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 COURSE 
 
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HIGH SCHOOL ALGEBRA 
 
 Elementary Course 
 
 BY 
 
 H. E. SLAUGHT, Ph.D. 
 
 ASSISTANT PROFESSOR OF MATHEMATICS IN THE UNIVERSITY 
 OF CHICAGO 
 
 AND 
 
 N. J. LENNES, M.S. 
 
 INSTRUCTOR IN MATHEMATICS IN THE WENDELL PHILLIPS 
 HIGH SCHOOL, CHICAGO 
 
 oXKc 
 
 Boston 
 
 ALLYN and BACON 
 1907 
 
COPYRIGHT, 1907, 
 BY H. E. SLAUGHT 
 AND N. J. LENNES. 
 
5<f3 
 
 PREFACE 
 
 The High School Algebra has been written in two parts, the 
 one designed solely for the first year and the other for a com- 
 plete review and advanced course at a later period. 
 
 The authors recognize the impossibility of combining in a 
 single treatment the qualities necessary for beginners with 
 the more mature point of view suitable for the third or fourth 
 year student. 
 
 The important features of the Elementary Course are : 
 
 1. Algebra is vitally and persistently connected with arithmetic. 
 
 Each principle in the book is first studied in its applica- 
 tion to numbers in the Arabic notation. The principles of 
 Algebra are thus connected with those already known in 
 arithmetic. Letters are introduced as abbreviations for " a 
 number" or "any number." Literal expressions are called 
 number expressions — the vague term "quantity" is not used. 
 In the exercises, Arabic figures are constantly involved in the 
 same manner as letters. Checking by the substitution of 
 parilcular numbers for letters occurs throughout. 
 
 2. TJie principles of Algebra used in the Elementary Course 
 are enunciated in a small number of short statements — eighteen 
 in all. 
 
 The purpose of these principles is to furnish in simple form 
 a codification of those operations of Algebra which are suffi- 
 ciently different from the ones already familiar in arithmetic 
 to require special emphasis. Such a codification has several 
 important advantages : 
 
 [5306144 
 
iv PREFACE 
 
 By constant reference to these few fundamental statements 
 they become an organic and hence permanent part of the 
 learner's mental equipment. 
 
 By their systematic use he is made to realize that the 
 processes of Algebra which seem so multifarious and hetero- 
 geneous are in reality few and simple. Invaluable training is 
 thus afforded in connecting things which are essentially, though 
 not apparently, related. 
 
 Such a body of principles furnishes a ready means for the 
 correction of erroneous notions, a constant incitement to effec- 
 tive review, and a definite basis upon which to proceed at each 
 stage of progress. 
 
 No attempt is made at formal demonstration of these prin- 
 ciples. It is believed that conviction as to their validity is 
 most effectively brought to a first year pupil by their proper 
 empirical exhibition. Formal argumentation is reserved for 
 the Advanced Course. 
 
 3. TJie mam purpose of the Elementary Course is the solution 
 of problems rather than the construction of a purely theoretical 
 doctrine as an end in itself 
 
 While the subject-matter of Algebra is developed in a logical 
 sequence around the principles, the attempt is made to connect 
 each principle in a vital manner with the learner's experience 
 by using it in the solution of a large number and great variety 
 of simple problems. 
 
 In making the problems the following criteria have been 
 observed : 
 
 The subject-matter should be easily within the comprehen- 
 sion of the pupil, and, so far as possible, the problems should be 
 such as one would actually need to solve in passing from known 
 to unknown data by means of given relations. Such are prob- 
 lems on physical relations in Chapter IV, and those on geomet- 
 rical relations in Chapters VI and VII. 
 
PREFACE V 
 
 However, with the pupil's meagre experience it is impossible 
 that all problems should be of this kind. Hence a considerable 
 body of problems has been introduced which involve artificial 
 relations imposed upon numbers. Such problems are of two 
 kinds : 
 
 (a) Those involving numbers related to concrete things, as, 
 for example, the problems on pages 44 to 46. The data used in 
 this class of problems (with very few exceptions) are such as 
 have a distinct value or real interest in themselves. Every 
 effort has been made to have the data entirely trustworthy. 
 
 (b) Those involving numbers not related to concrete things, 
 as, for example, problems 12 to 18, page 43. Such problems 
 are used as the basis for the development of formulas, pages 
 110 to 114, Chapter IV. 
 
 The utmost care has been taken in grading the problems, 
 both as to difficulty and with reference to the principles upon 
 which the solutions depend. 
 
 4. Tlie order of topics and the inclusion and exclusion of 
 subject-matter have been determined by the main purpose of the 
 Elementary Course as just stated. 
 
 The equation, therefore, as the instrument for the solution 
 of problems, occurjies the leading -place. New topics are intro- 
 duced only as they are needed in extending the use of the 
 equation. For example, the whole subject of linear equations, 
 including those with two and with three unknown quantities, 
 is treated before long division; and quadratic equatiens are 
 placed before the formal treatment of literal fractions. 
 
 The subject of factoring is introduced when quadratic equa- 
 tions are to be solved, and is immediately used for that 
 purpose in special cases. 
 
 Long division of polynomials is first found necessary as an 
 introduction to square root, which together with radicals is 
 essential to the solution of the general quadratic equation. 
 
vi PREFACE 
 
 The topics excluded are : (a) complicated factoring, (&) H. C.F. 
 by the long division process, (c) complicated fractions, (d) si- 
 multaneous equations in more than three unknowns, (e) cube 
 root, (/) fractional and negative exponents, (g) equations con- 
 taining complicated radicals, Qi) simultaneous quadratics except 
 the case of a quadratic and a linear equation. 
 
 These omissions have permitted the introduction of a much 
 greater number and larger variety of problems than would 
 otherwise be possible without diminishing the number of drill 
 exercises. 
 
 For example, instead of an extended and purely theoretical 
 discussion of radicals, only so much is given as is needed in 
 treating simple quadratic equations; and this is immediately 
 applied to the solution of a body of interesting problems in- 
 volving such concrete geometrical relations as are freely used 
 in modern grammar school arithmetics. 
 
 This larger space allotted to problems makes it possible to 
 pass by slow gradation from extremely simple to more compli- 
 cated cases which would otherwise be too difficult, and thus to 
 secure the facility and power in interpreting and solving prob- 
 lems which is demanded in physics and other sciences. 
 
 Attention is further called to the following features : 
 
 The simple and scientific treatment of the solution of the 
 equation as enunciated in Principle VIII, page 36. 
 
 The numerous illustrative solutions which have been given 
 in full for the purpose of exhibiting the best methods of attack 
 and the most effective arrangement of the work. 
 
 The development of negative numbers from concrete rela- 
 tions, immediately followed by a large number of problems 
 showing their practical use and interpretation. 
 
 The introduction of graphs at the beginning of the chapter 
 on simultaneous equations, thus making the graph the basis 
 of the study of simultaneous equations. 
 
PREFACE Vll 
 
 The notions of a formula and of substitution in a formula 
 are developed in connection with subjects already familiar, 
 such as interest, areas, volumes, etc. (See page 101.) These 
 notions are then applied to the study of other subjects. (See 
 page 115.) 
 
 Literal equations are in each case introduced as a generaliza- 
 tion of a series of concrete problems immediately preceding. 
 (See page 111.) Such equations are then to be solved for each 
 letter involved. (See page 127.) 
 
 Problems involving physical relations are in each case intro- 
 duced by means of a series of carefully graded problems lead- 
 ing up to the general case. (See page 120.) 
 
 If in any case the problems are found to be too numerous, 
 the later sections of Chapter IV may be omitted or postponed 
 until after graphs and simultaneous equations have been 
 studied. Many of the exercises are simple enough to be read 
 in class as mental drill exercises. 
 
 Keview questions and exercises are given at appropriate 
 intervals throughout the book. 
 
 Ratio and proportion are treated in the chapter on fractions 
 preparatory to their use in plane geometry. 
 
 The authors desire to express their thanks to many who 
 have given useful suggestions, and especially to Mr. J. A. 
 Dixon, Wendell Phillips High School, Chicago, and to Miss 
 Mabel Sykes, South Chicago High School, Chicago, who have 
 solved all the problems and exercises and whose criticisms 
 have been of special value. Thanks are also due to Mr. Edwin 
 Hand, Jr., for helping to prepare the cuts. 
 
 H. E. SLAUGHT. 
 N. J. LENNES. 
 Chicago, May, 1907. 
 
CONTENTS 
 
 CHAPTER I 
 INTRODUCTION TO THE EQUATION 
 
 PAGE 
 
 Peinciple I. Addition of Numbers having a Common Factor . 1 
 
 Peinciple II. Subtraction of Numbers having a Common Factor 9 
 
 Peinciple III. Multiplication of the Product of Several Factors 12 
 Pbinciple IV. Multiplication of the Sum or Difference of Two 
 
 Numbers 14 
 
 Peinciple V. Division of the Product of Several Factors . . 20 
 Peinciple VI. Division of the Sum or Difference of Two 
 
 Numbers 23 
 
 Peinciple VII. Number Expressions in Parentheses . 28 
 
 Peinciple VIII. Identities and Equations 34 
 
 Directions for Written Work 38 
 
 Equations involving Fractional Coefficients ... 41 
 
 CHAPTER II 
 POSITIVE AND NEGATIVE NUMBERS 
 
 A New Kind of Number .... 
 Peinciple IX. Addition of Signed Numbers 
 
 Addition by Counting 
 
 Averages of Signed Numbers 
 Peinciple X. Subtraction of Signed Numbers . 
 
 Peinciple XL Multiplication of Signed Numbers 
 Peinciple XII. Division of Signed Numbers 
 
 Interpretation and Use of Signed Numbers in Problems 
 
 47 
 49 
 53 
 55 
 58 
 62 
 65 
 67 
 
PAGE 
 
 72 
 74 
 82 
 88 
 92 
 
 X CONTENTS 
 
 CHAPTER III 
 
 INVOLVED NUMBER EXPRESSIONS 
 
 Double Use of the Signs + and — 
 Addition and Subtraction of Polynomials 
 Principle XIII. Multiplication of Polynomials 
 Squares of Binomials .... 
 Review Questions and Exercises . 
 
 CHAPTER IV 
 
 SOLUTION OF PROBLEMS 
 
 Problems involving Interest 100 
 
 Problems involving Areas 105 
 
 Problems involving Volumes 108 
 
 Problems involving Simple Number Relations . . . 110 
 
 Problems involving Motion 115 
 
 Problems involving the Simple Lever 120 
 
 Problems involving Densities 122 
 
 Problems involving Momentum 125 
 
 Problems involving Thermometer Readings . . . 128 
 
 Problems involving the Arrangement and Value of Digits . 130 
 
 Review Questions and Problems 131 
 
 CHAPTER V 
 
 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 
 Graphic Representation of Statistics 138 
 
 Graphic Representation of Motion 142 
 
 Graphic Representation of Equations 146 
 
 Solution of Simultaneous Equations by Substitution . . 153 
 Solution of Simultaneous Equations by Addition or Sub- 
 traction 156 
 
 Problems involving Two Linear Equations .... 160 
 
CONTENTS 
 
 XI 
 
 Linear Equations in Three Variables 
 Problems involving Three Unknowns 
 Keview Questions 
 
 PAGE 
 
 166 
 168 
 171 
 
 CHAPTER VI 
 
 SPECIAL PRODUCTS AND FACTORS 
 
 Principle XIV. Products of Powers of the same Base 
 Principle XV. Products of Monomials 
 
 Factors of Number Expressions 
 
 Monomial Factors 
 
 Trinomial Squares 
 
 The Difference of Two Squares 
 
 The Sum of Two Cubes 
 
 The Difference of Two Cubes 
 
 Trinomials of the Form x 2 + (a + 6) x + db 
 
 Trinomials of the Form ax 2 + bx + c 
 
 Factors found by Grouping . 
 
 Equations and Problems solved by Factoring 
 
 172 
 175 
 177 
 178 
 179 
 183 
 185 
 186 
 188 
 191 
 193 
 197 
 
 CHAPTER VII 
 
 QUOTIENTS AND SQUARE ROOTS 
 
 Principle XVI. Quotients of Powers of the same Base . . 210 
 
 Principle XVII . Division of Monomials ..... 212 
 
 Principle XVIII. Square Roots of Monomials .... 214 
 
 Division of Polynomials 216 
 
 Square Roots of Polynomials 221 
 
 Square Roots of Numbers in the Arabic Notation . . 225 
 
 Problems involving Square Roots 234 
 
 Solution of Quadratic Equations by Means of Square Root 240 
 
 Quadratic and Linear Equations 246 
 
 Problems involving Quadratics 247 
 
 Review Questions 253 
 
xii CONTENTS 
 
 CHAPTER VIII 
 FRACTIONS WITH LITERAL DENOMINATORS 
 
 PAGE 
 
 Common Factors 255 
 
 Common Multiples 257 
 
 Reduction of Fractions to Lowest Terms .... 260 
 
 Reduction of Fractions to a Common Denominator . . 262 
 
 Addition and Subtraction of Fractions .... 267 
 
 Multiplication and Division of Fractions .... 270 
 
 Complex Fractions 277 
 
 Ratio and Proportion 279 
 
 Equations involving Fractions 283 
 
 Simultaneous Equations involving Fractions . . . 289 
 
 Problems involving Fractions . . . . . . 291 
 
HIGH SCHOOL ALGEBRA 
 
 ELEMENTARY COURSE 
 
 CHAPTER I 
 
 INTRODUCTION TO THE EQUATION 
 
 ADDITION OF NUMBERS HAVING A COMMON FACTOR 
 
 1. Algebra like arithmetic deals with numbers. The num- 
 bers of algebra include those used in arithmetic, and also other 
 numbers which are defined as need arises for them. The sym- 
 bols employed to represent numbers, and the principles used 
 in operating upon them, are introduced by means of illustra- 
 tive problems as we proceed. 
 
 2. Illustrative Problem. The shortest railway route from 
 Chicago to New York is 912 miles. How long does it take a 
 train averaging 38 miles an hour to make the journey ? 
 
 Solution in words. The product of the average number of miles per 
 hour and the required number of hours equals the whole distance 
 traveled. That is, 38 multiplied by "the required number of hours " 
 equals 912. Hence " the required number of hours " is one thirty- 
 eighth of 912, or 24. 
 
 Solution using abbreviations. If, instead of the expression "the 
 required number of hours," we use the word "time," or simply the 
 abbreviation t, the solution may be written : 
 
 38 x t = 912. 
 Hence t = 912 + 38 = 24. 
 
 1 
 
2 INTRODUCTION TO THE EQUATION 
 
 Illustrative Problem. If in the above problem a train makes 
 the journey in 18 hours, find the average number of miles per 
 hour. 
 
 Solution. In words we have, as before, 18 multiplied by " the aver- 
 age number of miles per hour " equals 912. Hence " the average 
 number of miles per hour " equals one-eighteenth of 912, or 50§. 
 
 Using for the expression " the average number of miles per hour," 
 the word " rate," or simply the abbreviation r, the solution reads : 
 
 18 x r = 912. 
 
 r * 912 + 18 = 50$. 
 
 It should be clearly understood that t and r represent num- 
 bers whose values are unknown at the outset, but which become 
 known at the conclusion of the solution. 
 
 3. Abbreviations representing numbers and operations such as 
 are used in these problems occur constantly in algebra. In 
 fact, the systematic use of such abbreviations is one of the 
 chief distinctions between arithmetic and algebra. 
 
 4. The signs +, — , x , ■+-, and = are used in algebra with the 
 same meaning that they have in arithmetic. However, instead 
 of x a point written above the line is often used. Thus 2 • 3 
 means 2x3. 
 
 The product of two numbers represented by letters is gener- 
 ally indicated by writing the letters consecutively with no sign 
 between them. 
 
 Thus rt means the same as r • t or r x t. For example, in such prob- 
 lems as those just given we would write rt = d, meaning " the number of 
 miles per hour multiplied by the number of hours equals the distance 
 or number of miles traveled." 
 
 Similarly 2 r means 2 • r, but (as in arithmetic) 25 means 20 + 5, 
 not 2 • 5. 
 
 A fraction is often used to indicate division 
 
 Thus ? = 2-^3; - = r-<. 
 
 3 t 
 
ADDITION OF NUMBERS 3 
 
 PROBLEMS 
 
 In the same manner as above solve the following problems, 
 first by writing out the solution in words and then by using 
 such abbreviations as may be found convenient. 
 
 1. Five times a certain number equals 80. What is the 
 number ? Use n for the number. 
 
 2. Twelve times a number equals 132. What is the number ? 
 
 3. A tank holds 750 gallons. How long will it take a pipe 
 discharging 15 gallons per minute to fill the tank ? 
 
 4. The cost of paving a block on a certain street was $ 7 per 
 front foot. How long was the block if the total cost was $ 4620? 
 
 5. A city lot sold for $ 7500. What was the frontage if 
 the selling price was $ 225 per front foot ? 
 
 6. An encyclopedia contains 18,000 pages. How many 
 volumes are there if they average 750 pages to the volume ? 
 
 7. What is the cost of fencing per rod if it requires $ 256 
 to fence a quarter section of land ? 
 
 8. A square court 60 feet on a side is to be paved with 
 square tiles. How many square feet in each tile if the pave- 
 ment requires 1600 tiles ? 
 
 9. An excavation for a building is to be 130 feet long, 
 80 feet wide, and 9 feet deep. If 900 cubic feet are removed 
 each day, how long will it require to complete it ? 
 
 10. A railway embankment which contains 48,900 cubic 
 yards of earth was completed in 163 days. At what rate was 
 it filled in ? 
 
 11. Taking the length of the earth's orbit as 584 million miles, 
 find how far the earth travels in one day ; also in one hour. 
 
 5. Definition. If a number is the product of two or more num- 
 bers, then these numbers are called factors of the given number. 
 
 E.g. The integral factors of 10 are 1 and 10 or 2 and 5. 2\ and 
 4 are also factors of 10. 
 
4 INTRODUCTION TO THE EQUATION 
 
 6. Illustrative Problem. Divide the number 84 into three 
 parts such that the second part is five times the first and the 
 third part is eight times the first. 
 
 Solution. Using the abbreviation p for " the first part of the 
 number," we have 1 . porp = the first part> 
 
 5p = the second part, 
 8 p = the third part. 
 
 Since the sum of the three parts is 84, 
 p + 5p + &p = 84. 
 
 To complete the solution of this problem it is necessary to find the 
 sum of the numbers p, 5p, and 8p without knowing what number is 
 represented by p itself. If we suppose this sum to be lip, then 
 
 Up a 84 
 and p = 84 +■ 14 = 6. 
 
 This supposition leads to the correct result since, if p = 6, then 
 />+5j9 + 8jp = 6 + 5.6 + 8.6=6 + 30 + 48=84. Hence the three 
 parts into which 84 is divided are 6, 30, and 48. 
 
 7. The process here used for adding p, 5p, and 8p is a new 
 method of adding which is of very great importance in 
 algebra. This method is further exhibited by the following 
 examples : 
 
 (1) To add 18, 42, 54, and 30, we first factor these numbers so as 
 to show the common factor 6 and then add the remaining factors 3, 7, 
 9, and 5, and multiply this sum by the common factor 6. The result 
 is 24 • 6, or 144. The work may be arranged as follows: 
 
 18 a 3-6 Similarly (2) 
 
 42= 7-6 
 
 54= 9-6 
 
 30= 5-6 
 
 16 = 
 
 1 
 
 16 = 
 
 2 
 
 8 = 
 
 4 
 
 1 
 
 64 = 
 
 4 
 
 16 = 
 
 8 
 
 8 = 
 
 16 
 
 1 
 
 32 = 
 
 2 
 
 16 = 
 
 4 
 
 8 = 
 
 8 
 
 4 
 
 48 = 
 
 3 
 
 16 = 
 
 6 
 
 8 = 
 
 12 
 
 4 
 
 160 = 
 
 10 
 
 16 = 
 
 20 
 
 8 = 
 
 40 
 
 i 
 
 144 = 24 . 6 
 
 From example (2) we see that in case the numbers to be 
 added have two or more common factors it does not matter 
 
ADDITION OF NUMBERS 5 
 
 which one is selected. If each number is separated into tivo 
 factors one of which is a common factor, then it is evident 
 that the sum in any case is found by multiplying the common 
 factor by the sum of the other factors. 
 
 In the above manner add the following sets of numbers and 
 show in each case by adding in the ordinary way that the result 
 is correct : 
 
 14 
 
 36 
 
 32 
 
 17 
 
 39 
 
 32 
 
 28 
 
 72 
 
 64 
 
 34 
 
 52 
 
 48 
 
 35 
 
 48 
 
 128 
 
 68 
 
 78 
 
 40 
 
 70 
 
 12 
 
 256 
 
 85 
 
 91 
 
 72 
 
 8. Definition. If a number is the product of two factors, 
 then either of these factors is called the coefficient of the other 
 in that number. 
 
