for the term last found. EXERCISES II. Find the square root of each of the following expressions : 1. # 4 -4ar 3 4-8a; + 4. 2. 4w 4 -4m 3 +5m 2 -2m-f 1. 3. x*-2x 3 + 3x 2 -2x + l. 4. 5. 9x 4 +12or } -26z 2 -20a;+25. 6. 4x 4 - 7. x 4 y 4 4 x s y 3 + 6 x 2 y 2 4 xy -f- 1. 8. x 4 + fafy + 2a?y- 12 0^ + 90*. 9. a; 4 - 6 ax 3 + 13 oV - 12 a*x + 4 a 4 . 10. 4 a 2 + 9 6 2 + 16 c 2 - 12 a& + 16 ac - 24 be. 11. 49x 8 H-42iK 6 -19a; 4 -12 2 + 4. 12. 25 a* 4 - 30 ax 3 + 49 aV - 24 a ;3 aj + 16 a 4 . 13. a 4 2 14. 9 a 4 + 30 o 8 6 + 49 a?W + 40 a6 3 + 16 b\ 15. 89 a 2 6 2 - 70 ab 3 + 16 a 4 - 56 a 3 6 + 25 b 4 . 16. 4 a 6 - 12 a 4 6 - 28 a 3 6 3 + 9 a 2 6 2 + 42 ab 4 + 49 6 6 . 17. ^ 4 _i^ 2/ 4 2/ O ^ 18. ^ + ^ + a or 19. l + 2^-x 2 20. a 6 - 6 ax 5 + 15 a 2 x 4 - 20 aV + 15 a 4 ic 2 - 6 a d x + 21. 1 _ 4a + 64 a 6 - 64 a 5 - 32 a 3 + 48 a 4 + 12 a 2 . 22. 4 a 6 + 17 a 2 - 22 a 3 + 13 a 4 - 24 a - 4 a 5 + 16. 6 42 2 - 23. 9x 6 + 6 arty + 43 x 4 y 2 + 2&y + 24. - 2 x 4 or or 226 ALGEBRA. [Cn. XIV CUBE ROOTS OF MULTINOMIALS. 20. The process of finding the cube root of a multinomial is the inverse of the process of cubing the multinomial. Since (a + b) s = a 3 + 3 a 2 b + 3 ab 2 + W = a 5 + (3 a 2 + 3 ab + 6 s ) 6, (1) we have -^/(a 3 + 3 a 2 6 + 3 a& 2 + 6 3 ) = a + 6. (2) From the identity (2), we infer : (i.) The first term of the root is the cube root of the first term of the multinomial ; i.e., a=^/a 3 . (ii.) If the cube of the first term of the root be subtracted from the multinomial, the remainder will be 3a 2 b + 3ab 2 + b 3 , = (3a 2 + Sab + b-)b. Three times the square of the first term of the root, 3 a 2 , is called the Trial Divisor. (iii.) The second term of the root is obtained by dividing the first term of the remainder by the trial divisor ; i.e., b = - - 3 CL" The sum 3 a 2 + 3 ab -f b 2 is called the Complete Divisor. (iv.) If the product of the complete divisor by the second term of the root be subtracted from the first remainder, the second remainder will be 0. The work may be arranged as follows : 0*6 + 3 O 3 3a 2 b 3a 2 b+3ab 2 +b 3 a+b 3 a 2 trial divisor (1) 3 a 2 6^-3 a 2 =&, second term of root (2) 3 a 2 + 3 ab + 6 2 , complete divisor (3) b (4) 21. Ex.1. Find the cube root of 27 a^+54 x*y+ 36 xy 2 f 8/. The work, arranged as above, is : 27 x 3 + 54 ic 2 ?/ + 36 xy 2 + 8y* 27 y? 54 x*y 20-23] CUBE ROOTS OF MULTINOMIALS. 227 The cube root of 27 x? is 3 x, the first term of the root. The trial divisor is 3 (3 a-) 2 = 27 a 2 . *- I 9 The second term of the root is ^ x ^, = 2 ?/. The complete divisor is 2 + 3(3o?)(2 y) + (2y) 2 , = 27 x? + 18 a# + 4/, which is multiplied by the second term of the root, giving 22. The preceding method can be extended to find cube roots which are multinomials of any number of terms, as the method of finding square roots was extended. The work con- sists of repetitions of the following steps : After one or more terms of the root have been found, obtain each succeeding term by dividing the first term of the remainder at. that stage by three times the square of the first term of the root. Find the next remainder by subtracting Irom the last remainder the expression (3a 2 + 3a& + b z )b, wherein a stands for the part of the root already found, and b for the term last Jound. 23. The given multinomial should be arranged to powers of a letter of arrangement. Ex. 27-27z+90z 2 -55 27 -27 x -27 x+ 9x*- 81 z 2 - 3(3)2+3(3) (- EXERCISES III. Find the cube root of each of the following expressions : 2. 1 6 a; + 12 a 2 So 8 . 3. 64 a 3 + 240 a~b + 300 ab 2 + 125 b 3 . 4. x 6 - 6 x 5 + 15 x 4 - 20 x* + 15 y? - 6 x + 1. 228 ALGEBRA. [Cn. XIV 5. 8 x 6 - 36 .x 5 + 66 x 4 - 63 or 3 + 33 x 2 - 9 x + 1. 6. 156 a 4 - 144 a 5 - 99 a 3 + 64 a 6 + 39 a 2 - 9 a + 1. 7. l + 3x 8. l-6x 9. 8-12 8 x 3 10. 27 aV 5 + 54 aV + 9 aV - 28 aV _ 3 aV + 6 ax - 1. 11. 8 a 6 + 48 a 5 b + 60 a 4 6 2 - 80 a 3 6 3 - 90 a 2 6 4 + 108 6 5 - 27 6 6 . 12. x 3 + 3 x 1 - 9 a 11 - 27 x 15 - 6 tf - 54 x 13 + 28 x 9 . 13. 108 a 5 - 48 a 4 + 8 a 3 + 54 a 7 - 12 a 8 + a 9 - 112 a 6 . 14. 8 a 6 - 48 a 5 x + 60 aV - 27 x 6 - 108 ax 5 - 90 aV + 80 a 3 ar. 15. i + 3 x _ 8 x 3 - 6 x 4 + 6 or 5 + 8 x 6 - 3 x 8 - x 9 . 125 / 150 .y 5 165 y 4 172 y 3 99 y 2 54 y 27 x 6 x 5 x 4 x s x 2 ' x ROOTS OF ARITHMETICAL NUMBERS. Square Roots. 24. Since the squares of the numbers 1, 2, 3, , 9, 10, are 1, 4, 9, -, 81, 100, respectively, the square root of an integer of one or two digits is a number of one digit. Since the squares of the numbers 10, 11, , 100, are 100, 121, , 10000, the square root of an integer of three or four digits is a number of tivo digits ; and so on. Therefore, to find the number of digits in the square root of a given integer, we first mark off the digits from right to left hi groups of two. The number of digits in the square root will be equal to the number of groups, counting any one digit remaining on the left as a group. 25. The method of finding square roots of numbers is then derived from the identity (a + &) 2 = a 2 +(2 + &)6, (1) wherein a denotes tens and b denotes units, if the square root is a number of two digits. 23-26] ROOTS OF ARITHMETICAL NUMBERS. 229 26. Ex. 1. Find the square root of 1296. We see that the root is a number of two digits, since the given number divides into two groups. The digit in the tens* place is 3, the square root of 9, the square next less than 12. Therefore, in the identity (1), a denotes 3 tens, or 30. The work then proceeds as follows : 12' 96 9 00 3 96 3 96 30 + 6 = 36 2 a = 60, trial divisor (2ab + 6 2 ) -r- 2a = 396 -=- 60 = 6 + = (2 a + b) x 6 = (60 + 6) x 6 (1) (2) (3) The first remainder, 396, is equal to 2 ab -f b 2 , and cannot be separated into the sum of two terms, one of which is 2 ab. We cannot, therefore, determine b by dividing 2 ab by 2 a, as in finding square roots of algebraic expressions. Consequently step (2) suggests the value of 6, but does not definitely deter- mine it. As a rule, we take the integral part of the quotient, 6 in the above example, and test that value by step (3). This method may be extended to find roots which contain any number of digits. At any stage of the work a stands for the part of the root already found, and b for the digit to be- found. Ex. 2. Find the square root of 51529. The root is a number of three digits, since the given number divides into three groups. The digit in the hundreds' place is 2 y the square root of 4, the square next less than 5, Therefore in the identity (1), a denotes 2 hundreds, or 200, in the first stage- of the work. The work then proceeds as follows : 5' 15' 29 4 00 00 200 + 20 + 7 = 227 (1) (2) (3) (4) 2 a = 400, trial divisor (2 ab + fc 2 ) -*- 2 a = 11529 -- 400 = 20 + = (2 a + 6) 6 = (400 + 20) x 20 1 15 29 84 00 31 29 31 29 (2 a& + & 2 ) -- 2 a = 3129 -s- 440 = 7 + = (2 a + 6) 5 = (440 + 7) x 7 5' 15' 29 4 227 L = 2 11 + 4 42 1 15 84 31 29 31 29 312- 447 44 = 230 ALGEBRA. [Cn. XIV In the second stage of the work, a stands for the part of the root already found, 220, and b for the next figure of the root. In practice the work may be arranged more compactly, omitting unnecessary ciphers, and in each remainder writing only the next group of figures. Thus : (2) (4) Observe that the trial divisor at any stage is twice the part of the root already found, as in (2) and (4). 27. The abbreviated work in the last example illustrates the following method : After one or more figures of the root have been found, obtain the next figure of the root by dividing the remainder at that stage (omitting the last figure), by the trial divisor at that stage. See lines (2) and (4). Annex this quotient to the part of the root already found, and also to the trial divisor to form the complete divisor. Find the next remainder by subtracting from the last remainder the product of the complete divisor and the figure of the root last found. 28. Since the number of decimal places in the square of a decimal fraction is twice the number of decimal places in the fraction, the number of decimal places in the square root of a decimal fraction is one-half the number of decimal places in the fraction. Consequently, in finding the square root of a decimal frac- tion, the decimal places are divided into groups of two from the decimal point to the right, and the integral places from the decimal point to the left as before. 26-30] ROOTS OF ARITHMETICAL NUMBERS. 231 Ex. 14' 46.28' 09 9 38.03 68 5 46 5 44 2.28 09 2.28 09 76.03 In finding the second figure of the root, we have - 5 ^ = 9 ; but 69 x 9 = 621, which is greater than 546, from which it is to be subtracted. Hence we take the next less figure 8. EXERCISES IV. Find the square root of each of the following numbers : 1. 196. 2. 841. 3. 1296. 4. 65.61. 5. 7396. 6. 3481. 7. 667489. 8. 170569. 9. 1664.64. 10. 582169. 11. 1.737124. 12. 556.0164. 13. .00099225. Cube Roots. 29. Since the cubes of the numbers 1, 2, 3, , 9, 10, are 1, 8, 27, , 729, 1000, respectively, the cube root of any integer of one, two, or three digits is a number of one digit. The cube roots of such numbers can be found only by inspection. Since the cubes of 10, 11, , 100 are 1000, 1331, , 1000000, respectively, the cube root of any integer of four, Jive, or six digits is a number of two digits, and so on. Therefore, to find the number of digits in the cube root of a given integer, we first mark off the digits from right to left in groups of three. The number of digits in the cube root will be equal to the number of groups, counting one or two digits remaining on the left as a group. 30. The method of finding cube roots of numbers is derived from the identity (a + b) 3 = a 3 + (3 a 2 + 3 ab + b 2 ) b, (1) 232 ALGEBRA. [On. XIV wherein a denotes tens, and b denotes units, if the cube root is a number of two digits. Ex. Find the cube root of 59319. The digits in the tens 1 place of the root is 3, the cube root of 27, the cube next less than 59. Therefore in identity (1), a denotes 3 tens, or 30. The work may be arranged as follows : 59'319 27000 a + b 30 + 9 (1) (2) (3) 32 319 32319 3a 2 = 3(30) 2 = 2700 (3 a 2 b + 3 ab 2 + b 3 ) - 3 a 2 = 32319 + 2700 = 9 + 3a 2 = 3(30) 2 =2700 b- = 9 2 = 81 = (3 a 2 + 3 ab + b 2 ) xb= 3591 x 9 As in finding square roots of numbers, step (2) suggests the yalue of b, but does not definitely determine it. If the value of b makes (3 a 2 + 3 ab + b 2 ) x b greater than the number from which it is to be subtracted, we must try the next less number. In practice the work may be arranged more compactly, omitting unnecessary ciphers, and in each remainder writing only the next group of figures ; thus (1) (2) (3) 31. The preceding method may be extended to find roots that contain any number of digits. At any stage of the work a stands for the part of the root already found, and b for the digit to be found. The method consists of repetitions of the f ollowing^ steps : The trial divisor at any stage is three times the square of the part of the root already found ; as 27 in the preceding example. 59' 319 27 39 2700 810 81 32 319 32 319 3591 30-33] HIGHER ROOTS. 233 After one or more figures of the root have been found, obtain the next figure of the root by dividing the remainder at that stage {omitting the last two figures) by the trial divisor. In the last example, 9 + = 323 -r- 27. Annex this quotient to the part of the root already found. To obtain the complete divisor, add to the trial divisor (with two ciphers annexed) three times the product of the part of the root already found (with one cipher annexed) by the figure of the root just found, and also the square of the figure of the root just found. Find the next remainder by subtracting from the last remainder the product of the complete divisor and the figure of the root last found. 32. Evidently, in finding the cube root of a decimal fraction the decimal places are divided into groups of three figures from the decimal point to the right, and the integral places from the decimal point to the left as before. EXERCISES v. Find the cube root of each of the following numbers : 1. 2744. 2. 39304. 3. 110.592. 4. 328509. 5. 1.191016. 6. 74088000. 7. 340068392. 8. 426.957777. 9. 584067.412279. 10. 375601280.458951. 11. .041063625. HIGHER ROOTS. 33. Since -yVa 4 = a, and therefore, -yVa 4 = - Since -/a G = a, and -fy^/ a* = $/<$ = a, therefore, V a6 In general, since tya M = a, and therefore, **/a? q 234 ALGEBRA. [Cn. XIV That is, the pqth root of a number is the pth root of the qth root of the number. In particular, the fourth root is the square root of the square root, the sixth root is the cube root of the square root. EXERCISES VI. Find the fourth root of each of the following expressions : 2. + 16 a s b 5 + 10 a 2 b 6 + 4 ab 7 + b 8 . 3. 16 a 8 - 160 a 7 + 408 x 6 + 440 or 5 - 2111 x 4 - 1320 or 3 + 3672 x 2 + 4320 x + 1296. 4. 625 x 8 + 5500 x 7 + 17150 x + 20020 aj 5 + 721 x 4 - 8008 x 3 + 2744 x 2 - 352 x + 16. Find the sixth roots of each of the following expressions : 5. 64 x 12 - 192 x 10 + 240 x 8 - 160 x 6 + 60 x 4 - 12 x 2 + 1. 6. a 12 +6 o u 6+21 a 10 6 2 +50 a 9 +126 a*6 7 +90 a 4 6 8 +50 o 8 Find the value of each of the following indicated roots : 7. ^7279841. 8. ^3010936384. 9. ,J/164204746.7776. CHAPTER XV. SURDS. 1. In Ch. XIV. we considered only roots of powers whose exponents were multiples of the indices of the required roots. Such roots as -y/2, -y/a 2 , etc., were excluded. 2. It is proved in School Algebra, Ch. XVIII., that -y/ 2 , -^/a 2 , etc., cannot be expressed either as integers or as fractions. Thus, there is no integer or fraction whose square is 2. But it is there proved that the value of such a root can be found approximately to any degree of accuracy. E.g., approximate values of ^/2 are 1.4, 1.41, 1.414, etc. 3. It is also proved that these roots obey the fundamental laws of Algebra ; as ^/2 x ^/3 = ^/3 x -^2, etc. 4. An Irrational Number is a number which cannot be ex- pressed as an integer or as a fraction ; as ->/2, -^/a 2 . An Irrational Expression is an expression which involves an irrational number ; as -y/o, a 5. A Rational Number is a number which can be expressed as 2 x an integer or as a fraction; as 2, - , ^/(27a 6 ). 3y A Rational Expression is an expression which involves only rational numbers ; as -| a + ^ b, ab -f- ->/a 2 . 6. A Radical is an indicated root of a number or expression ; as y'7, V 9 > ^( a + 6). A Radical Expression is an expression which contains radi- cals; as 2V?, V^+V^ V( a + V & )- 7. A Surd is an irrational root of a rational number; as 236 ALGEBRA. [Cn. XV Observe that -^/(l + y'T) is not a surd, since 1 -f- -y"7 is not a rational number. 8. The Order of a surd is indicated by the index. Thus, y'a is surd of the second order, or a quadratic surd; -^/5 is a surd of the third order; and so on. Principles of Surds. 9. As in Ch. XIV., we limit the radicands to positive values, and the roots to positive roots. 10. The principles established in Ch. XIV., Art. 13, and their proofs, hold also for surds. For, any positive number is a power of either a rational or an irrational number. Thus, 4 = 2 2 , 3 = ( V3) 2 , a = (-tya)*. We have ^(ab) = VC(V a ) 2 (V 6 ) 2 ] = V a X V & 5 and so on - Therefore, (i.) ^/a" 8, therefore ^/9>^/8, or -^/3>V 2 - EXERCISES V. Reduce to equivalent surds of the same order: l. V 2, A/i 5 - 5 > -v/ 10 - 6 - 6 /4. 7. -/2 3. 8. /15 10. 9. 10. v Which is the greater, 13. 2V3 or 3V 2 ? 14. V 5 or A/ IO? 15 - iv" 25 or iV 11? 16. ^/'a 2 or y a, when a < 1 ? 17. -^Ac 3 or ^/x 4 , when x > 1 ? W^hich is the greatest, 18. V 3 > -v/ 5 ' or ^ 1()? 19 - Vf> ^/i or A/P Multiplication of Surds. 20. Multiplication of Monomial Surds. The converse of the principle of Art. 14 evidently holds. That is, ^/ax^/b=^/(ab). Ex. 1. 5^/4 x 2^/6 = 10^/24 = 20^/3. Ex. 2. V a x ->X - -\/ a3 X ^ a4 = ^ = a ^ a ' We thus have the following method : Reduce surds of different orders to equivalent surds of the same order. Multiply the product of the coefficients by the product of the surd factors. Simplify the result. 242 ALGEBRA. [On. XV Ex. 3. V 12 X v 36 = V( 4 X 3) x s^(4 x 9) = 2 V3 x ^(2 2 x 3 2 ) = 2^/3 3 x -4/(2< x 3 4 ) = 2^/(2 4 x 3 7 ) = 6^/(2 4 x 3) = 6-4/48. When the radicands contain numerical factors it is advisable to express them as powers of the smallest possible bases. 21. It is frequently desirable to introduce the coefficient of a surd under the radical sign. Ex. 4. 4 Vo = V 16 x V 5 = V 80 - Ex. 5. 3 a-ty(2 db) = ^(27 a 3 ) x EXERCISES VI. Multiply : 4. 4V15XV45. 5. Vt 7. 3^/45x5^/150. 8. 9^/54x3^/24. 9. ^/6 x 3^/36. 10. V a x V( 2 a )- u - 5 V m x 12. 7V(6^)x4V(lBic). 13. -^(a 2 x)x/a. 14. ^(5a?)x-4/(25ajy). 15. ^/(4 a 2 6) x 16. V( 1 +^)xV(^ + a )- 17 - ^/C 1 ~ x ? x 18. V 6 X -\/ 4 ' 19 ' A/ 50 X A/ 75 ' 20 ' V 21 X A/ 27 21. -^/20 X V 2 ' 22 ' A/ 72 X ^108. 23. -^'2 X V 3 - 27. -^/54 x 3 V6 X 5-4/2. 28. V 10 X -4/100 x r. That is, to raise a surd to any required power : Raise the radkand to the required power. Ex. 6. -/2 4 23. If the index of the root and exponent of the required power have a common factor, the work is simplified by Art. 14 : Ex.1. (^/5) 2 =-^/5. Ex.2. Ex. 3. [5 x EXERCISES VII. Simplify : 1. (V5) 2 . 2. (s 3 . 3. (A/^) 2 . 4. (-v/^) 5. (-\/2^) 6 . 6. (V3x) 3 . 7. (-v/Sa) 2 . 8. (3Va) 9. 2 2 . 10. A/aV 2 . 11. 3/2 5 . 12. 13. (Va 4 6) 2 . 14. (VSarV) 3 . 15. (V7a) 3 . 16. (2aV36) 244 ALGEBRA. [Cn. XV 24. Multiplication of Multinomial Surd Numbers. The work may be arranged as in multiplication of rational multinomials. Ex. Multiply 2 V5 + 3 V 2 by V 5 4 V 2 - We have 2^/5 + 3^/2 V5 - 4 V 2 10 + 3 V10 _ 8 ylO _ 24 10 - 5 V10 - 24 = - 14 - 5 V10. 25. Conjugate Surds. Two binomial quadratic surds which differ only in the sign of a surd term are called Conjugate Surds. E.g., V 3 + V 2 and ~ V 3 + V 2 5 1 -V 5 and * +V 5 - Either of two conjugate surds is the conjugate of the other. The product of two conjugate surds is a rational number. For, ( 26. Type-Forms. Many products are more easily obtained by using the type-forms given in Ch. V. Ex. (v 2 + V 3 ) 2 = (V 2 ) 2 + 2 V 2 x V 3 + (V 3 ) 2 EXERCISES VIII. 'Simplify each of the following expressions : 1. (v 3 + 3 V 6 - 5 V 8 ) x V 6 - 2. (V9 - 2 V45 + 5 V54) x V 3 - ; 3. (5+v 3 ) (l- 3 V 3 )- 4 - (V 10 - 2 ) (V 10 + 5 )- .5. (2 V7 - 5yi3) (V91 - 5). 6. (V6+HV 5 )(V 2 + 4 V 15 )- 9. jo. 24-26] SURDS. 245 11. 12. ( V 7 13. 14. 15. (5^/9 + 3^25) (-t/3 - 16. GJ/27- -^2) (2^/3 Find the value of each of the following expressions, without performing the actual multiplications : 17 - (V 5 -V 10 ) 2 - 18. (y6-4^/40) 2 . 19. 20. (V6-2^/2) 3 . 21. (1 + V2-V3) 2 . 22. 23. 24. 25- 25. 26. ^/(2 V2 - 3) x ^(2 V2 4- 3). 27- [ V( 7 + 2 V 1( >) - V( 7 - 2V 10 )] 2 - 28. [ Va + V( 2 - ^) + Va - 29. 30. 31- V(5 + V7) x V(2 - V 2 ) x V(5 - V7) x V(2 + V 2 )- 32. 33. Simplify each of the following expressions : id. t(rt* M\ v /** "*" u IK /(& rj- K\ v /^ / 3 s*. \r ^u / x -\ / * &^' ~\/ (y ~ ' v) x 36. X " /y-< / ~^2l-\l 2 xz + 4 z 2 . ,, + f +% lf -gV. + E-^-,). 246 ALGEBRA. . [Cii. XV Division of Surds. 27. Division of Monomial Surds. The converse of the prin- ciple of Art. 10 (iii.) evidently holds. That is, 2 3 ^ a2 3 -> 4 _ 3 6 / a 4 _ 3 6 /3 3 a _ , 6/f97 , ' 43 a ~ 4-/3 3 a 3 ~ 4\3V ~ 4\~3^" = ) We thus have the following method : Reduce surds of different orders to equivalent surds of the same order. Multiply the quotient of the coefficients by the quotient of the surd factors. Simplify the result. EXERCISES IX. Simplify each of the following expressions : i. v 60 -V 5 - 2. v 15 -Vt- 3- V-V-Vf 4- 5. V( 45 ^HV( 5 *)- 6 - 7. ^/x-*-tyx. 8. V^-*-^ 2 - 9 - 10. V 30 ^-^/ 45 - 1:L - 3V5^-^/15. 12. 13. 6V2^-^/9. 14. 2-^/6^-^/2. 15. 16. V( 14 ^)^-^( 28a2&2 )' 17 - ^/(15^)^-^/(25a:?/ 2 ). 18. (V6-5V14)-V 2 - 19 - (3V10-4V15HV 5 - 20. (V6-3^4)-j-^2. 21. 22. (3 V20 + 2 V15 - 4 V5)- V 10 - 23. (6-^4 -8^36 -15-^/48) - 24. y(6-a 2 )-i-V( a + 6 )- 25 - ^(a 2 6 26. x^+ 2 ^- a?2 - 27 - ^ 2 - 27-31] SURDS. 247 28. To Rationalize a surd expression is to free it from irra- tional numbers. Thus, -J/4 is rationalized by multiplying it by -J/2, since 29. The quotient of one surd divided by another, expressed as a fraction, may be simplified by rationalizing its denominator. Fv l V 5 - V 5 X V 3 - V 15 - EX<1> V3~V3^V3" 3 " We thus have the following method : Multiply the numerator and denominator by a factor which will rationalize the denominator. Ex 2 2 V a 2yaxs:/(2a) = 2-^/a 3 x #(4 a 2 ) a 2 ) x ^(2 a) ^ (8 a 3 ) 30. The Divisor a Binomial Quadratic Surd. We express the quotient as a fraction and rationalize the denominator. Ex.1. 5 V2 + 4 V3 (5 V2 + 4 V3) (5 V^ - 4 30-2y6-24 6-2y6 = 3 f = " V 50-48 We thus have the following method : Multiply the numerator and denominator by the conjugate of the denominator. Ex. 2. (l +aj )_(l_aj) a; 31. When the denominator contains three quadratic surds, a similar method may be employed. . 248 ALGEBRA. [Cu. XV Ex. 3. V2 V2(2V3-V2 2V3-V2+V5 [(2V3- _ 2y6-2-yio 12-4V6+2-5 " 9-4V6 (2ye-2- yiO)(9+4y6) EXERCISES X. Change each of the following fractions into an equivalent fraction with a rational denominator : 12 8 10 V2 ' V3 5 x . 6 ax 3^/4 7 ' 7^/25 a ' , x ' 3/( a V 9. : 10. i- 3 (Q L 4 / '/ /vT$ 12 3\ n / n'' ,\ 11 5 5+V21 2-V2* 3V5-2V2 ' 4+VH 5-2 V 6 15 5V2-4V3 5 V3 - 3 V? ' 5V2 + 4V3 10 1 2V5 - 3V 2 19 + 2V(2a-l) o 21 3 + 4V3 /1 n /'/* /^ -y/-L\J V ^"y t -' V6+V2-V 5 Surd Factors. 32. The expression x 2 + 2 ax + a 2 is evidently the square of x -f a. The third term of this ex- pression may be obtained as follows : 31-33] SURDS. 249 That is, the third term is the square of half the coefficient of x. Consequently, if to any binomial of the form x 2 + 2 ax, we add the square of half the coefficient of a?, the resulting tri- nomial will be the square of a binomial. This step is called completing the square. Thus, if to x 2 + 6 a?, we add (f ) 2 , = 9, we have x 2 -f 6 x + 9, =(x + 3) 2 . 33. By applying the principle of the preceding article, we can transform an expression of the second degree into the difference of two squares, and hence factor it. Ex. l. Factor x 2 + 6 x + 7. . We first complete x* + 6 x to the square of a binomial by adding (f) 2 , =9. In order 'that the value of the expression may remain unchanged, we also subtract 9 from it. We then have x * + 6 x + 9 _ 9 + 7) = (3. + 32 _ 2 Ex. 2. Factor x 2 + x 1. We have x 2 + x-l = x 2 + x + 2 - 2 - 1 Ex. 3. Factor 3 x 2 + 4 ^ - 2 f. Since the coefficient of x 2 is not 1, we first take out the fac- tor 3. We then have Completing x 2 -f- f %y to the square of a binomial by adding (| i/) 2 , =-|2/ 2 , to the expression within the parentheses, and also subtracting -| t/ 2 from it, we have = 3 (x + | y + i V10 y) (a + f y - i V 10 ?) 250 ALGEBRA. [Cn. XV We thus derive the following method : If the coefficient of x 2 is 1, add to, and subtract from, the given expression the square of half the coefficient of x. Write this result in the form a 2 b 2 and factor. If the coefficient of x 2 is not 1, factor out this coefficient, and treat the remaining factor as before. EXERCISES XI. Factor each of the following expressions : 1. oj 2 + 4aj + l. 2. x 2 -2x-ll. 3. 166 + 6x-ic 2 . 4. 9x 2 + 12^-1. 5. 4 an irrational number, is equal to - , a rational number. This is a contradiction of 2 b terms, and therefore the hypothesis ^/a = b + y'c is untenable. 37. If a+V* = *+V/> C 1 ) wherein ^/b and -^/y are surds, and a and x are rational, then a = x and b =y. For, if a = x, let a = x + m. Then (1) becomes x -|- m -f -y/& = a/' + -y/?