UNIVERSITY OF CALIFORNIA 
 AT LOS ANGELES 
 
 GIFT OF 
 
 Harriet E. Glazier
 
 ELEMENTARY ALGEBRA
 
 THE MACMILLAN COMPANY 
 
 NEW YORK BOSTON CHICAGO DALLAS 
 ATLANTA SAN FRANCISCO 
 
 MACMILLAN & CO., LIMITED 
 
 LONDON BOMBAY CALCUTTA 
 MELBOURNE 
 
 THE MACMILLAN CO. OF CANADA, LTD. 
 
 TORONTO
 
 ISAAC NEWTON 
 
 One of the greatest mathematicians of all time. As a young man he 
 invented the Binomial Theorem which is now studied in second year 
 courses in algebra. Newton wrote a Universal Arithmetic, which was 
 really a book on algebra. He was the first to use fractional and negative 
 exponents as we write them. He first used them in a letter dated 
 June 13, 1676.
 
 gear Course 
 
 BY 
 FLORIAN CAJORI 
 
 COLORADO COLLEGE 
 AND 
 
 LETITIA R. ODELL 
 
 NORTH SIDE HIGH SCHOOL, DENVER 
 
 Kefor ffork 
 THE MACMILLAN COMPANY 
 
 1916 
 
 All rights reserved
 
 COPYRIGHT, 1915, 
 BY THE MACMILLAN COMPANY. 
 
 Set up and electrotyped. Published June, 1915. Reprinted 
 June, 1916. 
 
 XortoooB lircss 
 
 J. 8. Cnshing Co. Berwick & Smith Co. 
 Norwood, Mass., U.S.A.
 
 152. 
 
 C\'> 
 
 V- I 
 
 Mathematical 
 
 Sciences 
 
 Liorary 
 
 PREFACE 
 
 IN this book algebra is presented to beginners in a simpler, 
 clearer, and more practical form than is usually found in 
 school texts. 
 
 The treatment is simplified by the omission of certain re- 
 
 dundant terms and notations, by maintaining an intimate 
 
 connection between algebra and arithmetic, and by starting 
 
 with definite assumptions of the laws of signs in subtraction 
 
 O and multiplication rather than with complicated and unsat- 
 
 NjT"' isfactory proofs. The part of algebra that deals with the 
 
 P- mechanical manipulation of algebraic expressions is reduced 
 
 in amount as much as is consistent with the acquirement of 
 
 accuracy. Simple fractions and easy radicals are introduced 
 
 (O early ; the discussion of fractions with binomial and trinomial 
 
 ~~7 denominators, and consideration of the more difficult parts of 
 
 v radicals, are postponed to a time near the close of the course. 
 
 N, Katio and proportion are brought into closer touch with the 
 
 equation and with graphs. 
 
 Clearness has been sought by careful definition, copious 
 
 f illustration, and by the use of language which recalls the 
 axiomatic processes involved. Such phrases as "clearing of 
 fractions " are objectionable because of the danger of their 
 . being applied by the pupil, not to the equation alone, but to 
 ^ algebraic expressions in general. " Canceling," as now used, 
 ^ is ambiguous, for it sometimes involves subtraction, at other 
 ^ times division. 
 
 The treatment is rendered more practical by a careful selec- 
 tion of problems. Mechanical problems involving the theory 
 
 348554
 
 Vi PREFACE 
 
 4 
 
 of the lever, specific mass, specific heat, the motion of falling 
 bodies, and technical terms in electricity are omitted at first, 
 and even near the end they are used only sparingly. Expe- 
 rience has shown that first year pupils in the high school 
 cannot cope successfully with the abstract concepts of me- 
 chanics. These topics belong more properly to the course in 
 physics usually given in the third or fourth year. In this 
 text stress is laid on practical applications to problems arising 
 in business. In such applications the pupil is merely continu- 
 ing on a somewhat higher plane the subjects first approached 
 in arithmetic. A distinguishing feature of this text is the 
 practical application of graphs. The ordinary procedure is 
 to use the graph merely in presenting to the eye the behavior 
 of two variables in an equation. In this text it serves, in 
 addition, for the determination by inspection of results de- 
 cidedly practical in character. 
 
 The authors are indebted for valuable criticisms and sug- 
 gestions to several teachers, but more particularly to Principal 
 E. L. Brown of the North Side High School in Denver. 
 
 FLORIAN CAJORI. 
 LETITIA R. ODELL.
 
 CONTENTS 
 
 I. INTRODUCTION 1 
 
 II. POSITIVE AND NEGATIVE NUMBERS ..... 14 
 
 III. MULTIPLICATION AND DIVISION ...... 41 
 
 IV. PROPORTION, GRAPHS ........ 58 
 
 V. EQUATIONS INVOLVING FRACTIONAL COEFFICIENTS . . 77 
 
 VI. SPECIAL PRODUCTS ........ 84 
 
 VII. FACTORING 89 
 
 VIII. EQUATIONS SOLVED BY FACTORING ..... 104 
 
 IX. MULTIPLICATION AND DIVISION OF POLYNOMIALS . . Ill 
 
 X. SQUARE ROOT ......... 114 
 
 XL GRAPHS OF SIMULTANEOUS EQUATIONS ..... 119 
 
 XII. SIMULTANEOUS LINEAR EQUATIONS ..... 126 
 
 XIII. QUADRATIC EQUATIONS ........ 141 
 
 XIV. FRACTIONS 162 
 
 XV. RADICALS AND GENERAL EXPONENTS 178 
 
 XVI. REVIEW OF ESSENTIALS OF ALGEBRA . . 192 
 
 vii
 
 CHAPTER I 
 
 INTRODUCTION 
 
 1. Algebra, like arithmetic, deals largely with the solution 
 of problems involving numbers. It is an extension of arith- 
 metic which enables one to solve more complicated problems. 
 It extends the field of arithmetic by means of three devices : 
 the use of letters as well as Hindu-Arabic numerals in the 
 study of numbers ; the introduction of new kinds of numbers 
 to be used in conjunction with those of arithmetic ; and the 
 development of simple methods of operation with those 
 numbers. 
 
 The signs of operation used in arithmetic are used also in 
 algebra. 
 
 Thus addition is indicated by -f-, subtraction by , multiplication by 
 x , division by -* . 
 
 The sign of multiplication ( x ) is often replaced in algebra by a dot (), 
 which is written above the line to distinguish it from a decimal point. 
 Thus 3 4 means 3x4. 
 
 The product of two numbers, one or both of which are" represented by 
 letters, is usually indicated by writing the numbers one after the other 
 without any sign between them. Thus ab means a x b or a 6, 4 c means 
 4 x c or 4 c. 
 
 When only one of two numbers is represented by a letter, it is custom- 
 ary to write the letter last. Thus we write 4 c, but not c 4. 
 
 In arithmetic, and also in algebra, division is frequently expressed in 
 the form of a fraction. Thus - = 4 -=- 5, - = a + b. 
 
 = is the sign of equality. 
 
 2. Any combination of numbers, letters, and symbols of opera- 
 tion, which represents a number, is called an algebraic expression. 
 
 Thus, a + 6, a ~ , 4 ab *- c are algebraic expressions. 
 2t C
 
 2 ELEMENTARY ALGEBRA 
 
 3. In arithmetic it is customary to use abbreviations, such, 
 as ft. for "foot," in. for "inch," A. for "acres," and so on. In 
 algebra the practice of using abbreviations is carried much 
 further. If we write i for " interest," p for " principal," r for 
 " rate," and t for " time," then the statement 
 
 " The interest is equal to the product of the principal, the 
 
 rate, and the time " 
 
 can be expressed briefly by the equation 
 
 i = prt. 
 
 By the use of such a system of shorthand much time and 
 labor is saved. The same letter may be used for different 
 things in different problems. 
 
 The letter m may be used for " miles " in one problem and 
 "minutes" in another. The letter x is used extensively to 
 designate a number which is unknown at the outset, but which 
 is to be determined by the solution of the problem in hand. 
 In one case x may mean the number of " dollars," in another 
 case the number of " pounds," in a third case the number of 
 " years." The meaning of a letter must be made plain in each 
 problem. To avoid confusion, no letter should stand for two 
 different things in the same problem. 
 
 If i means " inches " and / means " feet," then 
 .48 i = 4/ means "48 inches are equal to 4 feet." 
 
 15/+ 4 i means " 15 feet and 4 inches." If n stands for any 
 number, then 2 n -f 50 means " 2 times any number, increased 
 by 50," or " twice any number, plus 50." If n is taken equal to 
 3, then 2n + 50 = 2 3 + 50 = 56. If n = 25, then 2n + 50 
 = 2 25 + 50 = 100. 
 
 ORAL EXERCISES 
 
 4. Express in algebraic symbols a number which is 
 
 1. 5 more than x. 4. 10 less than five times c. 
 
 2. 5 less than x. 5. 20 less x. 
 
 3. .8 more than three times x. 6. 1.5 less five times a.
 
 INTRODUCTION 3 
 
 7. 6 less than y. 9. .18 less than ten times z. 
 
 8. 1.2 more than y. 10. 5 more than ^ of y. 
 
 11. 1 less than twice y. 
 
 Supplying for each letter a given number, express in words 
 exercises 12-25 : 
 
 12. 16 g. 19. 
 
 13. g + l. 20. 
 
 14. 2.T+31. 21. 7-4y. 
 
 15. 2x 5. 22. 4y ly 5. 
 
 16. 7z 13. 23. 15 | y. 
 
 17. 13-5z. 24. 17x + 5x = 22x. 
 
 18. i^-3. 25. 12% -3 a; = 9; x. 
 
 26. If t = interest, p = principal, t = time, r = rate, express 
 
 2 y y 
 
 in words i = jprt, p = , r = , t = . 
 rt pt ' pr 
 
 Find the values of the following expressions if a = 2, 6 = 3, 
 c = 5, d = 19 : 
 
 27.J* a -f- 2J>. 32. a+2b + 3c-d. 
 
 28. 3zT+~c. 33. d 5 a + 7 6 c. 
 
 29. 6c-&. 34. + a. 
 
 30. 9c + 4d. 35. 59 5c + 106. 
 
 31. a + b + c + d. 36. 2d + 199c-25a-19. 
 
 If m = 15, n = 12, p = 20, g = 0, what are the values of the 
 following algebraical expressions ? 
 
 37. m + n+p + q. 39. 12 n + 2p + 5m q. 
 
 38. g + 5p 6m. 40. 6p + 12 + 2w 3m. 
 (When g=0, then 12 g is equal to 12 times ; this gives the product 0.) 
 
 41. 2 m + ?i -(- 12 q. 43. 5 n + m 7 q. 
 
 42. 3p + 5 q + m. 44. 8 q + 5p 2 m.
 
 ELEMENTARY ALGEBRA 
 
 FIG. 1 
 
 EQUATIONS 
 
 5. An equation expresses an equality. In other words, an 
 equation is a statement that two expressions stand for the 
 same number. Thus x + 3 = 3x 1 is an equation. Equa- 
 tions are used in the solution of problems. Some number, 
 
 which at the outset is un- 
 known, is represented by 
 a letter; an equation is 
 formed, which enables one 
 to ascertain the value of 
 that unknown number. 
 
 An equation is like a 
 balance which is in equi- 
 librium when the weights 
 placed in one scale pan 
 are together equal to the 
 weights placed in the other 
 scale pan. 
 
 The equilibrium of a balance is not disturbed so long as 
 like changes in the weights are made simultaneously on both 
 sides. 
 
 So in equations, we may add the same number to both sides, 
 or subtract the same from both sides, or we may multiply or 
 divide both sides by the same number (except division by 
 zero) ; the equality is maintained during all these changes. 
 
 On the other hand, the equality is destroyed if more is 
 added to or subtracted from one side than the other, or if one 
 side is multiplied or divided by a larger number than is the 
 other side. 
 
 6. ILLUSTRATIVE PROBLEM. If 13 is added to twice a cer- 
 tain number, the sum is 41. Find the number. 
 
 Arithmetical Solution. Here we do not use equations. We subtract 13 
 from 41 and obtain 28. Then we divide 28 by 2 and obtain 14, which is the 
 required number. That is, we begin with 41, and go back to the required
 
 INTRODUCTION 5 
 
 number by subtraction and division, which are respectively the inverse of 
 the operations of addition and multiplication, named in the problem. 
 
 Algebraical Solution. Here we use equations. Let some letter stand 
 for the unknown number and then perform upon it the direct operations 
 named in the problem and obtain an equation. This direct way is usually 
 easier than the inverse. The solution is as follows : 
 
 Let x be the unknown number. Then 2 x is twice that number, and 
 2 x + 13 is the sum obtained by adding 13 to twice that number. 
 
 But this sum is equal to 41, as is stated in the problem. That is, 
 
 2 x + 18 = 41. 
 
 We want to find x. Remembering that an equality is like a balance, 
 that the equality is not destroyed if 13 is subtracted from both sides, we 
 
 obtain 2 x + 13 - 13 = 41 - 13, or 
 
 2 x = 28. 
 
 The equality remains true if both sides are divided by 2. Hence 
 
 ^ = ^,or 
 2 2 
 
 z = 14. 
 
 We have now solved the equation 2 x + 13 = 41 and have found the 
 previously unknown number x to be equal to 14. We see that the solu- 
 tion of the equation consists in making such changes in the equation that 
 x finally stands alone on one side of the equation. 
 
 7. SECOND ILLUSTRATIVE PROBLEM. If from ^ of a certain 
 number 17 is subtracted, the number resulting is 35. Find 
 the number. 
 
 Let x be the required number. Then 
 
 x 1 
 
 - is - of that number, and 
 3 3 
 
 x 1 
 
 17 is - of that number, less 17. 
 
 3 3 
 
 But this difference is equal to 35, as is stated in the problem. Hence 
 - 17 = 35. To solve this, add 17 to both sides. We get 
 
 o 
 
 I = 52. Multiplying both sides by 3, we obtain 
 
 o 
 
 x = 156, which is the answer. 
 
 This answer is correct, for | of 156 = 52, and 52 17 = 35, as stipu- 
 lated in the problem.
 
 6 ELEMENTARY ALGEBRA 
 
 PROBLEMS 
 
 8. 1. If 23 is added to three times a certain number, the sum 
 is 74. Find the number. 
 
 2. A wagon loaded with 12 sacks of flour weighs 1876 Ib. ; 
 the empty wagon weighs 700 Ib. Find the weight of 1 sack. 
 
 3. A train travels from Baltimore to New York, a distance 
 of 185 mi., in 4 hr. Find the average velocity of the train. 
 
 Work by arithmetic ; then by algebra. 
 
 In the latter case, let v be the velocity in miles per hour ; form an 
 equation which expresses the relation : 
 
 velocity x time = distance. 
 
 "4. The distance from New York to St. Louis is 1058 mi. 
 What is the velocity of a train which travels this distance in 
 23 hr. ? 
 
 5. How long will it take a train to travel the distance of 
 960 mi. from New York to Chicago at an average speed of 
 48 mi. an hour? 
 
 Let t = the no. of hours (" time "). 
 
 6. A lot sold for $2475. What was the frontage, if it sold 
 at $ 75 a front foot ? 
 
 7. The sura of two numbers is 31 ; the larger exceeds the 
 smaller by 7. Find the two numbers. 
 
 Let the smaller number = x. 
 
 Then the larger number = x + 1. 
 
 The sum of the two numbers = x + x + 7. 
 
 But the sum of the two numbers = 31. 
 
 Hence x + x + 7 = 31. 
 
 Since x + x = 2x, 22 + 7 = 31. 
 
 Subtract 7 from both sides, 2 = 24. 
 
 Divide both sides by 2, x = 12, the smaller number. 
 
 Whence x + 7 = 19, the larger number. 
 
 8. The sum of two numbers is 75; the larger exceeds the 
 smaller by 15. Find the two numbers. 
 
 9. Find two numbers whose difference is 34 and whose sum 
 is 126.
 
 INTRODUCTION 7 
 
 10. A father earns S 24 a week more than his son. Together 
 they earn $ 36 a week. What are the weekly wages of each ? 
 
 11. Divide $ 96 between father and son so that the son gets 
 j- of what the father gets. 
 
 12. Divide $124 between two brothers so that one receives 
 S 36 more than the other. 
 
 13. I propose to a boy the following puzzle: "Think of a 
 number, multiply it by 10, add 30, subtract 20." He gives 
 the result as 60. Find the number. 
 
 ^ / 14. To find the weight of a golf ball a man puts 10 golf 
 balls into the left scale pan of a balance and a 1-lb. weight into 
 the right ; he finds that too much, but the balance is restored 
 if he puts 1 oz. into the left scale pan. What was the weight 
 of a golf ball ? 
 
 15. A path half a mile long is to have a curb on both sides. 
 V/The curbstones used are 40 in. long, and 200 of them have 
 
 already been supplied. How many more are wanted ? 
 
 16. The cost of housekeeping for a family of n persons is 
 estimated at 7.5 + 2.5 n dollars per week. How many persons 
 are there in a family whose weekly expenses are $ 25 ? 
 
 17. A father leaves $ 14,000 to be divided among his three 
 children, so that the eldest child receives $ 1000 more than the 
 second, and twice as much as the third. What is the share of 
 each? 
 
 18. Divide $24,000 among A, B, and C, so that A's share 
 may be three times that of B, and C may have ^ of what A and 
 B have together. 
 
 - 19. A piece of silver is found to weigh a certain number of 
 ounces in a pair of scales ; on taking out a weight of 3 oz. from 
 one scale and placing it in the scale containing the silver, the 
 contents of one scale is twice as heavy as the contents of the 
 other. Determine the weight of the silver.
 
 8 ELEMENTARY ALGEBRA 
 
 20. Find the number whose excess over 2 is three times the 
 excess of its half over 4. 
 
 21. If the length of a circle is 3.1416 times its diameter, 
 find the radius of a circle (correct to two decimal places), when 
 the length of the circle is 7.56 ft. 
 
 To determine the nearest figure in the second decimal place, it is 
 necessary to compute the figure in the third decimal place. If the deci- 
 mal figures are, for example, .546, write .55 ; if the decimal figures are 
 .543, write .54 ; if the decimal figures are .545, write either .54 or .66. 
 
 22. If 6 be added to 7 times a certain number, the result is 
 equal to 144. Find the number, correct to two decimal places. 
 
 23. The sum of two numbers is 11.3; 8 times the greater 
 exceeds 11 times the smaller number by 2.5. Find the num- 
 bers, correct to three decimal places. 
 
 FACTOR, COEFFICIENT, TERM, EXPONENT 
 
 9. Each of the quantities which multiplied together form 
 a product is called a factor of the product. 
 
 Thus, each of the numbers 3, 6, x, y, is a factor of the product 15 xy. 
 Also, since 15 x times y gives the product 15 xy, 15 x is called a factor of 
 
 15 xy. 
 
 In general, the product of any two of the simple factors, 
 3, 5, x, y, is also called a factor of 15&?/; also, the product of 
 any three of the factors 3, 5, x, y, is called a factor of 15 xy. 
 
 10. In 4 c, 4 is called the coefficient of c. Similarly, 
 
 In 16 a, 16 is called the coefficient of a. 
 
 In 16 ab, 16 is called the coefficient of ab. 
 
 Here the word coefficient is applied to the factor which is 
 expressed in the Hindu-Arabic numerals. This is the most 
 common use of the word coefficient, but in a broader sense, 
 either of two factors which are multiplied together to form a 
 product is called a coefficient of the other factor. 
 
 For instance, 16 ab is the product of two factors 16 a and b ; hence 
 
 16 a is the coefficient of 6, and b is the coefficient of 16 a.
 
 INTRODUCTION 9 
 
 If no numerical coefficient is expressed, 1 is understood ; x 
 is the same as 1 x. Notice also that Ox, or x #, is equal to 0. 
 
 11. An algebraic expression may consist of parts which are 
 separated by the -f- or -- signs; these parts with the signs 
 immediately preceding them are called terms. 
 
 Thus, the expression 3a 4ft + 5c is separated by the -f or signs 
 into three parts ; it has the terms + 3 a, 4 6, + 5 c. / 
 
 12. An algebraic expression of one terra is called a monomial, 
 of two terms a binomial, of three terms a trinomial, and of 
 several terms & polynomial. 
 
 13. The product of two equal factors a a is called the square 
 of a. 
 
 a a is usually written a 2 and is read "a square" or "the second 
 power of a." 
 
 The product of three equal factors a a a is called the cube of a. 
 
 a -a -a is usually written a 3 , and is read "a cube" or "the third 
 power of a." 
 
 The product of four equal factors a a a a is called "a fourth " or 
 " the fourth power of a." 
 
 In a 2 , a 3 , a 4 , a 5 , a is called the base, the 2, 3, 4, 5, are called exponents. 
 
 14. An exponent is a number or letter placed a little above 
 and to the right of another number or letter, called the base. 
 
 When the exponent is a positive integer, it indicates how 
 many times the base is taken as a factor. 
 
 Avoid saying that a 2 means a multiplied by itself 2 times. This is not 
 true, a 2 means a a, where the base a is multiplied by itself only once. 
 
 Why is it incorrect to say that a 3 means a multiplied by itself 3 times ? 
 
 When no exponent is expressed, the exponent is regarded as 1. Thus, 
 a means the same as a 1 , 7 the same as 7 1 . 
 
 Later we shall extend the definition of an exponent, in order to give 
 meaning to such expressions as x*i or c*. 
 
 Care must be taken to distinguish between exponents and coefficients. 
 Notice that e* means e x e x e x e, but 4 e means e + e + e + e. If e = 
 3, then e 4 = 3 3 3 3 = 81, while 4e = 3 + 3 + 3 + 3=12.
 
 10 ELEMENTARY ALGEBRA 
 
 15. If the factors of a number are all equal, any one of 
 them is called a root of the number. 
 
 Nine has two equal factors, 3 and 3. We call 3 the square root of 9. 
 Similarly, 27 has three equal factors 3, 3, and 3 ; 3 is called the cube 
 root of 27. Again 256 has four equal factors, each being 4; hence 4 is 
 called the fourth root of 256. 
 
 Roots are indicated as follows: 3/9 = 3, v / 27=3, ^256 = 4. The 
 figure of the radical sign shows what root of the number is to be taken. 
 This figure is called the index of the root. If no figure is expressed, 
 square root is understood. 
 
 EXERCISES 
 
 16. Express in words the following : 
 
 1. a 5 . 6. 41m. 11. ^1 
 
 2. 6 3 . 7. 3y\ 12. 19 3 -5 4 . 
 
 3. a 5 + 6". 8. 5tf + c. 13. a + 
 
 4. e 2 4. 9. 6 h 9 x. 14. a? + y z + z mn. 
 
 5. 8 2 -29. 10. V36-V25. 15. ft 3 - 
 
 16. Compute the value of the expression in exercises 1-15, 
 if o = 2, 6 = 4, c = 50, e = 30, g = l, h = 5, Jfc=ll, m = 6, 
 TO = 10, p = 1, s = 12, u = 0, x = 3, y = 9, z = 29. 
 
 17. Find the value of 4 c 4 for each of the following values 
 if c = 1, 2, 5, 6, 4, 3, 10. 
 
 18. When s = 10, compute the values of s 2 2s, s 3 3s, 
 s 1 4 s, s 5 5 s. 
 
 19. When t = , compute the values of P, 1?, 5 P + 6 * 3 , 3 t P, 
 t-t*. 
 
 20. When 6 = 0, find the values of 6 2 , 6 3 . 
 
 PARENTHESES 
 
 17. When terms are to be grouped together, parentheses are 
 used. 
 
 Thus, (a; + y) 2 means that x + y, considered as a single num- 
 ber, is to be squared. That is, (x + y) 2 = (x + y) (x + y).
 
 INTRODUCTION 11 
 
 Again, 5 a (b 2 c + d) means that the entire expression 
 b 2 c + d is to be subtracted from 5 a. 
 
 When several parentheses are used in the same expression, 
 confusion may be avoided by using different forms. All these 
 forms go by the general name of " parentheses," but they are 
 designated by special names when it is desirable to distinguish 
 between them. Thus [ ] is called a " bracket," { f is called a 
 
 " brace," is called a " vinculum." But ( ) is always called 
 
 a " parenthesis." 
 
 That the sum of a and b is to be multiplied by c may be indicated in 
 four different ways as follows : (a + 6)c, [a + 6]c, {a 4- 6}c, a + b c. 
 
 EXERCISES 
 18. Read and tell the meaning of expressions 1-12 : 
 
 1. 20(5-3). 8, (2-*) + [10 -4]. 
 
 2. I0(x-2y). ?_/2_r 
 
 3. (x+y)(x-y). x \5 3, 
 
 5. 3 x(x + y). 
 
 6. 3x+( 
 
 7. (c + d) 5. 12. [a + & (c + d)] e. 
 
 13. Showthata 2 6 2 = (a 6)(a+ 6), whena = 12and6 = 10. 
 
 14. Showthat {a + 6p = a 2 + & 2 + 2a&, whena = 9and& = 5. 
 
 15. Show that (a - 6) 2 = a 2 + b 2 2 aft, when a = 12, 6 = 8. 
 
 19. Expressions like 19 +(7 + 3) may be worked out in 
 two ways: 
 
 (1) First simplify inside the parenthesis and then add. 
 
 (2) Add 7 to 19 and then add 3 to the result. That is, 
 
 19 +(7 + 3)= 19 + 10 = 29, 
 
 ' = 19 + 7 + 3 = 26 + 3 = 29.
 
 12 ELEMENTARY ALGEBRA 
 
 In the same way there are two ways of working 19 +(7 3) 
 or 19 - (7 + 3) or 9 x (7 + 3) or 9 X (7 3), as appears from 
 the following : 
 
 (1) 19 +(7 -3)= 19 + 4 = 23, 
 
 (2) = 19 + 7 - 3 = 23. 
 
 (1) 19 -(7 + 3)= 19 -10 = 9, 
 
 (2) = 19 - 7 - 3 = 9. 
 
 (1) 9 x (7 + 3)= 9x10 = 90, 
 
 (2) =9x7 + 9x3 = 90. 
 
 (1) 9 x(7-3)= 9x4=36, 
 
 (2) = 9 x 7 - 9 x 3 = 36. 
 
 EXERCISES 
 
 20. Perform each of the following exercises in two ways : 
 1. 8 +[6 + 3]. 6. 5x(5 + 5). 
 
 2. {10 + 51 + 122. + 5-3]x5. 
 
 3. 30 -[4 + 8]. 
 
 4. 16+(ll-6). 8 ' -. 
 
 5. 15 +[5 +6- 2]. 9- 30(10-4 + 3-1). 
 
 21. In expressions like 9-4 + 3 or 5-6-2- 10 + 2-3 or 
 3 + 3 6 H- 3 2 it is understood that the multiplications and 
 divisions (if there are any) are performed first in their order 
 from left to right; the additions and subtractions are carried 
 out afterwards in their order from left to right. 
 
 Notice that, in a term like 3 5 -5- 3 2, it is understood that 
 the divisor is 3, not 3 2. If we want 3 2 to be the divisor, 
 we must inclose it in a parenthesis and write the expression 
 thus, 3 5 -5- (3 2). Then 3 2 will be regarded as a single 
 number. Bearing these things in mind, we see that 
 
 5-4 + 3 = 1+3 = 4. 
 
 5-6-2- 10 + 2-3 = 30-20 + 6 = 16. 
 
 3 + 3-6-^-3-2 = 3 + 18 -=-3-2 = 3 + 6-2 = 15.
 
 INTRODUCTION 13 
 
 ORAL EXERCISES 
 
 22. Simplify the following : 
 
 1. 10-5 + 3-8. 5. 12 -4 -(2 -4)- 3. 
 
 2. 10-(5 + 3) + 8. 6. 12-4-5-2(4-3). 
 
 3. 10x2-9-3 + 5x2. 7. 10 -[12 -s- 3 -2]. 
 
 4. 12-4-2-4-3. 8. 10 12-5-3 2. 
 
 Of the following results, which are wrong ? 
 9. 50-10 + 8 = 50-18 = 32. 
 
 10. 20 + 24 -5- 3 4- 2 = 20 + 24 -=- 12 - 2 = 20 + 2 -2 = 20. 
 
 11. 15 + 36 -5- (3 + 9) + 5 = 15 + 36 - 3 + 9 + 5 = 41. 
 
 EVALUATION OF ALGEBRAIC EXPRESSIONS 
 
 23. When a = 5, 6 = 3, c = 2, d = 6, and e = 4, find the 
 values of : 
 
 1,1 a? + b 3 3 b 
 
 1. H 6. 
 
 be c + d 2 a 2 + 7 
 
 2. 
 
 3. 5a 3 -(2a6 + d). 
 
 d _ 6 8. (a + 6)(cx d 5). 
 
 e 9. 
 
 a _l_ c _(_ e 10. Va + 6 + c + d + e 4. 
 
 11. Does a? 2 + 5 a 36 = 0, when x = 4? Whencc = 5? 
 
 12. Does x 3 - x 2 + x - 10 = 18, when a? = 5 ? When a> = 3 ?
 
 CHAPTER II 
 
 POSITIVE AND NEGATIVE NUMBERS 
 
 24. The thermometer in your classroom indicates a tem- 
 perature of, perhaps, "67 above zero." In the northern states 
 the winter temperature out of doors sometimes drops 
 F - to " 15 below zero " or even lower. A shorter way 
 
 F. 
 
 212 
 
 212 of expressing this is as follows : 
 
 For "67 above zero" write " +67." 
 For " 15 below zero" write " - 15." 
 A + indicates that the temperature is " above zero" ; 
 100- 100 a ~~ i n dicates that it is "below zero." 
 
 But such plus and minus numbers are convenient 
 in other ways. A man who takes in and pays out 
 70-^-70 money may, for brevity, mark the sums taken in by 
 prefixing the + sign, and the sums paid out by pre- 
 32-f-32 fixing the sign. It is customary also to indicate 
 the amount a man owns by + and the amount a 
 man owes by . 
 
 In the same way we may write 1916 A.D. as +1916 
 -10" -10 an( ^ ^^ B-C- as ~ ^k These illustrations make it 
 -20- -20 plain how plus and minus numbers, or positive and 
 
 A negative numbers, as they are more usually called, 
 may be used in ordinary affairs of life. As we pro- 
 FIG. 2 cee< ^ further we shall see that the use of such num- 
 bers makes the solution of many problems much 
 easier and shorter. Such numbers can always be used when 
 there are pairs of quantities which are the exact opposites of 
 each other, as are the quantities in the above illustrations. 
 
 14
 
 POSITIVE AND NEGATIVE NUMBERS 15 
 
 ORAL EXERCISES 
 
 25. State the answers by prefixing + or to the numbers : 
 
 1. The temperature at noon is + 40 and falls 25 by night. 
 State the temperature at night. Give the answer if it falls 50 
 (that is, if it falls 40 and then 10 more). 
 
 2. A boy purchases a book which costs $ 1.75 and pays $ .50 ; 
 the balance he has charged. How may he indicate the amount 
 charged ? 
 
 3. His father gives him two dollars and tells him to pay the 
 balance on that book. How much has the boy left ? 
 
 26. One of the most common modes of representing positive 
 and negative numbers to the eye is by distances along a straight 
 line, as is done in the thermometer. It does not matter in 
 what direction the line is drawn. Usually it is most conven- 
 ient to draw the line horizontally^, 
 
 Some point is taken as the starting point (corresponding to 
 the zero in the thermometer). Distances to the right of are 
 usually indicated by the positive numbers, and distances to the 
 left of by the negative numbers. 
 
 f-t-1 
 
 -4 
 
 | 1 i 
 -3 
 
 -2 -1 
 FIG. 3 
 
 1 1 1 r 
 + 1 
 
 + 2 
 
 + 3 
 
 + 4 
 
 In the figure, the distance from to + 1 is taken as unit 
 distance or 1 space. The point marked + 2 is 2 spaces to the 
 right of 0; the point marked 2 is 2 spaces to the left of 
 0. 
 
 In the same way, a number + 2^ is indicated by a point 2\ 
 spaces to the right of ; 2^ is indicated by a point 2\ spaces 
 to the left of 0. 
 
 If to +1 we desire to add + 2, we start at the point marked 
 + 1, and go 2 units to the right, giving + 3 as the answer. 
 
 If from + 1 we desire to subtract + 2, we start at + 1 and go 
 2 units to the left, giving 1 as the answer.
 
 16 ELEMENTARY ALGEBRA 
 
 ADDITION 
 ORAL EXERCISES 
 
 27. 1. Locate on the line (Fig. 3) the following numbers : 
 
 + 4, -3, +H, -2|,3i,i. 
 
 2. To - 1 add 3. 
 
 We start at the point 1, and add 3 by going 3 units to the right, 
 giving + 2 as the answer. 
 
 3. From + 1 subtract 4. 
 
 4. Add 4 and - 3, - 4 and - 3, + 1| and 2, + 4| and - 2. 
 
 5. Add +5^ and 2, + 4i and 1, + 6 and -5, 15 and 25. 
 
 28. The student will have noticed that each of the symbols 
 + and is used in algebra in two senses. Thus + is used 
 sometimes to indicate addition and at other times it is used 
 to show that the number to which it is prefixed is positive. 
 
 Similarly, sometimes signifies subtraction and at other 
 times is used to represent a negative number. This double use 
 of these symbols may at first seem confusing, but we soon learn 
 how to interpret algebraic expressions in which they are used. 
 
 To express in algebraic symbols the addition of + 5 and 4 
 we inclose the numbers in parentheses and write thus : 
 
 The + between the two parentheses means addition; the + 
 in (+5) and the in ( 4) indicate whether the number is 
 positive or negative. 
 
 29. By the absolute value of a number is meant its value 
 without regard to the sign before it. Thus the absolute value 
 of both + 5 and 5 is 5. 
 
 Proceeding as in the examples given in 27, verify the 
 following: ( + 7) + (_|_5) = + i2. 
 
 -5)=+2. 
 -2.
 
 POSITIVE AND NEGATIVE NUMBERS 17 
 
 We see that these answers can be obtained by the following 
 rules which we shall find very useful in practice : 
 
 The sum of two numbers having the same sign is found by add- 
 ing their absolute values and writing their common sign before the 
 result. 
 
 The sum of two numbers having opposite signs is found by sub- 
 tracting the less absolute value from the greater and writing before 
 the result the sign of the number having the greater absolute value. 
 
 EXERCISES 
 
 30. Work exercises 1-9 by these rules and then verify the 
 answers by using the straight line as in the previous exercises. 
 
 1. (+3) + (-6). 4. (_7)4-(+17). 7. 
 
 2. (-3) + (+6). 5. (_17) + (-15). 8. 
 
 3. (-3) + (-6). 6. ( + 14) + (-10). 9. (- 
 
 10. Explain each of the above answers when the positive 
 and negative numbers represent assets and debts; also when 
 they represent temperatures above and below zero. 
 
 11. (+6) + (+10) + (+6) + (-9)=? 
 
 12. (+50) + (-30)-K+40) + (-10)=? 
 
 13. (_9) + (+20) + (-3) + (+20)= s ? 
 
 14. (+100) +(+900) + (-500)=? 
 
 SUBTRACTION 
 
 31. If the temperature is +10 in the morning and +40 
 at noon, we can find the rise in temperature by subtracting 
 + 10 from + 40. The rise is (+ 40) - ( + 10) or + 30. If 
 the temperature in the morning is 0, the rise is (+40) (0) 
 or + 40. If the morning temperature is 5, then the rise 
 must be still greater ; namely, + 45. 
 
 We have then (+40) - (-5) = + 45. 
 
 An easy way .to carry out this subtraction is to change the 
 sign of 5 to + 5 and then to add + 5 to 40. 
 c
 
 18 ELEMENTARY ALGEBRA 
 
 We assume the following general rule : 
 
 To subtract a number, change its sign and add. 
 
 This rule carries the operation of subtraction back to that of addition. 
 Notice that we have made no pretense of actually proving this rule to be 
 true in all cases. We take for granted that it is. In fact, assuming this 
 rule amounts to a definition of subtraction in algebra. To the question, 
 what is meant by "subtracting a number," the rule gives the answer : 
 " To subtract a number is to change its sign and add." 
 In subtracting a positive number from a larger positive number, as 
 7 from 12, it is easier to follow at once the familiar process used in arith- 
 metic. But the algebraic rule just given can be applied to this case 
 also. For we have 
 
 ( + 12) - ( + 7) = ( + 12) + (- 7) = + 5. 
 
 ORAL EXERCISES 
 
 32. 1. Verify the following subtractions : 
 
 + 7 +22 -7 -7 -20 
 
 -10 + 8 -10 + 8 -_5 
 
 + 17 +14 + 3 ^15 ^15 
 
 2- Check each answer in Ex. 1, by adding the remainder 
 to the subtrahend. What should the sum be equal to in each 
 case ? 
 
 3. Explain by the thermometer (or by assets and debts, or 
 by a straight line with a starting point ,0) how it is possible to 
 subtract 5 from a smaller number 3. Does the introduction of 
 negative numbers make subtraction always possible in algebra ? 
 
 Perform the following subtractions : 
 
 4. +20 5. +15 6. -25 7. -60 8. +30 
 + 30 -30 -40 +10 -50 
 
 9. 8 -(+12). 13. 0-(+10). 
 
 10. 5 -(-40). 14. O-(-lO). 
 
 11. 15- (+60). 15. (-20) -(+20). 
 
 12. (-30) -(-10). 16. 50 -(-30).
 
 POSITIVE AND NEGATIVE NUMBERS 19 
 
 17. 100 -[+30 + 10]. 19. 0-(+55 + 15). 
 
 18. (-5)- f20 + 40|. 20. (+20+30)-(+4o+25). 
 
 'Simplify the following: 
 
 21. 5 + (+ 10) - (+ 60). 24. (-30) -(+10) -(+40). 
 
 22. 6-[ + 9]-[-8]-[+7]. 25. 85 + {-90}-{ + 90|. 
 
 23. (-30)-(-10)-(-40). 26. 85- { + 90} + {-90}. 
 
 MULTIPLICATION INVOLVING NEGATIVE NUMBERS 
 
 33. In arithmetic we define 3 X 4 as meaning 4 + 4 + 4 ; 
 that is, 4 is to be taken as many times as there are units in 3. 
 In the multiplication of fractions, say, ^ X f , the above defini- 
 tion becomes inapplicable, for the reason that we cannot take 
 f a fractional number of times. We are therefore driven to a 
 different definition of multiplication ; we define the product as 
 the result obtained by multiplying the numerators together 
 for a new numerator and multiplying the denominators to- 
 
 1 2 1x2 
 
 gether for a new denominator. Thus, Q*^ = S ?' 
 
 2i O A X O 
 
 In algebra we are now studying a new type of numbers ; 
 namely, negative numbers. When we multiply one number by 
 another, and one or both numbers are negative, the question 
 arises : is the product a positive number or a negative number ? 
 We give the answer to the question in the following definition : 
 
 The product of two numbers having like signs is a positive 
 number and the product of two numbers having unlike signs is a 
 negative number. 
 
 In multiplying together two numbers, first find the product 
 of their absolute values and then write before it the proper 
 sign. 
 
 Thus we have, according to this definition, 
 
 ( + 3) x (+ 4) = + 12. (+ 3) x (- 4) = - 12. 
 
 (-3) x(+4) = -12. (-3)x(-4)=+12.
 
 20 ELEMENTARY ALGEBRA 
 
 ILLUSTRATION. If a man owes $ 10 to each of two persons, he owes them 
 $ 20 all together. Denote the $ 10 owed by - $ 10. 
 Then evidently, (+ 2)(- $10) = - $20. 
 SECOND ILLUSTRATION. Denote a profit of $ 10 by + $ 10. 
 Denote a loss of $ 10 by - $ 10. 
 Denote receive (add) by +. 
 Denote cancel (subtract) by . Then 
 
 a. To receive 5 profits of $ 10 each is to be worth $60 more. That is, 
 (+5)(+$10) = +$50. 
 
 b. To receive 5 losses of $ 10 each is to be worth $ 50 less. That is, 
 (+5)(-$10)=-$50. 
 
 c. To cancel 5 profits of $ 10 each is to be worth $ 50 less. That is, 
 
 d. To cancel 5 losses of $ 10 each is to be worth $ 50 more. That is, 
 (_6)(_$10) = +$50. 
 
 ORAL EXERCISES 
 
 34. State the following products : 
 
 1. + 5 2. -12 3. -70 4. +33 
 -11 -12 + 5 -11 
 
 5. (+ 5) x (- 5). 8. | x f 11. [+2*] X [-21]. 
 
 6. (_5)x(+5). 9. (_|) X (+*). 12. {-1.2} {-1.2}. 
 
 7. Ox (-5). 10. (-.5)x( + .61). 13. (+5)(-1.6). 
 
 PROBLEMS 
 
 35. 1. A man saved $175 a month for 4 months; then, 
 during a sickness of 3 months, he lost $200 each month. 
 How much did he have at the end of the 7 months ? 
 
 2. During 12 months a merchant expended $200 each 
 month, and took in $215 monthly. What was his profit 
 during the year? 
 
 3. A boy received from his father on July 4, $1.25; he 
 earned one dime, spent 85 $ for firecrackers, 25^ for a lunch, 
 and paid four 5^ car fares. How much did he have at the end 
 of the day ?
 
 POSITIVE AND NEGATIVE NUMBERS 21 
 
 4. A road from town A to town B leads over rolling 
 country ; 5 times the road passes over hillocks rising each 
 37 ft. and 6 times it descends from the hillocks 34 ft. How 
 much higher or lower is town B than town A ? 
 
 5. The water in a reservoir rises 3 in., then falls 6 in., 
 rises again 7 in., falls 9 in., and finally rises 13 in. How 
 much higher or lower is it than at first ? 
 
 6. An explorer 200 mi. south of the north pole travels north 
 47 mi., then 31 mi. south, and again 10 mi. farther south, then 
 86 mi. north. How far north from the starting point is he 
 finally ? 
 
 7. The boiling point of water is + 212. How much 
 above or below the boiling point are the following tempera- 
 tures : + 250, + 200, + 100, + 50, - 10 ? 
 
 8. Colorado Springs is 6000 ft. above sea level. How 
 much higher or lower are the following elevations above sea 
 level : 14,000 ft., 5000 ft., 400 ft., 100 ft., - 100 ft. ? How 
 do you interpret 100 ft. ? 
 
 9. The midday temperatures for one week were + 39, 
 -|- 48, + 60, + 57, + 39, + 55, + 60. Find the average 
 of these temperatures. 
 
 The average of a series of numbers is found by dividing their sum by 
 the number of them. In this case there are 7 numbers. Hence you 
 find their average by dividing their sum by 7. 
 
 10. The midnight temperatures for that same week were as 
 follows : + 10, - 3, - 12, + 8, + 12, - 20, - 5. Find 
 the average of these temperatures. 
 
 11. A merchant's monthly profits for five consecutive 
 months were +$400, +$200, -$100, -$300, + $ 500. 
 Find the average monthly profits. 
 
 12. The latitude of Key West, Fla., is 24 33' and of Chicago 
 is 41 50'. What is the latitude of a place halfway between 
 the two ?
 
 22 ELEMENTARY ALGEBRA 
 
 13. The latitude of the Cape of Good Hope is - 34 21' 
 (the indicating here south latitude) ; the latitude of Athens 
 in Greece is 37 58'! Find the latitude of a place halfway 
 between them. 
 
 14. At the seashore the rise and fall of the tides are 
 measured from a certain arbitrarily chosen level. A tide 
 which falls below that level is called a " minus " tide. If one 
 day the tide rises to 6' 3" and then falls to 1' 2", what is 
 the average water level for that day ? 
 
 15. The Greek philosopher, Plato, died 347. How many 
 years ago was that ? 
 
 16. The elevation above sea level of the top of Mont Blanc 
 is 16,050', and of the lowest part of the Atlantic Ocean is 
 27,800'. What is the difference in elevation between the two? 
 
 17. A man is rowing upstream. In still water his rate of 
 rowing is 6 mi. an hour. The rate of the stream is 2 mi. an 
 hour. How long will it take him to reach a place 12 mi. up- 
 stream ? 
 
 18. If a boat is steaming southward on a river at the rate 
 of 13 mi. an hour, while a man on the deck is walking toward 
 the stern at the rate of 3 mi. an hour, what is the man's 
 actual motion with respect to the shore ? 
 
 19. A balloon capable of exerting an upward pull of 395 Ib. 
 is attached to a car weighing 146 Ib. What is the net upward 
 or downward pull ? 
 
 DIVISION 
 
 36. From the rule of signs for multiplication we can ascer- 
 tain what the rule of signs should be for division. 
 
 In arithmetic, we test the correctness of a division by mul- 
 tiplying the quotient by the divisor ; their product should be 
 the dividend. That is, 
 
 Divisor x Quotient = Dividend. 
 We see that 15 -5- 3 = 5, because 3 x 5 = 15.
 