 E.g. In 2 • 3, 2 is the coefficient of 3, and 3 is the coefficient of 2. 
 In 9 rt, 9 is the coefficient of rt, r is the coefficient of 9 t, and t is the 
 coefficient of 9 r. In such expressions as 9 rt the factor represented by 
 Arabic figures is usually regarded as the coefficient. 
 
 The preceding examples illustrate the following principle : 
 
 9. Principle I. To add numbers having a common 
 factor, add the coefficients of the common factor and 
 multiply tJie sum by the common factor. 
 
 If n represents some number, then 4»i, in the same dis- 
 cussion, means four times that number, and 5 n means five 
 times that number. Hence, by Principle I, 4 n -f- 5 n = 9 n, 
 which is true no matter what number is represented by n. 
 
 By means of this principle perform the following additions, 
 understanding that each letter represents some number : 
 
 1. Sx + 7 a + 16 a +2 a. 4. 7b+8b +& + 6&. 
 
 2. 13 n + 8 n + 7 n + 9 n. 5. 8 £ + 7 t + 5t. 
 
 3. 3a + a + 2 a + 6 a. 6. 3r + 5r + llr. 
 
6 INTRODUCTION TO THE EQUATION 
 
 Show the correctness of the result in each of the above by 
 letting x = 2, n = 1, a = 4, 6 = 3, t = 6, and r = 7. Try also 
 other values for the letters. 
 
 Such a test for the correctness of an operation is called a check. 
 
 10. Definitions. Combinations of Arabic figures or letters, or 
 both, by means of the signs of operation, +, — , etc., are called 
 number expressions. 
 
 E.g. 38, 18 r, p + 5 p + 8p, are number expressions. 
 
 Two number expressions representing the same number, when 
 connected by the sign =, form an equality. 
 
 The expressions thus connected are called the members of the 
 equality and are distinguished as the right and left members. 
 
 Equalities, such as 8t-\-7 1 = 15t, in which the letters may 
 be any numbers whatever, are called identities. Equalities of 
 the type p + 5p + 8p = 84 are called equations. (See § § 29-34.) 
 
 EXERCISES 
 
 1. Add 5 1, 11 t, 20 t, and 47 t. Check the result by letting 
 t = 3 ; also t = 50, and t = 150. 
 
 2. Add 7 r, 23 r, 28 r, 52 r, and 117 r. Check the result for 
 r = 11, r = 20, and r = 1. 
 
 3. Add 3 rt, 7 rt, 65 rt, and 16 rt. Check for r = 1, t = 2. 
 
 4. Add 1^ w, 2| w, 3^ n, and check for n = 6. 
 
 5. Add 66, 88, 99, and 121 by Principle I. 
 
 6. Add 144, 96, 120, and 50 • 12 by Principle I. 
 
 7. Add 5 • 40, 8 . 60, and 6 • 20 by Principle I. 
 
 8. Add 5 ax, 3 ax, and 7 ax. Check for a = 2, x = 4. 
 
 9. Add 7 ax, 3 bx, and 12 ex. 
 
 In this case the common factor is x. Hence using Principle I, 
 
 7 ax + 3 bx + 12 ex = (7 a + 3 6 + 12 c)x. The parenthesis here indicates 
 that the numbers represented by x and the expression 7 a + 3 b + 12 c 
 are to be multiplied. 
 
ADDITION OF NUMBERS 
 
 SOLUTION OF PROBLEMS 
 
 11. One great object in the study of algebra is to simplify 
 the solution of problems. This is done by using letters to repre- 
 sent the unknown numbers, by stating the problem in the form 
 of an equation, and by arranging the successive steps of the 
 solution in an orderly manner. 
 
 Skill in translating problems into equations depends upon 
 attention to the following points : 
 
 (1) Read and understand clearly the statement of the prob- 
 lem, as it is given in words. 
 
 (2) Select the unknown number, and represent it by a suit- 
 able letter, say the initial letter of a word which will keep its 
 meaning in mind. If there are more unknown numbers than 
 one, try to express the others in terms of the one first selected. 
 
 (3) Find two number expressions which, according to the 
 problem, represent the same number, and set them equal to 
 each other, thus forming an equation. 
 
 These steps are exhibited in the following solution : 
 
 A tree 108 feet high was broken off by the wind so that the 
 
 part left above the first branch was three times as long as the 
 
 part broken off, and the part below the first branch was twice as 
 
 long as the part broken off. How long was the part broken off ? 
 
 Solution. Let 6 represent the number of feet broken off. 
 
 Then 36 is the number of feet left above the first branch, 
 and 2 6 is the number of feet below the first branch. 
 
 Hence, b + 36 + 26 and 108 are number expressions, each repre- 
 senting the total height of the tree. 
 
 Therefore 6 + 36 + 26 = 108. (1) 
 
 By Principle I, 66 = 108. (2) 
 
 Then 6 = one sixth of 108, or 18. (3) 
 
 Hence, the part broken off was 18 feet long. 
 
 Equation (3) is derived from (2) by dividing both members by 6. 
 
8 INTRODUCTION TO THE EQUATION 
 
 PROBLEMS 
 
 1. The greater of two numbers is 5 times the less, and 
 their sum is 180. What are the numbers ? 
 
 2. A number increased by twice itself, 4 times itself, 
 and 6 times itself, becomes 429. What is the number ? 
 
 3. A father is 3 times as old as his son, and the sum of 
 their ages is 48 years. How old is each ? 
 
 4. In a company there are 39 persons. The number of 
 children is twice the number of grown people. How many 
 are there of each ? 
 
 5. A and B receive $45 for doing a certain piece of work. 
 If A gets 4 times as much as B, how much does each receive ? 
 
 6. The population of Tokio is twice that of Canton, and the 
 sum of their populations is 2,700,000. How many inhabitants 
 in each city ? 
 
 7. Find two consecutive integers whose sum is 133. 
 
 8. The area of Louisiana is (nearly) 4 times that of 
 Maryland, and the sum of their areas is 60,930 square miles. 
 Find the (approximate) area of each state. 
 
 9. The horse-power of a certain steam yacht is 12 times 
 that of a motor boat. The sum of their horse-powers is 195. 
 Find the horse-power of each. 
 
 10. There are three circles on the blackboard. The circum- 
 ference of the second is 5 times that of the first, and the 
 circumference of the third is 10 "times that of the first. The 
 sum of their circumferences is 16 feet. Find the circumfer- 
 ence of each. 
 
 11. At a football game there were 2000 persons. The num- 
 ber of women was 3 times the number of children, and the 
 number of men was 6 times the number of children. How 
 many men, women, and children were there ? 
 
SUBTRACTION OF NUMBERS 9 
 
 12. The population of Portland, Oregon (estimate of the 
 Census Bureau, 1904), was twice that of Dallas, Texas, and the 
 population of Toledo was 3 times that of Dallas. The 
 three cities together had 300 thousand inhabitants. How 
 many were there in each city? 
 
 Let >i = number of thousands of inhabitants in Dallas. 
 Then 2ra = number of thousands of inhabitants in Portland, 
 and 3 n = number of thousands of inhabitants in Toledo. 
 Hence n + 2n + 3n = 300. 
 
 13. It is twice as far from New York to Syracuse as from 
 New York to Albany, and it is 4 times as far from New 
 York to Cleveland as from New York to Albany. The sum of 
 the three distances is 1015 miles. Find each distance. 
 
 14. In Maryland there were (census of 1900) 4 times as 
 many whites as negroes. The total population was 1185 thou- 
 sand. How many of each were there ? 
 
 If n equals the number of thousands of negroes, then the equation 
 is n + in = 1185. 
 
 SUBTRACTION OF NUMBERS HAVING A COMMON FACTOR 
 12. Numbers having a common factor may be subtracted in 
 a manner similar to the process exhibited under Principle I. 
 
 Thus, from 64 = 8 • 8 From 84 = 12 - 7 From 17 n 
 
 subtract 48 = 6-8 subtract 49 = 7-7 subtract 6n 
 Eemainder 16 = 2 . 8 35= 5-7 liw 
 
 In like manner perform the following subtractions : 
 
 1. 9.7-3-7. 5. 6.99-5.99. 9. Qn-2n. 
 
 2. 10-4-6-4. 6. 20-19-13-19. 10. 6-50-2-50. 
 
 3. 8.8-2-8. 7. 8a -3a. 11. 106-4 6. 
 
 4. 5-11-3-11. 8. 8-5-3-5. 12. 7a -4a. 
 
10 INTRODUCTION TO THE EQUATION 
 
 These examples illustrate the following principle : 
 
 13. Principle II. To find the difference of two numbers 
 having a common factor, subtract the coefficients of the 
 common factor and multiply the result by tlie common 
 factor. 
 
 Illustrative Problem. If thirteen times a certain number 
 diminished by eight times the number equals 75, what is the 
 number ? 
 
 Solution. Let n represent the required number. 
 Then 13 n — 8 n and 75 are expressions representing the same 
 number. 
 
 Hence, 13 n — 8 n =75. 
 
 By Principle II, 5n = 75. 
 
 Dividing each member by 5, n =15, the required number. 
 
 Check. 13 • 15 - 8 • 15 = 195 - 120 = 75. 
 
 EXERCISES AND PROBLEMS 
 
 1. By means of Principle II subtract 72 from 160 ; 
 50 from 300; 39 from 78; 34 from 85; 58 from 174; and 
 69 from 161. 
 
 2. Subtract 109 . 87 from 209 • 87 by Principle II. 
 Check by first finding the products and then subtracting as in 
 arithmetic. 
 
 Perform the following indicated operations and check 
 those in which letters are involved by substituting convenient 
 numbers : 
 
 3. 6St-llt. 7. 3.4n+5.4w+11.4w-7-4n. 
 
 4. 15w+25w-18n. 8. 13rt+16rt+3rt-20rt. 
 
 5. 70aj-15»+7aj-23a?. 9. 144-96+50.12-20-12. 
 
 6. 18- 7-3- 7- 2- 7+6- 7. 10. lla«-3aa;+4aa;. 
 
SUBTRACTION OF NUMBERS 11 
 
 11. 11 ax — dbx + kcx. 19. ar + br — cr. 
 
 12. 11- 9 -6 -9 + 3 -9. 20. 3ry-2sy-ty. 
 
 13. 20»-6n + 2w. 21. 11.17 + 47-17-8.17. 
 
 14. an — bn + en. 22. accy + bxy — 3 ay. 
 
 15. 5* + 2(H-3£. 23. 3a&c + 7a6c-2a&c. 
 
 16. 8s-3s + 20s. 24. 7 -5 a -3 -5^ + 8. 5 x. 
 
 17. 6a-4a + 3a-2a. 25. a«2r + 6-2r — c-2r. 
 
 18. llrs — 2rs + 4rs. 26. 2ar + 26r — 2cr. 
 
 27. Four times a certain number plus 3 times the number 
 minus 5 times the number equals 48. What is the number ? 
 
 28. One number is 4 times another, and their difference is 9. 
 What are the numbers ? 
 
 29. Find a number such that when 4 times the number is 
 subtracted from 12 times the number the remainder is 496. 
 
 30. The population of Ohio (1901) was twice that of Wis- 
 consin. The difference of their populations was 2100 thou- 
 sand. Find the population of each state. 
 
 31. The population of Illinois in 1903 was 5 times as great 
 as that of West Virginia. The difference between their popu- 
 lations was 4080 thousand. What was the population of each ? 
 
 32. There are three numbers such that the second is 11 times 
 the first and the third is 27 times the first. The difference 
 between the second and the third is 64. Find the numbers. 
 
 33. A cubic foot of asphaltum is twice as heavy as a cubic 
 foot of light anthracite coal. Seven cubic feet of coal weigh 390 
 pounds more than 1 cubic foot of asphaltum. Find the weight 
 per cubic foot of each. 
 
 34. Thirty-nine times a certain number, plus 19 times the 
 number, minus 56 times the number, plus 22 times the num- 
 ber, equals 12. Find the number. 
 
12 INTRODUCTION TO THE EQUATION 
 
 MULTIPLICATION OF A PRODUCT 
 
 14. Illustrative Problem. Three men, A, B, and C, invest 
 together $ 33,000. B puts in twice as much as A, and C 4 
 times as much as B. How much does each invest ? 
 
 Solution. Let d represent the number of dollars invested by A. 
 Then 2 d represents B's investment, and 4 (2 d) represents C's invest- 
 ment. 
 
 Hence, d + 2d + 4 (2d)= 33000. 
 
 To complete the solution of this problem it is necessary to multiply 
 2 c? by 4 without knowing what number is represented by d. If we 
 suppose the product to be 8 d, 
 then d + 2d + 8d = 33000. 
 
 By Principle I, 1 1 d = 33000. 
 
 Dividing each member by 11, d = 3000, A's investment ; 
 2c?= 6000, B's investment ; 
 4-26? = 24000, C's investment. 
 
 The supposition that 4 (2 d) = 8 d is justified by the fact that the 
 numbers thus found satisfy the conditions of the problem. 
 
 That is, 6?+2tf+4(2o?)= 3000 + 6000 + 24000 = 33000. 
 
 15. The process here used for multiplying 2 d by 4 is of great 
 importance in algebra. 
 
 The following examples further exhibit this new method of 
 multiplying : 
 
 1. 4(3- 5) =4. 15 = 60 2. 2(3-4.5) = 2-60 =120 
 
 Also 4(3-5) = 12- 5 = 60 Also 2(3 • 4 . 5)=6 -4 -5 =120 
 
 And 4(3 • 5) = 3 • 20 = 60 And 2(3 • 4 • 5) = 3 • 8 - 5 =120 
 
 And 2(3 -4- 5)= 3- 4- 10 = 120 
 
 In like manner find the following products in two or more 
 ways : 
 
 3. 6(2-3). 5. 2(5-149). 7. 20(5 a- 4). 9. 8(4*. 3). 
 
 4. 4(25-99). 6. 4(19-5). 8. 16 (4 s. 7). 10. 9(xy-5). 
 
MULTIPLICATION OF A PRODUCT 13 
 
 These examples illustrate the following principle : 
 
 16. Principle III. To multiply the product of several 
 factors by a given number, multiply any one of the factors 
 by that number, leaving the others unchanged. 
 
 EXERCISES AND PROBLEMS 
 
 Multiply as many as possible of the following in two or 
 more ways. Check where letters are involved. 
 
 7. 4(19-25). 13. 2(rs-16). 
 
 8. '5 (17- 20 -3). 14. 5(xy-5z). 
 
 9. 15(7a&). 15. 7(3-4a&). 
 
 10. 3 (4 ran). 16. 9(a-5-xy). 
 
 11. 5(abc). 17. 4(125-17). 
 
 12. 7(2xy). 18. 40(25-29). 
 
 19. There are three numbers whose sum is 80. The second 
 is 3 times the first and the third is twice the second. What are 
 the numbers ? 
 
 20. There are three numbers such that the second is 8 
 times the first and the third is 3 times the second. If the 
 second is subtracted from the third the remainder is 48. Find 
 the numbers. 
 
 21. The population of Bridgeport, Connecticut, is twice that 
 of Butte, Montana. Three times the population of Bridgeport 
 plus twice that of Butte equals 320 thousand. Find the popu- 
 lation of each city. 
 
 22. It is 4 times as far from New York City to Cincinnati 
 as from New York to Baltimore. Twice the distance from New 
 York to Cincinnati minus 5 times that from New York to Bal- 
 timore equals 567 miles. How far is it from New York to each 
 of the other cities ? 
 
 1. 
 
 7(3-4-5). 
 
 2. 
 
 8(7-2-3). 
 
 3. 
 
 9(2-3-4). 
 
 4. 
 
 5(2 ab). 
 
 5. 
 
 3(5 xy). 
 
 6. 
 
 12(8-4-20) 
 
14 INTRODUCTION TO THE EQUATION 
 
 23. The population of Hartford, Connecticut, is 3 times that 
 of Oshkosh, Wisconsin. Four times the population of Hart- 
 ford plus 5 times that of Oshkosh equals 510 thousand. Find 
 the population of each city. 
 
 24. One cubic inch of emery weighs twice as much as 1 
 cubic inch of ivory. The combined weight of 10 cubic inches 
 of each substance is 2.1 pounds. Find the weight per cubic 
 inch of each. 
 
 25. It is twice as far from Boston to Quebec as from Boston 
 to Albany and 3 times as far from Boston to Jacksonville, 
 Florida, as from Boston to Quebec. How far is it from Boston 
 to each of the other three cities, the sum of the distances being 
 1818 miles ? 
 
 26. A cubic inch of porcelain china is twice as heavy as 
 a cubic inch of ebony, and a cubic inch of rolled zinc is. 3 
 times as heavy as a cubic inch of porcelain. The combined 
 weight of 1 cubic inch of each substance is .387 pounds. Find 
 the weight per cubic inch of each. 
 
 MULTIPLICATION OF THE SUM OR DIFFERENCE OF TWO NUMBERS 
 
 17. Illustrative Problem. The length and width of a rectan- 
 gle together equal 58 inches. If the width were 5 inches 
 greater, the length of the rectangle would then be twice its 
 width. Find its dimensions. 
 
 Solution. Let 10 represent the number of inches in the width. If 
 it were 5 inches wider, it would then be w + 5 inches. Hence the 
 < length is 2 (w + 5), the parenthesis indicating that the sum of w and 5 
 is to be found and the result multiplied by 2. 
 Hence the sum of the length and width is 
 
 id + 2 (to + 5) = 58. 
 The solution of this problem involves the multiplication of the 
 number (w + 5) by 2 without knowing what number is represented 
 by w. 
 
MULTIPLYING A SUM OR DIFFERENCE 15 
 
 If we suppose that 2 (w + 5) = 2 w + 10, 
 
 then we have w + 2 w + 10 = 58. (1) 
 
 By Principle I, 3 w + 10 = 58. (2) 
 
 Hence 3w = 48, since 58 is 10 more than 3 w, (3) 
 
 and w = 16, the width. (4) 
 
 Then 58 - 16 =42, the length. 
 
 The correctness of the above process is shown by the fact that the 
 numbers 42 and 16 fulfill the conditions of the problem ; 
 that is, 2 (16 + 5) = 2 . 21 = 42, 
 
 and 16 + 42 = 58. 
 
 Equation (3) is derived from (2) by subtracting 10 from each 
 member. 
 
 18. The multiplication of the number (w + 5) by 2 without 
 knowing in advance what number is represented by w is a new 
 method of multiplying which is constantly used in algebra. The 
 following examples further exhibit this method : 
 
 (1) 4(2 + 7) = 4-9 = 36, 
 
 or 4(2 + 7)=4- 2 + 4- 7 = 8 + 28 = 36. 
 
 (2) 3(3 + 8 + 9) = 3- 20 = 60, 
 
 or 3(3 + 8 + 9) =3- 3 + 3 -8 +3- 9 = 9 + 24 + 27 = 60. 
 
 It is thus seen that in each case the same result is obtained 
 whether we first add the numbers in the parenthesis and then 
 multiply the sum or first multiply the numbers in the parenthe- 
 sis one by one and then add the products. 
 
 Multiply each of the following in two ways where possible : 
 
 1. 3(2 + 7). 5. 3(a + 6). 9. a(3 + 7 + 10). 
 
 2. 5(3 + 4 + 5). 6. ll(h + k). 10. 15(x + y + z). 
 
 3. 8(5 + 9+7). 7. 4(5a + 76 + c). 11. 20(ra+w+_p). 
 
 4. 7(6+11+9+9). 8. a(5 + 4 + 7). 12. 9(2r+3s+*). 
 
16 INTRODUCTION TO THE EQUATION 
 
 19. In a manner similar to the above the difference of two 
 numbers represented by Arabic figures may be multiplied by a 
 given number in either of two ways. ■ 
 
 E.g. 6(8-3) = 6 -5 = 30, 
 or 6(8-3) = 6-8-6- 3 = 48-18 = 30. 
 
 The same result is obtained whether we first perform the sub- 
 traction indicated in the parenthesis and then multiply the dif- 
 ference, or first multiply the numbers separately and then 
 subtract the products. In the case of numbers represented by 
 letters evidently the second process only is available. 
 
 E.g. 6(r-t) = 6r-6t. 
 
 Perforin as many as possible of the following multiplications 
 in two ways : 
 
 1. 7(9-2). 4. 17(18-11). 7. 5(3-1). 10. m{r-s). 
 
 2. 12(17-7). 5. 9 (a -2). 8. 3(y-2). 11. x{y-z). 
 
 3. 5(12-8). 6. 8 (ft -4). 9. a(c-d). 12. t(u~v). 
 