/, or m 252 ALGEBRA. [Cn. XV But, by Art. 36, this is impossible, unless m = 0. When m = 0, a =x, and therefore ^/b = ^/y. 38. If V(a + V A ) = V* + V/> then V( a - From y (a -f yft) = we obtain a -f- -^/b = x -f y + 2 Whence, by Art. 37, a = # + y, (1) and V& = Subtracting (2) from (1), Therefore V( a V 6 ) = V^ ~ V^- Square Roots of Simple Binomial Surds. 39. Ex. 1. Find a square root of 3 + 2^/2. Let V( 3 + 2 V 2 ) = V* + Vy- (!) Then, by Art. 38, V( 3 - 2 V 2 ) = V* ~ V2/- ( 2 ) Multiplying (1) by (2), V(9 - 8) = a? - y, or a; - y = 1. (3) Squaring (1), 3 + 2 V 2 = a + + 2 V(*2/) 5 whence, by Art- 37, x + y = 3. (4) Solving (3) and (4), we have x = 2, y = 1. Therefore V( 3 + 2 V 2 ) = V 2 + V 1 = V 2 + L This example could have been solved by inspection. We change 3 + 2y2 into the form m + 2-^/(mn) + n = (V w + V 71 ) 2 - We then have V(3 + 2 v 2 ) = V( 2 + 2 V 2 + 1) = V( V 2 + !) 2 = V 2 + 1- Ex. 2. Solve, by inspection, V(21 - 3 V 2 ^). We have V( 21 - 3 V 24 ) = V( 21 ~ 2 V 54 ) 37-40] SURDS. 253 In solving by inspection, first write the surd term of the given binomial surd in the form 2^/(ww), as 3^/24 = 2-^54. Then find by inspection two numbers whose sum is equal to the rational term of the given binomial surd, and whose product is equal to mn. EXERCISES XIII. Find a square root of each of the following expressions : 1. 7+V 48 - 2. 5-V24. 3. 2+V-3. 5. 3-V5- 8. 6 + 4V2. 11. 11+4V?. I* A-AV2- - fr 2 ). 17. n 2 6. 6+V 11 - 9. 7 + 2 ViO. 12. 30-10V5. 4. li+V2- 7. 8-V28. 10. 11-6V2. 13. ^ 16. 4< 18. 10 Approximate Values of Surd Numbers. 40. An approximate value of a surd number can be found to any degree of accuracy by the methods given in Oh. XIV. Ex. 1. Find an approximate value of -^/2 correct to three decimal places. The work proceeds as follows : 2.00' 00' 00' 00 1_ 100 96 4 00 2 81 1 19 00 1 12 96 60400 1.4142 24 281 2824 2828 The work is simplified by neglecting the decimal point, writing it only in the result. It is necessary to find the root to four decimal places in order to determine whether to take the figure found in the third place or the next greater figure, according to the well-known principle of Arithmetic. 254 ALGEBRA. [Cn. XV We now have y'2 = 1.4142 . This value lies between 1.4142, = yf^frf, and 1.4143, = |^f . It therefore differs from either of these fractions by less than they differ from each other. "Rut 14143 _ 141 42 _ 1 100TRT linnnF TTFOTTO' Consequently the error of taking either 1.4142 or 1.4143 as an approximate value of -^/2 is less than JOTTO^. By taking the root to more decimal places a still more accurate value can be found. It is therefore possible to find an approximate value such that the error will be less than any assigned num- ber, however small. Ex. 2. Find the value of -^/(l x) to three terms. The work proceeds as follows : l-x X 3xl 2 =3 3xl 2 +3xlx( An approximate value of a fractional surd is obtained most simply by rationalizing its denominator, then finding the required root of the numerator of the resulting fraction, and dividing this value by the denominator. Q Ex. 3. Find an approximate value of -, correct to three decimal places. v We have = 2^? = ? (1.4142) = 2.121. V 2 2 2 V EXERCISES XIV. Find an approximate value of each of the following expres- sions, correct to four figures : i. v 8 - 2 - W 2 - 5 - ' 3 - V 2 - 4 - IV 1 - 25 - 5. V 345 - 06 - 6. V 10862 - 321 - 7 - V 54 - 0001 - 8 -?-. 9 -?_. 10 1 11 JL, V 5 V 8 2 V 4 V 75 IRRATIONAL EQUATIONS. 255 13 ^ -i- -"-y . 14 V-*- 7 1-V3 ' 5-4VH V 2 - 5 +V 6 Find an approximate value of each of the following expres- sions, to include four terms : 18. A V(l + afl. 19. -vVfa 3 - 6 3 ). 20. IRRATIONAL EQUATIONS. 41. An Irrational Equation is an equation whose members are irrational in the unknown number or numbers ; as 42. To solve an irrational equation, we must first derive from it a rational, integral equation. This step, which is usually effected by raising both members of the equation to the same positive integral power one or more times, is called rationalizing the equation. Ex. l. Solve the equation ^/(36 -f- x' 2 ) x = 2. Transferring x, V( 36 + x *) = 2 + x - Equating squares of both members, Transferring and uniting terms, _4o; = -32. Dividing by 4, x = 8. Check : V( 36 + 64) = 2 + 8, or ^/WO = 10. Ex. 2. Solve the equation V(45 4- ) + V 37 9- Transferring -^/a;, V(^^ + a;) = 9 ^/x. Equating squares, 45 -f x 81 18 ^/x + a?. Transferring and uniting terms, Dividing by 18 and equating squares, tfa*4. 4 )+V 4:::::9 . or 7 + 2 = 9. 256 ALGEBRA. [Cii. XV The preceding examples illustrate the following method of solving irrational equations : Transform the given equation so that one radical stands by itself in one member of the equation. Equate equal powers of the two members when so transformed. Repeat this process until a rational equation is obtained. EXERCISES XV. Solve each of the following equations : 2. 2-x = 3. 3. ax = b. 4. V(a;-l) = 5. 5 - 7. 8-a:=4. 8. 9= 25. 26. 3 V(a? - 3) + V( 9 x 5 + 3 y(a? -7) 2 y(a? - 7) - 3 ' so. 31. v^ + 2tt )-V^ + 2 ^) = 2 V^ 32. V( + 4 ) + V(^ - 4 ) = V( 4 - 4 > 33. v( + 2 ) + V(* - 6 ) = 2 VC^ - 3 )- 34. V( ~ 5 ) - V( + 3 ) = -\/( x - 2 ) - CHAPTER XVI. IMAGINARY AND COMPLEX NUMBERS. 1. Attention was called in Ch. XIV., Art. 10, to the fact that -Y/ 16 cannot be expressed in terms of numbers with which we are, as yet, familiar. In general, since even powers of both positive and negative numbers are positive, even roots of negative numbers cannot be expressed in terms of either rational or irrational numbers. It is therefore necessary either to exclude from our consider- ation such roots as ^/ 1, and in general Sj/ a, or again to enlarge our ideas of number. 2. We will now define, that is, fix the meaning of, the num- bers V~~ 1 an( ^ A/~ a -> ky assuming that they obey the law (ya)' = a. This relation follows from the definition of a root, as was shown in Ch. XIV., Art. 5. We therefore have (V~ I) 2 = - 1, and ( 2 -/- a) 2n =-a. Whatever meaning and use these new numbers have must be derived from these relations. Imaginary Numbers. 3. The square root of a negative number is called an Imagi- nary Number; as -^3, ^/ 8. The study of these numbers is simplified by first considering the properties of ^/ 1, which is taken as the Imaginary Unit.* * The designation, imaginary, is unfortunate, since, as will be shown in Part II., Text-Book of Algebra, such numbers are no more imaginary (in the ordinary meaning of the word) than common fractions or negative numbers. Dr. George Bruce Halsted, Professor of Mathematics in the University of Texas, has suggested Neomon for the imaginary unit, and Weomonie for imaginary. 257 258 ALGEBRA. [Cn. XVI This new unit is commonly designated, by the letter i, and its opposite by i. We then have by definition (y- 1) = ( /)=-!. For the sake of distinction all numbers, rational and irra- tional, which have been used hitherto in this book are called Real Numbers. 4. The Fundamental Operations with the Imaginary Unit. It is proved in School Algebra, Ch. XX., that V 1 or ^ * s used like a real term or factor in the fundamental operations. Just as 3 = 1 + 1 + 1, and -3=-l-l-l; -=----~-? or =---. 5. We now have, in addition to the double series of real numbers, the double series of imaginary numbers : Si t -2 i, - i, 0, i, 2 i, 3 i, . 6. Powers of /. The following values of the positive in- tegral powers of -^/ 1, or i, follow directly from the definition of i and Art. 4 : V 1 = V 1> or i = i, (V- i) 3 =(V- i) 2 (V- 1) = - V- 1, (V- I) 4 - (V- 1) 2 ( V- 1) 2 = + 1, * = * 2 * 2 = 4- 1, (V- 1) 6 = (V- i) 4 ( V- 1) 2 = - 1, ^ = * -? = - 1- The preceding results give the following properties of powers of i: (i.) All even powers of i are real. (ii.) All odd powers of i are imaginary. 3-9] IMAGINARY AND COMPLEX NUMBERS. 259 The sign of any particular power of i is readily determined by expressing it as a power of i 2 if an even power, or of i 2 multiplied by i if an odd power. Ex. 1. t a =(i 2 ) 11 =(-l) 11 = -l. Ex.2. i*=(i s ) 18 =(-l) 18 = + l. Ex. 3. i 41 = i 40 x i = (i 2 ) 20 i = (- I) 20 i = + i. Ex. 4. i = i 38 x i = (i 2 ) 19 i = (- I) 19 <-. i. 7. Multiples of the Imaginary Unit. Since (V- a) 2 = - a, and (V X V~ I) 2 = (V a ) 2 (V~ I) 2 = - a, we have (V~ a) 2 = (V a X V - I) 2 - Whence V a = V a x V~ * Ex.i. v~ 9 = V 9 x V- 1 = 3 V- 1 = 3 *- Ex.2. v- 2 = V 2 xV- 1: =V 2 - i = i'V 2 - In all reductions involving imaginary terms or factors it is advisable thus to express them as multiples of -^/l or i. 8. Addition of Imaginary Numbers. Imaginary numbers are united by addition and subtraction just as real numbers are united. Ex. 1. V- 9 + V- 16 = 3 V- 1 + 4 V- 1 = 7V- 1 = 7*. Ex.2. Ex. 3. t" + i 15 = i + (- i} = 0. 9. Multiplication of Imaginary Numbers. The principle of Art. 7 is of importance in the multiplication of imaginary numbers. Ex. i. v- 9 x V 16 = V 9 x V- 1 x V 16 = 12 V- 1 =12 i. Ex.2, y-2 x V~ 8 =V 2 x V-l x V 8 x V-l = V 16 x(V-l) 2 =-4. 260 ALGEBRA. [Cn. XVI A point in Ex. 2 deserves special notice. Had we used the principle V x y & = as in surds, we should have obtained y[(_ 2) x (- 8)] = V16 = 4, and not - 4. But that principle was proved for positive roots of positive numbers, and therefore cannot be applied in this and similar examples. Ex. 3. V - 5 x V- 10 x V- 15 = V 5 x V 10 x V 15 x (V- 1) 3 x V- 1 = 5 i 10. Division of Imaginary Numbers. The following ex- amples illustrate all possible cases. Ex. 3. V6 ye = yexy-i /6 , _3 3 x~ 1 ^x-l 2 " 3 Ex V-4 V 4 xV-! V 4 EXERCISES I. Keduce each of the following expressions to the form a-y/ 1, or ai : i. v- 9 - 2 - ^V- 25 - 3 - V~ a2 - 4 - aV- 3 ^- 5. v- 12 - 6 - V- 10 - 7 - V-^- a (V 9. ^/(-Sa^ 3 ). 10. V(3-27). 11. -^/-64. 12. -^-a 12 . 0-11] IMAGINARY AND COMPLEX NUMBERS. 261 Simplify each of the following expressions : 13. i\ 14. i 29 . 15. i 54 . 16. 4 +i M . 17. i- 18. -i. 19. 4' 20- Add: 21. V-9+V-25. 22. V-16-V- 121 - 23. v-a 2 -V~& 2 - 24. 7V- 81 + 5V- 144 25. 5V- 8 -3V- 32. 26. 8V- 75 +V- 147. 27. 2V- 25 -3V- 49 + 4V- 100. 28. 2- 29. Perform the following indicated operations : so. v-^ 4 - 31 - (V-^) 4 - 32 - (V~ a ) 8 - 33 - V~ a8 - 34. v 3x V- 6 - 35 - V- 2x V- 8 - 36 - V~ 12x V 3 - 37. V-2xV~ 50 - 3 s - V- ax V(- 9a3 )- 39. V-^XV-^ 6 - 40. V~ 6 > V-^xV-2/ 4 - 43. V~ 2 xV~ 6 XV- 24. 44. V-5xV 8 xV- 20 - 45. vCi-^xvO- 1 )- 46 - v(^~ a2 )xV(- & )- 47. (V-5+V- 3 ) 2 - 48 - (2V-3+3V-2) 2 . 49. v-3-^-v- 3 - 5a V-3-^V 3 - 51 - V 3 ^-V~ 3 - 52. V-8-^-V- 2 - 53 - V-75n-v5. 54. V 12 -V- 3 - Complex Numbers. 11. A Complex Number is the algebraic sum of a real and an imaginary number ; as, 3 2 L The general form of a complex number is evidently a + bi, wherein a and b are real numbers. When 6 = 0, we have any real number. When a = 0, we have any imaginary number. 262 ALGEBRA. [Cn. XVI 12. Two complex numbers are said to be equal ivhen the real term of one is equal to the real term of the other, and the imagi- nary term of one is equal to the imaginary term of the other; as, 2 + 3 i = 2 + 3 i. That is, if a + bi = c + / 1) (4 4- 5^/ 1)= 4 2 (5-Y/ I) 2 = 16 4- 25 = 41. 18. Division of Complex Numbers. We express the quotient as a fraction, and simplify the result. Ex 1 1 + V- 2 = (1 + V- 2) (V- 3) = V- 3 - y6 2V- 3 2(V-3) 2 -6 Ex 2 1 _ . _ 2 ~V- 5 _ - 2-V-5 ' V-5) 2 2 -(V-5) 2 19. Any Even Root of a Negative Number. We have _ i)< = [(i 4V- 1) 2 ] 2 Therefore, -^/-4 = That is, the fourth root of 4 is a complex number. It will be proved in Text-book of Algebra, Part II, that any even root of a negative number is a complex number. Complex Factors. 20. A quadratic expression which is the product of two complex factors can be resolved into factors by the method used to resolve a quadratic expression into irrational factors. Ex. Factor x 2 -2x + 3. Completing x 2 2 x to the square of a binomial in a?, we have 264 ALGEBRA. [CH. XVI EXERCISES II. Simplify each of the following expressions : 1. (2 + 4i) + (2i-3). 2. (7-5z)-(3-4i). 3. (l+ V -9) + (4-V-4). 4. (6-V-l<3)-(5-V-36). 5. (1+ V -1)(1-V-1)- 6. 7. (2+3V-l)(3-4V-l). 8. 9. (3 + 5i)(V 12 - 3 0- 10. (v8 11. (J-iiV 3 ) 3 + 3 * 3 - 12. 5- is. ^ ^a Perform the following indicated divisions : 14. _ __ 15. -1 ___ 16. . . V- 2 2-V-3 2-3V-1 17 . is . 19 1 i 3 2 i 5 i V 3 a Factor each of the following expressions : 21. a5 8 -6aj + 26. 22. ^ + 4* + 68. 23. 0^-14 a? + 61. 24. 25. 4 2 + 4?/ + 3?/ 2 . 26. Make the indicated substitution in each of the following expressions, and simplify the results : 27. In x 2 - 6 a; + 14, let# = 3 + y-5. 28. In 3^2-50; + 7, let x = 2 - 3 V~ 2 - 29. In 0^ 22 i 2 let aj = 4 + 5i = 4 5t. CHAPTER XVII. DOCTRINE OP EXPONENTS. 1. We have already abbreviated such products as act, aaa, aaaa, , aaa n factors, by a 2 , a 3 , a 4 , , a n , respectively, and called them the second, third, fourth, , ntl^ powers of a. This definition of the sym- bol a n requires the exponent n to be a positive integer. Thus 2 5 means the product of 5 factors, each equal to 2. But 2 has, as yet, no meaning, since 2 cannot be taken times as a factor. For a similar reason 2~ 5 and 2^ are, as yet, mean- ingless. But, having introduced into Algebra the symbol a", it is natural to inquire what it may mean when n is 0, negative, or a fraction. We shall find that, by enlarging our conception of powers, quite clear and definite meanings can be given to such expres- sions as 2, 3~ 2 , and 4*. Positive Integral Powers. 2. The principle a m x a n = a m+n , wherein m and n are positive integers, was illustrated by par- ticular examples in Ch. III., Art. 24. In general, a m x a n = (aaa to m factors) (aaa to n factors) = aaa to m + n factors = a m+n . 3. The other principles upon which operations with positive integral powers depend have been proved in the preceding chapters. 265 206 ALGEBRA. [Cn. XVII For the sake of emphasis, and for convenience of reference, we restate them here : (i.) a m a" = a m+n . _/W jy Wl (ii.) = a m ~"j when m > n ; - = 1, when m = n ; a n a" m -I = - . when m < n. a" a"~ m (in.) (a m )" = a m ". (iv.) (ab) m = a m b m . m m Zeroth Powers. 4. The meaning of a symbol may be denned by assuming that it stands for the result of a definite operation, as was done in letting a n a a a n factors ; or by enlarging the meaning of some operation or law which was previously restricted in its application. In the latter way, negative numbers were introduced by extending the meaning of subtraction. 5. We now enlarge the meaning of powers by assuming that the principle ^=a m ~ n a n holds also when m = n. We then have = a m ~ m = a. a m But since = 1, a m it follows that a = 1. That is, the zeroth power of any base, except 0, is equal to 1. E.g., 1 = 1, 5=1, 99 = 1, (a + &) = l, etc. 3-7] DOCTRINE OF EXPONENTS. 267 6. Thus, by the assumption that the stated law holds when m = n, a definite value of the zeroth power of a number is obtained. Nevertheless, it will doubtless seem strange to the student that all numbers to the zeroth power have one and the same value, namely 1. But it should be distinctly noted that a is by definition a symbol for ; i.e., for the quotient of two like powers of the same base. Thus, 2 =|=|=|I= 1 - Negative Integral Powers. 7. We now still further enlarge the meaning of powers by assuming that the principle a n holds not only when m > n and m = n, but also when m < n. We then have, for example, ^ = a 2 - 5 = a~ 3 . a 5 But, cancelling as in fractions, a 5 a 3 Therefore, a~ s = - a 3 In general, since m < n, we may assume n = m + 7c. Then = = a w -< m+ *> = a~*. a n a m+ But J-= 1 = 1. a m+k a m+k-m tf Therefore, a-A = i. 0* That is, a negative power of a number is equal to the reciprocal of a positive power of the same number, the exponents being numeri- cally equal. 268 ALGEBRA. [On. XVII 8. We also have = = a k . a~ h !_ a k This relation and the relation which defined a negative integral power may be stated thus: Any power of a number may be transferred from the denomi- nator to the numerator, or from the numerator to the denominator, of a fraction, if the sign of its exponent be reversed. a( a) 4 a 5 This reciprocal relation between positive and negative powers is useful in reductions which involve negative powers. EXERCISES I. Find the value of each of the following expressions 1. 2- 3 . 2. 3- 2 . 3. (I)- 1 . 4. (3f)~ 3 . 5- tt)- 3 - 6. -^ 7. -L. 8. (2)- 6 . *lXt M0 Change each of the following expressions into an equivalent expression in which all the exponents are positive : 9. x?~ 4 . 10. 2c~*d. 11. 3~Wn- z . 12. 5or 3 - 13 2*r 3 M 4^ 5 " ' ' Sb~* In each of the following expressions transfer the factors from the denominator to the numerator : n 9 (-243)* and -fy 243, a q and ^a, are only different ways of representing the same numbers. Notice that the index of the root is the denominator of the exponent of the fractional power, and the radicand is the base. 270 ALGEBRA. [Cn. XVII 10. From the definition of a fractional power we have = (V9) 2 = 9, [(- 25)*] 3 = (j/- 25) 3 = - 25. In general, (a ? ~) 3 = (-f/a)* = a. i Also, (a 9 ) 3 = ^/a ? = a, if only positive roots be considered. Therefore, (a) 9 = (a 9 )?, for the positive root. 11. Meaning of a q , wherein ?? is a positive or a negative fraction. We may always assume q to be positive and p to have the sign of the fraction. P Whatever meaning a* may have must be derived by an application of the law a m a n = a m+n . By this law, 5* . 5* 5* = 5* + * + * = 5 2 . But, since 5% 5* - 5* = (5') 3 , we have (5^) 3 = 5 2 . That is, 5* is a number whose cube is 5 2 ; or 5^ = -%/5 2 - In general, tt^ 9. (af^fi 10. (a M )- 2 . 11. (a- M )- 3 . 12. (a~)". 13. a- 24 . 14. ~ f - 15 - 13] DOCTRINES OF EXPONENTS. 275 Powers of Products. (IV.) (ab) m = a m b m , for all rational values of m. Ex.1. (2 a)- = 2-**- = ^. Ex. 2. (3 of Vr 4 = 3-Vsr 8 = j 8 - (i.) m a negative integer. Let m = m 1? so that mj is positive. Then (ab} m = (ab}~ m i = -^ = i = a~ m ib-' n i = a m b m . (a6) OT i a ra i& ra i (ii.) m a fraction. Let m = , where j9 is a positive or nega- tive integer, and q is a positive integer. P If (6) ? be raised to the gth power, we have [(a&)*] ? =(a6) p , since q is an integer, = a p b p , by (i.). But (V>~) = (a~) (6 ? ") 9 = a^6 p . Therefore [(a&)*] =(a*6*) ff ; whence (aft)* = ab. Powers of Quotients. (V.) [ - ) = , for all rational values of m. o b m i We have - = (ab~ l ) m = q m b- m = EXERCISES VI. Simplify each of the following expressions : 1. (aV 1 )- 2 . 2. (la)-'. 3. (8a- 6 )i 4. a-^- 8 - 4 . 5. 2a*aj* 6. ajV*' 276 ALGEBRA. [Cn. XVII 7. ^-T a r " 9. W J \f ) 10. f "" J ' 11. f-^LV 5 - 12. 13. -^-1 - 14. ^-; 15. EXERCISES VII. MISCELLANEOUS EXAMPLES. Simplify each of the following expressions : a b _ a? b\ 2 a x a 4- x ^1/ri _L /it/r^ n w ^ _L 1 1 a^ ^ ~r a/ tC ^ A a; " ~r - . J- *. r" 1 ., ^1-5 -I a? 4- x* 4 1 x 1^-+ Find the square root of each of the following expressions 6. x? -f a;~^ + 2. 7. a- 4 x + 2 a~ V^ + aa;- 4 . 8. 4 it-- 4 - 12 or 3 + 13 or 2 - 6 or 1 4- 1. 9. 9x 2 + 10ar 2 -4ar 4 + ar 6 -12. 10. a 2 - f a^ - f a* + f 1 a 4 1. 11. f aj 8 - 5 a; V + -W- ^ - I * + A ^ 2 - Find the cube root of eacn of the following expressions : 12. ar 6 -6or 5 4l2ar 4 -8ar 3 . 13. 8^-36^-27^4 54 14. oj* - 3 a?* 4- 3 a?* + 2 x + 3 a?* - 3 * - 6 o?^ 4 CHAPTER XVIII. QUADRATIC EQUATIONS. 1. A Quadratic Equation is an equation of the second degree in the unknown number or numbers. E.g., x 2 =25, x 2 A Complete Quadratic Equation, in one unknown number, is one which contains a term (or terms) in x 2 , a term (or terms) in x, and a term (or terms) free from x, as x 2 2ax-\-b=cx d. A Pure Quadratic Equation is an incomplete quadratic equa- tion which has no term in x, as x 2 9 = 0. Pure Quadratic Equations. 2. Ex. l. Solve the equation 6 or 2 7 = 3 or 2 + 5. Transferring 3 x 2 to the first member, and 7 to the second member, 6^-3^ = 5+7, or 3 a 2 = 12. Dividing by 3, x 2 = 4. The value of x is a number whose square is 4. But 2 2 = 4, and (-2) 2 = 4. Therefore x = 2. 3. This example illustrates the following principle, which is proved in School Algebra, Ch. XXI. : The positive square root of the first member of an equation may be equated in turn to the positive and to the negative square root of the second member. 277 278 ALGEBRA. [Cn. XVIII Ex. 2. Solve the equation (a; 2) (x + 2) = 6. Simplifying, x 2 4 = 6. Transferring 4, cc 2 = 2. Equating square roots, a; = ->/ 2. These results are imaginary. Yet they satisfy the given equation, since In such a case the equation is said to have imaginary roots. The meaning of an imaginary result, when it arises in connec- tion with a problem, will be explained in Art. 16. 4. The methods used in Ch. VIII. for solving fractional equations which lead to linear equations apply also to frac- tional equations which lead to quadratic equations. Ex. 3. Solve the equation ^--^ + ^-^ = 0. b -f x x b Clearing of fractions, (a + x)(x - b) + (a; - a) (b + x) = 0, or, x 2 4- ax bx ab -f x 2 ax + bx ab = 0. Transferring and uniting terms, Dividing by 2 and equating square roots, x = y (a&). This equation therefore has irrational roots. EXERCISES I. Solve each of the following equations : 1. a 2 = 729. 2. x 2 - 25 = 144. 3. 5 a; 2 - 27-= 2 or 2 . 3_x ?_^_JL g a? 2 1_ " ~' ' ~ 10. 7aj 2 -8 = 9aj 8 -10. 11. 5 + 16 a 2 = 11 x 2 + 15. 3-4] QUADRATIC EQUATIONS. 279 12. 13. 5(3ar J + l) + 81 = 7(5a; 2 -16) + 18. 14 A JL = JL 15 2-x 2 7ar* + 9 37 ' 2x 2 3x 2 12 5 6 15* __ 15 x A . # + 10 16. f = D H 17. ar ar 18. (7 + 2 a;) (7 -2 a;) =13. 19. (a? + i)(a; - J) = . 20. 0-8)0+5)=3(3-#). 21. + 2 )0+3) =5 22. (x + 3) 2 = 49. 23. (3 x + 4) 2 49 = 24. 64 x 2 80 x + 25 = 9. 25. 26. ' = ' . 27. /y j_ P ^^99^^ */ J"~ *J t_/ *v # C O 28 ^5_ = 1^1. 29 3a?-4 = 7a;4-24. ic + 13 3a; + 18 ' 4^-1 8^ 19 .T + 3 10 1 6a; 14 + a^_o 3a s ^Ti = 2* 31 ' T"2^T7 = 32. (2 x - 3) (3 a; - 4) (a; - 13)0 ~ 4 )= 40 - 33. (5 x - 7) (3 x + 8) - - 10) (9 - a?) = 1634. 34. a?-l [ a? + l 35. 36. 07 + 1 # 1 X 2 1 5 3 132 x - 5 2 x + 3 77 (a; - 6) 64 11 6 81 x + 7 a; -8 a; + 2 a; + 12 37. (5 - a?) (3 - ) (1 4-aj)4-(54-aj)(3 + aj)(l - a?) = 16. 38. ax 2 = 6 4 . 39. (a bx) 2 = c 2 . 40 ax 2 -4- b 2 = bx 2 + a 2 41 (x 4- CL](X a] 42. m 2 x 2 4 mx + 4 = 9. 43 , _ , a b 44. -JL- + -?L- = 1. 45. " 2 * 2 a + x 6 + x ax b _ bx + a 47 a; + l __ a + bx + car^ a bx b + aa; a; 1 a to + car 1 280 ALGEBRA. [Cn. XVIII Solution by Factoring. 5. The principle on which the solution of an equation- by factoring depends was proved in Ch. VI, Art. 43. The methods given in Ch. VI, Arts. 9-13 ; Ch. XV, Art. 33 ; and Ch. XVI, Art. 20, enable us to factor any quadratic expression. The roots of the given quadratic equation are the roots of the equa- tions obtained by equating to each of its factors. Ex. 1. Solve the equation Dividing by 3, x 2 + f x - f = 0. Adding and subtracting (.jl^) 2 , = ff> we have or, (* + f) 2 -M = 0. Factoring, (x + f + f) (x + f - J) = 0, or, (x + 2)(x-=0, Equating each factor to 0, x + 2 = 0, whence x = 2 ; x i = 0, whence x = -|. Ex. 2. Solve the equation 2a 2 + 2#-l = 0. Dividing by 2, x* + x % = 0. Adding and subtracting (i) 2 , = J, or * + Factoring, (x + i + i V 3 ) + i - iV 3 ) = - Equating factors to 0, a + i + lV 3 = >^+-i- i V 3 = - Whence x = - | - ^/3, and - + Such roots are usually written J 5] QUADRATIC EQUATIONS. 