 POSITIVE AND NEGATIVE NUMBERS 23 
 
 Similarly with negative numbers : 
 (+ 15) -=- (- 3) = - 5, because (- 3) x (- 5) = + 15. 
 (- 15) - (+ 3) = - 5, because (+ 3) x (- 5) = - 15. 
 (- 15) -=- ( ; -3) = + 5, because (- 3) x (+ 5) = - 15. 
 From these examples we see that the rule of signs for divi- 
 sion is the same as for multiplication ; namely, 
 
 The quotient of two numbers having like signs is a positive 
 number. 
 
 The quotient of two numbers having unlike signs is a negative 
 number. 
 
 From the above it appears that division is the reverse of 
 multiplication. In multiplication, we are given two factors, to 
 find the product ; in division, we are given the product and one 
 of the factors, to find the other factor. 
 
 ORAL EXERCISES 
 
 37. Carry out the indicated divisions and check the 
 answers : 
 
 1. (+36)-K + 12). 4. (-42) -(-7). 7. (-51) -(+17). 
 
 2 . (_64)-(+16). 5. (7.5) -s- (1.5). 8. (+5.1)-!- (-1.7). 
 3- (-f)-K-f). 6. (-1.44)-K+1.2).9. (+#-!-(+ .5). 
 
 Perform the indicated operations : 
 
 10. (+45) -K-9) -(-6). 
 
 11. (+10) -(-7)- (+ 3). 
 
 12. (-4) -(-2) -(+5) +(+3). 
 
 13. (+9)-(-3)x(+12)x(-l). 
 
 14. (-18)- (+3) -(-6) -(-4). 
 15. 
 
 16. (_ 
 
 17. (-8). (-5) -(-10). 
 
 18. (_l 6 )-(+2)x(-9)-(-8). 
 
 19. (_24)-(-24) x (- 24) - (+ 24).
 
 24 ELEMENTARY ALGEBRA 
 
 PROBLEMS . 
 
 38. Use positive and negative numbers in the solutions of 
 the following problems : 
 
 1. A man has a contract to dig a well 56 ft. deep. If he 
 digs on an average 4 ft. a day, how many days will it take him 
 to dig the well ? 
 
 2. If the foundations of a wall are 5 ft. below the surface 
 and the wall is 41 ft. high above the surface, how many feet 
 are there from the foundations to the top of the wall ? How 
 many cubic feet of brick in it if it is 20 ft. long and 2 ft. 
 thick ? 
 
 A SIMPLIFIED NOTATION 
 
 39. We have seen that the expression 
 
 (+ 8) + (- 5) 
 
 means "to positive 8 add negative 5." The + between the 
 two parentheses means addition; the -f and in (+8) and 
 (5) serve the purpose of showing the quality of the num- 
 bers; namely, that the 8 is positive and the 5 is negative. 
 We know by our rules for addition and subtraction that 
 
 (+8)+(-5) = 8-5 = 3 and 
 
 (+8) -(+5) = 8 -5 = 3. 
 
 This shows that we can simplify the above notation by 
 writing, 8 _ 5 
 
 and that 8 5 may have two interpretations. It may signify 
 either (+8) _ (+5) or ( + 8 ) + (-5). 
 
 In the first interpretation the sign means an operation ; 
 in the second interpretation the sign means a quality. 
 The same argument shows that 
 
 = 7 and
 
 POSITIVE AND NEGATIVE NUMBERS 25 
 
 We can therefore simplify expressions as in the following 
 examples : 
 
 ( + 9) - ( - 2) + (+ 6) - ( + 7) = 9 + 2 + 6 - 7. 
 (-6) + (+9)- (-8) -(+3) =-6 + 9 + 8- 3. 
 The sign -f may be omitted in expressions involving multi- 
 plications and divisions like the following : 
 
 For (+ 27) ( + 9) we may write 27 9. 
 
 For ( 27) -=- (+9) we may write (- 27) -5-9. 
 
 For (-1- 15) -T- (-f 5) we may write 15 -=- 5. 
 
 EXERCISES 
 
 40. Simplify the following expressions : 
 
 1. (-5) + (+9) -(-12) -(5). 
 
 2. (+4)(+8) + (-6)-(+12). 
 
 3. (- 24) - (+ 30) + (+ 36) +- (+ 7). 
 
 4. (-35). (10) -(+8) -(-7). 
 
 5. (+18) - (+6) -(+5) + (+5). 
 
 6. (24 + 12) -(-42) + ( + 30)-*- (-28). 
 
 7. (4 + 3- 2) -(1 + 6) + (-1 + 8) -[-8-2 + 3]. 
 
 8. 3(3+l)-4(5+8) + (2-7)-(5 + 6-7) +9(8-5). 
 
 9. 10 {10 -6 + 1} + [7 + 3-12] .4 + 5-15+(3 + 4)5. 
 
 SIMILAR TERMS 
 
 41. Terms which have the same literal factors are called 
 similar. 
 
 Thus, 12 a and - 6 a are similar; so are 13 xy 2 and 25cey 2 . 
 
 On the other hand, terms which do not contain the same literal 
 factors are called dissimilar. 
 
 12 a and 15 b are dissimilar terms ; so are I3x 2 y and 23 xy*. 
 
 If /signifies "feet," we know from arithmetic that 
 10/+5/=15/.
 
 26 ELEMENTARY ALGEBRA 
 
 We find the sum by adding the numerical coefficients. This mode 
 of procedure is general. If /means "forks" or if / stands for 
 any abstract number, as 12, the same process of addition holds. 
 On the other hand, if the terms are dissimilar, as 5f and 
 10 i, then their sum is not 15/ nor is it 15 i ; all we can do in 
 such a case is to indicate the addition by writing 
 
 If one boy has 5 ducks + 6 hens + 3 rabbits, and another boy 
 has 3 ducks + 7 hens + 8 rabbits, then the two together have 
 8 ducks + 13 hens + 11 rabbits. Abbreviating, we write 
 
 (5 d + 6 h + 3 r) + (3 d + 7 h + 8 r) = 8 d + 13 h + 11 r. 
 This result is true, no matter what d, h, r may mean. 
 
 Similar remarks apply to the subtraction of similar and dis- 
 similar terms. 
 
 EXERCISES 
 
 42. Find the sums in exercises 1-11 : 
 
 1. 3a + 4a + 7a oa + 6a. 
 
 2. (+5c) (+3c) 5c. 
 
 3. 3aj + 3y + 4z + 9y + z + 2. 
 
 4. 3x + 6x + 8x 6x. 
 
 5. (+7) + (3aj + 4y) + 6y + 4as. 
 
 6. (20c-10d + 30e)+(4d-5c + 9e). 
 
 7. (3a + 5&-6c) + (10b + 4e-2a) + 6c-9a+20&. 
 
 8. 5 (a + 6) + 3 (a + 6) + 7 (a + 6) - (a + 6). 
 
 9. - 5 a 2 & 2 + 4 ab 2 - 2 a 2 6 2 - 6 a& 2 + 7 a& 2 c + a&c. 
 
 10. + 3fg, + 4 gh, - 5 gh, - 7 gh, + Qfg. 
 
 11. 2(a + &) + c, 3 a, 76, 9 c. 
 
 12. From 8 ax take 5 ax, then add to the result 6 ax.
 
 POSITIVE AND NEGATIVE NUMBERS 27 
 
 13. Add 9 mn to + 4?n?i ; from the result take 10 mn. 
 
 14. Subtract + 6 Jcl from - 24 kl, then add + 12 kl. 
 
 15. Subtract 3 be from 5 etc, then add 9 cd. 
 
 16. Simplify : 40 st - 5 st + 20 st + ty 10 ty. 
 
 17. From 6 (a + a) take 7 #, then again 4 a. 
 
 ADDITION AND SUBTRACTION OF POLYNOMIALS 
 
 43. When polynomials are to be added together, or when 
 one polynomial is to be subtracted from another, it is conven- 
 ient to write similar terms in the same column and to add or 
 subtract the terms in each column, proceeding from left to 
 right. 
 
 Add 2o + 46-5 c, -2a + 46 + 7c, 8a-56 + 9c. 
 Solution. 2 a + 4 6 5 e 
 -2a-M& + 7c 
 -8a-55 + 8c 
 
 8a + 36 + 10c, the required sum. 
 From Q x + 8y 5z take Qx 1 y + z. 
 Solution. 9x+ 8y5z 
 
 3 x + 15 y 9 z, the required difference. 
 
 EXERCISES 
 
 44. Add the following polynomials : 
 
 1. 2 ac 2 + 3 b 2 c - c 3 , Gac - 6 2 c + 9 ac 2 , ac + 6 2 c - 5 c 3 . 
 
 2. a 2 + 8 52 + 4 c . 2j 2 a 2 - 5 6 2 - 8 c 2 , 66 2 -f 7C 2 . 
 
 3. 6 .>; 7 ?/ + 4 xy, 4 a + 5 y + 10 xy, 20 x 9 xy + y. 
 
 4. 3 a + y 8z + 3w, 9a; 8 y 6z + 4w 3v. 
 
 5. a + 6, a + c, 6 + c, 2a 36 + 4~cT 
 
 6. a + 6+c d, 26 + 3c + 7d, 4a + 56-6d. 
 
 7. la + 66 + 3c-10d+/, a-5 6+8 c+2/, 26+3 c+4d 
 
 8. a 3 6 + 2 a& 3 - 6a6, 3 a 3 6 - 5 ab + 9 a& 3 .
 
 28 ELEMENTARY ALGEBRA 
 
 In exercises 9-11, subtract the second polynomial from the 
 first and check your answer by adding the remainder to the 
 subtrahend : 
 
 9. a + b + c + ci + e, 2a+3b 4 c 3e. 
 
 10. 10 x + 10 y 10 z + 10 w, 5 w + 3 x 8 y + 3 z v. 
 
 11. a + b x + y + 3z, a+b+x y 3z + 5x. 
 
 12. To a + c d add 4a + 5c + 6c and from the sum take 
 
 13. From 2m+4n-f9/) take 9 ra 8 n + 1 p and to the re- 
 sult add 4 m + 4 n 4- 4 p. 
 
 14. From 27/+ 240 + 23 ft take 19/ 15 h + 15 g and from 
 the result take 20 /+ 3 g 9 h. 
 
 15. Subtract 26 + 4c+7d + 6e from 56 4d + 6c + 9e 
 and check. 
 
 16. From 3 a + r + s take the sum of 10 r + 30 s and 5 r 
 7 s + 20 a. 
 
 17. From the sum of a + 3 6 and 2 a 6 6 take the sum of 
 - 3 a + 24 6 and a + b + c. 
 
 18. From x + y + z + w take a b c d, then add to the 
 result 3z + 4w-7c+3c?. 
 
 19. From 4 (a + ft) take 6 (a + b + 5 c). 
 
 20. Add 2 (a + 4 6 + 3 c), 3 (- 2 a + 3 b + 7 c), 4 (6 + c). 
 
 21. Add 4 a 4 + 4 a + a 4 &, a 4 - 3 a + 5 a 4 6, - 6 a 4 6, - 9 a. 
 
 Evaluate the following monomials and polynomials, when 
 a = - 2, b = - 3, c = - 1, d = 10 : 
 
 22. ab, abc, be, 2 aft, 2 ab, 3 be, 3 be, cd. 
 
 23. a + b + c + d, a b + c + 2d, 5a 3b + ac. 
 
 24. a 2 , 6 2 , c 2 , cP, a 2 6 2 . 
 
 25. a&cd, a?bcd, ab z cd, abc?d, abed?. 
 
 26. a 3 , 6 3 , c 3 , d 3 , a?b 3 , etc 3 , <?&. 
 
 27. aV, a 2 cP, 6cP, cd 3 , c 2 ^ 3 .
 
 POSITIVE AND NEGATIVE NUMBERS 29 
 
 CHECKING BY SUBSTITUTION OF NUMBERS 
 EXERCISES 
 
 45. 1. Add 3z 2 -4y, x 2 + 7y, 5x 2 128 y. 
 
 In the polynomials added and in their 3 x 2 4 ?/ = 3 8 
 
 sum, let each letter equal some simple x 2 + 1 y = 1 + 14 
 
 number and compare as indicated below. 5x 2 128 y 5 256 
 
 If a mistake has been made in addition, Sum 7 x 2 125 y 7 250 
 this test almost always reveals it. 7 x 2 125 y 243. 
 
 In each of the three binomials and 
 
 also in their sum, let x = 1, y = 2. We obtain the numbers on the right. 
 Adding the two columns of numbers, the sums are 7 250. This is the 
 very same result as that obtained by letting x = 1 and y = 2 in the sum 
 7 x 2 126 y. This checks the addition. 
 
 Add the following and check by letting x 2, y = 1 : 
 
 2. 5x* 6x + 4:y, z 2 + 7z 3y, 3x 2 5x 2y. 
 
 3. -x z 7x-2y, Wx* + 8x-4:y, -6z 2 + 4a; - Wy. 
 
 4. 10 x 2 x - y, 11 x 2 8 x 9 y, 20 x 1 + 1 x 5 y. 
 
 In exercises 5-7, subtract the second polynomial from the 
 first and check by letting a = 2, b = 3 : 
 
 5. 10 a 2 + a b, 5 a 2 4 a + b. 
 
 6. 25a 2 -10a + 26, 20a 2 -lla + 36. 
 
 7. 13 a 2 + 14 a + 15 b, 20 a 2 - 15 a + 6. 
 
 8. 10 a 3 + 2 6 2 , 7 a 3 - 3 6 2 . 
 
 9. 12 a 3 - 5 b 3 , 7 a 3 - 7 b\ 
 10. 7 a 3 - 24 6 4 , 8 a 3 - 20 6 4 . 
 
 REMOVAL OF PARENTHESES 
 
 46. In an example like 2 a +(a + 6 + c d), the polynomial 
 within the parenthesis is to be added to what precedes. This 
 may be accomplished by adding each term separately. We 
 may at first merely indicate this separate addition by rewriting 
 the expression with the parenthesis omitted ; thus, . 
 
 + c d.
 
 30 - ELEMENTARY ALGEBRA 
 
 Thereupon like terms are combined. In this instance we 
 obtain. 3 a + 5 + c d. 
 
 ^~ ** 
 
 In an example like 2a (a + 6 -f- c d), the polynomial 
 within the parenthesis is to be subtracted from what precedes. 
 This may be accomplished by subtracting each term separately. 
 We may at first merely indicate this subtraction by rewriting 
 the expression, but changing the sign of each term as it is 
 taken out of the parenthesis ; thus, 
 
 2 a (a + b + c d)=2a a 6 c + d. 
 Thereupon like terms are combined. In this instance we 
 obtain, a b c + d. 
 
 We have then the following rules for the removal of 
 parentheses : 
 
 A parenthesis preceded by the + sign may be removed without 
 changing the signs of the terms that were within the parenthesis. 
 
 A parenthesis preceded by the sign may be removed if we 
 change the sign of each term that was within the parenthesis. 
 
 When an expression contains a parenthesis within another 
 parenthesis, as (5 a [4 b 5 c]), remove one parenthesis 
 first, then the other. Thus, 
 
 It is usually found easier to remove the inner parentheses 
 first. 
 
 The student should form the habit of guarding carefully 
 against error in the application of these simple rules. 
 
 EXERCISES 
 47. Remove the parentheses and simplify : 
 
 1. a+(3a + 5 4c). 
 
 2. 6-(6 + 2c + 3d). 
 
 3. (a + 6 + 3c) + (a + &) (a &). 
 4. 
 
 5.
 
 POSITIVE AND NEGATIVE NUMBERS 31 
 
 .6. _ 
 
 7. j_91+8m 7%}+[ 71+6n+4m] 
 8. 
 9. 
 
 10. -(- 
 
 11. 6c- 
 
 12. 4a + 
 
 14. 5&-(4c 
 
 15. (m tt) + (3ra + 4n) J ra-f-(3ra n)f. 
 
 16. { (ra + 4w p) (p 2n) + (4p 3m) 7 raj. 
 
 INSERTION OF PARENTHESES 
 
 48. This process is the inverse of the preceding. Since in 
 removing a parenthesis with the + sign before it the signs of 
 the terms taken out are not changed, it follows that when the 
 parenthesis is restored, the signs of the terms inserted in 
 the parenthesis are not changed, provided the sign before 
 the parenthesis is +. 
 
 Since in removing a parenthesis with the sign before it the 
 sign of every term taken out is changed, it follows that when 
 the parenthesis is inserted, the sign of every term in the pa- 
 renthesis must be changed, provided the sign before the paren- 
 thesis is . 
 
 Accordingly a + b c + d = a + (b c + d), and 
 
 = a ( b + c d). 
 
 EXERCISES 
 
 49. Inclose in parentheses all the terms except the ones in- 
 volving the letters x and y ; in the odd-numbered exercises use 
 the + sign before the parentheses ; in the even-numbered exer- 
 cises use the sign. 
 
 1. x -f- a + 6 c + d + y ra + n s. 
 
 2. 2x + 3a 4
 
 32 
 
 ELEMENTARY ALGEBRA 
 
 3. 3a+2w-3e+x-r+5t+y+5t-6 
 
 6. _5&_a-_|_3?/ 4 c + 5 ci!. 
 
 7. 5m-f 6n + 7a; 3p 4^. 
 
 MAGIC SQUARES 
 
 + 7 1 a+a+6. 
 
 1. Figure 4 is a magic square. Add 
 the numbers along each of the three 
 horizontal dotted lines, each of the 
 three vertical lines, and each of the two 
 diagonal lines. What is there peculiar 
 about the sums? 
 
 FIG. 4 
 
 36 
 
 86 
 
 6 
 
 -2a 
 
 -la 
 
 
 26 
 
 46 
 
 66 
 
 a 
 
 -3a 
 
 5 a 
 
 76 
 
 
 56 
 
 -6a 
 
 a 
 
 -4a 
 
 FIG. 5 
 
 2. This magic 
 square (Fig. 5) 
 
 has algebraic expressions arranged 
 along the eight lines. Find the sum 
 of the three algebraical expressions 
 along each of these lines. What is 
 there peculiar about the sums ? 
 
 3. Take a = 1, 6=2, and compute 
 the resulting numerical magic square. 
 
 Select other integral values for a and 6, and construct other 
 
 magic squares. 
 
 4. If a = 2 and 6 = 1, then 
 a magic square is obtained, 
 most of whose numbers are 
 negative. Is the sum of the 
 numbers still the same for 
 each of the 8 different lines ? 
 
 5. Find the sum of the 
 expressions in each of the 10 
 lines in the larger magic 
 square (Fig. 6). 
 
 
 146 
 
 136 
 
 36 
 
 a 
 
 
 
 
 
 - 13 a 
 
 - 12 a 
 
 -2a 
 
 116 
 
 56 
 
 66 
 
 86 
 
 - 10 a 
 
 -4a 
 
 6 a 
 
 -7a 
 
 76 
 
 96 
 
 106 
 
 46 
 
 -6a 
 
 -8a 
 
 -9a 
 
 -3a 
 
 126 
 
 26 
 
 
 156 
 
 - 11 a 
 
 a 
 
 6 
 
 - 14rt 
 
 FIG. 6
 
 POSITIVE AND NEGATIVE NUMBERS 33 
 
 6. Answer questions 2 and 3, given above, for this larger 
 magic square. 
 
 ORAL EXERCISES 
 
 50. Represent a number : 
 
 1. Greater than n by 2. 6. 5 times x less 2. 
 
 2. Greater than x by 5. 7. 10 greater than a + 6. 
 
 3. Less than y by 3. 8. x greater than 9. 
 
 4. Less than r by 6. 9. a greater than n x. 
 
 5. Less than 7 n by a. 10. y less than 2 times a. 
 
 What is the other part if 
 
 11. One part of 15 is 6 ? 14. One part of y is x? 
 
 12. One part of n is 2 ? 15. One part of 24 is a ? 
 
 13. One part of x is y ? 16. One part of a + 6 is 10 ? 
 
 17. The difference between two numbers is 16; the greater 
 is 20. What is the smaller ? 
 
 18. The difference between two numbers is a; the greater 
 is x. What is the smaller ? 
 
 19. The smaller of two numbers is y ; their difference is 
 3. What is the greater ? 
 
 20. The sum of two numbers is 27 ; the smaller is n. What 
 is the greater ? 
 
 21. The sum of two numbers is a ; the greater is y. What 
 is the smaller ? 
 
 22. By how much does 28 exceed 15 ? 
 
 23. By how much does x exceed 11 ? 
 
 24. Express the excess of 52 over 45. 
 
 25. Express the excess of x oveivy- 
 
 26. Express the excess of a + b over a b. 
 
 27. If one book costs $ 2, what will 7 books cost ? 
 
 28. If one pencil costs 5 ^, what will b pencils cost ?
 
 34 ELEMENTARY ALGEBRA 
 
 29. If one coat cost x dollars, what will 3 coats cost ? 
 
 30. If 7 hats cost $ 171, w hat will 1 hat cost ? 
 
 31. If 2 oranges cost d cents, what will 1 orange cost? 
 
 32. If x apples cost 15^, what will 1 apple cost ? 
 
 33. What is 5 % of $200 ? 4 % of d dollars ? 
 
 34. What is the interest on $ 250 at 6 % ? 
 
 35. What is the interest on $ 300 at r % ' 
 
 36. What is the interest on p dollars at r % ? 
 
 37. If a man is 53 years old now, how can you represent his 
 age x years hence ? 
 
 38. If a man's age now is n years, how can you represent 
 his age 20 years ago ? 
 
 39. If a man is twice as old as his son, and his son is b 
 years old, how old is the father ? 
 
 40. The side of a square is 6 in. What is its perimeter ? 
 
 41. The side of a square is x in. What is its perimeter ? 
 
 42. The length of a rectangle is twice its width. Express 
 its length and its perimeter, if w represents its width. 
 
 43. The length of a rectangle is 4 ft. more than twice its 
 width. Express its length and its perimeter if x represents 
 its width. 
 
 44. If I have x dimes, how can I express the number of cents 
 I have ? 
 
 45. If n represents a certain number, how can I represent 
 the next greater number ? 
 
 46. If n represents a certain even number, how can I repre- 
 sent the next even number ? 
 
 47. Express two consecutive odd numbers.
 
 POSITIVE AND NEGATIVE NUMBERS 35 
 
 SOLUTION OF EQUATIONS 
 
 51. We have seen that an equation is like a balance ( 5) ; 
 the equality or balance is not disturbed as long as the same 
 changes are made simultaneously on both sides. 
 
 Thus, in solving 9 x 10 = 4 x . 
 Adding 10 to both sides, 9 x = 4x + 10. 
 
 Subtracting 4 x from both sides, 6 = 10. 
 
 Dividing both sides by 5, x = 2, the answer. 
 
 It will be noticed that adding the 10 to both sides has the 
 same effect as simply moving the 10 to the other side of the 
 equation and changing Us sign. 
 
 In the same way, subtracting 4 x from both sides had the same 
 effect as taking 4 x over to the left-hand side and changing its 
 sign. 
 
 This operation is called transposition. It consists in moving 
 a term from one side of the equation to the other and changing the 
 sign of the term. 
 
 Thus, in solving 4 6x = 2x 36. 
 
 Transposing 4 yields 6x = 2x 36 4. 
 
 Or - 6 x = 2 x - 40. 
 
 Transposing 2 x gives 8 x = 40. 
 
 Dividing both sides by 8, x = 6. 
 
 Check : Substitute 5 for x in the original equation. The result is 
 4 - 6 5 = 2 5 - 36, or 4 - 30 = 10 - 36, or 
 
 - 26 = -26 
 
 Since this is correct, we know that 5 is the correct value for x, provided 
 that no mistake was made in checking. 
 
 The method of solution which we have explained applies 
 equally well when the coefficients of the equation are numbers 
 represented by letters. 
 
 Consider, for example, mx + b c = d mx. 
 
 Transpose ma;, + 6, c, mx + ma; = d b + c. 
 
 Combine ma; + ma;, 2 mx d b + c. 
 
 Divide both sides of the equation by 2 m, x = , the answer.
 
 36 ELEMENTARY ALGEBRA 
 
 The answer obtained in the solution of an equation is called 
 a root of that equation. 
 
 A root of an equation is a quantity which, when substituted 
 for the unknown in the equation, " satisfies " the equation by 
 reducing both members to identical numbers. 
 
 EXERCISES 
 
 52. Solve the following equations, and check each answer : 
 
 1. 6x + o = 3x + 17. 11. 50^ + 2 
 
 2. 13x-5 = llx + 5. 4Q- 
 
 3. 4-5*= -6a; + 2. 12 - z + 10 + | 
 
 4. 4*-30 = 30-6, 13 2 , + 6 = a 
 
 5. 3y + S = 5y-5. 14 2x + b = x 
 
 6. 6</ + 45 = 7 2/ + 50. 15 ax + b = c 
 
 7. 40 + 4 = 90-6,. 16 5x = a + b 
 
 8. 1+X-6X + 30= 1? . ^y_ a= ; 
 
 x + 35-3x. i 8 . 03. + 4 = 5. 
 
 9. %x + 5 = 6. 19. C x-3c 
 
 10. ix-10 = 2a; + 20. 20. az + 2 6 = 3 c. 
 
 53. In a balance the equilibrium is not disturbed if we sub- 
 stitute on one side, say, a 10-pound weight in place of a 4- 
 pound weight and a 6-pound weight. So, in an equation, it is 
 permissible to leave one side unaltered and to make any change 
 on the other side we wish, provided this change does not alter 
 the value of that other side. 
 
 Only in an operation which changes the value of one side of 
 a balance or equation we must be careful that exactly the same 
 change in value is made on the other side. For example : 
 
 2 (4 x + 5) - 3 x - 4 - 3 (5 x - 9) + I3x. 
 8 x + 10 - 3 x = 4 - 15 x + 27 + 13 x. 
 6 a; + 10 = -2 a; + 31. 
 7se = 21. 
 a; = 3.
 
 POSITIVE AND NEGATIVE NUMBERS 37 
 
 EXERCISES 
 
 54. Solve: 
 
 2. 
 3. 
 
 4. \(x + 2) = x-20. 
 
 6. 3(z-5)-2(z + 2) = 5(z-7). 
 
 PROBLEMS 
 
 55. 1. The length of a rectangle exceeds its width by 10 ft. 
 Its perimeter is 52 ft. Find its length and width. 
 
 2. The length of a rectangle is 70 in. less than its perime- 
 ter. Find its length, when its width is 20 in. 
 
 3. A lot is 3 times as long as it is wide. If the perimeter 
 is 480 ft., what are its dimensions ? 
 
 4. A garden in the form of a rectangle is 70 yd. wide. 
 The perimeter is 4 times the length. Find the length. 
 
 5. A grocer desires to mix two grades of coffee costing 35^ 
 and 45^ a pound, so as to obtain a mixture weighing 100 Ib. 
 and worth 38^ a pound. How many pounds of each kind must 
 he take ? 
 
 6. How many pounds of 50^ coffee must be mixed with 
 60 Ib. of 40^ coffee to get a mixture worth 43^ a pound ? 
 
 7. Two grades of coffee costing 35^ and 30^ are to be 
 mixed so as to obtain a mixture weighing 100 Ib. and selling 
 for 50 ^ a pound at a profit of 50 % on the cost. How many 
 pounds of each kind are needed ? 
 
 8. The sum of two numbers is 206 ; the larger exceeds the 
 smaller by 46. Find the numbers. 
 
 9. What two numbers have 216 for their sum and 300 for 
 'their difference ? 
 
 54
 
 38 ELEMENTARY ALGEBRA 
 
 10. The difference between two numbers is 25 ; their sum is 
 0. Find the numbers. 
 
 11. If the difference between two numbers is 24, what must 
 each number be in order that their sum shall be 16 ? 
 
 12. Two bicyclists start in the same direction, one going 
 at the rate of 10 mi. an hour and the other at the rate of 12 mi. 
 an hour. How long before the second will be 1 mi. in advance 
 of the first ? 
 
 Let x = the no. of hr. required. 
 Then 10 x = the no. of mi. the first goes in x hr. 
 
 12 x = the no. of mi. the second goes in x hr. 
 
 In x hr. the second has gone 1 mi. farther than the first. Hence if 
 we add 1 mi. to the distance the first has traveled, we get 12 x, the dis- 
 tance traveled by the second. That is, 
 
 10 a; + 1 = 12 x. 
 Solving this, we get x = $. 
 
 13. Two boys start to run in the same direction. One runs 
 at the rate of 300 yd. a minute ; the other at a rate of 340 yd. 
 a minute. How long before the second will be half a mile in 
 advance of the first ? 
 
 14. If the two boys in problem 13 were to start at the same 
 moment and run in opposite directions on a two-mile circular 
 running track, in how many minutes would they meet ? 
 
 15. A boy has $ 1.35 in his pocket, all in dimes and nickels. 
 How many coins of each kind has he, if their total number is 
 19? 
 
 16. How many dimes and quarters, 16 coins in all, are 
 necessary to amount together to $ 2.65 ? 
 
 17. A newsboy has collected $ 1.99, the amount consisting 
 of nickels and pennies. How many has he of each, the total 
 number of coins being 103 ? 
 
 18. I have $ 7.45 in 19 coins, which are quarters, dimes, and 
 5 one-dollar pieces. Tell the number of coins of each kind.
 
 POSITIVE AND NEGATIVE NUMBERS 39 
 
 19. The average of the highest and the lowest temperature 
 of a winter's day is 17-| ; the difference between them is 55. 
 Find the extreme temperatures. 
 
 20. If the lowest temperature one day was 8, and the 
 average for that day was 20, find the maximum temperature. 
 
 21. A certain man is worth $ 3000 more than when he 
 started out in business ; one third of the sum of what he has 
 now and what he did have at first is $ 1915. How much does 
 he possess now ? 
 
 22. The sum of two investments is $ 2700 ; the first yields 
 annually at 5% $30 less than does the second investment at 
 6 lo Find each investment. 
 
 23. A man divides $5000 into two investments, of which 
 one at 6 % brings annually $40 more than the other at 7 %. 
 Into what sums was the $ 5000 divided ? 
 
 24. John can write twice as fast on a typewriter as James 
 can write by hand. How many words can each write per 
 minute, if together they write 162 words per minute ? 
 
 25. The sum of two numbers is 81 ; their difference is 11. 
 Find the two numbers. 
 
 26. The sum of two numbers is 4 a + b ; their difference is 
 a +- &. Find the numbers. 
 
 Let the first number = x. 
 
 Then the second number = 4 a + b x. 
 
 The first number subtracted from the second =4ta+ b x x 
 
 By the conditions of the problem, this difference = a + b. 
 Hence we have the equation 4 a + 5 2x = a + b. 
 
 Solve this equation. 
 
 27. The sum of two numbers is a ; their difference is b. Find 
 the two numbers. 
 
 28. The difference between two numbers is c; their sum is 
 0. Find the two numbers.
 
 40 ELEMENTARY ALGEBRA 
 
 29. How many dimes and quarters will make a sum of 
 10 a -f- 25 b cents, if the number of coins is a -}- 6 ? 
 
 30. If the lowest temperature one day was t, and the 
 average temperature was T, what was the maximum tem- 
 perature ? 
 
 31. A merchant owns now d dollars more than when he 
 started out in business ; one third of the sum of what he owns 
 now and what he did own at the beginning is D dollars. How 
 much does he own now ? 
 
 32. A father earns 4 times as much as his son. Together 
 they earn a dollars per day. How much does each earn ? 
 
 33. The perimeter of a rectangle is p ; its length exceeds its 
 width by q. Find the length and breadth. 
 
 34. A rectangle is 3 times as long as it is wide. Find its 
 dimensions, if the perimeter is P. 
 
 35. Find two consecutive integers whose sum is 25. 
 
 If the smaller integer is x, then the next higher integer is x + 1 . 
 
 36. Find three consecutive integers whose sum is 75. 
 
 37. What three consecutive integers produce the sum s ? 
 
 38. Find two consecutive even integers whose sum is 102. 
 
 If x is any integer, then we are certain that 2 a; is an even number, for 
 the reason that any integer, when doubled, gives an even number. 
 
 Letting 2 x stand for the smaller even integer, what must be added to 
 this to give the next higher even integer ?
 
 LEONHARD EULER 
 
 One of the greatest mathematicians of the eighteenth century. In 
 1770 he published an algebra which was translated from German into 
 French, English, Italian, and Latin, and was widely used. Euler was 
 a Swiss, but spent most of his time in Russia and Germany.
 
 
 MULTIPLICATION AND DIVISION 
 
 MULTIPLICATION OF MONOMIALS 
 
 56. By the definition of an exponent ( 14), we know that a 2 
 means a a, b 3 means b b b ; hence 
 
 a 2 x a 3 means a a x a a a or a 5 . Similarly, 
 c 3 x c 4 = c 7 , and a 3 6 2 c 4 x a 2 6 3 c 5 = a 5 6 5 c 9 . 
 
 Hence in the multiplication of terms containing like letters, the 
 exponent of any letter in the product is the sum of the exponents 
 of that letter in the factors. 
 
 The product of 12 abc and 5 aWc is 60 a 2 6 3 c 2 ; that is, the 
 coefficients are multiplied together. Bearing in mind the rule 
 of signs in multiplication, the following results are seen to hold : 
 + 3 xy x ( + 5 o?yz) = + 15 x 3 y z z. 
 - 4 xf x (+ 6 x 2 yz z ) = - 24 x 3 y 3 z 2 . 
 + 5 xyz 2 x (- 6 x 2 y 2 z) = - 30 tfff. 
 6 xyz x ( 7 xyz) = + 42 x z y 2 z 2 . 
 
 ORAL EXERCISES 
 
 57. Perform the following indicated multiplications : 
 
 1. +3a;( + 5ajy). 7. + 59 df 3 x - 9/V*. 
 
 2. + 5 xz (- 5 xy*z). 8. + 20 times + 30 e 3 . 
 
 3. - 7 a 3 6 2 c times - 8 &VeP. 9. - 30 acf X - 10 a6 2 cd/ 3 . 
 
 4. + a^V times 60. 10, ( 2a?#).(+4a; 2 ?/)-( lays). 
 
 5. (8 mVo 3 ) (- 7 HiWo*). 11. (- 2 a) (- 3 a 2 ) (- 2 aft). 
 
 6. -12h*x-7h*jk. 12. (+2a6)(-2a 2 6)(-3a5 3 ). 
 
 13. If two, four, or six negative terms are multiplied to- 
 gether, what is the sign of the product ? Why ? 
 
 41
 
 42 ELEMENTARY ALGEBRA 
 
 14. If three, five, or seven negative terms are multiplied 
 together, what is the sign of the product ? Why ? 
 
 15. If among the terms to be multiplied together a certain 
 number of terms is positive and the number of negative terms 
 is even, what is the sign of the product ? Does the number of 
 positive terms require special attention ? 
 
 16. If the number of negative terms is odd, what is the sign 
 of the product ? 
 
 MULTIPLICATION OF A POLYNOMIAL BY A MONOMIAL 
 
 58. It is seen that 4(2 + 3) =4 -5 = 20, 
 and that 4-2 + 4-3 = 20. 
 
 Hence 4(2 + 3)=4 -2 + 4 -3. 
 
 In general, a(b + c) = ab + ac, 
 
 and a(b + c d) =ab-{-ac ad. 
 
 Hence, in multiplication of a polynomial by a monomial, we 
 have the rule : 
 
 Multiply each term of the polynomial by the monomial, and 
 write down in succession the products with their respective signs. 
 
 Observe that the* foil owing results hold: 
 
 +2 ab\a* + b - c 3 ) = + 2 a 3 b 2 + 2ab 3 -2 ab*c*. 
 - 3 a 2 c[ - a + 2 be - 4 c] = + 3 a 3 c - 6 a^c 2 + 12 a V . 
 
 EXERCISES 
 
 59. 1. Multiply 2 x + ey + 6 z 5 w by 3 xy. 
 
 2. Multiply a 26 + ec 4d ebylO abc. 
 
 3. Find the product of 9 xy z z and xz + 4 y 3 z 5 xyz 2 
 
 Simplify 
 
 4. (3m+ 
 
 5. -7(5-2a+26-3c + 4 
 
 6. -l
 
 MULTIPLICATION AND DIVISION 43 
 
 7. (a-& s -2e-3d s )(-4a6cd)(-l). 
 
 8. (-2 ax 3 ) 
 
 9. (- 
 
 10. (_ 
 
 11. 12(1+2-3 + 4-5 + 6- 
 
 12. What are the two ways of working the last exercise ? 
 Which of those two ways is the shorter ? Why ? 
 
 MULTIPLICATION OF POLYNOMIALS 
 
 60. It is readily seen that (3 + 4)(2 + 6) = 56. It might be 
 worked also in this manner: 
 
 (3 + 4)(2 + 6)=3(2 + 6) + 4(2+6) = 3 2+3 - 6 + 4-2 + 4-6=56. 
 
 This second mode of procedure is longer in this particular 
 example. It is given here because it illustrates the process 
 which must be followed when the terms inside the parentheses 
 are not like terms. For instance, it can be used in a multiplica- 
 tion like the following : 
 
 We see from this that we can find the product of two poly- 
 nomials by multiplying one polynomial by each term of the other 
 and then adding the partial products. 
 
 In written exercises the work may be arranged as follows : 
 3x -4y 
 
 x +'2y 
 
 3 x 2 - 4 xy 
 
 6 xy - 8 y 2 
 3x* + 2xy-8y 2 
 
 Frequently a polynomial has some letter raised to different 
 powers in its terms. Thus, in the polynomial x 4 + 2 y? x 2 
 + 4 x 5, all the terms contain x, except the last term ;
 
 44 ELEMENTARY ALGEBRA 
 
 moreover, the term x 4 , which is the highest power of x in the 
 polynomial, is written first. Then follow the terms containing 
 the lower powers of x, arranged in descending order. This 
 polynomial is said to be arranged according to the descending 
 powers of x. 
 
 When written in reverse order, thus, 5 + 4 x x* + 2x 3 + x 4 , 
 we say that the polynomial is arranged according to the ascend- 
 ing powers of x. When polynomials of this kind are used, it 
 is convenient to arrange them first according to the descending 
 or the ascending powers of some letter. If in the multipli- 
 cation of two polynomials both are arranged according to the 
 ascending powers of some letter, or both according to the de- 
 scending powers, the partial products can be written down and 
 added with much greater ease. 
 
 Example. Find the product of 3 x 2 + 4 x 6 + 2 x 8 and 2 x + x 2 3. 
 Arranging both polynomials according to the descending powers of x, 
 the multiplication is as follows : 
 
 2a?-3a; l +4aj -5 
 
 x 3 
 
 9a; 2 -12a;+15 
 
 One method of checking a multiplication of this kind is to 
 arrange both polynomials according to the ascending powers 
 of the letters and then to find their product, thus : 
 
 - 3 -|- 2 x + 
 
 + 15-12*+ 9-6aj 
 -10*+ 8* 1 
 
 3a?*+ 2x* 
 
 The two products are the same, except that one is arranged 
 according to the ascending powers of *, and the other according
 
 MULTIPLICATION AND DIVISION 45 
 
 to the descending powers. The order in which the terms of a 
 polynomial are written does not affect its value, hence we say 
 that the two products are the same. The second multiplication 
 indicates that the first multiplication was correct. 
 
 A second method of checking multiplication is by assigning 
 some particular small number as the value of x. Let x = 2, 
 
 then 2x' s -3x 2 + 4 : x-5 = 7 
 
 a 2 + 2 x 3 = 5 
 
 2x 5 + x 4 -8 of + 12 x 2 -22 x + 15 35 
 Let x = 2, 64 + 16 - 64 + 48 - 44 + 15 = 35. 
 Hence we are reasonably certain that the multiplication is 
 correct. 
 
 EXERCISES 
 
 61. Multiply together the following polynomials, and check: 
 
 1. x + 1 and x + 2. 9. y 5 and y 3 2y z + 3yl. 
 
 2. 2 x + 1 and x + 3. 10. 2 z + 6 and - z 3 + 2 z 2 - 10. 
 
 3. 2 x 1 and x 3. 11. 3 x + y and 2 a 4 3 a 2 + 5. 
 
 4. 5 y?+ x+2 and x + 4. 12. x 2 + x and 5 x 2 + 4 x + 2. 
 
 5. 6a 2 -2a;-2 and -x + 3. 13. z 2 - 7 and 6z 3 -l. 
 
 6. y 3 2 y 2 + 4 and y 5. 14. a; 2 x + | and 2 a; + 4. 
 
 7. ^ 1 and ^/ 2 y 1. 15. a; 2 a; -f 3 and x -. 
 
 8. z 2 -2z +1 and z+-l. 16. z 4 + z 2 +1 and z 2 1. 
 
 Arrange the polynomials according to the descending powers 
 of the letter and multiply. Verify by arranging the polyno- 
 mials according to the ascending powers, and multiply : 
 
 17. x 1 and Sx 2 4 + 2 cc. 
 
 18. 6 + z 2 - 5 z and 2 - z 2 + 3 z. 
 
 19. - 1 + 3 f + 2 y and 6 + 5 y - y 2 .
 
 46 ELEMENTARY ALGEBRA 
 
 20. 
 
 21. x 3 + 2 x 4 - 3 x 2 - 4 and - 3 + 2 x + x 2 . 
 
 Perform the indicated multiplications : 
 
 22. (3x + y + z)(5x-2y-z + 5). 
 
 23. [o + 6-c][2a-26 + 3c]. 
 
 24. \x* + 2 xy + x\\a? xy + x + y\. 
 
 25. (a? + f + z 2 )(* + y + *) 
 
 26. (a-e + 3/-20) 2 . 
 
 27. (a + b - c + d + e) 2 . 
 
 28. (a + 6)(a - 6) (a 2 + 2 a6 - 6 2 ). 
 
 29. (x 2 y) 3 . 
 
 30. (x-y + z) 3 . . 31. (a-6-c) 3 . 
 
 HARDER EQUATIONS 
 
 62. 1. Solve (* + 2)(a?_ - 7) - ( + 5) 2 = 0. 
 Solution. ( X 2 _ 5 x - 14)- (x 2 + 10 x + 25) = 0. 
 
 Remove parentheses, x 2 5 x 14 x 2 10 x 25 = 0. 
 Transpose, x 2 - 5 x - x 2 - 10 x = 24 + 14. 
 
 Combine, - 15 z = 39. 
 
 Divide both sides by - 15, x = - f | = - 2f . 
 
 (ttecifc: (--tf + 2)(- V - 7)-(- + 5 ) 2 = - 
 
 = 0. 
 Solve : 
 
 2. (aj + l)(a;-l)-(aJ + 2)(aj-3)=-0. 
 
 3. a?-(*+4)(a? + l) = 0. 
 
 4. (2/ + 2)(y-3) + (i/-5)(i/-l)-2/ = 0. 
 
 5. (22/+l) 2 -4(2/ + l) 2 = 0. 
 
 6. [z + lJ[z-2]-[2!-3][2+4] = 0. 
 
 7. x 2 - 2 Z + 5 - (x + 2)(* - 3) = 0. 
 
 8- (y + 3)(y - 5) + (y + 2)(y + 1) - (y + 2)(y - 3)
 
 MULTIPLICATION AND DIVISION 
 
 47 
 
 9. + 1)0 - 1) - (a + 2)O - 2) + x + 10 = 0. 
 
 10. (3 - x)(2 - x) - (5 + a?)(l - x) - 2 x 2 = 6*. 
 
 11. = (1 - z)(z - 2) + (z + 4)(z - 2) + 8. 
 
 12. 3 = 5 + 0* - {6+}{a;+l} +5*. 
 
 13. (x - 5) 2 - (x -f 2) 2 = 5(*5 + 3). 
 
 14. 6 (of + 1) - (5 x + 6) (a? + 1) = + 2)( - 2). 
 
 15. (x 2 + x 4- 1)O - 5) - (a? - x - 7)(z - 3) = 50 + 7). 
 
 16. a; + 10 = - + 5)0 - 5) - (4 + a?)(5 SB). 
 
 17. (y + 3)(6y + 7) = (2y-3)(3y-4) + 6. 
 
 18. 2 - 5 = (z - 5)0 + 4) - (2 z + l)(z - 1) + z 2 . 
 
 19. 5 + (as + 1)0 - 1)0* + !) = (^ 2 + !)(* - 1) -f 
 
 20. O 2 + * + 2 )0 1) = (a? 1 !)( + 1). 
 
 63. Figure 7 shows that a rectangle 4 in. long and 3 in. 
 wide has an area of 12 sq. in. Similar relations hold when 
 the foot, yard, or some other unit of length 
 
 is taken. These considerations suggest 
 the following definitions : 
 
 in. X in. = sq. in. 
 ft. X ft. = sq. ft. 
 yd. x yd. x yd. = cu. yd., etc. FIG. 7 
 
 ORAL EXERCISES 
 
 64. Find the areas of rectangles whose dimensions are : 
 
 1. 7 ft. and 13 ft. 6. ab yd. and c yd. 
 