 The second of the above methods is needed in problems like 
 the following : 
 
 Illustrative Problem. The rate of an express train plus that 
 of a freight is 70 miles per hour. If the rate of the freight 
 were 7 miles less, the express would be going twice as fast as 
 the freight. Find the rate of each. 
 
 Solution. Suppose the rate of the freight is now r miles per hour. 
 
 Then r— 7 is the supposed rate of the freight, 
 
 and 2 (r — 7) is the present rate of the express. 
 
 Hence r + 2 (r - 7) = 70, the sum of the rates, (1) 
 
 and r + 2r- 14 = 70. (2) 
 
 By Principle I, 3 r - 14 = 70. (3) 
 
MULTIPLYING A SUM OR DIFFERENCE 17 
 
 Then 3 r = 84, since 70 is 14 less than 3 r. (4) 
 
 Hence r = 28, the rate of the freight, 
 
 and 2 (28 - 7 ) = 2 • 21 = 42, the rate of the express. 
 Check. 42 + 28 = 70. 
 
 Equation (4) is derived from (3) by adding 14 to both members 
 and observing that 14 — 14 = 0. 
 
 The foregoing examples illustrate the following principle : 
 
 20. Principle IV. To multiply the sum or difference of 
 two numbers by a given number, multiply each of the 
 numbers separately by the given number, and add or sub- 
 tract the products. 
 
 21. Principles III and IV should be carefully contrasted, as 
 in the following example : 
 
 2 (2 • 3 • 5).= 4- 3- 5 = 2- 6- 5 = 2- 3- 10, 
 
 but 2(2 + 3 + 5) =4 + 6 + 10. 
 
 In multiplying the product of several numbers we operate 
 upon any one of them, but in multiplying the sum or difference 
 of numbers we operate upon each of them. 
 
 EXERCISES AND PROBLEMS 
 
 1. Multiply 5 + 7 + 11 by 3 without first adding, and then 
 check by performing the addition before multiplying. 
 
 2. Multiply m + n by 4 and check for m = 5, n = 7. 
 
 4 (m + n) = 4 m + 4 n 
 Check. 4 (5 + 7) = 4 • 12 = 48, also 
 
 4 • 5 + 4 • 7 = 20 + 28 = 48. 
 
 3. Multiply r + s + x by a and check for r ==s = x = a = 2. 
 
 4. Multiply a + 6 + c by m and check for a = 1, b = 3, 
 c = 5, m = 4. 
 
 5. Multiply x + y by r and check for x = 2, y = 4, r = 6. 
 
18 INTRODUCTION TO THE EQUATION 
 
 Where letters are involved, check the results in the follow- 
 ing by substituting convenient values : 
 
 6. 8(13-5). 16. 8(2a-3&+4c). 
 
 7. 71(12 + 41-36). 17. 37(3x-2y-z). 
 
 8. 9(o-6 + 8). 18. 13(5r-3x + t). 
 
 9. 3(a + x + y -11). 19. 20(2a + 3x - 5y). 
 
 10. ra (a + & — c). 20. 7 (11 s — 2 s + 3 t). 
 
 11. a(18-7). 21. 35(x-2y + 3z). 
 
 12. 2 (13 + 8 + 9 - 21). 22. 78 (10 m + lln + 2 r). 
 
 13. 7(3 + 8 + 9 -a). 23. 4(25a; + 32a;-2/). 
 
 14. 5(17+ a -6). 24. 3(13 a? + 14 y - t). 
 
 15. 32 (a?— y-M). 25. 7(4a-36 + c). 
 
 26. Find two consecutive integers such that 3 times the first 
 plus 7 times the second equals 217. 
 
 27. Find two consecutive integers such that 7 times the first 
 plus 4 times the second equals 664. 
 
 28. There are three numbers such that the second is 17 less 
 than the first, and the third is 8 times the second. The sum of 
 the first and third is 89. What are the numbers ? 
 
 29. The number of representatives and senators together in 
 the United States Congress is 476. The number of represen- 
 tatives is 26 more than 4 times the number of senators. Find 
 the number of each. 
 
 30. The area of Illinois is 6750 square miles more than 10 
 times that of Connecticut. The sum of their areas is 61,640 
 square miles. Find the area of each state. 
 
 31. The sum of the horse-powers of the steamships Campania 
 and Mauritania is 102 thousand. The Mauritania has 12 
 thousand horse-power more than twice that of the Campania. 
 What is the horse-power of each ship? 
 
MULTIPLYING A SUM OB DIFFERENCE 19 
 
 32. The population of Wyoming (census of 1900) was 50 
 thousand more than that of Nevada, and the population of 
 Utah was 93 thousand more than twice that of Wyoming. 
 The population of Utah was 277 thousand. Find the popula- 
 tion of Nevada and Wyoming. 
 
 33. It is 73 miles farther from Newark to Philadelphia than 
 from New York to Newark, and it is one mile more than 10 
 times as far from Philadelphia to Chicago as from Newark to 
 Philadelphia. The sum of the distances from New York to 
 Newark and from Philadelphia to Chicago is 830 miles. Pind 
 each of the three distances, and the total distance from New 
 York to Chicago. 
 
 Let . d = number of miles from New York to Newark. 
 
 Then d + 73 = number of miles from Newark to Philadelphia, 
 
 and 10 (d + 73) + 1 = number of miles from Philadelphia to Chicago. 
 Hence d + 10 (d + 73) + 1 = 830. 
 
 34. In the championship season of 1906 the Chicago National 
 League baseball team lost 20 games less than New York, and 
 Pittsburg lost 12 less than twice as many games as Chicago. 
 Pittsburg and New York together lost 116 games. How many 
 did each of the three teams lose? 
 
 35. Pikes Peak is 3282 feet higher than Mt. ./Etna, and 
 Mt. Everest is 708 feet more than twice as high as Pikes Peak. 
 The sum of the altitudes of Mt. iEtna and Mt. Everest is 
 39,867 feet. Find the altitude of each of the three mountains. 
 
 36. The altitude of Chimborazo is 22,820 feet less than 
 3 times that of Mt. Shasta. What is the altitude of each 
 mountain if Chimborazo is 6060 feet higher than Mt. Shasta ? 
 
 Let x = the number of feet in the altitude of Mt. Shasta. Then 
 3x -22820 = altitude of Chimborazo, and 3a; - 22820 - x = 6060. 
 
 37. The distance of Mars from the sun is 39 million 
 miles less than 5 times as great as that of Mercury from the 
 sun. Mars is 105 million miles farther from the sun than 
 Mercury. What is the distance of each planet from the sun ? 
 
20 INTRODUCTION TO THE EQUATION 
 
 38. The population of New York City (estimate of Census 
 Bureau, 1904) was 5 thousand more than 11 times that of 
 Pittsburg. If 21 times the population of Pittsburg is sub- 
 tracted from twice that of New York, the remainder is 363 
 thousand. Find the population of each city. 
 
 39. The standing army of France (1906) was 383 thousand 
 less than twice that of Germany. If 7 times the number of 
 men in the German army is subtracted from 6 times the num- 
 ber of men in the French army, the remainder is 177 thousand. 
 Find the number of men in each army. 
 
 DIVISION OF A PRODUCT 
 
 22. The division of the product of several factors by a given 
 number may be performed in various ways : 
 
 E.g. (4- 6 -10) --2 = 240-=- 2 = 120. 
 
 Also (4.6.10)h-2 = 2-6.10 = 120, 
 
 (4.6.10)-2 = 4.3-10 = 120, 
 and (4- 6- 10)- 2 = 4- 6- 5 = 120. 
 
 In each case only one factor is divided, and since any factor 
 may be selected, we naturally choose one which exactly contains 
 the divisor if possible. 
 
 Perform each of the following divisions in more than one 
 way where possible : 
 
 1. (5 -8- 3) -=-2. 4. (11 • 20 • 16) -- 4. 7. (10 . 35 . 3) -5. 
 
 2. 20 abc-h 4. 5. 14 xyz '-*- 7. 8. 14 xyz -=- x. 
 
 3. 12a5e-i-3. 6. 12 abc + c. 9. (12 • 40 • 13) -s- 8. 
 
 These examples illustrate the following principle : 
 
 23. Principle V. To divide the product of several factors 
 by a given number divide any one of the factors by that 
 number, leaving the other factors unchanged. 
 
DIVISION OF A PRODUCT 21 
 
 Principle V is already known in arithmetic in the process 
 
 called cancellation. 
 
 2 • 6 • 9 
 Thus, in the fraction — , 3 may be canceled out of either 6 
 
 o 
 
 or 9, giving 2 ' 6 ' 9 = 2 . 2 . 9 or 2 • 6 . 3. 
 
 O 
 
 Principle V is necessary in the solution of problems like the 
 following : 
 
 Illustrative Problem. There are three numbers whose sum 
 is 26. The second is 8 times the first, and the third is one-half 
 the second. Find each of the numbers. 
 
 Solution. Let x represent the first number, 
 then 8 x represents the second number, 
 
 and — represents the third number. 
 
 8a; 
 Hence, x + 8 x -\ = 26, the sum of the numbers. 
 
 By Principle V, x + 8 x + 4 x = 26, since 8 x + 2 = 4 x. 
 
 By Principle I, 13 x = 26. 
 
 Hence, x = 2, the first number, 
 
 8 x = 16, the second number, 
 
 o_ 8.9 
 
 and — = — — = 8, the third number. 
 
 2 2 
 
 EXERCISES AND PROBLEMS 
 
 1. Multiply 2 x + 3 y by 5. Check for x = 7, y = 11. 
 
 2. Multiply 3 a + y by 8. Check f or a = 2, y = 3. 
 
 3. Divide 3 • 7 • 18 by 6 by means of Principle V. 
 
 4. Multiply 7 • 56 • 5 by 6 by means of Principle III. 
 
 5. Divide 21 • 36 • 42 by 7, leaving the result in two differ- 
 ent forms. 
 
 6. Multiply 5-13-27 by 3, leaving the result in three dif- 
 ferent forms. 
 
22 INTRODUCTION TO THE EQUATION 
 
 7. Divide 7 a • 14 b • 21 c by 7 in three different ways. 
 
 8. Add 5 a, 1 -^, %2, and 1|^, using Principles V and I. 
 
 9. From —p^- subtract x ^, using Principles V and II. 
 
 , _ ^ 14 a . 10 a i , ,6a 
 
 10. Irom — — H — — - subtract—— 
 
 — ■ o o 
 
 i , -pi- j j-v, ,. 16 a? 20 a; 16 a; „. •, Q 
 
 11. lind the sum of , , , 7 a, and 3 a:. 
 
 8 5 4 
 
 ,. • ja n 100 rs 90 rs , 25 rs 
 
 12. Find the sum of , , and — - — • 
 
 10 ' 9 5 
 
 13. From 25 xy subtract — — 2— 
 
 z 
 
 i-i a j a 18 abc OA , -, 30 be 
 
 14. Add , 20 be, and 
 
 a ' 10 
 
 15. Add , , and • 
 
 s a u 
 
 16. From i^M + 17a subtract ^^. 
 
 i M -c -(^n .39 mn , , . 25 am 
 
 17. From 179 m -\ ■ subtract 
 
 n a 
 
 18. From M(a + b) + U(a + b) subtract 7 (a + &) . 
 
 19. Cleveland had (estimate of Census Bureau, 1904) 8 times 
 as many inhabitants as Portland, Maine. If twice the popu- 
 lation of Portland is added to ^ that of Cleveland, the sum is 
 212 thousand. Find the population of each city. 
 
 20. One cubic foot of a certain kind of brick weighs as 
 much as 3 cubic feet of cedar. The combined weight of ^ of a 
 cubic foot of brick and 5 cubic feet of cedar is 187 pounds. 
 Find the weight per cubic foot of each. 
 
DIVISION OF A SUM OR DIFFERENCE 23 
 
 21. It is 8 times as far from Philadelphia to Louisville as 
 from Philadelphia to Baltimore. If \ the distance from Phila- 
 delphia to Louisville is added to 3 times that from Philadel- 
 phia to Baltimore, the sum is 485 miles. Find each of the two 
 distances. 
 
 22. Coinage silver weighs 4 times as much per cubic inch as 
 feldspar. The combined weight of \ a cubic inch of silver and 
 7 cubic inches of feldspar is .846 pound. What is the weight 
 of a cubic inch of each ? 
 
 23. It is 3 times as far from New York to Washington, D.C., 
 as from New York to New Haven, and it is 14 times as far 
 from New York to Seattle as from New York to Washington. 
 If \ the distance from New York to Seattle is added to 5 times 
 that from New York to New Haven, the sum is 836 miles. 
 Find the distance from New York to each of the other cities. 
 
 24. The population of Washington, D.C. (estimate of Cen- 
 sus Bureau, 1904), was 9 times that of Galveston, Texas, and 
 the population of Savannah, Georgia, was 33 thousand less 
 than \ that of Washington. The combined population of Gal- 
 veston and Savannah was 99 thousand. Find the popidation 
 of each city. 
 
 25. A cubic foot of steel weighs 17 times as much as a cubic 
 foot of yellow pine. The combined weight of 11 cubic feet of 
 pine and 3 cubic feet of steel is 1773.2 pounds. Find the 
 weight of 1 cubic foot of each. 
 
 DIVISION OF THE SUM OR DIFFERENCE OF TWO NUMBERS 
 24. In dividing the sum or difference of two numbers by a 
 given number, when these are represented by Arabic figures, 
 the process may be carried out in two ways. Thus, 
 
 (1) (12 + 8) --2=20-5- 2 = 10, 
 
 or (12 + 8) - 2 = 12 + 2 + 8 + 2 = 6 + 4 = 10. 
 
 (2) (20 - 12) + 4 = 8 - 4 = 2, 
 
 or (20 - 12) -s- 4 = 20 + 4 - 12 + 4 = 5 - 3 = 2. 
 
24 INTRODUCTION TO THE EQUATION 
 
 The same result is obtained in each case whether the num- 
 bers in the parenthesis are first added (or subtracted) and the 
 result then divided by the given number; or the numbers in 
 parenthesis are first divided separately and then the quotients 
 added (or subtracted). 
 
 If the numbers in the dividend are represented by letters, 
 the division can usually be carried out only in the second 
 manner shown above. 
 
 E.g. (r + t)+5 = r + 5+t + 5,or, r -±l = ^ + ±. 
 
 o 5 o 
 
 This is read : the sum of r and t divided by 5 equals r divided by 5 
 
 plus t divided by 5. 
 
 In this manner perform each of the following divisions in 
 two ways when possible : 
 
 1. (16 + 12) -h 4. 8. <» + $ + at) ■*- a 
 
 2. (20a- 10 6) - 5. 7. (15« - 3t + 18t)+3. 
 
 3. (36 x- 24^/) -s-6. 8. (18 a? + 36 y - 21 1) -s- 3. 
 
 4. (108y-72«)-s-12. 9. (m - n - r) -s- a. 
 
 5. (50 x + 75 x) -5- 25. 10. (m + n + r) -e- b. 
 
 These examples illustrate the following principle : 
 
 25. Principle VI. To divide the sum, or difference of 
 two numbers by a given number divide each number sep- 
 arately and find the sum or difference of the quotients. 
 
 Principle VI is necessary in problems such as the following : 
 
 Illustrative Problem. The population of Iceland is 40,500 less 
 than 10 times that of Greenland. The population of Green- 
 land plus \ that of Iceland is 27,600. Find the population of 
 each. 
 
DIVISION OF A SUM OR DIFFERENCE 25 
 
 Solution. Let x be the number of inhabitants in Greenland. 
 
 Then 10 x — 40500 is the number in Iceland. 
 
 u , 10 x - 40500 ___-. 
 
 Hence x + =27600. 
 
 5 
 
 By Principle VI, x + 2 x - 8100 = 27600. 
 
 By Principle I, 3 x - 8100 = 27600. 
 
 Adding 8100 to each member and observing that 8100 — 8100 = 0, 
 
 3 x =35700. 
 Therefore x = 11900, the population of Greenland, 
 
 and 10 x — 40500 = 78500, the population of Iceland. 
 
 Check. 1 1 900 + ^^ = 27600. 
 
 5 
 
 26. Principles V and VI should be carefully contrasted : 
 
 Thus : 12 * 1S ' 24 = 4- 18 • 24 = 12 • 6 • 24 = 12 • 18 • 8, 
 o 
 
 whi ,e 12 + ^ + 24=4 + 6 + 8. 
 
 That is, in dividing the product of several numbers we operate 
 upon any one of them as found convenient, but in dividing the 
 sum of several numbers we must operate upon each of them. 
 
 EXERCISES AND PROBLEMS 
 
 1. Divide 72 + 56 by 8 without first adding. 
 
 2. Divide 144 — 36 by 12 without first subtracting. 
 
 3. Divide r + t by 5 and check the quotient when r = 15, 
 t = 25; also when r = 60, t = 75. 
 
 4. Multiply 7 + 9 by 3 without first adding 7 and 9. 
 
 5. Multiply 25 — 8 by 5 without first subtracting. 
 
 6. Find the product of 12 and a + b, checking the result 
 when a = 5, b — 1. 
 
26 INTRODUCTION TO THE EQUATION 
 
 Perform the following indicated operations : 
 
 7. 3 (a + b + c + d). Check for a = 1, b = 2, c = 3, d = 4. 
 
 8. 7(r— s-\-t — x). Check for r = t = 5, s = x = 4. 
 
 9. (m + n + r) -h 4. Check for ra = 64, n = 32, »• = 8. 
 
 10. (x + y + z) -e- 5. Check for a? = 100, y = 50, and 2 = 25. 
 
 11. Add 6(o + b + c) and 6a + 126 + 24c . 
 
 o 
 
 12. From 25 (x + y + «) subtract 10 ° x + 50 ? + 25 z . 
 
 13 Add 6 ft6 + 7 ac +ad and 12 h + 15 c + 9 d 
 a 3 
 
 14. Perform the divisions: S3x ~^V and ™t-39s, 
 
 11 13 
 
 15. Divide 7 ma$ + 3 wa;?/ — 2 ram/ by y, and check for ra = 2, 
 
 n = 3, x = 4, y = 5. 
 
 16 Add 21 CTa? + 49 5y + 56 a& , 24 aa; + 56 by + 64 aft 
 7 n( 8 ' 
 
 and check for a=b = c = 2, x = y = 3. 
 
 17. If the sum of 4 times a number and 32 be divided by 2 
 the result is 30. Find the number. 
 
 18. The melting temperature of glass is 548 degrees (Centi- 
 grade) lower than 4 times that of zinc. One-half the number 
 of degrees at which glass melts plus 7 times the number at 
 which zinc melts equals 3434. Find the melting point of each. 
 
 19. The melting temperature of silver is 496 degrees (Centi- 
 grade) lower than that of nickel. Five times the number of 
 degrees at which nickel melts plus 7 times the number at 
 which silver melts equals 13,928. Find the melting point of 
 each metal. 
 
DIVISION OF A SUM OR DIFFERENCE 27 
 
 20. The population of Paris (1904) was 1360 thousand less 
 than twice that of Berlin. The sum of their populations was 
 4730 thousand. Find the population of each city. 
 
 21. A cubic foot of nickel weighs 1288 pounds less than 
 4 cubic feet of tin. One-half a cubic foot of nickel plus 1 cubic 
 foot of tin weighs 724 pounds. Find the weight per cubic foot 
 of each metal. 
 
 22. A cubic foot of gold weighs 2730 pounds less than 6 
 cubic feet of silver. One-third of a cubic foot of gold together 
 with 5 cubic feet of silver weighs 3675 pounds. Find the weight 
 per cubic foot of each metal. 
 
 23. The population of Japan (1904) was 106 million less 
 than 12 times that of Manchuria. If the population of Man- 
 churia be subtracted from \ that of Japan, the remainder is 12 
 million. Find the population of each country. 
 
 24. The population of Panama (1900) was 1160 thousand 
 less than 3 times that of Nicaragua (1905). Three times the 
 population of Panama plus twice that of Nicaragua is 2020 
 thousand. Find the population of each country. 
 
 25. The population of the Philippines (1903) was 400 thou- 
 sand less than 50 times that of Hawaii. One twenty-fifth of 
 the population of the Philippines plus the population of Hawaii 
 is 464 thousand. What was the population of each ? 
 
 26. One gallon of benzine weighs 45.4 pounds less than 8 
 gallons of alcohol. The weight of \ a gallon of benzine and 
 2 gallons of alcohol is 16.9 pounds. Find the weight of one 
 gallon of each liquid. 
 
 27. During the season 1906 the American League baseball 
 team of Chicago won 201 games less than 6 times as many as 
 Boston. One-third the number of games won by Chicago plus 
 7 times the number of games won by Boston equals 374. How 
 many games did each team win ? 
 