281 Ex. 3. Factor x 2 - 2 x + 19 = 0. Adding and subtracting ( I) 2 , = 1, x*-2x + l -1 + 19 = 0, or, since - 1 + 19 = 18 = - (- 18) = - (V~ 18) 2 , = -(3V-2) 2 , 0-1) 2 -(3V-2) 2 = 0. Factoring, (a; - 1 + 3 V- 2) (a; - 1 - 3 V- 2) = 0. Equating factors to 0, x - 1 + 3 V- 2 = 0, a> - 1 - 3 V- 2 = 0. Whence, x = 1 3 v ' - 2. EXERCISES II. Solve each of the following equations : 1. x 2 - 6#-f- 5 = 0. 2. 0^-7^ + 10 = 0. 3. ie 2 -4a;-21=0. 4. ^ = 11^ + 12. 5. 3z 2 + 4z + l = 0. 6. 9 a* -12 a? + 4 = 0. 7. 6x 2 + 13z-8 = 0. 8. 11 a 2 -7^-18 = 0. 9. 7a 2 -20a + 8 = 0. 10. 7-12a? 2 = 17aj. 11. 20^-79^ + 77 = 0. 12. 8x 2 + 13a;-82 = 0. 13. a,- 2 - 2 cc- 1 = 0. 14. or 8 6 a? -71 = 0. 15. x 2 -2x + 2 = 0. 16. a 2 - 4^ + 13 = 0. 17. (a? + 8)(aj + 3)=aj-6. 18. (x + 7) (x - 7) = 2(a? + 50). 19. (2a + l)0 + 2) = 3a 2 -4 20. (aj-l)(2a? + 3)=4aj 2 -22. 21. .T 2 -3 = i(x-3). 22. x(^ + 5) = 5(40-a^) + 27. 23 ^ 14 24 x + 7 = 3a;-5 x + 120 3 x - 10 2 2 + 3 x + 3 ' 25 5 = a? + ll. 26 5 4a? + 7 = 3, 4 a; 6 6 ' x x + 1 " 2 27 __ __ __ 28 ' 282 ALGEBRA. [Cn. XVIII 29. Qx-Sx 2 x x+ 3 30. X- -1 4 1 1 21-7x 21 x x-2 x 3 31. 1 x 2 - 106 1 32 X ~ - 2 + x 4- 2 2 (x + 3) x + 3 8^-72 8 x H -2 a? -2 x-3 33. x + 24 x-7 1 34 4 x- + 67 x 2 5 ar 2 - 5 3. + i 2 a? - 35. ar } + llaz + 28a 2 = 0. 36. x 2 - 14 mx + 33 m 2 = 0. 37. x 2 2 ax-{- a 2 b 2 = 0. 38. a? 2 3 a#-f 2 a 2 ab 6 2 =0. 39. 2 - (2 m - 1) a? + m 2 - m - 6 = 0. 40. x 2 - (3 a + 2 6) a; + 6 06. 41. ax 2 + (a + 2) a + 2 = 0. 42. bx 2 - 2 (b + c) x + 4 c = 0. 43. (a + 44. (a 2 -| 45. a 2 46. (m w) x 2 (m -f w) # + 2 n = 0. 47. _ + __ = 2. 48. . ox a 2 a b a ( a? a6 (6 1) 6 # + 4 >i x 4 ?i Solution by Completing the Square. 6. The following examples illustrate the solution of a quad- ratic equation by the method called Completing the Square. Ex. 1. Solve the equation a 2 5 x + 6 = 0. Transferring 6, x 2 5 x = 6. To complete the square in the first member, we add ( f) 2 , = -^-, to this member, and therefore also to the second. We then have Equating square roots, x -f = , by Art. 2. 5-6] QUADRATIC EQUATIONS. 283 Whence, x = f . Therefore the required roots are 3 and 2. Ex. 2. Solve the equation Transferring 1, 7 x 2 + 5 x = 1. Dividing by 7, # 2 + f x = f . Adding (^Y = AV, ^ + * * + T* = fWr - * = f Equating square roots, x -f T \ = T W~ 3- Whence, x = T \ Therefore the required roots are -3 and - - Ex. 3. Solve the equation Transferring a 2 , (a 2 - 6 2 ) x 2 - Dividingby^, rf- * ^ members > a 4 a 2 a^ Equating square roots, x 2 a = -^ Whence, * = ^^- Therefore the required roots are a and a b a + b The preceding examples illustrate the following method of procedure : Bring the terms in x and x 2 to the first member, and the terms free from x to the second member, uniting like terms. 284 ALGEBRA. [Cn. XVIII If the resulting coefficient of x 2 be not + 1, divide both members by this coefficient. Complete the square by adding to both members the square of half the coefficient of x. Equate the 2^ositive square root of the first member to the posi- tive and negative square roots of the second member. Solve the resulting equations. EXERCISES III. Solve each of the following equations : 0. 2. a*-5a; = -4. 3. a 2 + 2^ + 1 = 0. 4. 2 or 2 - 5. 3a 2 -53a + 34 = 0. 6. 7. x 2 -4z + 7 = 0. a 9. x 2 -2x + 6 = 0. 10. x 2 -l-^-x(x- 11. (3x-2)(x-l) = U. 12. (2a-l)(a-2) 13. # + - = 54-. 14. x--=U. x x 15. _!=:!?. 16. = a;-4. x x 17. -i- + J-=z-i. 18. a+i=7+i. 2x 3x 6 * 7 19. ^ =a; + 2. 20. 2 21. a? + 3 = a;-4 22 a?+l = 3a? + x + 9 cc 1 ' aj-ho 7^ 23 1Q | 27 =5 . 24. ^+j_2^~ 1 x 1 2ic re 5 x + 25. (2x-3) 2 = 8x. 26. (2 27. (5z-3) 2 -7=40a-47. 28. (x 29. (aj~7)(aj-4) + (2aj-3)(aj-5) 30. 10(2 a? + 3) (a? - 3) + (7 + 3) 2 = 20 (x + 3) (a; - 1). 31. (a? - 1) (a; - 3) -f- (x - 3) (or - 5) = 32. 32. (x - 1) (a? - 2) + (x - 3) (a? - 4) = (x - I) 2 - 2. 6-7] QUADRATIC EQUATIONS. 285 638 33. 34. ic 5 x 4 x 3 12 7 15 -1 6 x x-2 35. -^ + -6+^ = 5. 36 2x-7_ 7 11 9 ~ 1 "K "A o 7 3 ic , 6 38. -f 3 2 39. | !+* l-3a? 40. aj_ = -. 41. | - = . 6 a a? ?i a; 7i -f x n 2 x 2 42. a. = " (a 6) 2 x a 6 2 a; 2 H 2 no; ?i 1 a a-1 2 a a? a; a a -f 6 + x CKC + 1 ^ 2ic2 <* a 2 #H-ao? 48. a x 2 a x General Solution 7. The most general form of the quadratic equation in one unknown number is evidently as? + bx + c = 0. The coefficient a is assumed to be positive and not 0, but b and c may either or both be positive or negative, or 0. 286 ALGEBRA. [On. XVIII Dividing by a, x 2 + - x + - = 0. a a Transferring -, y? + - x = - -. a a a Adding (AT- = &' 4 a 2 Equating square roots, a + = V( &2 ~ 4ac ). Whence, x 2a 2a and a = - 2ci 2a 8. The roots of any quadratic equation can be obtained by substituting in the general solution the particular values of the coefficients a, b, and c. Ex. Solve the equation 3ot? + 7 x 10 = 0. We have a = 3, 6 = 7, c = - 10. Substituting these values in the general solution, we obtain * = - + iV[ 4 9-4x3(- 10)] = 1, and a = -i EXERCISES IV. Solve each of the following equations : 1. 2a 2 = 3o; + 2. 2. 5a^-6a; + l = 0. 3. 9o;(a? + l) = 28. 4. x 2 -b 2 = 2ax-a? 5. oj 2 + 6oa; + l=0. 6. ar 2 + l = 2i#. 7. (a-5) 2 +(>-10) 2 = 37. a 2tt(3n-4a;) = tt 2 9. n 2 ( 2 + l) = a 2 + 2n 2 a?. 10. x 2 + (a? + a) 2 = a 2 . 7-10] QUADRATIC EQUATIONS. 287 Relation between Roots and Coefficients. 9. If the roots of the quadratic equation ax 2 + bx + c = 0, or x 2 -}--a; + - = a a be designated by r and r. 2 , we have r = b V(6 2 -4ac) 2a 2a The sum of the roots is n+r,=-- (i) The product of the roots is b -4ac)-] r 6 T~ J L~^~ The relations (1) and (2) may be expressed thus : (i.) If the coefficient of the second power of the unknown num- ber be 1, the sum of the roots is equal to the coefficient of the first power of the unknown number, with sign reversed. (ii.) If the coefficient of the second power of the unknown num- ber be 1, the product of the roots is equal to the term free from the unknown number. E.g., the roots of the equation x 2 5# + 6 = are 2 and 3 ; their sum is 5 (the coefficient of x with sign reversed), and their product is 6 (the term free from x). The roots of the equation 6 x 2 x 2 = 0, or x 2 J x ^ = 0, are J and ^ ; their sum is -J, and the product is ^-. 10. Formation of an Equation from its Roots. The relations of the last article enable us to form an equation if its roots be given. We may always assume that the coefficient of the second power of the unknown number is 1. 288 ALGEBRA. [Cn. XVIII Ex. 1. Form the equation whose roots are 1, 2. We have r -f r 2 = 1 4- 2 = 1, the coefficient of x, with sign reversed ; and rfa = 1 x 2 = 2, the term free from x. Therefore the required equation is x 2 x 2 = 0. Ex. 2. Form the equation whose roots are 1 -f 2^/3, 1 2^/3. We have i\ + r 2 = (1 + 2 V 3 ) + (1 - 2 V 3 ) - 2 ; and rfa = (1 4- 2 V 3 ) (1 - 2 y3) = 1 - 12 = - 11. Therefore the required equation is cc 2 2 x 11 = 0. 11. It follows from Art. 9, that the quadratic equation may be written in the form ^-Oi + r^aj + r^^O, or (x r x ) (x r 2 ) = 0. Ex. Form the equation whose roots are 1, 2. We have (x -f 1) (a; 2) = 0, or x 2 - x - 2 = 0. When the roots are irrational or imaginary, the method of the preceding article is to be preferred. EXERCISES V. Form the equations whose roots are : 1. 8, 2. 2. - 5, - 3. 3. 10, 10. 4. 7, - 3. 5. 4, - 10 6. 2, If. 7. - j, - 1 J. 8. - J, 8. 9. 2, 0. 10. a, b. 11. - a, - 1. 12. a 2 , - 4 a 2 . is. v 2 > -V 2 - 14 - W~ 3 > -4V- 3 - 15. 1 -}- V7, 1 - V7. 16. i - i V 11 * * + t V 11 - 17. 3--5 3+-5. 18. - Nature of the Roots. 12. In many applications it is important to know, without having to solve an equation, the nature of its roots, i.e., whether they are both reed and unequal, whether they are both real and equal, whether they are imaginary. 10-12] QUADRATIC EQUATIONS. 289 In the general solution b y(6 2 -4ac) b y(& 2 -4ac) 2a + 2a 2a ~~2^~ of the equation aa? 2 + b% + c = 0, a, 6, and c are limited to real, rational values. (i.) The tivo roots are real and unequal when b 2 4 ac is positive, i.e., when b 2 4 ac > 0. E.g., in x 2 + 4# - 12 = 0, a = 1, 6 = 4, c = 12 ; and since 6 2 4 ac, = 16 + 48, is positive, the roots of this equation are real and unequal. (ii.) The two roots are real and equal when b 2 4 ac is equal to ; i.e., when b 2 = 4 ac. E.g.,m or j -4x + 4 = 0, a = 1, b = 4, c = 4 ; and since 6 2 = 4 ac, the roots of this equation are real and equal. (iii.) The two roots are conjugate complex numbers when b 2 4 ac is negative ; i.e., when b 2 4 ac < 0. E.g., in ^-2^ + 3 = 0, a = 1, b = - 2, c = 3; and since b 2 - 4 ac, = 4 - 12, = - 8, is negative, the roots of this equation are complex numbers. EXERCISES VI. Without solving the following equations, determine the nature of the roots of each one : =0. 2. a^+12 z= -40. 3. 4. x 2 -x = 12. 5. x 2 - Sx + 25 = 0. 6. aj 2 -8 = 16. 7. 8. 8^-2^-25 = 0. 9. 10. 10 a 2 - 21 a; -10 = 0. 290 ALGEBRA. [Cii. XVIII For what values of m are the roots of each of the following equations equal ? For what values of m are the roots irra- tional? And for what values of m are the roots complex numbers ? 11. wa; 2 -h4a,' + l=:0. 12. 2ar> -f mz + 1 = 0. 13. 3# 2 + 6o; + m = 0. 14. mo? + wa + 1 = 0. IRRATIONAL EQUATIONS. 13. An irrational equation may lead to a quadratic equation when rationalized. Ex. l. Solve the equation x + ^/(25 x 2 ) = 7. Transferring x, V( 25 ^ 2 ) = 7 ~ ^ 0-) Squaring, 25 - or 2 = 49 - 14 oj + x 2 . (2) The roots of this equation are 3, 4. Both roots of (2) satisfy the given equation, since 3 + V(25 - 9) = 7, and 4 -f- V( 25 ~ 16 ) = 7. Ex. 2. Solve the equation a? V( 25 a 2 ) = 1. Transferring #, -^(2o x 2 ) = 1 #. (1) Squaring, 25 -x 2 = 1 - 2x + x 2 . (2) The roots of this equation are 4 and 3. The number 4 is a root of the given equation, since 4-V(25-16) = l; but the number 3 is not a root of the given equation, since - 3 - V(25 - 9) = - 7, not 1. Therefore the root 3 was introduced by squaring. Now observe that the same rational equation (2) would have been obtained, if the given equation had been that is, if the surd term had been of opposite sign. The root 3 satisfies equation (3), since - 3 + V( 25 -9) = -3 + 4 = 1. 12-13] IRRATIONAL EQUATIONS. 291 Therefore equation (2) is equivalent to equations (1) and (3) jointly. It frequently happens that no root can be found to satisfy an equation obtained by giving to the square root either its positive or its negative value. In Ex. 1, the equation thus derived is and is not satisfied by either of the roots obtained. The equa- tion is then said to be impossible. Ex. 3. Solve the equation V(2* + 3) -V(7 -*) = !. If both positive and negative square roots be admitted, the given equation is equivalent to the four equations : -s) = l CO, ^(2x+3)-^/(7-x) = l (2), -x) = l (3), -V( 2 a+3)- V(7-) = 1 (4). The same rational integral equation will evidently be de- rived by rationalizing any one of these equations. In (1) transferring -^/(7 x), Squaring, 2 x + 3 = 1 - or 3o; 5 = Again squaring, 9 x 2 30 x + 25 = 28 4 x, or 9aj 8 -26a?-3 = 0. The roots of this equation are 3 and i-. By substitution we find that equation (2) is satisfied by the root 3, and equa- tion (3) by the root ^. The other two equations are impossible. Consequently, in solving an irrational equation, we must expect to obtain not only its roots, but also the roots of the other equations obtained by changing the signs of the radicals in all possible ways. Some of these equations will be impos- sible. The roots of the other irrational equations will be the roots of the rational equation. 292 ALGEBRA. [Cn. XVIII 14. Ex. Solve the equation V(3 x z - 2 x + 4) - 3 ar + 2 x = - 16. Since - 3 y? + 2 a? = -(3 or - 2 a; + 4)+ 4, we may take -^/(S x 2 2 a; + 4) as the unknown number, replac- ing it temporarily by y. We then have the quadratic equation y - i/ 2 + 4 = - 16. The roots of this equation are 5, and 4. Equating ^/(3 y? 2 x -f 4) to each of these roots, we have -^(3 x 2 2 x + 4) = 5, whence x = 3, J. V(3 y? - 2 x + 4) = -4, whence a; = J (1 V37). The numbers 3, J satisfy the given equation, and are therefore roots of that equation. The numbers -V(1 V^7) do not satisfy the given equation. But if the value of the radical is not restricted to the posi- tive root, the given equation comprises the two equations V(3 2 - 2 x + 4) - 3 a? + 2 x = - 16, (1) -V(3 2 -2aj + 4)-3oj a + 2a; = -16. (2) Then -|(1 ^/37) are roots of (2). The given equation is said to be in quadratic form. EXERCISES VII. Solve each of the following equations, and check the results. If a result does not satisfy an equation as written, determine what signs the radical terms must have in order that the result may satisfy the equation. 2. 4z 3. 3-3^-4^ + 9=0. 4. 5^ 7 14-15] HIGHER EQUATIONS. 293 17 25 11. r ?^/x=3x 2 -\-3x-53. 12. 13. (5- 15. 16. 20 ' 21. x 2 -x + 2V(^ --!!) = 14. 22. 23. 24. 27. 28. * __ HIGHER EQUATIONS. 15. Certain equations of higher degree than the second can be solved by means of quadratic equations. Ex. 1. Solve the equation x 3 1 = 0. Factoring, (a; 1) (x 2 + x + 1) = 0. This equation is equivalent to the two equations x 1 = 0, whence x = 1 ; and x 2 + x + 1 = 0, whence x = 294 ALGEBRA. [Cn. XVIII This example gives the three cube roots of 1, since x 3 1=0 is equivalent to x 3 = 1, or x = -J-l. Therefore the three cube roots of 1 are In general, the three cube roots of any number can be found by multiplying the arithmetical cube root of the number in turn by the three algebraic cube roots of 1. ^8 = 2^/1=2, -lV-3. Ex. 2. Solve^the equation a 4 - 9 = 2 x 2 - 1. Since x 4 = (x 2 ) 2 , we may take x 2 as the unknown number and solve this equation as a quadratic in x 2 . We then have (x 2 ) 2 -2x 2 -8, = 0. Factoring, (or 2 - 4) (x 2 + 2) = 0. Whence, x 2 - 4 = 0, or x = 2 ; and x 2 + 2 = 0, or a; = V~ 2. In general, any equation containing only two powers of the unknown number, one of which is the square of the other, can be solved as a quadratic equation. Ex. 3. Solve the equation (x 2 3x+l) 2 =6+5(x 2 3a;+l). In this example x 2 3 x -f- 1 is regarded as the unknown number, and may temporarily be represented by the letter y. The equation then becomes y 2 = 6 + 5 y ; whence y = 6, and 1. We therefore have the two equations x 2 -3x + l = 6, whence x = f iV 29 5 x 2 3a? + l= 1, whence x = 2, a? = 1. Therefore the roots of the given equation are f %^/2$, 2, 1. Attention is called to the fact that, in each example, we have obtained as many roots as there are units in the degree of the equation. 15-16] PROBLEMS. 295 EXERCISES VIII. Solve each of the following equations : 1. 3^ + 1 = 0. 2. z 4 -l = 0. 3. a 6 + 1=0. 4. a 6 - 1 = 0. 5. (a; -I) 3 = 8. 6. x? = (2a-x) 3 . 7. (^ + 1) 4 =16. 8. 4 + 9 = 10r J . 9. x 4 -6x 2 = -l. 10. o; 6 -65ar J = -64. 11. 12. 13. 14. q a -a.o 16 ' (a + ) 8 + (a - a?) 8 a 4 - 6ar> + 1 2 x 2 a 2 x 2 -\- a? _ 34: '' PROBLEMS. 16. Pr. 1. The sum of two numbers is 15, and their product is 56. What are the numbers ? Let x stand for one of the numbers ; then, by the first con- dition, 15 x stands for the other number. By the second condition x (15 x) = 56 ; whence x = 7, and 8. Therefore x = 7, one of the numbers, and 15 x = 8, the other number. Observe- that if we take x = 8, then 15 x = 7. That is, the two required numbers are the two roots of the quadratic equation. Pr. 2. Divide 100 into two parts whose product is 2600. Let x stand for the less part, and 100 x for the greater. By the second condition, x (100 - x) = 2600. The roots of this equation are 50 + 10^/ 1 and 50 10 V~ 1. 296 ALGEBRA. [Cn. XVIII An imaginary result always indicates inconsistent conditions in the problem. The inconsistency of these conditions may be shown as follows : Let d stand for the difference between the two parts of 100. Then 50 -f- d stands for the greater part, and 50 d for the less. The product of the two parts is (50 + J d) (50 - d), = 2500 - Q cZ) 2 = 2500 - J d 2 . Since d 2 is positive for all real values of d, the product 2500 \ d 2 must be less than 2500. Consequently 100 cannot be divided into two parts whose product is greater than 2500. 17. When the solution of a problem leads to a quadratic equation, it is necessary to determine whether either or both of the roots of the equation satisfy the conditions expressed and implied in the problem. Positive results, in general, satisfy all the conditions of the problem. A negative result, as a rule, satisfies the conditions of the problem, when they refer to abstract numbers. When the required numbers refer to quantities which can be understood in opposite senses, as opposite directions, etc., an intelligible meaning can usually be given to a negative result. An imaginary result always implies inconsistent conditions. 18. The interpretation of a negative result is often facili- tated by the following principle: If a given quadratic equation have a negative root, then the equation obtained by changing the sign of x has a positive root of the same absolute value. E.g., the roots of the equation y? 5 # + 6 = are 2 and 3 ; and the roots of the equation or a^-f 5 x + 6 = 0, are 2 and 3. 16-18] PROBLEMS. 297 Pr. 3. A man bought muslin for $ 3.00. If he had bought 3 yards more for the same money, each yard would have cost him 5 cents less. How many yards did he buy ? Let x stand for the number of yards the man bought. Then 800 1 yard cost - - cents. If he had bought x + 3 yards for the x 300 same money, each yard would have cost - - cents. ^nn ^nn x -\- o Therefore ^--^-=5; whence x = 12 and -15. x x + 3 The root 12 satisfies the equation and also the conditions of the problem ; the root 15 has no meaning. But if x be replaced by x in the equation, we obtain a new equation, 300 300 - 300 300 K = 5, or = 5, (2) x ic + 3 a; 3 x whose roots are 12 and -f 15. Equation (2) evidently corresponds to the problem : A -man bought muslin for $ 3.00. If he had bought 3 yards less for the same money, each yard would have cost him 5 cents more. Notice that the intelligible result, 12, of the first statement has become 12 and is meaningless in the second statement. EXERCISES IX. 1. If 1 be added to the square of a number, the sum will be 50. What is the number ? 2. If 5 be subtracted from a number, and 1 be added to the square of the remainder, the sum will be 10. What is the number ? 3. One of two numbers exceeds 50 by as much as the other is less than 50, and their product is 2400. What are the numbers ? 4. The product of two consecutive integers exceeds the smaller by 17,424. What are the numbers ? 5. If 27 be divided by a certain number, and the same num- ber be divided by 3, the results will be equal. What is the number ? 298 ALGEBRA. [Cn. XVIII 6. What number, added to its reciprocal, gives 2.9 ? 7. What number, subtracted from its reciprocal, gives n ? Let n = 6.09. 8. If n be divided by a certain number, the result will be the same as if the number were subtracted from n. What is the number ? Let n = 4. 9. If the product of two numbers be 176, and their differ- ence be 5, what are the numbers ? 10. A certain number was to be added to , but by mistake -J- was divided by the number. Nevertheless, the correct result was obtained. What was the number ? 11. If 100 marbles be so divided among a certain number of boys that each boy shall receive four times as many marbles as there are boys, how many boys are there ? 12. The area of a rectangle, one of whose sides is 7 inches longer than the other, is 494 square inches. How long is each side? 13. The difference between the squares of two consecutive numbers is equal to three times the square of the less number. What are the numbers ? 14. A merchant received $ 48 for a number of yards of cloth. If the number of dollars a yard be equal to three-sixteenths of the number of yards, how many yards did he sell ? 15. In a company of 14 persons, men and women, the men spent $ 24 and the women $ 24. If each man spent $ 1 more than each woman, how many men and how many women were in the company ? 16. A pupil was to add a certain number to 4, then to sub- tract the same number from 9, and finally to multiply the results. But he added the number to 9, then subtracted 4 from the number, and multiplied these results. Nevertheless he obtained ,the correct product. What was the number ? 17. A man paid $ 80 for wine. If he had received 4 gallons less for the same money, he would have paid $ 1 more a gallon. How many gallons did he buy ? 18] PROBLEMS. 299 18. A man left $ 31,500 to be divided equally among his children. But since 3 of the children died, each remaining child received $ 3375 more. How many children survived ? 19. Two bodies move from the vertex of a right angle along its sides at the rate of 12 feet and 16 feet a second respectively. After how many seconds will they be 90 feet apart ? 20. A tank can be filled by two pipes, by the one in two hours less time than by the other. If both pipes be open 1J hours, the tank will be filled. How long does it take each pipe to fill the tank ? 21. From a thread, whose length is equal to the perimeter of a square, 36 inches are cut off, and the remainder is equal in length to the perimeter of another square whose area is four-ninths of that of the first. What is the length of the thread ? 22. A number of coins can be arranged in a square, each side containing 51 coins. If the same number of coins be arranged in two squares, the side of one square will contain 21 more coins than the side of the other. How many coins does the side of each of the latter squares contain ? 23. A farmer wished to receive $ 2.88 for a certain number of eggs. But he broke 6 eggs, and in order to receive the desired amount he increased the price of the remaining eggs by 2|- cents a dozen. How many eggs had he originally ? 24. Two bodies move toward each other from A and B re- spectively, and meet after 35 seconds. If it takes the one 24 seconds longer than the other to move from A to B, how long does it take each one to move that distance ? 25. It takes a boat's crew 4 hours and 12 minutes to row 12 miles down a river with the current, and back again against the current. If the speed of the current be 3 miles an hour, at what rate can the crew row in still water ? 26. A man paid $ 300 for a drove of sheep. By selling all but 10 of them at a profit of $ 2.50 each, he received the amount he paid for all the sheep. How many sheep did he buy ? CHAPTER XIX. SIMULTANEOUS QUADRATIC AND HIGHER EQUATIONS. 1. The solution of a system of quadratic or higher equa- tions in general involves the solution of an equation of higher degree than the second, and therefore cannot be effected by the methods for solving quadratic equations. But there are many special systems whose solutions can be made to depend upon the solutions of quadratic equations. The proofs of the following methods are given in School Algebra, Ch. XXIV. 2. Elimination by Substitution. When one equation of a system of two equations is of the first degree, the solution can be obtained by the method of substitution. Ex. Solve the system y + 2 x = 5, j (1) rf-jf = -8.J (2) Solving (1) for y } y = 5 - 2 x. (3) Substituting 5 2 x for y in (2), From this equation we obtain x = 1, and x = 5J. Substituting 1 for x in (3), y = 3. Substituting 5| for x in (3), y = 6^-. It is proved in School Algebra, Ch. XXIV., that the above method is based upon equivalent equations. Therefore the solutions of the given system are 1, 3 ; 5f , 6J, the first number of each pair being the value of x, and the second the corresponding value of y. 300 1-3] SIMULTANEOUS QUADRATIC EQUATIONS. 301 Had we substituted 1 for x in (2), we should have obtained y = 3. But the solution 1, 3 does not satisfy equation (1). Therefore, always substitute in the linear equation the value of the unknown number obtained by elimination. 