 2. 6 in. and 9^ in. 7. 3 x mi. and 4 y mi. 
 
 3. 3| ft. and 4| ft. 8. 2 ab in. and 7 abc in. 
 
 4. 5 yd. and x yd. 9. x 2 yd. and xz yd. 
 
 5. 6 ft. and 9 ft.
 
 48 ELEMENTARY ALGEBRA 
 
 Find the volumes of rectangular solids whose dimensions 
 are : 
 
 10. 3 in., 4 in., 5 in. 14. ab yd., abc yd., and a?b yd. 
 
 11. 5 ft., 6 ft., n ft. 15. 4x in., 5 y in., and 2 z in. 
 
 12. 2 yd., b yd., c yd. 16. 5 y ft., 2 ?/ 2 ft., and 2 */z ft. 
 
 13. J in., w in., & in. 17. x rd., y rd., and xyz rd. 
 
 Find the area of a square whose edges are : 
 
 18. 6 s ft. 20. 4 a 2 in. 22. (a + 5) yd. 
 
 19. 4.5 a in. 21. ran 2 f t. 23. (x + 2 y) in. 
 
 Find the volume of a cube whose edges are : 
 
 24. 4 a. 25. 2 ft 2 . 26. (c + d)ft. 
 
 27. If the box shown in the adjoining figure has I 4 in., 
 
 w = 3 in., h = 2 in., 
 
 (a) What is the area of the bottom ? 
 
 (&) What is the area of the bottom and top 
 taken together ? 
 
 (c) What is the area of one end of the box ? 
 Of both ends ? 
 
 (d) What is the total surface of the box ? 
 (e) What is the volume of the box ? 
 
 28. Answer the questions in Ex. 27 when I = 1, w = 2, and 
 ft is any number. 
 
 29. Answer the questions in Ex. 27 when I = 10 and w and 
 h are any numbers. 
 
 30. Answer the questions in Ex. 27 when I, w, h are any 
 numbers. 
 
 Find the area of a rectangle whose dimensions are as fol- 
 lows: 
 
 31. 9, (a -|- b). 34. (3 - x), (4 + x). 
 
 32. ( x - 6), 5. 35. (a + 2 6), (a - 2 b). 
 
 33. (a + 6), (a - 6). 36. [5 a - 2 6], [2 a + 3 6].
 
 MULTIPLICATION AND DIVISION 49 
 
 37. Find the numerical value of the area in Ex. 36 when 
 a = 10, b = 5. 
 
 DIVISION BY MONOMIALS 
 65. Just as in arithmetic, 3 -=- 4 may be written f ; 
 
 71 
 
 so in algebra, n + d may be written - 
 
 a 
 
 In algebra, as in arithmetic, ^ may be interpreted in two 
 ways : 
 
 (1) As an indicated division. 
 
 (2) As a regular fraction, showing that 3 of the 4 equal 
 parts of a unit are taken. 
 
 71 
 
 The same is true of - . It means " n divided by d," or " a 
 
 unit is divided into d equal parts and n such parts are taken." 
 We can adopt either interpretation. 
 
 We have learned in arithmetic that the value of an expres- 
 sion (quotient or fraction) is not changed, when the dividend 
 and divisor in a division (or the numerator and denominator 
 of a fraction) are both multiplied or both divided by the same 
 number. 
 
 That is, | = |=3. 
 
 This principle is used in simplifying the dividend and 
 divisor, or in reducing a fraction to its lowest terms. 
 
 Thus, dividing by 5 a both terms of the fraction , we obtain 
 
 36c 5ad 
 
 its equal, 
 a 
 
 The rule of signs must be followed carefully. In division, as in multi- 
 plication, " like signs give plus, unlike signs give minus." 
 
 We know ( 13) that 2 5 = 2 . 2 . 2 . 2 2, and 2 3 = 2 . 2 - 2. 
 
 05 2 2 2 2 
 Hence | = 2>2>g " = 2' or 2-. 
 
 . ., . a 5 a-a-a a a M 
 
 Similarly, = = a 2 or a^ 3 - 
 
 a 3 a- a- a
 
 50 ELEMENTARY ALGEBRA 
 
 This gives the following very important rule : 
 The exponent of any letter in the quotient is equal to its ex- 
 ponent in the dividend minus its exponent in the divisor. 
 
 It is thus seen that in division we subtract the exponents of any letter, 
 while in multiplication ( 56) we add its exponents. 
 
 In general, in multiplication, a m a" = a m+ " ; 
 
 ,. . . a m 
 
 in division, = a m ~ B - 
 a" 
 
 ORAL EXERCISES 
 
 66. Perform the indicated divisions : 
 
 l 12abc 24afy 36 aby 3 Scdx* 
 4 ab 6 xy 12 aby' 12 dx 
 a 2 c 3 +Z 4 3m 3 + 5 x* 
 
 ' ' 9 ' I O ' 1 
 
 a c z 2 + 3m or 
 
 
 a 5 b ab*c 4 x*y* + 9 m 2 n 5 
 ~ab' -2a& 3 ' + 4 afy 2 ' - SraV* 
 
 67. EXERCISES 
 
 1. - 124 aW -f- 12 aW. 
 
 2. - 25 rffz 9 ^- 15 ajyz 5 . 
 
 3. 236 m 6 nV ^- 24 m^o 8 . 
 
 4. + 50 ftc 4 ^ 6 -*- ( - 60 6 3 c 5 dV) . 
 
 Reduce the following fractions to the lowest terms : 
 -28/> 5 gV + 72 aW 2 
 
 + 56 pY^s 2 ' - 44 a Wef ' 
 
 Let A be the area of a rectangle, L its length, W its width. 
 Find the second dimension in the following rectangles : 
 
 8. A = 56 a 2 sq. ft., L = 8 a f t. 
 
 9. A = 136 ofy 2 sq. in., L = 12xy in. 
 
 10. ^l = 
 
 11. ^ = 
 
 12. .4 = 1.
 
 MULTIPLICATION AND DIVISION 51 
 
 MULTIPLICATION OF FRACTIONS 
 
 68. Figure 9 shows that | of f of an inch is f of an inch ; 
 this recalls to mind the arithmetical rule for the multiplication 
 of fractions : Multiply the numerators to- 
 gether for a new numerator, and the denomi- 
 nators together for a new denominator. 
 
 The process is the same in algebra. 
 
 It is easiest, as a rule, at first merely to 
 
 indicate the multiplication, then to divide both numerator and 
 denominator by every factor common to them. 
 I a 2 6' 2 
 
 Thus, -Mutinies 
 
 3c 
 
 EXERCISES 
 69. Eind the product of the following fractions : 
 
 Sab 12 ed ^ -2ab - 3 a 
 
 4cd' 15 a' 6xy ' 8 a*V 
 
 16 zw* 85 a 2 * 5 , + 8 * 5 
 
 
 
 15 at ' 64 zw -24 
 
 1 64 ttyV 6 - 24 
 
 ' 
 
 -36mV 14 aV 
 
 Since in division "unlike signs give minus," it is evident that the 
 following equalities hold : 
 
 That is, the sign may be in front of the fraction, or before a factor 
 of the numerator, or before a factor of the denominator, and there is no 
 change in the value of the fraction. 
 
 7 ~ 25 q, 2 c 4 - 
 
 ' 
 
 5a 4 c 2 ' - aW - 
 
 27 kWm 3 r*s 2 t 6 8/y/i 4 5 t 
 
 " . _ r~: X - _ _ __ . ~. 10.
 
 52 ELEMENTARY ALGEBRA 
 
 If a fraction - is to be multiplied by an integer c, it is easiest to write 
 b 
 
 for c its equal -, and then to multiply - by -, by the rule for the 
 1 61 
 
 multiplication of fractions. 
 
 11. 24 aW times - ab<icd \ 13. 32 fcVd 6 times 
 
 30^% .. 20 h*fk* ,. 
 
 12. - - times 3x 2 z*w 2 . 14. - ? tunes 
 
 -3m 2 n 4 o 2 
 
 DIVISION OF FRACTIONS 
 
 70. In algebra, as in arithmetic, a fraction is divided by 
 another " by inverting the divisor and multiplying." 
 
 This rule may be proved as follows : Let the quotient of - -s- - be x. 
 
 o a 
 
 The dividend is - : the divisor is - . Since 
 6 d 
 
 divisor x quotient = dividend, we have, 
 
 = -. Multiplying both sides of the 
 d b 
 
 equation by bd gives ocdx_abd^ QT ^ re( j ucm g tne fractions to 
 
 d b 
 the lowest terms, bcx = ad. Dividing both sides by the 
 
 coefficient of x, we get x = . 
 
 be 
 
 This value of x is the required quotient. One sees at once that a quick 
 way of getting this answer is to invert the divisor and multiply. 
 
 ORAL, EXERCISES 
 71. Perform orally the following divisions : 
 
 7. +. 
 
 4. l} + 2t. 6. + !. 8. 
 
 b
 
 MULTIPLICATION AND DIVISION 53 
 
 In divisions like a -=- - or - -=- a, it is easiest to write - in place of a ; 
 c c 1 
 
 then proceed as in the previous exercises. 
 
 11. 5---. 
 
 13. - + -. 
 
 n 
 
 de 4 
 
 12. | + 2. 
 
 2 6 
 
 " H-I- 
 
 9. e-i-5. 
 
 10. *+. 
 
 72. WRITTEN EXERCISES 
 
 1 UasW . 28a 2 s 2 d 72 m 
 
 25 a% ' 125 xf' 
 256 eW . 128 eWy 
 36a6V ' 144 ' 
 
 ORAL PROBLEMS 
 73. 1. How many 25^ pieces make $4? 
 
 2. How many 25 ^ pieces make $ d ? 
 
 3. A man "steps off" the length of a playground and finds it 
 to be 125 paces. How many feet is this, if his pace is 3 ft. ? 
 
 4. If a man's pace is 3 ft., what distance do p such paces 
 measure ? 
 
 5. A man finds the width of a lot to be a lengths of a cer- 
 tain rod which is 6 ft. long. How wide is the lot ? 
 
 6. The water in a canal flows at the rate of - mi. an hour. 
 
 h 
 
 In what time will it flow - mi ? 
 
 k 
 
 1. An aviator flies at the rate of 95 mi. an hour. How long 
 
 will it take him to go - of a mile ? 
 s 
 
 8. A girl cuts off two-thirds of a ribbon 21 yd. long. How 
 many feet of ribbon does she cut off ? 
 
 9. A woman gets the - part of a supply of coffee weighing - 
 
 ct c 
 
 pounds. How many pounds does she receive ?
 
 54 ELEMENTARY ALGEBRA 
 
 DIVISION OF POLYNOMIALS BY MONOMIALS 
 
 74. This is easily done by dividing each term of the poly- 
 nomial by the monomial and connecting these partial results by the 
 proper signs. 
 
 Thus, (15 x 2 y z -6xy 3 + l2xy)^-(3 xy) 
 
 , 12xy K 
 + * = 5 xy - 2 y* + 4, the quotient. 
 
 3 xy 3xy 
 Again, (a 2 6 3 - 3 a 2 6* - 5 a 3 6 3 ) -4- 5 a 2 6 2 
 
 a 2 6 3 3a 2 6 4 5a 3 6 3 6 36 2 
 
 = = - ab, the answer. 
 
 o cro 2 5 a 2 o 5 a 2 o 2 5 5 
 
 The correctness of the answer can be tested by multiplying the quotient 
 by the divisor. The resulting product should be equal to the dividend. 
 
 EXERCISES 
 75. Perform the indicated divisions and check your results : 
 
 1 _ **^ v **-Vf rr 
 
 2x 
 
 3a 2 6 2 
 
 3. . 9. 
 
 15 &3 C 2 _ 20 Wed 1A 
 
 4. 10. 
 
 15 b z c 4 ioV 
 
 5a6c-6a 2 6c 2 -|-8a 2 6c 
 abc 
 
 
 6. M -^ 2 < 12. 
 
 24sV + 12sy 
 
 
 -12x*y 
 18 aftV - 12 a 2 6 4 c 3 + 24 aW
 
 MULTIPLICATION AND DIVISION 55 
 
 DIVISION OF POLYNOMIALS BY BINOMIALS 
 
 76. An inspection of the following multiplication will sug- 
 gest some of the steps to be taken in the inverse operation 
 Division. 4 a? + 3 a; +5 
 
 x-2 
 
 43? + 3 a? + 5 a; 
 
 - 8 x* - 6 x - 10 
 4 o - 5 a? - a- 10 
 
 If we are required to divide 4 a? 5 x 9 - x 10 by x 2, 
 then the quotient which must be the other factor, 4 x 2 -f 3 x + 5, 
 can be found as follows: 
 
 Dividend 4 a^ 5 x 2 x 10 
 1st partial product 4 3? $ a; 2 
 
 a? -2 Divisor 
 
 3 x + 5 Quotient 
 
 1st remainder + 3 # 2 x 10 
 
 2d partial product + 3 x 1 6 x 
 
 2d remainder + 5 x 10 
 
 3d partial product + 5 x 10 
 
 Last remainder 
 
 Notice that both the given dividend and divisor are arranged accord- 
 ing to the descending powers of the leading letter x. The ascending 
 powers would have answered equally well. It is a very great conven- 
 ience to have the dividend and divisor arranged either both according to 
 the ascending or both according to the descending powers of a leading 
 letter. If they are not so arranged, then they should be, before the 
 division is begun. 
 
 In the division above, notice the following : 
 
 (1) The first term in the quotient 4 a; 2 is obtained by dividing the 
 first term in the dividend 4 x 3 by the first term in the divisor x. 
 
 (2) The first partial product is the product of the first term in 
 the quotient 4 x 2 and the divisor x 2. 
 
 (3) The first remainder 3 x 2 x 10 is obtained by subtracting 
 this partial product from the dividend. 
 
 The 10 is not needed in the second step of the division 
 and may therefore be omitted until the third step.
 
 56 ELEMENTARY ALGEBRA 
 
 The remaining steps of the division, should there be any, are 
 repetitions of steps above. 
 
 77. Explain: 
 
 (1) How is the term + 3 x of the quotient found ? 
 
 (2) How is the second remainder found ? 
 
 (3) How are the + 5 in the quotient and the third partial 
 product found ? 
 
 (4) How can you check your division ? 
 
 In many cases the division does not come out exact and there will be 
 a final remainder. Ju exercises 1-23 which follow, the divisions will 
 come out exact. 
 
 Divide 9 - 9 x + 8 a? - 7 x 3 + 2 a; 4 by 3 -'2 x. 
 
 ,3 2x 
 9-6 
 
 + 6 a; 2 - 4 a 3 
 
 EXERCISES 
 78. Divide: 
 
 1. 0^ + 203 + 1 by x + 1. 5. x*-2x*+2x-lbyx-l. 
 
 2. t/ 2 -f 4 y + 4 by y + 2. 6. w 3 + 1 by w + 1. 
 
 3. x 2 - x 6 by x - 3. 7. c 3 - 1 by c - 1. 
 
 4. z 2 - 7 2 -f 10 by z - 5. 8. x* - 16 by x -2. 
 
 9. Qx 4 + x 3 ~l5x 2 + 38x-3Qby x 1. 
 10. 4 + 28 a; + 29 x 2 - 30 y? + 25 x 4 by 5 x + 2. 
 11. 
 
 12. a 2 6 2 by a - 1). t 14. a 2 + 2 a^ + y 2 by x + y. 
 
 13. ic 2 y 2 by sc + y. 15. as 2 2 xi/ + t/ 2 by x y.
 
 MULTIPLICATION AND DIVISION 
 
 57 
 
 16. m 3 -fn 3 by m + n. 
 
 17. m 3 n 3 by m n. 
 
 18. x & t/ 6 by a; y. 
 
 19. a 6 6 6 by a 2 b*. 
 
 20. c 6 d 6 by c 3 d 3 . 
 
 21. a 6 6 6 by a 3 + b 3 . 
 
 22. of y 6 loy x y. 
 
 23. a; 5 + y 5 by a; + y. 
 
 DIVISIONS INVOLVING REMAINDERS 
 
 79. 11 H- 3 gives the quotient 3 and the remainder 2; the division is 
 not exact ; the answer is 3 and f . We see that 
 
 remainder 
 
 dividend -=- divisor = quotient + 
 
 divisor 
 
 Divide x 5 
 
 a 5 
 
 - 2 y* 
 
 + 3x 3 
 + 3 x 3 
 
 - 2 
 
 - 6 
 
 -)- x + 4, the remainder. 
 x ~4- 4 
 
 Hence the quotient is x 2 + 3 + -. 
 
 X s 2 
 
 Check : dividend = quotient x divisor + remainder. 
 EXERCISES 
 
 80. Divide: 
 
 1. or 3 + 2 x z + 1 x 5 by x 9. 
 
 2. t/ 4 - 4 y 3 + 5 f - 3 y + 12 by / + 4. 
 
 3. z 5 - 16 z 3 - 6 2 -f 3 z + 5 by z 2 - 4 z. 
 
 4. to 8 + 1 by w 1 + 1. 
 
 5. a 4 + 3 a 2 + 9 a - 10 by a 3 - 7. 
 
 6. m 5 5 m 4 8 ?/i 3 + 4 m + 8 by m 3 + 1. 
 
 7. r 16 -(- 1 by r 4 + 1. 
 
 8. s 16 - 1 by s 5 - 2.
 
 CHAPTER IV 
 
 PROPORTION 
 
 81. A proportion expresses the equality of two common 
 fractions. 
 
 Thus, |f and f are equal to each other ; hence, |f = f is a proportion. 
 
 A common fraction may be said to indicate the division of the number 
 
 in the numerator by the number in the denominator. When so con- 
 
 sidered, the fraction is frequently called a ratio. The ratio - is often 
 written a : b. 
 
 Since a proportion is a special type of equation, it may be treated like 
 an equation. If one of the terms of either fraction in the proportion is 
 unknown, the rules for the solution of equations enable us to find the 
 value of the unknown term. 
 
 x 3 
 
 1. Solve the proportion = -. 
 ] o 
 
 Multiplying both sides of the equation by numbers which will remove 
 the denominators, that is, multiplying both sides by 12 and by 5, we get 
 6 . yi x _ ft 12 - 3 
 
 ^ % 
 
 Or 5 x = 36. 
 
 Divide both sides by 5, x = S J = 7, the answer. 
 
 2. Solve the proportion a ~ = -. 
 
 c a 
 
 Multiplying both sides by c, and also by d~, 
 cd(a 6) _ cdx 
 ~~T~ = ~d" 
 
 Or d(a b) =cx. 
 
 Change sides, for convenience, ex = d(a &). 
 
 = ' a 
 
 58
 
 PROPORTION 59 
 
 EXERCISES 
 
 82. Solve the following proportions : 
 . x 23 x mn 
 
 6 = y 8 ' 12 = ^" 1 
 
 11. - = - 18. 
 c d 4 
 
 12 - - = 19. 5.5: x= 10.7: 25.4. 
 
 an 
 
 _ 10.5 x t x 
 
 6. _ = _-. 13. =- 20. -x:a = m:n, 
 
 o 8 20 m 
 
 7. ^ = ^- 14. -^ = 2- 21. _r:-6.8 = y:s. 
 5 15 6s 
 
 PROBLEMS IN PROPORTION 
 
 83. 1. If 5 acres of land are worth $875, what are 7 acres 
 of land of the same quality worth ? 
 
 Here the number of acres is said to be in the "same ratio" as or 
 " directly proportional " to the cost. That is, 
 5 acres _$ 875 
 7 acres <$x 
 
 Since 5 acres = and ^^ = , the original proportion can be written 
 7 acres 7 $ x a; 
 
 in the simplified form, 5 _ 875 
 
 7~~aT 
 We get x = 1225. Hence, the answer is -f 1225. 
 
 2. If 11 men can build a wall in 80 days, how long would it 
 take 14 men to do the same work ? 
 
 Here the larger the number of men, the less the time for doing a given 
 piece of work; the number of men is not in the "same ratio" as the
 
 60 ELEMENTARY ALGEBRA 
 
 times needed to complete the work, but in the " inverse ratio." That is, 
 the number of men is "inversely proportional" to the time required to 
 finish the work. We have, therefore, 
 
 , or more simply, 
 
 13 men 80 days 
 U = x_ 
 13 80' 
 Solve this proportion. 
 
 3. If, in a broad jump, a boy can jump 15 ft. when he takes 
 a running start of 20 ft., how far can he jump with a running 
 start of 400 ft. ? 
 
 Is the distance a boy can jump either "directly" or "inversely" 
 proportional to the running start ? 
 
 Can this example be worked by proportion ? 
 
 In working problems we must consider the following : 
 
 (1) Can the problem be worked by proportion ; that is, do the quan- 
 tities named in the problem vary in either direct or in inverse ratio ? 
 
 (2) If they do, determine whether the variation is in "direct ratio" 
 or in "inverse ratio." 
 
 It should be observed that, while a variation may not be strictly in 
 "direct" or in "inverse ratio," the variation may be so nearly in that 
 ratio that the method of proportion yields answers that are sufficiently 
 accurate for practical purposes. 
 
 For example, it does not necessarily follow that 4 men can do a piece 
 of work in half the time that 2 men can do the work. The 4 men may 
 not be equally efficient or 4 men may be in each other's way, so that 4 
 men could not work as efficiently as 2 men can. But in many such cases 
 the differences are so slight as to be negligible. 
 
 4. Find two numbers in the ratio of 4 to 5 whose sum is 
 207. 
 
 Let the two numbers be 4 a: and 5 x. 
 
 5. Find two numbers in the ratio of 3 to 8 whose sum is 
 -275. 
 
 6. Find two numbers in the ratio 2 to 9 whose sum is 
 -217. 
 
 7. Find two numbers in the ratio of 2:5, such that the 
 first number minus the second yields 133.
 
 PROPORTION 
 
 61 
 
 8. Two sums of money bear interests which are in the 
 ratio of 6 to 7. Find each interest, if the two interest sums 
 together amount to 650 dollars. 
 
 9. Two sums of money, at the same time and rate, bear 
 interests which are in the ratio of 3 to 4. What is the larger 
 sum, if the smaller is $ 3360 ? 
 
 10. Two equal sums of money, both at the same rate, bear 
 interests which are to each other as 10 is to 13. If the time for 
 the lesser interest is 5 years and 5 months, what is the time for 
 the other ? 
 
 11. What is the height of a flagstaff which casts upon level 
 ground a shadow of 84 ft., if at the same moment a vertical rod, 
 10 ft. long, casts a shadow 
 
 7 ft. long ? 
 
 As shown by Fig. 10, two 
 triangles are formed which 
 have their sides parallel, re- 
 spectively. Such triangles 
 have the same shape, but not 
 the same size. It is shown 
 in geometry that such tri- 
 angles have their correspond- 
 ing sides proportional. In 
 this case the horizontal sides 
 are to each other as the verti- 
 cal sides and as the slanting 
 sides. In our problem the slanting sides are not used in forming the 
 proportion. We form the proportion thus: 
 
 Horiz. side of small tri. _ Vert, side of small tri. 
 Horiz. side of large tri. ~~ Vert, side of large tri. 
 
 _10 
 
 84 ~ ' 
 x being the required height of the flagstaff. 
 
 12. A pole for wireless telegraphy casts a shadow of 105 ft. 
 upon level ground at the same time that a vertical rod, 12 ft. 
 long, casts a shadow 7 ft. long. Compute the length of the pole. 
 
 7 ft.
 
 62 ELEMENTARY ALGEBRA 
 
 13. By the method just explained determine the height of 
 the flagpole of your school, or of some similar high object. 
 
 14. At the moment when the sun sets behind a mountain 
 peak, a vertical rod 12 ft. high casts a shadow of 27 ft. on level 
 ground. If the mountain is known to tower 7000 ft. above the 
 plain, what is the distance in a horizontal line to the center of 
 the mountain ? 
 
 15. Concrete is made by mixing cement, sand, gravel, and 
 water. A 1 : 2 : 4 concrete is made up of one volume of cement, 
 twice that volume of sand, and 4 times that volume of stone. 
 The volume of the mixed cement is less than the sum of the 
 volumes of the parts, because of the open spaces between the 
 stones which must be filled in making the concrete. One cubic 
 foot of concrete contains : 
 
 for 1:2:4 concrete : .22 cu. ft. cement, .44 cu. ft. sand, .88 
 cu. ft. gravel. 
 
 f or 1 : 2| : 5 concrete : .19 cu. ft. cement, .475 cu. ft. sand, 
 .95 cu. ft. gravel. 
 
 In the building of a silo, 1125 cu. ft. of concrete is needed, 
 of which must be 1 : 2 : 4 concrete, the rest 1 : 2|- : 5 concrete. 
 How many cubic feet of cement, of sand, and of gravel are 
 needed for making the 1:2:4 cement ? 
 
 Let the required number of cubic feet of cement, sand, and gravel be, 
 respectively, 22 x, 44 x, and 88 x. 
 
 16. In Ex. 15, how many cubic feet of cement, of sand, and 
 of gravel are needed for making the 1 : 2|- : 5 cement ? 
 
 17. A contributes $ 2000 to a speculation, B $ 3500, C $ 1500. 
 The total profits are $ 725. What amount of the gain ought 
 each to receive ? 
 
 18. Find a number such that its excess over 8 bears the same 
 ratio to its excess over 9 as the number itself does to its excess 
 over 2.
 
 PROPORTION 63 
 
 19. Two pounds of tea cost as much as 3 Ib. of coffee, and as 
 much as 5 Ib. of butter. The cost of 1 Ib. of tea, 1 Ib. of coffee, 
 and i Ib. of butter is $ 1.40. Find the cost per pound of each 
 article. 
 
 GRAPHIC REPRESENTATION OF TEMPERATURE 
 
 84. If we desire information on the changes of temperature 
 for any period of time, we may take readings of a thermometer 
 at certain intervals of time. We may take hourly readings of 
 the outdoor temperature during the afternoon and evening of 
 a winter day and obtain data, in degrees Fahrenheit, perhaps 
 as follows : 
 
 Noon, 20. 7 P.M., 15. 
 
 1 P.M., 21. 8 P.M., 9. 
 
 2 P.M., 22. 9 P.M., 5. 
 
 3 P.M., 21.5. 10 P.M., 0. 
 
 4 P.M., 20. 11 P.M., - 2. 
 
 5 P.M., 17. 12 P.M., - 3. 
 
 6 P.M., 16. 
 
 These readings convey an idea of the changes of tempera- 
 ture during the 12 hours following midday. We see that the 
 maximum temperature was at 2 P.M., the minimum at mid- 
 night. The temperature dropped most rapidly between 7 and 
 8 P.M. 
 
 These relations can be seen much more easily, if the changes 
 of temperature are shown graphically. Draw a horizontal line, 
 as in Fig. 11, and let successive equal intervals from o along 
 the line ox indicate the successive hours after midday. We 
 mark in this way the hours 1, 2, 3, etc. 
 
 At each of these points we draw a line perpendicular to ox 
 to represent the temperature at that particular time. If the 
 temperature is +, we measure the distance from ox upward, 
 but if the temperature is , we measure the distance from ox 
 downward. If 1 is represented by the length of one of the
 
 64 
 
 ELEMENTARY ALGEBRA 
 
 divisions up or down, then the midday temperature is repre- 
 sented by 20 spaces from ox up, while 3 at midnight is 
 measured from ox by 3 spaces down. If we connect the ends 
 
 
 V 
 
 
 S 
 
 S 
 
 
 ~~, 
 
 s 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ,x* 
 
 S 
 
 
 
 
 
 ^ 
 
 ^ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 y 
 
 k 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 . 
 
 
 
 
 
 
 v. 
 
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 12 
 
 M 
 
 
 
 t 
 
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 \ 
 
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 [ 
 
 
 
 f 
 
 
 
 
 
 
 ^ 
 
 < 
 
 > 
 
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 K 
 
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 TS 
 
 
 
 
 
 
 
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 FIG. 11 
 
 of the lines thus drawn, we obtain a broken line which shows 
 to the eye the changes in temperature. This broken line is 
 the graphic representation of temperature. It is called a graph 
 or a diagram. Drawing such a graph or diagram is called 
 plotting it ; locating a point is called plotting the point. 
 
 How can you tell from the graph the time when the tempera- 
 ture was highest ? The time when it was lowest ? The time 
 when the temperature changed least? The time when the 
 temperature changed most rapidly ? What was the drop in 
 temperature between 2 P.M. and 9 P.M. ? Between 7 P.M. and 
 
 10 P.M. ? Between 8 P.M. and 12 P.M. ? Between 1 P.M. and 
 
 11 P.M.?
 
 PROPORTION 65 
 
 Care should be taken, in plotting, to use a scale not too 
 small and yet small enough so that all the statistics can be 
 used. 
 
 EXERCISES 
 
 85. 1. Plot the curve of temperatures for the next 12 hours 
 the observed temperatures being as follows : 
 
 12 P.M., - 3. 7 A.M., - 4. 
 
 1 A.M., - 4. 8 A.M., 0. 
 
 2 A.M., - 4. 9 A.M., + 3. 
 
 3 A.M., - 5. 10 A.M. + 7. 
 
 4 A.M., - 5. 11 A.M. + 10. 
 
 5 A.M., - 6. 12 Noon, + 15. 
 
 6 A.M., - 6. 
 
 2. Find the rise in temperature between 5 A.M. and 11 A'M. 
 
 3. At what time was the rise in temperature most rapid? 
 
 4. What was the difference in temperature between 12 P.M. 
 and 12 noon ? 
 
 5. Plot the curve of mean temperatures for 1913, Denver, 
 
 Col. 
 
 Jan. 30. July 72. 
 
 Feb. 22. Aug. 73. 
 
 Mch. 38. Sept. 59. 
 
 Apr. 49. Oct. 46. 
 
 May 58. Nov. 44. 
 
 June 67 Dec. 23. 
 
 6. Plot the rise and fall of the tide on the coast of Southern 
 California, Aug. 2, 1913. 
 
 At 3 : 20 A.M., - 1.2 ft. 
 9 : 50 A.M., - 4.2 ft. 
 2 : 50 P.M., 1.7 ft. 
 9: 00 P.M., 7 ft.
 
 66 
 
 ELEMENTARY ALGEBRA 
 
 FLUCTUATIONS IN THE PRICE OF COAL 
 ORAL EXERCISES 
 
 86. Figure 12 shows the changes in the price of anthracite 
 coal in the United States from 1860 to 1914. 
 
 The years are shown on a horizontal line, the cost on a ver- 
 
 c $7.00 
 
 $6.00 
 
 > 15.00 
 
 $4.00 
 
 $3.00 
 
 Year 
 FIG. 12
 
 PROPORTION 67 
 
 tical line. For instance, in 1890 the price per ton is seen to 
 be $ 3.90 ; in 1910, $ 4.80. 
 
 1. What effect had the Civil War upon the price of coal ? 
 
 2. In what year was the price $4.50? 
 
 3. !STame the highest price shown. 
 
 4. Xame the lowest price shown. 
 
 5. How many tons of coal could be bought in 1893 for 
 $57.20? 
 
 6. What was the difference in the price per ton in 1874 and 
 1861? 
 
 7. What was the difference in the price of 100 tons in 1866 
 and 1899 ? 
 
 DRAWING EXERCISE 
 
 87. 1. Draw a diagram showing the fluctuations in the price 
 of iron per ton during the years 1870 to 1910, the data being 
 as follows : 
 
 1870, $33. 1885, $18. 1900, $20. 
 
 1872, $44. 1887, $21. 1902, $16. 
 
 1874, $30. 1889, $18.' 1903, $21. 
 
 1877; $19. 1890, $18.50 1904, $19. 
 
 1878, $18. 1894, $13. 1905, $13. 
 
 1880, $28. 1895, $13. 1906, $20. 
 
 1881, $25. 1898, $12. 1907, $23. 
 
 1882, $26. 1899, $19. 1910, $17. 
 
 2. Make a table of varying quantities in your own experience 
 and draw the graph. Select the number of pupils in your 
 school, the amount of money you have spent in successive 
 weeks or months, the amount of rainfall in successive months, 
 the number of immigrants coming to A_merica in different 
 years, or some similar data which you are able to obtain. 
 
 88. From these graphs we can see, at a glance, the variation 
 in one thing as another changes. For instance in Fig. 12 there 
 is a change in the price of coal per ton as the time increases.
 
 68 
 
 ELEMENTARY ALGEBRA 
 
 These quantities are called variables ; the value of one 
 variable depends upon the value of the other. 
 
 89. This idea can be carried over to the relation between 
 two unknown quantities expressed by an equation. 
 For instance, in x + y = 1 : 
 
 if x = 3, 
 if x = 2, 
 if x = 1, 
 if x = 0, 
 
 then 
 then 
 then 
 then 
 
 3 + y = 7, 
 
 y = 4 ; 
 
 hence 
 
 2 + y = 7, hence y = 5 ; 
 
 1 -f- y = 7, hence y = 6 ; 
 
 + y = 7, hence y = 7 ; 
 
 ifa; = 1, then l + */=7, hence y = 8; 
 
 if a; = 4, then 4 + y = 7, hence y = 9. 
 
 As a; increases, ?/ decreases ; and as x decreases, y increases. 
 
 Thus x and y are two variables, the value of y depending 
 upon what the value of x may be, and the value of x depending 
 upon the value of y; thus x is a, function of y and y is & function 
 of x. 
 
 90. Let us make a table, assuming ten values of x and com- 
 puting the corresponding values for y in the equatiqn x y = 5. 
 The result is the following table : 
 
 x y = 5 
 
 X 
 
 2/ 
 
 Point 
 
 6 
 
 1 
 
 ^ 
 
 6 
 
 
 
 B 
 
 4 
 
 - 1 
 
 c. 
 
 3 
 
 -2 
 
 D 
 
 2 
 
 -3 
 
 E 
 
 1| 
 
 -sj 
 
 F 
 
 1 
 
 -4 
 
 G 
 
 
 
 -5 
 
 H 
 
 _ i 
 
 -6 
 
 I 
 
 -2 
 
 -7 
 
 J 
 
 We count off values of x on the horizontal line 
 OX (called the x-oxis) and the corresponding 
 
 FIG. 13
 
 PROPORTION 69 
 
 values of y on the vertical line OY (called the y-axis), counting to the 
 right of the origin for + values of x and to the left for values of x ; 
 up for -f values of y, and down for values of y. 
 
 This becomes plainer, if we plot each of the points in Fig. 13. The 
 first point, x = 6 and y = 1, is the point A, 6 spaces to the right and 
 1 space up. The second point, x = 5, y = 0, is the point B, 5 spaces to 
 the right, on the x-axis. The third point, z = 4, y = 1, is the point C, 
 4 spaces to the right and 1 space down, and so on. Points represented by 
 fractional values of x and y are located in the same way. Thus x = 1, 
 y = 3|, locate the point F. 
 
 The values of x and y which satisfy the equation x y = 5 
 are represented by points which form a straight line. Further- 
 more, every point in the line corresponds to values of x and of y 
 which satisfy the equation. 
 
 Thus the graph of a linear equation in two unknowns is a 
 straight line. 
 
 EXERCISES 
 
 91. Tabulate five pairs of values of x and y which satisfy 
 each of the following equations. Draw a separate pair of 
 axes for each equation and plot the points. 
 
 1. 2z + 3y = 6. 3. x y = 2. 5. x = 2y. 
 
 2. 3z-4?/ = 12. 4. x + y = Q. 6. y = 2x. 
 
 92. In the preceding exercises the student has observed that 
 a linear equation in two unknowns produces a graph which is 
 a straight line. As two points determine the position of any 
 straight line, it is necessary and sufficient to find only two points 
 on the graph of a linear equation in order to fix the position of 
 the line. The two points most easily found are x = 0, y = ? 
 and x = ?, y = 0. 
 
 If x = and y = satisfy the equation, it will be necessary 
 to let x have some value other than and to find the cor- 
 responding value of y.
 
 70 
 
 ELEMENTARY ALGEBRA 
 
 For instance, in2x 5y = 0, ifaj = 0, y = 0, and if y = 0, x = 0. 
 
 But if x = 5, y = 2. 
 
 Plot (0, 0), and (5, 2), and draw the line, as in Fig. 14. 
 
 FIG. 14 
 
 EXERCISES 
 
 93. Kepresent the fol- 
 lowing equations graph- 
 ically : 
 
 1. x + y = 3. 
 
 2. x y = 5. 
 
 3. 2x + y = 4. 
 
 4. x 2y = 6. 
 
 5. 2x-y = 10. 
 
 6. x + 2y = Q. 
 
 7. x 3y = 0. 
 
 8. 3 a- y = Q. 
 
 9. 5 a; = 4 y. 
 
 PRACTICAL APPLICATIONS OF GRAPHS 1 
 
 94. One of the simplest applications of graphs is in the re- 
 duction of denominate numbers from one unit to another with- 
 out the labor of numerical computation. Thus, Fig. 15 enables 
 us by inspection to reduce miles to Kilometers, or Kilometers 
 to miles. The following example shows how the graph may be 
 drawn. 
 
 95. If 10 miles are equal to 16.1 Kilometers, how many Kilo- 
 meters are equal to x miles ? 
 
 Let y be the required number of Kilometers. 
 
 Then by proportion, 
 
 Omitting the names of units, 
 
 Multiply both sides by 16.1, 
 Or 
 
 x mi. y Km. 
 10 mi." 16.1 Km.' 
 
 a? _ y 
 
 10 16.1 ' 
 1.61 x = y. 
 y = 1.61 a. 
 1 May be postponed for the present.
 
 PROPORTION 
 
 71 
 
 y 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 f 
 
 
 m 
 ^10 
 
 
 
 B 
 ( 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ' t 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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 / 
 
 s 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 t 
 
 S 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 S 
 
 ** 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 s 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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 4 
 
 s 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 . 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 y 
 
 * 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 s 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 (j 
 
 ^ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ,' 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 f 
 
 Y/ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 S 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ^ 
 
 ' 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ) 10 D 20 30 x 
 
 Kilometers 
 
 FIG. 15 
 
 The graph of this equation is a straight line. 
 To draw this graph, carefully locate two points on it and 
 then draw a straight Hue through the two points. We obtain 
 
 X 
 
 y 
 
 Point 
 
 
 
 
 
 
 
 10 
 
 16.1 
 
 A 
 
 In the graph, Fig. 15, Kilometers are measured along the 
 x-axis, miles are measured along the y-axis. 
 
 To determine how many Kilometers are equal to 8 miles, we find the 
 point B on the ?/-axis which indicates 8 miles, then proceed to the right 
 (in a direction parallel to the x-axis) to the point C on the graph, then 
 proceed downward to the point D on the x-axis. The number of Kilo- 
 meters equivalent to 8 miles is seen to be approximately 13. 
 
 Results that are absolutely accurate cannot be obtained for three 
 reasons : .Fi'rsZ, the relation " 10 miles = 16.1 Kilometers " is correct only
 
 72 ELEMENTARY ALGEBRA 
 
 to tenths of Kilometers ; secondly, even if this relation were accurate, the 
 graph could not be drawn with absolute accuracy ; third, to estimate the 
 fraction of a Kilometer at the point (7, without some slight error, is hardly 
 possible. Of necessity all measurements are only approximate. 
 
 ORAL EXERCISES 
 
 96. Estimate, to the first decimal, the number of Kilometers in 
 
 1. 7 mi. 4. 13 mi. 7. 17 mi. 
 
 2. 9 mi. 5. 14 mi. 8. 18 mi. 
 
 3. 11 mi. 6. 16 mi. 9. 20 mi. 
 
 Estimate, to the first decimal, the number of miles in 
 
 10. 8 Km. 13. 17 Km. 16. 23 Km. 
 
 11. 12 Kin. 14. 19 Km. 17. 27 Km. 
 
 12. 13 Km. 15. 21 Km. 18. 32 Km. 
 
 CONSTRUCTION AND USE OF GRAPHS 
 
 97. 1. 10 Kilograms are approximately equal to 22 pounds. 
 Draw a graph for reducing at sight, Kilograms to pounds, and 
 pounds to Kilograms. Change at sight, 6 lb., 7 lb., 9 Ib. to 
 Kilograms ; change also 21 Kg., 17 Kg., 13 Kg. to pounds. 
 
 2. 5|- yards = 1 rod. Construct a graph for converting 
 yards to rods and rods to yards. At sight change to yards, 
 2.2 rd., 1.7 rd., 1.3 rd., .8 rd. 
 
 Let 1 in. along the x-axis stand for 1 rd., and ^ in. along the /-axis 
 stand for 1 yd. This gives a graph which is convenient for reducing a 
 small number of rods to yards. 
 
 3. Draw a graph for finding the lengths of circles when 
 the diameters are given, and vice versa. 
 
 Take the length of a circle equal to ty times its diameter. 
 
 4. Draw a graph for converting temperatures on the 
 Fahrenheit scale to temperatures on the Centigrade scale, and 
 vice versa. 
 
 We know that a temperature of 0C. is the same as one of 32 F. ; 
 also that a temperature of 20 C. is the same as one of 68 F. From these
 
 PROPORTION 
 
 73 
 
 data we locate the points A and jB, in Fig. 16, which determine the line. 
 This graph is applicable to temperatures below zero, or negative tem- 
 peratures. 
 
 30 
 
 -21 
 53 
 
 ilO 
 
 eit 
 
 FIG. 16 
 
 5. In Fig. 16 change the following to temperatures on the 
 Fahrenheit scale : 30 C., 25 C., 17 C., C., - 10 C., - 20 C. 
 
 6. In Fig. 16 change the following to temperatures on the 
 Centigrade scale: 80 F., 65 F., 52 F., 32 F., 15 F., 8F., 
 0F., -3F. 
 
 7. A wholesale dealer's profit when he sells to a retail 
 merchant is 20%. Draw a graph for ascertaining the cost 
 price when the selling price is known, and vice versa. 
 
 When there is a profit of 20%, the selling price is to the cost, as 120 
 is to 100. 
 
 Let x be the selling price and y the corresponding cost. 
 
 Then, by proportion , 120 = 100 ' 
 
 Multiply both sides by 100, x = y. 
 Or y = x- 
 
 The graph of this equation is a straight line.
 
 74 ELEMENTARY ALGEBRA 
 
 From this equation the following data are obtained : 
 
 X 
 
 y 
 
 Point 
 
 
 
 
 
 
 
 30 
 
 25 
 
 A 
 
 
 $20 
 
 no 
 
 Through the points O and A, in Fig. 17, draw a straight line, which is 
 the graph required. It shows the relation between the cost and the sell- 
 ing price. 
 
 To determine the cost of 
 an article that sells for $ 36, 
 take the point B 36 divisions 
 from O on the sc-axis, then 
 OB represents the selling 
 price. From B pass verti- 
 cally to C on the " wholesale 
 to retail " graph, thence 
 horizontally to D. Then 
 OD, or 30, is the cost to 
 the wholesale dealer. 
 
 O 
 
 $10 
 
 B$40 
 
 $20 $30 
 Selling Price 
 
 Fzo.17 8 - In Fi S' 17 find 
 
 the cost of articles sold 
 
 to the retail dealer for $10, $14, $17, $24, $28, $32. 
 
 9. In Fig. 17, find the selling price to the retail dealer of 
 articles which cost $ 5, $ 15, $ 20, $ 23, $ 25. 
 
 10. A wholesale dealer makes a profit of 15 % when he sells 
 to a retail dealer. The retail dealer makes a profit of 65 % 
 when he sells to a consumer. Draw a graph for finding at 
 sight the cost to the wholesale dealer of articles whose selling 
 price to the consumer is known. 
 
 When there is a profit of 15%, the selling price is to the cost as 115 is 
 to 100. 
 
 Let x be the selling price to the retail dealer and y the cost to the 
 wholesale dealer. 
 
 Then, by proportion, -2- = -^-. 
 115 100
 
 PROPORTION 
 
 75 
 
 x = y. 
 
 Multiply both sides of the equation by 100, 
 
 Or y = ftx. 
 
 When x 0, y = 0, hence the graph passes through 0. 
 
 When x = 46, y = 40, hence the graph passes through A. 
 
 Draw the line OA. 
 
 The retail dealer sells at a profit of 65 %. That is, his selling price to 
 the consumer is to his buying price as 165 is to 100. 
 
 Let x be the selling price to the consumer and y the cost to the retail 
 dealer. 
 
 Then, 
 
 Or 
 
 165 100 
 y = fjx. 
 
 When x = 0, y = 0, hence the graph passes through O in Fig. 18. 
 When x = 66, y = 40, hence the graph passes through B. 
 Draw the line OB. 
 