28 INTRODUCTION TO THE EQUATION 
 
 28. The altitude of Aconcagua is 70,800 feet less than 6 
 times that of Mt. Blanc. One-sixth the altitude of Aconcagua 
 added to that of Mt. Blanc equals 19,760 feet. Find the alti- 
 tude of each mountain. 
 
 29. The area of Great Britain is 30,387 square miles less 
 than 12 times that of the Netherlands, and the area of Japan 
 is 42,065 square miles less than 15 times that of the Nether- 
 lands. One-third the area of Great Britain plus £ the area 
 of Japan is 69,994 square miles. Find the area of each 
 country. 
 
 30. The diameter of the earth is 1918 miles more than 
 twice that of Mercury, and the diameter of Venus is 1700 
 miles more than twice that of Mercury. The diameter of the 
 earth plus \ that of Venus equals 11,768 miles. Find the 
 diameter of each planet. 
 
 31. The diameter of Jupiter is 500 miles more than 20 
 times that of Mars, and the diameter of Saturn is 4200 miles 
 more than 16 times that of Mars. One-tenth the diameter of 
 Jupiter plus \ that of Saturn is 45,150 miles. Find the 
 diameter of each planet. 
 
 32. The diameter of Neptune is 29,000 miles less than 
 twice that of Uranus. One-half the diameter of Neptune plus 
 4 times that of Uranus is 145,000 miles. Find the diameter 
 of each planet. 
 
 From the last three problems make a table of the diameters 
 of all the planets. 
 
 NUMBER EXPRESSIONS IN PARENTHESES 
 27. When a number expression is inclosed in a parenthesis, 
 it is sometimes possible to perform the operations indicated 
 within the parenthesis and thus to remove it. 
 
 E.g. 6 + (7 + 10 - 9) = 6 + 8 = 14. 
 
REMOVAL OF PARENTHESES 29 
 
 When, however, the operations within the parenthesis can- 
 not be carried out, the parenthesis may sometimes be removed 
 by Principle IV or VI. 
 
 E.g. 5(2 n + 7 r) = 10 n + 35 r, and (20 x - 16 y) + 4 = 5 x - 4 y. 
 
 In these examples the operations upon the parentheses are 
 multiplication or division. The cases in which the operations 
 on the parentheses are addition or subtraction are considered 
 in the following examples : 
 
 (1) 5 + (3 + 4) = 5 + 7 = 12, 
 
 also 5 + (3 + 4) = 5 + 3 + 4 = 8 + 4 = 12. 
 
 (2) 10 + (7-3) = 10 + 4 = 14, 
 
 also 10 + (7-3)=10 + 7-3 = 17-3 = 14. 
 
 (3) 20- (7 +2) =20-9 = 11, 
 
 also 20 - (7 + 2) = 20 - 7 - 2 = 13 - 2 = 11. 
 
 (4) 20-(7-2)=20-5 = 15. 
 
 also 20-(7-2)=20-7+2 = 13 + 2 = 15. 
 
 From (1) it appears that the expression, 3 + 4, may be added 
 to 5 by first adding 3 and then adding 4. 
 
 From (2) it is seen that the expression, 7 — 3, may be added 
 to 10 by first adding 7 and then subtracting 3. 
 
 From (3) it is seen that the expression, 7 + 2, may be sub- 
 tracted from 20 by first subtracting 7 and then subtracting 2. 
 
 From (4) it is evident that the expression, 7 — 2, may 
 be subtracted from 20 by first subtracting 7 and then adding 2, 
 since 7 is 2 more than the number which was to be subtracted. 
 
 In like manner perform each of the following operations in 
 two ways : 
 
 1. 18 + (5 + 2 + 3). 4. 40 -(6 + 7+3). 7. 25n+(22w-3rc). 
 
 2. 28 + (7 -3 + 4). 5. 55 -(16 -7). 8. 18*-(6a;-2aj). 
 
 3. 32 -(5 + 3). 6. 101 + (73 -22). 9. 16r-(7r + 2r). 
 
30 INTRODUCTION TO THE EQUATION 
 
 These examples illustrate the following principle : 
 28. Principle VII. A number expression consisting of 
 two or more numbers connected by the signs + or — may 
 be added to anotiier given number by adding each number 
 with the plus sign and subtracting each number with the 
 minus sign. Such a number expression may be sub- 
 tracted from a given number by subtracting each number 
 with the plus sign and adding each number with the 
 minus sign. 
 
 Principles IV, VI, and VII are useful in removing parenthe- 
 ses from number expressions when the values of the numbers 
 involved are not known. 
 
 Instead of a parenthesis, a bracket [ ], or a brace { }, may be used. 
 Thus, 2 (6 + 4) = 2 [6 + 4] = 2 {6 + 4}. 
 
 In expressions involving parentheses the operations within 
 the parentheses should be performed first if possible. Then 
 perform the indicated multiplications and divisions, and finally 
 the remaining additions and subtractions. 
 
 E.g. Given 7 + 15 - (9 - 4) - 4 [7 + 11] + (29 - 20). 
 
 Performing operations within the parentheses, this reduces to : 
 7 + 15-5-4-18^9. 
 
 Performing multiplications and divisions, we have 7 + 3 — 8. 
 
 Performing the remaining additions and subtractions, the given 
 expression reduces to 2. 
 
 EXERCISES 
 
 In performing the following indicated operations, state in 
 each case what principles are used. 
 
 1.8(7+4 + 8 + 2). 8. 6(7-3)+2. 
 
 3. 3(2a-4)-4(a-6). 4. 14 + 4 (8 + 4) -=- (32-20). 
 
 The minus sign preceding 4 (a — 6) indicates that this whole 
 
 product is to be subtracted. Hence, using Principle IV, we have 
 
 6a— 12 — (4a —24). Then, using Principle VII, this becomes 
 
 6a-12-4a + 24 = 2a + 12. 
 
REMOVAL OF PARENTHESES 31 
 
 5. 8-3(4-2)+6 + 3[6 + l]. 
 
 6. 16 -5- [3 + 1]- [12 -3j -4-3. 
 
 7. 5 (17 - 5) + 18 -=- (8 - 2) - (21 + 14) -=- 7. 
 
 8. 7(a + 6)+6(a + 6). 
 
 9. 5(a+6)+ 3 (a + b + c). 
 
 10. 16(r + *) + H {r4 0- 11. 15s-3(r + s). 
 
 12. 15 (r + s) -3(r-s). 13. 20-(cc + w), 
 
 14. 50x-25(x-y). 15. 12 m - 6(* - 2m). 
 
 16. ll< + 5(2* -1) -3(2 + 2). Check fori = 2. 
 
 17. (12* -6m) -r-2 + 3« — 2m. Check for t = 1, u = 1. 
 
 18. 3(5a-7y) + (21a-28M) -e- 7. Check for a; = 2, m = 1. 
 
 19. 8(r — s) + 5(2r + s) — 3 (r + s). Check for r = 1, 8 = 1. 
 
 20. 10(a; + y) -r- 5 + 6 (a? - y) -*- 3. Check f or cc = 2, m = 1. 
 
 21. 5(h+k) + 2(h+k)+3(h + 7c). 
 
 Use Principle I, then IV ; also IV, then I. 
 
 22. 5(7x-4:y)+9(9x-5y). 
 
 23. 8(r-s)-2(r-s). 24. 9(3p- q)-S(3p-q). 
 Use Principle II, then IV ; also reverse this order. 
 
 25. (16 x -12 x) -f- 4. 26. (18 a6 — 12 a6a?) -s- a. 
 
 Use first Principle VI, then II ; also II, and then V. 
 
 27. 9(6 + 4) -3 (6 -4). 28. (5 axy - 3 ex) -=- x. 
 
 29. 9 (6 abc + 4: xyz) — 3(6 abc — 4 an/z). 
 
 30. 8(5a;-M + 2z)-ll(3a; + 2M-7z). 
 
 31. 3[2(a + 6 + c)-3(a-6 + c)]. 
 
 First remove the parentheses, then the bracket. 
 
 32. 7\Sx-(2y-3x)+(2x-Ay)' i . 
 
32 INTRODUCTION TO THE EQUATION 
 
 PROBLEMS 
 
 1. There are three numbers whose sum is 80. The second 
 is 3 times the first, and the third twice the second. What are 
 the numbers ? 
 
 2. A man has three buildings whose total value is $46,800. 
 The second building cost $ 800 less than the first, and the third 
 cost twice as much as the second. What is the cost of each 
 building ? 
 
 3. The population of Connecticut (Census of 1900) was 50 
 thousand more than twice that of Rhode Island, and the popu- 
 lation of Massachusetts- was 81 thousand more than 3 times 
 that of Connecticut. The population of Massachusetts minus 
 that of Connecticut was 1897 thousand. Find the population 
 of each state. 
 
 4. The population of Indiana (Census of 1900) exceeded that 
 of Iowa by 284 thousand, and the population of Illinois was 
 210 thousand less than twice that of Indiana. The population 
 of Illinois minus that of Indiana was 2306 thousand. Find 
 the population of each state. 
 
 5. The melting point of copper is 250 degrees (Centigrade) 
 lower than 4 times that of lead. Ten times the number of 
 degrees at which lead melts minus twice the number at which 
 copper melts equals 1152. What is the melting point of each 
 metal ? 
 
 6. The melting point of iron is 450 degrees higher than 5 
 times that of tin. Three times the number of degrees at 
 which iron melts plus 7 times the number at which tin melts 
 equals 6410. Find the melting point of each metal. 
 
 7. In 1904 the gold product of Africa was 11 million dol- 
 lars more than 3 times that of Russia, and the gold product of 
 Australia was 84 million less than twice that of Africa. 
 Russia and Australia together produced 113 million. How 
 much did each country produce ? 
 
REMOVAL OF PARENTHESES 33 
 
 8. In 1904 the value of the silver produced in the United 
 States was 5 million dollars more than 14 times as much as 
 that of Canada, and the product of Mexico was 1 million less 
 than 16 times that of Canada. Twice the product of the 
 United States minus that of Mexico equals 71 million. How 
 much did each country produce ? 
 
 9. The Nile is 100 miles more than twice as long as the 
 Danube. Ten times the length of the Danube plus 4 times the 
 length of the Nile equals 3400 miles. How long is each river ? 
 
 10. The number of first-class battleships in the United States 
 navy (1906) was 22 less than 5 times the number of protected 
 cruisers. Twelve times the number of cruisers less twice the 
 number of battleships equals 60. Find the number of each. 
 
 11. The number of torpedo boats in the United States navy 
 (1906) was 77 less than 7 times .as great as the number of 
 torpedo boat destroyers. Three times the number of torpedo 
 boats minus 4 times the number of destroyers equals 41. Find 
 the number of each. 
 
 12. The weight of a cubic foot of spruce is 16 pounds more 
 than that of a cubic foot of cork, and the weight of a cubic 
 foot of dry live oak is 5 pounds more than twice that of spruce. 
 One cubic foot of oak weighs 36 pounds more than one cubic 
 foot of spruce. Find the weight of a cubic foot of each. 
 
 13. In 1904 Canada produced 3 million dollars more of 
 gold than Mexico, and the United States produced 17 million 
 more than 4 times as much as Canada. The combined product 
 of Mexico and the United States was 94 million. How much 
 did each country produce ? 
 
 14. The value of the copper produced in the United States 
 in 1904 was 5 million dollars more than the value of the crude 
 petroleum, and the value of the bituminous coal was 94 million 
 more than twice the value of the copper. Nine times the value 
 of the copper minus twice the value of the coal equals 342 
 million. Find the value of each. 
 
34 INTRODUCTION TO THE EQUATION 
 
 15. The pressure developed in the chamber of a modern 
 twelve-inch gun is 21 thousand pounds per square inch more 
 than that developed in an ordinary hunting rifle. The 
 maximum pressure which gunpowder can develop is 18 
 thousand pounds per square inch less than 3 times as great 
 as that developed in the twelve-inch gun. The sum of 
 the three pressures is 151 thousand pounds. Find each 
 pressure. 
 
 16. The standing army of Australia (1906) was 14 thousand 
 more than that of Canada, and the standing army of Great 
 Britain was 47 thousand more than 4 times that of Australia. 
 Great Britain's army contained 287 thousand men. How many 
 men were there in the armies of Canada and Australia ? 
 
 IDENTITIES AND EQUATIONS 
 
 29. In the preceding pages equations have been freely 
 used, and certain simple methods have been employed in 
 solving problems by means of them. We now proceed to a 
 more detailed study of these methods and of the equation 
 itself. 
 
 30. Equalities in which letters are used as number symbols 
 are of two kinds as shown by the following examples: 
 
 (1) 3n + 4n = 7»isa true statement no matter what number is 
 represented by n. 3 (5 x + 6y) = 15 x + 18 ?/ holds for all values 
 which may be assigned to x and y. 
 
 (2) to + 2 (w + 5) = 58 is a true statement if, and only if, w = 16. If 
 w is replaced by any number less than 16, the number expression on 
 the left is less than 58, and if w is replaced by any number greater 
 than 16, the result is greater than 58. 
 
 31. The equality w -f 2 (w -f 5) = 58 is said to be satisfied by 
 w=16, because this value of w reduces both members to the 
 same number, 58. 
 
IDENTITIES AND EQUATIONS 35 
 
 32. Definition. An equality which is satisfied, no matter 
 what numbers are substituted for one or more of its letters, 
 is called an identity with respect to those letters. 
 
 E.g. 3n + 4n = 7nisan identity with respect to n. 
 
 a (b + c) = ab + ac is an identity with respect to a, b, and c. 
 
 When it is especially desired to distinguish an equality as an 
 identity, the equality sign is written =. 
 
 Each of the Principles I to VII may be thus stated in symbols as 
 identities. Thus, 
 
 I, 4n + 6n = 10n; II, 12 n - 5n=7n; 
 III, 5 (4 ab)=20 ab ; IV, 5 (a ± b)=5 a±5b; 
 V, 30a&^-6 = 5aZ>; VI, (16 a ± 206) + 4=4 a ± 56; 
 VII, x + (a - b) - (c - d)=x + a-b-c + d. 
 
 33. Definition. An equality which is satisfied only when 
 certain particular values are given to one or more of its letters 
 is called a conditional equality with respect to those letters. 
 
 E.g. 3 x + 5 = 35 is an equality only on the condition that x = 10. 
 x + y = 10 is an equality for certain pairs of letters like 1 and 9, 2 and 8, 
 3 and 7, 5 and 5, but certainly not for all values of x and y ; for instance, 
 not for 3 and 8. 
 
 34. A conditional equality is called an equation, and a letter 
 whose particular value is sought to satisfy an equation is called 
 an unknown number in that equation, or simply an unknown. 
 
 The equations at present considered contain only one un- 
 known. 
 
 35. To solve an equation in one unknown is to find the value 
 or values of the unknown which satisfy it. 
 
 The following example exhibits the process of solving an 
 equation which contains one unknown and which is satisfied 
 by one value only of the unknown : 
 
% 
 
 36 INTRODUCTION TO THE EQUATION 
 
 Given w + 2 (w + 5) = 58. (1) 
 
 By Principle III, to + 2 to + 10 = 58. (2) 
 
 By Principle I, 3 to + 10 = 58. (3) 
 
 Subtracting 10 from both members, 3 to = 48. (4) 
 
 Dividing both sides by 3, w = 16. (5) 
 
 Check. Putting w = 16 in (1), 16 + 2 (16 + 5) = 16 + 2 . 21 = 58. 
 
 Explanation. Any value of to which satisfies (1) also satisfies (2) 
 
 and (3), since the expressions in the left members of (1), (2), and (3) 
 
 represent the same number expressed in different forms. Any value 
 
 of to which satisfies (3) also satisfies (4), for if 10 more than 3 to is 
 
 58, then 3 to must be 48. Any value of to which satisfies (4) also 
 
 satisfies (5), for if 3 to is 48, then to is | of 48. 
 
 By similar considerations the value of to which satisfies (5) may be 
 shown to satisfy (4), (3), (2), and (1). Hence to = 16 is the solution 
 of the given equation. 
 
 The above solution illustrates the following principle : 
 36. Principle VIII. An equation may be changed into 
 another equation such that any value of the unknown 
 which satisfies one also satisfies the other, by means of any 
 of the following operations : 
 
 (1) Adding the same number to both members. 
 
 (2) Subtracting the same number from both members. 
 
 (3) Multiplying both members by the same number. 
 
 (4) Dividing both members by the same number. 
 
 (5) Changing the form of either member in any way 
 which leaves its value unaltered. 
 
 The operations under Principle VIII are hereafter referred 
 to in detail by means of the initial letters, A for addition, S for 
 subtraction, M for multiplication, D for division, and F for 
 form changes. 
 
 Note. — It is not permissible to multiply or divide the members of 
 an equation by any expression which is equal to zero. See Advanced 
 Course. 
 
SOLUTION OF EQUATION H 37 
 
 37. The operations involved in Principles I to VII are 
 all form changes which leave the value of the expression 
 unaltered. 
 
 E.g. 4 (vi + n) by Principle IV has the same value as 4 m + 4 n. 
 
 There are other form changes which are already familiar in 
 arithmetic. 
 
 E.g. 2 + 4 = 4 + 2, or in general a + b = b + a. Likewise 2-7 = 7-2, 
 or in general ab = ba. That is, the order in which numbers are added 
 or multiplied is immaterial. 
 
 Again, 3 + 4 + 6 = (3 + 4) + 6 = 3 + (4 + 6), or in general a + b + c 
 = (a + 6) + c = a + (6 + c), and likewise a -b • c = (a • b) • c = a • (6 • c) ; 
 i.e. numbers to be added or multiplied may be grouped in any manner 
 desired. Still other form changes will be learned later. 
 
 38. Principle VIII is further illustrated as follows : 
 
 On the scale pans of a common balance are placed objects of 
 uniform weight, say tenpenny nails. The scales balance only 
 when the weights are the same in both pans ; that is, when the 
 number of nails is the same. 
 
 If now the scales are in balance, they will remain so under 
 two kinds of changes in the weights : 
 
 (a) When the number of nails in the two pans is in- 
 creased or diminished by the same number ; correspond- 
 ing to the operations A, S, M, D, on the members of an 
 equation. 
 
 (6) When the number in each pan is left unaltered but 
 the nails are rearranged in groups or piles in any manner ; 
 corresponding to the operations F on the members of an 
 equation. 
 
 The equation, then, is like a balance, and its members are 
 to be operated upon only in such ways as to preserve the 
 balance. 
 
38 INTRODUCTION TO THE EQUATION 
 
 DIRECTIONS FOR WRITTEN WORK 
 
 39. The solution of an equation consists in deducing, by 
 means of Principles I to VIII, another equation whose first 
 member contains the unknown only and whose second member 
 does not contain the unknown. The successive steps should 
 be written down as in the following example : 
 
 Solve 6 (4 n - 3) + 25 (n + 1) = 50 + 31 ft + 2 (3 - ft)- 9. (1) 
 
 By F, using III and IV, we obtain from (1) 
 
 24 n -18 + 25 n + 25 = 50 + 31n + 6-2 n- 9. (2) 
 
 By F, I and II, we obtain from (2) 
 
 49 » + 7 = 29 n + 47. (3) 
 
 Subtracting 7 and 29 n from each member of (3) and using Prin- 
 ciple II, we have 20n = 40. (4) 
 
 Dividing each member of (4) by 20, 
 
 n=2. (5) 
 
 Check. Substitute n = 2 in Equation (1). 
 
 For convenience this work can be abbreviated as follows: 
 
 6(4n - 3) + 25 (n + 1) = 50 + 31 n + 2(3- n) - 9. (1) 
 
 By F, III, IV, 24 n - 18 + 25 n + 25 = 50 + 31 n + 6 - 2 n - 9. (2) 
 By F, I, II, 49n + 7 = 29n + 47. (3) 
 
 By S 1 7, 29 n, 20n = 40. (4) 
 
 ByD|20, n = 2. (5) 
 
 S | 7, 29 n means that 7 and 29 n are each to be subtracted from 
 both members of the preceding equation. D 1 20 means that the 
 members of the preceding equation are to be divided by 20. 
 
 Similarly, in case we wish to indicate that 6 is to be added to each 
 member of an equation, we would write A I 6, and if each member is 
 to be multiplied by 8, we would write M\ 8. It is important that the 
 nature of each step be recorded in some such manner. 
 
+ 4 n - 6 = 20 + 5n - 5 + 3 n. 
 
 (2) 
 
 21 n + 2 = 15 + 8 n. 
 
 (3) 
 
 21n-8n = 15-2. 
 
 (4) 
 
 13 n = 13. 
 
 (5) 
 
 n = 1. 
 