3. Elimination by Addition and Subtraction. This method can frequently be applied. Ex. Solve the system x 2 + 3 y = 18, 1 (1) 2x 2 -5y = 3. J (2) We will first eliminate y. Multiplying (1) by 5, 5 x 2 + 15 y = 90. (3) Multiplying (2) by 3, 6 x? - 15 y = 9. (4) Adding (3) and (4), 11 x 2 = 99. Whence, x = 3, and x 3. Substituting 3 for x in (1), y = 3. Substituting 3 for x in (1), y = 3. The given system has the two solutions 3, 3 ; 3, 3. Notice that this example could also have been solved by the method of substitution. EXERCISES I. Solve each of the following systems : 2. < 3. 4 , " U2/ = 96. '' (3x-2y = l, '' * + 6. 302 ALGEBRA. Sx-J + tf-l, r -L3. 1 _ ~ - 1*. i [Cn. XIX 15 ' {lx-y = 7 +f 16 ' { or 2 + 5 xy + / = 43, 17. 19. 4 3 * / 17 7* x y 18. 20. 5a;-2 = 4. Homogeneous Equations. When all the terms which con- tain the unknown numbers in both equations of the system are of the second degree, a system can always be derived whose solution is obtained by the method of Art. 2. Ex. Solve the system Multiplying (1) by 73, 73 x 2 + 73 xy + 146 y 2 = 74 x 73. Multiplying (2) by 74, 148 x 2 + 148 xy + 74 y 2 = 74 x 73. Subtracting (3) from (4), 75 x 2 + 75 xy - 72 y 2 = 0, (1) (2) (3) (4) (5 x - 3 y) (5 x + 8 y) = 0. Therefore the given system is equivalent to a* + xy + 22/ 2 = 74,j (a) ' x> + xy + 2y* = The solutions of these systems, and hence of the given system, are respectively 3, 5 ; 3, 5 ; 8, 5 ; 8, 5. In applying this method to such systems, we must first derive from the given equations a homogeneous equation in which there is no term free from the unknown numbers. 3-6] SIMULTANEOUS QUADRATIC EQUATIONS. 303 5. Such examples can also be solved by a special device. Ex. Solve the system x 2 + 4 f = 13, (1) ^ + 2^ = 5. (2) In both equations, let y = tx. (3) Then from (1), y? + 4 aft 2 = 13, whence a 8 = ^ 13 , 9 ; (4) 1 + 4 r and from (2), x 2 t + 2tft 2 = 5, whence y? = 5__- (5) Equating values of x 2 , 13 = - -- (6) 1 + 4* 2 + 2 2 Whence t = 1, and = f . When = i, x- = 13 9 = 9, whence a; = 3. When t = %, x 2 = 1, whence a; = y^- When a; = 3, y = tx = %( 3)= 1. When a = vi, y = - 1( Vi)= * IVi- EXERCISES II. Solve each of the following systems : f aj - a ' ' - 2xy + 15 = 0. ' ' g = 61 - 3xy, ' -y 2 = 31 - 2xy. 6. Symmetrical Equations. A Symmetrical Equation is one which remains the same when the unknown numbers are interchanged. A system of two symmetrical equations can be solved by first finding the values of x -{-y and x y. 304 ALGEBRA. [Cn. XIX Ex. 1. Solve the system x 2 + y 2 = 13, j (1) xy = G. J (2) Multiplying (2) by 2, 2 o;y = 12. (3) Adding (3) to (1), x* + 2 xy + f = 25. (4) Subtracting (3) from (1), x 2 2xy + y 2 = l. (5) Equating square roots of (4), x + y = 5. (6) Equating square roots of (5), x y = 1. (7) It is proved in School Algebra, Ch. XXIV., that equations (6) and (7) are equivalent to a; + y = 5, * + # = 5, # + ?/ = 5, 1 a; + y = 5, * + # = 5, j # + ?/ = 5, 1 a; + y = 5, -y = -l,J x-y = +l,) a-2/ = -l. The solutions of these four systems are respectively 3, 2 ; 2,3; -2, -3; -3, -2. The solutions of (6) .and (7) should be obtained mentally, without writing the equivalent systems. Each sign of the second member of (6) should be taken in turn with each sign of the second member of (7). Notice that these solutions differ only in having the values or x and y interchanged. This we should expect from the definition of symmetrical equations. When the equations are symmetrical, except for sign, the solution can be obtained by a similar method. Ex. 2. Solve the system r-f-4 (1) x 2 + f = 29. (2) Squaring (1), tf - 2 xy + f = 9, (3) Subtracting (3) from (2), 2 xy = 20, or xy = 10. (4) The solutions of (1) and (4) are 5, 2 ; - 2, - 5. Notice that the solutions in this case differ not only in hav- ing the values of x and y interchanged, but also in sign. 6] SIMULTANEOUS QUADRATIC EQUATIONS. 305 EXERCISES III. Solve each of the following systems : L I aw = 32. xy = 15. 2. 5. . xy = b. xy = 28. 3 xy = - 60. ay = 12. io |9x 2 + 2/ 2 = 37a 2 , L ay = - 90. r I cc = 2 a5. 12. \ ' 13. la? + y = 15. (x 2 -y 2 = 28, f x 2 -4 : y 2 = -3. 15. i 16. j 17. I x?/ = 48. I xy = 1. x y = 5. 16 ^+49 ?/ 2 =113, (x y = Z. (6xy = w. l f a# = 80, a? + y = 16, 21. 1 1 _ 1 22. 1 1_1 23. 24. 25. ^ 'a? y 16 y a? ~ 15' 26. =2. x-y = 2. f x + ccw + v = 29, f 27. 9 28. ^ + ?y 2 - f x 2 H- w 2 - (aj - v) = 20, f 29. 30. ^ *+f ?/- - . = 6. 32 " + /-a^ = 21. 306 ALGEBRA. [Cn. XIX 7. Higher Equations. The solutions of certain equations of higher degree than the second can be made to depend upon the solutions of quadratic equations. Ex. l. Solve the system x? + y s = 9, (1) a? + y = 3. (2) Dividing (1) by (2), x 2 xy+y 2 = 3. (3) Subtracting (3) from the square of (2), 3 xy = 6, or xy = 2. (4) The solutions of (2) and (4), and therefore of the given system, are 1, 2, and 2, 1. Ex. 2. Solve the system x 4 + f = 17, (1) x + y = 3. (2) We first find the value of xy. Let xy = z. (3) Squaring (2), x 2 + 2 xy + y 2 = 9, (4) or cu 2 + 2/ 2 = 9 2 z. (5) Squaring (5), x 4 + 2 x 2 y 2 + i/ 4 = 81 - 36 z + 4 z 2 , (6) .Since # 4 + y 4 = 17, we have from (7), 2z 2 - 36z + 81 = 17. (8) Whence z = 16, and 2. (9) Therefore, from (3) and (9), xy = 16, (10) and . xy = 2. (11) The solutions of (2) and (10) and of (2) and (11) are readily found, and should be checked by substitution. EXERCISES IV. Solve each of the following systems : 2 ' (x s -y* = 7. - 2) 3 = 28, i ^ 4. ' x ' ' ^ ' l32(x 3 + 2 / 3 ) = 2285. 7-8] SIMULTANEOUS QUADRATIC EQUATIONS. 307 Problems. 8. Pr. The front wheel of a carriage makes 6 more revolu- tions than the hind wheel in travelling 360 feet. But if the circumference of each wheel were 3 feet greater, the front wheel would make only 4 revolutions more than the hind wheel in travelling the same distance as before. What are the circumferences of the two wheels ? Let x stand for the number of feet in the circumference of front wheel, and y for the number of feet in the circumference of hind wheel. Then in travelling 360 feet the front wheel makes 360 360 revolutions, and the hind wheel makes revolutions. x y By the first condition, = + 6. (1) x y If 3 feet were added to the circumference of each wheel, the front wheel would make revolutions, and the hind wheel 360 x -\- o revolutions. By the second condition, - = - + 4. (2) x+3 y +o Whence x = 12, the circumference of the front wheel, and y = 15, the circumference of the hind wheel. 308 ALGEBRA. [Cn. XIX EXERCISES V. 1. The square of one number increased by ten times a second number is 84, and is equal to the square of the second number increased by ten times the first. 2. The sum of two numbers is 20, and the sum of the square of the one diminished by 13 and the square of the other in- creased by 13 is 272. What are the numbers ? 3. Find two numbers such that their difference added to the difference of their squares shall be 150, and their sum added to the sum of their squares shall be 330. 4. Find two numbers whose sum is equal to their product and also to the difference of their squares. 5. The sum of the fourth powers of two numbers is 1921, and the sum of their squares is 61. What are the numbers ? 6. If a number of two digits be multiplied by its tens' digit, the product will be 390. If the digits be interchanged and the resulting number be multiplied by its tens' digit, the product will be 280. What is the number ? 7. If a number of two digits be divided by the product of its digits, the quotient will be 2. If 27 be added to the number, the sum will be equal to the number obtained by interchanging the digits. What is the number ? 8. The product of the two digits of a number is equal to one-half of the number. If the number be subtracted from the number obtained by interchanging the digits, the remainder will be equal to three-halves of the product of the digits of the number. What is the number ? 9. If the difference of the squares of two numbers be divided by the first number, the quotient and the remainder will each be 5. If the difference of the squares be divided by the second number, the quotient will be 13 and the re- mainder 1. What are the numbers ? 8] SIMULTANEOUS QUADRATIC EQUATIONS. 309 10. The sum of the three digits of a number is 9. If the digits be written in reverse order, the resulting number wil} exceed the original number by 396. The square of the middle digit exceeds the product of the first and third digit by 4. What is the number ? 11. A rectangular field is 119 yards long and 19 yards wide. How many yards must be added to its width and how many yards must be taken from its length, in order that its area may remain the same while its perimeter is increased by 24 yards ? 12. The floor of a room contains 30J square yards; one wall contains 21 square yards, and an adjacent wall contains 13 square yards. What are the dimensions of the room ? 13. A. merchant bought a number of pieces of cloth of two different kinds. He bought of each kind as many pieces and paid for each yard half as many dollars as that kind contained yards. He bought altogether 19 pieces and paid for them $ 921.50. How many pieces of each kind did he buy ? 14. The diagonal of a rectangle is 20| feet. If the lengtk of one side be increased by 14 feet and the length of the other side be diminished by 2f feet, the diagonal will be increased by 124 feet. What are the lengths of the sides of the rectangle ? 15. A certain number of coins can be arranged in the form of one square, and also in the form of two squares. In the first arrangement each side of the square contains 29 coins, and in the second arrangement one square contains 41 more coins than the other. How many coins are there in a side of each square of the second arrangement ? 16. A piece of cloth after being wet shrinks in length by one-eighth and in breadth by one-sixteenth. The piece con- tains after shrinking 3.68 fewer square yards than before shrinking, and the length and breadth together shrink 1.7 yards. What was the length and breadth of the piece ? 17. A merchant paid $ 125 for two kinds of goods. He sold the one kind for $ 91 and the other for $ 36. He thereby 310 ALGEBRA. [Cn. XIX gained as much per cent on the first kind as ho. lost on the second. How much did he pay for each kind ? 18. Two workmen can do a piece of work in 6 days. How long will it take each of them to do the work, if it takes one 5 days longer than the other ? 19. Two men, A and B, receive different wages. A earns $ 42, and B $ 40. If A had received B's wages a day, and B had received A's wages, they would have earned together $ 4 more. How many days does each work, if A works 8 days more than B, and what wages does each receive ? 20. It takes a number of workmen 8 hours to remove a pile of stones from one place to another. Had there been 8 more workmen, and had each one carried 5 pounds less at each trip, they would have completed the work in 7 hours. Had there been 8 fewer workmen and had each one carried 11 pounds more at each trip, they would have completed the work in 9 hours. How many workmen were there and how many pounds -did each one carry at every trip ? 21. A tank can be filled by one pipe and emptied by another. If, when the tank is half full of water, both pipes be left open 12 hours, the tank will be emptied. If the pipes be made smaller, so that it will take the one p'ipe one hour longer to fill the tank and the other one hour longer to empty it, the tank, when half full of water, will then be emptied in 15} Iiours. In what time will the empty tank be filled by the one rpipe, and the full tank be emptied by the other ? CHAPTER XX. RATIO, PROPORTION, AND VARIATION. RATIO. 1. The Ratio of one number to another is the relation between; the numbers which is expressed by the quotient of the first, divided by the second. E.g., the ratio of 6 to 4 is expressed by f , = f . The ratio of one number to another is frequently expressed by placing a colon between them ; as 5 : 7. The first number in a ratio is called the First Term, or the Antecedent of the ratio, and the second number the Second Term, or the Consequent of the ratio. Thus, in the ratio a : b, a is the first term, and b the second. 2. Since, by definition, a ratio is a fraction, all the proper- ties of fractions are true of ratios ; as a : b = ma : mb. 3. The definition given in Art. 1 has reference to the ratio of one number to another. But it is frequently necessary to compare concrete quantities, as the length of one line with the length of another line, etc. If two concrete quantities of the same kind can be expressed by two rational numbers in terms of the same unit, then the ratio of the one quantity to the other is defined as the ratio of the one- number to the other. 2i 35 E.g., the ratio of 2 L yards to 11 yards is 2-J-: 1^, =^- = . 1^- 16 Observe that by this definition the ratio of two concrete quantities is a number. Also that the quantities to be com- pared must be of the same kind. Dollars cannot be compared, with pounds, etc. 311 312 ALGEBRA. [Cn. XX 4. If two concrete quantities cannot be expressed by two rational numbers, integers or fractions, in terms of the same unit, they are said to be Incommensurable one to the other. Thus, if the lengths of the two sides of a right triangle be equal, the length of the hypothenuse cannot be expressed by a rational number in terms of a side as a unit, or any fraction of a side as a unit. If a side be taken as the unit, the hypothenuse is expressed by -Y/2, an irrational number. And the ratio of the hypothe- nuse to a side is >/2 : 1, = -^/2. But as was shown in Ch. XV, Art. 40, an approximate value of -y/2 can be found to any required degree of accuracy. 5. In general let P and Q be two incommensurable quan- tities. It is proved in School Algebra, Ch. XXV, that two rational numbers and m can be found, between which n n the value of the ratio P : Q lies. These two fractions differ by -. Therefore, the ratio P: Q, which lies between them, differs n .j from either of them by less than -. By taking n sufficiently :great we can make - as small as we please, that is, less than n in which r = i> and 1 -- -J- f -r + > in which r = f , are descending progressions. The flth Term of a Geometrical Progression. 21. By the definition of a geometrical progression a l = a 1? a 2 = c^r, a 3 = a> 2 r = a^, a 4 = a 3 r = c^r 3 , etc. The law expressed by the relations for these first four terms is evidently general, and since the exponent of r in each is one less than the number of the corresponding term, we have * = a ^~ 1 ' ( L ) That is, to find the nth term of a geometrical progression : Raise the ratio to a power one less. than the number of the term, and multiply the result by the first term. Ex. 1. If a l = i, r = 3, n = 5, then a 5 = \ - 3 4 = - 8 J-. This relation may also be used to find not only a n , when a 1? r, and n are given, but also to find the value of any one of the four numbers when the other three are given. Ex. 2. If % = 4, a 6 = i, n = 6, then -J- = 4 r 5 , whence r = . The Sum of a Geometrical Progression. 22. We have S n =a 1 + a 1 r+a l r 2 -\ ----- ha 1 r n - 2 +a 1 r n - 1 , (1) and rJS n = a x r -f a^ 2 H ----- f- a^"- 2 -f a^"- 1 -f- a^ n . (2) Consequently, subtracting (2) from (1), # n (l - r) = ! - a^", whence S n = - = -. (IL) 1 r r 1 20-24] GEOMETRICAL PROGRESSION. 333 Substituting a n for a^ M ~ l in (II.), we have Sn = 1 ~ !/* = g^-fll. (HI.) 1 / /* 1 The first forms of (II.) and (III.) are to be used when r < 1, the second when ? > 1. 23. Ex. 1. Given x = 3, r = 2, w = 6, to find 6 . From (II.), # 6 = 3( |j" l 1 ) = 189. Formulae (II.) and (III.) may be used not only to find S n when Oi, r, and n, or a l5 a w , and r are given, but also to find the value of any one of the four numbers when the other three are given. Ex. 2. Given S n = - 63}, a = - }, a B = 32, to find r. By (III.), - 63} = ~ + 32r , whence r = 2. 24. Formulae (I.) and (II.), or (I.) and (III.), may be used simultaneously to determine any two of the five elements, !, , r, n , n, when the three other elements are given. Ex. Given r = 2, a n = 16, S n = 31}, to find ^ and n. From (III.), 31} = 16 * 2 ~ ai , whence % = }. ^ 1 From (I.), 16 =} 2 n ~ 1 , whence n = 6. EXERCISES IV. Find the last term and the sum of the terms of each of the following geometrical progressions : 1. 3 + 6 + to six terms. 2. 2 4 -f- - to ten terms. 3. 32 1.6 + to seven terms. 4. If -f 2|-f to six terms. 2 1 5. 2 2 2 + to eleven terms. 6. -^ +^ + to n terms. 7. 1 + (1 + ) + to four terms, to n terms. 334 ALGEBRA. [Cn. XXI In each of the following geometrical progressions find the values of the elements not given : 8. % = 1, r = 4, n = 5. 9. a n = 10, r = 2, n = 4. 10. a n =96, w=4, S n =127.5. 11. r=10, n=7, # B =3,333,333. 12. Oi = 74|, n = 6, a n = 21. 13. a : = 7, r = 10, a n = 700. 14. ! = !, a B =512, # B = 1023. 15. a n = 3125, r = 5, n = 3905. 16. a! =4, r=3, n =118,096. 17. a x = 100, w = 3, = 700. 25. The Sum of an Infinite Geometrical Progression. If the number of terms in a geometrical progression is unlimited, the exact value of the sum of the series cannot be obtained. Thus, in the series 1+i+i+iH ---- without end, the sum continually increases as more and more terms are included in it. =2 -ay- 1 . And S, = 1, S 2 = 11 S 3 = 1}, ^ 4 = 1J, ^ 1000 = 2 (I) 999 ; and so on. We thus see that, although the sum of this series grows larger and larger, it does not increase without limit, but ap- proaches the value 2 more and more nearly as more and more terms are included in the sum. Evidently the sum can be made to differ from 2 by as little as we please, by taking a sufficient number of terms. We therefore call 2 the limit of the sum of the series, or more briefly, the sum of the series. The exact sum 2, however, can never be obtained. 26. In general, when r < 1, the term 0$* in the formula 24-26] GEOMETRICAL PROGRESSION. 335 decreases as n increases. It can be proved, as in the particular example, that this term can be made as small as we please, by taking n sufficient!} 7 great. Therefore, when r < 1, we take as the sum of the infinite geometrical progression. This theory can be applied to find the value of a repeating (recurring) decimal. Ex. Verify that .6 = |. We have .666 ... = T % + T ^ + ^ + -, a geometrical progression whose first term is $ and whose ratio is -^. Consequently Q _ TU" _ 6 _ 2 - 1_ 1 - TT - 3 ' *- 10 EXERCISES V. Find the sum of the following infinite geometrical progres- sions : 1. 6 + 4 + ..-. 2. 60 + 15 + ..-. 3. 10-6 + .... 4 - i + i + -- 5 - l-i-H- 6. 5-1 + .... 7. f-f + .... 8. Vf_ f . V 2 + ... t 10. l + + & i2 + .-, when 11. l+- + iH ---- , when x x 2 Find the value of each of the following repeating decimals : 12. .44.... 13. .99..-. 14. .2727-... 15. .015015-... 16. .199.... 17. 1.0909- ... 18. .122323-... 19. .2014T5475-... Verify each of the following identities : 20. 336 ALGEBRA. [Cii. XXI Geometrical Means. 27. A Geometrical Mean between two numbers is a number, in value between the two, which forms with them a geometrical progression. E.g., + 2, or 2, is a geometrical mean between 1 and 4. Let G be the geometrical mean between a and b. Then by definition of a geometrical progression, ^= -|; whence G = V(6). TJiat is, the geometrical mean between two numbers is the square root of their product. Ex. Find the geometrical mean between 1 and -J. We have 28. Geometrical Means between two numbers are numbers, in value between the two, which form with them a geometrical progression. E.g., 4 and 16 are two geometrical means between 1 and 64; and 2, 4, 8, 16, 32 are five geometrical means between 1 and 64. Ex. Insert five geometrical means between 1 and 729. We have a l = 1, n = 7, a, = 729. Therefore 729 = r 6 , or r = 3. The required means are : 3, 9, 27, 81, 243. EXERCISES VI. Insert a geometrical mean between 1. 2 and 8. 2. 12 and 3. 3. -J- and yfy. 4. ^/a and V (2 a). 5. 75 m 3 and 3 mn*. 6. and . q p 7. (a-6) 2 and(a+6) 2 . 8. (a 2 +1) (a 2 -!)- 1 and \ (a 4 -!). 9. Insert five geometrical means between 2 and 1458. 10. Insert seven geometrical means between 2 and 512. 27-30] GEOMETRICAL PROGRESSION. 337 11. Insert six geometrical means between 3 and 384. 12. Insert six geometrical means between 5 and 640. 13. Insert nine geometrical means between 1 and Problems. 29. Pr. A farmer agrees to sell 12 sheep on the following terms : he is to receive 2 cents for the first sheep, 4 cents for the second, 8 cents for the third, and so on. How much does he receive for the twelfth sheep, and how much for the 12 sheep, and what is the average price ? We have a x = 2,n = 12, r = 2. Then a 12 == 2 x 2 11 = 2 12 = 4096. And # 12 = ~ = 2 x 4095 = 8190. That is, he receives 4096 cents, or $ 40.96, for the twelfth sheep, and 8190 cents, or $ 81.90, for the 12 sheep. The average price is ^&, = $ 6.82 J. 1^ 30. In many examples the elements necessary for determin- ing the element or elements directly from (I.)-(IIL) are not given, but in their place equivalent data. Ex. The fifth term of a G. P. is 48, and the eighth term is 384. Find the first term and the ratio. From (I.), 48 = a,r 4 , and 384 = a,r 7 ; whence r 8 = 8, or r = 2. Therefore a x = 3. Or, we could have regarded 48 as the first term and 384 as the last term of a progression of four terms. Then by (I.), 384 = 48 r 3 , whence r = 2 as before. EXERCISES VII. 1. .The first term of a G. P. of six terms is 768, and the last term is one-sixteenth of the fourth term. What is the sum of the six terms of the progression? 338 ALGEBRA. [Cn. XXI 2. The first term of a G. P. of ten terms is 3, and the sum of the first three terms is one-eighth of the sum of the next three terms. Find the elements of the progression. 3. The twelfth term of a G. P. is 1536, and the fourth term is 6. What is the ratio, and the sum of the first eleven terms ? 4. In a G. P. of eight terms, the sum of the first seven terms is 4441, and is to the sum of the last seven terms as 1 to 2. Find the elements of the progression. 5. The sum of the first four terms of a G. P. is 15, and the sum of the terms from the second to the fifth inclusive is 30. What is the first term, and the ratio ? 6. Find the elements of a G. P. of six terms whose first term is 1, and the sum of whose first six terms is 28 times the sum of the first three terms. 7. The sum of the first three terms of a G. P. is 21, and the sum of their squares is 189. What is the first term ? 8. The product of the first three terms of a G. P. is 216, and the sum of their cubes is 1971. What is the first term, and the ratio ? 9. If the numbers 1, 1, 3, 9 be added to the first four terms of an A. P., respectively, the resulting terms will form a G. P. What is the first term, and the common difference of the A. P. ? 10. A G. P. and an A. P. have a common first term 3, the difference between their second terms is 6, and their third terms are equal. What is the ratio of the G. P., and the com- mon difference of the A. P. ? 11. Show that, if all the terms of a G. P. be multiplied by the same number, the resulting series will form a G. P. 12. Show that the series whose terms are the reciprocals of the terms of a G. P. is a G. P. 13. Show that the product of the first and last terms of a O. P. is equal to the product of any two terms which are .equally distant from the first and last terms respectively. 