 O $10 $20 $30 $40 $50 $60 $70 
 
 Selling Price 
 
 FIG. 18 
 
 To determine the cost to the wholesale dealer of an article for which 
 the consumer pays $60, two steps are necessary. 
 
 The first step is to find its cost to the retail dealer. Take the point C 
 60 divisions on the z-axis, then OC represents the price paid by the con- 
 sumer. From C pass vertically to D on the "retail to consumers" graph,
 
 76 ELEMENTARY ALGEBRA 
 
 thence horizontally to E. Then OE, or about $52.20, is the cost to the 
 retail dealer. 
 
 The second step is to find the cost to the wholesale dealer. Take the 
 distance OF equal to OE ; from .Fpass vertically to G on the "wholesale 
 to retail " graph, thence horizontally to H. Clearly OH is the cost to the 
 wholesale dealer, about $31 . This is approximately the required answer. 
 
 11. What is the cost to the wholesale dealer of articles 
 which are sold to the consumer for $ 20 ? $ 30 ? $ 50 ? $ 75 ? 
 
 12. What does the consumer pay for articles which cost to 
 the wholesale dealer $ 50 ? $ 40 ? $ 15 ? 
 
 ^'- ^ V / 
 
 rro ? L -r 
 
 V 
 
 3- 
 
 $^ 

 
 
 CHAPTER V 
 
 EQUATIONS INVOLVING FRACTIONAL COEFFICIENTS 
 
 98. Solve x + \x = 15. 
 
 ^- * / 
 
 Solution. . i + i = t- 
 
 Hence we have fa: = 15. 
 
 -ir o 
 
 Divide both sides by f , x = -- = 18. Ans. 
 
 o 
 
 Second Solution. $x + x = 15. . 
 
 Multiply both sides by 6, 2x + 3 x - 90. 
 
 5* = 90. 
 
 x = 18. Ans. 
 
 Check: | (18) + (18) = 15. 
 
 6 + 9 = 15. 
 
 15 = 15. 
 
 In the second solution we multiply both sides of the equation by some 
 number which will remove the denominators. It is usually easiest to 
 select the least number into which each denominator will go without a 
 remainder. In the example above, that number is 6. 
 
 EXERCISES 
 99. Solve and check : 
 
 2. fy + |y = l. 8 
 
 363' y-1 y-2 = 2 y-3 
 
 n 5n 3n 2 334 
 
 O O 4: C ^~" O , t ~f" <^ v H~ ^ j 
 
 10. - + - - = 4. 
 
 5. 
 
 11. f(6 -a?) = 5. 
 
 77
 
 78 ELEMENTARY ALGEBRA 
 
 13. $(* + !) =2. 16. .2z = 48-.04z. 
 
 14. i (s + 4) - i s - 8 = 0. 17. .2 w + 3 = 3.8 + .04 o. 
 
 15. .5x = 3. 
 
 18. 7(z + 2) + f(z-12) = (z+9). 
 
 19. -l--2 = -n-3. 
 20. 
 
 PROBLEMS 
 
 100. 1. One half a certain number plus one fourth that 
 number increased by one third the number equals 26. Find 
 the number. 
 
 2. The sum of two numbers is 29. One half the first plus 
 one third the second is 12. What are the numbers ? 
 
 3. The difference between two numbers is 6. One ninth 
 of the first plus one third of the second is 6. Find the 
 numbers. 
 
 4. Separate 48 into two parts so that ^ the larger minus ^ 
 the smaller equals ^ the number itself. 
 
 5. One fourth of a certain number increased by 1^, and the 
 result diminished by the quotient obtained by dividing the sum 
 of twice the number and 4 by 9, equals 1. Find the number. 
 
 6. The width of a rectangle is f of its length and its pe- 
 rimeter is 54 in. Find its dimensions. 
 
 7. The length of a rectangle is 12 in. more than the width. 
 The sum of the length and width is twice their difference. 
 Find the area. 
 
 8. The width of a rectangle is 5 ft. less than the length. 
 If the length be decreased 2 ft. and the width increased 3 ft., 
 the area will be increased 11 sq. ft. Find the area. 
 
 9. A certain square has the same area as a rectangle whose 
 dimensions are 3 ft. longer and 2 ft. shorter than those of the 
 square. What is the area of each ?
 
 FRACTIONAL COEFFICIENTS 79 
 
 10. Find two consecutive integers such that ^ the first plus 
 | the second equals 7 less than the first. 
 
 11. Find three consecutive even integers such that ^ of the 
 first and | of the second, plus ^ of the third, equals the third. 
 
 12. What three consecutive integers have their sum equal 
 to 48 ? How does the sum compare with the second integer ? 
 Is this true of any three consecutive integers ? 
 
 13. The sum of 1 of a certain even integer, ^ of the next 
 odd one, and -| of the next even one equals the even integer 
 just before the one first mentioned. Find the series of in- 
 tegers. 
 
 14. A is 15 years old ; B is 25 years old. In how many 
 years will A be f as old as B ? 
 
 15. A is twice as old as B ; five years ago he was 2\ times 
 as old. How old is each now ? 
 
 16. A man's age 6 years ago was ^ of what his age will be 
 30 years hence. Find his age now. 
 
 17. A father is 3 times as old as his son ; in 5 years the 
 father will be 21 times as old as the son. What is the age of 
 each? 
 
 18. A man who is 35 years old has a son 10 years old. In 
 how many years will the boy be | as old as his father ? 
 
 19. A man gave to one son ^ of his money, to a second ^ of 
 his money. He had left $10 less than the number of dollars 
 he gave to both. How much had he at first ? 
 
 20. I spent $1.75 for 1^ and 2^ stamps, buying 25 more of 
 the former than of the latter. How many of each kind did I 
 buy? 
 
 21. I took a trip of 105 miles, partly by trolley, p*artly by 
 train. If I went \ as far again by train as by trolley, how 
 far did I go by each ?
 
 80 ELEMENTARY ALGEBRA 
 
 22. A collection of 5^ pieces and quarters amounts to $ 1.60. 
 There are 2 more 5^ pieces than quarters. How many are 
 there of each ? 
 
 23. I have three more half dollars than quarters. The 
 value of the half dollars exceeds the value of the quarters by 
 $ 4.50. How much money have I ? 
 
 24. Thirty coins dimes, nickels, and quarters amount 
 to $5.30. There are ^ as many nickels as quarters. How 
 many are there of each ? 
 
 25. Three boys have 150 marbles. If the third has f as 
 many as the second, and the second f as many as the first, 
 how many marbles has each boy ? 
 
 26. The same number is subtracted from 60 and from 45. 
 One third of the first remainder equals ^ of the second. What 
 is the number ? 
 
 27. In going a certain distance, a train traveling at the rate 
 of 40 miles an hour takes 21 hours longer than one traveling 
 50 miles an hour. What is the distance ? 
 
 Use d = rt. 
 
 28. I have 13 hours at my disposal. How far may I go at 
 the rate of 3 miles an hour, so as to return home in time, corn- 
 ing back at the rate of 3^ miles an hour ? 
 
 29. A man sold a home for $4944, gaining 3% of the cost. 
 How much did the home cost ? 
 
 Hint, x + .03 x = 4944. 
 
 30. An agent deducted 5 J commission for the sale of 
 property, remitting $ 6222.50. Find the selling price. 
 
 31. At 6 /o interest for a certain time, $ 350 amounted to 
 $ 444.50. What was the length of time ? 
 
 Use i = prt. 
 
 32. The interest of $365 for 2 years 3 months at a certain 
 rate was $32.85. What was the rate ?
 
 FRACTIONAL COEFFICIENTS 81 
 
 33. The amount of a certain principal at 6 f in 5^- years is 
 $ 7714. What is the principal ? 
 
 34. A man invests a part of $ 1500 at 5 % and the re- 
 mainder at 4 %. The 5 % investment yields $39 more a year 
 than the 4 f - How much has he invested at 4 % ? 
 
 35. A part of $ 3680 is invested at 5| % and the remainder 
 at 6%. The total interest in 2 years amounted to $411.60. 
 How many dollars are there in each investment ? 
 
 36. If the valuation of a certain property is $17,000, and 
 the annual tax thereon is $ 161.50, find the tax rate in mills 
 on a dollar. 
 
 Let x mills = the tax on a dollar. Then 17000x = 161.50. 
 
 1000 
 
 Find the tax rate in mills on a dollar : 
 
 VALUATION TAX VALUATION TAX 
 
 37. $6,560 $55.76 40. $125,500 $909.875 
 
 38. $29,800 $186.25 41. $375,400 $2909.35 
 
 39. $90,700 $589.55 42. $875,500 $6566.25 
 
 43. I wish to get $ 7280 from a bank. For what sum must 
 I make out a 60-day note, to obtain this sum, bank discount 
 being at the rate of 6 % per annum ? 
 
 Bank discount is computed on the face of the note. 
 
 Let x = the required sum. The bank discount for 60 da. is 1 % of 
 $ x, or $ .01 x. Subtracting $ .01 x from $ x, leaves $ .99 x, the proceeds. 
 Hence, .99 x = 7280. 
 
 44. A man needs $ 2975. For what sum must he make out 
 a 60-day note to obtain the $ 2975 as proceeds, if the bank dis- 
 counts the note at 6 % per annum ? 
 
 45. Mr. Murray makes out a 90-day note, which is dis- 
 counted by the bank at 2 J for the 90 days. What must be 
 the face of the note, to yield $ 9700 as proceeds ? 
 
 46. John Allen issued a note to a bank which secured for 
 him, after it was discounted at 3 f for the term of 4 months, 
 the sum of $ 29,750. Find the face of the note.
 
 82 ELEMENTARY ALGEBRA 
 
 The federal income tax in the United States requires that every unmar- 
 ried person shall pay an annual tax of 1 % on his net income in excess of 
 $ 3000, and that married persons living together shall pay an annual tax of 
 1% on their joint income in excess of $4000. When the net income ex- 
 ceeds $ 20,000, an additional tax thereon is levied as follows : 
 
 1 % on part of income over $ 20,000 and not above $ 50,000. 
 
 2% on part of income over $ 50,000 and not above $ 75,000, etc. 
 
 Thus, an unmarried man pays on an income of $ 70,000 the following 
 tax : 1 % on $ 67,000 + 1 % on $ 30,000 + 2 % on $ 20,000. 
 
 47. An unmarried person whose net annual income is be- 
 low $20,000 pays an income tax of $145.70. Find his in- 
 come. 
 
 48. A man and wife have a joint income that is below 
 $ 20,000 ; their income tax is $ 135.45. Find their income. 
 
 49. A man and wife pay an income tax of $200. Does 
 their annual income exceed $ 20,000 ? Find their income. 
 
 Let $ x = income. Then .01 (x - 4000) + .01(x - 20,000) = 200. 
 
 50. The annual income tax of an unmarried man is $450. 
 Is his income above $20,000? Above $50,000? Find it. 
 
 51. The joint income tax of a man and his wife is $1360. Is 
 the annual income above $50,000? Above $75,000? Com- 
 pute their income. 
 
 52. What is the annual net income of an unmarried man 
 whose income tax is $ 850 ? 
 
 53. A boy on a farm receives 2% ft a dozen for delivering 98 
 dozen eggs and agrees to have 27^ deducted from his pay for 
 every dozen eggs that he breaks and does not deliver. He is 
 paid $ 1.55. How many dozen eggs did he break ? 
 
 54. Two persons traveling toward each other set out at the 
 same time from towns 198 miles apart. One person walks 10 
 miles a day, the other 12 miles. In what time will they meet, 
 and how many miles will each have walked ? 
 
 55. At an election of a constituency abstained from voting, 
 and | of the constituency voted for the successful candidate.
 
 FRACTIONAL COEFFICIENTS 83 
 
 At a subsequent election f abstained from voting, and \ voted 
 for the successful candidate. The minority in the latter elec- 
 tion was 120 less than in the former. Find the number of 
 voters in the constituency. 
 
 56. A rectangular tank is 10 ft. 6 in. long and 7 ft. 3 in. 
 wide, and contains 400 gallons of water. Find the depth of 
 the water, correct to one tenth of an inch, having given that a 
 gallon contains 231 cu. in. 
 
 57. A train is 5 minutes late when it performs its journey 
 at the rate of 29^- miles an hour, and is 2 minutes late when it 
 travels at 30 miles an hour. What is the number of miles in 
 the journey ? 
 
 58. There are two lawns in one garden. One is square ; the 
 second is 3 yd. narrower and 4 yd. longer than the first, but 
 of equal area. Find the area of each.
 
 CHAPTER VI 
 
 SPECIAL PRODUCTS 
 
 101. I. (a +- 6) 2 . 
 
 a + b 
 a + b 
 
 a? + 2 ab + b* 
 
 II. (a-&) 2 . 
 
 a 6 . 
 
 a b 
 
 ab 
 
 a?-2ab + b z 
 .-. (a -H 6) 2 = a 2 + 2 ab + W. 
 (a - b) 2 = a 2 - 2 ab + ft 2 . 
 
 RULE. T7ie square of the \ sum \ of two numbers is equal 
 
 ( difference J 
 
 to the square of the first, I [ twice the product of the first 
 
 ( minus j 
 
 and the second, plus the square of the second. 
 
 ORAL EXERCISES 
 
 102. Find, by inspection, the values of : 
 
 1. (m + w) 2 . 5. (4 + ) 2 . 
 
 2. (c-d) 2 . 6. (6-7) 2 . 
 
 3. (a-26) 2 . 7. (a 2 -!) 2 . 
 
 4. (3z + y) 2 . 8. (2a 2 -36) 2 . 
 
 84
 
 SPECIAL PRODUCTS 
 
 85 
 
 10. (7a&-2) 2 . 
 
 11. (9 x 2 - 2 ?/ 2 ) 2 . 
 
 12. (5 a 3 -f- 4 a 2 ) 2 . 
 
 13. (1 + z) 2 . 
 
 14. (ia-3) 2 . 
 
 15. c 
 
 9. (4 c + 3 d 3 ) 2 . 16. (o + 2/*) 2 . 
 
 17. (3cr-2&") 2 . 
 
 18. (7 a 6) 2 . 
 
 19. (5 d 3 e) 2 . 
 
 20. (m + 12) 2 . 
 
 21. (9 a; 3y) z . 
 
 22. (2a-76) 2 . 
 
 23. What must be added to a 2 + 10 a to make it the square 
 of a + 5 ? 
 
 24. What must be added to x 2 xy + y 1 to make it the 
 square of x y ? 
 
 25. What must be added to 36 12 a 2 to make it a perfect 
 square ? 
 
 In the following, supply terms necessary to form perfect 
 squares : 
 
 34. 64z/ 2 + 32y +( ). 
 
 35. 121m 6 -22m 3 +( ). 
 
 37. ( )-66 2 + 9. 
 
 38. ( 
 
 39. ( r 
 
 40. ( )-24s 5 + 144. 
 
 41. ( )- 182?n 2 +169. 
 Find the binomials whose squares are : 
 
 42. x 2 + 6 a + 9. 
 
 43. 4 a 2 4 c 
 
 44. x 2 + 22 xy + 121 y 2 . 
 
 46. m 2 + 40 m + 400. 
 
 47. w 2 4n + 4. 
 
 26. c 2 +( 
 
 27. ra 2 +( 
 
 28. z 2 +( 
 
 29. 2 -( )+16. 
 
 30. 4a 2 -( )+l. 
 
 31. 9z 2 -f( ) + 36. 
 
 32. 166 2 -( )+49. 
 33. 
 
 48. 
 
 49. 25 c 4 + 16003 + 256 c 2 . 
 
 50. 9 a 2 - 30 axy* + 25 ofy 4 . 
 
 51. 36 6 2 + 60 a 2 6c 3 + 25 aV. 
 
 52. 64 c 8 - 16 c 4 d 5 + d 10 . 
 
 53. 4 x 2 y 2 + 28 xy + 49.
 
 86 ELEMENTARY ALGEBRA 
 
 103. III. (a + 6) (a -6). 
 
 a + b 
 a-b 
 
 -ab-b* 
 
 RULE. 7%e product of the sum of two numbers and their differ- 
 ence is equal to the square of the first number minus the square 
 of the second. 
 
 ORAL EXERCISES 
 
 104. Find, by inspection, the values of : 
 
 1. (k + y)(Tc-y). 11. (a 2 - 9 6 2 ) (a 2 + 9 ft 2 ). 
 
 2. (l-3a)(l+3a). 12. (13 x 5 + y)(13 x 5 - y). 
 
 3. (2h + 7)(2h-7). 13. (4 ab - c)(4 ab + c). 
 
 4. (5 -&)(& + 5). 14. [( ft + 6)+c][(a+6)-c]. 
 
 5. (10 -3 a) (3 a + 10). 15. (x + b + 2)(x + b -2). 
 
 6. (3rc + 7)(3w-7). 16. [(m-w)-3][(m-7t)+3]. 
 
 7. (3a-a)(3a; + a). 17. [+(> + w)p-(> + u)]. ' 
 
 8. (5a6 + 3)(5a6-3). 18. 
 
 9. (2n + 7)(7-2n). 19. 
 
 10. (66-l)(66 + l). 20. (2 + c-d)(2-c + 
 
 Find two binomials whose product is : 
 
 21. a^ - 4 f. 28. 121 c 2 cP - 144 x 4 . 
 
 22. 16 -a 4 . 29. 2256-l. 
 
 23. 4 a; 2 - 16. 30. 1 - 81 a 10 . 
 
 24. 9a 2 -646 4 . 31. 100 6 2 - 36. 
 
 25. 25m 4 -49n 6 . 32. ^-225. 
 
 26. a 4 -9. 33. ff -a 6 
 
 27. 9a 2 -25. 34. 16m 2 -25w 2 .
 
 SPECIAL PRODUCTS 87 
 
 36. (x ?/) 2 z 2 . 38. a 2 (b + c) 2 . 
 
 36. (c + d) 2 - e 2 . 39. t* - (s - w) 2 . 
 
 37. (ra-?i) 2 -64. 40. 81-(a-6) 2 . 
 
 In each of these exercises (21-40), the two binomials found are called 
 factors of the given expression. Thus, in Ex. 21, x 3 + 2 y 3 and a; 3 2 y 3 
 are said to be factors of x 6 4 j/ 6 . 
 
 105. IV. (a? + a )(* + &). 
 
 x + a 
 x + b 
 
 x 2 + ax 
 
 ab 
 .: (x + a) (or + ft) = jr* + (a + '&)* + a&. 
 
 RULE. The product of ttvo binomials having a common term 
 is equal to the square of the common term, and the sum of the tm-> 
 like terms times the common term, plus the product of the unlike 
 terms. 
 
 ORAL EXERCISES 
 
 106. Find, by inspection, the values of : 
 
 1. (z + 3)(a + 4). 11. (a + 4c)(a c). 
 
 2. (a; - 3)( - 4). 12. (2 1 + 3)(2 1 - 5). 
 
 3. (* + 3)(* 4). 13. (a? - 16) (a; + 20). 
 
 4. (aj-3)(a; + 4). 14. (3fc- l)(3fc- 19). 
 
 5. ( <y +2)(y-7). 15. (m + 17)(m-2). 
 
 6. (a 9)(o 3). 16. (c-x)(c + 3x). 
 
 7. (6-5)(6 + 6). 17. (s-8)(*-3). 
 
 8. (a* + l)(aaj-2). 18. (7 fc - 15)(7 h + 2). 
 
 9. (v - a)(v - 2 a). 19. (4^ - a 2 )(4gr + 5a 2 ). 
 10. (n + &)O + 46 )- 20 - (5c 2 -9)(5c 2 -4).
 
 88 ELEMENTARY ALGEBRA 
 
 Find two binomials (two factors) whose product is : 
 
 21. a 2 + 7 a + 12. 31. 100 m 2 + 10 m - 12. 
 
 22. x 2 - 9 a + 20. 32. ato? + 12 xy - 64. 
 
 23. 6 2 + 26-35. 33. 16s 4 + 16s 2 -5. 
 
 24. a 2 + 11 a - 12. 34. 15 - 8 m + m 2 . 
 
 25. c 2 - 20 c + 51. 35. 8 - 42 a + 49 a 2 . 
 
 26. m 2 - 25 m + 100. 36. m 2 - 33 m + 90. 
 
 27. 2 25 + 66. 37. 121 1 2 - 77 t - 18. 
 
 28. 4^ + 22^ + 30. 38. 96 2 -96-40. 
 
 29. 25s 2 -30s + 8. 39. 100 z 2 - 40 x - 12. 
 
 30. 9 & - 60 d + 19. 40. / 2 - 13/- 30.
 
 CHAPTER VII 
 
 FACTORING 
 
 107. In the preceding chapter we found the products of 
 certain binomials. We performed also the inverse operation : 
 given the product, to find the binomials. 
 
 For instance, starting with the product x 2 4 y' 2 , we found two bi- 
 nomials, x + 2 y and x 2 y, whose product is x' 2 4 y' 2 . 
 
 This inverse operation is called factoring ; x + 2 y and x 2 y are 
 called factors of x 2 - 4 y 2 . 
 
 In this chapter we enter upon a fuller treatment of factoring. 
 
 Factoring an expression is the process of finding two or more 
 expressions which, multiplied together, produce the given ex- 
 pression. 
 
 In factoring it is frequently necessary to find roots of mono- 
 mials. 
 
 108. The square root of a monomial is one of the two equal 
 factors whose product is the monomial. 
 
 Since 5 5 = + 25, and (- 5)( 5) = + 25. it follows that there are two 
 square roots of 25 ; namely, 4- 5 and 5. 
 
 Similarly there are two square roots of a- ; namely, -f- a and a. 
 
 The two square roots of 25 are usually written in the form 5 ; sim- 
 ilarly the two square roots of a 2 are written a. 
 
 The positive square root is called the principal square root. 
 
 We write V25 = 5. When V has no sign before it, or has 
 the + sign before it, the principal root is always understood. 
 When we write V we mean the negative root. 
 
 Thus, V25 stands for + 5, V25 does not stand for 5. 
 V25 stands for 5, V25 does not stand for + 5. 
 
 V25 = 5. 
 
 89
 
 90 ELEMENTARY ALGEBRA 
 
 Since x 3 x 3 = of, and ( a?)( a^)= x 6 , we have 
 
 _ Va?= - , Va? = 
 
 _ 
 
 Similarly, Vl6 ic 6 = + 4 or 5 , V16 z 6 = - 4 z 3 , 
 
 Also, V4 a 6 6 4 = 2 a 3 6 2 , V36 a 8 6 10 = 6 o. 4 b 5 . 
 How is the exponent of a letter in the square root found ? 
 
 RULE. The two square roots of a monomial are found by 
 writing the square root of the numerical coefficient, times the 
 letters of the monomial, each with an exponent that is half of its 
 exponent in the given monomial. 
 
 Similar results hold for the fourth or sixth roots, or higher 
 even roots of monomials. 
 
 Thus, \/16=+2, -\/i6=-2, vl6=2, 
 =-3x 2 , \/81 x s = 3 x 2 , 
 
 109. The cube root of a monomial is one of the three equal 
 factors whose product is the monomial. 
 
 Thus, \/8 = 2,_because 2-2-2 = 8. 
 
 Notice that v/S'is not - 2, for (- 2)( 2)( 2) is not 8. 
 
 4^8 =-2, because (_2)(- 2)(-2) = - 8. 
 
 Observe that a negative number can be the cube root of a negative 
 number, but a negative number cannot be the square root of a negative 
 number. 
 
 That is, -4 is v/ 64, because ( 4)( 4)(- 4) = - 64, but Sis 
 not V 64, because ( 8)( 8) is not 64. 
 
 Nor indeed can V 64 be + 8, because 8 8 is not 64. 
 
 In other words, neither a positive number nor a negative number can 
 be the square root of a negative number. 
 
 From -\/S = 2 and \/ 8 = 2 we see that the cube root of 
 a monomial has the same sign as the monomial. 
 
 Since (3a 2 ) (3 a 2 ) (3 a 2 ) = 27 a 6 , we have V27 a 6 = 3 a 2 . 
 
 110. The square root of 2 cannot be exactly expressed by 
 the Hindu-Arabic numerals. One can only approximate its
 
 FACTORING 
 
 91 
 
 value^ by extracting the square root to three or four, or more, 
 decimal places. The radical V2~, and other radicals of the 
 same kind, like V3, Vo, etc., whose values cannot be found 
 exactly, but only approximately, represent numbers called 
 irrational numbers. 
 
 ORAL EXERCISES 
 111. Give the values of: 
 
 1. V121. 5. Vl6zy. 
 
 2. -V12T. 6. -V25/2 6 . 
 
 3. VOm 2 . 7. +VlOOa 2 6 2 c 2 . 
 
 8. Vo^wV. 
 
 9. _ 
 
 10. V4 a 
 
 11. 
 
 4. V9 
 
 m. 
 
 Give both square roots of : 
 
 13. 144 a 12 . 15. 
 
 14. 169 a 6 6 8 . 16. 100 r*s 16 . 
 
 Give the values of : 
 
 19. ^125. 23. ^32. 
 
 20. ^/-125. 24. -v/a 4 ^^ 12 . 
 
 21. -\/16a 4 b*. 25. -^216 a 9 . 
 
 22. \/64a^. 26. 
 
 12. V49 m 6 n*o 2 p*. 
 
 17. 400a 16 6c 20 . 
 18. 
 
 31. Simplify V32. 
 
 32 is equal to 16 2, where 16 is a perfect square. 
 
 Hence V32 = VlQ 2 =4\/2. We take the square root of 16, which is 
 4, and write the 4 as a factor before the v x . The factor 2 is not a perfect 
 square and is kept under the radical sign. 
 
 32. Simplify VoOa 2 6 5 . 
 
 Here 50 a 2 6 5 = 25 a 2 6* times 2 6. Since 25 o 2 6* is a perfect square, we 
 have V50 a 2 6 5 =
 
 92 ELEMENTARY ALGEBRA 
 
 Simplify the following : 
 
 5. Vl2a 2 6 2 c 5 . 
 
 6. V300m 2 np 2 . 
 7. 
 
 8. Vl28a6 2 c 3 d 4 . 
 
 112. An integral number or expression is one that is free 
 from fractions: 
 
 For example, 125 and 7 a + 4 6 2 are integral numbers or expressions. 
 
 A prime factor of an integral number is an integral factor 
 which is exactly divisible only by itself and one. 
 Thus 3 and 7 are prime factors of 63. 
 
 A prime factor of an integral algebraic expression is an 
 integral factor which cannot itself be resolved into rational 
 factors other than itself and one. 
 
 Since (a 2 + 2 ft 2 ) (a 2 2 ft 2 ) = a 4 - 4 6 4 , we know that a 2 + 2 6 2 and 
 a 2 2 b' 2 are factors. Moreover, these are prime factors. By multiply- 
 ing we see that (a + V2 6) (a - V2 ft) = a 2 - 2 ft 2 . But (a + V2 6) and 
 (a V2 6) are not prime factors, because they contain the irrational V2. 
 
 In this chapter we shall find the prime factors of expres- 
 sions ; we shall not search for factors involving irrationals. 
 
 Factors of monomials are easily found by inspection. Thus 
 the monomial 30 a 2 x?y = 2-3-5 a 2 x*y = 2 3 5 a -a x -x-x-y. 
 
 If a polynomial contains a monomial factor, it may be found 
 easily by inspection. In factoring a polynomial, proceed as 
 follows : 
 
 (1) Divide the polynomial by the highest monomial factor 
 common to all the terms. 
 
 (2) When possible, factor the quotient thus obtained. 
 
 Thus, in 32 a 8 6 3 + 28 a 2 6 2 12 aft, we see that each term contains the 
 factor 4 ab. Divide the polynomial by 4 ab ; the quotient is 8 a 2 ^ 2 + 
 7 ab 3, which (as we shall see later) cannot itself be resolved into 
 factors free from inexact roots. 
 
 Therefore, 32 a s b s + 28 a 2 6 2 - 12 ab = 4 ab(8 a a & 2 + 7 ab - 3).
 
 FACTORING 93 
 
 113. To check an example in factoring either 
 
 (I) Multiply the factors together, or 
 
 (II) Substitute numbers for the letters and simplify. 
 
 In 32 a 3 6 3 + 28 a 2 6 2 - 12 aft = 4 a&(8a 2 6 2 + 7 ab - 3), let a = b = 1. 
 Then 32 a 3 6 3 + 28 a 2 ^ - 12 ab = 48 j 4 ab = 4, 8 a 2 6 2 + 7 a& - 3 = 12, 
 and 4 12 = 48.* 
 
 114. Since factoring is the inverse of multiplication, certain 
 type forms can always be factored. 
 
 I. Type form : ay + by eg. 
 
 ay + by - cy = y(a + b - c), 
 
 wherein y stands for the highest monomial factor common 
 to all the terms. 
 
 Hence 15 Wx - 10 b*y - 5 6 3 z = 5 b\3 x-2y-z). 
 
 EXERCISES f 
 
 115. Factor: 
 
 1. 4 a 8. 6. 2 m*n 6 raV + 8 ra 2 3 . 
 
 2. a 2 -a 3 . 7. 15 a 4 - 10 a 3 - 25 a 2 . 
 
 3. 3a6-6ac. 8. 16 c 3 + 8 c z d - 64 c 4 ^ 2 . 
 
 4. 2afy 6 a 2 *. 9. x a 2 + ar s + :c 4 . 
 
 5. a 2 6 + a 2 c - a 2 d. 10. 12 a 6 - 24 a 7 + 6 a 5 . 
 
 11. a^ + 2 xy x xz. 
 
 12. 18d 6 -81cf + 27d 2 -9cP. 
 
 13. IT ay - 34 ay 4- 68 a 3 . 
 
 * If an error exists, checking by the substitution of particular numbers for 
 the letters usually reveals the error, but not always. There is the possibility 
 that some particular numbers will be hit upon which fail to reveal the error. 
 For instance, a 2 b*=(a 2 6) (a + 6) (where =^= means " is not equal to "), 
 yet a = 1, b = 2 yields 1 4 = (1 2) 3, or 3 = 3. 
 
 Again, a = 2, 6 = 2 yields = 2-0, or = 0. Again a = 0, 6 = 10 yields 
 100 = 100. If the substitution yields an inequality, the factoring is cer- 
 tainly wrong ; if the substitution yields an equality, the factoring is very 
 probably correct, but the test is not infallible in this case.
 
 94 ELEMENTARY ALGEBRA 
 
 14. 3 afyz + 3 xfz + 3 xyz\ 
 
 15. 44 m 3 w 2 55 m?n + 66 m?r. 
 
 16. aW - aW + aW. 
 
 17. 125 c 2 d 2 + 100 c 2 cP - 150 c 3 ^. 
 
 18. 15 & 3 0/i 2 + 17 bgW-Wtfh - 
 
 19. a n & a"c a n d. 
 
 20. 2 a?" 6 &ic + 4 cof. 
 
 21. 3 a"6 n c 9 a n b n cd + 6 a n b"ce. 
 
 22. cy^-dy^ + gy*". 
 
 23. 5 a 2 6 n + 10 a s &" 15 a 4 &". 
 24. 
 
 116. II. Type form : or + ay + bx + &y. 
 
 Group the first two terms together and the last two terms 
 together. Then factor each group. Notice that the factoring 
 should reveal the same binomial or polynomial factor in all 
 the groups. 
 
 We obtain 
 
 (ax + ay) + (bx + by) = a(x + y ) + 6(z + y). 
 
 Divide by x + y and we obtain the quotient a + b. 
 Taking the divisor x + y as one factor, and the quotient a + b as the 
 other factor, we have 
 
 ox + ay + bx + by = (x +y)(a + b). 
 
 Factor 2 a - 66 - ad + 3bd. 
 
 Grouping, (2 a - 6 6) - (ad - 3 bd). 
 Factoring each group, 2 (a 3 6) d(a 3 b). 
 The factor a 3 b is seen to occur in both groups. 
 Dividing by a 3 6, we obtain 2 d. 
 
 Taking the divisor a 3 b as one factor, and the quotient 2 d as an- 
 other factor, we have 
 
 2 a - 6 b - ad + 3 bd =(a - 3 6)(2 - d). 
 
 The groups may contain two or more terms, but all the groups 
 must have the same polynomial factor. In grouping, the terms
 
 FACTORING 95 
 
 may be taken in any order which promises to reveal a common 
 polynomial factor. 
 
 For example, ab c be + a may be written ab be + a c. 
 Then ab - be + a - c = b(a - c)+(a c) = (a c)(6 + 1). 
 
 EXERCISES 
 
 117. Factor : 
 
 1. a(x y}+b(x y). 3. x(c - a) + y(c - a). 
 
 2. a(x + 2 y) - b(x + 2 y}. 4. 3(c d) 2(0 - d). 
 
 5. a(3 c + d) - 6(3 c + d) + c(3 c + d). 
 
 6. 2(a-f-#) + a(a-f #) e(a4-#). 
 
 7. 6(a c) c(c a). 
 
 8. r(z-j/)+<z-.y) + %-z). 15. 03 + ^ + 0+ 1. 
 
 9. ax ay bx + by. 16. 2 a b + ab. 
 
 10. 3 a - 3 y - dx + dy. 17. 6 6a? -12 by- 50 c?/+25 ex. 
 
 11. ac + ad - 6c - 6d. 18. 3 6 5 6t/ 3 - 6 &y + 10 ft?/ 4 . 
 
 12. y - bx ay + ab. 19. 49 ar lat + 21 6?- 6 bt. 
 
 13. cd + 3c-2d 6. 20. 1+a + b + ab. 
 
 14. 6rs + 4r 9s 6. 21. ax + a?/ + 02 + 60; +by+bz. 
 
 22. a a ex + c bx + b. 
 
 23. ax ay + 2 6x 2 by -\- ex cy. 
 
 24. 2 a 2 6 + 2 c + fl 6^ + ex + a?/ by -\- cy. 
 
 118. III. Type form : a 2 + 2 ab + b 1 . 
 
 a 2 2 aft + 2 = (a 6) 2 . 
 1 . Factor 25 x 4 - 60 zV + 36 ?/ 6 . 
 Here a 2 = 25x 4 , 6 2 = 36?/ 6 . 
 
 Hence a = 5 a; 2 , 6 = 6 y 3 , 
 
 and - 2 a& = - 2 . 5 x 2 6 f- = - 60 a: 2 ?/ 8 - 
 
 This product 60 x 2 y s is the same as the middle term of the trinomial 
 which we are factoring. 
 
 Hence that trinomial is factored by type form III, and we obtain 
 25 x 4 - 60 x' 2 y 3 + 36 j/ 6 = (5 x 2 - 6 y 3 ) 2 .
 
 96 ELEMENTARY ALGEBRA 
 
 2. Factor 4 & + 2 x + 1. 
 Here a 2 = 4 z 2 , 6 2 = 1. 
 
 We obtain a = 2x, b = 1, + 2 ab = 4a:. 
 
 The product 4x is nof the same as the middle term of the given 
 trinomial. 
 
 Hence this trinomial cannot be factored by type form III. 
 
 The type form here considered enables one to factor any 
 trinomial which is a perfect square. 
 
 EXERCISES 
 119. Factor and check by multiplying the factors together : 
 
 1. m 2 + 2mn + w 2 . 12. 100 + 20ft + ft 2 . 
 
 2. a 4 - 4 a 2 6 + 4 6 2 . 13. 9 w 2 - 18 w + 9. 
 
 3. 4 -6a 2 6 2 + 96 4 . 14. 1-26& + 1696 2 . 
 
 4. a 2 + 24 a; +144. 15. 25 + 40a + 16a 2 . 
 
 5. &6 + 16 + 86 3 . 16. 9 c 2 + 54 cd?e 3 + 81 d 6 e 6 . 
 
 6. d 2 + 16d + 64. 17. 16 n* - 56 np + 49 p 2 . 
 
 7. i. + y2_ 2/ . 18. 225 P 
 
 8. ^a^ + laj + y 1 ^. 19. 4m 2 + 6 
 
 9. 36 a 2 - 12 aft 3 + 6 6 . 20. 225 s 2 - 60 s 5 + 4 i 10 . 
 
 10. 121e 8 -22e 4 /+/ 2 . 21. (a+ 6) 2 - 6(a + 6) + 9. 
 
 11. 9 2 + 6.v + l. 22. a;- 
 
 23. 36(r - s) 2 - 60(r - s) + 25. 
 
 24. 9( 
 
 25. 4-12(a-)+9(a a;) 2 . 
 
 26. a 2 + 2 ay 2 zz + y 1 2 yz + z 2 . 
 
 27. c 2 + 2 cd + 2 ce + d 2 + 2 de + e 2 . 
 
 28. 2 - 
 
 29. 4 
 
 30. 2
 
 FACTORING 97 
 
 120. IV. Type form : a 2 - W. 
 
 a 2 - b 2 = (a + 6)(a - b). 
 
 Factor 81 x*y 2 100 z 4 . 
 
 Here a 2 = 81 ^y 2 , b' 2 = 100 2*. 
 
 We obtain a = V81 xy 2 = 9 x 3 ?/, ft = v/100 z* = 10 z 2 . 
 
 Hence 81 zy 2 - 100 z* = (9 .rfy + 10 z 2 ) (9 3 y - 10 z 2 ) . 
 
 We see that every binomial which is the difference of two 
 perfect squares is composed of two binomial factors. 
 
 RULE. (1) Find the square root of each perfect square. 
 
 (2) Take the sum of the square roots as one factor, and 
 
 (3) The first square root minus the second square root as the 
 other factor. 
 
 EXERCISES 
 
 121. Factor: 
 
 1. &-y*. 1. 25c 6 -49d 4 . 13. ft 2 -169. 
 
 2. a 2 -9. 8. 81 -121s 12 . 14. 4 a 2 -!. 
 
 3. 1-6 2 . 9. 9r 2 -16* 2 . 15. 16a 2 6 2 -25c 2 . 
 
 4. 4-7M 4 . 10. 100/>y Slto 2 . 16. 225 g 2 - 144 z 2 . 
 
 5. ^-25. 11. 36x 10 -49y 12 . 17. 100 w 2 - 256V 4 . 
 
 6. 64c 2 -z 2 . 12. 144 1^ 25 x 4 . 18. 9zV 121. 
 
 19. 49c 6 -64 8 . 29. 6 2 - 18 6 + 81 -49s 2 . 
 
 20. 289 a 2 6 2 c 2 25 d 2 . 30. d 2 12d + 36 25 J 2 . 
 
 21. (x+y) 2 1. 31. 1 (a + 6)". 
 
 22. (a-j-36) 2 -c 2 . 32. 16 -(x-y}\ 
 
 23. (x-2yy z 2 . 33. 9 (m 3w) 2 . 
 
 24. (m + 6n) 2 -4r 2 . 34. k z -(h + 10) 2 . 
 
 25. (7 d - r) 2 - 25 s 2 . 35. a 2 -6 2 -26c-c 2 
 
 26. a 2 + 2 06 + b 2 4c 2 . =a 2 (6 2 + 26c + c 2 ) 
 
 27. z 2 2xy + y z z 2 . =a 2 -(6 + c) 2 . 
 
 28. c 2 - 6 cd + 9 d 2 e 2 . 36. < 2 - m 2 + 2 mra n 2 .
 
 98 ELEMENTARY ALGEBRA 
 
 37. 49-c 2 -6cd-9d 2 . 
 
 38. 
 
 39. 
 
 40. 
 
 = (a -|- 6 + c -f d)(a + 6 - c - d). 
 
 41. (a? + 2y) (TO n). 44. 144(m -) 2 121 (w + t/) 2 . 
 
 42. 9(c - d) 2 - 4( - y) 2 . 45. 36(# - z) 2 - 196(u> + z) 2 . 
 
 43. 64(fc 4- Z) 2 - 49(r - t) 2 . 46. c*+2cd+d 2 -m 2 +2mr-r 2 . 
 
 122. V. Type form : j? + gx + h. 
 
 x 2 + gx + h=(x + a)(x + 6), 
 wherein g = a + b aiid h =ab. 
 
 1. Factor a 2 5 cc + 6. 
 
 Here g = 5, A = 6. .-. a + 6 = 5, aft =6. 
 
 By trial, find two numbers whose sum is 5, and whose product is 6. 
 These numbers are 3 and 2. 
 
 Hence x 2 - 5 x + 6 = (x 2)(x 3). 
 
 2. Factor x t -2x-35. 
 
 Here = 2, A = - 35. 
 
 That is, a + 6 = 2, aft = 35. 
 
 By trial, find two numbers whose sum is + 2 and whose product is 
 -35. 
 
 These numbers are + 7 and 5. 
 
 Hence x 2 + 2z 35 =( + 7)( 5). 
 
 3. Factor a? + 5 x 7. 
 
 Here = 5, h = 7. 
 
 a + 6 = 5, ab = 7. 
 
 We cannot find two integers whose sum is 5 and whose product is 7. 
 Hence this trinomial cannot be factored by this type form. 
 
 ORAL EXERCISES 
 
 123. Factor : 
 
 1. 3*4- 5* + 6. 3. x 2 + x -6. 
 
 2. x 2 - x 6. 4. ic 2 + 9 x + 20.
 
 FACTORING 99 
 
 5. af-x-20. 21. d 2 + 7d+10. 
 
 6. or* -9 a +20. 22. d 2 -14d-15. 
 
 7. or + a; -20. 23. /*- 15/+ 26. 
 
 8. a; 2 _ a _ 12. 24. # 2 + 9 + 8. 
 
 9. a^ + 9a; + 14. 25. tf + Wh-39. 
 
 10. x 2 -3o:-18. 26. ft 2 -11 ft -60. 
 
 11. z? + Ux + 33. 27. fc 2 - 12 Jc + 32. 
 
 12. x 2 - 16 a; - 36. 28. fc 2 - 3 k - 54. 
 
 13. 0^ + 140;+ 45. 29. fc 8 + 27 A; - 90. 
 
 14. 2^ + 14 a + 24. 30. Z 2 - 25 Z + 100. 
 
 15. or z 72. 31. m 2 + 9m 112. 
 
 16. y? - 16 x + 28. 32. r*s* + 12 rs -|- 20. 
 
 17. a 2 -3a-10. 33. a 2 6 2 + 16 ab + 55. 
 
 18. m 2 + 7 ?7i + 12. 34. y?y*-xy- 380. 
 
 19. a 3 +3a + 2. 35. c 2 ^ 2 - 12 cd - 13. 
 
 20. a 2 -a- 2. 36. (x+y)* l(x + y)-18. 
 
 124. VI. Type form : />^ 2 + qx + r. 
 
 px 2 + qx + r=(ax + b)(cx -\- d), 
 wherein p = ac, q = be -f ad, r = bd. 
 
 1. Factor 6 tf - x - 12. 
 
 Here p = 6, q = - 1, r = 12. 
 
 If possible, we must find numbers a, 6, c, and d, such that ac = 6, 
 6d=-12, 6c + ad=- 1. 
 
 Here, ac = 6 suggests the values a = 3, c = 2, or a = 1, c = 6, etc., 
 and 6d = 12 suggests the values 6=4, d 3, or 6 = 6, d = 2, etc. 
 
 Try the various sets of values of a and c, whose product is 6, with each 
 pair of values of b and d, whose product is 12, until a combination is 
 found which satisfies be + ad = 1 ; that is, a combination which gives 
 x as the middle term of the product. 
 
 Try the sets of values, a = 3, c = 2, b = 4, d - - 3.
 
 100 ELEMENTARY ALGEBRA 
 
 +8x 
 Examine (3 x + 4) (2 x - 3) . 
 
 We see that the sum of the two cross products + 8 x and 9 x is x, 
 which is the middle term of 6 x 2 x 12. 
 
 Hence 6 x' 2 - x - 12 = (3z + 4)(2 x - 3). 
 
 2. Factor 8 x 2 - 34 x + 35. 
 Try a = 8, c = 1, 6 = 7, d =-5. 
 
 -7* 
 
 Examine (8 x 7 ) (x 5) . 
 
 i i 
 
 -40 x 
 
 The sum of the cross products 40 x and 7 x is 47 x. 
 But the middle term in 8 x 2 34 x + 35 is not 47 x. 
 Hence (8 x 7)(x 5) are not factors of 8 x* 34 x + 35. 
 Try a = 4, c = 2, b = - 7, d = - 5. 
 
 -14x 
 Examine (4x - 7)(2x 5). 
 
 -20 x 
 
 The sum of the cross products 20 x and 14 x is 34 x. 
 34 x is the middle term of 8 x 2 34 x -f 36. 
 Hence 8x 2 - 34 x + 35 =(4x- 7)(2x- 5). 
 
 EXERCISES 
 125. Factor: 
 
 1. 6 x 2 + 7 x + 2. 9. 10 6 2 - 37 6 + 30. 
 