 (6) 
 
 SOLUTION OF EQUATIONS 39 
 
 (2) Solve 17 n + 4(2 + ft) - 6 = 5(4 + n) - 5 + 3 w. (1) 
 
 By F, IV, 17 n 
 
 By F, I, 
 By S | 2, 8 n, 
 By F, II, 
 By Z> 1 13, 
 
 Check. Substitute n = 1 in equation (1); 
 then 17 + 12 - 6 = 25 - 5 + 3, 
 
 or 23 = 23. 
 
 An equation may be translated into a problem. For example, 
 the equation 21a; + 2 = 15 + 8ic may be interpreted as follows : 
 Find a number such that 21 times the number plus 2 is 15 
 greater than 8 times the number. 
 
 EXERCISES AND PROBLEMS 
 
 Solve the following equations, putting the work in a form 
 similar to the above and checking each result. Translate the 
 first twenty into problems. 
 
 1. 13a-40-tt = 8. 
 
 2. 3z + 9 + 2a; + 6 = 18 + 4: C . 
 
 3. 5 x + 3 — x = x-^-18. 
 
 4. 13y + 12 + 5y=32+8y. 
 
 5. 4m + 6ra + 4 = 9m + 6. 
 
 6. 7m-18 + 3m = 12 + 2m + 2. 
 
 7. 3y-4; + 2y-6 = y + 7 + y + 3 + 10. 
 
 8. 5« + 3 + 2a; + 3 = 2a; + 5 + 3a; + 3 + a;. 
 
 9. 2a; + 4a;4-9 + a; + 6 = 20 + 3a; + 5 + 2£c. 
 
 10. 18 + 6m + 30 + 6m = 4m + 8+12 + 3m + 3 + m + 29. 
 
40 INTRODUCTION TO THE EQUATION 
 
 11. y + 72 + 45y = 106 + 12y. 
 
 12. 42 a -56 = 20 a: + 10. 
 
 13. 6x + 8-3x + 4: + 5x = 7x + 32 + x-20. 
 
 14. 32x-4 + 7x = B8 + 3x + 5x. 
 
 15. 12m-3-3ra = 32 + 2m. 
 
 16. 15ra + 3-2m + 7 = 3m + 60. 
 
 17. a + 7+3a = 2a + 45. 
 
 18. 5 6-30 + 66 = 36 + 90. 
 
 19. 3c + 18+14c = 6c + 51. 
 
 20. 17x- + 4 + 3x = 7a; + 30. 
 
 21. 7(m + 6) + 10 m = 42 - 8(2 m + 2) + 181. 
 
 22. 20 - 3(x - 4) + 2 x = 2 x + 17. 
 
 23. (8x + 4)-=-2 + 7a; = 4z + 9. 
 
 24. 6 a + 4(4 x + 2) = 85 - 3(2 x + 7). 
 
 25. 8 + 7(6 + 6 n) + 2 w = 2(4 n + 5) + 18 n + 49. 
 
 26. 5(9a: + 3) + 6a: = 24a:-4(3a; + 2) + 36. 
 
 27 7(12 «; + 8 ) + 13 + 5 a; - 6 = 47. Apply V and VI. 
 28 . 12(5 + 4o ; )_5(6 + 4 a; ) +5 - ()==a; + 18j 
 
 29 15 , 21(3 + *) 2(6 + 18^) ^3(9 ^+12) 2S 
 7 3 3 "*" 
 
 __ ll(5z + 25), 3(6^-2) 7(4a; + 8),12 a; +36 jW 
 30. ^ + — — -J = r -^ + _ +35. 
 
SOLUTION OF EQUATIONS 41 
 
 EQUATIONS INVOLVING FRACTIONS 
 
 40. Solve w + ^ + ^ = 88. (1) 
 
 First Solution. The coefficients of n are 1, f, and J. 
 Applying Principle I, If n = 88. (2) 
 
 By D I If , n = 88 + If = 48. (3) 
 
 Second Solution. Multiply both members of (1) by 6. 
 
 That is, by 1/ 1 6, 6n+^ + ^= 528. (2) 
 
 By F, V, 6n+3n+2n= 528. (3) 
 
 By F, I, 11 n = 528. (4) 
 
 By D | 11, n = 48, as before. (5) 
 
 The object is to multiply both members of the equation by such a 
 number as will cancel_each denominator. Hence the multiplier must 
 contain each denominator as a factor. 
 
 Evidently 12 or 18 might have been chosen for this purpose, but not 
 8 or 10. 6 is the smallest number which will cancel both 2 and 3, and 
 hence this was chosen as the multiplier. 
 
 41. The process explained in the second solution above is 
 called clearing of fractions. 
 
 As another illustration solve the equation 
 
 Here the smallest multiplier available is 36. 
 
 Hence by M|36, §^L* + ?<L£ + Hl&£ - 89 • 65i (2) 
 
 By V, each denominator is cancelled by the factor 36, 
 
 9 • 3 x + 18 x + 4 • 5 x = 36 • 65. (3) 
 
 By III, 27 x + 18 x + 20 x = 36 • 65. (4) 
 
 By I, 65 x = 36 . 65. (5) 
 
 ByD,V, x = §6^65 = 36< (6) 
 
 o5 
 Check. Substitute x = 36 in (1). 
 
42 INTRODUCTION TO THE EQUATION 
 
 EXERCISES AND PROBLEMS 
 
 Solve the following equations, indicating the principles 
 used at each step. Check each solution by substituting in 
 the original equation the value obtained. Translate each 
 into a problem. 
 
 For instance, from equation 2: Find a number such that when 
 increased by its half, its third, and its fourth, the sum is 25. 
 
 1. V- = 5. 
 2^3 
 
 3. ^ + 4n-^=^ + 26. 
 
 4 3 2 
 
 2 3 4 10 
 
 lb 5 10 
 
 6. 4n + ^ = 3ft + 2 
 
 7 2 
 
 7 . 12 | 4 ( 9a; + 6) 2(3 + lla;) _ 5(4 a; + 4) | 11 
 
 o o o 
 
 8 . l3 & + 7-^±^ = ?A±9 + 3 8 . 
 
 7 5 
 
 5 a -f 7 . 2a + 4 = 3a + 9 ^ . 
 
 2 3 4 
 
 10 17a ~ 5 10a + 2 = 5a + 7 g 
 
 3 4 " 2 
 
 11. ^±3 + 5(2a ; -fl0) = 2a; + 24 
 — 5 
 
SOLUTION OF EQUATIONS 43 
 
 12. The sum of two numbers is 12, and the first number is 
 \ as great as the second. What are the numbers ? 
 
 13. The smaller of two numbers is f of the larger. If their 
 sum is 66, what are the numbers ? 
 
 14. Find two consecutive integers such that 4 times the 
 first minus 3 times the second equals 9. 
 
 15. Find three consecutive integers such that \ of the first 
 plus the second minus ^ the third equals 5. 
 
 16. Find three consecutive integers such that 3 times 
 the first plus 9 times the second minus 4 times the third 
 equals 73. 
 
 17. There are three numbers such that the second is 4 more 
 than 9 times the first, and the third is 2 more than 6 times 
 the first. If -|- of the third is subtracted from \ of the second, 
 the remainder is 3. Find the numbers. 
 
 18. There are three numbers such that the second is 2 more 
 than 9 times the first and the third is 7 more than 11 times the 
 first. The remainder when 4 times the third is subtracted from 
 13 times the second is 144. Find the numbers. 
 
 19. The population of Philadelphia (estimate of Census 
 Bureau, 1904) was 508 thousand less than 20 times that of 
 Dayton, Ohio. The population of St. Louis was 245 thousand 
 more than 4 times that of Dayton. One-half the population of 
 Philadelphia plus ^ that of St. Louis was 851 thousand. Find 
 the population of each city. 
 
 20. The shortest railway route from Boston to Chicago 
 is 166 miles more than 4 times that from Boston to New 
 York ; and the shortest route from Boston to Atlanta is 
 196 miles less than 6 times that from Boston to New York. 
 The distance from Boston to Chicago is 481 miles more than 
 \ the distance from Boston to Atlanta. Find each of the three 
 distances. 
 
44 INTRODUCTION TO THE EQUATION 
 
 Solution. Let d — distance in miles from Boston to New York ; 
 then 4 d + 166 = distance in miles from Boston to Chicago, 
 
 and 6 d — 196 = distance in miles from Boston to Atlanta, 
 
 and 4rf + 166 = 6J ~ 196 + 481. 
 
 By VI, 4 d + 166 = 3 d - 98 + 481. 
 
 By S, II, d = 481 - 98 - 166 = 217 = distance from Boston 
 
 to New York. 
 
 4 d + 166 = 1034 = distance from Boston to Chicago, 
 
 and 6 d — 196 = 1106 = distance from Boston to Atlanta. 
 
 21. The railway mileage in the United States in 1900 was 
 8 thousand greater than twice that of 1880, and in 1904 it was 
 66 thousand less than 3 times that of 1880. One-half the mile- 
 age of 1900 plus \ that of 1904 equals 168 thousand. Find 
 the railway mileage in 1880, 1900, and 1904. 
 
 22. The number of newspapers in the United States in 1880 
 was 381 less than 4 times that in 1850. The number in 1905 
 was 700 more than twice that of 1880 and 7990 more than 
 6 times that of 1850. Find the number of newspapers in 1850, 
 1880, and 1905. 
 
 23. The number of telephones in the United States in 1900 
 was 40 thousand less than 30 times as great as the number in 
 1880. Half the number in 1905 was 660 thousand more than 
 that in 1900 and 80 thousand more than 40 times that in 1880. 
 Find the number of telephones in use in 1880, 1900, and 1905. 
 
 24. The displacement of the battleship Kearsarge is 1232 
 tons greater than that of the Oregon, and the displacement 
 of the Connecticut is 4480 tons greater than that of the 
 Kearsarge. One-third the displacement of the Kearsarge 
 minus \ that of the Connecticut equals 640. Find the dis- 
 placement of each ship. 
 
 25. The distance of the fixed star Vega from the sun is 11.7 
 light-years greater than the distance from the sun to Sirius. 
 The distance of Regulus is 8 light-years less than twice that of 
 
SOLUTION OF EQUATIONS 45 
 
 Vega. What is the distance of each star from the sun if \ the 
 distance of Vega minus \ that of Regulus is 2 light-years ? 
 
 Light travels at the rate of 186,000 miles in one second. A light- 
 year is the distance traveled by light in one year. 
 
 26. The distance from the sun to the fixed star 61 Cygni is 
 4 light-years more than that from the sun to Alpha Centauri 
 (the star nearest the sun), and the distance from the sun to 
 Polaris (the pole star) is 6 times that of 61 Cygni. One- 
 eighth the distance of Polaris minus \ the sum of the dis- 
 tances of the other two equals 2 light-years. How many 
 light-years is each star from the sun? 
 
 27. In 1905 the wheat crop of Russia was 47 million bushels 
 less than 4 times that of Argentina, and the crop of the United 
 States was 77 million bushels less than 5 times that of Argentina. 
 The crop of the United States exceeded that of Russia by 124 
 million bushels. What was the wheat crop of each country ? 
 
 28. The total imports of the United States in 1900 were 60 
 million dollars less than 10 times the imports in 1800. The 
 imports in 1906 were 473 million less than twice the imports of 
 1900, and 135 million more than 12 times the imports of 1800. 
 Find the imports of the United States in 1800, 1900, and 1906. 
 
 29. The total exports of the United States in 1900 were 26 
 million dollars less than 20 times the exports in 1800. The 
 exports in 1906 exceeded those of 1900 by 350 million and were 
 31 million less than 25 times those of 1800. Pind the exports 
 of 1800, 1900, and 1906. 
 
 30. The gross tonnage of the German merchant marine was 
 (1906) 548 thousand more than that of the United States, and 
 the tonnage of the British marine was 2662 thousand greater 
 than 4 times that of the German marine. The British marine 
 was greater by 930 thousand tons than twice that of Germany 
 and 3 times that of the United States combined. Find the 
 tonnage of each marine. 
 
46 INTRODUCTION TO THE EQUATION 
 
 31. The population of San Francisco (estimate of Census 
 Bureau, 1904) was 289 thousand more than that of Springfield, 
 Massachusetts, and the population of Chicago was 132 thousand 
 more than 5 times that of San Francisco. The population of 
 Springfield minus -£-$ that of Chicago was 2 thousand. Find 
 the population of each city. 
 
 32. The per capita tax of New York City (1905) was $ .49 
 less than twice that of Chicago and the per capita tax of Boston 
 was $ 3.56 less than 3 times that of Chicago. The combined 
 per capita tax of Boston and New York was $ 52.15. What 
 was the per capita tax of each city ? 
 
 33. The number of pupils attending public high schools in 
 Chicago (1905) was 1237 more than twice as great as in Phila- 
 delphia. One-third the number of pupils in Chicago minus \ 
 the number in Philadelphia was 2524. How many pupils 
 attended public high schools in each city? 
 
 34. The number of pupils in public high schools in Boston 
 (1905) was 1574 less than 4 times as many as in Baltimore. 
 Three times the number in Boston minus 6 times the number 
 in Baltimore equals 5268. How many pupils in each city ? 
 
 35. The distance from San Francisco to Yokohama is 2031 
 statute miles less than 3 times that from San Francisco to 
 Honolulu, and the distance from San Francisco to Hongkong is 
 2219 more than twice that from San Francisco to Honolulu. 
 One-fifth the distance to Hongkong plus \ the distance to 
 Yokohama equals 3152. Find the distance from San Francisco 
 to each of the other three cities. 
 
 36. The distance from New York to Singapore is 542 statute 
 miles more than 3 times that from New York to London, and 
 the distance from New York to Manila is 5342 miles less than 
 5 times that from New York to London. Three times the dis- 
 tance from New York to Manila plus 6974 miles equals 4 times 
 the distance from New York to Singapore. Find the distance 
 from New York to each of the other three cities. 
 
CHAPTER II 
 POSITIVE AND NEGATIVE NUMBERS 
 
 42. Thus far the numbers used have been precisely the same 
 as in arithmetic, though their representation by means of 
 letters and some of the methods used in operating upon them 
 are peculiar to algebra. 
 
 43. A new kind of number will now be studied in connection 
 with problems of the following type. 
 
 Illustrative Problems. 1. If a man gains' $1500, and then 
 loses $ 800, what is the net result ? Answer, $ 700 gain. 
 
 2. The assets of a commercial house are $ 250,000, and the 
 liabilities are $ 275,000. What is the net financial status of 
 the house ? Answer, $ 25,000 net liabilities. 
 
 3. The thermometer rises 18 degrees and then falls 28 
 degrees. What direct change in temperature would produce 
 the same result ? Answer, 10 degrees fall. 
 
 4. A man travels 700 miles east and then 400 miles west. 
 What direct journey would bring him to the same final desti- 
 nation ? Answer, 300 miles east. 
 
 In the statement of each of these problems the numbers are 
 applied to things which are opposite in quality ; namely, gain 
 and loss, assets and liabilities, rise and fall, distances measured 
 in two opposite directions, as east and west. 
 
 The answer in each case not only gives the proper arithmeti- 
 cal number, but also connects with this number one of the two 
 opposite qualities involved in the problem. 
 
 E.g. The answer in (1) is $700 gain and not simply f 700; in (2) 
 it is not $25,000, but $25,000 liabilities; in (3) it is not 10°, but 10° 
 fall ; and in (4) it is not 300 miles, but 300 miles east. 
 
 47 
 
48 POSITIVE AND NEGATIVE NUMBERS 
 
 These problems are illustrations of the sense in which the 
 word " opposite " is here used. 
 
 E.g. Gain and loss are opposite in that they annul each other 
 when taken together ; rise and fall of temperature are opposite in that 
 one counteracts the other. 
 
 44. In all problems involving one or both of two qualities 
 which are opposite in this sense there is constant need to dis- 
 tinguish which one is meant. 
 
 E.g. On a day in winter it is not sufficient to say the temperature 
 is 5° ; we must specify whether it is above or below zero. 
 
 In the case of the thermometer we call the temperature positive if 
 it is above zero and negative if it is below zero. 
 
 These words positive and negative are used to describe all pairs of 
 qualities which are opposite in the sense here understood. 
 
 The signs + and ~ stand respectively for the words "posi- 
 tive " and " negative." 
 
 E.g. $ + 700 is read positive $ 700, and means either $700 gain or $700 
 assets, according to the problem in which it occurs. Likewise $~700 
 is read negative $700, and means either $700 loss or $700 liability. 
 
 Referring to the thermometer, +18° is read positive 18°, and means 
 either a temperature of 18° above zero or a rise of the mercury 18° from 
 any point. The latter is the meaning in Problem 3. Similarly ~28° 
 means either a temperature 28° below zero or a, fall of the mercury 28° 
 from any point. 
 
 It is commonly agreed to call gain positive and loss negative, 
 assets positive and liabilities negative, above zero positive and 
 below zero negative, motion upward positive and downward 
 negative, motion to the right positive and to the left negative. 
 
 45. Numbers marked with the sign + are called positive and 
 those marked with the sign ~ are called negative. 
 
 The positive sign maybe omitted; that is, a number with 
 neither sign written is understood to be positive. 
 
ADDITION OF SIGNED NUMBERS 49 
 
 ADDITION OF POSITIVE AND NEGATIVE NUMBERS 
 
 46. While in each of the problems on page 47 the result 
 was obtained by subtracting one number from the other, 
 yet they are not properly subtraction problems, but addition 
 problems. 
 
 E.g. In Problem 1, we are not asking for the difference between 
 $ 1500 gain and $800 loss, but for the net result when the gain and the 
 loss are taken together ; that is, the sum of the profit and loss. Hence 
 we say $1500 gain + $800 loss = $700 gain, or, using positive and 
 negative signs, + 1500 + ~ 800 = +700. . 
 
 Similarly in Problem 2, 
 
 $250,000 assets + $275,000 liabilities = $25,000 net liabilities, 
 Or + 250,000 + - 275,000 = - 25,000. 
 
 In Problem 3, 18° rise + 28° fall = 10° fall, 
 
 Or +18 +-28 = -10. 
 
 In Problem 4, 
 
 700 miles east + 400 miles west = 300 miles east, 
 Or +700 + -400 = +300. 
 
 Note. — In the last case east is called + and west -, since, as a map 
 is usually held, east is to the right and west to the left. Likewise 
 north or up is called + and down or south is called - . 
 
 PROBLEMS 
 
 1. A balloon which exerts an upward pull of 460 pounds is 
 attached to a car weighing 175 pounds. What is the net 
 upward or downward pull? Express this as a problem in 
 addition, using positive and negative numbers. 
 
 Solution. 460 lb. upward pull + 175 lb. downward pull = 285 lb. net 
 upward pull. Using positive numbers to represent upward pull and 
 negative numbers to represent downward pull, this equation becomes 
 
 + 460 + -175 = +285. 
 
50 POSITIVE AND NEGATIVE NUMBERS 
 
 2. How would the morning paper print the following 
 thermometer readings from the weather report ? Jacksonville 
 38° above zero, Seattle 5° below zero, Nashville 28° above zero, 
 Chicago 13° below zero. 
 
 3. The temperature rises 15° and then falls 24°. What 
 direct change in temperature would produce the same final 
 result ? Express this as an example in addition. 
 
 4. A vessel on the equator sailed north 3° and was then 
 forced south 5° by a hurricane. It then resumed its course 
 northward 8°. Express this as a sum and show the final posi- 
 tion with reference to the equator. (See note, § 46.) 
 
 In each of the following translate the solution into the language of 
 algehra by means of positive and negative numbers, as in Problem 1. 
 
 5. A 450-pound weight is attached to a balloon which 
 exerts an upward pull of 600 pounds. What is the net up- 
 ward or downward pull ? 
 
 6. A man's property amounts to $45,000 and his debts to 
 $ 52,000. What is his net debt or property ? 
 
 7. The assets of a bankrupt firm amount to $245,000 and 
 the liabilities to $ 325,000. What are the net assets or liabili- 
 ties ? 
 
 8. A weight exerting a downward pull of 280 pounds 
 when submerged in water is attached to a float which will just 
 support a weight of 240 pounds. What is the net pressure 
 upward or downward when both are submerged ? 
 
 9. A man on the deck of a steamer is walking at the rate 
 of 4 miles an hour toward the stern. If the boat is sailing 
 eastward in a river at the rate of fifteen miles per hour, what 
 is the actual motion of the man with respect to the bank of 
 the river ? 
 
 10. A man can row a boat at the rate of 6 miles per hour. 
 How fast can he proceed against a stream flowing at the rate 
 of 2^ miles per hour ; 7 miles per hour ? 
 
ADDITION OF SIGNED NUMBERS 51 
 
 11. A steamer which can make 12 miles per hour in still 
 water is running against a current flowing 15 miles per 
 hour. How fast and in what direction does the steamer 
 move ? 
 