30-34] HARMONICAL PROGRESSION. 339 14. A merchant's investment yields him each year after the first, three times as much as the preceding year. If his investment paid him $ 9720 in four years, how much did he realize the first year and the fourth year ? 15, Given a square whose side is 2 a. The middle points of its adjacent sides are joined by lines forming a second square inscribed in the first. In the same manner a third square is inscribed in the second, a fourth in the third, and so on indefinitely. Find the sum of the perimeters of all the squares. HARMONICAL PROGRESSION. 31. A Harmonical Progression (H. P.) is a series the recipro- cals of whose terms form an arithmetical progression. E-9-, l + i + i + i+-- is a harmonical progression, since 1 + 2 + 3 + 4+.. is an arithmetical progression. Consequently to every harmonical progression there corre- sponds an arithmetical progression, and vice versa. 32. Any term of a harmonical progression is obtained by finding the same term of the corresponding arithmetical pro- gression and taking its reciprocal. Ex. Find the eleventh term of the harmonical progression 4,2,|,.... The corresponding arithmetical progression is i *> i, -, and its eleventh term is J ^-. Therefore the eleventh term of the given progression is T 4 T . 33. No formula has been derived for the sum of n terms of a harmonical progression. 34. A Harmonical Mean between two numbers is a number,, in value between the two, which forms with them a harmonical progression. 340 ALGEBRA. [Cn. XXL E.g., f is a harmonical mean between ^ and f . Let H stand for the harmonical mean between a and ft, then is an arithmetical mean between - and Consequently M 1 a b 2ab -, or H= -. -Ll Ex. Insert a harmonical mean between 2 and 5. We have g = 2 *2x5 = 20. 2 + 5 7 35. Harmonical Means between two numbers are numbers, in value between the two, which form with them a harmonical progression. E.g., f, 1, J, |, i are five harmonical means between 3 and f . Ex. Insert four harmonical means between 1 and 10. We have first to insert four arithmetical means between 1 and Y 1 ^, and obtain TTP ft? If? if The required harmonical means are therefore 50 50 50 50 if* fib ft> ft- Problems. 36. Pr. 1. The geometrical mean between two numbers is -^, and the harmonical mean is fi What are the numbers? Let x and y represent the two numbers. Then V or y = i > (1) and = , or 5 xy = x + y. (2) x -\- y o Solving (1) and (2), we obtain x = 1 , y = J, and # = ^, y = 1. EXERCISES VIII. Find the last term of each of the following harmonical pro- gressions : 1. 1+ + 1+.. . to 8 terms. 2. ++ + ... to 15 terms. 3. 2-2-f ---- to 11 terms. 4. 8-f- T * T ---- to 16 terms. 34-36] HARMONICAL PROGRESSION. 341 5. i + -f H ---- to 25 terms. a 2 a 3 a Find the harmonical mean between 7. 2 and 4. 8. - 3 and 4. 9. } and 1. 10. _J_ and -- LJ 11. H* and a; 1 x + 1 a-j-6 a 6 12. Insert 5 harmonical means between 5 arid i. 13. Insert 10 harmonical means between 3 and 1. 14. Insert 4 harmonical means between 7 and ^. 15. If a be the harmonical mean between b and c, prove that a 6_g 6 c c 16. The arithmetical mean between two numbers is 6, and the harmonical mean is - 3 ^. What are the numbers ? 17. If one number exceeds another by two, and if the arith- metical mean exceeds the harmonical mean by ^, what are the numbers ? 18. The seventh term of a harmonical progression is y 1 ^-, and the twelfth term is ^ What is the twentieth term ? 19. The tenth term of a harmonical progression is J, and the twentieth term is . What is the first term ? CHAPTER XXII. THE BINOMIAL THEOREM FOR POSITIVE INTEGRAL EXPONENTS. 1. The expansions of the powers of a binomial, from .the third to the fourth inclusive, were given in Ch. XIII., Arts. 7-8, and the laws governing the expansion of these powers were stated. As yet, however, we cannot infer that these laws hold for the fifth power without multiplying the expansion of the fourth power by a + b ; nor for the sixth power without next multiplying the expansion of the fifth power by a + b; and so on. If, however, we prove that, provided the laws hold for any particular power, they hold for the next higher power, we can infer, without further proof, that because the laws hold for the fourth power, they hold also for the fifth; then that because they hold for the fifth, they hold also for the sixth, and so on to any higher power. 2. If the laws (i.)-(vi.) hold for the rth power, we have 1 2i 3 Notice that only the first four terms of the expansion are written. But it is often necessary to write any term (the fcth, say) without having written all the preceding terms. To derive this term, observe that the following laws hold for each term of the expansion : (i.) The exponent of b is one less than the number of the term (counting from, the left). Thus in the first term we have b l ~ l = 6 = 1 ; in the second, 6 2 - 1 = b ; in the tenth, b l ~ l = b 9 ; and in the &th term, ft*- 1 . 342 1-3] BINOMIAL THEOREM. 343 (ii.) The exponent of a is equal to the binomial exponent less the exponent of b. Thus, in the first term we have a r ~ = a r , in the second, a r ~ l ; in the tenth, a r ~ 9 ; and in the &th term, a r ~ (k ~ l) , = a r ~ k+l . (iii.) The number of factors (beginning with 1 and increasing by 1) in the denominator of each coefficient, and the number of factors (beginning with r and decreasing by 1) in the numerator of each coefficient, is equal to the exponent ofb in that term. Thus, in the coefficient of the second term the denominator is 1 and the numerator is r; in that of the third term the denominator is 1 2 and the numerator is r(r 1); in the tenth term the denominator is 1 2 9 and the numerator is r(r 1) ... (r 8); and in the &th term the denominator is 1 2 3 (k 1), and the numerator is Therefore the kth term in the expansion of (a + b) r is r ( r _ 1) ( r _ 2) ... (r - k + 2) ^-fc+i^-i. In like manner, any other term can be written. Thus, the (k l)th term is -a'- k+2 b k - 2 . 1.2-3 (fc-2) 3. We can now prove that, if the laws (i.)-(vi.) hold for (a + b) r , they also hold for (a + b) r+l ; that is, if they hold for any power they hold for the next higher power. Assuming, then, that the laws hold for (a -f 6) r , we have -i , . 1. 2 V 3 ..(* -2) (&-,!) The first three terms of the expansion are written, then all terms are omitted, except the (k l)th and the Jfcth. 344 ALGEBRA. [Cn. XXII Multiplying the expansion of (a -f- b) r by (a + 6), we obtain : (a -f- b) r+l = a r+l + ra r b + r ^~ 1 ) a r ~ l b 2 -f f-r(r-l)...(r-fc + 2) r(r-l)...(r-fe+3)-| 1. 2 ...(A;-1) 1.2...(fc-2) But and - fe + 3) 1.2-..(A;-1) 1-2- (fc -2) = r(r - 1) -. (r - ft + 2) + r(r - 1) (r - k + 3) (ft - 1) 1.2...(/b-l) r ( r - 1) ... ( r - fe + 3) (r - fe + 2 + ft - 1) Therefore, d and ra 2 > m lt or when hm t < d and m 2 < m^ A positive result means that the problem is possible with the assumption made; i.e., that the couriers meet at a point to the right of B. (ii.) A Negative Result. The result will be negative either when hm l > d and m 2 < m 1} or when hm^ < d and m 2 > ra^ Such ALGEBRA. [Cn. XXIII a, result shows that the assumption that the couriers meet to the right of B is untenable, since, as we have seen, in that case the result is positive. That under the assumed conditions the couriers can meet only at some point to the left of B can also be inferred from the following considerations, which are independent of the negative result : If lim^ > d, the first courier has passed B when the second courier is seen at that station ; that is, the second courier is behind the first at that time. And since also w 2 < nil) the first courier is travelling the faster, and must therefore have overtaken the second, and at some point to the left of B. On the other hand, if hm^ > d, the first courier has not yet reached B when the second is seen at that station ; that is, the first courier is behind the second at that time. And since also w 2 > m 1? the second courier is travelling the faster, and must therefore have overtaken the first, at some point to the left of B. Similar reasoning could have been applied in (i.). (iii.) A Zero Result. A zero result is obtained when /iWj = d, and m 2 is not equal to ?% ; that is, the meeting takes place at B. This is also evident from the assumed conditions. For the first courier reaches B h hours after he was seen at A ; and since the second courier is seen at B h hours after the first was seen at A< the meeting must take place at B. (iv.) Indeterminate Result. An indeterminate result is ob- tained if hnii = d, and m 2 = m^ In this case every point of the road can be regarded as their place of meeting. For the first courier evidently reaches B at the time at which the second courier is seen at that station ; and since they are trav- elling at the same rate, they must be together all the time. The problem under these conditions becomes indeterminate. (v.) An Infinite Result. An infinite result is obtained when hm l ^ d, and m 2 = ?%. In this case a meeting of the couriers is impossible, since both travel at the same rate, and when the second is seen at B the first either has not yet reached B or has already passed that station. 18] LIMITS. 357 An infinite result also means that the more nearly equal mi and m. 2 are, the further removed is the place of meeting. EXERCISES II. Solve the following problems, and interpret the results : 1. In a number of two digits, the digit in the tens 7 place exceeds the digit in the units' place by 5. If the digits be interchanged, the resulting number will be less than the origi- nal number by 45. What is the number ? 2. The sum of the first and third of three consecutive even numbers is equal to twice the second. What are the numbers ? 3. A father is 26 years older than his son, and the sum of their ages is 26 years less than twice the father's age. How old is the son ? 4. In a number of two digits, the digit in the units' place exceeds the digit in the tens' place by 4. If the sum of the digits be divided by 2, the quotient will be less than the first digit by 2. What is the number? Discuss the solutions of the following general problems : 5. What number, added to the denominators of the fractions - and -, will make the resulting fractions equal ? b d 6. Having two kinds of wine worth a and b dollars a gallon, respectively, how many gallons of each kind must be taken to make a mixture of n gallons worth c dollars a gallon ? 7. Two couriers, A and B, start at the same time from two stations, distant d miles from each other, and travel in the same direction. A travels n times as fast as B. Where will A overtake B ? CHAPTER XXIV. UNDETERMINED COEFFICIENTS. CONVERGENT AND DIVERGENT SERIES. 1. The infinite series is a decreasing geometrical progression, whose ratio is -| . It follows from Ch. XXI., Art. 26, that the sum of this series approaches a definite finite value as the number of terms is indefinitely increased. Let S n = 1 + | + | + Jy H ---- to n terms. Then, by Ch. XXI., Art. 26, as n increases indefinitely. By actual computation, we obtain S, = 1, S, = If, 3 = 2i S 4 = 2^, etc. These sums approach 3 more and more nearly, as more and more terms are included. This infinite series may therefore be regarded as having the finite sum 3. But the sum of the series increases beyond any finite number, as the number of terms increases indefinitely. 2. The examples of the preceding article illustrate the fol- lowing definitions : 358 1-4] CONVERGENT AND DIVERGENT SERIES. 359 Any infinite series is said to be Convergent, when the sum of the first n terms approaches a definite finite limit, as n increases indefinitely. An infinite series is said to be Divergent when the sum of the first n terms increases numerically beyond any assigned num- ber, however great, as n increases indefinitely. 3. It was shown in Ch. XXI., Art. 26, that, when r < 1, the sum of the series a + ar -f- cut* + approaches the definite finite value as the number of 1 r terms is indefinitely increased. Therefore, any decreasing geometrical progression is a con- vergent series. 4. Infinite series arise in connection with many mathematical operations. Thus, for example, if the division of 1 by 1 x be continued indefinitely, we obtain as a quotient the infinite series When x is numerically less than 1, this series is a decreasing geometrical progression, as in Art. 1. Therefore, by the preced- ing article it is convergent. When x = 1, the series becomes and is evidently divergent. When x 1, we have 1 _ 1 + 1 _! + .... The sum of n terms of this series is -1-1 or 1, according as n is odd or even. The series is said to oscillate and is neither convergent nor divergent. When x is numerically greater than 1, we have, by Ch. XXI., Art, 22 (II.), o 1 aJ" 360 ALGEBRA. [Cn. XXIV By taking n sufficiently great this expression can be made to exceed numerically any number, however great. Therefore the series is divergent. Thus, when x = 2, the series becomes The sum of this series can evidently be made greater than any assigned number, however great. But when x = 2, the value of the fraction - is , = 1. We therefore conclude that the infinite series approaches in value the fraction - - for all values of x be- 1 x tween 1 and + 1. Conversely, we may look upon the series as the expansion of the fraction for all values of x between these limits, but for no other values of x. In general, an infinite series, no matter how obtained from a given expression, can be regarded as the expansion of that expression only when the series is convergent. This fact should be kept in mind, without further emphasis, in all the expansions that we shall derive in this chapter. 5. If an infinite series a + a^x + a>$? + a$? + be con- vergent for values of x greater than 0, the sum of the series approaches a , as x approaches 0. Let a + a^ + a#? + a 3 or 3 -\ ---- = a + xS wherein S 1 = a l + a& + a 3 x 2 + . Evidently, if the given series is convergent, that is, if a + xSi is finite, then Si is finite. Therefore, by Ch. XXIII. , Art. 13, xSi = 0, when x = 0. Consequently a + aix + a$? + , a + X S, = o> when x = 0. 6. If two integral series, arranged to ascending powers of x, be equal for all values of x which make them both convergent, the coefficients of like powers of x are equal. 4-8] EXPANSION OF FRACTIONS. 361 Let a + a^x -f- a.-p? -f- = b Q 4- b& for all values of # which make the two series convergent. Then the sums of the two series approach equal limits when x = 0. But, by the preceding article, the sum of the one series approaches a , that of the other b Q ; consequently a 6 , and ! + 0*1?+ "- = b]X + b$? + -. Since these two series are convergent for all values of x for which the original series are convergent, they are equal for values of x other than zero, and the last equation may be divided by x. Hence % + <*>& -f a$? H ---- = &i + b 2 x + b s x 2 -\ ---- ; and, as before, a^ = b^ and a^x + a^x 2 -\ ---- = b& + b$? + ~>. In like manner, we can prove a 2 = b 2 , a s = b 3 , etc. 7. Evidently the principle of the preceding article holds with greater reason if .either or both of the series be finite, i.e., have a limited number of terms. There is, in this case, no question of convergence of the finite series. The series must be equal for all values of x, if they be both finite ; or, if one be infinite, for all values of x which make that series convergent. EXPANSIONS OF RATIONAL FRACTIONS. 8. We shall now give a method of expanding a fraction in an infinite series, without performing the actual division. Ex. 1. Expand 2 ~ x > 1 + x x 2 in a series, to ascending powers of x. We equate the fraction to a series of the required form, in which the coefficients of the different powers of x are un- known, or undetermined. Assume 2 ~ x = A + Bx + Cx 2 + Dx* + Ex 4 , 1 + x x* whence A, B, (7, D, E, are constants to be determined. 362 ALGEBRA. [Cn. XXIV learing the equation of fractions, we obtain 2-x=A+B A x+C + B -B - C In this work the powers of x in the terms of the second and third partial products are' omitted, it being understood that the letters remaining are the coefficients of the powers of x just above in the first partial product. Thus the coefficient of x is A + B, etc. The series on the right is infinite ; that on the left may be regarded as an infinite series with zero coefficients of all powers of x higher than the first. By Art. 6, we have B + A = ^-1, whence B^ 3; C+B-A = 0, whence (7-5; C-B = Q, whence Z> = -8; D-C=Q, whence E = 13; etc., etc. Hence, substituting these values of A, B, (7, D, assumed series, we have 2-x in the 1+x-x 2 We can assume that this series is equivalent to the fraction only when x has such values as make it convergent. Let the student compare tliis result with that obtained by division. In fact, the latter method of expanding a fraction is to be preferred when only a few terms are wanted. But the successive coefficients, after a certain stage, may be computed with great facility by the method of undetermined coefficients. A moment's inspection of the preceding work will convince the student that the coefficient Z>, and all which follow it, are each connected with the two immediately preceding coefficients by a definite relation. Thus, D+C- B = Q, E + D- C=0, F+ E-D= 0, etc. 8] EXPANSION OF FRACTIONS. 363 In assuming as the expansion of a rational fraction an infinite series of ascending powers of x, it is usually necessary tirst to determine with, what power the series should com- mence. This is done by division, when both numerator and denominator are arranged to ascending powers of x. In fact, this step also determines completely the first term of the series. Ex.2. Expand f in a series to ascending powers of x. The first term in the expansion, obtained by division, is evidently ^x~ 2 . We therefore assume ~ x = i x~ 2 -\-Bx~ 1 4- C -f- Dx - Clearing of fractions, we obtain B 3C -B -C By Art. 6, we have 1 = 1, 3B ^ = 1, whence B = 3C-B= 0, whence C= 3 D C = 0, whence D= etc., etc. Hence ' EXERCISES I. Expand the following fractions in series, to ascending powers of x, to four terms : 1 1 2 3 3 6 l2x ' l + 3x ' 3-x 2 4 . 5 ' l-x ' 7. . a ~ 9. 364 10. x _ 3* 11. ALGEBRA. 2-x [Cii. XXIV 13. 2+*-3< 14. ^=- 3 -^ 3 - x + 3 a? 12. 15. EXPANSION OF SURDS. 9. Ex. Expand V( 1 -^ 2 + 2 ^ in a series, to ascending powers of x. Assume ^/(l _ or 2 + 2 a 3 ) = 1 + Bx + Os 2 + Dx 3 + Squaring both sides of the equation, we have 2BC + 2BD Equating coefficients, 1 = 1. 25 = 0, whence 5 = 0; 2 C + B 2 = - 1, whence C = - J ; 2 Z) + 2 5(7 = 2, whence Z) = + 1 ; 2 .E + 2 BD + C* = 0, whence E = %\ etc. Hence V(l - *? + 2a^)= 1 - EXERCISES II. Expand the following expressions in series, to ascending powers of x, to four terms : i. v( 1 + a )- 2 - V( 2 - 2 ^)- 3 - 4. ^(4 - 2 a; + a 2 ). 5. V( 5 + 3a; + 9 ^)- 6 - PARTIAL FRACTIONS. 10. It is frequently desirable to separate a rational alge- braical fraction into the simpler (partial) fractions of which it is the algebraical sum. 2x 1 1 E.g., 1-x 2 1-x 8-9] PARTIAL FRACTIONS. 365 The process of separating a given fraction into its partial fractions is, therefore, the converse of addition (including sub- traction) of fractions ; and this fact must guide us in assuming the forms of the partial fractions. We shall also assume that the degree of the numerator is at least one less than that of the denominator. A fraction whose numerator is of a degree equal to or greater than that of its denominator can be first reduced by division to the sum of an integral expression and a fraction satisfying the above condi- tion. The latter fraction will then be decomposed. The denominators of the partial fractions can be definitely assumed. For they are evidently. those factors whose lowest common multiple is the denominator of the given fraction. But there is one case of doubt ; namely, when a prime factor is repeated in the denominator of the given fraction. E.g., 6-2X 2 = 3 2_ 1 (1 - x) 2 (1 + x) 1 - x (1 - x) 2 1 + x ' 3 + x 2 2 1 We could not have decided, in advance, whether either of the two given fractions is the sum of two or of three partial fractions. There must necessarily be a partial fraction having (1 x) 2 as a denominator, since, otherwise, the L. C. M. of the denominators would not contain the prime factor 1 x to the second power. But it cannot be determined, in advance, whether there is a partial fraction having 1 x as a denomi- nator. In such cases, therefore, it is advisable to make provision for all possible partial fractions by assuming as denominators all repeated factors to the first power, second power, etc. The numerators of partial fractions thereby assumed, which should not have been included, will acquire the value zero from the subsequent work, so that those fractions drop out of the result. 366 ALGEBRA. [Cn. XXIV The numerators of the partial fractions must be assumed with undetermined coefficients. Since the numerator of the given fraction is, by the hypothesis, of degree at least one less than the denominator, the same must be true of each partial fraction. We therefore assume, for each numerator, a complete rational integral expression with undetermined coefficients of degree one lower than the corresponding denominator. If any term in the assumed form of the numerator should not have been included, its coefficient will prove to be zero. An exception to this principle occurs when the denominator of the partial fraction is the second or higher power of a prime factor, as, (1 x) 2 . In that case the numerator is assumed as it would be according to the above principle if the prime factor occurred to the first power only. We may briefly restate the above principles : Separate the denominator of the given fraction into its prime factors. Assume as the denominator of a partial fraction each prime factor ; in particular, when a prime factor enters to the nth power, assume that factor to the first power, second power, and so on, to the nth power, as a denominator. Assume for each numerator a rational integral expression, with undetermined coefficients, of degree one lower than the prime factor in the corresponding denominator. Let us first decompose the two fractions which we have used to illustrate the theory. (1 - x) 2 (l + x) 1 - x (1 - xf 1 + x Since the prime factor in the denominator of each partial fraction is of the first degree, each numerator is assumed to be of the zeroth degree. Clearing the equation of fractions, we have 6 - 2 or 2 = A(l - x)(l + x) + B(l + ) + C(l - x) 2 9] PARTIAL FRACTIONS. 367 Since this equation must be true for all values of x, we have B-2C= 0, 1 Whence^l = 3, 5 = 2, O^l. A + B+C= 6. j Ex.2. 3 + f. =-A_ + 7 -A_ H _ # (1 - x)\l + x) 1 - x (1 - x) 2 1 + a The forms of the partial fractions are assumed the same as in Ex. 1. ' We have 3 + x* = (-A+C)x* + (B-2C)x + A + B+C, and then A + C = 1, 5-20=0, Therefore 3 -f Whence ^1 = 0, B = 2 } (7 = 1. When the factors of the denominator of the given fraction are of the first degree, as in Exs. 1 and 2, the work may be shortened. Begin with the equation 6 - 2 tf = A(l - x) (1 + x) + B (1 + x) + C (1 - x) 2 , of Ex. 1. Since this equation is true for all values of x, we may substitute in it for x any value we please. Let us take such a value as will make one of the prime factors zero. Substituting 1 for x, we obtain 4 = 25, whence B = 2. Next, letting x = 1, we have 4 = 40, whence O=l. There is no other value of x which will make a prime factor zero, but any other value, the smaller the better, will give an equation in which we may substitute the values of B and Q already obtained. * ALGEBRA. [Cn. XXIV Letting x = 0, we obtain 6 = A + B + (7, whence A = 3. The same method can be applied to Ex. 2. Ex 3 a?2 a? + 3 = or* -# + 3 = .4 a 8 -! a-laj' + aj + l o?