 2. 6 x 2 + 13 x + 5. 10. 6 y 2 - 31 y + 35. 
 
 3. 12z 2 + 7z-5. 11. 6 2 -19J + 10. 
 
 4. 7 x 2 + 11 x + 4. 12. 3 s 2 + 13 s - 10. 
 
 5. 3 a; 2 + 7 a + 2. 13. 24 s 2 + s - 3. 
 
 6. 20 x z + 33 aj + 7. 14. 30 ?/ 2 + 19 y - 5. 
 
 7. 2 a 2 - 27 a + 13. 15. 8 m 2 - 22 m - 21. 
 
 8. 8 m 2 - 31 m + 30. 16. 3 m* - m - 2.
 
 FACTORING 101 
 
 17. 15m 2 + 14m-8. 21. 11 g* - 21 g - 2. 
 
 18. 18a 2 + 9a-2. 22. 2g 2 + gh-15W. 
 
 19. 6c 2 + 35c-6. 23. 3a 2 -f-lla6-4& 2 . 
 
 20. 6 d 2 - 13d -5. 24. 3a 2 -7a6-66 2 . 
 
 126. VII. Type forms : a 3 ft 3 . 
 
 (a 3 + 6 3 ) = (a + 6) (a 2 - ab + 6 2 ). 
 (a 3 - 6 3 ) = (a - 6) (a 2 + ab + 6 2 ). 
 
 Factor ar 5 + 27. 
 
 a 3 = x 3 , 6 3 = 27 ; a = x, 6 = 3. 
 .-. 3 + 27 = (z + S)(z*-3 
 
 Again, factor 125 m s 8. 
 
 Here a 3 = 125 m 3 , 6 3 = 8 ; a = 5 m, 6=2. 
 
 .-. 125 m 3 - 8 = (5 m - 2) (25m 2 + 10 m + 4). 
 
 Factor : 
 
 1. tf + y 3 . 11. y? + f. 
 
 2. x 3 - y 3 . 12. 8 a 6 - & 6 . 
 
 3. m 8 + l. 13. 1 - 27 a 9 . 
 
 4. m 3 1. 14. m l2 + n n . 
 
 5. o 3 6 3 + 8. 15. 27 c 15 + 1. 
 
 6. c 3 ^ 3 - 27. 16. 8 a 3 " - 1. 
 
 7. 64 a 3 -!. 17. 6 9 " + c 6n . 
 
 8. 1255 3 + 216. 18. x 9 + y 9 . 
 
 9. 8^ 3 -27/i 3 . 19. 64 a 3 - 125 b 3 . 
 10. zy + 343. 20. 216-27ic 18 . 
 
 The foregoing are the simpler type forms of factoring in 
 algebra; more complicated ones are given in more advanced 
 texts on algebra. 
 
 127. Skill in factoring depends upon an easy recognition of 
 the type forms and a facility in applying correctly the method 
 of each one.
 
 102 ELEMENTARY ALGEBRA 
 
 Some suggestions will be helpful. 
 
 I. See if the terms contain a common monomial factor. 
 If so, divide the expression by it, and keep it as one factor; the 
 quotient will be the other factor. 
 
 II. Determine to which type form the quotient thus 
 obtained belongs and factor accordingly. 
 
 III. Inspect each factor and, if possible, resolve it into new 
 factors. 
 
 TYPE FORMS AND FACTORS 
 128. I. ay+by-cy = y(a + b-c). 
 II. 
 
 III. a 2 2fl&+6 2 =(a6) 2 . 
 
 a 2 - ft 2 = (a + ft)(a - ft). 
 
 i a 2 + 2aft + & 2 -c 2 =(a+&) 2 -c 2 =(a+6+c)(a+&-c). 
 a 2 + 2a& + ft 2 -c 2 -2a?-d 2 = (a+ft) 2 -(c+<f) 2 
 
 V. x* + gx+h = (x + a)(x+b); g=a + b,h = ab. 
 VI. px z +qx+r=(ax+b)(cx+d) ; p=ac, q=bc+ad, r=bd. 
 
 ' 
 
 REVIEW EXAMPLES 
 129. Factor: 
 
 1. a 3 + 2a 2 + a, 7. a 2 + 16 a + 15. 
 
 2. 2s* + 12a; 2 + 18. 8. ax - 2x- ay + 2y. 
 
 3. 9 y* 24 y 6 + 16 y 8 . 9. a 2 , & 2 m crto + fe 
 
 4. & 3 -&. 10. c 3 + c 2 c 1. 
 
 5. a^-16. 11. a; 6 -6x 3 -7. 
 
 6. a 2 + 6a-7. 12. V - 17 A? + 72.
 
 FACTORING 103 
 
 13. 64-o. 17. 3m 4 -8m 2 -35. 
 
 14. h 3 1 W 2 h + 14. 18. y 3 64. 
 
 15. 81 h* k*. 19. L l + L + |. 
 
 21. (a + 6)(a 2 + & 2 ) + (a + 6)(2a6). 
 
 22. 2 a 3 -14 a 2 + 70 -10 a. 36. c 16 -216c 7 . 
 
 23. (a + 6) 2 + 2(a + 6) + l. 37. a 2 - 4 6 2 + 46c- c 2 . 
 
 24. a 2 + 6 2 - 2 ab 9. 38. a 2 + m 2 > 
 25 g2 ci 2 + 2 d& 6 2 39 3 x 2 3 CLX 
 
 26. 1 + x u . 40. 7 c 2 + 20 c - 
 
 27. r 2 + 3r-154. 41. e 3 + 512/ 3 . 
 
 28. x 2 + 11 x - 210. 42. 6y 
 
 29. a 12 - 6 12 - 43. a 2 
 
 30. 2s 2 -5s + 2. 44. 1- 
 
 31. 7 a 2 + 175 -70 a. 45. 9 c 4 16 c 2 ^ 2 + 7 d 4 . 
 
 32. 18-6x-12a; 2 . 46. 9 m 4 - 13 m 2 w 2 + 4 n*. 
 
 33. 17 w* + 25 10 - 18. 47. a 2 - b 2 - a - b. 
 
 34. 9x 2a -25. 48. a 3 + 6 3 + a 2 -6 2 . 
 
 35. a 2 ac + 2 ab 2 be. 
 
 49. 4 a 2 - 4 a& + 6 2 - 9 c 2 24 cd 16 d?. 
 
 50. (x + y) 2 - o(x + y) + 6. 54. 25^ - 61 x 3 y 3 + 36 y 6 . 
 
 51. a & xy ab*xy. 55. a^ + 729?/ 6 . 
 
 52. 4 a 5 " - a-. 56. 196 r% - z) 2 - 225 s 2 . 
 
 53. 49 a 21 - 84 a'b* + 36 & 2 ". 
 
 58. 21 c - 5 d + 3 cd - 2 de - 14 e - 35. 
 
 59. (x 2 - I) 2 - <y - I) 2 . 60. 49y 2 -70yz + 252 2 .
 
 CHAPTER VIII 
 
 EQUATIONS SOLVED BY FACTORING 
 
 130. The degree of an equation in one unknown is the same 
 as the highest power of the unknown in the equation. 
 
 An equation of the first degree is a linear equation ; one of 
 the second degree is a quadratic equation ; one of the third 
 degree a cubic equation ; one of the fourth degree a quartic 
 equation; etc. 
 
 ax + b is a linear equation. 
 ay? + bx -f c = is a quadratic equation. 
 aa? + bx 2 + ex + d = is a cubic equation. 
 a* 4 + bx 2 + c = is a quartic equation. 
 
 An equation of second, third, fourth, or higher degree may 
 sometimes be solved by the methods of factoring which we 
 have learned. 
 
 The solution of such equations by the method of factoring 
 depends upon the principle that if the product of two or more 
 factors equals zero, , at least one of the factors must equal 
 zero. 
 
 131. 1. Solve x* - 7 x = - 12. 
 
 Transposing 12, so that the right-hand member of the equation may 
 
 be x 2 - 7 x + 12 = 0. 
 
 Factoring, (x - 4) (x - 3) = 0. 
 
 Since the product of the factors (x 4)(x 3) is equal to zero, one or 
 the other of the factors must be equal to zero. 
 
 When x - 4 = 0, 
 
 we obtain the root x = 4. 
 
 104
 
 EQUATIONS SOLVED BY FACTORING 105 
 
 When x 3 = 0, 
 
 we obtain the other root, x = 3. 
 
 Hence the given quadratic equation has the two roots, 3 and 4. 
 
 Check : If x = 4, x 2 7 x = 12 becomes 
 
 16-28 =- 12. 
 
 - 12 = - 12. 
 
 If x = 3, x 2 7 x = 12 becomes 
 
 9-21 =-12. 
 
 - 12 = - 12. 
 
 2. Solve ar f -x = 2z 2 -2. 
 
 Transpose, x 3 x 2 x 2 + 2 = 0. 
 
 Factor, a:(x 2 - 1) - 2(x 2 - 1) = 0. 
 
 (x 2 -l)(x-2)=0. 
 
 Ifx + l = 0, x=-l. 
 
 If x- 1=0, x = l. 
 
 If x - 2 = 0, x = 2. .4ns. - 1, 1, 2. 
 
 Check : If x = - 1, - 1 + 1 = 2 - 2, = 0. 
 
 If x = 1, 1 - 1 = 2 - 2, = 0. 
 
 If x = 2, 8 - 2 = 8 - 2, 6 = 6. 
 
 RULE. Transpose all the terms to the left side so that the right 
 side may be zero., Factor the left side. Make each factor con- 
 taining the unknown equal to zero and solve each resulting 
 equation. 
 
 If there is a factor that does not contain the unknown, divide both 
 members of the equation by it. Thus, 7 x 2 + 21 x 63 = has the factor 
 7. Dividing both sides of the equation by it, the simplified equation is 
 x 2 + 3 x - 9 = 0. 
 
 9 
 
 132. Observe that a second-degree equation will produce two 
 factors, a third-degree equation three factors, and so on. There 
 will be as many roots as linear factors containing the unknown, 
 and hence an equation has as many roots as the highest power 
 of the unknown in it.
 
 106 ELEMENTARY ALGEBRA 
 
 Some of the roots may be alike, as in 
 a 2 4 x = 4. 
 
 x-2 = 0,x = 2. Ans. 2, 2. 
 
 This is called a double root. 
 
 In solving an equation care must be taken not to discard a 
 factor which contains the unknown quantity ; by so doing, a 
 root will be lost. 
 
 Solve x--2x = 0. 
 
 Divide both sides by x, x 2 = ; x = 2, one root. 
 But by factoring, x(x 2) = 0. 
 
 Whence x = ; x = 2, two roots. 
 
 EXERCISES 
 
 133. Solve and check: 
 
 1. a; 2 25 = 0. 13. cz 2 = a*c. 
 
 2. a 2 = 36. 14. 2a? + 6 = 7*. 
 
 3. x* = 4:X. 15. 12 x 2 + x = 6. 
 
 4. a 2 = ax. 16. 25 a; 2 70 x = 49. 
 
 5. x 2 - 7 a? = - 10. 17. 25 x 2 - 25 a; + 6 = 0. 
 
 6. x 2 3 x = 10. 18. z 2 - aa; bx + ab = 0. 
 
 7. 2 a 2 14 x = 0. 19. a; 2 ax bx = 0. 
 
 8. 5 or* = 15 a?. 20. Qx 2 $ax 2cx= ac. 
 
 9. x* + x = 72. 21. ^ = 40;. 
 
 10. a 2 2 aa = a 2 . 22. 0^ + 8 = 20? + 4 a. 
 
 11. 2 a? = 3 a. 23. a 3 - 3x 2 + 2 a = 0. 
 
 12. bx* ax = 0. 24. x 3 + 20 x = 9 or 5 .
 
 EQUATIONS SOLVED BY FACTORING 107 
 
 PROBLEMS 
 
 134. Solve the following problems and see which roots of 
 the equations satisfy the conditions of the problem : 
 
 1. What number is it whose square plus the number itself 
 equals 42 ? 
 
 2. The product of a certain number decreased by 2 and 
 the same number decreased by 3 equals 20. Find the number. 
 
 3. If to the square of a certain number, the number itself 
 and 5 be added, the result will be 115. Find the number. 
 
 4. If a certain number be subtracted from 19 and 21, in 
 turn, the product of the remainders will be 80. What is the 
 number ? 
 
 5. The difference between two numbers is 3 and the sum 
 of their squares is 89. What are the numbers ? 
 
 6. A rectangular field is 5 rd. longer than it is wide. If 
 each dimension be increased by 1 rd., its area would be 50 sq. 
 rd. What is its present area ? 
 
 Let x rd. = the width. 
 
 Then (x + 5) rd. = the length. 
 
 (x + 1) rd. = the width increased by 1 rd. 
 (x + 6) rd. = the length increased by 1 rd. 
 We obtain the equation (x + l)(x + 6) = 50. 
 x 2 + 7 x + 6 = 50. 
 x 2 + 7 x - 44 = 0. 
 
 The roots of the equation are + 4 and 11. 
 
 We discard the root 11 as being inapplicable to our problem, on the 
 ground that we do not speak of a rectangle having a negative length and 
 a negative breadth ; the problem involves the condition that only the 
 positive roots shall be used. 
 
 7. If the length of a certain square be increased 3 ft. and 
 the width decreased 2 ft., the area would not be changed. 
 What are the dimensions of the square ?
 
 108 ELEMENTARY ALGEBRA 
 
 8. By how many inches must the length, 38 in., and the 
 width, 22 in., of a rectangle be increased, in order to increase 
 the area by 189 sq. in. ? 
 
 9. How may $ 84 be divided among a number of persons 
 so that each person shall receive $ 5 more than the number of 
 persons ? 
 
 10. Is the difference of the squares of two consecutive num- 
 bers even or odd ? Prove your answer. 
 
 11. The sum of the squares of two consecutive numbers is 
 85. What are the numbers ? 
 
 12. The sum of the squares of two consecutive odd numbers 
 is 202. What are the numbers ? 
 
 13. A schoolroom contains 48 seats, and the number in a 
 row is 4 less than twice the number of rows. How many seats 
 in a row ? 
 
 14. If a cubical box has its dimensions increased by 1, 2, 3 
 inches respectively, the capacity of the box has been increased 
 93 cu. in. What are the dimensions ? 
 
 15. A rectangular yard has a fence around it measuring 
 100 ft. ; the area of the yard is 600 sq. ft. How does its 
 width compare with its length ? 
 
 16. The surface of a cube contains 384 sq. in. Find its di- 
 mensions. 
 
 The altitude of a triangle is its height, or the length of a perpendicular 
 
 line from its vertex to its base. 
 
 The area of a triangle equals one 
 half its altitude multiplied by its base. 
 Area = | ab. 
 
 17. If the area of a triangle is 
 FlG 19 36 sq. in. and the base is 12 in., 
 
 find the altitude. 
 
 18. The base of a triangle is 3 times the altitude and the 
 area is 24 sq. ft. Find the base.
 
 EQUATIONS SOLVED BY FACTORING 
 
 109 
 
 19. The area of a triangle is 35 sq. ft. and the base is 3 ft. 
 more than the altitude. Find the altitude and the base. 
 
 The square on 
 
 leg 
 FIG. 20 
 
 If the perimeter is 72 in., 
 
 The hypotenuse of a right triangle is its longest side, 
 the hypotenuse is equal to the sum of 
 the squares on the legs. 
 
 20. One leg of a right triangle 
 is 3 ft. longer than the other, and 
 the area is 54 sq. ft. Find each 
 leg and the hypotenuse. 
 
 21. The sides of a certain right 
 triangle are in the ratio 3:4:5. 
 what is its area ? 
 
 22. The diagonal of a square is 50 in. What is its perim- 
 eter ? 
 
 23. The diagonal of a rectangle is 640 in. If the length of 
 the rectangle is three times its width, what is its area ? 
 
 24. The area of a rectangle is 48 sq. ft. If its length is 
 2 ft. more than the width, what will its diagonal be ? 
 
 The area of a trape- 
 zoid is equal to one 
 half the altitude times 
 the sum of the bases. 
 
 b' 
 FIG. 21 
 
 a 
 
 25. The alti- 
 tude of a trapezoid 
 is 12 in., one base 
 is 3 in. shorter than 
 
 the other; the area is 162 sq. in. How long is the other 
 
 base? 
 
 26. One base of a trapezoid is 2 ft. longer than the other 
 and the altitude is 1 ft. longer than the longer base ; the area 
 is 120 sq. ft. Find the bases and the altitude.
 
 110 ELEMENTARY ALGEBRA 
 
 The length of a circle (circumference), C = 2 irR (wherein ir(pl) is a 
 Greek letter standing for 3.1416, or ^-, approximately). 
 The area included by a circle A = vR 2 . 
 
 27. If a circle is 88 in. long, what is its radius ? 
 
 28. Over how much ground may a horse graze if tied to a 
 stake by a 30-foot rope ?
 
 FRANCOIS VIETA 
 
 Vieta's Notation bfodern Notation 
 
 lC-8Q + Wy aequ. 40 ; x 3 - 8 x 2 + 16 x = 40. 
 
 C stands for cubus or " cube " of the unknown. 
 Q stands for quadratus or "square " of the unknown. 
 N stands for numerus or the unknown itself. 
 aequ. stands for aequalis or equal.
 
 CHAPTER IX 
 
 MULTIPLICATION AND DIVISION OF POLYNOMIALS 
 REVIEW OF MULTIPLICATION 
 
 135. Arrange both polynomials that are to be multiplied to- 
 gether, according to the descending or ascending power of 
 some letter. Multiply each term of one polynomial by each 
 term of the other, and add the partial products. 
 
 Thus, find the product of : 
 
 1. 2 a 2 + 3 a - 5 and 3 a 2 2 a + 1. 
 
 2. 3 a 3 - a 2 + 4 a - 2 and a 2 + a - 3. 
 
 3. 3 x* - x 4- 4 and 2 a? + 3 * 5. 
 
 4. ;c 2 + ay + y- and x y. 
 
 5. 2 x 2 - 5 a;y + 2 y 2 and 2 x 3 y. 
 
 6. x 4 a 2 + 1 and x 4 + cc 2 + 1. 
 
 7. 1 - a - 2 a 2 + a 4 and 1 - 2 a + a 2 . 
 
 8. c 6 - 2 c 3 + 1 and c 2 - 2 c + 1. 
 
 9. 05 ?/ and afy + a/'i/ 3 + y 4 + a% 2 + a^ 4 - 
 
 10. 1 + 5 a 3 - 6 a 4 and 3 a 2 + 1 - a. 
 
 Simplify : 
 
 11. (ra 4 -9m 2 + 12ra-4)(ra 2 + 3m-2). 
 
 12. (3r 3 -2r 
 
 13. (a 3 + a 2 6 
 
 14. (x 2 + y 2 + z 2 xy xz yz)(x + y -\-z). 
 
 15. (x 2 -|- 3 x - 2)(4 x z - 5 x + 1)(2 z 2 - a - 3). 
 
 ill
 
 112 ELEMENTARY ALGEBRA 
 
 DIVISION OF POLYNOMIALS 
 
 136. We learned ( 76) to divide a polynomial by a bino- 
 mial. This process may be extended to cases in which the 
 divisor is itself a polynomial. 
 
 (1) Divide 6x*-5x?-7x z + Sx-2}>y2x z -3x + l. 
 g/p4 5 ic 3 7 # 2 -4- 8 x 2 2 cc 2 3 x 
 
 f) nt __ O /y3 ]_ Q /y2 
 
 3 x* + 2 x 2 
 
 4 a; 3 
 
 . 4a; 2 + 6 x 2 
 
 There is no remainder ; the division is exact. The quotient 
 x-2. 
 
 (2) Divide 16 + 5a 4 - 3a 3 -32a + 36 a 2 by a 2 + 4-4a. 
 
 16 - 32 a + 36 a 2 - 
 16 - 16 a + 4 a 2 
 
 3 a 3 + 5 a 4 
 
 4 
 
 4 a + a 2 
 
 4 
 
 - 4 a + 4 a 2 
 
 - 16 a + 32 a 2 - 
 - 16 a + 16 a 2 - 
 
 3 a 3 + 5 a 4 
 4 a 3 
 
 
 16 a 2 + a 3 + 5 a 4 
 16a 2 -16a !> 
 
 17 a 3 + a 4 
 
 There is a remainder, and the division is "not exact." The 
 quotient is ^ 3 , 4 
 
 4 4 a + a 2 ' 
 
 137. Emphasis must be placed upon the necessity of arrang- 
 ing both polynomials, and each remainder, according to the as- 
 cending or the descending power of some letter. 
 
 Divide the first term of the dividend by the first term of the 
 divisor for the first term of the quotient. 
 
 Multiply the entire divisor by the first term of the quotient. 
 
 Subtract the product from the dividend. 
 
 Treat the remainder as a new dividend and proceed as before.
 
 POLYNOMIALS 113 
 
 To check an example in division : 
 
 I. Multiply the r quotient by the divisor, add the remainder 
 to this product. The result should be the dividend ; or 
 
 II. Substitute particular numbers for the letters and divide 
 the resulting values. 
 
 To illustrate the second method of checking : in our first division, let 
 
 6 x* - 5 X s - 1 z 2 + 8 x - 2 = 42. 
 
 2 x 2 - 3 x + 1 = 3. 
 
 3 x 2 + 2 x - 2 = 14. 
 Then, 42 -*. 3 = 14. 
 
 Hence the division is correct. In this mode of checking, care must be 
 taken to select numbers for the letters which will not reduce the divisor 
 to zero. 
 
 EXERCISES 
 138. Divide and check : 
 
 1. x 4 - 9 a? + 19 x 2 - 25 x + 6 by x 2 - 2 x + 3. 
 
 2. 6 a 4 - a 3 b - 10 a 2 6 2 + 31 ab 3 - 20 6 4 by 3 a 2 + 4 ab - 5 6 2 . 
 
 3. 8 a 4 + 8 a 3 b - 28a 2 6 2 + 53 ab 3 - 21 6 4 by 4a 2 - 6 ab + 76 2 . 
 
 4. x 5 - x* + 4 a? 3 x- + 5 x 6 by 3? + x 2. 
 
 5. a 3 - 3 a 2 6 -f- 3 a& 2 - 6 3 by a 2 - 2 ab + 6 2 . 
 
 6. 10a 5 +lla 4 -39a 3 + 42a 2 -17a+5by 2a 3 +3a?-7 a+5. 
 
 7. r 8 + r 7 r 6 r 5 + r 3 + r 2 1 by r 4 r 2 + 1. 
 
 8. a 4 + a 2 6 2 + & 4 by a 2 + b 2 - ab. 
 
 9. x 5 y 5 by a; y. 
 
 10. a 3 + 6 3 + c 3 - 3 a&c by a + 6 + c. 
 
 Simplify : 
 
 11. (z 2 + 2 x + 2)(z 2 - 4 x - 5) -s- (x - 5). 
 
 12. (a; + t/) 2 (x 2 xy + y 2 ). 
 
 13. [(a 3 + 6 3 ) -r- (a 2 - 06 4- 6 2 )] [a 2 + ab + 6 s ]. 
 
 14. (a - 6) 3 (a + &) 2 *- (a 2 - & 2 ). 
 
 15. [(x + 2/) s - (x - 1/) 3 ] -s- (3 x 2 + 2/ 2 ).
 
 CHAPTER X 
 
 SQUARE ROOT 
 
 139. The square root of an expression is one of the two 
 equal factors of the expression. 
 
 Since (a + 6) 2 = a 2 + 2 ab + 6 2 , 
 
 Va 2 + 2 ab + & 2 = a + 6. 
 
 Thus it is a simple matter, by inspection, to extract the 
 square root of a trinomial, which is seen to be of the form 
 
 Or we may find the square root in this way : 
 
 a 2 + 2 ab + b 2 \a + b 
 a 2 
 
 Trial divisor, 2 a 
 
 Complete divisor, 2 a + b 
 
 Thus the square root is a + &. 
 
 The study of this simple case enables one to devise a rule 
 which is applicable to more complicated cases. The procedure 
 is as follows : 
 
 I. Extract the square root of the first term and subtract its 
 square from the polynomial, leaving 2 ab + 6 2 . 
 
 II. We see in 2 ab the factor b which we know is the second 
 term of the root. In cases when & is not known, we see that 6 
 may be obtained by dividing 2 ab by 2 a. We call 2 a the trial 
 divisor. Since a is the part of the root already found, we see 
 that the trial divisor is double the root already found. After 6 
 has been found, add it to the trial divisor and we have 2 a + b, 
 the complete divisor. 
 
 114
 
 SQUARE ROOT 115 
 
 III. Multiply 2 a + b by 6, and subtract. 
 This process may now be extended to finding the square root 
 of any polynomial. 
 
 140. The extraction of the square root of 49 a; 4 42ar 3 +79x 2 
 30 x + 25 is as follows : 
 
 49x 4 -42x 8 +79x 2 -30a;-|-25 |7a 2 -3x+6 
 49 x 4 
 
 -42x 8 + 79x 2 30x+25, 1st remainder 
 -42z+ 9x 2 
 
 +70x 2 30x+25, 2d remainder 
 +70x 2 -30x+25 
 
 1st trial divisor, 2(7 x 2 ) = 14x 2 
 
 1st complete divisor, 14 x 2 3 x 
 
 2d trial divisor, 2(7 x 2 3 x) =14 x' 2 6 x 
 
 2d complete divisor, 14 x 2 6x+5 
 
 3d remainder 
 
 The process is as follows : 
 
 (1) The square root of 49 x 4 is 7 x 2 . Here a = 7 x 2 ; 2 a = 14 x 2 , the 
 first trial divisor. 
 
 (2) The first term in the first remainder is 42 x 3 . Take 42 x 3 =2 ab. 
 
 (3) To find 6, divide - 42 x 3 by 14 x 2 . We obtain 6 = - 3x. 
 
 (4) Multiply the first complete divisor by 3 x and subtract. The 
 second remainder is what is left after subtracting (7x 2 3x) 2 from the 
 given polynomial. 
 
 (5) Proceed with the second remainder as you did with the first. Let 
 7 x 2 3 x be the new value of a. 
 
 (6) Divide the first term of the second remainder by the first term of 
 the second trial divisor ; you obtain + 5. 
 
 (7) Multiply the second complete divisor by + 5, and subtract. The 
 third remainder is what is left over after subtracting (7x 2 3x + 5) 2 
 from the given polynomial. As this remainder is zero, (7 x 2 3 x + 6) 2 
 is equal to that polynomial. 
 
 Hence 7 x 2 3 x + 5 is the required square root. 
 
 EXERCISES 
 141. Find the square root of : 
 
 1. a 4 -4 a 3 + 6 a 2 - 4a + l. 
 
 2. x* - 6 y? + 13 x 2 - 12 x + 4. 
 
 3. c 6 + c 4 - 26 c 3 + 6 c 5 + 8 c 4- 10 c 2 + 1. 
 4.
 
 116 ELEMENTARY ALGEBRA 
 
 5. 250* - 20afy + 34oty 2 - 32^ + 17 aty 4 
 
 6. 9 c 4 ^ - 18 c 3 ^ - 3 cW + 12 cd 5 + 4 d 6 . 
 
 7. m 4 n, 4 + 16 r 4 + mVr 2 - 6 m 3 n?r + 24 mm*. 
 
 10. 4a?-20a:-- 
 
 a; 
 
 SQUARE ROOT OF ARITHMETICAL NUMBERS 
 
 142. 1 2 = 1, 2 2 = 4, .- 9 2 = 81, 10 2 = 100, 90 2 = 8100, 100 J 
 = 10,000, 1000 2 = 1,000,000. 
 
 This is sufficient to show that a perfect square consisting of 
 two digits has one digit in the root ; one consisting of three or 
 four digits has two digits in the root; one consisting of Jive or 
 six digits has three digits in the root ; and so on. In other 
 words, the square of a number has twice as many digits, or 
 one less than twice as many, as the number itself. 
 
 Hence, to find the square root we must separate the number 
 into periods of two digits each, towards the left and right from 
 the decimal point. The period farthest to the left may have 
 one or two digits, whereas the one farthest to the right must 
 have two, a cipher being annexed, if necessary, to complete it. 
 
 The square root has therefore one digit corresponding to every 
 period in its square. Separating the square number into 
 periods enables one to find one digit in the root at a time. 
 
 This is seen more clearly if we consider that the square of 
 any number of tens, say 4 tens (40), ends in two ciphers 
 (1600) ; hence the two digits on the right are not needed to 
 find the tens' digit (4), and are set aside until the units' digit 
 is to be found. 
 
 Likewise, the square of any number of hundreds, say 4
 
 SQUARE ROOT 117 
 
 hundreds (400), ends in four ciphers (160,000), and all of these 
 may be set aside until the hundreds' digit is found and they 
 are needed for the tens' and units' digits. 
 
 After pointing off the number into periods, the method is 
 similar to the one used for polynomials. The procedure is 
 based on the formula a 2 -f- 2 ab + b 2 = (a + &) 2 . 
 
 When dividing by the trial divisor, exclude, for brevity, the right-hand 
 digit in the remainder. 
 
 The reason for this is seen in the example which follows ; the trial 
 divisor 2 is equal to 20 of the next lower units. Now 12-^-2 gives the 
 same digit in the root as 127 -=- 20. 
 
 Find the square root of 2272254.76. 
 
 2272254. 76 1 1507 A 
 1 
 
 1st trial divisor = 2(1) =2 
 
 1st complete divisor = 26 [125 
 
 2d trial divisor = 2(15) =30 
 
 3d trial divisor = 2(150) = 300 
 
 3d complete divisor = 3007 J31049 
 
 4th trial divisor = 2(1507) = 3014 
 
 4th complete divisor = 30144 
 
 127 1st remainder 
 
 22254 2d remainder 
 
 120576 3d remainder 
 120576 
 
 Ans. 1507.4. 
 
 When the 2d trial divisor (30) will not go into 22, a cipher is placed 
 in the root ; the next trial divisor becomes 2(150) or 300 and the next 
 period is brought down. Then 2225 is divided by 300. 
 
 143. The square root of a fraction may be found by taking 
 the square root of the numerator and of the denominator 
 separately, or by reducing the fraction to an equivalent decimal 
 and taking the square root of the decimal. Unless the 
 denominator of the fraction is a perfect square, the second 
 method of procedure is far better. 
 
 EXERCISES 
 
 144. Find the square root of : 
 
 1. 3249. 3. 51529. 5. 248004. 
 
 2. 6084. 4. 93636. 6. 118.1569.
 
 118 ELEMENTARY ALGEBRA 
 
 7. 1010025. 11. .01522756. 
 
 8. 10.4976. 12. 1459.24. 
 
 9. .00564001. 13. 420.496036. 
 10. 64.1601. 14. 7753607.3209. 
 
 Find, to three decimal places, the square root of : 
 
 15. 2. 16. 3. 17. 5. 18. f. 19. f. 20. f.
 
 CHAPTER XI 
 
 GRAPHS OF SIMULTANEOUS EQUATIONS 
 
 145. The plotting of linear equations of the form ax + by 
 = c was explained in 90. 
 
 In the equation x + Oy = 2, y may have any value ; but x 
 must equal 2. Consequently such points as (2, 0), (2, 3), 
 (2, 2), (2, 5), (2, - 1), (2, - 2) would lie on the graph. Thus 
 the graph must be a line parallel to the y-axis, two units to the 
 right. 
 
 EXERCISES 
 
 146. Represent equations 1-8 graphically : 
 
 1. x = 4. 3. x = 7. 5. x = 0. 7. y = 2. 
 
 2. y = 1. 4. y = 3. 6. ?/ = 0. 8. x = 3. 
 
 9. Is the point (3, 4) on the graph of the equation 2 x y 
 = 7 ? Can you tell without making a graph of the equation ? 
 
 10. Can you tell without making a graph of an equation 
 whether the graph goes through the origin or not ? 
 
 11. Which of the equations 1-8 above go through the 
 origin ? 
 
 12. What would you note about the graphs of the equations 
 
 13. Do x + y = 7 and 2 # + 2^ = 14 stand for two distinct 
 graphs or only one ? Why ? 
 
 14. Can you tell by looking at a linear equation whether its 
 graph is parallel to the oj-axis, or to the y-axis, or parallel to 
 neither of them ? How ? 
 
 15. Why are the equations in Ex. 12 called "inconsistent"? 
 Will the same finite values of x and y satisfy both of them ? 
 
 119
 
 120 
 
 ELEMENTARY ALGEBRA 
 
 16. Will the same values of x and y satisfy both equations 
 in Ex. 13 ? 
 
 TWO EQUATIONS IN X AND Y 
 
 147. The solution of a pair of linear equations may be 
 obtained by drawing their graphs, referred to the same pair of 
 axes. 
 
 For instance: x + 2y = 4, (1) 
 3 x - y = 9. (2) 
 In (l),if z = 0, y = -2. 
 x = - 4, y = 0. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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 GRAPHS OF SIMULTANEOUS EQUATIONS 121 
 
 Plot these points and draw the graph. 
 
 In (2), if a;=0, y = -9. 
 a; = 3, y = Q. 
 
 Plot these points and draw the graph. 
 
 The graphs intersect at point P, whose coordinates are 2, 3. Since 
 two straight lines can intersect in only one point, there can be only one 
 pair of values of x and y which will satisfy two linear equations. 
 
 Test these coordinates by substituting them in both equations. 
 
 Let a; = 2, y = 3 ; x + 2y = 4 becomes 4 = 4 ; 3x y = 9 
 becomes 9 = 9. 
 
 Hence x = 2, y = 3 satisfy both equations. 
 
 What would you infer if the graphs were parallel ? 
 What would you infer if the graphs were identical ? 
 
 EXERCISES 
 
 148. Plot the following pairs of equations, each pair on the 
 same set of axes ; determine the coordinates of the intersection, 
 and test the result in each pair of equations : 
 
 1. x + y = 3, 7. x + y = 7, 
 aj + 4?/ = 9. x = y. 
 
 2. x y = 4, 8. x 2y = l, 
 
 2x+y = U. x = 2y. 
 
 3- x-y = 5, 9. x -y = l, 
 
 x + y = 9. y = -2. 
 
 4. x + 2y + 3 = Q, 10. 3x-2y = Q, 
 2x-3y-S = Q. 2x + 3y = Q. 
 
 5. 3x 4y = 18, 11. a; = 4, 
 
 l. = 3. 
 
 6. 2x-y = 3, 12. 4x-5y=2Q, 
 
 4=x-2y = Q. 4x 5y = 4Q. 
 
 13. What is the equation of the #-axis ? 
 
 14. What is the equation of the v/-axis ?
 
 122 
 
 ELEMENTARY ALGEBRA 
 
 PRACTICAL APPLICATIONS OF GRAPHS* 
 
 149. Draw a graph for changing prices expressed in dollars- 
 per-yard to prices expressed in francs-per-meter. 
 
 1 yard = .915 meter, 1 dollar = 5.18 francs. 
 Hence $ 1 per yard = 5.18 fr. per .915 meter. 
 
 If .915 meter cost 5.18 fr., then 1 meter costs 5.18 fr. -=- .915 = 5.66+fr. 
 Hence 1 dollar-per-yard = 5.66 fr. per meter, nearly. 
 Let x = francs-per-meter, 
 
 and y = dollars-per-yard. 
 
 Then x = 5.66 y. 
 
 When y = 0, then x = ; these values determine the point O in Fig. 23. 
 When y = 5, then x = 28.3 ; these values determine the point A. 
 
 j6 jjjjiijjljj;jlj3 H 
 
 ^5 : 
 
 * hR II 
 
 
 
 i !l 1 N 1 1 Illllllll 
 
 O 5 10 15 20 25 
 Francs - per - meter 
 
 FIG. 23 
 
 The line OA is the required graph. To reduce 4 dollars-per-yard to 
 francs-per-meter, pass from B to C, and from C to D, The answer is 
 22.6 francs-per-meter, nearly. 
 
 EXERCISES 
 150. By inspection, change the following to francs-per-meter : 
 
 1. $2 per yard. 3. $ 6 per yard. 5. $4| per yard. 
 
 2. $3 per yard. 4. $5^ per yard. 6. $3| per yard. 
 
 * May be postponed for the present.
 
 GRAPHS OF SIMULTANEOUS EQUATIONS 123 
 
 Change the following to dollars-per-yard : 
 
 7. 5 fr. per meter. 10. 20 fr. per meter. 
 
 8. 10 fr. per meter. 11. 25 fr. per meter. 
 
 9. 15 fr. per metet. 12. 27 fr. per meter. 
 
 13. Given 1 mile = 1.61 Kilometers, and 1 dollar = 5.18 
 francs, construct a graph for changing dollars-per-mile to francs- 
 per-Kilometer. 
 
 14. Draw a graph for changing miles-per-hour to feet-per- 
 minute. 
 
 Hint. I mile-per-hour = 5280 feet per 60 minutes = 5280 -4- 60 feet- 
 per-minute. 
 
 15. Draw a graph for changing rods-per-minute to feet-per- 
 second. 
 
 16. A freight train starts from Denver at noon, running at 
 the rate of 20 miles an hour. Four hours later a passenger 
 train starts in the same direction at the rate of 40 miles per 
 hour. Draw a 
 
 graph showing the 
 distances of the 
 trains from Den- 
 ver during the 
 first ten hours. 
 
 Let x = the num- 
 ber of hours after 
 12 M, and y = num- 
 ber of miles traveled. 
 
 (a) For the freight 
 train, y = 20 x. 
 
 When x = 0, then 
 y = ; these values 
 determine the point 
 O in Fig. 24. 
 
 When x = 10, then 
 y = 200 ; these values 
 determine the point A. 
 
 200 
 
 180 
 160 
 
 b 
 V 
 
 > 140 
 
 Q 120 
 S 
 
 2 100 
 
 M 
 01 
 
 80 
 
 3 
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 345678 
 Hours after 12 M. 
 FIG. 24 
 
 9 10
 
 124 
 
 ELEMENTARY ALGEBRA 
 
 The line OA shows the distances traveled in different times. 
 (6) For the passenger train, y 40 (x 4). 
 When x = 4, then y = ; these values determine the point B. 
 When x 7, then y = 120 ; these values determine the point C. 
 The line BC shows distances traveled by the passenger train. 
 
 (c) How far from Denver is each train after 4| hr. ? 5 hr. ? 
 8J hr. ? 
 
 (d) How far from Denver does the passenger train overtake the freight 
 train ? 
 
 (c) At what time does it overtake the freight train ? 
 
 17. A starts off on a bicycle at 7 miles an hour. Two hours 
 later B rides in the same direction at 10 miles an hour. Draw 
 a graph and tell from it how far apart the two men are after 
 3 hr., 5 hr., 6^ hr. Where and when does the second man 
 overtake the first ? 
 
 18. A train leaves Denver for Colorado Springs at 8 A.M., 
 traveling 30 miles an hour. At 8 : 30 A.M. a second train leaves 
 Colorado Springs for Denver, traveling 40 miles an hour. The 
 two stations are 80 miles apart. Draw a graph and tell from it 
 how far apart the trains are at 9 o'clock, 9 : 30 o'clock, 10 o'clock. 
 
 Where and when do the 
 trains meet ? 
 
 Let x number of hours 
 after 8 A.M., and y = num- 
 ber of miles from Denver. 
 
 (a) The point O marks 
 the time and place where 
 the first train starts. Ex- 
 plain. 
 
 (6) The point A marks 
 the time and place where the 
 second train starts. Ex- 
 plain. 
 
 (c) Explain how the line 
 OB is drawn. 
 
 (d) Explain how the line 
 AC is drawn. 
 
 
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 Hours after 8 A.M. 
 
 FIG. 25
 
 OF SIMULTANEOUS EQUATIONS 125 
 
 19. A man cycles from A to B at 8 miles an hour, and re- 
 turns at 10 miles an hour. If he takes 4 hr. to go there and 
 back, find the distance from A to B. Draw a graph showing how 
 far he was from B after 1 hr., 2 hr., 3 hr. When was he at B ? 
 
 20. A boy begins work with a weekly wage of $9 and 
 receives an increase of 25^ every week. Another boy starts 
 with a weekly wage of only $6, but receives an increase of 50^ 
 every week. Draw a graph which shows the wage of each at 
 the beginning of every week, for 20 weeks. When will their 
 wages be the same ? 
 
 Let 
 and 
 
 x weeks = the time, 
 y dollars = the wage. 
 
 (a) After x weeks the wage of the first boy is y = 9 +-. 
 
 When z=0, y=9; 
 these fix the point A. 
 
 When x = 20, 
 y 14 ; these fix the 
 point B. 
 
 The line AB shows 
 the wage of the first 
 boy for every week. 
 
 (6) After x weeks 
 the wage of the sec- 
 
 ond boy is y = 6 + - 
 
 Dollars per Week 
 
 O > W 
 
 y 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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 10 20 
 Weeks 
 FIG. 26 
 
 When x=0, y=Q ; 
 these fix point C. 
 
 When x = 20, 
 y = 16 ; these fix D. 
 
 The line CD shows 
 the wage of the second 
 boy for every week. 
 
 (c) During which week are the wages the same ? 
 
 (d) State the difference in wage after 7, 10, 13, and 19 weeks. 
 
 21. Fred has $7 in the bank and adds to this 75^ a day. 
 David has $5 in the bank and adds to this $1.25 a day. 
 Draw a graph showing the amounts for the first 7 days.
 
 CHAPTER XII 
 
 SIMULTANEOUS LINEAR EQUATIONS 
 
 151. In a linear equation in one unknown, that unknown 
 can have just one value. Thus 2 x = 8, x = 4. 
 
 In a linear equation in two unknowns, one unknown may 
 vary at pleasure ; the other unknown varies also, but it is de- 
 pendent upon the values of the first. In 2 x y = 4, if x = 0, 
 y = 4 ; if a; = 2, y = ; if x = 3, y = 2, etc. 
 
 But if besides 2 x y = 4, we have a second relation between 
 the variables, wholly independent of the first relation, as 
 2 x + y = 12, then we have what is called a system of linear 
 equations in which x and y are subjected to two conditions 
 which restrict x and y to a single value each. 
 
 The solution of equations by graphs is not always easily ac- 
 complished and may not be accurate. 
 
 To solve the system without graphs we must reduce the two 
 equations in two unknowns to one equation in one unknown. 
 This process is called elimination. 
 
 The elimination may be accomplished in either of two ways : 
 I. addition, or subtraction, or II. substitution. 
 
 SOLUTION BY ADDITION OR SUBTRACTION 
 
 152. I. Solve x + y = 7. (1) 
 
 2 x - 3 y = 4. (2) 
 
 Multiply (1) by 2, to make the coefficients of x alike in both equations. 
 2x + 2y = l4. (8) 
 
 2x-3y = 4 (2) 
 
 Subtract, 5 y = 10 
 
 y = 2. 
 Substitute 2 for y in (1), x + 2 = 7. 
 
 x = 5. Ans. 5, 2. 
 126
 
 SIMULTANEOUS LINEAR EQUATIONS 127 
 
 Check: Substituting 5 for x and 2 for y in (1), and also in (2), we 
 obtain : 
 
 5 + 2 = 7, 7 = 7. 
 10 _ 6 = 4 T 4 = 4. 
 
 II. Solve 2 x - 3 y = 4. (1) 
 
 7 + 4 y = _ 15. (2) 
 
 Multiply (1) by 4, 8 x - 12 y = 16 (3) 
 
 Multiply (2) by 3, 21 x + 12 y = - 46 (4) 
 
 Add, 29 x = - 29 
 
 x =- 1. 
 
 Substitute 1 fora; in (1), 2 - 3 y = 4. 
 - 3 y = 6. 
 
 y=2. Ans. - 1, 2. 
 
 Check : Substituting in (1), -2 + 6 = 4; 4=4. 
 
 Substituting in (2), - 7 - 8 = - 15 ; 15 = - 15. 
 
 Notice that the coefficients of x or of y in the two equations may be 
 made alike in absolute value by multiplying both sides of one equation or 
 of each equation by some appropriate number. Then either add or sub- 
 tract to eliminate one letter. By substitution, find the value of the other 
 letter. 
 
 EXERCISES 
 153. Solve and check : 
 
 1. 2x-3y = -l, 7. 3t-10s = 32, 
 5x + 2y = 45. 6 * - 20 s = 64. 
 
 2. 7x + 2y = l, 8. 2m = 3n, 
 
 4 x y = 7. 9 m 17 n = 7. 
 
 3. x = 2 y, 9. 3 a + 6 b = 7, 
 
 3 x + 7 y = 130. 6 a + 12& = 15. 
 
 4. 6x-3y = 3, 10. 
 
 5. 3x 2y = l, 11. 3x-5y = 23, 
 4 x + 3 y = 2|. 7 x + y = - 35. 
 
 6. 5a + 76 = 14, 12.
 