 12. A dove capable of flying 40 miles per hour in calm 
 weather is flying against a hurricane blowing at the rate of 60 
 miles per hour. How fast and in what direction is the dove 
 moving ? 
 
 13. If of two partners, one loses $ 1400 and the other gains 
 $ 3700, what is the net result to the firm ? 
 
 14. A man's income is $ 2400 and his expenses $ 1500 per 
 year. What is his saving ? 
 
 15. A man loses $ 800 and then loses $ 600 more. What is 
 the combined loss ? Indicate the result as the sum of two 
 negative numbers. 
 
 16. A man gains $ 500 and then gains $ 700 more. What 
 is the combined gain ? Express the result as the sum of two 
 positive numbers. 
 
 17. A tug which can steam 9 miles per hour in still water is 
 going down a stream whose current is 6 miles per hour. How 
 fast is the tug moving ? 
 
 18. The thermometer falls 8° and then 17° more. Express 
 the combined result of the two changes as a negative number. 
 
 47. Definitions. Positive and negative numbers are some- 
 times called signed numbers, because each such number consists 
 of an arithmetic part, together with a sign of quality. 
 
 The arithmetic part of a signed number is called its absolute 
 value. 
 
 Thus, the absolute value of + 3 and also of ~ 3 is 3. Two 
 signed numbers are of like quality when they have the same 
 signs, and of opposite quality when one is positive and the 
 other negative. 
 
52 POSITIVE AND NEGATIVE NUMBERS 
 
 In the one case we say they have like signs, in the other op- 
 posite or unlike signs. The preceding exercises illustrate the 
 following principle : 
 
 48. Principle IX. To add two signed numbers of like 
 quality, find the sum of their absolute values, and 
 prefix to this the common sign of quality. 
 
 To add two signed numbers of opposite quality, find 
 the difference of their absolute values, and prefix to 
 this the sign of that one whose absolute value is the 
 greater. 
 
 In case their absolute values are equal, their sum is 
 zero. 
 
 The sum of two signed numbers thus obtained is called their 
 algebraic sum. 
 
 49. This principle may be further illustrated by means of 
 the following graphic representation of signed numbers. 
 
 On an unlimited straight line call some starting point zero, 
 and lay off from this point equal divisions of the line indefi- 
 nitely both to the right and to the left, as shown in the figure. 
 
 ■i o + i 
 
 -HH- 
 
 In order to describe the position of any one of these division 
 points, we require not only an integer of arithmetic, to specify 
 how far to count from the starting point (the point marked 
 zero) in order to reach the given point, but also a quality sign, 
 to indicate the direction of the counting. 
 
 E.g. + 7 marks the division point 7 units to the right of zero, and 
 - 5 marks the point 5 units to the left of zero. Such a diagram is 
 called the scale of signed numbers. 
 
 The part of the scale to the right, taken alone, would need no 
 signs, and would picture the integral numbers of arithmetic, while the 
 
ADDITION BY COUNTING 53 
 
 two parts together require the distinguishing signs of quality, and 
 picture the integral numbers of algebra. 
 
 Fractions would of course be pictured at points between the inte- 
 gral division points, on the right or the left of the scale, according 
 as the fractions are positive or negative. 
 
 ADDITION BY COUNTING 
 
 50. In arithmetic two numbers are added by starting with 
 either and counting forward as many units as there are in the 
 other. 
 
 E.g. To add 3 and 5 we start with 5 on the number scale and 
 count 6, 7, 8 ; or start with 3 and count 4, 5, 6, 7, 8. 
 
 51. In algebra two signed numbers are added in the same 
 manner except that the direction, forward or backward, in which 
 we count, is determined by the sign, + or ~, of the number 
 which we are adding. 
 
 Thus, to add + 7 and +5, begin at 7 to the right of the zero point in 
 the scale of signed numbers and count 5 more toward the right, or 
 begin at 5 to the right and count 7 more to the right, in either case 
 arriving at + 12. 
 
 To add _ 7 and ~5, we may begin at 7 to the left and count 5 more 
 toward the left, or begin at 5 to the left and count 7 more in that di- 
 rection, in either case arriving at _ 12. 
 
 To add + 7 and _ 5, begin at 7 to the right and count 5 toward the 
 left, or begin at 5 to the left and count 7 toward the right, in either 
 case arriving at +2. 
 
 To add _ 7 and + 5, begin at 7 to the left and count 5 toward the 
 right, or begin at 5 to the right and count 7 toward the left, in either 
 case reaching - 2. 
 
 I.e. +7 + +5 = +12, -7 + -5 = ~12, +7 + "5 = +2, ~7 + +5 = -2. 
 
 52. In adding several signed numbers, we reach the same 
 result whether we add them in the order in which they happen 
 to be given, or in any other order. Thus we may add first all 
 
54 POSITIVE AND NEGATIVE NUMBERS 
 
 the positive numbers and then all the negative numbers, and 
 finally combine these two results. 
 
 E.g. + 5 + ~8 + +7 + -6, taken in order from left to right, gives - 3 
 as the sum of the first two, +4 as the sum of the first three, and ~2 as 
 the final result. But +5 + +7 s + 12, -8 + "6 = "14, and +12 + -14 = 
 -2, the same result as before. 
 
 53. Addition by counting makes it clear that two numbers 
 of opposite signs tend to cancel each other when added. 
 
 E.g. In adding ~o to +8 we start with +8, which we reach by count- 
 ing from zero eight units to the right, and then add ~5 by counting 
 five units to the left, thus retracing five of the units just counted. 
 That is, - 5 annuls five of the +8 units, leaving the sum +3. 
 
 In case two numbers have opposite signs and equal absolute 
 values one completely cancels the other. 
 E.g. +8 +-8 = 0. 
 
 EXERCISES AND PROBLEMS 
 
 Perform the following indicated additions : 
 
 1. -31 + +42. 3. +68 + ~46. 5. +104 + ~245. 
 
 2. "17 + -13. 4. +34 + "46. 6. -llf + +8£. 
 7. +16 + ~3 + +7 + +4 + -19. 8. +42 + + 74 + -92 + -7 + -3. 
 9. -13n + + 7n = + 7n + -13n = -6n. 
 
 Solution. By Principle IX, to add ~13 n to +7 n or to add +7 n to 
 -13 n is to "find the difference of their absolute values and prefix to 
 this the sign of that one whose absolute value is the greater." That is, 
 -13n + +7n = -(13 n— 7n) = _ 6n, since by Principle II, 13/i — 7n = 6n. 
 
 10. +14* + -8* 4- "3* + ~45*. 11. + 68x+ + 3±x+-16x+-3x. 
 12. 296r* + -367r* + -119rt. 13. ~3 (a + 6) + +4 (a + 6). 
 14. +7(x — y) + -5(x — y). 15. "81 (r + s) + ~91 (r + s). 
 
 16. A man travels 10 miles east, then 23 miles west, and 
 finally 5 miles east. Express his final distance and direction 
 from the starting point as the sum of three signed numbers. 
 
AVERAGES OF SIGNED NUMBERS 55 
 
 17. The temperature falls 35°, rises 24°, falls 3°, rises 17°, 
 and finally falls 5°. What is the net change in temperature 
 between the last reading and the original reading ? 
 
 18. A real estate firm gains $5000 on one transaction, loses 
 $2500 on a second, loses $1400 on a third, and gains $200 on a 
 fourth. What is the aggregate result of the four transactions ? 
 
 AVERAGES OF SIGNED NUMBERS 
 
 54. Half the sum of two numbers is called their average. 
 
 Thus 6 is the average of 4 and 8. Similarly, the average of 
 
 three numbers is one-third of their sum, and in general the 
 
 average of n numbers is the sum of the numbers divided by n. 
 
 Find the average of each of the following sets : 
 
 1. 10, 12, 14, 16, 18. 4. 7, 9, 11, 13, 15. 
 
 2. 5, 9,20,30, 3. 5. 7,10,21,29,30. 
 
 3. 11, 10, 4, 6, 5. 6. 13, 8, 9, 10, 4. 
 
 The average gain or loss per year for a given number of 
 years is the algebraic sum of the yearly gains and losses di- 
 vided by the number of years. 
 
 Illustrative Problem. A man lost $400 the first year, gained 
 $300 the second, and gained $1000 the third. What was the 
 average loss or gain ? 
 
 Solution. -400 +. + 300 + +1000 = +900 = +30Q 
 
 3 3 
 
 That is, the average gain is $ 300. 
 
 Illustrative Problem. During four years a certain business 
 shows an average annual gain of $ 5000. What was the loss 
 or gain the first year, if for the remaining years there were 
 gains of $8000, $9000, and $7500 respectively? 
 
 Solution. Let x represent the number of dollars gained or lost the 
 first year. Then the average for the four years is 
 x + +8000 + +9000 + +7500 
 
56 POSITIVE AND NEGATIVE NUMBERS 
 
 But the average for the four years is given as $ 5000. 
 
 Hence « + + 8000 + +9000 + + 75Q0 = +5000. 
 
 4 
 
 By If, x + + 8000 + +9000 + +7500 = +20000. 
 
 By IX, x + + 24500 = + 20000. 
 
 By A, x + + 24500 + - 24500 = + 20000 + -24500. 
 
 By IX, a; =-4500. 
 
 Hence there was a loss of S$ 4500 the first year. 
 
 PROBLEMS 
 
 1. Find the average of $1800 loss, $3100 loss, $6800 gain, 
 $10,800 loss, and $31,700 gain. 
 
 2. Find the average of $180 gain, $360 loss, $480 loss, 
 $100 gain, $700 gain, $400 gain, $1300 loss, $300 gain, $4840 
 gain, and $ 12,000 gain. 
 
 3. A merchant gained an average of $ 2800 per year for 5 
 years. The first year he gained $3000, the second $1500, the 
 third $4000, and the fourth $2400. Did he gain or lose and 
 how much during the fifth year ? 
 
 4. A certain business shows an average gain of $4000 
 per year for 6 years. During the first 5 years the results 
 were, $8000 loss, $10,000 gain, $7000 gain, $3000 gain, 
 and $12,000 gain. Find the loss or gain during the sixth 
 year. 
 
 5. Find the average of the following temperatures : 7 a.m., 
 "4°; 8 a.m., -2°; 9 a.m., -1°; 10 a.m., +1°; 11 a.m., +5°; 
 12 m., +7°. 
 
 6. During the 12 hours ending at 6 a.m., January 19, 1892, 
 the U. S. Weather Bureau at Helena, Montana, recorded the 
 following temperatures: "9°, "8°, "8°, "9°, "9°, ~9°, "8°, +36°, 
 + 37°, +40°, +20°, +16°. Find the average temperature for the 
 12 hours. 
 
AVERAGES OF SIGNED NUMBERS 57 
 
 Find the average yearly temperatures at the following 
 places, the monthly averages having been recorded as given 
 below : 
 
 7. For New York City: +29°, +33°, +39°, +46°, +53°, +63°, 
 + 67°, +67°, +61°, +52°, +47°, +41°. 
 
 8. For Singapore, Straits Settlement : + 81°, + 84°, * 85°, + 85°, 
 +85°, +86°, +86°, +87°, +85°, +84°, +83°, +81°. 
 
 9. For St. Vincent, Minnesota: "5°, 0°, +15°, +35°, +55°, 
 + 60°, +66°, +63°, +55°, +40°, +22°, +5°. 
 
 10. For Nerchinsk, Siberia: "23°, -13°, "10°, +35°, +55°, 
 + 70°, +70°, +64°, +50, +30°, +5°, ~15°. 
 
 If the latitude of a place is midway between the latitudes of 
 two given places, then its latitude equals one-half the algebraic 
 sum of the two given latitudes. 
 
 Thus, if the latitude of one place is +16° and that of the 
 other " 56°, then the latitude of the place midway between them 
 
 is" 56 ° + +16O = ^ == -20°. 
 2 2 
 
 The data used in some of the following problems vary 
 
 slightly from the published records, but in no case more 
 
 than 5'. 
 
 11. The latitude of New Orleans, Louisiana, is + 30°, and of 
 Toronto, Canada, +43° 40'. Find the latitude of Norfolk, Vir- 
 ginia, which is midway between these latitudes. (Positive 
 numbers represent north latitude and negative numbers repre- 
 sent south latitude.) 
 
 12. The latitude of Alexandria, Egypt, is +31° 10', and of 
 Christiania, Norway, + 59° 50'. That of Venice, Italy, is mid- 
 way between these latitudes. Find the latitude of Venice. 
 
 13. The longitude of Edinburgh, Scotland, is "3° 10', and of 
 Warsaw, Poland, +21°. Find the longitude of Bremen, Ger- 
 many, which is midway between these. (Positive numbers 
 represent east longitude and negative numbers west longitude.) 
 
58 POSITIVE AND NEGATIVE NUMBERS 
 
 14. The longitude of Vienna, Austria, is + 16° 20', and of 
 Splugen Pass, Switzerland, +9° 20'. The longitude of Splilgen 
 Pass is midway between the longitudes of Vienna and Paris, 
 France. Pind the longitude of Paris. 
 
 15. The longitude of Brussels, Belgium, is + 4° 20', and of 
 Jena, Germany, + 11° 40'. The longitude of Brussels is mid- 
 way between those of Jena, and Liverpool, England. Find 
 the longitude of Liverpool. 
 
 16. The longitude of Berlin, Germany, is +13° 20', and of 
 St. Petersburg, Russia, +30° 20'. The longitude of Berlin is 
 midway between those of St. Petersburg and Madrid, Spain. 
 Find the longitude of Madrid. 
 
 17. The latitude of Eome, Italy, is +41° 55', and of Cux- 
 haven, Germany, + 53° 55'. The latitude of Rome is midway 
 between those of Cuxhaven and Cairo, Egypt. Find the 
 latitude of Cairo. 
 
 18. The longitude of Calais, France, is + 1° 55' and of Luck- 
 now, India, + 80° 55'. The longitude of Calais is midway be- 
 tween those of Lucknow and Washington, D.C. What is 
 the longitude of Washington? 
 
 SUBTRACTION OP SIGNED NUMBERS 
 55. In arithmetic the accuracy of subtraction is tested by 
 
 showing that the remainder added to the subtrahend equals 
 
 the minuend. 
 
 E.g. We say 8 - 5 = 3 because 5 + 3 = 8. 
 
 Indeed, we may, and often do perform subtraction by start- 
 ing with the subtrahend and counting until we reach the 
 minuend. 
 
 E.g. A clerk in changing a dollar bill after a purchase of 63 cents 
 might count out two pennies, a dime, and a quarter, saying "65, 75, 
 one dollar." That is, he counts from 63 cents (the subtrahend) up 
 to 100 cents (the minuend). 
 
SUBTRACTION OF SIGNED NUMBERS 59 
 
 56. In like manner, signed numbers also may be subtracted, 
 but we must find not only the number of units, but also the 
 direction from the subtrahend to the minuend. This is most 
 easily done by reference to the number scale, page 52. 
 
 Ex. 1. +8 — + 5 = +3, since in passing from + 5 to +8 on the 
 number scale we count 3 units in the positive direction. 
 
 Ex. 2. - 8 — - 5 = ~3, since from ~5 to "8 we count 3 units in 
 the negative direction. 
 
 Ex. 3. "8 — + 5 = "13, since from +5 to "8 we count 13 units 
 in the negative direction. 
 
 Ex. 4. +8 — ~5 = + 13, since from ~5 to + 8 we count 13 units 
 in the j>ositive direction. 
 
 57. In each of these examples we see that the result fulfills 
 the test of arithmetic ; namely, 
 
 Subtrahend + Difference = Minuend 
 
 Hence _ 8 — ~5 = - 3 is checked by finding that _ 5 + -3 = ~8. 
 
 +8 - -5 = +13 is checked by finding that "5 + +13 = +8. 
 
 EXERCISES AND PROBLEMS 
 
 Perform the following indicated subtractions by finding the 
 distance and direction on the number scale from subtrahend 
 to minuend, and apply the check to each result. 
 
 1. -10- "5 6. "17 --20 11. +93 -+22 
 
 +6 --14 12. +17 --13 
 
 +7 - -9 13. -78 - -37 
 
 11- +6 14. +57 -+84 
 
 -21- "6 15. "48 --31 
 
 16. How many degrees, and in what direction, must the 
 temperature change in order to vary from 12° below zero to 
 38° above zero ? This is an example in subtraction, since we 
 are required to find a number which added to "12° gives +38°. 
 Hence we write it +38° - ~12° = what ? 
 
 2. 
 
 -15- +5 
 
 7. 
 
 3. 
 
 +20 - -15 
 
 8. 
 
 4. 
 
 +11- +3 
 
 9. 
 
 5. 
 
 -11- +5 
 
 10. 
 
60 POSITIVE AND NEGATIVE NUMBERS 
 
 17. How many degrees, and in what direction, does the ther- 
 mometer change in passing from 27° above zero to 3° below 
 zero ? That is, ~3° - +27° = what ? 
 
 18. What must be added to $35 loss to make the sum $30 
 gain ? That is, +30 - -35 = what ? 
 
 19. What must be added to $ 15 gain to make the sum $ 8 
 loss ? That is, "8 - +15 = what ? 
 
 58. Subtraction always possible. In arithmetic subtraction is 
 possible only when the subtrahend is less than or equal to the 
 minuend. 
 
 E.g. 5 from 8 leaves 3, 5 from 5 leaves ; but we cannot take 5 
 from 2, since when we have subtracted 2 units from 2 there are no 
 more to take away. 
 
 However, by means of negative numbers we can as easily 
 perform the subtraction, 2 minus 5, as 5 minus 2. 
 
 Thus, 2 - 5 = -3, since -3 + 5 = 2; and 5 - 2 = +3, since, +3 + 2 
 = 5. 
 
 It thus appears that, in terms of signed numbers, a — b has 
 a meaning no matter what numbers are represented by a and 
 b ; that is, a — b means the number which added to b gives a. 
 
 59. A short rule for subtraction. Since +8 — ~5 = + 13, and 
 since + 8 +• + 5 = + 13, it follows that subtracting ~5 from +8 
 gives the same result as adding + 5 to + 8. Similarly ~8 — _ 5 = 
 -3 and "8+- + 5 = -3. 
 
 Hence subtracting a negative number is equivalent to adding 
 a jiositive number of the same absolute value. 
 
 Since +8 -+5 = +3, and since +8 + -5 = +3, it follows that 
 subtracting + 5 from + 8 gives the same result as adding ~5 to + 8. 
 Similarly ~8 - +5 = ~13 and "8 + ~5 = "13. 
 
 Hence subtracting a positive number is equivalent to adding 
 a negative number of the same absolute value. 
 
SUBTRACTION OF SIGNED NUMBERS 61 
 
 These statements are illustrated by such facts as : Removing 
 a debt is equivalent to adding property and removing property is 
 equivalent to adding debt. 
 
 Perform the following subtractions as explained in this 
 paragraph, by changing the sign of the subtrahend and adding: 
 
 1. "5 - -2. 5. +57 - -32. 9. +37 -+50. 
 
 2. "4 - +1. 6. "32 - +34. 10. -23 - +57. 
 
 3. -5- +2. 7. -52 -"32. 11. -16 a- +4 a 
 
 4. +3 _ -5. 8. "16 - "12. 12. +13 * - ~20 1 
 
 The preceding exercises illustrate the following principle : 
 
 60. Principle X. To subtract one signed number from 
 another signed number, add the subtrahend with its sign 
 changed to the minuend. 
 
 The change in the sign of the subtrahend may be made men- 
 tally without re-writing the problem. The results are to be 
 checked by showing that the difference added to the subtrahend 
 equals the minuend. 
 
 EXERCISES 
 
 1. Prom +6 + ~2 subtract "14. 
 
 2. From "6 a -f + 2a subtract "14 a. 
 
 3. From 17 ab+ 8 ab subtract "35 ab. 
 
 4. From 5 ax + 4 ax subtract 7 ax + 2 ax. 
 
 5. From 54 abc +~47 abc -f 36 abc subtract 80 abc. 
 
 6. From 54 . 13 +~47 .13 + 36-13 subtract 80 . 13. 
 
 7. From 29 • 3 • 11 + 37 • 3 • 11 subtract "34 . 3 • 11. 
 
 8. From 29 xy + 37 xy subtract ~34 xy. 
 
 9. Solve x + 8 = 4. 
 
 Solution. Subtract +8 from each member (which is equivalent to 
 adding -8). 
 
62 POSITIVE AND NEGATIVE NUMBERS 
 
 Then x = +4 - +8 = ~4, which is correct, since ~4 + +8 = +4. This 
 is a problem in subtraction, since one of two numbers, 8, and their 
 sum, 4, are given, and we are to find the second number, which is 
 represented by x. 
 
 Solve the following equations : 
 
 
 
 10. x + -'3 = 7. 
 