-l In this example, the one prime factor being of the second degree, we assume the corresponding numerator to be a com- plete linear expression. Clearing of fractions, we have x 2 - x + 3 = A (x 2 + x + 1) + (Bx + C) (aj - 1) = (A + J3)aj* + (A - B + C)x + -4 - C. Equating coefficients of like powers of x, we obtain -4 + ^ = 1, A-B+C = -1, A-C = 3; whence, ^1 = 1, 5-0, C = - 2. Or, we might have used the second method, beginning with a- 2 - a; + 3 = 4 (a 2 + a; + 1) + (jB.e + 0) (o> - 1). Letting a; = 1, we obtain 3 = 3-4, whence A = l. Since no other value of x will make a factor vanish, we take any simple values. When x =0, we have 3 = A - C, whence G = -2. Finally, letting x = 1, we have 5 = A-2B-2C, whence B = 0. * - x + 3 1 2 Therefore x-1 or' 2-2a + 4ft 2 = Ax + B Cx + D E " /-i O\ O s -I \ -4 . > I /- . O\ O (1 + x 2 ) 2 (l -x) 1 + x 2 (1 + x 2 ) 2 1 - a; The prime factors in the denominator of the first two partial fractions being of the second degree, expressions of the first degree are assumed as numerators. 9] PARTIAL TRACTIONS. 369 Clearing of fractions, we have 2_2# + 4cc 2 = (Ax + B) (I + x 2 ) (1 - a?) + (Cx + D) (1 - x) + E (1 + z 2 ) 2 - ( - A + jBJ) a?* + (A - B) x 3 + ( - ^L + - C + 2 J0) x 2 + (A-B+C-U)x+(B + D + E). Equating coefficients of like powers of x, we obtain whence, A = l, B=l, C=-2, D = 0, E = l. Therefore EXERCISES III. Separate the following fractions into partial fractions : 1 6 7 (a>-2)(l-2aO 3x 1 " (5 + 3x)(, + 4) 1 l- (, + 3)(*-2) ^ ^ fi 7 T-y 8 ar 4 9 or 9 x -\- ^ x 1^ 10 3 a? + ^ 9a^-16 ' (* 2 -l)(*-2) U< 6(aj-9)(aj-3) 13 , 2 + 5^ + 10 L " (* + !)(* -I) 2 o a; (a; + o) (^ + l)(x + 2)(x + 3) 15 3 ~ X 1 1 2 17 . 1R > ^ (.-I) 3 ' 19 ^ + 1 ' 3?-l ' X*+l r 21 19> ?~1 CHAPTER XXV. THE BINOMIAL THEOREM FOR ANY RATIONAL EXPONENT. 1. From Ch. XXII., Art. 4, we have when n is a positive integer. In this case, as we have seen, the series ends with the n -f- 1th term. But if n be not a posi- tive integer, the expression on the right of (1) will continue without end, since no factor of the form n k can reduce to 0. Therefore the series will have no meaning unless it be con- vergent. 2. It is proved in Elements of Algebra, Ch. XXXI., that this series is convergent when x lies between 1 and + 1 ; and, in Ch. XXXII., that when the series is convergent, it is the expansion of (1 -f x) n . Ex. This infinite series can be taken as the expansion of (1 -f a?)^, =V(1 + ic ) on ty when x is numerically less than 1. 3. Expansion of (a + 6)". We have (1) and ( a + ft). = i += i + (2) 370 1-4] THE BINOMIAL THEOREM. 371 When b is numerically less than a, a and, by (1) above, .. (3) In a similar way it can be shown that, when a is numerically less than 6, (a + b) n = b n + nb*~ l a + M ^~ 9 1 V-*a* + .... (4) Notice that when n is a fraction or negative, formula (3) or (4) must be used according as a is numerically greater or less than 6. 4. Ex. 1. Expand - - -to four terms. S/(a-46 2 ) If we assume a > 4 6 2 , by (3), Art. 3, we have = 32 5 4 896 6 6 9 a 2 ^/a 81 a?^/a If a < 4 b 2 , we should have used (4), Art. 3. Any particular term can be written as in Ch. XXII., Art. 9. / 1 \ 2 Ex. 2. The 6th term in the expansion of f x \ is \ ^J (-2) (-3) (-4) (-5) (-6) 2 _ 5 1.2.3.4.5 372 ALGEBRA. [Cn. XXV 5. Extraction of Roots of Numbers. Ex. Find ^/VI to four decimal places. We have = 4 (1 + .03125 - .00048 + .00001 ---- ) = 4 x 1.03078 = 4.12312. Therefore -^/17 = 4.1231, to four decimal places. EXERCISES. Expand to four terms : 1. (1 + a)*. 2. (1-x)- 1 . 3. (1 - )- 3 . 4. (1 + x^. 5. (1 + a)- 4 . 6. (1 ?/ 2 )~ 2 . 7. (x* + y)-i. 8. (x y 2 )-*. 9. (27 + 5z)i 10. (8a 3 -36)i 11. (3 + 2 a)*. 19 /'^ /-2 Q 7i3\"3 A<&. ( O Ct O ) . 13 1 14. 1 C 2 -6 2 ) ^(a 3 -6) v \ Find the 16. 4th term of (1-2 z)i 17. 6th term of (1 18. 5th term of (a;* - ar 1 /)"^. 19. 8th term of (a 3 ^/b 2 20. fc 5th term of (1 + 21. 2 fcth term of T^ 2 - Find to four places of decimals the values of : 22. ^/5. 23. ^/27. 24. CHAPTER XXVI. LOGARITHMS. 1. It is proved in Elements of Algebra, Ch. XXXV., that a, value of x can always be found to satisfy an equation of the form 10* = n, wherein n is any real positive number. E.g., when n = 10, x = 1, when n = 100, x = 2, when n = 1000, x = 3, etc. When n is not an integral power of 10, the value of x is irra- tional, and can be expressed only approximately. Thus, when n = 24, the corresponding value of x has been found to be 1.38021*", to five decimal places; or 10 1.38021- = 24. A value of x is called the logarithm of the corresponding value of n, and 10 is called the base. In general, a value of x which satisfies the equation b* = n, is called the logarithm of n to the base b. E.g., since 2 3 = 8, 3 is the logarithm of 8 to the base 2 ;. since 10 2 = 100, 2 is the logarithm of 100 to the base 10. The Logarithm of a given number n to a given base b is, there- fore, the exponent of the power to which the base b must be raised to produce the number n. 2. The relation b x a is also written x = Iog 6 a, read x is the logarithm of a to the base b. Thus, 2 3 = 8 and 3 = log 2 8, 10 2 = 100 and 2 = log w 100, are equivalent ways of expressing one and the same relation. 373 374 ALGEBRA. [Cn. XXVI 3. The theory of logarithms is based upon the idea of representing all positive numbers, in their natural order, as powers of one and the same base. Thus, 4, 8, 16, 32, 64, etc., can all be expressed as powers of a common base 2 ; as 4 = 2 2 , 8 = 2 3 , 16 = 2 4 , etc. Since, also, all the numbers intermediate between those given above can be expressed as powers of 2, the exponents of these powers are the logarithms of the corresponding numbers. The logarithms of all positive numbers to a given base form what is called a System of Logarithms. The base is then called the base of the system. It follows from Art. 1, that any positive number except 1 may be taken as the base of a system of logarithms. EXERCISES I. Express the following relations in the language of logarithms : 1. 5 2 = 25. 2. 2 5 = 32. 3. 7 3 = 343. 4. 3 7 = 2187. Express the following relations in terms of powers : 5. Iog 3 81 = 4. 6. Iog 9 81 = 2. 7. Iog 4 64 = 3. 8. Iog 2 64 = 6. Determine the values of the following logarithms : 9. Iog 2 32. 10. Iog^l28. 11. Iog 2 .5. 12. Iog 2 .25. 13. Iog 4 64. 14. log^S. 15. Iog 2 .125. 16. Iog 5 .04. To the base 16, what numbers have the following logarithms ? 17. 0. 18. i. 19. - 2. 20. f . 21. - 1. Principles of Logarithms. 4. The logarithm of 1 to any base is 0. For 6 = 1, or log, 1 = 0. 5. The logarithm of the base itself is 1. For b l = b, or log* 6 = 1. 6. The logarithm of a product is equal to the sum of the loga- rithms of its factors ; or, log& (m x ri) = Iog6 m -f log& n. 3-7] LOGARITHMS. 375 Let Iog 6 m = x and Iog 6 n = y ; then b x = m and 6 y = n, and therefore, mn = b x b y = 6 I+y . Translated into the language of logarithms, this result reads Iog 6 (rari) = x + y. But x = Iog 6 m and y = Iog 6 ?i, and consequently Iog 6 (mw) = Iog 6 m + Iog 6 n, for all positive values of b. This result may be readily extended to a product of any number of factors. For, log b (mnp) = log b (mn) + log s p = Iog 6 m + Iog 5 n + log b p. And, in like manner, for any number of factors. E.g. Given Iog 2 32 = 5, and Iog 2 64 6 ; what is the loga- rithm of 2048 to the base 2 ? Since 2048 = 32 . 64, we have Iog 2 2048 = Iog 2 32 + Iog 2 64 = 5 + 6 = 11 7. The logarithm of a quotient is equal to the logarithm of the dividend minus the logarithm of the divisor; or, Iog 6 (m + n)= log b m - Iog 6 n. Let Iog 6 m = x and Iog 6 n = y; then b x = m and b y = n, and therefore m -i- n = b x -5- b y = b x ~ y . In the language of logarithms the last equation is Iog 6 (m-r-ri)=x y=. Iog 6 m Iog 5 n, for all positive values of b. E.g. Given Iog 3 3 = 1 and Iog 3 21 87 = 7, what is the loga- rithm of 729 to the base 3 ? Since 729 = ^-^, we have Iog 3 729 = Iog 3 2187 - Iog 3 3 = 7 - 1 = 6. 376 ALGEBRA. [Cn. XXVI 8. Both m and n may be products, or the quotient of two numbers. E.g., lo glo |^-f = lo glo (4 x 5) - lo glo (9 x 8) / X o = Iog 10 4 + Iog 10 5 - Iog 10 9 - Iog 10 8. 9. The logarithm of the reciprocal of any number is the oppo- site of the logarithm of the number. For, Iog 5 - = Iog 6 1 - Iog 6 n n = Iog 6 n, since Iog 6 1 = 0. E.g., Iog 2 4 = 2, and Iog 2 J = - 2. 10. TJie logarithm of any power, integral or fractional, of a number is equal to the logarithm of the number multiplied by the exponent of the power ; or log(m^) = p logm. Let Iog 6 m = x, then b x = m. Raising both sides of the last equation to the pfh power, we have b px = m p , or Iog 6 (m p ) = px =p Iog 6 m. E.g., If Iog 5 25 = 2, what is Iog 5 (25) 3 ? We have Iog 5 (25) 3 = 3 log, 25 = 3 x 2 = 6. 11. When the exponent is a positive fraction whose numera- tor is 1, this principle may be conveniently stated thus : The logarithm of a root of a number is the logarithm of tlic number divided by the index of the root. For, log, (m) = - log m = l2iL^:. q q E.g., If Iog 7 2401 = 4, what is Iog 7 V2401 ? We have Iog 7 V2401 = - Iog 7 2401 = -.4 = 2. 8-12] LOGARITHMS. 377 EXERCISES II. Express the following logarithms in terms of log a, log 6, log c, and log d : Express the following sums of logarithms as logarithms of products and quotients. 8. log a -f log b log c. 9. log a (log b -f log c). 10. 31oga-^log(6 + c). 11. i log (1 - a) + flog (! + *) 12. 21og- + 31og- 13. 21oga-f logfc-f-ilogc. b a Given Iog 10 2 = .30103, Iog 10 3 = :47712, Iog 10 5 = .69897, Iog 10 7 = .84510, find the values of the following logarithms, to the base 10 : 14. log 50. 15. log 6. 16. log 8. 17. log 9. 18. log 12. 19. log 36. 20. log 108. 21. log-4J-. 22. log 2}. 23. logSf. 24. Iog5-f. 25. log 360. 26. log 3072. 27. log 3500. 28. log 5880. 29. log ^72. 30. log VI 80 - 31. log ^1715. 32 1o^ 490 33 lo ^^ 9 ^XVl05 j 32. log. 33. logVV. 34. log (4J) , Systems of Logarithms. 12. The two most important systems of logarithms are : (i.) The system whose base is 10. This system was intro- duced, in 1615, by the Englishman, Henry Briggs. Logarithms to the base 10 are called Common, or Briggs 's Logarithms. 378 ALGEBRA. [Cn. XXVI (ii.) The system whose base is the sum of the following infinite series, The value of this sum, which to seven places of decimals is 2.7182818, is denoted by the letter e. Logarithms to the base e are called Natural Logarithms; sometimes also Napierian Logarithms, in honor of the inventor of logarithms, the Scotch Baron Napier, a contemporary of Briggs. Napier himself did not, however, introduce this sys- tem of logarithms. These two systems are the only ones which have been gen- erally adopted ; the common system is used in practical calcu- lations, the natural system in theoretical investigations. The reason that in all practical calculations the common system of logarithms is superior to other systems is because its base 10 is also the base of our decimal system of numeration. The logarithms of most numbers are irrational, and thus approximate values are used. Properties of Common Logarithms. 13. In the following articles the subscript denoting the base 10 will be omitted. We now have 10= 1, orlogl -0; 10' = 10, or log 10 = 1; 10 2 = 100, or log 100 = 2 ; 10 3 = 1000, or log 1000 = 3 ; (a) 10- 1 = .1, or log .1 = - 1 ; 10- 2 '= .01, or log .01 =-2; 10- 3 = .001, or log .001 =-3; 10 4 = .0001, or log .0001 = - 4 ; 12-15] LOGARITHMS. 379 Evidently the logarithms of all positive numbers, except positive and negative integral powers of 10, consist of an inte- gral and a decimal part. Thus, since 10 1 < 85 < 10 2 , we have 1 < log 85 < 2, or log 85 = 1 -f a decimal. 14. The integral part of a logarithm is called its Character- istic. The decimal part of a logarithm is called its Mantissa. 15. Since a number having one digit in its integral part, as 7.3, lies between 10 and 10 1 , it follows from table (a) that its logarithm lies between and 1, i.e., is + a decimal. Since any number having two digits in its integral part, as 76.4, lies between 10 1 and 10 2 , its logarithm lies between 1 and 2, that is, is 1 + a decimal In general, since any number having n digits in its integral part lies between 10"" 1 and 10 n , its logarithm lies between n 1 and n, i.e., is n 1 + a decimal. We there- fore have : (i.) The characteristic of the logarithm of a number greater than unity is positive, and is one less than the number of digits in its integral part. E.g., log 2756.3 = 3 + a decimal. Since a number less than 1 having no cipher immediately following the decimal point lies between 10 and lO" 1 , it follows from table (6) that its logarithm lies between and 1, i.e., is 1 4- a positive decimal. Since a number less than 1 having one cipher immediately following the decimal point lies between 10- 1 and 10- 2 , its logarithm lies between 1 and 2, i.e., is 2 -f a positive decimal. In general, since a number less than 1 having n ciphers immediately following the decimal point lies between W~ n and 10- (M+1) , its logarithm lies between - n and (n -f 1), i.e., is (n + 1) + a positive decimal. We therefore have : (ii.) The characteristic of the logarithm of a number less than 1 is negative, and is numerically one greater than the number of ciphers immediately following the decimal point. E.g., log .00035 = 4 + a decimal fraction. 380 ALGEBRA. [Cn. XXVI It follows conversely from (i.) and (ii.) : (iii.) If the characteristic of a logarithm be -\-n, there are n-f-1 digits in the integral part of the corresponding number. (iv.) If the characteristic of a logarithm be n, there are n 1 ciphers immediately following the decimal point of the correspond- ing number. 16. It lias been found that 538 = 10 2 - 73078 to four decimal places, or log 538 = 2.73078. We also have log .0538 = log ySf for = log 538 - log 10000 = 2.73078 - 4 = .73078 - 2 ; log 5.38 = log fff = log 538 - log 100 = 2.73078 - 2 = .73078; log 53800 = log (538 x 100) = log 538 + log 100 = 2.73078 + 2 - 4.73078. These examples illustrate the following principle : If two numbers differ only in the position of their decimal points, their logarithms have different characteristics but the same positive mantissa. 17. The characteristic and the mantissa of a number less than 1 may be connected by the decimal point, if the sign ( ) be written over the characteristic to indicate that the character- istic only is negative, and not the entire number. Thus, instead of log .00709 = .85065 - 3 = - 3 + .85065, we may write 3.85065 ; this must be distinguished from the ex- pression 3.85065, in which the integer and the decimal are both negative. Similarly, log .082 = 2.91381, while log .820 = 2.91381. Five-Place Table of Logarithms. 18. The logarithms, to the base 10, of a set of consecutive integers have been computed. In tabulating these logarithms, compactness is important. 15-20] LOGARITHMS. 381 For this reason, all unnecessary detail is omitted. Since the characteristic of the logarithm of any number can, as we have seen, be determined by inspection, it is unnecessary to write it with the mantissa in the table. Consequently, only the man- tissas, without the decimal points, are there given. Neither is it necessary to give the logarithms of decimal fractions, since their mantissas are the same as the mantissas of the numbers obtained by omitting the decimal point. The logarithms may be carried to any number of decimal places, and the extent to which they are carried depends upon the degree of accuracy required in their use. 19. The accompanying five-place table gives the mantissas of the logarithms of all consecutive integers from 1 to 9999 inclusive. In this table the first three figures of each number are given in the column headed N, and the fourth figure in the horizontal line over the table. The first figure, which is the same for all numbers in a given column, is printed in every tenth number only. The columns headed 0, 1, 2, 3, etc., contain the mantissas, with decimal points omitted. In the column headed 0, when the first two figures are not printed, they are to be taken from the last mantissa above which is printed in full. In the columns headed 1, 2, 3, etc., the last three figures only are printed ; the first two are to be taken from the column headed in the same horizontal line. When a star is prefixed to the last three figures of a man- tissa, the first two figures are to be taken from the line below. To Find the Logarithm of a Given Number. ' 20. When the Number consists of Four or Fewer Figures. Take the mantissa that is in the horizontal line with the first three figures and in the column under the fourth figure of the given number. Determine the characteristic by Art. 15. 382 ALGEBRA. [Cn. XXVI E.g., log 2583 = 3.41212, log 46.32 = 1.66577. In writing logarithms with negative characteristics it is customary to modify the characteristics so that 10 is uniformly subtracted from the logarithms. Thus, 2.45926 = .45926 - 2 = 8.45926 - 10 ; 4.37062 = .37062 - 4 = 6.37062 - 10. That is, we add 10 to the negative characteristic, and write 10 after the logarithm. log .5757 = 9.76020 - 10, log .02768 = 8.44217 - 10. Observe that the first two figures of the mantissa of log .5757 are taken from the line below, in accordance with the directions in Art. 19. If the given number consists of fewer than four figures, annex ciphers until it has four figures, in taking the mantissa from the table. E.g., mantissa of log 78 = mantissa of log 7800 = .89209, and log 78 = 1.89209. In like manner, log 583 = 2.76567, log .02 = 8.30103 - 10. 21. When the Number consists of more than Four Significant Figures. The method used is called interpolation, and depends upon the following property of logarithms: The difference between two logarithms is very nearly propor- tional to the difference between the corresponding numbers when this difference is small. . The error made by assuming that these differences are exactly proportional will be so small that it may be neglected. Ex. 1. Find log 27845. ' Omitting, for the moment, the decimal points from the mantissas, we have mantissa of log 27850 = 44483, mantissa of log 27840 = 44467, difference of mantissas = 16. 20-21] LOGARITHMS. 383 Let x stand for the difference between the mantissas of log 27845 and log 27840 ; that is, for the correction to be added to the smaller mantissa to give the required mantissa. Then, by the above property, x = 27845 - 27840 = 5 - 16 27850 - 27840 ~~ 10 " Whence x = .5 x 16 = 8. Therefore, mantissa of log 27845 = 44467 + 8 = 44475, and log 27845 = 4.44475. Observe that, by Art. 16, the mantissa of log 27850 is the same as the mantissa of log 2785. In subsequent work such ciphers will be omitted. The method can now be stated more concisely for practical work : /Subtract the mantissa corresponding to the first four figures of the given number from the next mantissa in the table ; multiply this difference by the remaining figure or figures of the given number, treated as a decimal ; add the product to the first (and smaller) mantissa. Prefix finally the proper characteristic. In thus finding the mantissa, a decimal point in the given number is ignored, in accordance with Art. 16. The difference between two consecutive mantissas in the table is called the Tabular Difference. Ex. 2. Find log 78.1283. We have mantissa of log 7813 = 89282, mantissa of log 7812 = 89276, tabular difference = 6, correction == .83 x 6 = 4.98, mantissa of log 781283 = 89276 + 5 = 89281. Therefore log 78.1283 = 1.89281 Observe that the correction added to the mantissa of log 7812 is 5, the nearest integer to 4.98. 384 ALGEBRA. [Cn. XXVI 22. In the table of logarithms a column containing the required corrections (head Pp. Pts., i.e., proportional parts) is given. In this column there are several small tables, each con- taining two columns of numbers. One of these columns consists of the consecutive numbers 1 to 9 ; the other, headed by a tabu- lar difference, contains the correction corresponding to each one of the figures 1 to 9, when it is the fifth figure of the number whose logarithm is required. When it is the sixth figure, the corresponding tabular correction must evidently be divided by 10 ; when it is the seventh figure, by 100 ; and so on. Thus, in Ex. 1 of the preceding article, we take the correc- tion opposite 5, under the tabular difference 16, and obtain 8, as before. In Ex. 2, we take the following corrections from the column headed by the tabular difference 6 : for 8, correction = 4.8 for 3, correction = 0.18 final correction = 4.98, as before. Observe that the correction for the sixth figure of the given number does not affect the result. Ex. 3. Find the log .0128546. We have mantissa of log 1286 = 10924, mantissa of log 1285 = 10890, tabular difference = 34. From the column of proportional parts headed by 34, we obtain : correction for fifth figure 4 = 13.6 correction for sixth figure 6 = 2.04 total correction = 15.64 Therefore, mantissa of log 128546 = 10890 + 16 = 10906, and log .0128546 = 8.1090.6 - 10. Observe that in this example the correction for the sixth figure does affect the result. 22-24] LOGARITHMS. 385 EXERCISES III. Verify the following statements : 1. log 13 = 1.11394. 2. log 14.84 = 1.17143. 3. log 73000 = 4.86332. 4. log 5884.4 = 3.76970 5. log .031586 = 8.49949 - 10. 6. log .00391857 = 7.59S13 - 10. Find the logarithms of each of the following numbers : 7. 5. 8. 18. 9. 540. 10. 3876. 11. 2076. 12. 59.80. 13. 1.87. 14. .01832. 15. .0004129. 16. 63072. 17. 59.836. 18. 4376.4. 19. .070518. 20. 185462. 21. .00103987. To find a Number from its Logarithm. 23. Mantissa given in the Table. If the mantissa of the given logarithm is found in the table, the first three figures of the required number will be in the same line with it in the column headed N, and the fourth figure over the column in which the given mantissa stands. The characteristic is determined by Art. 15 (iii.) and (iv.). Ex. 1. Find the number whose logarithm is 4.82099. The mantissa .82099 corresponds to the number 6622; but since the given characteristic is 4, the required number must have five integral places. Consequently 4.82099 = log 66220. Ex. 2. Find the number whose logarithm is 8.78625 10. The mantissa .78625 corresponds to the number 6113; but since the characteristic is 2, the required number must be a decimal having its first significant figure in the second deci- mal place. Consequently 8.78625 - 10 = log .06113. 24. Mantissa not given in the Table. The method employed is the converse of that used in Art. 21 to find the logarithms of numbers that consist of more than four significant figures. 