 128 ELEMENTARY ALGEBRA 
 
 SOLUTION BY SUBSTITUTION 
 
 154. Solve 2x-3y = 27. (1) 
 
 5x + 2y = l. (2) 
 
 From (1), 2z = 27+3j/. 
 
 x = 27+3 y. (3) 
 
 2 
 
 Substitute in (2), 136 + 15y + 2 y = 1. (4) 
 
 i| 
 
 Multiply both sides by 2, 
 
 /=:2. (6) 
 
 Whence I9y = -153. 
 
 y = -l. 
 
 07 _ 21 
 
 Substitute in (3), x - = 3. ^4ras. 3, 7. 
 
 8 
 
 . Cftedfc : In (1), 6 + 21 = 27 ; 27 = 27. 
 In (2), 15 - 14 = 1 ; 1 = 1. 
 
 Notice that the value of x was found in terms of y in one equation and 
 substituted in the other ; the resulting equation was solved, and the value 
 of the other letter found by substitution. 
 
 Either letter may be eliminated, its value in terms of the other letter 
 being found in either equation and substituted in the other equation. 
 
 EXERCISES 
 
 155. Solve, by substitution, and check. Which two equa- 
 tions are inconsistent ( 146, Ex. 15) ? Which are not inde- 
 pendent ? 
 
 1. 4 x + 5 y = 10, 5. x + 21 y = 2, 
 
 2. 3x + 2y = 60, 6. 
 
 3. 4 cc 6y = 96, 7. 8m + 5n= 1, 
 
 4. 3x+5y = 5Q, 8. k = 5l + 3, 
 
 x-7y = S. I = 2k - 24.
 
 SIMULTANEOUS LINEAR EQUATIONS 129 
 
 9. 4r + 3s = 5, 11. x + 1 
 
 10. a -26 = 2, 
 
 26-6a = 3. 6z-8y = 7. 
 
 EQUATIONS CONTAINING FRACTIONS 
 
 156. Solve + ^ = 4. (1) 
 
 Multiply (1 ) by 3, m + = 12. (3) 
 
 4 
 
 Multiply (2) by 6, m - 3 ?t = - 18. (4) 
 
 Subtract (4) from 3, 3 n + = 30. (5) 
 
 4 
 
 Multiply (5) by 4, 12 n+ 3 n = 120. 
 
 15 n = 120. 
 
 n = 8. 
 Substitute 8 for n in (4), m 24 = 18. 
 
 m = 6. ^ins. m = 6, n = 8. 
 Check this answer by substitution in (1) and (2). 
 
 EXERCISES 
 157. Solve, and check: 
 
 i + ^ = ^ 4 *_y = _l 
 
 323' 34 12' 
 
 a 6_7 a; 3y_l 
 
 2 3~6' 2 16 ~2* 
 
 m n o
 
 130 
 
 7. 
 8. 
 9. 
 
 r + 3 
 
 ELEMENTARY 
 
 ALGEBRA 
 
 i n 
 
 :0, 
 
 b 10 
 
 2 
 s + 9 
 
 r-2 ft 
 
 3 3 
 2 a - 7 13 - 
 
 10 
 
 u + 2 
 
 3 
 v-3 
 
 3 6 
 
 5 + 3a 5y- 
 
 2 o 
 
 9 
 
 y + 7 
 
 2 
 u 4 -. 
 
 7 4 
 60; + 8 y =108. 
 a + b 5 
 
 = 20f 
 
 11 
 
 10 
 2 -3 
 
 x-y - 
 
 2 ' 
 2 1 
 
 ' b-a 9' 
 3a 3& + 11. 
 
 x + y 
 
 2 5 
 
 
 13. a; = 2 y, 
 
 x + y a; y_3a: 2y 3y a; 31 
 5 5 3 2 10* 
 
 . " ^ = i' - * 
 
 10 3 =2/ ~3' 
 
 LITERAL SIMULTANEOUS EQUATIONS 
 
 158. Solve ax + y=b, (1) 
 
 ex dy = e. (2) 
 
 Multiply (1) by d, adx + dy = 6d, (3) 
 
 ex dy = e. (2) 
 
 Add (2) and (3), (d + c)x = bd + e. (4) 
 
 Divide both sides of (4) by (ad + c), 
 
 ad + c 
 
 By substituting this value of x in (1) or in (2) we can find y. 
 But it is easier, in this case, to find y by the same method which gave 
 x. Accordingly, 
 
 Multiply (1) by c, acx + c,y = be. (6) 
 
 Multiply (2) by a, acx ady = ae. (6)
 
 SIMULTANEOUS LINEAR EQUATIONS 
 
 131 
 
 Subtract (6) from (5), (c + ad)y = be, ae. 
 
 _ be ae 
 ~ c + ad 
 
 EXERCISES 
 
 10. m + cm = 0, 
 bm + w = 1. 
 
 159. Solve: 
 
 1. x + y = a, 
 x y = b. 
 
 2. ax + by = c, 
 x y =d. 
 
 3. ex dy = m, 
 dx + cy = n. 
 
 4. x = ay, 
 
 bx + dy = bed. 
 
 5. mx -+- ny = 1, 
 cx dy = 1. 
 
 6. 3x 2y = 
 
 2 x 3 y = m n. 
 
 7. x y = l, 
 
 (a + b)x (a 6)y = 0. 
 
 8. (a b)x y = 0, 
 + (a - 6)y = 0. 
 
 9. r + ds = 3, 
 drs = 3d. 
 
 15. Solve for R : (7=2 TT.R. 
 
 16. Solve for a and b : A --. 
 
 M 
 
 17. Solve for a and ^ : A = \(a 
 
 18. Solve for r and t : d = rt. 
 
 19. Solve for M: F = Ma. 
 
 3 bx 4- 2 ay = d. 
 
 5-2 
 
 a 6 
 
 13. + 
 
 a o 
 
 ? + - 
 & a 
 
 , . r 
 14. -
 
 132 ELEMENTARY ALGEBRA 
 
 20. Solve for &, I, and n : S = ^ (tt + f). 
 
 21. Solve for r and t : A = P + Prt. 
 
 22. Solve for d and n : I = a + (n 
 
 23. Solvefor^: C = %(F-32). 
 
 24. Solve for s and a : = -5L . 
 
 160. Solve + = 20, . (1) 
 
 a; y 
 
 --- = 17. (2) 
 
 a 2/ 
 
 If we multiply both sides of each equation by xy, the resulting 
 equations are much more difficult to solve. Hence do not multiply by xy, 
 but eliminate the fraction containing either x or y. 
 
 Multiply (1) by 3, and (2) by 2, Check : 
 
 M = 60 (3) 14 + 6 = 20. (1) 
 
 x y 20 = 20. 
 
 0-4=34 (4) 
 
 21 - 4 = 17. (2) 
 
 13 17 = 17. 
 
 Subtract, = 26 
 
 In (1) substitute \ for y, - + 6 = 20, 
 x 
 
 2 = 14x, 
 
 x = }. Ans. \, \. 
 
 EXERCISES 
 161. Solve and check : 
 1 5 + 6 - 27 2 5 _ 4 - 1 
 
 1. -- 1 -- 41, 6. - - -, 
 
 x y x y 2 
 
 ^_ 3 =-39. ^ + ^ = 5.
 
 SIMULTANEOUS LINEAR EQUATIONS 133 
 
 o 2 3_ , 2 m_ 
 
 I.-) = O, /. = tt, 
 
 x y x y 
 
 8 _ 5 _ 31 3 _ 6 
 
 x y a; i/ 
 
 4 . 1_ 1^ 8 . i_=_i, 
 
 2 a; 3 y ax by 
 
 _5_ + JL = 44 
 
 3 a; 2?/ bx ay 
 
 1,1 6 5 
 
 5. - 4- - = & 5 9- = 
 
 x y ax by 
 
 11 9 ^ 
 
 i ^ O 
 
 x y ax by 
 
 x y nx my 
 
 c 1 , 7i , 9?i . . . 
 
 = a. -| = m 2 4- n 2 . 
 
 THREE SIMULTANEOUS LINEAR EQUATIONS 
 
 162. If the system consists of three independent equations 
 containing three unknowns, one of the unknowns must be 
 eliminated between two pairs of the equations. The resulting 
 equations may then be solved and the third unknown found by 
 substitution. 
 
 Solve x + y z = 0, (1) 
 
 = 9, (2) 
 
 = 10. (3) 
 
 Eliminate y between (1) and (2), 
 
 x + y- 2 = (1) 
 
 2x-y + 3z_9 (2) 
 
 Add, 3 x + 2 z - 9 (4) 
 
 Eliminate y between (2) and (3), by multiplying (2) by 2, 
 
 4z-2y+6z = 18 (5) 
 
 3 a; + 2 y + g = 10 (3) 
 
 Add, lx+ 7e = 28 (6)
 
 134 ELEMENTARY ALGEBRA 
 
 Divide by 7, x -f z = 4. (7) 
 
 Eliminate z between (4) and (7), by multiplying (7) by 2, 
 
 2sc + 2z= 8 (8) 
 
 3x + 2z= 9 (4) 
 
 Subtract, x 
 
 In (7) substitute 1 for a;, l+z = 4;z = 3. 
 
 In (1) substitute 1 fora:, and 3 forz, l+y 3 = 0; y = 2. 
 Ans. 1, 2, 3. 
 
 Be sure that after eliminating one unknown between two pairs of the 
 equations, you have two equations containing the same two unknowns. 
 
 Should there be more than three unknowns, reduce the system to a 
 system containing one less equation and one less unknown. Continue 
 this process until you have one equation with one unknown. 
 
 EXERCISES 
 163. Solve and check : 
 
 1. 2z-3y + 4z = 20, 7. 2r-3s = 5, 
 
 3x + y-z = S, 3r + t = W, 
 
 5x- + 3z = 28. s-2t = 5. 
 
 2. x + y + z = 2, 8 . 
 
 x y + z = 16, a 2c = 9, 
 
 x + y-z=-2. 7c-46 = 44. 
 
 3. 2x-5y-4:z= -4, 1,1,1 
 z +2 , + z = 0, 9. - + - + - = 6, 
 
 4a-3y-22=-4. 1_1 + 1 = 2 
 
 4. 4a-36 + 3c = 20, a~~ b c~ 
 
 -7a + 26 + 6c = 5, 1 + 1 _ 1 _ Q 
 
 8a & + 2c = 45. a 6 c 
 
 5. 2a-3& + 4c = 2, 1 1_ 
 3a-3&-15 = 0, 10 ' x + y~ ' 
 7a-4c-31 = 0. 1 
 
 6. 
 
 10 n +3p = -2.
 
 SIMULTANEOUS LINEAR EQUATIONS 135 
 
 , -l9 - ., 
 
 11. , n - L "j , -, o 
 
 a o c a o C+ a= .d, 
 
 5 6 7 _ _ 2 g a 
 
 a 6 c~ a + 6-4c+3d=-.8. 
 
 1 1,8 
 
 ---- - - = 43. 
 
 PROBLEMS 
 
 164. 1. The sum of two numbers is 78; their difference is 
 4. Find the numbers. 
 
 2. The sum of two numbers is 76; their difference is 26. 
 Find the numbers. 
 
 3. If the difference of two numbers is 12 and their sum is 
 45, what are the numbers ? 
 
 4. There are two numbers whose sum is 25. If the larger 
 be subtracted from the smaller, the remainder is 6. Find 
 the numbers. 
 
 5. Twice a certain number plus another number is 35. The 
 difference between the two numbers is 1. What are the 
 numbers ? 
 
 6. The sum of two numbers is 72. If the second be sub- 
 tracted from twice the first, the remainder is 3. Find the 
 numbers. 
 
 7. The difference between two numbers is 10 ; their sum 
 is 98. What are the numbers ? 
 
 8. The sum of two numbers is 26 ; their difference is 2. 
 Find them. 
 
 9. A father earns $45 more a month than his son. 
 Together they earn $125 a month. What is the monthly 
 salary of each ? 
 
 10. A real estate dealer purchases two houses for $11,250; 
 one house is worth $250 more than the other. Find the 
 value of each.
 
 136 ELEMENTARY ALGEBRA 
 
 11. A mother is 21 years older than her daughter ; their 
 combined ages are 71 years. Ascertain the age of each. 
 
 12. Five pounds of sugar and three dozen of eggs cost $ 1.20 ; 
 seven pounds of sugar and five dozen of eggs cost $1.92. 
 What will one dozen of eggs and one pound of sugar cost ? 
 
 13. I bought 10 yards of silk and 5 yards of cambric for 
 $14. If I had bought 2 yards less of silk and 3 yards more 
 of cambric, my bill would have been $ 1.60 less. What is the 
 price of each per yard ? 
 
 14. Find two numbers such that the greater exceeds the 
 less by 11, and one fourth the less plus one third the greater 
 is 13. 
 
 15. Find the fraction in which if 1 be added to the 
 numerator its value is 1, but if 6 be added to the denominator 
 the value is . 
 
 16. The sum of three numbers is 29 ; twice the first plus 3 
 equals the second ; twice the third minus 1 equals the second. 
 Find the numbers. 
 
 17. A man hired 3 boys for a day for $4.25. The second 
 boy received half again as much as the first, and the third 
 received 25^ more than the second. How much did each 
 receive ? 
 
 18. Find two numbers such that three times the greater 
 exceeds twice the less by 61, and twice the greater exceeds 3 
 times the less by 4. 
 
 19. A and B have equal sums of money. A spent half of 
 his and B earns $ 10 ; then B has 3 times as much as A. How 
 much had each at first ? 
 
 20. If A gives B $5, A will have ^ as much as B; but if B 
 gives A $25, B will have the same amount as A. How much 
 has each ?
 
 SIMULTANEOUS LINEAR EQUATIONS 137 
 
 21. A can do a piece of work in 3 days, B can do the same 
 work in 4 days. In what time will both together do the work? 
 
 22. A and B together do some work in 4 days, A and C 
 together in 5 days, B and C in 10 days. How long would it 
 take each alone ? 
 
 23. A company of boys bought a boat. If there had been 5 
 boys more, each would have paid $ 1 less ; but if there had 
 been 3 boys fewer, each would have paid $1 more. How 
 many boys were there ? What did each pay ? What did the 
 boat cost ? 
 
 24. A two-digit number has the sum of its digits equal to 9. 
 If 9 be added to the number, the digits will be interchanged. 
 Find the number. 
 
 Let t the digit in tens' place, u = the digit in units' place. Then 
 10 1 + u = the number. 
 
 25. The tens' digit is twice the units' digit. If 36 be 
 subtracted from the number, the digits will be reversed. What 
 is the number ? 
 
 26. The sum of the digits of a three-digit number is 11. 
 The units' digit is twice the tens'. If the digits are reversed, 
 the number will be 396 more than it is. Find the number. 
 
 27. Three cities are located at the vertices of a triangle. 
 The distance from A to B by way of C is 150 miles ; from C 
 to A by way of B is 140 miles ; from B to C by way of A is 
 110 miles. How far apart are the cities ? 
 
 28. If a rectangle were 1 inch longer and 2 inches narrower, 
 it would contain 20 square inches less; if it were 3 inches 
 wider, it would become a square. Find its dimensions. 
 
 29. A train maintains a uniform speed for a certain dis- 
 tance. If the speed had been 10 miles an hour less, the time 
 would have been 1\- hours more ; if the speed had been 10 
 miles an hour more, the time would have been If hours less. 
 Find time, rate, distance.
 
 138 ELEMENTARY ALGEBRA 
 
 30. A man invests a certain sum of money at 4 <f and a 
 second sura at 3\%. His income from the two investments is 
 $445. Had the first sum been invested at 5%, his yearly 
 income would have been $50 more. How much had he in 
 each investment ? 
 
 31. B's yearly income is f of A's, and A spends $500 per 
 annum more than B. At the end of 3 years A has saved 
 $ 4500 and B $ 2250. What is the yearly income of each ? 
 
 32. A person had $ 1750 invested so as to bring in an annual 
 income of $ 77. Part was bearing 4% interest, the rest 5%. 
 How much was invested at 4 % ? 
 
 33. A merchant mixes two qualities of tea in the ratio of 
 two parts of the cheaper to one part of the dearer, and, by sell- 
 ing the mixture at 60 ^ per pound, he gains a profit of 25 % ; 
 on mixing them in the ratio of three parts of the cheaper to 
 two parts of the dearer, and selling the mixture at 65^ per 
 pound, he gains the same rate of profit. Find the prices at 
 which he bought. 
 
 THE BEGINNINGS OF ALGEBRA 
 
 165. The earliest traces of algebra are found in a very old Egyptian 
 papyrus, written by Ahmes about 1700 or 2000 B.C. This papyrus is a 
 copy of a still older document. The Ahmes papyrus was found in 1868 
 and is the most important source of our knowledge of ancient Egyptian 
 mathematics. It contains interesting matters on fractions, geometry, 
 algebra, and even the rudiments of trigonometry. It gives solutions of 
 linear equations. It might be expected that the symbols used 3000 years 
 ago would be as different from ours as their language and speech were 
 different from ours. To give an idea of the writing in the Ahmes papy- 
 rus, we reproduce one equation and place beneath it the English transla- 
 tion and the equation written in modern symbols. 
 
 nnn 
 
 Heap, its , its i. its i. its whole, it makes : 37 
 
 That is, x (f + i + t + *) = 37 -
 
 SIMULTANEOUS LINEAR EQUATIONS 139 
 
 Notice particularly that efl> means |, / means , dZ> is the 
 
 symbol to indicate that 7 is the denominator of a unit fraction, p means 
 10. 
 
 No further progress was made in algebra until the time of the Greek 
 Diophantus (about 300 B.C.), who was very skillful in solving problems. 
 He had a symbol of his own for the unknown x and for x 2 . He had 
 also a symbol to express subtraction and equality. Juxtaposition, which 
 with us represents multiplication, meant addition with him. A most 
 momentous impulse to the development of algebra was given by the 
 Hindus, after the fifth century A.D. They displayed wonderful skill in 
 developing arithmetic and algebra. The Hindu and Greek knowledge of 
 algebra was later transmitted to the Arabs who, in turn, became the 
 teachers of European nations at the close of the dark ages. The early 
 European writers on algebra used hardly any symbols ; everything except 
 arithmetical numbers was fully written out in words. In the sixteenth 
 century the modern symbols began to be invented. These symbols con- 
 stitute a sort of shorthand method of indicating algebraic relations and 
 processes, whereby the power and usefulness of algebra in the solution of 
 problems is very greatly increased. How important they are can be 
 realized when the attempt is made to solve difficult problems without their 
 aid. The sixteenth-century symbols are more clumsy and less convenient 
 than the modern. The great French algebraist, Vieta, wrote in 1595 the 
 polynomial 
 
 2-16 x 2 + 20x* - 8 x 6 + x 8 , thus : 2 - 16 Q + 20 QQ - 8 CO + QCC. 
 
 Here Q (quadratus) stands for x 2 , C (cubus) stands for x 3 . The Portu- 
 guese mathematician, Pedro Nunez, in 1567, wrote 
 
 VV756 + 27 - V V756 - 27 thus : 
 
 R. V. cu. R. 756 p"27 m R. V. cu. R. 756 m 27. 
 
 Here R. V. cu. means radix vniversalis cubica or general cube root, 
 p means plus, m means minus. The notations of Vieta and Descartes 
 will be illustrated by other examples which we place under their portraits 
 opposite pages 110 and 140. The notation of Descartes was nearly the 
 same as the modern notation. 
 
 The sign of equality ( = ) was first used for this purpose by an English 
 writer on arithmetic and algebra, Robert Recorde. The following photo- 
 graph of part of a page in his algebra, entitled the Whetstone of Witte,
 
 140 
 
 ELEMENTARY ALGEBRA 
 
 published in London in 1657. shows the place in that book where the 
 sign = is first introduced. 
 
 ,foj eaffc alteratfo of ff *tf w.3l Ml p;o 
 pounoe a fc IDC era ple0, bkaufe the cr traction of tljetr 
 rootcsjtnaie tije mo;c sptlp bet tojougijte. 3nD 
 uoiae tl)e tcoioufe repetition of tbcfc UJOO:QCS : 
 quallc to : 3 tuill fette as 31 Hoc often in luoo;fec 
 pa( re of pacallelcs,o; Ccmotue lines of one lengtbe, 
 tljns^^^jbtcaurc noe.2. tinges, can be moare 
 equalle. flno no ID marlic thcfe iiombcrs. 
 
 2i3.?; 
 
 9.5- 
 ,n t6c firflc tftcrc appearetfj. 2 , nombcrf , tfjat is 
 
 From Recorde's Whetstone of WUU, 1557. 
 FIG. 27
 
 RENE DESCARTES 
 
 Modern Notation 
 
 Descartes' Notation 
 x 8 * +px + q ooO; 
 
 Descartes used oo as the sign of equality. The * after x 3 signified that 
 the next lower power of x was missing from the expression.
 
 CHAPTER XIII 
 
 QUADRATIC EQUATIONS 
 
 166. The equation x 2 + ax + b = 
 
 is called a complete quadratic equation. It is a quadratic equa- 
 tion because the highest power of the unknown x is the 
 second ; it is complete because it contains also a term involv- 
 ing the unknown x to the first power and a term b (often 
 called the absolute term} which is free from x. 
 Quadratic equations of the forms 
 
 x* = b, 
 
 x 1 + ax = 0, 
 
 are called incomplete quadratic equations, because either the 
 term involving the unknown x to the first power or the abso- 
 lute term b is absent. 
 
 INCOMPLETE QUADRATIC EQUATIONS 
 
 167. Equations of the form x 2 + ax = can be solved by 
 the method of " completing the square," to be explained later. 
 Much easier is the solution by the method of factoring, treated 
 in a previous chapter. 
 
 Factoring, we obtain x(x + a) = 0. 
 
 Making the first factor equal to zero, x = 0. 
 
 Making the second factor equal to zero, x + a = 0. 
 
 x = a. 
 
 Very simple is the solution of x- = b. For greater conven- 
 ience, write it x 2 = c 2 . 
 
 Extracting the square root of both sides, x = c. 
 
 Since the square root of both sides has been extracted, it might be 
 claimed that the sign should be written on both sides, giving 
 
 x = c. 
 141
 
 142 ELEMENTARY ALGEBRA 
 
 But this result is the same as when we write x = c. 
 For the equation x = c means here 
 
 (1) +x = + c. (3) +x=-c. 
 
 (2) - x = - c. (4) - x = + c. 
 
 Of these four sets, the first two are the same, and the last two are the 
 same. Hence, x = c gives all the values of x. 
 
 EXERCISES 
 
 168. 1. Solve - 3 = - + 5. 
 2 2 
 
 Transposing and combining, 
 
 4*2 = 8. 
 
 Hence x 2 = 2, 
 
 and x = \/2. 
 
 Check : ^ - 3 = ? + 5, or 6 = 6. 
 
 Solve the following : 
 2. 4z 2 - = 
 
 3 
 
 11. ^ + 3^=13. 
 
 12. 2 x z + 5.78 x = 0. 
 
 10 7 ~.2 1 ^ ;r 
 
 5. 4 ?/ 2 = 3 ?/ 2 + ^. 
 
 6. 7 a 2 -63 = 7. 14. 3z-A = o. 
 
 2y 2 -8 = 2 T" 
 
 16. Ax '- = 0. 
 
 a; 
 
 Q <-'*"' I -M-i-r ^^ o 
 
 13 17. .01 x 2 - .08 a? = 0. 
 
 COMPLETE QUADRATIC EQUATIONS 
 
 169. In 131 we solved complete quadratic equations by 
 the method of factoring. That method is the best when the
 
 QUADRATIC EQUATIONS 143 
 
 factors can be found by inspection. But there are equations, 
 like x 2 -\-3x + 1 = 0, which cannot be solved by any method 
 of factoring thus far studied. Let us consider another method 
 of solving complete quadratic equations. 
 
 Of what expression is the trinomial a 2 -|- 2 ab + b 2 a perfect 
 square ? Of what is x 2 + 2 bx + ft 2 a perfect square ? What 
 must be added to x 2 + 2 bx, to make the resulting trinomial a 
 perfect square ? How can b 2 be found from x 2 + 2 bx ? The 
 process of finding this third term, 6 2 , when the two terms 
 3? 4- 2 bx are given is called completing the square. 
 
 KULE. To complete the square of x 2 -f- 2 bx, add to this the 
 square of half the coefficient of x. 
 
 ORAL EXERCISES 
 
 170. Complete the square of the following : 
 
 1. x 2 -f 2 x. 3. x 2 4 x. 5. z 2 + 3 z. 7. y 2 + 16 y. 
 
 2. x 2 + 6 a. ( 4. # 2 8 y. 6. x 2 + 5 a. 8. z 2 + 11 2. 
 
 SOLUTION OF QUADRATIC EQUATIONS BY" COMPLETING THE 
 
 SQUARE 
 
 171. 1. Solve the equation x 1 -f 6 x 7 = 0. 
 
 (1) Transpose the term 7 which does not 
 
 involve x or a; 2 . x 2 + 6 x 7 = 0. 
 
 (2) Complete the square in the left member, 
 
 by adding 3 2 to both sides. x 2 + 6 x + 9 = 16. 
 
 (3) Take the square root of both sides. x + 3 = 4. 
 
 (4) Transpose the 3 to the right side. x = 4 3. 
 Simplify. x = + 1 or 7. 
 Checking in the original equation, x 2 + 6 x 7 = 0. 
 
 The value x = 1 gives : The value x = 7 gives : 
 
 1 + 6(1)-7=0. (_7)S + 6(_7)-7=0. 
 1 + 6 - 7 = 0. 49 - 42 - 7 = 0. 
 
 Hence 1 is a root. 7 is a root. 
 
 The reason for substituting the values of x in the original equation, 
 rather than in some equation derived from it, is evident. An error may 
 have been committed in getting the derived equation, but this derived
 
 144 ELEMENTARY ALGEBRA 
 
 equation may be correctly solved. Substituting the values of x in the 
 derived equation would not reveal the error. 
 
 2. Solve 9z 2 -5=-4z. 
 
 (1) Transpose 4 x to the left side and 5 to the right, 
 
 9 x 2 + 4 x = 6. 
 
 (2) Divide both sides by 9, the coefficient of x 2 , 
 
 x 2 + i x = f . 
 
 (3) Complete the square by adding (f) 2 to both sides, 
 
 (4) Find the square root of both sides, x + f = . 
 
 (5) Transposing and combining, x = $ $. 
 
 Checking in the original equation, 9x 2 5 = 4 x. 
 The value x = 1 gives : The value x = f gives : 
 
 9(- I) 2 - 5 = - 4(- 1). 9(*) - 5 = - 4(f ). 
 
 9-6 = 4. ^_6 = _^. 
 
 4 = 4. _^,e_y. 
 
 Hence 1 is a root. Hence | is a root. 
 
 172. To solve complete quadratic equations : 
 
 (1) Transpose the terms in y? and in x to the left side, and the 
 terms free of x to the right. 
 
 (2) Divide both sides by the coefficient of x 2 . 
 
 (3) Complete the square by adding to each side the square of 
 half the coefficient of x. 
 
 (4) When possible, simplify the right side by combining its terms. 
 
 (5) Extract the square root of both sides, and use both signs 
 on the right. 
 
 (6) Find the two values of x. 
 
 EXERCISES 
 
 173. Solve the following quadratic equations by completing 
 the square, and check : 
 
 1. x 2 + 6z-16 = 0. 4. 25 x 2 + 20 x = 21. 
 
 2. cc 2 + 10z = 24. 5. 
 
 3. 4x 2 + 12a; = 27. 6.
 
 QUADRATIC EQUATIONS 145 
 
 7. Solve 9 x 2 + 36 x - 2 = 0. 
 
 (1) Transpose -2, 9 x 2 + 36 x = 2. 
 
 (2) Divide both sides by 9, x 2 + 4 x = - . 
 
 7 
 
 (3) Complete the square by adding 4 to both sides, 
 
 x 2 
 
 (4) Find the square root of both sides, 
 
 8 
 (6) Solve for X, a:=-2 
 
 o 
 
 The value of the radical V38 cannot be computed accurately. Taking 
 \/38 = 6.16 +, we obtain the following approximate values for x : 
 
 x = +6 ' 16 -2 = 2.05 - 2 = .06. 
 
 x = - 2.05 - 2 = - 4.05. 
 
 Except when the approximate values are needed in some computation, 
 it is customary to retain the radical in the expression for x ; thus, 
 
 Solve the following : 
 
 8. 4 y 2 + 20 y = 15. 20. 4 x 2 - 20 x + 5 = 0. 
 
 9. 9y 2 -12?/ = 10. 21. 10 - 7 a? = 25 x\ 
 
 10. 49 z 2 - 28 x = 16. 22. 16 a; 2 -5 a; = 4. 
 
 11. 10z-l = z 2 . 23. 3x + 5 = 9a; 2 . 
 
 12. 81 x 2 = 13 - 54 x. 24. 3 x = 10 - 16 x*. 
 
 13. 21 - 200 x = 100 z 2 . 25. 36 a 2 + 24 a = 5. 
 
 14. 100 f = 25 + 50 y. 26. 2/ 2 + 7y = 13. 
 
 15. 121 z 2 - 22 2 = 8. 27. 13 = a 2 + 14 a. 
 
 16. a* + fa; = l. 28. 21 - 5 a; - 2^ = 0. 
 
 17. 4 a; 2 - 6 a? = 0. 29. 3 a 2 + 8 x = 2. 
 
 18. 17?/ 2 + 34y=0. 30. 5# 2 + 9?/-l = 0. 
 
 19. 2 2 5z = 1. 31. 4 a? 1 = a?. 
 
 L
 
 146 ELEMENTARY ALGEBRA 
 
 32. 2x + 5 = 3x\ 38. z 2 + 2a? + f = 0. 
 
 33. a; 2 - a; + f- = 0. 39. 4 a? 2 + 4 z - 15 = 0. 
 
 34. 16 = 11 x 4- 3 z 2 . 40. 9 a; 2 - 12 a? -5 = 0. 
 
 35. 15 a; = 18 -2 a,- 2 . 41. a; 2 -3a; = 5. 
 
 36. 4a? = 6a? + l. 42. 3 z 2 - 5 = 5z - a? + 2. 
 
 37. 4a 2 5 = 9a. 43. 6 a; 2 7aj 5 = 3z + 5. 
 
 44. Solve a? - 2 a; - 31 = 0. 
 We have x 2 2 x + 1 = 32. 
 
 a; - 1 = \/32. 
 
 The radical V32 can be simplified as follows ( 111): As 32 = 2.16, 
 and 16 is a perfect square, we have V32 = x/16.2 = 4 V2. Proceeding in 
 the same way, show that V36 = 5V2, V82 = 2*V2, V32o 8 = 4aV2a. 
 
 45. Simplify the following radicals : 
 
 V18, V8, 
 
 Solve the following: 
 
 46. cc 2 - 4 & - 4 = 0. 49. 9 x 2 - 12 x 46 = 0. 
 
 47. z 2 - 6z 9 = 10. 50. 16 a; 2 -40 =175. 
 
 48. 4 cc 2 - 12 x = 23. 51. 0^-60; -11 = 0. 
 
 52 . Solve 
 
 05 1 
 
 When x appears in the equation in the denominators of fractions, 
 multiply both sides of the equation by a number which will remove the 
 denominators. In this case multiply by (a; !)(;+ 1). We obtain 
 
 (a; -l)(g + l)_2g(g -!)(+!)_ 
 a; 1 x+l 
 
 Dividing both terms of each fraction by the factor common to those 
 terms, we get
 
 QUADRATIC EQUATIONS 147 
 
 Or x + 1 = 2x(x-l). 
 
 2x 2 -3x = 1. 
 
 4 4 
 
 x = 3 
 
 44 
 
 Solve the following : 
 
 53. -- f = . 
 
 a; 1 2 
 
 54. 
 
 2z-l 
 
 55. ?LJ: = X , 
 
 rf 1 O 
 
 v ^"^ X J 
 
 56. -? =,!. 
 
 x 1 x 
 
 57 =- +- 
 3^ y' 
 
 
 x 4 
 
 X 
 
 
 t><) 
 
 1 
 
 1 
 
 3 
 
 
 * + l 
 
 1 # 4 
 
 10' 
 
 60 
 
 2 
 
 iB 
 
 
 fil 
 
 X + l 
 
 2 
 
 
 
 y + 5 
 
 y + 2' 
 
 
 
 
 
 % 
 
 
 fi2 
 
 *-^ 
 
 
 
 
 2-5 
 
 ~2-4' 
 
 
 PROBLEMS 
 
 174. 1. The sum of two numbers is 10, their product is 24. 
 Find the numbers. 
 
 Let x = one of the numbers, then 
 
 10 x = the other number. 
 
 x(10 x) = their product, which must be equal to 24. 
 Hence x(10 x) = 24. 
 
 x 2 - 10z = - 24. 
 x 2 - 10x + 25= 1. 
 
 x = 1 + 5. 
 x = 6 or 4. 
 10 - x = 4 or 6. 
 Hence the two numbers are 6 and 4. 
 
 2. The product of two numbers is 143; their sum is 24. 
 Find the numbers.
 
 148 ELEMENTARY ALGEBRA 
 
 3. The area of a rectangle is 112 sq. in. ; the sum of its 
 length and breadth is 23 in. Find its length and breadth. 
 
 4. Find the length and breadth of a rectangle whose area 
 is 4 sq. in. and whose perimeter is 8 T ^ in. 
 
 5. The perimeter of a window frame is 18 ft. ; the area of 
 the window is 19^ sq. ft. Find the dimensions of the window. 
 
 6. If a rectangular table is 7 ft. longer than wide, and it 
 contains 36f sq. ft., what are its dimensions ? 
 
 Are both roots of the quadratic equation to which the problem leads 
 applicable to the problem ? 
 
 7. A tinner wants to make a square box 2 inches deep, 
 with a capacity of 72 cubic inches. From each corner of a 
 
 square sheet of tin a 2-inch square is cut ; 
 the four rectangular strips thus made are 
 turned up. What must be the dimensions 
 of the sheet of tin ? 
 
 8. The longer side of a rectangular box 
 exceeds the shorter side 2 in. The box is 
 3 in. deep and has a capacity of 105 cu. in. 
 
 "Pin 28 
 
 If made out of a single piece of cardboard, 
 what were the dimensions of the cardboard, supposing that 
 there is no other waste than the 3-inch squares cut from each 
 corner ? 
 
 9. A flower bed is 12' x 15'. How wide a walk must sur- 
 round the bed, to increase the total area by 160 sq. ft. ? 
 
 10. The area of a rectangle exceeds that of a square by 
 24 sq. in. If the side of the square is equal to half a shorter 
 side of the rectangle, and the longer side of the rectangle ex- 
 ceeds the shorter side by 3 in., find the dimensions of the 
 rectangle. 
 
 11. In a right triangle the hypotenuse is 10 ft. longer than 
 the longer leg, and the shorter leg is 10 ft. shorter than the 
 longer leg. Find all three sides.
 
 QUADRATIC EQUATIONS 149 
 
 12. Two legs of a right triangle are equal. The hypotenuse 
 exceeds the length of a leg by 3 in. Find the three sides. 
 
 13. The product of two consecutive integers is 3540. Find 
 
 them. 
 
 14. The longer leg of a right triangle exceeds the shorter 
 by 5 in. The area of the triangle is 88 sq. in. Find the 
 length of each leg. 
 
 15. Find a number such that, if 20 is subtracted from it, 
 and the remainder is multiplied by the number itself, the 
 product is 189. 
 
 16. A school board has $ 22 to spend on geographies. If 
 the bookseller reduces the price of each book 5 ^, 4 more books 
 can be purchased than at the original price. How many books 
 can be bought at the reduced rate ? 
 
 17. Had a man's daily wage been $ 1 less, he would have 
 been obliged to work 7 days longer to earn $ 140. How many 
 days did the man work ? 
 
 Do both roots apply to the problem ? Why ? 
 
 18. A picture 8" x 12" is placed in a frame of uniform 
 width. If the area of the frame is equal to that of the pic- 
 ture, how wide is the frame ? Draw a figure. 
 
 19. In a trapezoid, AB exceeds CD by 4 in., and CD exceeds 
 the altitude CE by 1 in. The area of the trapezoid, com- 
 puted by taking the product of the 
 
 altitude and one half the sum of 
 the parallel sides, is 28 sq. in. Find 
 the lengths of the parallel sides and 
 
 of the altitude. A B 
 
 FIG. 29 
 
 20. The difference between the 
 
 parallel sides of a trapezoid is 20 ft. ; the altitude is equal to 
 the shorter of the parallel sides ; the area is 119 sq. ft. Find 
 the lengths of the parallel sides. 
 
 21. Find two consecutive integers whose product is 30,800.
 
 150 ELEMENTARY ALGEBRA 
 
 22. Find two consecutive even numbers whose product is 
 1088. 
 
 Hint. Let x be an integer ; then 2 x must be an even integer. How 
 much must be added to 2 x to get the next higher even integer ? 
 
 23. Find two consecutive odd numbers whose product is 
 783. 
 
 Hint. If 2 x is an even integer, what must be added to 2 x to get the 
 next higher odd integer ? 
 
 24. The edges of a rectangular box (with cover) are in the 
 ratio 2:3:6. If the entire surface is 648 sq. ft., find the 
 lengths of the edges. 
 
 Hint. Let the number of feet in the edges be 2 x, 3 x, 6 x, respectively. 
 
 SOLUTION OF THE GENERAL QUADRATIC EQUATION 
 
 175. The equation ax+bx + c = is called the general 
 quadratic equation, in which the coefficients a, b, c may be 
 any numbers. 
 
 Let us solve it by completing the square. 
 
 (1) Divide both sides by a 
 
 C i& J_^ C 
 
 and transpose - to the right or -\--x- 
 
 n 
 side. 
 
 (2) Complete the square by & / 6 V & 2 c 
 adding fY to both sides. a \W 4 2 a 
 
 (3) Perform the indicated r2 i & / & y_6 2 -4ac 
 
 ' * 1 o~~ ) i 2 
 subtraction on the right side. \* a J 
 
 (4) Find the square root of b . V& 2 4 ac 
 
 x -\ = x ' 
 
 both sides. 2 a 2 a 
 
 (5) Transpose to the b , V& 2 - 4 ac 
 
 v / o n x = x " 
 
 right side. 2 a 2 a 
 
 (6) Another way of writing _ b V& 2 4 ac < ^ j 
 the answer. 2 a
 
 QUADRATIC EQUATIONS 151 
 
 SOLUTION OF QUADRATIC EQUATIONS BY FORMULA 
 EXERCISES 
 
 176. 1. Solve 3z 2 + 5z+2=0. 
 
 Here a = 3, b = 5, c = 2. Substitute these numbers in the formula 
 I above. We obtain 
 
 = 2 
 
 6 63 
 
 Check : (1) 3(- f) 2 + 5(- f ) + 2 = 0. 
 
 | - -^ + 2 = 0. 
 = 0. 
 
 (2) 3(-l) 2 + 5(-l)+2=0. 
 
 3 _ 5 + 2 = 0. 
 = 0. 
 Solve: 
 
 2. 5z 2 -7a + 2 = 0. 8. iz 
 
 3. 
 
 4. 3 x 2 - a? -10 = 0. 
 
 10. .3a^ + .2x .5 = 0. 
 
 5. 5x 2 - 6 x 3 = 0. 
 
 6. x*- 
 
 n - 
 
 7. fic2-a;-2=0. 12. x 2 lOlx + 7 = 0. 
 
 MISCELLANEOUS QUADRATIC EQUATIONS 
 EXERCISES 
 
 177. Solve by the method of factoring, whenever the trino- 
 mial can be factored at sight. Otherwise solve by the formula 
 or by completing the square. 
 
 1. z 2 - 7 a; -18 = 7. x* + 24 = 10 x. 
 
 2. x 2 - 6 x - 8 = 0. x 7 
 
 8. 
 
 3. x* + 32 = -lSx. a + 60 3 a; -5 
 
 4. 2a^ + 5x + 2 = 0. 9. x* + 3x = 18. 
 
 5. .2 x 2 + x .5 = 0. 10. x* + 7 x = 0. 
 
 6. 1.5 ^ + 50: + 1.2 = 0. 11. 40 = z 2 + 3z.
 
 152 ELEMENTARY ALGEBRA 
 
 12. X* - 5 x + 6 = 0. 14. .1 x 2 - .1 x = 10. 
 
 13 3+12. x ^26 15. (*- 3) = (aj -!)(* + 1) 
 
 x x + 12 5 16. a; 2 -8 a; + 15 = 0. 
 
 17. 4 x 2 1 x - 2 = 0. 
 
 18. (aj-2)(*-3) = (a; 5)(aj + l). 
 
 19. (x f)(* + 2) = 0. 
 
 20. x(x + iQ) = (a: l)(a? 2). 
 
 21. 6z 2 - 19 a: + 10 = 0. 
 
 22. a; (a? a? 1) = (a? 1) (a? + 1). 
 
 
 7 n& 
 
 
 
 -J Cfi 
 
 97 
 
 
 1 
 
 
 , 2 
 
 r 
 
 358. 
 
 
 
 
 
 
 rr 3 
 
 D 
 
 i 
 
 1 
 
 
 I" 
 
 t)A 
 
 1 
 
 in 
 
 Q 
 
 
 OQ 
 
 
 a; 
 
 
 1 1 
 
 
 
 X 
 
 
 
 
 
 a; 
 
 - 
 
 1 
 
 * 3 
 
 
 OK 
 
 1 
 
 i 
 
 1 
 
 _ i 
 
 29 
 
 X 
 
 
 
 1 
 
 1 
 
 Q 
 
 
 X + 
 
 1 ' 
 
 X 
 
 1 
 
 
 
 x 
 
 
 <v 1 
 
 v "^ X 
 
 
 no 
 
 1 
 
 i 
 
 1 
 
 o 
 
 OA 
 
 1 
 
 
 
 
 
 
 X 2 
 
 1 ' 
 
 a; 
 
 
 
 a; 
 
 
 
 
 
 IMAGINARY ROOTS 
 
 178. Solve x* + 2 x + 5 = 0. 
 
 Complete the square, a; 2 + 2a; + l = 4. 
 
 Extract the square root, a; + l = V 4. 
 
 x = -1 V^l. 
 
 We cannot express V 4 as a number of the type studied 
 thus far : 
 
 V- 4 is not + 2, because + 2)(+ 2) = +4. 
 
 V^l is not - 2, because (- 2)( 2) = -f- 4. 
 
 V 4 is a new type of number; it is called an imaginary 
 number.
 
 QUADRATIC EQUATIONS 153 
 
 Since 4 =(+ 4)( 1), we may write 
 
 The roots of the quadratic equation may be written 
 
 We call V 1 the imaginary unit. The geometrical mean- 
 ing of imaginary numbers will be explained in the advanced 
 course. Whenever a quadratic equation yields roots involving 
 the square root of a negative number, it is evident that some 
 condition is called for which cannot be satisfied by the ordi- 
 nary positive and negative numbers. 
 
 For example, find two numbers whose sum is 2 and whose 
 product is 3. 
 
 Let x = one of the numbers. 
 
 Then 2 x = the other number, 
 
 and os(2 - a;) = 3. 
 
 Solving, x 2 - 2 x = - 3, 
 
 _ 
 
 x - 1 = V- 2, 
 x = lV^2. 
 
 The solution involves V 2 ; it shows that no two numbers of the 
 kind thus far studied can satisfy the conditions. Only these new num- 
 bers, these imaginary numbers, satisfy the conditions. 
 
 
 EXERCISES 
 
 Solve : 
 
 1. x 2 + x + 1 = 0. 3. 2 x 5 = tf. 
 
 2. x 2 + 2 z + 6 = 0. 4. 2^ + 33 + 4 = 0. 
 
 GRAPHIC SOLUTION OF QUADRATIC EQUATIONS 
 179. 1. Solve graphically j+2z-i = 0. 
 
 We draw the graph of y = x + 2z 1. Assume different values for 
 4 
 
 x 2 
 
 x and compute 2x, , and y, as shown in the following table : 
 4
 
 154 
 
 ELEMENTARY ALGEBRA 
 
 X 
 
 2x 
 
 x 2 
 4 
 
 - 1 
 
 y 
 
 2 
 
 4 
 
 1 
 
 -1 
 
 4 
 
 1 
 
 2 
 
 I 
 
 -1 
 
 H 
 
 
 
 
 
 
 
 -1 
 
 - 1 
 
 -1 
 
 -2 
 
 i 
 
 - 1 
 
 -2| 
 
 -2 
 
 -4 
 
 1 
 
 -1 
 
 -4 
 
 -3 
 
 -6 
 
 2 i 
 
 -1 
 
 -4| 
 
 -4 
 
 -8 
 
 4 
 
 -1 
 
 -5 
 
 -5 
 
 - 10 
 
 6J 
 
 - 1 
 
 -4f 
 
 -6 
 
 -12 
 
 9 
 
 - 1 
 
 -4 
 
 -7 
 
 - 14 
 
 121 
 
 - 1 
 
 -21 
 
 -8 
 
 -16 
 
 16 
 
 - 1 
 
 -1 
 
 -9 
 
 - 18 
 
 20} 
 
 - 1 
 
 11 
 
 -10 
 
 -20 
 
 25 
 
 -1 
 
 4 
 
 FIG. 30 
 
 When y = 0, we have *- + 2 x 1=0. Hence values of x in the graph 
 4 
 
 which correspond to y = furnish roots of the equation ?- + 2x 1=0. 
 