 16. 
 
 x+ 9= 3. 
 
 11. a; + -9 = 1. 
 
 17. 
 
 -4 + x = 7. 
 
 12. 3+ x = 0. 
 
 18. 
 
 -5 + a = 4. 
 
 13. x + -1 = 2. 
 
 19. 
 
 -20+ * = -12. 
 
 14. a: +13 =7. 
 
 20. 
 
 8+ n=-l& 
 
 15. x + 4=2. 
 
 21. 
 
 ft + -40 = -65. 
 
 MULTIPLICATION OF SIGNED NUMBERS 
 61. The multiplication of signed numbers is illustrated by 
 the following problems. 
 
 Illustrative Problem. A balloonist, just before starting, makes 
 the following preparations : (a) He adds 9000 cubic feet of gas 
 with a lifting power of 75 pounds per thousand cubic feet ; (b) 
 He takes on 8 bags of sand, each weighing 15 pounds. How 
 does each of these operations affect the buoyancy of the 
 balloon ? 
 
 Solution, (a) A lifting power of 75 lbs. is indicated by + 75, and 
 adding such a power 9 times is indicated by + 9. Hence + 9 • +75 = 
 + 675, the total lifting power added. 
 
 (b) A weight of 15 lbs. is indicated by ~15, and adding 8 such 
 weights is indicated by + 8. Since the' total weight added is 120 lbs., 
 we have + 8 • "15 = - 120. 
 
 Illustrative Problem. During the course of his journey this 
 balloonist opens the valve and allows 2000 cubic feet of gas to 
 escape, and later throws overboard 4 bags of sand. What 
 effect does each of these operations produce on the balloon ? 
 
MULTIPLICATION OF SIGNED NUMBERS 63 
 
 Solution, (a) The gas, being a lifting power, is positive, but the 
 removal of 2000 cubic feet of it is indicated by ~2, and the result is a 
 depression of the balloon by 150 lbs. ; that is, _ 2 • + 75 = _ 150. 
 
 (b) The removal of 4 weights is indicated by ~4, but the weights 
 themselves have the negative quality of downward pull. Hence to 
 remove 4 weights of 15 lbs. each is equivalent to increasing the 
 buoyancy of the balloon by 60 lbs; that is, ~4 • _ 15 a +60. 
 
 62. These illustrations of multiplying signed numbers are 
 natural extensions of the process of multiplication in arith- 
 metic. 
 
 E.g. Just as 3 • 4 = 4 + 4 + 4 = 12, so 3 . -4 = -4 + -4 +~4 = -12, 
 and since 3 • 4 is the same as + 3 • + 4, we write + 3.- + 4 = +12. 
 
 Again, just as we take the multiplicand additively when the 
 multiplier is a positive integer, so we take it subtractively when 
 the multiplier is negative integer. 
 
 E.g. - 3-+4 means to subtract +4 three times; that is, to sub- 
 tract + 12. But to subtract +12 is the same as to add ~12. Hence 
 -3- +4 =~12. Again, -3 • _ 4 means to subtract - 4 three times; 
 that is, to subtract -12. But to subtract -12 is the same as to add 
 +12. Hence -3- -4 = +12. 
 
 EXERCISES AND PROBLEMS 
 
 Explain the following indicated multiplications and find the 
 product in each case : 
 
 1. 
 
 -3 • -10. 
 
 5. 
 
 43 • -192. 
 
 9. 
 
 71 • -x. 
 
 2. 
 
 -3- +10. 
 
 6. 
 
 -27 • -235. 
 
 10. 
 
 -112 • ~t. 
 
 3. 
 
 -5 • +50. 
 
 7. 
 
 ~5-+r. 
 
 11. 
 
 -U-y. 
 
 4. 
 
 -75 • -89. 
 
 8. 
 
 + 16- -r. 
 
 12. 
 
 -20 • -v. 
 
 13. A man gained $212 each month for 5 months, then lost 
 $175 per month for 3 months. Express his net gain or loss 
 as the sum of two products. 
 
64 POSITIVE AND NEGATIVE NUMBERS 
 
 14. A man gained $2100 during a certain year. For the 
 first 4 months he lost $125 per month. During the next 5 
 months he gained $500 per month. Find his gain or loss dur- 
 ing the remaining 3 months of the year. Express the net gain 
 as the sum of two products. 
 
 15. A raft is made of cork and iron. What effects are pro- 
 duced upon its floating qualities by the following changes ? 
 
 (a) Adding 4 braces, each weighing (under water) 5 lbs. (b) Ee- 
 moving 3 pieces of cork, each capable of sustaining 3 lbs. 
 (c) Adding 10 pieces of cork, each capable of sustaining 7 lbs. 
 
 16. What are the effects on a shipwrecked man's ability to 
 float ? (a) If he holds fast to 3 bags of gold, each weighing 1 lbs. 
 
 (b) If he ties on two life preservers, each capable of supporting 
 15 lbs. (c) If he throws away his two boots, each weighing 2 lbs. 
 
 The preceding exercises illustrate the following principle : 
 
 63. Principle XI. If two signed numbers are of the same 
 quality, their product is positive; if they are of opposite 
 quality, their product is negative. The absolute value of 
 the product is the product of the absolute values of the 
 factors. 
 
 In applying this principle observe that the sign of the prod- 
 uct is obtained quite independently of the absolute value of 
 the two factors. 
 
 E.g. f • -5 = -('£) = -3f ; "12 • ~3.5 =+42. 
 
 64. Principle XI is also stated in symbols as follows : 
 +a • +6 = + ab, ~a • ~b — + ab, + a • ~b =~ab, ~a • + b =~ab. 
 
 The product of several signed numbers is found as illustrated 
 in the following : 
 
 -2 • +5 • -3 • -4 • +6 = -10 . -3 • -4 • +6 = +30 . ~4 • +6 = 
 _ 120 • + 6 = ~720. That is, any two factors are multiplied 
 together, then this product by another factor, etc., until all the 
 factors are multiplied. 
 
DIVISION OF SIGNED NUMBERS 65 
 
 65. Evidently the factors in such a product may be taken in 
 any desired order. Let the student try other orders in the 
 above example. 
 
 Since the product of all positive factors is positive, the final 
 sign depends upon the number of negative factors. If this 
 number is even, the product is positive ; if it is odd, the product 
 is negative. 
 
 E.g. If there are 5 negative factors, the product is negative; if 
 there are 6, it is positive. 
 
 EXERCISES 
 
 In the following exercises determine the sign of the product 
 before finding its absolute value. State each principle used in 
 the reduction to the final form. 
 
 1. -4- +3- ~6 .~7. 9. «(3a+"4a- + 5a). 
 
 2. -50 • -20 • "30 • -40. 10. 8 (16 x + "20 x) -s- 4. 
 
 3. ~a- ~b -+c- + d- + e. 11. 5x — 3("2a; + + 3:»— -4a>). 
 
 4. a>-b- + c--d--x. 12. 6r + 4(3r — "5r+-7r). 
 
 5. -5 ("3 +"7). 13. -5 ("4 • +3 . -2). 
 
 6. a(-b — ~c). 14. 6x— "14 x— (~5 x +~1 x). 
 
 7. -c{x-~y). 15. 8 y -16 y +(4 y + "11 y). 
 
 8. 1 a{x+-y—~z). 16. lit — 20 (* +' - 3 • -5 *). 
 
 DIVISION OF SIGNED NUMBERS 
 
 66. In arithmetic we test the correctness of division, by 
 showing that the quotient multiplied by the divisor equals the 
 dividend. 
 
 E.g. 27 -r- 9 = 3, because 9 -3 = 27. 
 
 Hence division may be defined as the process of finding one 
 of two factors when their product and the other factor are given. 
 
66 
 
 POSITIVE AND NEGATIVE NUMBERS 
 
 This definition also applies to the division of signed numbers. 
 In dividing signed numbers, however, we must determine the 
 sign of the quotient as well as its absolute value. 
 
 E.g. -42- +6 = -7, because "7- + 6 = -42; 
 
 also -42 -=- -6 = +7, because +7- -6 = -42. 
 
 So in every case the test is : 
 
 Quotient x Divisor 
 
 Dividend. 
 
 In like manner perform the following : 
 
 1. -26 + 5. 4. -9rs-=-+3. 
 
 2. -ab + a. 5. + 75y+~15. 
 
 3. + 5xy+~x. 6. -121 x + +11. 
 
 The preceding exercises illustrate the following principle : 
 
 67. Principle XII. The quotient of two signed numbers 
 is positive if the dividend and divisor have the same sign, 
 negative if they have opposite signs. The absolute value of 
 the quotient is the quotient of the absolute values of divi- 
 dend and divisor. 
 
 EXERCISES 
 
 Perform the following indicated divisions. Check by multi- 
 plying quotient by divisor. 
 
 -99 x -=--25. 
 • 87 y ■+■ -400. 
 8*-f-4y)-f--2. 
 16 a + -20 b) -=- -4. 
 6r + 9s--120n--3. 
 7 aa;+-14 ay-~21 &z)-=- + 7. 
 12 xy — ~3 ax) •+■ ~x. 
 27 ab - 36 ac) -=- "9. 
 3-4?/ + 6-8a)-f--3. 
 x — 3 y — z) -f- a. 
 
 1. 
 
 -28 -s- + 7. 
 
 
 
 11. 1 
 
 2. 
 
 -42 -s- -6. 
 
 
 
 12. 1 
 
 3. 
 
 51 -=--17. 
 
 
 
 13. (< 
 
 4. 
 
 21xy + 3. 
 
 
 
 14. (' 
 
 5. 
 
 "16a&-s--4. 
 
 
 
 15. ( 
 
 6. 
 
 " 15 ax -=- cc. 
 
 
 
 16. ( 
 
 7. 
 
 -32 (a- 6) + 
 
 _ (a- 
 
 -&). 
 
 17. C 
 
 8. 
 
 21(x + y) + 
 
 9. 
 
 
 18. (! 
 
 9. 
 
 4.-9z-=--3. 
 
 
 
 19. (. 
 
 10. 
 
 3-8*-s--4 
 
 
 
 20. 2 
 
INTERPRETATION OF SIGNED NUMBERS 67 
 
 68. While Principles I-VIII were studied in connection 
 with unsigned, or arithmetic numbers only, it is now very 
 important to note that they all apply to signed numbers as 
 well. The form changes described in § 37 also apply to signed 
 numbers just the same as to arithmetic numbers. 
 
 In the statement of these principles the word number will 
 from now on be understood to refer either to the ordinary num- 
 bers of arithmetic or to the signed numbers, as occasion may 
 require. It should also be noticed that the numbers of arith- 
 metic are used as freely in algebra as in arithmetic. It is only 
 when we wish to distinguish them from negative numbers that 
 they are called positive numbers. 
 
 The number system of algebra, as far as we have studied it, 
 consists of the numbers of arithmetic together with the negative 
 numbers. 
 
 INTERPRETATION AND USE OF NEGATIVE NUMBERS 
 
 69. Illustrative Problem. Divide 34 units into two parts such 
 that one part is equal to the remainder when 3 times the other 
 part is subtracted from 46. 
 
 Solution. Let the two numbers be represented by x and 34 — x. 
 
 Then 34-x = 46-3x. (1) 
 
 S|34, -x = 12-3z. (2) 
 
 ,4|3ar, 8*-x=12. (3) 
 
 Principle II, 2 x = 12. (4) 
 
 D\2, x = Q. (5) 
 
 Substituting x = 6 in (1), 34 - 6 = 46 - 18, or 28 = 28. 
 
 In the above solution, equation (2) was obtained from (1) by sub- 
 tracting 34 from each member. This would clearly be impossible 
 without the use of negative numbers. 
 
 In this case the problem itself does not involve negative numbers, 
 but in the course of its solution they naturally occur. If the nega- 
 tive number could not be used, we should be compelled to keep the 
 
68 POSITIVE AND NEGATIVE NUMBERS 
 
 members of each equation positive or zero. This would be impossible, 
 since we do not know what numbers are represented by the letters 
 involved, and hence cannot tell by inspection whether a given term 
 is positive or not. 
 
 70. We have seen how naturally the use of signed numbers 
 has arisen in problems where things of opposite qualities have 
 to be distinguished. 
 
 In solving a problem, a negative result may have a natural 
 interpretation or it may indicate that the conditions of the 
 problem are impossible. 
 
 A similar statement holds in reference to fractional answers in 
 arithmetic. For example, if we say there are twice as many girls as 
 boys in a schoolroom and 35 pupils in all, the number of boys would 
 be 35 -^- 3 = llf, which indicates that the conditions of the problem 
 are impossible. 
 
 71. Illustrative Problem. The crews on three steamers to- 
 gether number 94 men. The second has 40 more than the first, 
 and the third 20 more than the second. How many men in 
 each crew ? 
 
 Solution. Let n = number of men in first crew. 
 
 Then n + 40 = number of men in second crew, 
 
 and n + 40 + 20 = number of men in third crew. 
 
 Hence n + n + 40 + n + 40 + 20 = 94, 
 
 and 3 n + 100 = 94. 
 
 n s= -2. 
 
 Here the negative result indicates that the conditions of the 
 problem are impossible. 
 
 72. Illustrative Problem. A real estate agent gained $ 8400 on 
 four transactions. On the first he gained $6400, on the second 
 he lost $ 2100, on the third he gained $ 5000. Did he lose or 
 gain on the fourth transaction ? 
 
INTERPRETATION OF SIGNED NUMBERS 69 
 
 Solution. Since we do not know whether he gained or lost on that 
 transaction, we represent the unknown number by n, which may be 
 positive or negative, as will be determined by the solution of the 
 problem. 
 
 Thus we have 6400 + -2100 + 5000 + n = 8400. (1) 
 
 Hence by IX, F, + 9300 + n = 8400. (2) 
 
 By S, n = 8400 - 9300. (3) 
 
 ByX, n = -900. (4) 
 
 In this case the negative result indicates that there was a loss on 
 the fourth transaction. 
 
 PROBLEMS 
 
 In the following problems give the solutions in full and state all 
 principles used, together with the interpretation of the results : 
 
 1. A man gains $2100 during one year. During the first 
 three months he loses $125 per month, then gains $500 per 
 month during the next five months. What is the gain or loss 
 per month during the remaining four months ? 
 
 2. A man gained $ 1250 during four months. During the 
 second month he gained $600 more than the first month, the 
 third month he gained $300 less than the second, and the 
 fourth he gained $200 more than the third. Find the gain 
 or loss for the first month. 
 
 3. A box containing a Christmas toy weighed 25 oz. When 
 the toy and the packing were removed, the box weighed 20 oz. 
 The packing weighed 7 oz. What kind of a toy was it ? 
 
 4. A man rowing against a swift current rows 8 miles in 5 
 hours. The second hour he rows one mile less than the first, 
 the third two, miles more than the second, and the fourth and 
 fifth one mile more each than he rowed the third hour. How 
 many miles did he row each hour ? 
 
 5. There are three trees the sum of whose heights is 108 
 feet. The second is 40 feet taller than the first, and the 
 third is 30 feet taller than the second. How tall is each tree ? 
 
70 POSITIVE AND NEGATIVE NUMBERS 
 
 Find the average yearly temperature at each of the follow- 
 ing places, the average monthly temperatures being as here 
 given : 
 
 6. Port Conger, off the northwest coast of Greenland ; 37°, 
 -43°, "32°, -15°, +14°, +18°, +35°, +34°, +25°, +4°, -17°, - 30°. 
 
 7. Franz Joseph's Land ; "20°, "20°, "10°, 0°, 15°, 30°, 35°, 
 30°, 20°, 10°, 0°, "10°. 
 
 8. Western Baffin Land ; "30°, "30°, ~20°, 0°, 20°, 35°, 40°, 
 35°, 25°, 10°, -10°, -20°. 
 
 Find the average yearly loss or gain in each of the following : 
 
 9. $1600 gain, $8000 loss, $24,000 gain, $40,000 loss. 
 
 10. $32,000 gain, $45,000 loss, $24,000 gain, $42,000 loss. 
 
 11. The average yearly temperature of north central Siberia 
 is ~5°. The average monthly temperatures beginning in Febru- 
 ary are: "60°, "30°, 0°, 15°, 40°, 40°, 35°, 30°, 0°, "30°, -50°. 
 Find the temperature for January. 
 
 12. The business transactions of a certain firm averaged 
 $ 1500 loss for 4 years. For the first year there was a gain of 
 $ 800, the second year a loss of $ 1800, the third year a loss of 
 $300. What was the loss or gain for the fourth year ? 
 
 13. A commercial house averaged $15,000 gain for 5 
 years. What was the loss or gain the first year if the remain- 
 ing years show : $8000 gain, $24,000 gain, $2000 loss, $20,000 
 gain, and $ 50,000 gain, respectively ? 
 
 14. The longitude of Boston, Massachusetts, is "71° 10', and 
 that of Chicago, Illinois, is ~87° 35'. Find the longitude of 
 Lake Chautauqua, which is midway between these. 
 
 15. The longitude of New Haven, Connecticut, is ~72° 58', 
 and that of Bombay, India, is +72° 48'. The longitude of St. 
 Paul's Cathedral, London, is midway between these. Find the 
 longitude of the cathedral. 
 
INTERPRETATION OF SIGNED NUMBERS 71 
 
 16. The longitude of Cincinnati, Ohio, is ~84° 30', and that 
 of Indianapolis, Indiana, is "86° 5'. The longitude of Cincin- 
 nati is midway between those of Indianapolis, and Columbus, 
 Ohio. Find the longitude of Columbus. 
 
 17. The longitude of Bristol, England, is ~2° 30', and that of 
 Minneapolis, Minnesota, is ~93° 20'. The longitude of Bristol 
 is midway between those of Minneapolis and Calcutta, India. 
 Find the longitude of Calcutta. 
 
 18. The latitude of Columbus, Ohio, is + 40° and that of 
 Winnipeg, Canada, is + 50°. The latitude of Columbus is mid- 
 way between those of Winnipeg and Houston, Texas. Find 
 the latitude of Houston. 
 
 19. The longitude of Montreal, Canada, is "73° 40' and that 
 of Baltimore, Maryland, "76° 40'. Find the latitude of Phila- 
 delphia, which is midway between these. 
 
 20. The latitude of Lima, Peru, is ~12° and that of Buenos 
 Ay res, Argentina, "34° 35'. The latitude of Lima is midway 
 between those of Buenos Ayres and Caracas, Venezuela. Find 
 the latitude of Caracas. 
 
 21. The longitude of Providence, Rhode Island, is _ 71° 75' 
 and that of Fargo, North Dakota, -96° 50'. The longitude of 
 Fargo is midway between those of Providence and Seattle, 
 Washington. Find the longitude of Seattle. 
 
I 
 
 CHAPTER III 
 INVOLVED NUMBER EXPRESSIONS 
 
 73. Double Use of the Signs + and—. In the preceding 
 chapter it has been found that the negative quality may be 
 regarded as implying subtraction and the positive quality as 
 implying addition. It was for this reason that + and ~ were 
 selected as symbols for the words " positive " and " negative." 
 
 74. It is now possible to dispense with these special signs 
 of quality. For, by Principle X, a — + b and a + ~ b are the same 
 in effect, and likewise a—~b and a-\- + b. Hence, omitting 
 the positive signs (§ 45), we may write 
 
 a -f- ~& = a — b and a — ~b = a + b. 
 
 One set of signs is, therefore, sufficient as symbols both of 
 operation and of quality. 
 
 E.g. 5 — 7 means either 5 +~7 or 5 — +7, and in either case equals 
 ~2, which we now write — 2. Thus 5 — 7 equals — 2, which is read, 
 5 minus 7 equals negative 2. 
 
 Ex. 1. "5.-4 = + 20 is now written - 5 • - 4, or (— 5)(-4), 
 = + 20, V_ ' 
 
 and -5- + 4 = -20is written -5 • +4 or, (-5) ( + 4),= -20. 
 
 Ex.2. 5(9 a +"2 a) = 5(9 a— 2 a). 
 
 By IV, =45 a- 10a. 
 
 By II, = 35 a. 
 
 Ex.3. 5a+-4(-3a- + 76) = 5a-4(-3a-7&). 
 By IV, XI, =5a + 12a + 28&. 
 
 By I, =17 a +28 6. 
 
 72 
 
POLYNOMIALS 73 
 
 EXERCISES 
 
 Rewrite the following, using one set of signs, and then per- 
 form the indicated operations. 
 
 1. 7- +3 + "8. 5. 27 abc + ~35abc- 2 abc. 
 
 2. -9 --3 + -12. 6. 5 ("2 + -3) + 4(5 -~7). 
 
 3. -4a + -5a + "6a. 7. '8 (7 - 2) - 8(6 + "9). 
 