386 ALGEBRA. [Cn. XXVI Ex. 1. Find the number whose logarithm is 2.81727. We have given mantissa = 81727 ; next smaller mantissa = 81723, corresponding number = 6565 ; next larger mantissa = 81730, corresponding number = 6566. Let x stand for the difference between 6565 and the required number ; that is, for the correction to be added to 6565. We then have x = 81727 - 81723 Qr x = = 6 6566 - 6565 ' 81730 - 81723' ' T 1 " 7 " corrected for the first decimal place. Notice that the signifi- cance of the decimal point in the result is that the correction is to be annexed as an additional figure to the smaller number. Therefore, the figures in the required number are 65656 ; and since the characteristic of the given logarithm is 2, there are only three integral places. Hence 2.81727 = log 656.56. This process may also be stated concisely for practical work : Take the mantissa next smaller and the mantissa next larger than the given mantissa, and note the numbers corresponding; next divide the difference between the given mantissa and the next smaller by the difference between the next larger and the next smaller. Annex the quotient to the number corresponding to the smaller mantissa, neglecting the decimal point of the quotient. Place the decimal point in the number thus obtained as it is determined by the given characteristic. Ex. 2. Find the number whose logarithm is 7.18281 10. We have given mantissa = 18281 ; next smaller mantissa = 18270, corresponding number = 1523 ; next larger mantissa = 18298, corresponding number = 1524. Hence the correction to be annexed to 1523 is 18281 - 18270 = 11 = 39 , 18298-18270' 28' 24-25] LOGARITHMS. 387 Therefore the figures of the required number are 152339 ; and since the characteristic of the given logarithm is 3, there must be two ciphers between the decimal point and the first significant figure. Consequently 7.18281 - 10 = log .00152339. In general, in using a five-place table, the numbers corre- sponding to given mantissas should be carried to only Jive significant figures, as in Ex. 1. But with mantissas in the first two pages of the table, the corresponding numbers may be carried to six figures. The reason being that the tabular differences later become so small that the correction for a sixth figure will not in general affect the result. See Exx. 2-^3, Art. 22. 25. The correction to be added to the number corresponding to the next smaller mantissa may also be taken from the column of proportional parts. In this column turn to the table headed by the number which is equal to the difference between the next larger and the next smaller mantissa. As the first figure of the correc- tion take the figure in this table which is opposite the pro- portional part nearest to the difference between the given mantissa and the next smaller mantissa. If a second figure in the correction is to be found, we should take as the first figure that figure which is opposite the pro- portional part next smaller than the difference between the given mantissa and the next smaller. Multiply by 10 the difference between the proportional part already used and the difference between the given mantissa and the next smaller, and take the product as a proportional part in determining the second figure of the correction ; and so on. Thus, in Ex. 1 of the preceding article, we turn to the column headed by the tabular difference 7. The proportional part in this table that is nearest to 4 (the difference between the given mantissa and the next smaller) is 4.2 ; the number opposite 4.2 is 6, the correction previously obtained. 388 ALGEBRA. [Cn. XXVI In Ex. 2, we turn to the column headed by the tabular differ- ence 28. The proportional part next smaller than 11 (the difference between the given mantissa and the next smaller) is 8.4; the figure opposite 8.4 is 3, the first figure of the correction. We next multiply 2.6 (=11 8.4) by 10, and take the product 26 as a proportional part. The figure opposite 25.2 (nearest to 26) in the column headed by 28 is 9, the second figure of the correction. Therefore, the required correction is found to be 39, as before. EXERCISES IV. Verify the following statements : 1. log x = 3.14926, x = 1410.13. 2. log x = 1.59187, x = 39.073. 3. log x = . 34159, x = 2.1958. 4. log x = 9.57187 - 10, x = .37314. 5. log x = 7.83957 -10, x = .0069115. . log x = 6.18953 - 10, x = .00015471. Find the numbers whose logarithms are : 7. 2.26150. 8. .59726. 9. 8.94655-10. 10. 3.88825. 11. 6.19815. 12. 6.72576-10. 13. 4.98880. 14. 1.68417. 15. 9.23360-10. Cologarithms. 26. The Cologarithm of a number, or, as it is sometimes called, the Arithmetical Complement of the logarithm, is de- fined as the logarithm of the reciprocal of the number. That is, colog n log - = log 1 log n = log n. n We thus see that the cologarithm of a number is obtained by subtracting its logarithm from 0. But this step would leave the mantissa as well as the characteristic negative. To avoid a negative mantissa, therefore, we subtract the logarithm from 10 - 10, = 0. 25-27] LOGARITHMS. 389 Ex. 1. Find the colog 3. Subtracting log 3, = .47712, from 10 10, we have 10. - 10 .47712 9.62288 - 10 Therefore colog 3 = 9.62288 - 10. Ex. 2. Find colog .0054. Subtracting log .0054, = 7.73239 - 10, from 10 - 10, we have 10. - 10 7.73239 - 10 2.26761 Therefore colog .0054 = 2.26761. EXERCISES V. Verify the following statements : 1. colog 543 =7.26520-10. 2. colog 72.318 =8.14075-10. 3. colog 8.9134 =9.04996-10. 4. colog .38145 =.41856. 5. colog .051984 =1.28413. 6. colog .0091437 = 2.03887. Find the cologarithm of each of the following numbers : 7. 5817. 8. .6305. 9. .009812. 10. 763.85. 11. 15.482. 12. 7.00386. 13. .000594. 14. 32581.9 Applications. 27. Ex. 1. Compute the value of a?, when x = 53.847 x .0085965. . log x = log 53.847 + log .0085965. log 53.847 = 1.73117 log .0085965= 7.93433 - 10 log x = 9.66550 10 x = .46291. 390 ALGEBRA. [Cn. XXVI Ex. 2. Compute the value of x, when x = 8.4394 -- .31416. log x = log 8.4394 + colog .31416. log 8.4394= .92631 colog .31416 = .50285 log i x = 1.42916 x = 26.863. Ex. 3. Compute the value of a?, when = 6.4319 x .59218 7.9254 x .062547* log x = log 6.4319 + log .59218 + colog 7.9254 + colog .062547. log 6.4319 = .80834 log .59218= 9.77246-10 colog 7.9254 = 9.10098 - 10 colog .062547 = 1.20379 log x = 20.88557 -20 = .88557. x = 7.6837. Ex. 4. Find the value of a?, when x = .5318 4 . log x = 4 log .5318 = 4(9.72575-10) = 38.90300 - 40 = 8.90300 - 10. x = .079983. Ex.5. Find the value of ty-. 031459. Since a negative number cannot be expressed as a power of + 10, such a number does not have a logarithm. In this example, therefore, and in all similar examples, we first deter- mine the sign of the result. We then find the value of the expression obtained by changing each sign to +, and to that result prefix the sign previously determined. 27] LOGARITHMS. 391 The sign of the result of this example is Let a; = y. 031459. Then log x = 1 log .031459 = 1(28.49775-30) = 9.49925 - 10, and x = .31568. Therefore, the required result is .31568. Observe that in dividing log .031459 by 3, we first modified the characteristic so that the number, 30, which is subtracted from the logarithm is 10 times the divisor ; that is, so that the quotient obtained by dividing this number by 3 is 10. Ex. 6. Compute the value of x, when = 4.5921 x -C/.021946 = .059318 x .41587 3 ' log x = log 4.5921 + 1 log .021946 + colog .059318 + 3 colog .41587. log 4.5921 = .66201 i log .021946 = i (28.34135 - 30) = 9.44712 - 10 colog .059318 = 1.22681 3 colog .41587 = 3 x .38104 = 1.14312 log x = 12.47906 -10 = 2.47906. x = 301.34. Ex. 7. Compute the value of x, when 3/5.4318 x y.31459 7.1938 x .2934 2 For convenience in arranging the logarithmic work, we first cube both members of this equation, and obtain ^ = 5.4318 x y.31459 m 7.1938 x .2934 s 392 ALGEBRA. [Cn. XXVI Taking logarithms, we have 3 log x = log 5.4318 + i log .31459 + colog 7.1938 + 2 colog .2934. (2) In practice, step (1) should be performed mentally, and the result (2) be at once written. log 5.4318= .73494 i log .31459 = 1(19.49775 - 20) = 9.74887 - 10 colog 7.1938 = 9.14304 - 10 2 colog .2934 = 2 x 0.53254 = 1.06508 3 log x = 20.69193 -20 = .69193. log x=. 23064. x = 1.70076. EXERCISES VI. Find the values of each of the following expressions : 1. 31.834 x 185.592. 2. 8.0043 x .5319. 3. .004893 x 6.5942. 4. (- .0514) x .123857. 5 ' 78 . 6 1539 19.7939 ' 347 380.14 x (- .0576) 78395 9. 11 3892.7 (- 9.7408) x .000395 7.3792 5.83 x 91.358 36.937 57.13 x 9.0047 .00479 5.382 x .07235 12 4.9 x (-306) x 48.3 .79 x 891.3 x. 00099 100.088 x 2.9 x .081 ' ' (-10.236) x .07 x .0031 14. 7.0435 3 . 15. .31844 4 . 16. 2.3817 5 . 17. (3.68 x .97) 4 . 18. (.7918 x 3.17) 5 . 19. [.034 x (- 4.973S)] 4 . 20. (17.19 x .00001986) 5 . 27*28] LOGARITHMS. 393 21. ^/13. 22. J/-251. 23. -^39.837, 24. ^163.4 3 . 25. sX-31492 2 ; 26. -/1.0031. 27. ^/||. 28. -4/11 29 30. l^/f. 31. 21^/1 32. 33. (.74sy8.21) 4 . 34. (5.21-^/.3817) 6 . 35. f-^'- 5x^/17. 36. 3f ^/.38 x .{/7.3815. 37. ^(- 25 V 3 )- 38 - ^/(H2.34-^/.003914). 39. ^/(17.2^/.718). 40. ty(- 23^7.18943). 41. 5.341^/(27.39^/.1439). 42. 23.491 2 -^(.l 8-^/17.3). c /3.1 \ .5 - 9.2614 .519VH7.38 Exponential Equations. 28. An Exponential Equation is an equation in which the unknown number appears as an exponent of a known or an unknown number, as a* = b. Ex. 1. Solve the equation 3 X = 9. Taking logarithms, x log 3 = log 9 = 2 log 3. Hence x = 2. This result could have been obtained by inspection, by writing the given equation 3 X = 3 2 . Ex. 2. Find the value of x in 3* = 5. taking logarithms, x log 3 = log 5 ; log 5 T~ ~ log 3 .47712 log 5 .69897 whence X = ~ ~ = ~ = 1.46497. 394 ALGEBRA. [Cn. XXVI Ex. 3. Find the value of x in the following equation taking logarithms, (3 x + 1) log 2 = (2 x - 1) log 7. Removing parenthesis, 3 x log 2 + log 2 = 2 x log 7 log 7, or o?(3 log 2 - 2 log 7) = - log 7 - log 2; when ce g log 7 + log 2 2 log 7 - 3 log 2 .84510 4- .30103 1.69020 - .90309 1.14613 .78711 = 1.4561. EXERCISES VII. Solve the following exponential equations : 1. 2* = 64. 2. 3' = SI. 3. 2*~ l = .&*-*. 4. (_8)-*=16. 5. 4 8 - 1 = .5- 8 . 6. 4* = 8. 7. 8 X =32. 8. 5 X = (V5)- 1 . 9. 4 I+1 = 8-2' +2 . 10. 25 3 *- 1 = 625 5*+ 3 . 11. 7^ (I - 3) = 343- 1 49v"*- 3 >. 12. 27^ (I - 3) = ( V3) V(x+8) . 13. V all ~ x = a8 ~ z - 14. ^/a^ 2 -V aX ~ 3 - 15 - V 3 ~ 4j: ^^ <5 ~ 7a: xal=l. 16. (i) z =25. 17. (i)- 7 =64. 18. (fj) li;! - fi =(^) 78 - 8 . 19- (|) 4 *- 7 = .75 2 - 3 *. 20. 4^-6.2^ + 8 = 0. 21. 9* + 243 = 36 3*. 22. 3's z = 9. 23. 5 log2 * = 625. 24. 16 log3x = 32 IO x . 25 ;V = 10. 26. 16* = 45. 27. 11" = 310. 28. 25* =10. 29. 7* = 300. 30. 3.594* = 359600. 31. ^/9.8926 = 1.29. 32. 5* = 7 3 - 14 . 33. aV 2 = ^/3. 34. r>* +3 =1000. 35. 7 X+1 =5. 36. 1.58*- 5 = 9.847. 37. 5* +1 = II- 1 . 38. 3* +7 =7*+ 3 . 39. 31* +3 = 2tV +4 . 40. 35 qe+2 = 40 *. 28-32] LOGARITHMS. 395 Compound Interest and Annuities. 29. To find the compound interest, /, and the amount, A, of a given principal, P, for n years at r per cent. If the interest is payable annually, the amount of $ 1 at the end of one year will be 1 -f r dollars, and the amount of P dollars will be P(l + r) dollars. This amount, P(l + r), be- comes the principal at the beginning the second year. There- fore, at the end of the second year the amount will be P(l + r) X (1 4- ?), = P(l 4- rf dollars, and so on. Therefore, at the end of n years the amount will be P(l 4- ?*)" dollars, or *=/(! +/)". 30. This formula can be used not only to find A, but also to find P, r, or n, when the three other quantities are given. Thus, P = (i + 31. An Annuity is a fixed sum of money, payable yearly, or at other fixed intervals, as half-yearly, once in two years, etc. 32. To find the present value, P, of an annuity of A dollars, payable yearly for n years, at r per cent. The present *vorth of the first payment is dollars, of A IT? the second payment is dollars, and, in general, of the nth payment is dollars. (l + r) n Therefore the present worth of all the payments is A + , .** + Multiplying numerator and denominator by 1 4- r, we have ~r|_' (1 + /)"_] 396 ALGEBRA. [Cn. XXVI Ex. 1. Find the amount of $500 for 8 years at 5% com- pound interest. A = P (I + r) n = 500 x 1.05 8 . log A = log 500 + 8 log 1.05. log 500 = 2.69897 8 log 1.05 = .16952 log A = 2.86849 A = 738.73. Therefore the require'd amount is $ 738.73. Ex. 2. Find the present value of an annuity of $ 1000 for 6 years, if the current rate of interest is 5%. (l-fr) We will first compute 1.06 6 , log(1.05) 6 = 6. log 1.05 = 6 x .02119 = .12714. (1.05) 6 = 1.34012. We then have 20000 X" 84012 X .05 1.34012 log P = log 20000 + log .34012 + colog 1.34012. log 20000= 4.30103 log .34012= 9.53163-10 colog 1.34012 = 9.87286 - 10 log P= 23.70552 -20 = 3.70552. P=5076. Therefore the present value of the annuity is $ 5076. 32] LOGARITHMS. 397 EXERCISES VIII. Find the amount at compound interest : 1. Of $3600 for 5 years at 41%. 2. Of $1875.50 for 8 years at 5%. 3. Of $ 12,350 for 6 years at 3J%. 4. Of $21,580 for 7 years 4 months at 4%. Find the principal that will amount to : 5. $7913 in 5 years at 5% compound interest. 6. $14,770 in 10 years at k\% compound interest. 7. $11,290 in 8 years at 4% compound interest. 8. $11,090 in 6 years 6 months at 3% compound intere'st. 9. In what time, at 4%, will $8010 amount to $11,400 at compound interest ? 10. In what time, at 4J%, will $3530 amount to $5987, if the interest is compounded semi-annually ? Find the rate of compound interest : 11. If $ lllO.amounts to $ 1640 in 8 years. 12. If $ 3750 amounts to $6070 in 14 years. Find the present value of an annuity : 13. Of $ 1000 for 10 years, if the current rate of interest is 4%. 14. Of $ 1250 for 8 years, if the current rate of interest is 41%. 15. Of $ 2500 for 10 years, if the current rate of interest is 5%. 16. Of $3000 for 12 years, if the current rate of interest is 6%. 100-149 N. 1 2 I 3 4 5 6 7 8 9 Pp. 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Pts. 11 150-199 N. 1 2 3 4 5 6 7 8 9 Pp. Pts. 150 17609 638 667 696 725 754 * 782 811 840 869 9.O ?R 5 1 898 926 955 984 *OI 3 *0 4 I *O99 *I2 7 *I 5 6 ~ y 2.8 S 2 18 184 213 241 270 298 327 355 384 412 441 2 5- 6 53 469 498 526 554 583 61 i 639 667 6 9 6 * 724 7 i ' 54 752 780 808 837 865 893 921 949 977 J 4 ii .6 II. 2 55 19033 06 1 089 117 145 173 201 229 257 285 5 14-5 14.0 56 312 340 368 396 424 45 i 479 57 53? 562 6 17.4 16.8 57 590 618 645 673 700 728 J56 783 811 838 7 20.3 19.6 58 866 893 921 948 976 *00 3 "058 *o85 *II2 8 23.2 22.4 59 20 140 167 194 222 249 276 33 330 358 385 9 26.1 25.2 160 412 439 466 493 520 548 575 602 629 656 27 o.fi 61 683 710 * 737 763 790 817 844 871 898 925 i / 2- 1 2.6 62 952 978 "032 *59 +085 *II2 *i 39 "165 *I92 2 63 21 219 245 272 299 325 352 378 405 431 458 2 < .1 7.8 64 484 5" 537 5 6 4 590 617 643 669 696 722 J 4 10.8 / * y 1.9 22 635 655 674 6 94 7*3 733 753 772 792 811 2 3.8 23 830 850 869 889 908 928 947 967 986 *oo5 7 O c.7 24 35025 044 064 083 102 122 141 1 60 1 80 199 O 4 7-6 2 5 218 238 257 2 7 6 295 315 334 353 372 392 5 9-5 26 411 43 449 468 488 57 526 545 564 583 6 ii 4 27 603 622 641 660 679 698 717 736 755 774 7 13 3 28 793 813 832 *o 5 o 870 889 908 927 946 965 8 15 .2 29 984 *oo3 *02I "059 *078 *097 *n6 *i35 *i 54 9 17 .1 230 36173 192 211 229 248 267 286 305 324 342 18 3 1 36i 380 399 418 436 455 474 493 5 11 53 i j .8 32 549 568 586 605 624 642 661 680 698 717 2 3-6 33 736 754 773 791 810 829 847 866 884 903 7 O c.4 34 922 940 959 977 996 *oi4 *033 "051 "070 *o88 O 4 D r 7.2 35 37 I0 7 125 144 162 181 199 218 236 254 273 5 9.0 36 291 310 328 346 365 383 401 420 438 457 6 10.8 37 475 493 5 11 530 548 566 585 603 621 639 7 12.6 38 658 676 694 712 749 767 785 803 822 8 14.4 39 840 858 876 894 912 949 967 985 *oo3 9 16.2 240 38021 039 057 075 093 112 130 148 1 66 184 17 41 202 220 238 256 274 292 310 328 346 364 I 1.7 42 382 399 417 435 453 471 489 57 525 543 2 3.4 43 561 578 596 614 632 650 668 686 703 721 3 O *T 5.1 44 739 757 775 792 810 828 846 863 88 1 899 4 .8 45 917 934 952 970 987 *oo5 *023 ""041 *058 *076 5 8-5 46 39094 in 129 146 164 182 199 217 235 252 6 IO.2 47 270 287 30? 322 340 358 375 393 410 428 7 II 9 48 445 463 480 498 515 533 55 568 585 602 8 13.6 49 620 637 655 672 690 707 724 742 759 777 9 '5 3 N. 1 2 3 4 5 <5 7 8 9 Pp. Pts. IV 250-299 N. O 1 2 3 4 5 6 7 . 8 9 Pp. Pts. 250 39 794 811 829 846 863 88 1 898 915 933 95 18 5 1 967 985 *002 *oi9 -037 *054 *07i *o88 *io6 *I2 3 j 1.8 52 40 140 157 175 192 209 226 243 261 278 291 2 3.6 53 312 329 346 364 38i 398 415 432 449 466 2 j* v c.4 54 483 500 5 I8 535 552 569 586 603 620 637 O 4 D'T- 7-2 55 654 671 688 75 722 739 75 6 773 790 807 9.0 56 824 841 858 875 892 909 926 943 960 976 6 10.8 57 993 *OIO ="027 *044 *o6i *078 *095 *ni *I28 *I45 7 12.6 58 41 162 179 196 212 229 246 263 280 296 3 J 3 8 14.4 59 330 347 363 380 397 414 43 447 464 481 9 16.2 200 61 497 664 5*4 68 1 547 714 564 58i 747 597 764 614 780 631 797 647 814 i7 1.7 62 830 847 863 880 896 9U 929 946 963 979 2 / 3 4 63 996 *OI2 *O29 *o 45 *062 "=078 *O95 *m *i2 7 1*144 3 O T 1 c.l 64 42 1 60 I 77 193 2IO 226 243 259 275 292 308 4 6.8 65 3 ? 341 357 374 390 406 423 439 455 472 8.5 66 488 54 521 537 553 57 586 602 619 635 6 IO.2 3 651 813 667 830 684 846 700 862 716 878 732 894 749 911 765 927 781 943 797 959 I "i 13.6 69 975 991 *oo8 *O24 *O4O *056 *072 *o88 *IO4 *I2O 9 15-3 270 43 136 J 52 169 185 20 1 217 233 249 265 28l 16 7i 297 3 I 3 329 345 361 377 393 409 425 441 i 1.6 72 457 473 489 505 52i 537 553 569 584 600 2 3.2 73 616 632 648 664 680 696 712 727 743 759 3 4.8 74 775 791 807 823 838 854 8 7 o 886 902 917 O 4 6.4 75 933 949 965 981 996 *OI2 *028 *044 *o 59 *75 5 8.0 76 44091 107 122 138 154 I 7 185 20 1 217 232 6 9.6 77 248 264 279 295 3" 326 342 358 373 389 7 II. 2 78 404 420 436 45 i 467 483 498 5 J 4 5 2 9 545 8 12.8 79 560 576 592 607 623 6 3 8 654 669 685 700 9 14.4 280 716 73i 747 762 778 793 809 824 840 855 JC 81 871 886 902 917 932 948 963 979 994 *OIO i *j I.C 82 45025 040 056 071 086 IO2 117 '33 148 163 2 J 3.0 o 3 179 194 209 225 240 255 271 286 301 3i7 4.c 84 332 347 362 378 393 408 423 439 454 469 4 "0 6.0 85 484 500 5*5 530 545 561 576 59i 606 621 5 7-5 86 637 652 667 682 697 712 728 743 758 773 6 ! 9.0 87 788 803 818 834 849 864 i 879 894 909 924 7 10.5 88 939 954 969 984 *ooo 015 *030 "045 *o6o *75 8 12.0 89 46 090 105 120 135 150 165 180 195 210 225 9 J 3-5 290 240 255 2 7 285 300 3i5 330 345 359 374 14 9i 389 404 419 434 449 464 479 494 59 523 i 1.4 92 538 553 568 583 598 613 627 642 657 672 2 2.8 93 687 702 7 l6 73i 746 761 776 790 805 820 -2 4.2 94 835 850 864 879 894 909 923 938 953 967 O 4 5.6 95 982 997 *OI2 *026 *O4i 056 *07o "085 *IOO *ii4 5 7- 96 47 I2 9 144 1 59 J 73 188 202 217 232 246 261 6| 8.4 97 276 290 305 3i9 334 349 363 378 392 407 7 9-8 98 422 436 45 i 465 480 494 59 524 538 553 8 II. 2 99 567 582 596 611 625 640 654 669 683 698 9 12.6 N. O 1 2 3 4 5 6 7 8 9 Pp. Pts. 300-349 N. 1 2 3 4 5 6 7 8 9 Pp. Pts. 300 47 7 12 727 741 756 770 784 799 813 828 842 OI 857 871 885 900 914 929 943 958 972 986 02 48001 015 029 044 058 073 087 101 116 130 3 144 1 73 187 202 216 230 244 259 273 15 04 287 302 330 344 359 373 387 401 416 i i-5 05 430 444 458 473 487 501 530 544 558 2 3-o 06 572 586 601 615 629 643 657 671 686 700 6.0 07 714 728 742 75 6 77 785 799 813 827 841 4 5" 7r 08 855 869 883 897 * 9 " 926 940 * 954 968 982 O f\ f\ 09 996 *OIO *024 ="038 *o66 *o8o *io8 *I22 7 y.u 10.5 310 49 136 150 164 178 192 206 220 234 248 262 8 12.0 ii 276 290 304 332 346 360 374 388 4O2 9 13-5 12 415 429 443 457 471 485 499 513 527 541 13 554 568 582 596 610 624 638 651 665 679 14 693 707 721 734 748 762 776 790 803 817 15 831 841 859 872 886 900 914 927 941 * 955 i 14 16 969 982 996 *OIO *024 *037 ="051 "065 *o 79 2 2.8 17 50 1 06 1 20 I 33 '47 161 174 1 88 202 215 229 40 18 243 256 270 284 297 3" 325 338 352 365 ,4 5.6 19 379 393 406 420 433 447 461 474 488 501 5 7.0 320 515 5 2 9 542 556 569 583 596 610 623 637 6 8.4 21 ^S 1 664 678 691 70? 718 732 745 759 772 7 9.8 22 786 799 813 826 840 853 866 880 893 * 9 7 8 1 1.2 23 920 934 947 961 974 987 *OOI *oi4 *028 9 12.6 24 51055 068 08 1 093 108 121 135. 148 162 173 2 5 26 1 88 322 202 335 $ 228 362 242 375 388 268 402 282 415 295 428 308 44 i 27 28 455 587 468 60 1 481 614 495 627 508 640 521 654 667 548 680 g 574 706 i 2 2.6 29 720 733 746 759 772 7 86 799 812 825 838 3 3-9 330 851 865 878 891 * 9 4 917 93 943 957 97 4 5-2 31 983 996 *oo9 *022 "048 *o6i *075 *o88 *IOI 5 6.5 32 5 2 114 127 140 153 166 179 192 205 218 231 6 7.8 33 244 257 270 284 297 310 323 336 349 362 7 9.1 34 375 388 401 414 427 440 453 466 479 492 8 10.4 35 54 517 530 543 556 569 582 595 608 621 9 11.7 36 634 647 660 673 686 699 711 724 737 75 37 763 776 789 802 815 827 840 853 866 * 879 38 892 901 917 930 943 95 6 969 982 994 12 39 53020 033 046 058 071 084 097 no 122 135 i 1.2 340 148 161 173 1 86 199 212 224 237 250 263 2 2.4 41 275 288 301 314 326 339 352 364 377 390 3 3-6 42 43 403 529 415 542 428 555 441 567 466 593 479 605 491 618 54 631 III 4 4.8 5 6.0 44 656 668 694 706 719 732 744 757 769 6 7.2 45 782 794 807 1 820 832 843 857 870 882 895 7 8.4 ft ' /-> A 46 908 920 933 945 958 970 983 995 *oo8 *O2O o 9.0 4% 1 *-v Q 47 54033 045 058 070 083 095 108 1 20 133 145 9 io.o 48 158 170 183 195 208 220 233 245 258 270 49 283 295 307 320 332 345 357 370 382 394 N. 1 2 3 4 5 6 7 8 9 Pp. Pts. VI 350-399 N. 1 2 3 4 5 6 7 8 9 Pp. Pts. 350 54407 419 432 444 45 6 469 481 494 506 518 5 1 53i 543 555 568 580 593 605 617 630 642 52 654 667 679 691 704 716 728 741 753 765 53 777 790 802 814 827 839 851 864 876 888 13 54 900 913 92? 937 949 962 974 986 998 *on i 2 *-3 2.6 55 55 2 3 035 047 060 072 084 096 1 08 121 133 7 2.Q 56 143 157 169 182 194 206 218 230 242 255 O 4 O s C.2 57 267 279 291 303 315 328 340 352 364 376 6. 