 4 
 
 We see that y = at the points A and B. At ^4, x = 8, nearly ; at B, 
 x = , nearly. Hence 8 and -f are approximations to the roots of 
 the quadratic equation. 
 
 Solve graphically: 
 
 3. a 8 
 
 TWO SIMULTANEOUS EQUATIONS: ONE LINEAR, THE OTHER 
 QUADRATIC 
 
 180. I. Solve the two equations y = x"-, 
 
 To find the values of x and y that will satisfy both these equations, 
 eliminate either x or y. The elimination of y is easier. Writing in the 
 first equation 2 x + 3 in place of y we obtain
 
 QUADRATIC EQUATIONS 
 
 155 
 
 2 x + 3 = x 2 . 
 x 2 - 2 x + I = 4. 
 x - 1 = 2. 
 
 x = 3 or - 1. 
 
 From the equation y = 2 x + 3, we see that, when 
 
 x = 3, y = 9, 
 and when x = 1, y = 1. 
 
 The same values of y are obtained with even greater ease from the 
 equation y j.2 
 
 f x=-l 
 
 The answers are 
 
 and 
 
 It is important to study the geometrical meaning of problems 
 like this. 
 
 In the equation 
 
 give various values to x and 
 compute the corresponding 
 values of y. We obtain, 
 say, 
 
 x=2, 1, 0, -1, -2, -3. 
 y = 7, 5, 3, 1, -1, -3. 
 
 Plotting the pairs of 
 values of x and y, as is 
 done in Fig. 31, we see 
 that all of them lie on a 
 straight line. 
 
 In the equation y = x\ 
 give different values to x 
 and compute the corre- 
 sponding values of y, we 
 obtain pairs of values, as 
 follows : 
 
 x = 3, 2, 1, 
 y = 9, 4, 1, 
 
 (3 
 
 FIG. 31 
 
 0, -1, -2, -3. 
 0, 1, 4, 9.
 
 156 ELEMENTARY ALGEBRA 
 
 These points lie on the curve shown in Fig. 31. This curve 
 is called a parabola. 
 
 We see that the straight line cuts the parabola in the two 
 points A and B, having the coordinates x = 1 and y = 1, 
 x = 3 and y = 9. These points of intersection represent the 
 values ofx and y which satisfy both of the two given equations. 
 
 II. Solve x = 2y z . 
 x + 2y = . 
 
 Here x is easier to eliminate. In the second equation write 2 y 2 for x. 
 We obtain 2 y 2 + 2 y = 4. 
 
 (1) Divide both sides by 2, t/ 2 + y = 2. 
 
 (2) Complete the square, 2/ 2 + J/+i = f. 
 
 (3) Take the square root of both sides, y + = |. 
 
 (4) Transposing, / = $ = 1 or 2. 
 
 (5) From x+2 y=4, we get se= 2 when y = 1, and a; = 8 when y = 2. 
 
 The answers are I x = 2 and x = 8 
 
 = 1 y = - 2. 
 
 Solve the following: 
 
 1. / = 4a;, 6. 2a+3 y 0, 
 # + 2^=5. a 2 4y = 0. 
 
 2. 3y 2 = 4z, 7. y 2 + 3z=0, 
 
 2 a; + 4 y = 14. 6 a + y - 15. 
 
 3. x* = 5y, 6. ?/ 2 +4o; = l, 
 
 5 y 60 = 4 a. x- 5y = 23. 
 
 4. a 2 = 10 y, 9. y 2 - 2 a? + 7 = 0, 
 
 5. 3x = 7y 2 , 10. z 2 + 3 y + 5 = 0, 
 
 PROBLEMS 
 
 181. 1. The sum of two numbers is 18; the square of one 
 number is equal to 3 times the other. Find the two numbers. 
 
 Let x = one number, 
 
 and y the other number.
 
 QUADRATIC EQUATIONS 157 
 
 Then x + y = 18, 
 
 and x 2 = 3 y. 
 
 Solving, we get from the first equation, 
 
 y = 18 x. 
 
 Eliminate y, x 2 = 3(18 x) = 54 - 3 x. 
 
 Transpose, x 2 + 3 x = 64. 
 
 Complete the square, x 2 + 3 x + (|) 2 = 54 + f = *$&. 
 Extract the square root, x + f = -^. 
 
 x = -^ - f = 6 or - 9. 
 
 / = 18 x = 12 or 27. 
 
 The answers are I x ~ 6 and I X = ~ 9 
 [ 3, = 12 If = 27. 
 
 This problem could have been solved by the use of only one unknown 
 quantity. 
 
 Using both x and y solve the following : 
 
 2. Find two numbers such that 10 times the square of one 
 is equal to 3 times the other, and their sum is 33. 
 
 3. Find two numbers whose difference is |, the square of 
 the larger being equal to f times the smaller. 
 
 4. Five times one number is equal to 3 times another. The 
 square of the smaller is equal to 18 times the other. Find the 
 numbers. 
 
 EXERCISES 
 
 182. 1. Solveajy = 6. 
 
 x y = 5. 
 
 The second equation gives x = 5 + y. 
 
 Substitute in the first, (5 + y~)y = 6. 
 
 Or 2/2 + 5 y _ e _ o. 
 
 Factor, (y + 6) (y 1) = 0. 
 
 y=-6, +1. 
 Substitute in x = 5 + y, x = 1, when y =6. 
 
 x = 6, when y = 1. 
 
 The answers are I * ~ and J x ~ 
 
 \y=-Q U = l. 
 
 Let us plot the equation x y = 5 ; compute values as follows : 
 y = '2, 1, 0, -1, -2, -6. 
 z = 7, 6, 5, +4, +3, -1.
 
 158 
 
 ELEMENTARY ALGEBRA 
 
 *y: 
 
 \ 
 
 (6,-1-V 
 
 The graph is a straight line, 
 as shown in Fig. 32. 
 
 Plot next xy = 6. We have 
 
 y = 6, 3, 2, 6, -1, 
 -2, -3, -6. 
 
 x = 1, 2, 3, 1, - 6, 
 - 3, - 2, - 1. 
 
 The pairs of points which 
 are here computed lie on 
 a curve composed of two 
 branches, called hyperbola. 
 
 The straight line and the 
 hyperbola intersect in two 
 
 points, 
 
 FIG. 32 
 
 x = 6 and y = 1, 
 
 x = 1 and y = 6. 
 These pairs of values are the answers to the simultaneous equations. 
 
 Solve the following : 
 
 2. z + ?/ = 9, 
 xy = - 36. 
 
 3. x + y = 23, 
 xy = 132. 
 
 4. x + = 
 
 6. o#+y = 40, 
 x-y = 2. 
 
 7. xy + x = 36, 
 a*/ =-30. 
 
 8. 
 
 5. xy 18.75, 
 
 9. 
 
 x = 9, 
 + y = 5. 
 
 10. Solve a? + y 2 = 25. 
 
 a + y = 7. 
 
 (1) The second equation gives x = 7 y. 
 
 (2) Write (7 y) in place of x in the first equation. 
 
 (7 - j/) 2 + j/ 2 = 25. 
 49 14 y + y 2 + y 2 = 26. 
 
 2 y* - 14 y = - 24.
 
 QUADRATIC EQUATIONS 
 
 159 
 
 The answers are 
 
 x = 7 ?/ = 3 or 4. 
 and 
 
 fx = S 
 < . 
 \ f = 4 
 
 = 3. 
 
 The graph of x + y = 1 can be drawn from the following values : 
 
 x = 3, 1, 0, -1, -2. 
 
 y = 4, 6, 7, 8, 9. 
 These pairs of values are points lying on the line in Fig. 33. 
 
 (3;4 
 
 (4,3) 
 
 FIG. 33 
 
 To find the graph of x 2 + y 2 = 25, assume values of x and compute 
 corresponding values of y as follows : 
 
 z = 5, 4, 3, 2, 1, 0, _-!,_ -2, -3, -4, -5. 
 
 2/=0, 3, 4, 4.6, 4.9, 5, 4.9, 4.6, 4, 3, 0. 
 
 Notice that for every value of x (except x = 5 and 5) there are two 
 corresponding values of y. The pairs of points all lie on a circle. 
 
 11. 
 
 12. 
 
 x - y = 1. 
 
 2x -y =3.
 
 160 ELEMENTARY ALGEBRA 
 
 13. tf+ y z = 29, 20. x 2 + f = 2, 
 a; 2y =1. 5x # = 4. 
 
 14. 3z + 2y =37; 21. 4z + y = 50, 
 x 2 + t/ 2 = 113. z 2 + y 1 = 200. 
 
 15. a?-y 2 = 0, 22. 7 a; + y = 17, 
 a + 2y =15. a5* + 3f = 13. 
 
 16. a? - 2 y* = - 17, 23. 3 x 2 - f = 242, 
 a -f y = 4. # y = 0. 
 
 17. z 2 + 2</ 2 =36, 24. 2x z + 3y* = 5, 
 a? + y = 6. a; + y = 0. 
 
 18. 2s 2 -y 2 = 14, 25. 3z 2 -2?/ 2 = 43, 
 x y = 5. x y =1. 
 
 19. a? + t/ 2 = 221, 26. 2 y 2 - 40? = 82, 
 2x-3y= -13. 3 + 6 y = 41. 
 
 HISTORY OF QUADRATIC EQUATIONS 
 
 183. While the earliest known solution of equations of the first degree 
 is found in the Ahmes papyrus, about 2000 B.C., the earliest known solu- 
 tions of quadratic equations occur in Euclid's Elements of geometry, 
 about 300 B.C., about 1700 years later. Euclid solved quadratic equa- 
 tions by geometric construction. In proposition 11, of Book II, of the 
 Elements, Euclid solved by drawing lines the problem : To divide a 
 given straight line into two parts, so that the rectangle contained by the 
 whole line and one part of it may be equal to the square on the other part. 
 In algebra, this problem demands the solution of the quadratic equation 
 a(a x) a; 2 , where x is the length of the required line. About two 
 centuries after Euclid, a Greek writer, Heron, gives a root of a quadratic 
 equation, showing that he knew how to solve the equation algebraically, 
 but he nowhere tells how he did it. Equally non-committal is Diophan- 
 tus, the celebrated Greek writer on arithmetic and algebra, of the fourth 
 century A.D. Diophantus gives correct answers to several quadratic 
 equations. Part of his book is lost ; the parts that are extant nowhere 
 explain his mode of solution. Diophantus does not give more than one 
 root of a quadratic equation, even if both are positive, nor does he ever 
 recognize a negative root. He nowhere uses negative numbers. In 
 that respect his algebra was like our school arithmetics.
 
 QUADRATIC EQUATIONS 161 
 
 Marked advance in algebra was made by the Hindus. In the fifth 
 and sixth centuries they solved quadratics by a process which can be ex- 
 plained easiest by using our modern notation. They wrote the roots of 
 
 ax 2 + bx = c ; thus : x = - - -- Somewhat later Qridhara, an- 
 
 V4 ac + 6' 2 b 
 
 other Hindu mathematician, uses the form of x = Toob- 
 
 2a 
 
 tain the first form of x one begins by multiplying both sides of ax 2 + bx = c 
 by a. To obtain the second form of x, one begins by multiplying both 
 sides of the equation by 4 a. This last process is now called the " Hindu 
 method" of solution. It has the advantage of excluding fractions under 
 the radical sign. 
 
 The Hindu mathematician Bhaskara (twelfth century A.D.) recognized 
 that a quadratic equation has two roots, when both of them were positive. 
 He gives x = 50 or 6 as the roots of x 2 45 x = 250, and remarks : 
 " But the second value is in this case not to be taken, for it is inadequate ; 
 people do not approve of negative roots." Thus negative roots were seen, 
 but not admitted. The recognition of negative roots came much later. 
 The Italian mathematician, Cardan (1539), and the Belgian mathematician, 
 Simon Stevin (1585), admitted negative roots. Soon after, Albert Girard 
 (1629) took the still more advanced position of admitting the possibility of 
 imaginary roots, thereby recognizing the theorem that every quadratic 
 equation has two roots or, more generally, that every equation of the nth 
 degree has n roots. But opposition to the acceptance of imaginary roots 
 continued for over a century after Girard. Not until the beginning of 
 the nineteenth century did all opposition to them vanish. This opposition 
 to imaginary roots grew mainly out of the fact that they could not be 
 explained geometrically. A positive root could be pictured to the eye by 
 the length of a line drawn in a given direction ; a negative root could be 
 interpreted by the length of a line drawn in the opposite direction. But 
 how could imaginary roots be consistently represented by lines ? Such 
 geometric representation is, however, possible. How this can be done 
 was first shown by the Danish engineer, C. Wessel, in 1797. Most influ- 
 ential in securing the general recognition of imaginary numbers was the 
 great German mathematician, Gauss.
 
 CHAPTER XIV 
 
 FRACTIONS 
 
 184. We have seen in previous exercises that fractions in 
 algebra are subject to the same general rules as fractions 
 in arithmetic. In algebra, as in arithmetic, the value of a 
 fraction is not changed, if both of its terms are multiplied by the 
 same factor or if both of its terms are divided by the same factor. 
 
 2 4 a ac ax + bx a + b 
 
 Thus, 
 
 3 6 6 be ax bx a b 
 
 The most common mistake made by beginners is to say that - - is 
 
 x z + 6 
 
 equal to . But this answer cannot be obtained by multiplying or 
 6 
 
 dividing both terms of the given fraction by any number. 
 
 EXERCISES 
 
 185. 1. Is ^^ equal to when x = 1 ? When x = 2? 
 
 ^ + 6 6 
 
 2. Show by an example that subtracting a number from the 
 numerator and denominator changes the value of a fraction. 
 
 3. Which of the following equalities are correct ? 
 
 3 + l_ 1 (a + 6)(a -6) __ a + b a + a 2 __ a 
 3-2 -2' (a + 26)(a 6) a + 26' b - a 2 b' 
 
 4. Reduce to the lowest terms, 
 
 ' 
 
 Factor numerator and denominator, then divide both terms by their 
 common factors. We get 
 
 3 
 
 (a 2 - 4 y 2 ) _J2-a^-(a: + 2y)(j J -J>-y)' = 3(x + 2 y) 
 - 2 y) ltt^(v-~-2rfj 4 x 
 
 162
 
 FRACTIONS 163 
 
 Reduce to the lowest terms : 
 
 * _i56 19 6(g - 6)(6 - c) 
 
 (36-3a)(46-4c) 
 
 6. 20 a2 - 2a ~ 3 . 
 wy a-3 
 
 7. mx . 21 a 2 +a6-66 2 
 *y 5(o-26) ' 
 
 8 a(x + y) 22 2a 2 6(a 3 -63) 
 
 a(* + y) ' 6 ab(a 2 + ab + 6 2 ) ' 
 
 9 . *<? + y c \ 23 m 2 -! 
 
 *r ye (l-m)(m 2 
 
 10. . ./fot 1 )' .. 24. ( 2 y~ 1 ) 2 
 
 1X da; + dy 25 n 3 - n 2 - 12 n 
 4a; + 42/ 4n 2 -16n 
 
 12. MaP. 26 a-ay. 
 abx (y _ a-)2 
 
 13 X + y? - 27 5a ~ 106 
 2/ + ^ 26-a 
 
 \2 
 
 14. C^-y) . 28 ma; + my 
 y x nx + ny 
 
 15 a 2 -*/ 2 . 5a + 56 
 
 *' T^; T^ ' 
 
 y x I5a 2 b + 15ab 
 
 rtS. W 2 ^,2 Ki t A 
 
 16. ^ 2L. 30. 
 
 (x + y) 2 "V-y 
 
 17 12 cc 2 ^ + 2)(cc + 4) 4 cc 2 4 xy + y 2 
 
 9 a;y 2 (a; + 4) 2 ay 2 ax 
 
 18 (.V-a;)(.^ 2 + a; 2 ). 32 ^3y-a;) 2 
 
 Qu 11 *C O V
 
 164 ELEMENTARY ALGEBRA 
 
 MULTIPLICATION AND DIVISION OF FRACTIONS 
 
 186. In algebra we have the same rules of operation as in 
 arithmetic ; namely, 
 
 1. In finding the product, multiply the numerators together for 
 a new numerator, and the denominators together for a new de- 
 nominator. 
 
 2. In finding the quotient, invert the divisor and then proceed 
 as in the multiplication of a fraction by a fraction. 
 
 Thus, $-*.$ = jx$ = tt. 
 
 The correctness of these rules is shown in arithmetic. How- 
 ever, it is easy to establish the truth of them by means of the 
 equation. 
 
 For instance, the second rule may be proved as follows : 
 
 Let ... >l=x, J = y, 
 
 then a = bx, c = dy, 
 
 and ad = bdx, be = bdy. 
 
 Dividing equals by equals, 
 
 ad _ bdx _ x 
 be bdy y 
 
 -D * x a c 
 
 But - = --*- _. 
 
 y b d 
 
 TT a c ad 
 
 Hence - -5- - = 
 
 6 d be 
 
 This proves the rule for the division of one fraction by 
 another. 
 
 Notice that the rules for the multiplication and division of 
 fractions are still applicable when in place of one of the frac- 
 tions we have an integral expression ; we need only write the 
 integral expression with the denominator 1. 
 
 Thus, .c=U.5 = ?. 
 
 b bib 
 
 --5-C = --*--__- 1_ <L. 
 
 b b I b c be
 
 FRACTIONS 165 
 
 Perform the indicated operations and simplify by reducing 
 the resulting fraction to its lowest terms : 
 
 1 120 . 1 5.5 
 
 * T5T T5TP 
 
 Here 
 
 3 31 
 
 aer.^s _ 31 _ 31 
 
 S5r.ieCF 4-17 68' 
 
 17 
 
 x + y 
 
 x 2 -4/ ' x- 1y 
 Inverting the divisor and multiplying, we obtain 
 
 (x 2 -4j/*)(z + y) 
 
 It is usually best not to carry out the indicated multiplications in the 
 numerator and denominator. On the contrary, it is usually better, if 
 possible, to factor still further, with the purpose of exposing to view all 
 the factors that are common to the numerator and denominator. 
 
 Factoring and then dividing both terms by their common factors, we 
 obtain 
 
 x + 2y 
 _ a b . a 2 - ab , . 15 m 30 3m 
 
 O. r- 1U. 
 
 a 2 + ab a- b 2 2m 5m 10 
 
 4 a 3 2 a 2 y z x-. 
 
 4. 
 
 b 2 b x xy x 2 
 
 g _6f^.36_ 10 7a;-14y . l(x- 
 
 6. 
 7. 
 8. 
 
 9 ' 3 ' a - 1
 
 166 ELEMENTARY ALGEBRA 
 
 2x 2 + 5x + 3 2x 2 + x-, 
 
 22. 
 
 24 
 
 - 1 x + 1 
 
 056 a + x 
 
 ax + x 2 2x 2 +7x-15 
 
 3 x*y + 6 y? ' 6 ?/ 2 + 15 y + 6 
 a*-! 
 
 x y 
 
 COMPLEX FRACTIONS 
 
 187. A complex fraction can usually be simplified most 
 easily by multiplying both its numerator and denominator by 
 the lowest common denominator of the fractions occurring in 
 the numerator and denominator of the complex fraction. 
 
 a 6 
 
 1. Simplify - _ Z. 
 a_b 
 
 x y 
 
 The 1. c. d. of the minor fractions - and - is xy. Multiply both the 
 
 x y 
 
 numerator and the denominator of the complex fraction by xy. We 
 obtain 
 
 a. 
 
 
 
 a ,b 
 x y 
 
 ay + bx 
 
 
 
 a _ b 
 
 ay bx 
 
 
 
 x y 
 
 
 4 
 
 5 
 
 _6 
 
 
 a 
 
 b 
 
 c 
 
 
 1 
 
 3 
 
 _4 
 
 
 a 
 
 b 
 
 c 

 
 FRACTIONS 167 
 
 Multiply both terms of the complex fraction by abc. 
 
 4 5_6 
 
 Th en a b c _ 4 be + 5 gc 6 ab 
 
 1 _ 3 _ 4 ~~ be - 3 ac - 4 aft ' 
 a 6 c 
 
 25 
 
 49 
 3. Simplify 
 
 42 
 
 In this case the process explained in arithmetic is easier : 
 Invert the divisor and multiply. 
 
 25 5 6 
 
 42 7 7 
 Simplify: 
 
 _2 1_ _3 1_ 22 
 
 4 5 a; ly 4m 3n 63 
 
 ^+i' 7TX" W 
 
 5x ly mn 72 
 
 5. J2_J5. 8. +2m + 1 . 11. tl. 
 
 w , m ^ _ 1 _o* 
 
 m n 2m + l 6 2 
 
 1 ^ ^ 9 
 
 H . -L O . O 4 l 
 
 K ' a 
 
 e. ;. 9. J-4- 12. 
 
 o J- mi i 
 
 ~m 60~20 
 
 13. ^^ 1 +y 
 
 1x 1+y
 
 168 ELEMENTARY ALGEBRA 
 
 THE HIGHEST COMMON FACTOR 
 
 188. Only expressions which are free of fractions and of 
 radicals will be considered here. 
 
 The highest common factor (h. c. f.) of two or more expres- 
 sions is the product of such prime factors as occur in each ex- 
 pression, every factor being taken the least number of times it 
 occurs in any one expression. 
 
 This definition indicates the process of finding the h. c. f. 
 
 1. Find the h. c. f. of 12a 2 6 3 c, 24a 3 6c 2 , 36 aW. 
 Resolving into prime factors, we obtain 
 
 =2-2-3-a-a-6-&.6-c. 
 = 2-2.2-3-a-a-a-&.c-c. 
 = 2.2-3.3-a-a-a-a-6-6-c-c-c. 
 The h. c. f . = 2 2 3 a a b c = 12 o 2 6c. 
 
 2. Find the h. c. f . of a 2 (a - 2) 2 , ab(a - 2) 3 , 3 a(a - 2) 4 . 
 
 We may write 
 
 a 2 (a-2) 2 =a-a(a-2)(a-2). 
 ab(a - 2)3 = a 6(a - 2)(a - 2)(a - 2). 
 3 a(a - 2)* = 3 a(a - 2)(a - 2)(a - 2)(a - 2). 
 .% theh.c. f. = a(a-2)(a-2) =a(a-2) 2 . 
 
 EXERCISES 
 
 189. Find the h. c. f . of : 
 
 1. a 2 6, a 2 c, a 2 d. 8. 2 a + 2 6, a 2 - b 2 . 
 
 2. aorz/, &##, cxy. 9. a; ?/, cc 2 y 2 . 
 
 3. 10 a 2 , 15 a 3 , 25 a 4 . 10. a 2 - 1, z 2 - 2 a: -f 1. 
 
 4. 6 a 2 6c, 12 aW, 18 a 2 6 3 c2. 11. 2 a 2 + 4 a + 2, 4 a 2 - 4. 
 
 5. a 2 a&, a 2 ac, a 2 + ad. 12. 3 a& + 3 b 2 , a 2 + 2 ab + 6 2 . 
 
 6. mxmy, nxny, pxpy. 13. cc 2 2 a;?/ + y 2 , x* y 3 . 
 
 7. 2 a? + 2t/ 2 , ax 2 + ay 2 . 14. x 2 - 4, z 2 + a; - 6. 
 
 15. x 3 + y 3 ,x 2 -y 2 , tf + 2 xy + y 2 . 
 
 16. m 2 9, m 2 + 5 m + 6, m 2 m 12. 
 
 17. a 2 - 1, a 3 - 1, a& - & - 2 a + 2.
 
 FRACTIONS 169 
 
 18. 3 a 6 - 48 a 2 , 3 a 4 + 13 a 3 + 14 a 2 . 
 
 19. (a? + </) 4 - y) 2 (a + 6), (x + y}\x - y)\a - b). 
 
 20. r 2 6 rs + 8 s 2 , r 2 + 2 rs 8 s 2 , r 2 5 rs + 6 s 2 . 
 
 LOWEST COMMON MULTIPLE 
 
 190. The lowest common multiple (1. c. in.) of two or more 
 expressions is the product of all the prime factors that occur 
 in the expressions, every factor being taken the greatest number 
 of times it occurs in any one expression. 
 
 This definition indicates the process of finding the 1. c. m. 
 
 1. Find the 1. c. m. of 3 a 2 , 6 a 3 , 5 oft 2 . 
 
 3 a 2 = 3 a a ; 6 a 3 = 2- 3- a- a- a; 5 ab 2 = 5 a 6-6 
 Hence the 1. c. m. = 2 3 5 a a a . b b = 30 a 8 6 2 . 
 
 2. Find the 1. c. m. of w 2 + 2 ab + b 2 , a 2 -b 2 , 2 a 2 + 2 ab. 
 
 a 2 -6 2 = Ca + 6)(o-6). 
 2a 2 + 2a6 = 2o(a + 6). 
 Hence the 1. c. m. = 2a(a+6) 2 (a 6). 
 
 EXERCISES 
 191. Find the 1. c. m. of: 
 
 1. x, y, z. 5. a 2 , a 2 + a. 
 
 2. a 3 &, a& 3 . 6. a 2 + a, a 2 a. 
 
 3. b 2 cd, bc*d, bed?. 7. c 2 + cd, cd + d 2 . 
 
 4. 5 , 7 d, 10 e. 8. a 2 + ab, ab b 2 . 
 
 9. a 2 4, a 2 + a 6. 
 
 10. x>-f,x 2 + 2xy + y 2 , (x - y) 3 . 
 
 11. a 2 - b 2 , a 3 - b 3 , a 2 - 2 ab + b 2 . 
 
 12. m + n, m 2 + n 2 , m 2 + 2 mn -\- n 2 . 
 
 13. 3m 3n, m 3 n 3 , m 2 + mn + n 2 . 
 
 14. x 2 + 6 a? + 9, x 2 + 5 x + 6.
 
 170 ELEMENTARY ALGEBRA 
 
 16. a 2 -6a + 8, a 2 -a -12. 
 
 16. 1 - a 2 , 1 - of, 1 + x. 
 
 17. xyz, x*y xy z , tfy xy 3 . 
 
 18. x* 9, x + 3, a? + 10* + 21. 
 
 19. a 2 1, a 3 +'a 2 + a + 1, a 3 - a 2 + a - 1. 
 
 20. x 2 - 4 xy + 4 t/ 2 , a? + 2 ay - 8 y 2 , x 2 - xy 2 y 2 . 
 
 192. Find the h. c. f . and 1. c. m. of : 
 
 1. 24, 36. 6. 25, 75 cd 4 , 100 acU 
 
 2. 96, 144. 7. 3 a, a 2 - 06. 
 
 3. 49, 98, 147. 8. 4 ax, 4 ate - 4 aa?. 
 
 4. 252,216,180. 9. ^-9, a 2 +5z + 6. 
 
 5. 8 xy 3 , 4 afyz, 10 afy 2 z 2 . 10. ace + ab, bx + 6 2 . 
 
 11. x 2 - y\ a? + 2 a*/ + y 2 , + y 3 . 
 
 12. 27 - a 3 , (3 - a) 2 , 4 (3 - a). 
 
 13. a 2 - 2 ab, a 2 - 6 aft + 8 6 2 , a 2 + ab - 6 6 2 . 
 
 14. c 2 cd* 2 a" 2 , cm + dm en d"w. 
 
 ADDITION AND SUBTRACTION OF FRACTIONS 
 
 193. The process is the same as in arithmetic. If the 
 fractions do not all have the same denominator, they must be 
 reduced to fractions having a common denominator. The 
 lowest common denominator (1. c. d.) is obtained by finding the 
 lowest common multiple (1. c. m.) of the given denominators. 
 
 1. Add --- + 
 3a 2 6 
 
 The 1. c. d. of 2 ab 8 , 3 a 2 6, 4 a 2 6 2 is 12 a 2 6 8 . 
 
 x 6a 
 
 -T- 3 a 2 6 = 4 ft 2 , = 
 
 ~ " 4 ft 2 
 
 2 aft 3 6 a 12a 2 6 8 
 
 X 2 4 6 2 
 
 12a 2 & 3 ^4a 2 & 2 = 36, * 36 3ftx " 
 
 4a 2 6 2 3 ft 12a 2 6 3
 
 FRACTIONS 171 
 
 We obtain^ ___ gg_ + 6 as - 
 
 12a 2 6 3 
 
 In practice, much of this work can be done mentally and need not be 
 written down. 
 
 2 2s 63 
 
 (a; - y}\x + y) (x - y)(x + y) 2 ' x 
 The 1. c. d. = (x - y) 2 (x + y) 2 . 
 
 (x - y) 2 (x + y} 2 divided by (x - y) 2 (x + y) = x + y. 
 
 (x - yy 2 (x + y) 2 divided by (x y)(x + y) 2 = x y. 
 
 Hence 2x + Qy 
 
 (x- y) 2 (x + y) (x-y)(x + y) 2 (x- y) 2 (x + y) 2 
 
 194. EXERCISES 
 
 1. Add f, I, T V 3. Add A, 
 
 z 
 
 __ __ 
 3 xy 2 15 x z y 2 
 
 2. Add 5, -., ^. 4. Add ^, -^, 
 
 ' ' ' 
 
 , -., . . , , . 
 
 5 15' 25 5 ' 25' 50 
 
 Change to a single fraction and reduce that to its lowest 
 terms : 
 
 5. "" y * y. 12. i . 
 
 5 10 x + y x y 
 
 6. _? *_. 13. -J L. 
 
 x+y x-y x-y x+y 
 
 7. i -* 14. [ 
 
 (a; y) 2 x y' x(m x) m x 
 
 r* rg* ->/ <K 
 
 8. x 1 15. x 
 
 x + y (x + y) 2 ' x 2 + xy x 2 -y 2 ' 
 
 5 
 
 n 1 g ^ I 
 
 ' x 2 -y 2 x + y' ' 6 x + 15 8 a; + 20' 
 
 10 y \- 1 17 y- 1 y- 2 
 
 (4-V) + / 15y-5 18y-6*
 
 172 ELEMENTARY ALGEBRA 
 
 19 30* | 4 5 
 
 -l 3x-l 
 20. 
 
 - __ -- -- 
 
 (z-x)(x-y} (x~y)(y-z) (y - z )(z - x) 
 
 (a + 6)(o + c) (a + &)(& + c) (6 + c)(o + c) 
 
 EQUATIONS CONTAINING FRACTIONS 
 
 195. In solving equations containing fractions it is usually 
 best to multiply both sides of the equation by an expression 
 which removes all fractions. 
 
 \ a i X . X 2 7 
 
 1. Solve -+ -. 
 
 Multiplying both sides by 6, x + 2 x 4 = 14. 
 
 3z = 18. 
 a; = 6. 
 Check : Substituting 6 for x in the original equation, 
 
 1+1=4- 
 l-f 
 
 2. Solve -+ 
 
 X X + 1 # 1 
 
 The 1. c. d. = x(x + l)(x 1). It is better to retain the 1. c.d. in 
 the factored form. 
 
 Multiplying both sides by x(x + l)(x 1), we obtain. 
 (x + 1)( Z - 1)+ x(x - 1) = 2x(x + 1). 
 x* - 1 + x 2 - x = 2 a; 2 + 2 x. 
 
 <> * , 
 3. Solve 
 
 -!=-! 
 3 4 11 
 
 The 1. c.d. =6a;(a:+l)(x-l).
 
 FRACTIONS 173 
 
 Multiply both sides by the 1. c. d. 
 
 18 x - 18 + 24 2 _ 24 x = 11 & + 11 x. 
 13 z 2 - 17 x - 18 = 0. 
 (z-2)(13z + 9) = 0. 
 
 x = 2 or - ft. 
 
 EXERCISES 
 196. Solve the following : 
 
 1. 
 
 2. 
 3. 
 
 4. 
 
 5. 
 6. 
 
 7. 
 8. 
 9. 
 10. 
 
 12. 
 13. 
 
 1 7 
 
 14 x-1 3a;-4 
 
 x 8 
 
 2 5 
 3 2x' 
 
 2x-l x-2 2 
 
 -j *C O U *C O 
 
 Q />2 /^ 
 1C A/ U^ ^/ 
 
 17 2 * 6 
 
 2 33 
 x + 1 x 13 
 
 a; 1 2 a; a; 
 18 1 I 2 - 5 . 
 
 344 
 | x + 5 x = 7. 
 6 a? 4 4 
 
 x-1 ' a;-2 6 
 
 9 3 
 19. ^ + ^ --0. 
 
 3-x 3-x 3 
 
 0.6 7 
 a; ~12' 
 
 3a; 1 
 
 x-1 x-2 
 
 9O it _|_ 9 
 
 ~y + 2' 
 
 7-1 10 
 
 o-i ^ x i 
 
 /. I 1 /v. 1 
 
 13 x 14 ar 
 
 a?+l 3 
 
 -J ff Q 
 
 a; + 4 a; + 5 
 .001 .025 
 
 *-5 4 3a?-l 
 
 4 5-a; 4 
 
 a: + 2 24 
 11.5 x + 3 1.75 
 
 25. a; -j- 12 = . 
 a; 
 
 o 6 1 *. 
 
 4 6 
 
 _1_ 7 K 
 -j i O 
 
 x-4~ 4
 
 174 ELEMENTARY ALGEBRA 
 
 3 2 
 
 39. 
 
 x-l x 2 90 z + 1 z + 2 2+3 
 
 28. X "*".+ ~ X = - 30. 
 
 x 1 3 + x x x1 x+1 
 
 PROBLEMS 
 
 197. 1. By what number must 72 be divided, that the 
 quotient may exceed the divisor by 6 ? 
 
 2. What number, when divided by 13, yields a quotient 
 that is less than the number by 168 ? 
 
 3. A is 8 years old, B is 33 years old. When will A be 
 as old as B ? 
 
 4. A is 36 years old, B is 48 years old. How many years 
 ago was A ^ as old as B ? 
 
 5. What fraction, equal to ^, becomes equal to f when 15 is 
 added to its numerator and denominator ? 
 
 Hint. Let be the fraction. 
 
 6. What must be the dimensions of a rectangle containing 
 192 sq. in., in order that the perimeter be 56 in. ? 
 
 7. A rectangle whose area is ^ sq. ft. has a perimeter of If 
 ft. Find its dimensions. 
 
 8. Divide 84 into two parts, such that the fraction formed 
 by these parts is ^. 
 
 9. A number exceeds the sum of its one third, one fourth, 
 and one fifth by 26. Find the number. 
 
 10. The rate of one train running between two towns exceeds 
 the rate of another by 10 miles an hour. The difference in 
 time is 1 hour. Find the rate of each train, if the towns are 
 200 miles apart. 
 
 Negative answers for the rates may be rejected as not permissible in 
 our problem.
 
 FRACTIONS 175 
 
 11. An oil tank can be filled by one pump in 6 hours, or by 
 another pump in 12 hours. How long will it take to fill the 
 tank when both pumps are working ? 
 
 Let x = time in hours, when both pumps are working. 
 
 Then - = part of tank filled by both pumps in 1 hr. 
 
 X 
 
 I = part of tank filled by the first pump in 1 hr. 
 ^ = part of tank filled by the second pump in 1 hr 
 
 Hence - + - - Solve. 
 
 6 12 x 
 
 12. How long will it take two pumps to fill an oil tank, 
 when one pump alone can fill it in 6 hr., the other pump alone 
 can fill it in 8 hr. ? 
 
 13. Two pumps working at the same time can fill an oil 
 tank in 5 hr. One pump working alone can fill it in 8 hr. 
 How long will it take the second pump alone to fill it ? 
 
 14. In a debating society a motion was carried 4 to 3 ; on a 
 reconsideration, 4 affirmative votes changed over to the nega- 
 tive, and the motion was lost 10 to 11. Find the number that 
 voted in the affirmative on the first ballot. 
 
 Hint. Let the affirmative and negative vote on the first ballot be 4 a; 
 and 3 x, respectively. 
 
 15. Find the world's production of rubber in a year, when 
 South America produced |, Africa | of the total amount, and 
 the rest of the world produced 2800 tons. 
 
 16. A man spends -\ of his income in board, y 1 ^ in clothes, 
 in sundries, and has $ 630 left. What is his income ? 
 
 17. A man placed at interest S6000 at 5%, and 6 months 
 later $5500 at 6 %. When will the interest on the two sums 
 be equal ? 
 
 18. Two ball teams in a league have the following record: 
 A won 50, lost 24, B won 56, lost 24 games. The two teams 
 play a final series of 10 games together. How many must A 
 win in order that the ratio of games won to games lost be 
 equal for both teams ?
 
 176 ELEMENTARY ALGEBRA 
 
 19. The height of a bin is 1 foot less than the width and 4 
 feet less than the length. Find the capacity of the bin if the 
 height is equal to half the length. 
 
 HISTORY OF FRACTIONS 
 
 198. Fractions occur in some of the earliest historic records, but 
 ancient peoples experienced great difficulty in computing with fractions. 
 At the present time fractions may have any number as numerator and 
 any number as denominator. Not so in olden times. The Babylonians, 
 as early as 2000 B.C., perhaps even earlier, used fractions with the 
 denominator 60. In measuring time and angles, they used subdivisions 
 on the scale of 60. The subdivisions of the hour into 60 minutes and the 
 minute into 60 seconds, as also of a degree (angular measure) into 60 
 minutes and a minute into 60 seconds, are of Babylonian origin. Every 
 time we measure angles and every time we consult our watches we are 
 using units of measure invented by astronomers near the banks of the 
 Ganges over 3000 years ago ! When the Babylonians took 60 as the de- 
 nominator of all fractions, there was no need of actually writing the 
 denominator. They wrote simply the numerator. To distinguish it from 
 a whole number, they placed it a little to the right of the ordinary position 
 for a word or number. 
 
 A fractional system involving the same denominator was in vogue also 
 among the Romans. They usually took 12 as the denominator; with 
 them 12 was also the number of subdivisions of weights and measures ; 
 the coin called as was subdivided into 12 uncice, the foot into 12 inches. 
 
 While the Babylonians and Romans preferred the use of fractions with 
 the same denominator and different numerators, the Egyptians and 
 Greeks had, as a rule, fractions with constant numerators and different 
 denominators. In the Ahmes papyrus, an old Egyptian mathematical 
 treatise found about fifty years ago, much attention is given fractions. 
 All fractions in that papyrus, except the fraction f , have the numerator 
 unity. In this case there was no need of writing the numerators ; a 
 fraction was designated by writing the denominator and then placing 
 over it a dot or another simple mark. Two ninths was indicated in the 
 Ahmes papyrus as | T , the sum of J and ^ being f . 
 
 The more modern point of view in computation with fractions appeared 
 among the Hindus after the fifth century of our era and, somewhat later, 
 among certain Arabic scholars, and among European writers of the 
 renaissance. The Hindus wrote the numerator over the denominator,
 
 FRACTIONS 177 
 
 but did not separate the two by a fractional line. Among the first to use 
 this line was the Italian mathematician Leonard of Pisa (thirteenth 
 century). At the beginning of the sixteenth century the fractional line 
 had come into general use. 
 
 Even as late as the fifteenth century certain European writers of 
 prominence experienced difficulty in explaining the multiplication of 
 fractions. If to " multiply " means to "increase," how can the product 
 of two proper fractions be smaller than either of the two fractions ? The 
 earliest algebraist to make extensive use of letters as the representatives 
 of numbers was the Frenchman, Vieta (1540-1603). The use of fractions 
 involving letters was introduced since his day. 
 
 The most conspicuous figure in the invention of decimal fractions is the 
 Belgian mathematician, Simon Stevin. He explained them in a booklet, 
 La Disme, published in 1585. He did not use the decimal point. He 
 
 0123 
 
 had a clumsy notation; he wrote the fraction 5.912 thus 5 9 1 2 or thus 
 5912. The small digit above, or inside a circle, indicated the 
 decimal place of the digit affected. Among the first, if not the first, to 
 simplify the notation of decimals by the use of the decimal point, or 
 comma, was John Napier, of Scotland, in publications of 1616 and 1617. 
 
 Thus it is seen that the modern fraction and the modern methods of 
 computing with fractions are the result of many stages of evolution, 
 reaching over a period of not less than 4000 years, and that Asiatic, 
 African, and European nations shared in effecting this evolution.
 
 CHAPTER XV 
 
 RADICALS AND GENERAL EXPONENTS 
 
 199. All numbers of elementary algebra can be placed in 
 one of two groups, real numbers or imaginary numbers. 
 
 "^j | 7, 4.675 are real numbers. 
 
 V 3, 2V 1, V 1 are imaginary numbers. 
 
 Imaginary numbers or expressions are numbers which in- 
 volve the square roots of negative numbers. Real numbers 
 are. rational or irrational. Rational numbers are positive or 
 negative integers, and numbers which are the quotients of 
 such integers, such as 9, 5, , .125. 
 
 Irrational numbers or irrationals are numbers that are real 
 but not rational, such as V2, V9, V5. 
 
 A radical is a root of any algebraic expression, indicated by 
 the use of a radical sign, such as V9, V6, V^ 2 x 6. 
 
 From this definition it is evident that a radical may be a 
 rational or an irrational number. 
 
 The integer of the rational sign is called its index. Thus 
 the index of the radical sign -\/ is 4. If no index appears, 
 the index 2 is understood. 
 
 A number by which a radical is multiplied is called its 
 coefficient. 
 
 Thus, in 7 \/6, 7 is the coefficient of the radical v/6. 
 
 200. The value of an irrational number, like V2~, cannot be 
 expressed exactly in Hindu-Arabic numerals, but we may 
 represent it accurately by means of lines. For instance, the 
 diagonal of a square whose side is 1 is equal to V2. We 
 
 178
 
 RADICALS AND GENERAL EXPONENTS 
 
 179 
 
 cannot find V2 exactly, but we can draw the square and its 
 diagonal. 
 
 1 
 
 FIG. 34 
 
 Likewise, in the right triangle whose legs are 1 and 2, the 
 hypotenuse is Vl 2 + 2 2 = Vo. 
 
 In this semicircle whose diameter is 4, the perpendicular 
 erected at the distance 1 from one end equals V3, as is shown 
 in geometry. 
 
 FRACTIONS AS EXPONENTS 
 
 201. All the exponents used thus far have been positive 
 integers. We defined a 5 as a a a a a. See 14. That is, 
 the exponent 5 indicates that a is taken as a factor five times. 
 
 What is the meaning of a 2 ? It would be absurd to say 
 
 that a* signifies a taken as a factor one half times. A number 
 can be taken as a factor only a whole number of times; for 
 example, three times or ten times. 
 
 For the purpose of attaching a meaning to a 2 , we stipulate 
 
 that we shall be able to multiply a? by a* according to the 
 same rule by which a 2 is multiplied by a 2 . The product of a 2 
 and a 2 is found by adding the exponents. That is, a 2 a 2 = a 4 . 
 
 If we add the exponents in the multiplication of a* by a*, we 
 obtain a$ + ^ or a 1 . That is, 
 
 at . a? = a.
 
 180 ELEMENTARY ALGEBRA 
 
 It is seen that CM is one of the two equal factors of a, or the 
 square root of a. 
 
 Hence a* is another way* of writing the square root of a. 
 That is, a* = Va. 
 
 In the same way, since 
 
 a? a^ = al + $ = a 3 , . 
 a* is another way of writing the square root of a 3 . 
 
 That is, a^ = Va 1 - 
 
 Similarly, since 
 
 | | f _ | 
 
 iu ^^v,,^ .^^ . = 
 
 and x* = Va; 5 , etc. 
 
 Thus the numerator of a fractional exponent indicates the 
 power of the base and the denominator indicates the root* 
 
 202. In finding the value of 8 7 , we may take the cube root 
 of 8, which is 2, and square 2, getting 4 as a result. 
 
 Or we may square 8, which gives us 64, and then take the 
 cube root of 64, which is 4. 
 
 That is, 8* = V8* = ( V8) 2 . 
 
 All irrational numbers which are expressed by the use of 
 radical signs can be expressed by the use of fractional expo- 
 nents. In fact, the simplification of expressions can be effected 
 more easily by the latter, as will appear later. 
 
 Thus, Vl6 aPb = 16?a*&^ = 2 a^&i. 
 
 i 
 
 * It can be shown that v'a or a n has n different root values, but in ele- 
 mentary algebra only one of these values is usually considered, the so-called 
 principal value. If a is a positive number, then the principal value of tya is 
 its positive root ; if a is negative and the index n is odd, there is no positive 
 root, and the negative root is taken as the principal root. If a is negative 
 and the index n is even, then all the roots are imaginary.
 
 RADICALS AND GENERAL EXPONENTS 
 
 181 
 
 ORAL EXERCISES 
 
 203. Express with fractional exponents and simplify : 
 
 11. 
 
 12. - 
 13. 
 
 14. Vwi Vwi Vra. 
 15. 
 
 16. 3^/|. 
 
 17. -- 7 \l 
 
 18. )~v c . 
 
 19. Va + 6. 
 20. 
 
 EXERCISES 
 
 204. Write with radical signs : 
 
 3 
 
 1. a*. 9. 6 
 
 2. &i 10. 7 
 
 3. a;%l 11. 
 
 4. (7a)i 12. 3a 
 
 5. 2(5aj)^. 13. ( + y 
 
 14. 4 (c 
 15- 5a^o; 
 
 6. 2 ai 
 
 7. (2 a)*. 
 
 8. 2 
 
 16. a&.
 
 182 ELEMENTARY ALGEBRA 
 
 MEANING OF A ZERO EXPONENT 
 
 205. Thus far, no meaning has been assigned to a. Can 
 we not discover a meaning ? By the law of exponents in di- 
 vision (i.e. subtracting the exponent in the divisor from that 
 in the dividend) we obtain a 3 -5- a 3 = a. But when a is not 
 
 a 3 
 
 zero, a 3 -=- a 3 = = 1. . . a = 1. 
 a? 
 
 Therefore any number except with a zero exponent is equal 
 to 1. 
 
 MEANING OF A NEGATIVE EXPONENT 
 
 206. Reasoning as in 205, we obtain 
 
 a 5 -5- a 8 = or 3 , also a 5 -j- a 8 = = . .*. a~ 3 = 
 
 a 8 a 3 a 3 
 
 Therefore, any quantity with a negative exponent is equal to 1 
 divided by that quantity with the exponent positive. 
 
 - V - 
 Further, ' 
 
 a~ 3 b-*c ! c a 2 c 
 
 Therefore, a factor may be moved from the numerator to the 
 denominator of a fraction provided the sign of its exponent be 
 changed. 
 
 Care must be taken to transfer only factors in - i- , a~ 2 is not a 
 
 b 
 
 - ~*" Since 
 
 IIIV lAVl <L lilH 1 1 lllll I 1 tljl'l 
 
 AC xo ifiuii crv^uai UL> . 
 
 o a 2 6 
 
 (j-2 * fa follows that 
 
 a~ 2 + 1 a 2 1 + a 2 
 
 a' 2 
 
 6 6 a 2 6 
 
 
 EXERCISES 
 
 207. Simplify: 
 
 
 1. 36~*. 
 
 4. 7.16-t. 7. (&)-*. 
 
 2. 64*. 
 
 5. 9*. 8. (^)*. 
 
 3. 125*. 
 
 6. (-27)*. 9. 36^-34
 
 RADICALS AND GENERAL EXPONENTS 183 
 
 10. (- 8)-* + (jh)*. 13- (- 2)~ 2 (- 3). 
 
 \ / \ A i U/ 
 
 i-s A. 1 
 
 12. (|)^-2 . 2- 1 *3* 
 
 16 ' ' 20 ' I' 24 ' 4 5^' 
 
 17. m -A- 21. 2-^. 
 
 25. 
 
 
 22. 
 
 7 V-*x~ l y-* 26. 
 
 19. 1. 23. -<. 
 
 3 2/- 3 z 
 
 a- 1 - 6- 
 
 REDUCTION OF RADICALS 
 
 208. A radical is in its simplest form when the following 
 
 three conditions are all satisfied : 
 / 
 
 (1) There are no fractions under the radical sign. 
 
 VJ is not in its simplest form. 
 
 (2) The number under the radical sign contains no factor 
 raised to a power equal to or exceeding the index of the radical 
 sign. 
 
 Vo* is not in its simplest form, because the exponent 4 exceeds the 
 index 3 of the radical sign. 
 
 (3) In a radical of the form Va", m, and n must have no 
 common factor other than 1. 
 
 v^ i s no t j n t ne simplest form, because 2 and 4 have the common 
 factor 2. 
 
 Simplify \/64 a 8 6 2 . We obtain 
 
 = 2* a . & = 2 cflb = 2
 
 184 
 
 ELEMENTARY ALGEBRA 
 
 ORAL, EXERCISES 
 
 9. 
 
 14. V729ary 2 - 
 15. 
 
 16. 
 
 210. 1. Simplify V98. 
 
 \/98 = >/49 2, wherein 49 is the largest factor of 98 which is a perfect 
 
 square. V49-2 = 7V2. .-. v / 98 = 7V2. 
 
 Again 5V/54 = Sv/27 2, wherein 27 is the largest factor of 54 which is 
 a perfect cube. 
 
 5v / 2TT2 = 5.3^2 = 16^2. 
 
 2. Simplify Vf. 
 
 Multiply both numerator and denominator by some number which wiU 
 make the denominator a perfect square ; multiplying by 3, Vl = Vf 
 = Vi 6, wherein 1 is the largest factor of 6 which is a square, and 9 is 
 used for its denominator. Since VI = J* Vfc 8 = i V^- ' Vf = 
 
 Again 7 Vi? = 7-y^il ' I = ^ Vrfy 10 = 
 
 EXERCISES 
 
 211. Simplify: 
 
 
 
 1. V8. 
 
 3. V20. 
 
 5. V28. 
 
 2. V12. 
 
 4. V24. 
 
 6. V40.
 
 RADICALS AND GENERAL EXPONENTS 185 
 
 7. 
 
 3 A/44. 
 
 27. 10V10000 
 
 47. V^T-- 
 
 8. 
 
 4V18. 
 
 28. A/32. 
 
 48. 3 A/|. 
 
 9. 
 
 -6A/27. 
 
 29. 3^48. 
 
 49. </f 
 
 10. 
 
 5 A/48. 
 
 30. 7 A/80. 
 
 50. A/|. 
 
 11. 
 
 2V56. 
 
 31. A/224. 
 
 51. 3A/^V 
 
 12. 
 
 7 V121. 
 
 32. A/2 3 . 
 
 52. A/i- 
 
 13. 
 
 -8V160. 
 
 33. A/C^ 
 
 - 
 
 14. 
 
 2V75. 
 
 34. V2p. 
 
 53. -J-- 
 
 15. 
 16. 
 17. 
 18. 
 19. 
 
 3A/125. 
 -9V300. 
 3V162. 
 2V320. 
 
 -fVl47. 
 
 35. ttA/C 3 . 
 
 o c 7^ -/ /- ft 
 
 37 m~\/?n 3 n 
 38. 5V}. 
 39. A/|. 
 
 54. "V/-^' 
 
 c c ^ / 
 
 x / 
 
 56. ^. 
 
 y ' or 
 
 20. 
 21. 
 22. 
 23. 
 24. 
 25. 
 26. 
 
 |V108. 
 
 2 A/54. 
 5^32. 
 7 A/56. 
 3A/88. 
 
 40. V|. 
 41. A/|. 
 42. V^V- 
 43. V^. 
 44. ^VJ^. 
 45. v'- 
 46. A^f. 
 
 c rf / ^ 01 1 4 1 * 
 
 O I . y^o/ t/y A/ 
 
 CQ -VWWt. / 
 
 OO> //f /f % 
 * 7/fc Ti 
 
 V(g + 6) 2 
 
 
 
 ^ , a o 
 
 rr~ 
 
 fiO ^7* 2 7/ 2 ^ A / " 
 
 fA/250. 
 
 Vxy 
 
 ADDITION AND SUBTRACTION OF RADICALS 
 
 212. Similar radicals are radicals having the same radical 
 sign and the same number under the radical sign. 
 
 4 -\/a, 2Va, b\/a have the same radical sign v' , and the same num- 
 ber a under the radical sign. Hence they are similar.
 
 186 
 
 ELEMENTARY ALGEBRA 
 
 Add 2 V8, 3 VlO, 5V50, - 7V40. 
 
 - 7V40 
 = 4V2 + 3 VlO + 25 V2 - 14 VlO = 29V2 - llVlO. 
 
 EXERCISES 
 
 213. Simplify: 
 1. V3-2V3. 
 
 2. 5V2 + 3V2-V2. 
 3. 8VTI-5VII. 
 4. 9Va-2V. 
 5. aVb + bv'b. 
 6. 2V2-3V|. 
 7. 5V3+9V|. 
 8- Vf + VJ. 
 9. Vf- V|. 
 
 10 . ^+\/?- 
 
 11. V20+V45-V5. 
 12. 2V2-V18 + 3V50. 
 13. 5V28 + 3V63-VII2. 
 14. VJ-2V2 + V}. 
 15. Vct 2 & + V6 3 V9 6. 
 
 16. 2 Va 3 6 Vci6c 2 + V4 a6. 
 
 19. Vmw + \/ \/ 
 ^ /i ^ mw 
 
 20. VC^ 2 2/ 2 ) 
 
 (a: y) V4 cc 3 ?/ 2 -(- 4 ar 2 ?/ 3 . 
 
 21. \Vl6 + \ / 54+Vl28. 
 
 23. \V8 x - 16 + V27 x - 54. 
 
 22. -\/5 -\/7 + A/-^-. 
 
 24. v / (-3) 2 +V9a-27. 
 
 MULTIPLICATION OF RADICALS HAVING THE SAME INDEX 
 
 214. Multiply 5Va by 7V6 and we have 
 5 . 7 Va . V6 = 35 Va6. 
 Multiply 3V2 by 4 VlO and we have 
 
 - 3 . 4 V2 VlO = - 12V20 = - 24 V5. 
 
 RULE. Multiply the rational factors and the radical factors, 
 and simplify. 
 
 It will be observed that the multiplication of two radicals, 
 Va; and Vy, rests on the identity -\/xy = V# Vy.
 
 RADICALS AND GENERAL EXPONENTS 187 
 ORAL EXERCISES 
 
 215. Find the indicated products : 
 
 216. Multiply V2 + V3 - V5 by V2 - V3 + V5. 
 
 V2+V3-V5 
 V2-V3+V5 
 
 -6 +2V15 
 
 RULE. Multiply each term of one polynomial by each term 
 of the other, simplify, and combine. 
 
 EXERCISES 
 
 217. Multiply: 
 
 1. V6 - VY - 2 V2 by V3. 
 
 2. V7+V3-VI1 by V5. 
 
 3. 3V2-2V3+V5 by VB.
 
 188 ELEMENTARY ALGEBRA 
 
 4. V5-4V2+VK) by VlO. 
 
 5. V3 + 7V6-5VI2 by V6. 
 
 6. 3 + V2 by 7 - V3. 
 
 7. 8 - V6 by V5 - 4. 
 
 8. 3 V2 + 4 V3 by 2V2 - 5V3. 
 
 9. 2 V7 + 7 V2 by 2 V7 - 7V2. 
 
 10. 5V3 + 3V5 by 5V3-3V5. 
 
 11. V6-V7+V8 by V8+V7-V6. 
 
 12. 2 V5 - 3 V6 - 4 V7 by VlO - V5 - V2. 
 13.' Vi+Vj-V| by V5+V6-V7. 
 
 14. x Vy + V#y y Va? by 
 
 15. V2--\/3+Vl by ^ 
 
 DIVISION OF RADICALS HAVING THE SAME INDEX 
 
 218. 1. Divide 2 V2 by V3. 
 
 (a) 2V2-^V3 =2vf = fV6, or 
 
 (b) 2V2-,V3=^.^=2V6 = V6. 
 
 V3 V3 3 3. 
 
 Unless the division of the quantities under the radical signs gives a 
 whole number, the second method is preferable. 
 
 In (&) we express the division in the form of a fraction, then multiply 
 both numerator and denominator by some number which will make the 
 denominator rational. 
 
 2. Divide 3 by 2 + V3. 
 
 2-V3 4-3 
 
 Again, (3V2 + 4 V3) -s- (3 V2 - 4 V3) 
 
 _3\/2+W3 _ 3v / 2 + 4\/3_ 18 + 24 V6+48 _ 66 + 24V / 6_ 33 + 12V6 
 ~3V2-4V3 ' 3V2+4V3~ 18-48 -30 16
 
 RADICALS AND GENERAL EXPONENTS 189 
 
 If the denominator is of the form Va Vft, multiply both terms of the 
 fraction by Va T Vb. This is called rationalizing the denominator. 
 
 To find the numerical value of 2 V2 -f- V'3 in its present form requires the 
 extracting of two square roots and a long division. It is much less work 
 to rationalize the denominator, then in the resulting expression, |V6, 
 find V6, and take f of it. 
 
 EXERCISES 
 219. Divide: 
 
 1. V2 by V5. 8. V# -f- y by Va; y. 
 
 2. 1 by V3. 9. 5 by V5 + 1. 
 
 3. 24 by V6. 10. 3 V7 by V7 - 4. 
 
 4. 2by\/2. 11. V2+V3by V5+V61 
 
 5. 36 by V48. 12. V3 + V5 by V2 - V6. 
 
 6. 6by2-V3. 13. VIO-V2by V3-V5. 
 
 7. a by Va"+6. 14. Vl5 - V3 by V5 + V6. 
 
 15. vn - vi2 by vn + vi2. 
 
 16. 2 V2 + 3 V3 by 2 V2 - 3 V2. 
 
 17. 2 V3 - 3 V2 by 2 V3 + 3 V2. 
 
 18. a V6 + & Va by a V6 6 Va. 
 
 19. a Va + 6 by Va + b. 
 
 20. Va; + y Vic y by VaH- y+^/x y. 
 
 RADICAL EQUATIONS 
 
 220. 1. Solve VE - 2 = 5. 
 
 Square both sides, x 2 = 25 ; x = 27. ^4ns. 27. 
 
 CAedfc ; V27 - 2 = 5 ; 5 = 5.
 
 190 ELEMENTARY ALGEBRA 
 
 2. Solve V* + 6 + V* 1 = 3. 
 
 Transpose, Vx + 5 = 3 Vx 1. 
 
 Square both sides, x + 5 =9 6 Vx 1 + x 1. 
 
 Transpose and combine, 
 
 Square both sides, 
 Check : 
 
 221. RULE. If the equation contains but one radical, trans- 
 pose so that it will stand alone. 
 
 If the equation contains two radicals, transpose so that there is 
 one radical on each side and so that the rational terms are com- 
 bined on one side. 
 
 Raise both sides to a power corresponding to the degree of the 
 radicals. Solve the resulting equation. 
 
 In squaring both sides of an equation, a value of x may be obtained 
 which will not satisfy the original equation. Such a value is called an 
 extraneous root. For instance, squaring both sides of the equation 
 
 Vx + 5 = 1 - Vx, 
 
 there results x + 5 = 1 2Vx + x. 
 
 Transpose and simplify, 2 = Vx. 
 
 Square both sides, 4 = x. 
 
 Substitute in \/x +5 = 1 Vx", 3=1 2, which is absurd. 
 
 Hence x = 4 is not a root of the original equation, but an extraneous 
 root. In fact, there is no value of x that will satisfy Vx + 5 = 1 Vx. 
 In other words, this equation has no roots. 
 
 Only by substitution in the original equation involving radicals can we 
 tell whether a value obtained for x is a root of that equation or whether 
 it is an extraneous root.
 
 RADICALS AND GENERAL EXPONENTS 
 EXERCISES 
 
 222. Solve and check : 
 
 191 
 
 2. 
 3. 
 4. 
 
 6. x = 5 Vo? x 6. 
 
 7. ^/4z-16=2. 
 
 5 = 
 
 + 2. 
 
 8. 8 - -Vx 1 = 5 + Vx + 2. 
 
 9. x = 13 - VaT- 13. 
 
 12. 
 
 2 
 
 + 4 
 
 5. x 3 = Va 2 6. 10. 3 Vx = Vz + 3. 
 
 11. -Vx*-7x + 6= Vz 2 - x - 12. 
 
 V;e 4- 1 VaT - 3 
 
 15. Vz + 3=^- 
 
 aj-f-3 
 
 5 
 
 13. Vo; + 17 = x 3. 
 
 14. V x 5 x 1. 
 
 15. Vz + 3 = 
 
 16. Vo; 2 
 
 V-2 
 
 = 0.
 
 CHAPTER XVI 
 
 REVIEW OF THE ESSENTIALS OF ALGEBRA 
 DEFINITIONS 
 
 223. Define the following : 
 
 Coefficient Monomial Linear equation 
 
 Exponent Binomial Quadratic equation 
 
 Term Polynomial Root of an equation 
 Factoring 
 
 EXERCISES ON PARENTHESES 
 
 224. Remove the parentheses and simplify : 
 
 1. (5 _ 6) - (7 - 8) + (5 - 6 - 7) - (- 5 - 4 + 6). 
 
 2. ll_(3-[4-5]) + [4-(3-5)+(4-l)]. 
 
 3. 4(3 - 2) + 4(3 -=- 2) - 4 -- (4 - 2). 
 
 4. (4 a + 6) - (8 b - c) - (a - 7 &). 
 
 5. 4 x* - 7 y 2 - [7 x 2 - 6 x + 8 y] - [5 x - 3 y]. 
 
 6. 3a? 5xy-\2xy 4:a? 9y*} \7a? 5xy\. 
 
 7. a - [6 + (c + 2 d) - (- c - d)] - [2 a - 3 &]. 
 
 8. -(a - c + d) + (2 a - 3 6 + 4 c) - (4 a - [5 b - 6 c]). 
 
 Group all terms with a single letter in a parenthesis with 
 + before it, and all terms with two different letters in a 
 parenthesis with before it : 
 
 9. a b + 4 c db + 2 oc 4 be + o?b. 
 
 10. x + 5 y 4 2 -}- xy 3 xz + 7 ?/z. 
 
 11. 2ar J 
 
 12. cde 4cd 3de+f d + 2c -3e. 
 
 192
 
 REVIEW OF THE ESSENTIALS OF ALGEBRA 193 
 
 EXERCISES IN ADDITION, SUBTRACTION, MULTIPLICATION, 
 AND DIVISION 
 
 225. Simplify: 
 
 1. 2 z + O -I) 2 . 6. 2m(w-4)-ra 2 . 
 
 2. 4*+(2* + l) 2 . 7. 3( + y)-2(aj-y). 
 
 3. 4+(2-:f). 2 8. (z + 2/) 2 +(a;-2/) 2 . 
 
 4. 4y-(2?/ + l) 2 . 9. (a-6) 2 +(a-6) 2 . 
 
 5. (ce-?/) 2 +-4a*/. 10. a(a 
 
 Find the values of the following at sight : 
 
 11. 4 3 , 3 4 , 4 1 , I 4 , 2, 8, 25*. 
 
 12. a 2 a 3 , a 4 a, 
 
 ^ , f 
 
 a_ _ <8 5 6 
 
 a 9 ' 6 10 '5c 5 ' 
 
 14 4a 2 6 3 12 a 3 5 4 15 x 5 y 5 
 6 a 3 6 ' 4 a& 2 ' 25 x b y ' 
 
 Simplify : 
 
 15. (a+6) 2 +(a-6) 2 . 18. (a + 2 ft) 2 -(a-f 6)(a-&). 
 
 16. (a-26) 2 +(a + 6) 2 . 19. (2 a - 6) 2 + (a - 6) 2 . 
 
 17. (a + 6-2c) 2 . 20. (a-6) 2 - 
 
 Multiply orally : 
 
 21. (x + 3)(z + 9). 25. (6 - 5)(& - 4). 
 
 22. (2, + 7)(y + 4). 26. (2- Z )(3-). 
 
 23. (7-2)(5-z). 27. (a? + 7)(> 2 + 5). 
 
 24. (a-6)(a + 8). 28. (ay + l)(ay + 4). 
 
 Perform the indicated divisions : 
 
 29. 
 
 30. 
 
 31. 
 
 o
 
 194 ELEMENTARY ALGEBRA 
 
 EXERCISES IN FACTORING 
 
 226. Factor: 
 
 1. 10 xf - 15 tff. 8. 3 x* (x - 1) - (1 - a?). 
 
 2. (2x + y)a + (2x + y)b. 9. 3ra(a 2 + 6 2 ) - 2 w (6 2 + a 2 ). 
 
 3. 5 *(a + 6) - 3 x(a + 6). 10. (m - ?) 2 - (m - w). 
 
 4. o^ 5 x 5 - 3 x* + 3 a; 1 . 11. 06 as + 6y sy. 
 
 5. x 6 - 5 or 5 + a 4 5 a 3 . 12. a 3 - a 2 6 + a& 2 - 6 3 . 
 
 6. (a-&)a 2 -(a-6)6 2 . 13. y* + Wy + 25. 
 
 7. (a? - 1) m + (a? - 1) w. 14. v? + 8 x + 16. 
 
 15. a 2 6 2 + 12 ab + 36. 
 
 16. a 2 + & 2 + c 2 2 a& 2 ac + 2 6c. 
 
 17. m 2 -49. 32. 6 2 -12&-45. 
 
 18. 64 - 2 . 33. c 2 + 14c-32. 
 
 19. 81^-25/. 34. x* + 10 a -75. 
 
 20. (a-6) 2 -c 2 . 35. x 2 -10x-75. 
 
 21. x* - (5 aj + yY- 36. a; 2 - 7 ajy - 8 y 2 . 
 
 22. a 2 + 2a6 + 6 2 -c 2 . 37. x* + 15 afy + 14 a^/ 2 . 
 
 23. a* 5 a; + 4. 38. 70 a 2 + 17 a + 1. 
 
 24. x 2 + 5 a; + 4. 39. 2 a 2 + 9 a? + 9. 
 
 25. y* + 6y 16. 40. 2-y-6y*. 
 
 26. y* Qy- 16. 41. 6 - m - 15 m 2 . 
 
 27. a 2 -? a + 10. 42. a 3 + 86 3 . 
 
 28. cc 2 + 7a; + 10. 43. a 3 - 8 6 3 . 
 
 29. z 2 + 3a-10. 44. 8^-27^. 
 
 30. x 2 -3a;-10. 45. 8 a? + 27 y 3 . 
 
 31. a 2 + 7a-18. 46. a^-1.
 
 REVIEW OF THE ESSENTIALS OF ALGEBRA 195 
 EXERCISES ON FRACTIONS 
 
 227. 1. 
 
 h 6 b +b b 
 
 2. Show that ^ = .=-2 = _^ = _?. 
 
 3. -J- -L- -1_ ? 
 
 x) (x-y}(x-z) 
 1 ? 
 
 Reduce each fraction to its lowest terms : 
 6x + 3-2c 
 
 3(x + y) ' 9 4c 2 
 
 6 5 fo + y) 3 . 10 bx + ax 
 20(x + y)* ' no? nb 2 
 
 7 a-6 
 2a-2&' 
 
 8. " ~ l ~" 
 
 * 
 
 a 2 - a& (a? - 2) 3 
 
 Multiply : 
 
 13. by. 15. gng by 10a; 
 
 3 16a; 2 2a: J ? 2 - oj 
 
 , 34^ gy-y , 
 
 17 z 2 J 5a;^ ' x 2 + xy 
 
 Divide : 
 
 17. *=* by ^- 27 - 20. (a 2 - rf) by 
 
 J 
 
 a 
 
 18. ^-Z^ by ^^L. 21 . "~ by (a _ 2 6). 
 cn m+n a + b 
 
 19. 6m3 by 3m 22. ^~g+A by *^- 
 n?-m 3 J m-n a; 2 -2 a; -35 & + 5
 
 196 ELEMENTARY ALGEBRA 
 
 Perform the indicated additions and subtractions of 
 fractions : 
 
 23 . I_*= 26. 
 
 a 3 a 5 (a -I) 3 (1-a) 8 
 
 24. - + _. 27. 
 
 2y 3y Sy x y x + y 
 
 25. - ___ _. 28 . _ __ _'. 
 
 (x y) 3 (y a;) 2 a 3 a 4 
 
 Reduce each expression to an equivalent fraction : 
 
 29. 3x + l^. 32. m-^Zll. 
 
 4 3 m 
 
 30. 2y + 5x ~ 3y . 33. a; 2 + aw + y 2 + ~^ . 
 
 4 x y 
 
 9 n* 7T 3 
 
 31. 80 z - * -- 2 
 
 a + 6. + y 
 
 EXERCISES IN LINEAR EQUATIONS 
 
 228. Solve: 
 
 1. 8 x + 19 = 11 -5 x + 13. 3. #=12 -a. 
 
 2. |y + 7 = |y + 8. 4. fa 2 fa = 4aj. 
 
 5. (16 * + 5) (9 a? + 31) = (4 x + 12)(36 a; + 10). 
 
 6. x (x + 1) + x (x + 2) = (a; + 3)(2 a? + 4). 
 
 7. mx = nx + 3. a; + 4 _ - 
 
 K 
 
 8. ay + by = c. 
 
 9. 2 ay 2 6y = 16 
 10 ax + bx ex = d. 
 
 9. 2 ay 2 6y = 16 + 2 a. 15. 5 - - = 3. 
 
 11. m# nx x = 
 
 ifi 
 
 x -\-25 
 12. (z + a) 2 =.(*-&)'. 
 
 
 13. (a;-4) 2 -(a;-5) 2 =4a;+5. a; -4 a; -2
 
 18. 
 19. 
 20. 
 21. 
 22. 
 23. 
 
 25. 
 
 REVIEW OF THE ESSENTIALS OF ALGEBRA 197 
 x-1 x-3 
 
 2x-l~2x-2 
 
 x + 1 x 
 
 X 1 
 
 a: 1 X + 1 
 
 3x-5 = _2 
 
 X-4: ~~ 3 
 
 4 x 4- J 
 
 + 1. 
 
 2a; 2 
 
 2 -f y + 3 _ 
 2y + 1 
 
 26 
 
 27 
 
 28 
 
 29. 
 
 30. 
 
 31. 
 
 I2a? *=3& 
 
 _ Q K 
 O JU t>. 
 
 a; 2y 5 = 0. 
 
 2 x + 1 a 4- 7 
 
 3 5 
 
 3. y __ J 
 
 3a?-|-8?/ = 5, 
 
 a; + w = 2 
 
 3 a; + cy = 3 c 2 
 3 x cy = c 2 . 
 
 EXERCISES IN QUADRATIC EQUATIONS 
 
 229. Solve: 
 1. x 2 6x + 5 = 0. 
 
 3. a; 2 -5 a; = -4. 
 
 12. 15+- = 11. 
 a; 
 
 4. (3aj-2)(aj-l)=14. 
 
 5. (3aj-2)(aj-l)=200. 
 
 6. (2a;-l) 2 =_8a;. 
 
 7. x 2 2ax = b 2 . 
 
 8. x 2 -(a - l)a? + a = 0. 
 
 9. a; 2 + mx + n = 0. 
 
 10. px 2 + qx + r = 0. 
 
 11. - A_ = 8x + 2 
 x-1 x 2 - 1 
 
 13. 
 
 14. _3_ + _ 3 _ = 8. 
 
 1 + a; 1 x 
 
 x l_ d 
 
 "C a; + l 
 
 x + ~L x + 2 
 
 15. 
 16. 
 
 x 1 a; 2 
 
 __ IT 
 
 17.
 
 198 
 
 ELEMENTARY ALGEBRA 
 
 18. 
 
 19. 
 
 20. 
 
 21. 
 
 22. 
 
 x + y = 15. 23. 
 
 24. 
 
 1-2=2. 
 x y 
 
 - y* = 7. 
 
 PROBLEMS 
 
 230. 1. What number added to the numerator and denomi- 
 nator of ^ will give a fraction equal to f ? 
 
 2. A train runs 300 miles in a certain time. If the train 
 were to run 10 miles an hour faster, it would take it 1 hour 
 less to travel the same distance. Find the time and rate. 
 
 3. The length of the rim of one of the hind wheels of a 
 carriage exceeds the length of the rim of one of the front 
 wheels by 3 ft. The front wheel makes the same number of 
 revolutions in running 200 ft. that the hind wheel makes in 
 running 260 ft. Find the length of the rim of each wheel. 
 
 4. A farmer found that the supply of feed for his cows 
 would last only 20 weeks. He therefore sold 50 cows, and his 
 supply lasted 30 weeks. How many cows had he at first ? 
 
 5. Two trains start together and run in opposite directions, 
 one at the rate of 47 miles an hour, the other at the rate of 34 
 miles an hour. After how many hours will they be 1053 
 miles apart ? 
 
 6. The area of a rectangle is 1850 sq. in. Its length 
 exceeds its width by 13 in. How long is each side ?
 
 REVIEW OF THE ESSENTIALS OF ALGEBRA 199 
 
 7. A farmer expected to receive $5.52 for his eggs. He 
 broke 6 eggs, but by charging 2 ^ a dozen more for the remain- 
 ing eggs, he received the desired amount. How many dozen 
 eggs had he originally ? 
 
 8. Weighed on a defective balance a body appears to 
 weigh P pounds when the true weight is a + bP pounds, where 
 a and b are numbers that are the same for all weights. If the 
 balance shows 10 pounds for a real weight of 11.2 pounds, and 
 30 pounds for a real weight of 33.2 pounds, find a and b. 
 What will the balance show for a real weight of 40 pounds ? 
 
 9. According to tradition, the following problem was 
 assigned by Euclid of Alexandria to his pupils, about 3 cen- 
 turies B.C. : A mule and a donkey were going to market laden 
 with wheat. The mule said to the donkey, " If you were to 
 give me 1 measure, I would carry twice as much as you, if I 
 were to give you 1 measure, our burdens would be equal." 
 What was the burden of each ? 
 
 a 
 
 10. A room shaped as in Fig. 37 has a floor 
 area of 320 sq. ft. If the lengths marked a 
 are all alike, and b is 22 ft., find a. 
 
 11. To pass from one corner of a rectangu- 
 lar park to the opposite corner I must go 700 
 
 j IG. oT 
 
 feet, if I go around the sides; if I walk 
 
 diagonally across, I save 200 feet. What are the dimensions 
 
 of the park ? 
 
 12. It takes a cyclist 18 minutes longer to go a distance of 
 27 miles from A to B than it takes to return by the same 
 route. If he rides up hill 6 miles an hour and down hill 15 
 miles an hour, how many miles of his trip going is up-hill 
 travel, and how many miles is down hill ? 
 
 13. A dealer sells bicycles at a 20 % profit. A rival dealer 
 obtains the same kind of bicycles $ 3 cheaper and sells them 
 $3 cheaper, thereby realizing a profit of 21| %. What price 
 does the first dealer pay for his bicycles ?
 
 200 ELEMENTARY ALGEBRA 
 
 14. If speculum metal contains 67 % of copper and 33 % of 
 tin (by weight) and gunmetal contains 90 f of copper and 
 10 / of tin, how many pounds of gunmetal should be melted 
 with 300 pounds of speculum metal, to obtain an alloy in 
 which there is 5 times as much copper as tin ? 
 
 15. Water in freezing expands 10 % of its volume. How 
 much water when frozen will fill a 5-gallon freezer ? 
 
 16. In a price list, the cost of sewer pipe, per foot of length, 
 is given by the formula c = .4 d? + 14, where d is the diame- 
 ter of the pipe in inches and c is the cost in cents. What will 
 400 ft. of 15 in. pipe cost ? 
 
 PROBLEMS ON THE LEVER 
 
 231. 1. What force must be applied by the hand to balance 
 or raise a known weight W= 748 pounds ? 
 
 f w 
 
 FIG. 38 
 
 By careful measurement it has been ascertained that the force multi- 
 plied by the force arm of the lever is equal to the weight multiplied by the 
 weight arm. Notice that the weight arm is measured from the point of 
 application of the weight to the prop or fulcrum, A, and the force arm is 
 measured from the point of application of the force to the prop. In this 
 case / = 4, w = 2. 
 
 Let F = the required force in pounds, then 
 
 4 F = 2 x 748. 
 F = 374 pounds, the answer. 
 
 2. How large a weight can be raised with a force of 500 lb., 
 if the force arm is 7.5 ft. and the weight arm .5 ft. ? Draw 
 a figure. 
 
 3. The force arm is 16 times greater than the weight arm. 
 What force will balance a weight of 732 pounds ?
 
 REVIEW OF THE ESSENTIALS OF ALGEBRA 201 
 
 4. The force arm is 2 ft. longer than the weight arm. How 
 long is the weight arm, if a force of 12 pounds balances a 
 weight of 35 pounds ? 
 
 5. A lever is 75 in. long. Where must the prop be placed 
 in order that a force of 2 pounds at one end may move 5 
 pounds at the other end ? 
 
 6. Two men carry a load of 212 pounds on a pole between 
 them. If the load is 3.5 ft. from one man and 5.4 ft. from the 
 other, how many pounds does each man carry ? 
 
 7. A man weighing 170 pounds has a crowbar 6 ft. long. 
 What pressure can he exert toward moving a rock, if the prop 
 is 3 in. from the lower end of the crowbar ? 
 
 8. On an untrue balance a weight of 13 oz. appears to 
 weigh 13.2 oz. If the beam is 6 in. long, and the error is due 
 to a displacement of the fulcrum, how much longer is one arm 
 than the other ? 
 
 9. A team is hitched to a doubletree 4 ft. long. At what 
 point must the doubletree be 
 
 attached to the plow, so that 
 one horse will pull li times as 
 much as the other ? 
 
 c r?^ D 
 
 10. Each of two horses LXJ Plow-Beam 
 
 hitched to the doubletree A in FlG 39 
 
 Fig. 39 pulls 4 as much as a 
 
 horse at B. At what point should a plow beam be attached to 
 the doubletree evener CD, which is 5 ft. long ? 
 
 PROBLEMS ON FALLING BODIES 
 
 232. Bodies falling from a state of rest obey laws expressed by the 
 following formulas : v _ 32.2 t 
 
 s = 16.1 1*, 
 
 v = V64.4 s, 
 
 where v means the velocity, t the time in seconds, s the space or distance 
 (in feet) through which the body falls.
 
 202 ELEMENTARY ALGEBRA 
 
 1. Express each of the three formulas in words. 
 
 2. Eliminate t from the first two formulas, and thereby de- 
 rive the third formula. 
 
 I 
 
 3. Eliminate s from the last two formulas, and thereby de- 
 rive the first. 
 
 4. Eliminate v from the first and third, and thereby de- 
 rive the second. 
 
 5. A stone dropped into a mine shaft is heard to strike 
 bottom after 5 seconds. Neglecting the time it takes sound 
 to travel, estimate the depth of the shaft. 
 
 6. With what velocity will a body strike the bottom of a 
 mining shaft 1000 ft. deep ? 
 
 7. A stone is heard to strike the bottom of a mine shaft 
 after ten seconds, sound traveling, in this instance, at a rate 
 of 1126 ft. a second. What is the depth of the shaft ?
 
 INDEX 
 
 Numbers refer to sections 
 
 Absolute, term, 166. 
 
 value, 29. 
 Addition, algebraic, 26, 28, 29. 
 
 checking, 45. 
 
 fractions, 193. 
 
 monomials, 41. 
 
 polynomials, 43. 
 
 radicals, 212. 
 Ahmes, 165. 
 Algebraic expression, 2. 
 
 evaluation of, 4, 23. 
 Area, circle, 134. 
 
 rectangle, 63. 
 
 trapezoid, 134, 174. 
 
 triangle, 134. 
 Arithmetical operations, order 
 
 21. 
 
 Average, 35. 
 Axis, x-axis, 90 ; y-axis, 90. 
 
 Bank discount, 100. 
 Base, 13, 14. 
 Binomial, 12. 
 Braces, 17. 
 Brackets, 17. 
 
 Cardan, 183. 
 Circle, 134, 182. 
 
 area of, 134. 
 
 length of, 8, 134. 
 Checking, addition, 45. 
 
 division, 137. 
 
 equations, 51. 
 
 factoring, 113. 
 
 multiplication, 60. 
 
 subtraction, 45. 
 Coefficient, 10. 
 
 of radical, 199. 
 Concrete mixture, 83. 
 Coordinates, 147. 
 
 of, 
 
 Degree of an equation, 130. 
 Descartes, 165. 
 Dissimilar terms, 41. 
 Division, 36. 
 
 by zero, 5, 137. 
 
 checking, 36, 74, 137. 
 
 law of exponents, 65. 
 
 of fractions, 70, 186. 
 
 of monomials, 65. 
 
 of polynomials by binomials, 
 76. 
 
 of polynomials by monomials, 
 74.7 
 
 of polynomials by polynomials, 
 135; 
 
 of radicals, 218. 
 
 with remainders, 79. 
 
 Elimination, 151, 162. 
 
 by addition or subtraction, 152. 
 
 by substitution, 154. 
 Equation, 5. 
 
 complete quadratic, 166, 169. 
 
 containing fractions, 98, 156, 
 160, 195. 
 
 cubic, 130. 
 
 degree of, 130. 
 
 double root, 132. 
 
 fractional coefficients, 98. 
 
 general quadratic, 175. 
 
 graphical solution of quad- 
 ratic, 179. 
 
 graph of, 89, 90. 
 
 history of quadratic, 183. 
 
 incomplete quadratic, 167. 
 
 involving parentheses, 54, 62. 
 
 linear, 90, 130. 
 
 literal, 51. 
 
 plotting, 89, 90. 
 
 quadratic, 130, 166. 
 
 quartic, 130. 
 203
 
 204 
 
 INDEX 
 
 Equation, radical, 220. 
 
 root of, 51, 132. 
 
 solution of, 6, 51. 
 Equations, inconsistent, 146. 
 
 literal, 158. 
 
 one linear, one quadratic, 180. 
 
 simultaneous linear, 147, 151, 
 
 162. 
 Exponent, 13, 14. 
 
 fractional, 201. 
 
 law for division, 65. 
 
 law for multiplication, 56. 
 
 negative, 206. 
 
 zero, 205. 
 Extraneous roots, 221. 
 
 Factor, 9. 
 
 highest common factor, 188. 
 
 prime, 112. 
 
 rationalizing, 218. 
 Factoring, 107. 
 
 checking, 113. 
 Formulas, distance, 8, 100. 
 
 interest, 3, 100. 
 Fractions, 65, 184. 
 
 addition, 193. 
 
 complex, 187. 
 
 division, 70, 186. 
 
 history of, 198. 
 
 multiplication, 68, 186. 
 
 multiplication by integer, 69. 
 
 reduction, 65, 185. 
 
 signs of terms, 69. 
 
 subtraction, 193. 
 Function, 89. 
 
 Gauss, 183. 
 
 Graphic representation, 84. 
 
 Graphs, construction and use of 
 
 graphs, 97. 
 identical, 147. 
 of linear equations, 90. / 
 of quadratic equations, / 179. 
 of simultaneous linear equa- 
 tions, 145. 
 parallel, 147. 
 
 practical application of, 94, 
 149. 
 
 Highest common factor, 188. 
 Hyperbola, 182. 
 
 Imaginaries, 199. 
 Imaginary unit, 178. 
 Income tax, 100. 
 Inconsistent equations, 146. 
 Index, of radical, 199. 
 
 of root, 15. 
 Interest, 100. 
 
 Least common denominator, 193. 
 Least common multiple, 190. 
 
 Magic squares, 49. 
 Monomial, 12. 
 
 addition, 41. 
 
 division, 65. 
 
 multiplication, 56. 
 
 subtraction, 41. 
 Multiplication, 33. 
 
 checking, 60. 
 
 fractions, 68, 186. 
 
 law of exponents, 56. 
 
 monomials, 56. 
 
 numbers, 33. 
 
 polynomials, 58, 60, 135. 
 
 radicals, 214. 
 
 Negative numbers, 24, 26. 
 
 Two uses of + and -, 28, 39. 
 Newton, frontispiece. 
 Number, absolute value of, 29. 
 
 integral, 112. 
 
 unknown, 5, 6. 
 Numbers, addition of, 26, 28, 29. 
 
 division of, 36. 
 
 imaginary, 178, 199. 
 
 irrational, 110, 199, 200. 
 
 multiplication of, 33. 
 
 negative, 24, 26. 
 
 positive, 24, 26. 
 
 rational, 199. 
 
 real, 199. 
 
 subtraction, 31. 
 
 Origin, 90. 
 
 Parabola, 180. 
 Parentheses, 17. 
 
 equations involving, 53, 62. 
 
 insertion of, 48. 
 
 removal of, 46. 
 PI, 134.
 
 INDEX 
 
 205 
 
 Plotting, 84, 90, 145, 146. 
 Polynomials, 12. 
 
 addition, 43. 
 
 division, 74, 135. 
 
 multiplication, 58, 60, 135. 
 
 subtraction, 43. 
 Portraits, Descartes, page 140. 
 
 Euler, page 40. 
 
 Newton, frontispiece. 
 
 Vieta, page 110. 
 Positive numbers, 24, 26. 
 
 uses of + and -, 28, 39. 
 Power, 13. 
 
 cube, 13. 
 
 fourth, 13. 
 
 square, 13. 
 Problems, algebraic solution, 6. 
 
 arithmetical solution, 6. 
 
 bank discount, 100. 
 
 distance, 100. 
 
 falling bodies, 232. 
 
 income tax, 100. 
 
 interest, 100. 
 
 lever, 231. 
 
 concrete mixture, 83. 
 
 taxes, 100. 
 
 Products, special, 101. 
 Proportion, 81. 
 
 direct, 83. 
 
 inverse, 83. 
 
 Quadratic equation, 130, 166. 
 complete, 166, 169. 
 general, 175. 
 graph of, 179. 
 history of, 183. 
 incomplete, 166, 167. 
 solved by completing 
 
 square, 171. 
 
 solved by factoring, 130. 
 solved by formula, 176. 
 
 Radical, 199. 
 
 coefficient of, 199. 
 
 equation, 220. 
 
 index of, 199. 
 
 sign, 15. 
 Radicals, 199. 
 
 addition, 212. 
 
 division, 218. 
 
 multiplication, 214. 
 
 the 
 
 Radicals, rationalization, 218. 
 
 reduction of, 208. 
 
 similar, 212. 
 
 simplify, 111. 
 
 subtraction, 212. 
 Ratio, 81. 
 
 Rational number, 199. 
 Recorde, 165. 
 Rectangle, area of, 63. 
 Root, cube, 15, 109. 
 
 fourth, 15. 
 
 index, 15 
 
 of an equation, 51. 
 
 principal, 108, 201. 
 
 square, 15, 108, 139, 142. 
 Roots, extraneous, 221. 
 
 imaginary, 178. 
 
 Sign, change of, in fractions, 69. 
 
 not equal to, 113, note. 
 
 of equality, 1. 
 
 of operation, 39. 
 
 of quality, 39. 
 
 radical, 15. 
 Similar, radicals, 212. 
 
 terms, 41. 
 Simultaneous equations, 151, 162, 
 
 180. 
 Square root, 15, 108, 139. 
 
 of fractions, 143. 
 
 of numbers, 142 
 Subtraction, 26, 28, 31. 
 
 checking, 32, 45. 
 
 fractions, 193. 
 
 monomials, 41. 
 
 polynomials, 43. 
 
 radicals, 212. 
 
 Term, 11. 
 
 Terms, dissimilar, 41. 
 
 similar, 41. 
 'Transposition, 51. 
 Trapczoid, 134. 
 
 area of, 134, 174. 
 Triangle, altitude, 134. 
 
 area of, 134. 
 
 hypothenuse, 134. 
 
 leg, 134. 
 Trinomial, 12. 
 
 Unknown, 5, 6.
 
 206 
 
 INDEX 
 
 Variables, 88, 89. 
 Vieta, 165. 
 Vinculum, 17. 
 
 Volume of rectangular solid, 64. 
 of cube, 64. 
 
 Whetstone of Witte, 165. 
 
 X-axis, 90. 
 F-axis, 90. 
 
 Zero, division by, 137. 
 exponent, 205. 
 operations with, 130. 
 
 Printed in the United States of America
 
 T 
 
 HE following pages contain advertisements of a 
 few of the Macmillan books on kindred subjects
 
 GEOMETRY 
 
 By Professor W. B. FORD, of the University of Michigan, and CHARLES 
 AMMERMAN, of the William McKinley High School, St. Louis, with the 
 editorial cooperation of Professor E. R. HEDRICK, of the University of 
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 Plane and Solid Geometry. Cloth, 12, ill., ix and 321 pages $1.25 
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 Solid Geometry. With Syllabus of Plane Geometry, 106 pages .80 
 
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 The editor of the series to which this book belongs was a member 
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 MATHEMATICS TEXTS BY PROFESSOR CAJORI 
 
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