5 58 388 400 4i3 425 437 449 461 473 485 497 ^ 59 509 522 534 546 558 570 582 594 606 618 7 9.1 360 630 642 654 666 678 691 73 7i5 727 739 8 IO.4 61 75 i 763 775 787 799 811 823 831 847 859 9 II.7 62 871 883 895 907 919 93 i 943 955 967 979 63 991 *oo3 *oi5 *O27 *o 3 8 *O5O *062 *074 *o86 *098 64 56 no 122 134 146 158 170 182 194 205 217 12 65 229 241 253 265 277 289 301 312 324 336 i 1.2 66 348 360 372 384 396 407 419 43i 443 451 2 2.4 67 467 478 490 502 5'4 526 538 549 561 573 1 T^ 3.6 68 585 597 608 620 632 644 656 667 679 691 O 4 i8 69 73 714 726 738 750 761 773 785 797 808 5 6.0 370 820 832 844 855 867 879 891 902 914 926 6 7-2 7i 937 949 961 972 984 996 *oo8 *oi9 *03i *043 7 8.4 72 57054 066 078 089 101 "3 124 136 148 I 59 8 9 ' 73 171 183 194 206 217 229 241 252 .264 276 9 10.8 74 287 299 310 322 334 345 357 368 380 392 75 403 415 426 438 449 461 473 484 496 57 76 519 530 542 553 565 576 588 600 611 623 ii 77 634 646 657 669 680 692 73 7 J 5 726 738 i I.I 78 749 761 772 784 795 807 818 830 841 852 2 2.2 79 864 875 887 898 910 921 933 944 955 967 3 3-3 380 978 990 *OOI *oi3 *024 *035 *047 "=058 "070 *o8i 4 4-4 81 58092 104 "5 127 138 149 161 172 184 195 5 5-5 82 206 218 229 240 252 263 2 74 286 297 309 6 6.6 83 320 331 343 354 3 6 5 377 388 399 410 422 7 7'7 84 433 444 456 467 478 490 5oi 5 12 524 535 8 8.8 85 546 557 569 580 591 602 614 625 636 647 9 9.9 86 659 670 68 1 692 704 7i5 726 737 749 760 87 782 794 805 816 827 838 850 86 1 872 88 883 894 906 917 928 939 95 961 973 984 10 89 991 *oo6 *oi7 *028 *040 *o5i *062 *073 *o84 *095 i I.O 390 59 106 118 129 140 J5 1 162 J 73 184 J 95 207 2 2.0 91 218 229 240 25 1 262 273 284 295 306 3i8 3 3-o 92 329 340 35 1 362 373 384 395 406 417 428 4 4.0 93 439 45 461 472 483 494 506 5 J 7 528 539 5 5-o 94 550 561 572 583 594 605 616 627 638 649 6 6.0 95 660 671 682 693 704 7*5 726 737 748 759 g 7.0 8.0 96 770 780 791 802 813 824 83i 846 857 868 o.o 97 879 890 901 912 923 934 945 95 6 966 977 ^w 98 988 999 *OIO *O2I *O32 *Q43 *Q54 *o65 *076 *o86 99 60 097 108 119 I 3 141 152 163 173 184 195 N. O 1 2 3 4 5 6 7 8 9 Pp. Pts. 400-449 VII N. 1 2 3 4 5 6 7 8 9 Pp. Pts. 400 60 206 217 228 239 249 260 271 282 293 304 01 3H 325 336 347 358 369 379 390 401 412 02 423 433 444 455 466 477 487 498 59 520 03 4 53i 638 649 IS 563 670 574 681 595 73 606 713 617 724 627 735 05 746 756 767 778 788 799 810 821 831 842 06 853 863 874 885 895 906 917 927 938 949 ii 07 959 97 981 991 *OO2 013 *02 3 *034 *o 4 5 *055 i i.i 08 6 1 066 077 087 098 109 119 130 140 1 S 1 162 2 2.2 09 172 183 194 204 215 225 236 247 257 268 3 3-3 410 278 289 300 310 3 2I 331 342 352 363 374 4 4.4 ii 384 395 405 416 426 437 448 458 469 479 I 66 12 490 500 511 5 21 S3 2 542 553 563 574 584 o o.o 13 595 606 616 627 637 648 658 669 679 690 I 7-7 8.8 14 700 711 721 73i 742 752 763 773 784 794 15 805 815 826 836 847 857 868 878 888 899 9 9.9 16 909 920 930 941 95 * 962 972 982 993 *cx>3 17 62 014 024 034 045 55 066 076 086 097 107 18 118 128 138 149 1 S9 170 1 80 190 20 1 211 19 221 232 242 252 263 273 284 294 34 315 420 321 335 346 356 366 377 387 397 408 418 10 21 428 439 449 459 469 480 490 500 5" 521 I(-V 22 531 542 55 2 562 572 583 593 603 613 624 2 ,\J 2 O 23 634 644 653 665 675 685 696 706 716 7 26 7 /-k 24 737 747 757 767 778 788 798 808 818 829 4 3- 4.0 2 5 839 849 859 870 880 890 900 910 921 931 5- 26 941 95 * 961 972 982 992 *002 *OI2 *O22 * 033 6 6.0 27 63043 053 063 073 083 094 104 114 124 134 7 7.0 28 144 155 165 175 185 195 205 215 225 *2 3 6 8 8.0 2 9 246 256 266 276 286 296 306 317 327 337 9 9.0 430 347 357 367 377 387 397 407 417 428 438 3i 448 458 468 478 488 498 5 08 5*8 528 538 3 2 548 558 568 579 589 599 609 619 629 639 33 649 659 669 679 689 699 709 719 729 739 34 749 759 769 779 789 799 80 9 819 82 9 839 35 849 859 869 879 889 899 909 919 929 939 9 36 949 959 969 979 988 998 *oo8 *oi8 *028 *038 i 0.9 37 64048 058 068 078 088 098 108 118 128 137 2 1.8 3^ 147 : 57 167 177 187 197 207 217 227 237 3 2.7 39 246 256 266 276 286 296 306 3i6 326 335 4 3-6 440 345 355 365 375 385 395 404 414 424 434 5 4-5 r A 4i 444 454 464 473 483 493 503 5i3 523 S3 2 iv4 42 542 552 562 572 582 59i 60 1 611 621 631 7 6'3 hy /> 43 640 650 660 670 680 689 699 709 719 729 81 44 738 748 758 768 777 787 797 807 816 826 9 0. 1 45 836 846 856 865 875 885 895 904 914 924 46 .933 943 953 963 972 982 992 *OO2 *OII *O2I 47 6503 1 040 050 060 070 079 089 099 108 118 48 128 137 147 157 167 176 1 86 196 205 215 49 225 234 244 254 263 273 283 292 302 312 N. 1 2 3 4 5 6 7 8 9 Pp. Pts. 450-499 N. 1 2 3 4 5 <5 7 8 9 Pp. Pts. 450 65321 33 1 341 350 360 369 379 389 398 408 5 1 418 427 437 447 456 466 475 485 495 54 52 5H 523 533 543 552 562 57 1 58i 59i 600 53 610 619 629 639 648 658 667 677 686 696 54 706 7*5 725 734 744 753 763 772 782 792 55 801 811 820 830 839 849 858 868 877 887 10 56 896 906 916 925 935 944 954 963 973 982 i I.O 57 992 *OOI *on *020 *O3O *039 *o 4 9 *o 5 8 *o68 *77 2 2.O 58 66087 096 106 5 124 134 H3 J 53 162 172 3 3-0 59 181 191 200 2IO 219 229 238 247 257 266 4 4.0 460 276 285 2 9 5 34 3H 323 332 342 35i 36i | 5- 6.0 61 370 380 389 398 408 4i7 427 436 445 455 7 7.0 62 464 474 483 492 502 5" 521 530 539 549 8 8.0 63 64 558 652 567 661 577 671 586 680 596 689 605 699 614 708 624 717 633 727 642 736 9 9.0 65 745 755 764 773 783 792 801 811 820 829 66 839 848 857 867 876 885 894 904 913 922 67 932 941 95 960 969 978 987 997 *oo6 *oi5 68 67025 034 043 052 062 071 080 089 099 108 69 117 127 136 H5 '54 164 173 182 j 191 20 1 470 210 219 228 237 247 256 265 274 284! 293 7i 3 02 3ii 321 33 339 348 357 367 376 3^5 i 0.9 72 394 403 4i3 422 43i 440 449 459 468 477 2 1.8 73 486 495 504 5*4 523 532 54i 55: 56o 569 7 2.7 74 578 587 596 605 614 624 633 642 i 651 660 O 4 / 3-6 75 669 679 688 697 706 7i5 724 733 742 752 5 4-5 76 761 770 779 788 797 806 815 825 834 843 6 5-4 77 852 861 870 879 888 897 906 916 925 934 7 6-3 78 943 95 2 961 97 979 988 997 *oo6 *oi5 *O24 8 7- 2 79 68034 043 052 06 1 070 079 088 097 1 06 "5 9 8.1 480 124 133 142 151 1 60 169 178 187 196 205 81 215 224 233 242 251 260 269 278 : 287 296 82 305 3H 323 332 34i 350 359 368 I 377 386 83 395 404 4i3 422 43 1 440 449 458 i 467 476 84 485 494 502 5" 520 529 538 547 556: 565 85 574 583 592 60 1 610 619 628 637 646 655 86 664 673 681 690 699 708 717 726 735 i 744 8 87 753 762 771 780 789 797 806 815 824 833 i 0.8 88 842 851 860 869 878 886 895 904 9i3 922 2 1.6 89 93 1 940 949 958 966 975 984 993 *OO2 *OII 3 2.4 490 69 020 028 037 046 55 064 73 082 090 099 4 3-2 9i 108 117 126 i35 144 1 S 2 161 170 I 79 1 88 5 4.0 . o 92 197 205 214 223 232 241 249 258 267 276 4.0 e f\ 93 285 294 302 3" 320 329 338 346 355 364 7 I 6 A 94 373 38i 390 399 408 4^7 425 434 443 452 9 0.4 7.2 95 461 469 478 487 496 54 5'3 522 53i 539 96 548 557 566 574 583 592 601 609 618 627 97 636 644 653 662 671 679 688 697 75 714 98 723 732 740 749 758 767 775 784 793 801 99 810 819 827 836 84! 854 862 871 880 888 N. 1 2 3 4 5 6 7 8 9 Pp. Pts. 500-549 IX N. O 1 2 3 4 5 6 7 | 8 9 Pp. Pts. 500 69897 906 914 923 932 940 949 958 966 975 01 984 992 *OOI *OIO *oi8 *02 7 "036 *044 *53 *062 02 70 070 079 088 096 105 114 122 131 140 148 3 J 57 165 *74 183 191 2OO 209 217 226 234 04 243 252 260 269 278 286 293 33 312 321 05 329 338 346 353 364 372 38i 389 398 406 9 06 4i5 424 432 441 449 458 467 475 484 492 i ?8 07 08 5 01 586 59 593 518 603 526 612 535 621 544 629. 1% 561 646 % 578 663 2 3 I.o 2 '7 09 672 680 689 697 706 714 723 73i 740 749 4 3-6 510 757 766 774 783 791 800 808 817 825 834 i 4-5 c.4 ii 842 851 859 868 876 885 893 902 910 919 J i 6.3 12 927 935 944 952 961 969 978 986 993 *00 3 g O 7.2 13 71 OI2 020 029 037 046 054 063 071 079 088 9 8.1 14 9 6 103 H3 122 130 39 147 '55 164 172 15 181 189 198 206 214 223 231 240 248 2 57 16 265 273 282 290 299 307 315 324 332 34i J 7 349 357 366 374 383 39i 399 408 416 423 18 433 441 4jo 458 466 475 483 492 500 508 19 5 J 7 525 533 542 550 559 567 575 584 592 520 600 609 617 625 634 642 650 659 667 675 8 21 684 692 700 709 717 725 734 742 75 759 j 0.8 22 767 784 792 800 809 817 825 834 842 2 1.6 23 850 858 867 875 883 892 900 908 917 923 2 2.4 24 933 941 95 958 966 975 983 991 999 *oo8 O 4 **T 3-2 25 72 016 024 032 041 049 57 066 074 082 090 5 4.0 26 099 107 115 123 132 140 148 156 165 173 6 4-8 2 7 181 189 198 206 214 222 230 ^39 247 255 7 5.6 28 263 272 280 288 296 34 313 321 329 337 8 6-4 29 346 354 362 370 378 387 395 403 411 419 9 7.2 530 428 436 444 452 460 469 477 485 493 5oi 3i 59 518 526 534 542 550 558 567 575 583 32 59i 599 607 616 624 632 640 648 656 665 33 673 681 689 697 705 713 722 730 738 746 34. 754 762 770 779 787 795 803 811 819 827 35 835 843 852 860 868 876 884 892 900 908 36 916 925 933 941 949 957 965 973 981 989 7 37 997 *oo6 *oi4 *022 *O3O ="038 *o 4 6 *054 *062 *O7O i 0.7 3 3 4 5 6 7 8 9 Pp. Pts. Xll 650-699 N. 1 | 2 3 4 5 6 7 8 9 Pp. Pts. 650 81 291 298 30? 3 11 3i8 32? 33i 338 34? 35 1 5 1 358 36? 378 38? 398 405 411 418 52 425 431 438 44? 458 471 478 48? 53 491 498 505 5" 518 52? 53' 538 544 551 54 558 564 578 584 598 604 611 617 55 624 631 637 644 651 657 664 671 677 684 56 690 697 704 710 717 723 730 737 743 75 57 757 763 770 776 783 790 796 803 809 816 58 823 829 836 842 849 856 862 869 875 882 59 889 895 902 908 9i? 921 928 93? 941 948 660 954 961 968 974 981 987 994 *ooo *oo7 *oi4 7 r\ *r 61 82 020 027 033 040 046 053 060 066 073 079 0.7 62 086 092 099 I0 5 112 119 I2 5 132 138 H5 1.4 63 151 158 164 171 I 7 8 184 191 197 204 2IO 3 2. 1 64 217 223 230 236 243 249 256 263 269 2 7 6 5 3-5 65 282 289 295 302 308 315 321 328 334 341 6 4.2 66 347 354 360 373 380 387 393 400 406 7 4.9 67 413 419 426 432 439 445 452 458 46? 471 8 5-6 68 478 484 491 497 54 5*> 5 X 7 523 53 536 9 6.3 69 543 549 556 562 569 575 582 588 59? 60 1 670 607 614 620 627 633 640 646 653 659 666 7 1 672 679 685 692 698 70? 711 718 724 73 72 737 743 75 756 763 769 776 782 789 795 73 802 808 814 821 827 834 840 847 853 860 74 866 872 879 885 892 898 905 911 918 924 75 93 937 943 95 956 963 969 * 975 982 988 76 995 *OOI *oo8 *oi4 *020 *O27 *033 *046 *O52 77 83059 065 072 078 08? 091 097 104 I 10 117 78 123 129 136 142 149 155 161 1 68 174 181 79 187 193 200 206 2I 3 219 225 232 238 24? 680 25 1 257 264 270 2 7 6 283 289 296 302 308 81 3 I 5 321 327 334 340 347 353 359 366 372 6 82 378 385 398 404 410 423 429 43 6 0.6 83 442 448 455 461 467 474 480 487 493 499 2 1.2 84 506 5 12 518 52? 531 537 544 550 556 563 3 1.8 85 569 575 582 588 594 601 607 613 620 626 4 2.4 86 632 639 651 658 664 670 677 683 689 5 3.0 87 696 702 708 721 727 734 740 746 753 6 3-6 oo 55 759 765 771 778 784 790 797 803 809 816 7 4.2 89 822 828 835 841 847 853 860 866 872 879 8 4.8 690 88? 891 897 904 910 916 923 929 935 * 942 9 54 91 948 954 960 967 973 979 985 992 998 92 84011 017 023 029 036 042 048 05? 061 067 93 73 080 086 092 098 10? in 117 123 130 94 136 142 148 155 161 167 173 180 1 86 192 95 198 205 211 217 223 230 236 242 248 25? 96 261 267 273 280 286 292 298 30? 3" 317 97 323 330 336 342 348 354 361 367 373 379 98 386 392 398 404 410 417 423 429 435 442 99 448 454 460 466 473 479 48? 491 497 504 N. O 1 2 3 4 5 | 6 7 8 9 Pp. Pts. 700-749 Xlll N. O 1 2 3 4 5 6 7 8 9 Pp. Pts. 700 84 510 516 522 528 533 54 1 547 553 559 566 OI 572 578 584 590 597 603 609 615 621 628 02 634 640 646 652 658 665 671 677 683 689 03 696 702 708 714 720 726 733 739 743 75 i - 04 757 763 770 776 782 788 794 800 807 05 819 825 831 837 844 850 856 862 868 874 7 06 880 887 893 899 905 911 917 924 930 936 I 0.7 07 942 948 954 960 967 973 979 985 991 997 2 1.4 08 85 003 009 016 022 028 034 040 046 052 058 3 2.1 _ Q 09 065 071 077 083 089 095 IOI 107 114 1 20 4 2.8 710 126 132 138 144 150 156 163 169 173 181 | 3-5 4.2 ii 187 193 199 205 211 217 224 230 236 242 7 4.9 12 248 254 260 266 272 278 283 291 297 303 8 5.6 14 309 370 376 321 382 388 333 394 339 400 406 352 412 358 418 364 423 9 6-3 15 43 1 437 443 449 455 461 467 473 479 485 16 491 497 53 59 5 l6 522 528 534 540 546 17 55 2 558 5 6 4 570 576 582 588 594 600 606 18 612 618 625 631 637 643 649 653 66 1 667 19 673 679 685 691 697 73 709 721 727 720 733 739 745 751 757 763 769 775 781 788 6 21 794 800 806 812 818 824 830 836 842 848 0.6 22 854 860 866 872 878 884 890 896 902 908 2 1.2 23 914 920 926 932 938 * 944 93o 95 6 962 968 7 1.8 24 974 980 986 992 998 *010 *oi6 *022 *028 j 4 2.4 25 86034 040 046 052 058 064 070 076 082 088 5 3-0 26 094 IOO 1 06 112 118 124 130 136 141 147 6 3-6 2 7 '53 J 59 165 171 177 183 189 195 201 207 7 4.2 28 213 219 225 231 237 243 249 253 26l 267 8 4.8 29 273 279 285 291 297 33 308 320 326 9 54 730 332 338 .344 350 356 362 368 374 380 386 31 392 398 404 410 421 427 433 439 445 32 33 457 5i6 463 522 469 528 473 534 481 540 487 546 493 55 2 499 558 54 564 34 570 576 581 587 593 599 605 611 617 623 35 629 633 641 646 652 658 664 670 676 682 36 688 694 700 705 711 717 723 729 733 741 5 11 747 806 753 812 759 817 764 823 77 829 776 835 782 841 788 847 794 853 800 859 2 I.O 39 864 870 876 882 888 894 900 906 911 917 3 1.5 740 923 929 933 941 947 953 958 964 97 976 4 2.0 41 982 988 994 999 *oo5 *OII *oi7 *O23 *029 *33 5 2-5 42 87 040 046 052 058 064 070 75 08 1 087 093 6 3- 43 099 io3 in 116 122 128 134 140 146 151 I 3-5 44 '57 163 !6 9 173 181 1 86 192 198 204 2IO O 4.0 45 216 221 227 233 239 243 251 256 262 268 9 4-5 46 274 280 286 291 297 303 309 313 320 326 47 332 338 344 349 355 361 367 373 379 384 48 390 396 402 408 419 425 437 442 49 448 454 460 466 47i 477 483 489 495 5 00 N. O 1 2 3 4 5 6 7 8 9 Pp. Pts. XIV 750-799 N. O 1 2 3 4 5 6 7 c* 9 Pp. Pts. 750 87506 5^2 518 523 529 535 54i 547 SS 2 558 5 1 564 57 576 581 587 593 599 604 610 616 52 622 628 633 639 645 651 656 662 668 674 53 679 685 691 697 703 708 714 720 726 73i 54 737 743 749 754 760 766 772 777 783 789 55 791 800 806 812 818 823 829 835 841 846 5 6 852 858 864 869 875 88 1 887 892 898 904 57 910 915 921 927 933 938 944 950 955 961 58 967 973 978 984 990 996 *OOI *oo7 *oi3 *oi8 59 88024 030 036 041 047 053 058 064 070 076 760 081 087 093 098 104 no 116 121 127 133 6 _. 61 138 144 150 156 161 167 173 178 184 190 i o.o 1/7 62 '95 20 1 207 213 218 224 230 235 241 247 4fi T R 63 252 258 264 270 275 281 287 292 298 304 3 I.o 64 309 3i5 321 326 332 338 343 349 355 360 4 5 2.4 3-0 8 366 423 372 429 377 434 383 440 389 446 395 45i 400 .457 406 463 412 468 417 474 6 7 3-6 4.2 67 480 485 491 497 502 508 5i3 519 525 530 8 4.8 68 536 542 547 553 559 5 6 4 570 576 581 587 9 5.4 69 593 598 604 610 615 621 627 632 638 643 770 649 655 660 666 672 677 683 689 694 700 7i 705 711 717 722 728 734 739 745 750 756 72 762 767 773 779 784 790 795 801 807 812 73 818 824 829 835 840 846 852 857 863 868 74 874 880 885 891 897 902 908 913 919 925 75 930 936 941 947 953 958 964 969 971 981 76 986 992 997 *oo3 *oo9 *oi4 *O2O *O25 *03i *037 77 89 042 048 053 059 064 070 076 08 1 087 092 78 098 104 109 "5 1 20 126 131 137 H3 148 79 154 '59 165 170 176 182 I8 7 193 198 204 780 209 215 221 226 232 237 243 248 254 260 81 265 271 2 7 6 282 287 293 298 304 310 3i5 82 321 326 332 337 343 348 354 360 365 37 1 i o c 83 376 382 387 393 398 404 409 415 421 426 2 o I.O 84 432 437 443 448 454 459 465 470 476 481 3 i-5 85 487 492 498 54 509 5*3 520 526 53i 537 4 2.0 86 542 548 553 559 5 6 4 57 575 581 586 592 5 2-5 87 597 603 609 614 620 625 631 636 642 647 6 3-0 88 653 658 664 669 675 680 686 691 697 702 7 3-5 89 708 713 719 724 730 735 741 746 752 757 8 4.0 790 763 768 774 779 785 790 796 801 807 812 9 4-5 9i 818 823 829 -834 840 845 851 856 862 867 92 873 878 883 889 894 900 905 911 916 922 93 927 933 938 944 949 955 960 966 971 977 94 982 988 993 998 *oo4 *oo9 "015 *020 *026 *Q3i 95 90037 042 048 053 059 064 069 7 5 080 086 96 091 097 IO2 108 H3 119 124 129 135 140 97 146 IS 1 J 57 162 1 68 173 179 184 189 195 98 200 206 211 217 222 227 233 238 244 249 99 255 260 266 271 2 7 6 282 287 293 298 34 N. 1 2 3 4 5 | 6 7 8 9 Pp. Pts. 800-849 XV N. 1 2 3 4 5 6 7 8 9 Pp. Pts. 800 90309 3H 320 325 33 i 336 342 347 352 358 01 363 369 374 380 385 390 396 401 407 412 02 417 423 428 434 439 445 450 455 461 466 03 472 477 482 488 493 499 54 509 515 520 04 526 536 542 547 553 558 563 569 574 5 580 585 590 596 601 607 612 617 623 628 06 634 639 644 630 655 660 666 671 677 682 07 687 693 698 73 709 714 720 725 730 736 08 741 747 75 2 757 763 768 773 779 784 789 09 795 800 806 811 816 822 827 832 838 843 6 810 849 854 859 865 870 875 881 886 891 897 ! 0.6 n 902 907 913 918 924 929 934 940 945 * 95 2 1.2 12 95 6 961 966 972 977 982 988 993 998 3 1.8 13 91 009 014 020 025 030 036 041 046 052 57 4 2.4 H 062 068 73 078 084 089 094 100 105 no 5 3- 15 116 121 126 132 137 142 148 153 158 164 6 3-6 16 169 174 1 80 190 196 201 206 212 217 7 4.2 *7 222 228 233 238 243 249 254 259 263 270 8 4.8 18 275 28l 286 291 297 302 307 312 318 323 9 6.4 19 328 334 339 344 350 355 360 365 371 376 820 381 387 392 397 403 408 413 418 424 429 21 434 440 445 45 455 461 466 47 i 477 482 22 487 492 498 503 508 5'4 5 I 9 524 529 535 23 540 545 55 1 556 561 566 572 577 582 587 24 593 598 603 609 614 619 624 630 631 640 2 5 645 651 656 66 1 666 672 677 682 687 693 26 698 703 709 714 719 724 730 735 740 745 27 75 1 756 761 766 772 777 782 787 793 798 28 803 808 814 819 824 829 834 840 845 850 29 855 861 866 871 876 882 887 892 897 903 830 908 913 918 924 929 934 939 944 91 ^955 3 1 960 965 971 976 981 986 991 997 *OO2 32 92 012 018 023 028 033 038 044 049 054 059 f\ K 33 06^ 070 075 080 085 091 096 101 106 in 2 o I O 34 122 127 132 137 H3 148 '53 158 163 3 35 169 174 179 184 189 195 200 205 210 215 4 2.0 36 221 226 231 236 241 247 252 2 57 262 267 2-5 37 273 2 7 8 283 288 293 298 34 309 314 3i9 6 3-0 38 324 330 335 340 345 35 355 3 66 7 3-5 39 376 381 387 392 397 402 407 412 418 423 8 4.0 840 428 433 438 443 449 454 459 464 46 9 474 9 4-5 41 480 490 495. 500 5 11 516 521 526 42 531 536 542 547 552 557 562 567 572 578 43 583 588 593 598 603 609 614 619 624 629 44 634 639 64? 650 655 660 665 670 675 68 1 45 686 691 696 701 706 711 716 722 727 732 46 47 737 788 742 793 747 799 752 804 % 763 814 768 819 773 824 77 8 829 783 834 48 840 845 850 855 860 865 870 875 881 886 49 891 896 901 906 911 916 921 927 932 937 N. 1 2 3 * 5 6 7 8 9 Pp. Pts. XVI 850-899 N. 1 | 2 3 4 5 6 7 8 9 Pp. Pts. 850 92 942 947 952 957 962 967 973 978 * 983 988 Si 993 998 "003 *oo8 *oi3 *oi8 *024 *029 *39 52 9344 049 054 059 064 069 075 080 *o1t 090 53 095 IOO I0 5 no "5 1 20 I2 5 131 136 141 54 146 151 156 161 1 66 171 176 181 186 192 55 197 202 207 212 217 222 227 232 237 242 56 247 252 258 263 268 273 278 283 288 293 6 r (\ 57 298 303 308 313 3*8 323 328 334 339 344 i O.O 58 349 354 359 364 369 374 379 384 389 394 2 1.2 59 399 404 409 414 420 425 430 435 440 445 3 4 2-4 860 450 455 460 465 47 475 480 485 490 495 ^ 3' 61 500 55 5 10 5*5 520 526 53 1 536 54i 546 ft 36 62 551 556 561 566 57 1 576 581 586 596 7 O 4.2 63 601 606 611 616 621 626 631 636 641 646 8 4-8 64 651 656 661 666 671 .676 682 687 692 697 9 T^ 54 65 702 707 712 717 722 727 732 737 742 747 66 75 2 757 762 767 77 2 777 787 792 797 67 802 807 812 817 822 827 832 837 842 847 68 852 857 862 867 872 877 882 887 892 897 69 902 907 912 917 922 927 932 937 942 947 870 952 957 962 967 972 977 982 987 992 997 ! s 7 1 94 002 007 012 017 022 027 032 037 042 047 I 0.5 72 052 057 062 067 7 2 077 082 086 091 096 2 I.O 73 101 106 III 116 121 126 *3* 136 141 146 I.B 74 I 5 I 156 161 1 66 171 176 181 1 86 191 196 4 J 2.0 75 201 | 206 211 216 221 226 231 236 240 245 5 2.5 76 250 255 260 265 2 7 275 280 285 290 295 o 3- 78 3 00 349 305 354 310 359 315 364 320 369 325 374 330 335 340 379 384 389 345 394 1 3-5 4.0 79 399 404 409 414 419 424 429 433 438 443 9 4-5 880 448 453 458 463 468 473 478 j 483 488 493 81 498 503 507 5^2 SJ7 522 527 532 537 542 82 83 547 596 552 60 1 557 606 562 6n 567 616 571 621 III I 81 630 586 635 640 84 645 650 655 660 665 670 675 680 685 689 85 694 699 704 709 714 719 724 729 734 738 4 86 743 748 753 763 768 773 778 783 787 l 0.4 87 792 797 802 807 812 817 822 827 832 836 2 0.8 88 841 846 851 856 861 866 871 876 880 885 3 1.2 89 890 895 900 90S 910 915 919 924 929 934 4 1.6 890 939 944 949 954 * 959 963 968 973 978 983 5 2.O 9 1 988 993 998 *002 *OI2 *oi7 *O22 *O27 *O32 X 92 95 O 36 041 046 051 056 06 1 066 7 I 075 080 ^ 3*2 93 085 090 095 IOO 105 109 114 119 124 129 94 134 139 143 148 I 5 8 163 1 68 173 177 95 182 187 192 197 202 207 211 216 221 226 96 231 236 240 245 250 255 260 265 2 7 274 97 279 284 289 294 299 33 308 313 318 323 98 328 332 337 342 347 35 2 357 361 366 371 99 376 386 390 395 400 405 410 415 419 N. O 1 2 3 4 5 6 7 8 9 Pp. Pts. 900-949 N. O 1 2 3 4 5 6 7 8 9 Pp. Pts. 900 95424 429 434 439 444 448 453 458 463 468 01 472 477 482 487 492 497 501 506 5" 516 02 52i 525 53 535 540 545 550 554 559 5 6 4 03 569 574 578 583 588 593 598 602 607 612 04 617 622 626 631 636 641 646 650 655 660 5 665 670 674 679 684 689 694 698 73 708 06 7U 718 722 727 732 737 742 746 75i 756 07 761 766 77 775 780 785 789 794 799 804 08 809 813 818 823 828 832 837 842 847 852 09 856 861 866 871 875 880 885 890 895 899 910 904 909 914 918 923 928 933 938 942 947 i 0.5 ii 95 2 957 961 966 971 976 980 985 990 995 2 I.O 12 999 *oo4 *oo9 *oi4 *oi9 *023 *028 *0 33 *o 3 8 *O42 3 1.5 U 96047 052 057 061 066 071 076 080 085 090 4 2.O H 095 099 104 109 114 118 123 128 133 J 37 5 2-5 IS 142 H7 152 156 161 1 66 171 175 180 185 6 3-o 16 190 194 199 204 209 213 218 223 227 232 7 3-5 1 7 237 242 246 2 5 l 256 261 265 270 275 280 8 4.0 18 284 289 294 298 303 308 3U 3 1 7 322 327 9 4-5 19 332 336 34i 346 350 355 360 365 369 374 920 379 384 388 393 398 402 407 412 417 421 21 426 43i 435 440 445 450 454 459 464 468 22 473 478 483 487 492 497 5 01 506 5" 5 J 5 23 520 525 530 534 539 544 548 553 558 562 24 567 572 577 58i 586 59i 595 600 605 609 2 5 614 619 624 628 633 638 642 647 652 656 26 661 666 670 ^75 680 685 689 694 699 73 27 708 7 1 3 717 722 727 73i 736 741 745 750 28 755 759 764 769 774 778 783 788 792 797 29 802 806 811 816 820 825 830 834 839 844 930 848 853 858 862 867 872 876 881 886 890 3 1 895 900 904 909 914 918 923 928 932 937 32 942 946 95 * 95 6 960 965 970 974 979 984 4 33 988 993 997 *002 *oo 7 *OII *oi6 *02I *O25 *o^o 0.4 o 8 34 97035 039 044 049 053 058 063 067 072 077 3 LJ.O 1.2 35 081 086 090 95 IOO 104 109 114 118 123 4 1.6 36 128 132 *37 142 146 !5* J 55 1 60 165 169 5 2.0 37 174 179 183 1 88 192 197 202 206 211 216 6 2.4 38 220 225 230 234 239 243 248 253 257 262 7 2.8 39 267 271 276 280 285 290 294 299 34 308 8 3-2 940 3i3 317 322 327 331 336 340 345 350 354 9 3-6 41 359 364 368 373 377 382 387 39i 396 400 42 405 410 414 419 424 428 433 437 442 447 43 45i 456 460 465 470 474 479 483 488 493 44 497 502 506 5 11 516 520 525 529 534 539 45 543 548 552 557 562 566 57i 575 580 585 46 589 594 598 603 607 612 617 621 626 630 47 635 640 644 649 653 658 663 667 672 676 48 681 685 690 695 699 704 708 7U 717 722 49 727 73i 736 740 745 749 754 759 763 768 N. 1 2 3 4 5 6 7 8 9 Pp. Pts. XV111 950-999 N. 1 2 3 4 5 6 rr 8 9 Pp. Pts. 950 97 772 777 782 786 791 795 800 804 809 813 5 1 818 823 827 832 836 841 845 850 855 859 52 864 868 873 877 882 886 891 896 900 905 53 909 914 918 923 928 932 937 941 946 95 54 955 959 964 968 973 978 982 987 991 996 55 98 ooo 005 009 014 019 023 028 032 037 041 56 046 050 055 059 064 068 073 078 082 087 57 091 096 IOO 105 109 114 118 123 127 132 58 137 141 146 150 155 J 59 164 1 68 J 73 177 59 182 1 86 191 '95 200 204 209 214 218 223 960 227 232 236 241 245 250 254 259 263 268 61 272 277 281 286 290 295 299 34 308 313 62 318 322 327 33i 336 340 345 349 354 358 63 363 367 372 376 381 385 390 394 399 403 5 f\ f 64 408 412 417 421 426 430 435 439 444 448 2 -5 I.O 65 453 457 462 466 471 475 480 484 489 493 3 1.5 66 498 502 57 5" 5 l6 520 525 529 534 538 4 2.Q 67 68 543 588 547 592 552 597 it? f 605 f5 610 570 614- 574 619 579 623 583 628 I 2-5 3-Q 69 632 637 641 646 650 655 659 664 66.8 673 7 3-5 970 677 682 686 691 695 700 704 709 713 717 8 4.0 7 1 722 726 73i 735 740 744 749 753 758 762 9 4-5 72 767 771 776 780 784 789 793 798 802 807 73 811 816 820 825 82 9 834 838 843 847 851 74 856 860 865 869 874 878 883 887 892 896 75 900 905 909 914 9l8 923 927 932 936 941 76 941 949 954 958 963 967 972 976 981 985 77 989 994 998 *oo3 *oo7 *OI2 *or6 *02I *O25 *029 78 99034 038 043 047 052 056 06 1 065 069 074 79 078 083 087 092 096 IOO I0 5 IO9 114 118 980 123 127 131 136 140 H5 149 *54 158 162 81 167 171 176 1 80 185 I8 9 193 198 202 207 4 82 211 216 220 224 229 233 238 242 247 2 5 * i 0.4 83 255 260 264 269 273 277 282 286 291 295 2 0.8 84 300 34 308 3*3 3 1 7 322 326 330 335 339 3 1.2 85 344 348 352 357 36i 366 370 374 379 383 4 5 1.6 2.O 86 388 392 396 401 405 410 414 419 423 427 6 2 4. 87 432 436 441 445 449 454 458 463 467 471 *T 2.8 88 476 480 484 489 493 498 502 506 5" 5 X 5 8 2 2 89 520 524 528 533 537 542 546 550 555 559 9 J** 3-6 990 564 568 572 577 58i 585 590 594 599 603 9i 607 612 616 621 625 629 634 638 642 647 92 651 656 660 664 669 673 677 682 686 691 93 695 699 704 708 712 717 721 726 730 734 94 739 743 747 752 756 760 765 769 774 778 95 782 787 791 795 800 804 808 813 817 822 96 826 830 835 839 843 848 852 856 86 1 865 97 870 874 878 883 887 891 896 900 904 909 98 9i3 917 922 926 .93 935 939 944 948 95 2 99 957 961 965 970 974 978 983 987 991 996 N. O 1 2 3 4 5 6 7 8 9 Pp. Pts. THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO SO CENTS ON THE FOURTH DAY AND TO $1.OO ON THE SEVENTH DAY OVERDUE. 6 1*48 ,**; ., LD 21-100m-12, '43 (8796s) VB 17267 M306CH5 QA THE UNIVERSITY OF CALIFORNIA LIBRARY