 4. -3-5a; + 4.7a;-8.8;K. 8. 3(4a-"5 b) -11(~2 a+~36) 
 
 Rewrite the following expressions, using special signs of 
 quality so that all signs of operation shall indicate addition : 
 
 9. 5-8-14 = 5 +-8 +-14. 13. 56 ay — 72 ay + 7ay. 
 
 10. -7 + 8-18. 14. 3(2-6) + 6(3-7). 
 
 11. -4a + 5a-17a. 15. -3(8 - 6) -4(6-9). 
 
 12. -7-4x+7-4x-- 8 -4a:. 16. 8(4* -5?t)-5(— t + 4n). 
 
 POLYNOMIALS 
 
 75. We have found that the solution of problems leads 
 us to build involved number expressions out of single number 
 symbols. 
 
 E.g. If a; is a number representing my age in years, then 2 (x — 10) 
 is double the number representing my age 10 years ago, and 
 2 l(x — 10) + (x+ 15)] is the number representing twice the sum of my 
 ages 10 years ago and 15 years hence. 
 
 Number expressions are now to be studied more in detail. 
 
 76. Definition. A number expression composed of parts con- 
 nected by the signs + and — is called a polynomial. Each of 
 the parts thus connected together with the sign preceding it is 
 called a term. 
 
 E.g. 5 a - 3 xy — f rt + 99 is a polynomial whose terms are 5 a, 
 -Zxy, - | rt, and + 99. The sign + is understood before 5 a. 
 
74 INVOLVED NUMBER EXPRESSIONS 
 
 77. Definitions. A polynomial of two terms is called a bino- 
 mial, one of three terms is called a trinomial. A term taken 
 by itself is called a monomial. 
 
 E.g. 5 a — 3 xy is a binomial; 5 a — 3 xy — § rt is a trinomial ; 5 a, 
 — 3 xy, —%rt are monomials. 
 
 According to the above definition x + (6 + c) may be called 
 a binomial notwithstanding it is equivalent to the trinomial 
 x + b + c. 
 
 In this case x is called a simple term and (b + c) a compound 
 term. Likewise we may call 3t + £x — 5(a-\- b)y a trinomial 
 having the simple terms 3 t, -f 4 x, and the compound term 
 -5(a + b)y. 
 
 It should be clearly understood that a negative or positive 
 sign before a compound term (as well as before a simple term) 
 applies to the number represented by the whole term. 
 
 78. Definition. Two terms which have a factor in common 
 are called similar with respect to that factor. 
 
 E.g. 5 a and — 3 a are similar with respect to a ; — 3 xy and — 7 x 
 are similar with respect to x ; 5 a and — 5 b are similar with respect 
 to 5 ; 7 abc and — f abc are similar with respect to abc. 
 
 Similar terms may be combined by Principles I, II, and IX. 
 
 E.g. 5a-3a = (5-3)a = 2a; -3xy-7x= -x (3 y+7) ; 5a-5b = 
 5(a-b). 
 
 ADDITION AND SUBTRACTION OF POLYNOMIALS 
 
 79. In adding or subtracting polynomials the work may be 
 conveniently arranged by placing the terms in columns, each 
 column consisting of terms which are similar. 
 
 Ex. Add 5x-6y + 4:z + 5at, — 3 x + 11 y -16z- 96*, 
 and —7y-\-8z. 
 
ADDITION OF POLYNOMIALS 75 
 
 Arranging as suggested and applying Principles I, II, and IX, 
 
 we have 
 
 5x— 6y + 4 2 + 5 at 
 
 -3x + lly-16z-9bt 
 
 - iy+ 8z 
 
 2x — 2y- 4:Z + t(5a-9b) 
 
 5 x and — 3 x are similar with respect to their common factor 
 x. Hence by Principle I we add the other factors 5 and — 3, 
 obtaining (5 — S)x = 2x. 
 
 Likewise we add + 5 at and — 9 bt with respect to the com- 
 mon factor t, obtaining (5 a — 9 b)t. In the second column the 
 sum is (— 6 + 11 — l)y= — 2y, and in the third column the 
 sum is ( + 4 — 16 + 8) z = — 4 z. 
 
 Check by giving convenient values to x, y, z, t, a, and b. 
 
 EXERCISES 
 
 1. Add76-3c + 2o'; _26 + 8c-13d\ 
 
 2. Add 6x— 3 y + ±t — 7 z; x — 5 y-3t; 4 x-±y + 8 t. 
 
 3. Add 5 ac + 3 6c -4c + 86; 26 + 3c-2 6c -3 ac; 46 + 
 4 c + 6c — ac; 2 6c + 4ac + c; 36— 4c. 
 
 4. Add 3-4.7 — 5x + 5abx; 3aby + 3x — 5 -4 ■ 7; 7 x + 
 2-4-7; baby — 3x — 5-4- 7; 45# + a6cc — 4 -7. 
 
 5. Add 3(x — 5) + a(c + 6) + 6(a-?/); 6(c + 6) — a(x — y) 
 + 8(a>-5); 7(c + 6)-4(a- </); 3(a>-y) + (a-5). 
 
 6. Add 16(a+6-c)-3(x-2/)+2(a-6); 2{x-y)- 
 3(a-6) + (a+ 6— c); 7 (a — 6) + a(x — y) — 6(a + 6 — c). 
 
 7. Add a(a — 6) — c(# + ?/) + d(x — z)— 4a6c; c(# — 2)— 
 d(x+y) + (a— 6)+2a6c; e(a — 6)+ma6c+3(a — z)+8(a+y). 
 
 8. Add 7a — 4a+12 2; 6a — 3 x + C2 ; 2 6a + 4 a — 3 cz. 
 
76 INVOLVED NUMBER EXPRESSIONS 
 
 9. Add (a — &) — 3(c — d)+m(a + d) ; c(a—b)+a(c — d). 
 
 10. Add 34 ax + 4 by — 3 z; 2by + 5z; 3bx — 7y + 5dz. 
 
 11. Add 3b + 4:cd—2ae; ab—3cd + 3ae; 3cd — 2ab. 
 
 12. Add 7ax-13by + 5; 9ax + 8by-4; 3 b -12 ax. 
 
 13. Add5(a + 6)-3(c-d); 3(c-d)-8(a + 6); — 2(a + 6). 
 
 14. Add 3 + 4(c -d)-5(a-6-c) ; 4 (a— 6 - c) + 5(c-d); 
 3( a _6_c)-9(c-d) + 12. 
 
 15. Add U(c — 9) + 3(x + y) + 21ivu; —71wu — 5(x + y) 
 -13(c-9). 
 
 16. Add 5a& - 3- 7 -9 + 5(a-l); 5- 9- 7 + 3a&- 2 (z-1); 
 3(»-l)-4.7.9 + 2a&. 
 
 17. Add 31 • 50 - 43 • 74 + 2 • 18 ; 21 • 74 + 7 • 18 - 56 • 50 ; 
 - 12 • 18 + 42 • 50 - 6 • 74. 
 
 18. Add 7(a5-y)-4(aj + y) + 4-7; 9 (a; 4- 2/) + 3(x-y) - 
 9-7; 6(^-^)4-2.7-3(^ + 2/). 
 
 19. Add 16 xy - 13 • 64 ; 15 a& - 2 a# ; 34 • 64 - 3 xy 4- 2 a& ; 
 14- 64 -3a?/- 2 a&. 
 
 80. The subtraction of polynomials is illustrated by the 
 following example : 
 
 From 15 ab — 17 xy + 11 rt subtract — 5 ab 4- 4 xy — 5 nt. 
 
 Arranging as on page 75 and applying Principles II and X, 
 
 15 ab —17xy + 11 rt 
 — 5ab + 4 xy — 5 n< 
 20 ab -21 xy+ t(Ur + 5 n) 
 
 As suggested in § 60, it is sufficient to change the signs of the sub- 
 trahend mentally, rather than to rewrite them before adding to the 
 minuend. 
 
SUBTRACTION OF POLYNOMIALS 77 
 
 EXERCISES 
 
 1. From 9x+3y— 11 z subtract — 5x+8y— 3z. 
 
 2. From 12 ab — 3 cd + 12 xy subtract 3 ab + 2 cd — 11 xy. 
 
 3. From 9 xc + 4 ad — 3 cz + 2 ?/ subtract 3 ?/ — 3 ad + 5 cz. 
 
 4. From 13 t + 5 mx — 5 cu subtract 2 £ — 4 ma + 3 cv. 
 
 5. From 3 v — 2 w + 5 ?nn — 4 asz subtract — -y + 5 w — 3 raw. 
 
 6. From 31 b + 4 ary -f 10 a.c — 4 subtract 8 6 — 5 xy — 3 ax. 
 
 7. From 4 — 3 a — 5xz— 3vy-x subtract 7 a + 2 az 4- 4 vy. 
 
 8. From 8 .ry — 3 x + 4 y subtract — 2xy + 13w + 4x—2y. 
 
 9. From 2 a& — 5 + 7 v + 13 abc subtract 3 ab + v + 8 a&c. 
 
 10. From 8 ca;a — 4yb—3yc subtract 4 fom + 2 yb + 4 yc — 49. 
 
 11. From 31 • ir> — 7xy subtract 12 • 45 + 9 a*/. 
 
 12. From 3 abc - 4 -28 + 2 (x + y) -3 xy subtract 6 • 28 
 + 4 xy — 3 (x + y) + 8 a&c. 
 
 13. From 7 • 3 • 5 + 9 (»y — z) -f 4 • 3 (a + 6) subtract 8(a# — z) 
 -8-3(a + b)+8-3-5. 
 
 14. From 5 ax — 3 by + 4 ax+ 5 by subtract 5 by— 3ax+ 7 by. 
 
 15. From 19 (r - 5 s) + 13 (5 x — 4) + 7 (a? — y) subtract 
 17(5 x _ 4) - 5 (x- y) - 11 (r - 5 s). 
 
 16. From 16 - 15 • 30 + 14(* - 5 yz) — 13 (5 y — z) subtract 
 32-16- 30 + 8(5y-*). 
 
 17. From -41.3 + 13-4-16 subtract 7- 4- 16-8-3. 
 
 18. From a (b + c) + 4 (ra + ») —16 c subtract 9 (ra + n) 
 + 31c-d(6 + c). 
 
 19. From 5(7a;-4) + 3(5t/-3a;) + 5 - 7 subtract 8-7 
 -9(7 x - 4) + 8(5 y— 3*). 
 
 20. From 15 • 48 + 8 a& + 49 a; subtract 7 • 48 - 9 a& - 14 x. 
 
78 INVOLVED NUMBER EXPRESSIONS 
 
 EXERCISES IN ADDITION AND SUBTRACTION 
 
 1. Add 5x — 3y — 7 r + 8t, —7 x + 18y — ±r — 7t, —20 a; 
 
 — 24 y + 18 r - 15 1, and 13 x + 15 y + 11 r + 6 t. 
 
 Check the sum by substituting a; = 1, y = 1, r = 1, f as 1. 
 
 2. Add 17a — 9 6, 3c + 14a, 6 — 3 a, a — 17c, and a — 36 
 -f 4 c. Check for a = 1, 6 = 2, c = 3. 
 
 3. Add 2x + 3y — t, —6y + 8t, —x + y — t, — 4* + 7z,and 
 3 y. Check f or x = 2, y = 3, £ = 1. 
 
 4. Add 17r + 4s-£, 2t + 3u, 2r-3s-f4£, 5w-6«, 7r 
 
 — 3s + 8u, and 8r — 2f + 6w. Check by putting each letter 
 equal to 1 ; also equal to 2. 
 
 5. Add 3 ft + 2 £ + 4 u and ft + 3 1 + 3 w. Check by putting 
 ft = 100, t = 10, rt = 1 ; i.e. 324 + 133 = 457. 
 
 6. Add 4 ft + 3 £ + u and 3 ft + 2 £ -f 7 u. Check as in 5. 
 
 7. Write 247, 323, 647, 239, and 41, as number expressions 
 like those in 5 and 6 and then add them. 
 
 8. Add At — u, bt — u, 6t — u, 7 t — u, and 8 1 — u. Check 
 for t = 10, u = 1 ; also t = 1, u = 1. 
 
 9. Add 647, 391, 276, and 444 as in example 7. 
 
 10. Simplify: 3 xyz — 2 xyz + 5 xyz — 4 xyz + xyz — xyz. 
 
 11. Subtract 5 a — 3 6 + 6 c from — 8a-f 76 — lie and check. 
 
 12. From 7 xy + 8 xz + 9 yz take 17 a;y — 19 xz — 20 yz. 
 
 13. From 6 a; — 3 y take 8 y — 3 z. 
 
 14. From 3p — 4 g + 8 r take 7j9 — 11 r + 11 5- 
 
 15. From a + b + c take a; — y + z. Suggestion : By Principle 
 VII, a + 6 + c-(a; — y + z)=za + b + c-x + y — z. 
 
 16. From 2a; — 3y take 5. r + 7 y + 2 a — 3 6. 
 
 17. From the sum of 18 a6c — 27 xyz + 13 rst and — 11 a6c 
 -f 16 xyz — 52 rst take 67 rst — 39 a6c. 
 
ADDITION AND SUBTRACTION 79 
 
 18. To the difference between the subtrahend 15 a; — 18 y 
 + 27 z and the minuend 117 x + 07y — 81 z add 4 a: — 6 y + 3z. 
 
 19. Add ll(a?-y) + 15 (a -6) and -20(a5-y)-37(o-6) 
 and from the sum subtract 135 (a; — y)— 213 (a — b). 
 
 20. Add 6ax +7 bx — 8cx, —11 ax — 18 &# + 25 ex, and 19 ax 
 -16cx + 24:bx. 
 
 21. From 13 mn — 25 mp + 36 mq subtract 18 mn + 23 mp. 
 
 22. Add by Principle I, 6 • 3 • 9 - 11 • 5 • 7 + 16 ■ 9 • 11 and 
 - 8 • 3 . 9 + 24 • 5 • 7 - 23 • 9 . 11. 
 
 23. From 83-9 + 78 -13 subtract 57-9-93.13+85.17. 
 
 24. From 3 h + 4 1 + 2 u subtract h + 5t+3u. Check. 
 
 25. Subtract 7(a— x)— 10 (b — y) from 13(a — x)+5(b — y). 
 
 81. In solving problems it is often unnecessary to arrange the 
 work of addition and subtraction in columns as above. In most 
 cases the operations can be readily indicated by means of paren- 
 theses as illustrated in the following example : 
 
 From the sum of 3 a + 4 b and 5 a — Sb subtract 7 a — 6 b. 
 
 Indicate these operations thus : 3 a + 4 b + (5 a — 8 6) — (7 a — 6 b). 
 Applying Principle VII, we have 3a + 46 + 5a — 86 — 7 a + Qb. 
 Collecting similar terms, 3a + 5a — 7a + 46 — 86 + 6 6. 
 Finally, applying Principles I and II, we obtain a + 2 6. 
 After a little practice the last two steps can be taken at once. 
 
 EXERCISES AND PROBLEMS 
 
 Perform the following indicated operations by collecting 
 similar terms at once without arranging in columns : 
 
 1. 15 + (7 -9 a;)- (-7* -9) +9. 
 
 2. 7 + 5y-(3y + 2) + (8-4:y). 
 
 3. 2a + 3 + (4a-5)-(lla-14). 
 
 4. 326-(176-12)-(46-13). 
 
80 INVOLVED NUMBER EXPRESSIONS 
 
 5. 16c-(41-7c) + (15-8c). 
 
 6. - (5 a - 3 c) - (2 c - 8 a) + 3 a. 
 
 7. _(_ 12 a _7 ?/ -15 a:) -(-9 # + 8 a + 3?/). 
 
 8. (19 x + 4 y - 32 x - 17 x) - 12 a - (49 y + 18 a> - 70 a). 
 
 9. 17a-3-(7a-2) + (6a-5). 
 
 10. 5a- (8-4x + 7y) + (5x + 3)-(5y + 3x-99). 
 
 11. -(3a + 56-7c) + (8a-4c)-(9c-46 + 4a)-91a. 
 
 12. 7-(4-4c + 2d-2a)+31c-(4-2a-5d)-(-8c). 
 
 13. (41a&-21c+4)-(36c + 15-78a6) + (13c-90ao-8). 
 
 14. 9 by - (4 c - 8 by - 13) - 2c - 16 - (34 by -12 c + 8 by). 
 
 15. 6 mn+(— 9m — 7 w +14) -8 n + (13 mn—11 m)+34 raw. 
 
 16. 34 ax— (— 17 a# + 42) + 8 a -(14 a + 24 ax — 7). 
 
 17. 19 - (+ 2 - 7a - 4 b + 11 oft) - (- 2 6 + 8 ab + 4 a). 
 
 18. 41 % - (4 b - 13 y + 17 by) - (- 5 b - 17 6y + 13 y). 
 
 19. 39 rs - 20 s - 19 r - (7 »•* + 8 s - 19 r) - (15 r - 5 s - 56). 
 
 20. a(3 x — 2 y — z) — (5 ax — ay + 3 z) + az. 
 
 21. 5(4 h + 3bk-7br)- 6(15 & - 35 r) - 20 h. 
 
 22. The altitude of Popocatepetl is 1716 feet less than that 
 of Mt. Logan, and the altitude of Mt. St. Elias is 316 feet 
 greater than that of Popocatepetl. Find the altitude of each 
 mountain, the sum of their altitudes being 55,384 feet. 
 
 23. The Ganges River is 1800 miles shorter than the Ama- 
 zon, and the Orinoco is 300 miles shorter than the Ganges. 
 The sum of their lengths is 6900 miles. How long is each ? 
 
 24. A cubic foot of red oak weighs 35 pounds less than 2 
 cubic feet of cherry wood, and 21 pounds more than a cubic 
 foot of chestnut ; while a cubic foot of chestnut weighs 100 
 pounds less than 3 cubic feet of cherry. Find the weight of 
 each kind of wood per cubic foot. 
 
ADDITION AND SUBTRACTION 81 
 
 25. Lead weighs 259 pounds more per cubic foot than cast 
 iron, and 166 pounds more than bronze ; while a cubic foot of 
 bronze weighs 807 pounds less than 3 cubic feet of iron. Find 
 the weight per cubic foot of each metal. 
 
 26. Green glass weighs 60 pounds per cubic foot less than 
 dense flint glass, and 8 pounds more than crown glass ; while a 
 cubic foot of crown glass weighs 293 pounds less than 2 cubic 
 feet of flint glass. Find the weight per cubic foot of each. 
 
 27. Europe has 12 million inhabitants less than 10 times as 
 many as South America, and North America has 29 million 
 more than twice as many as South America. If 3 times the 
 population of North America be subtracted from that of Eu- 
 rope, the remainder is 65 million. How many inhabitants has 
 each continent ? 
 
 28. The length of the Rio Grande River is f that of the 
 Volga, and the Mississippi is 600 miles less than twice as long 
 as the Volga. If \ the length of the Mississippi be subtracted 
 from that of the Rio Grande, the remainder is 400 miles. 
 Find the length of each river. 
 
 29. In 1900 the total wealth of the United States was 1532 
 million dollars more than 13 times as great as in 1850, and 
 9016 million more than twice as great as in 1880. The total 
 wealth in 1880 was 174 million less than 6 times as great as 
 in 1850. What was the wealth in each of the three years ? 
 
 30. The money circulation of the United States in 1880 
 was 13 million dollars more than 60 times that in 1800 and 
 in 1905 it was 188 million more than 150 times that in 1800. 
 One-seventh of the amount in 1880 plus \ the amount in 1900 
 was 786 million. Find the circulation for each year. 
 
 31. The total bank deposits in the United States in 1905 
 were 3127 million dollars less than twice as great as in 1900 
 and 681 million more than 5 times as great as in 1880. The 
 deposits in 1880 were 5105 million less than in 1900. Find 
 the deposits for each of the three years. 
 
82 INVOLVED NUMBER EXPRESSIONS 
 
 32. The amount of deposits in savings banks in the United 
 States in 1905 was 703 million dollars greater than in 1900, 
 and 183 million less than 4 times that in 1880. The amount 
 in 1900 was 67 million less than 3 times as great as in 1880. 
 Find the deposits for each of the three years. 
 
 33. The total value of the farms in the United States in 
 1880 was 280 million dollars more than 3 times their value in 
 1850, and 8333 million less than their value in 1900. The 
 value of the farms in 1880 was 1924 million less than ^ their 
 value in 1900. Find the value in each of the three years. 
 
 34. The value of the manufactures in the United States in 
 1900 was 811 million dollars more than 12 times that in 1850, 
 and 3071 million less than 3 times that of 1880; while the 
 value in 1880 was 744 million less than 6 times that in 1850. 
 Find the value of the manufactures for each year. 
 
 MULTIPLICATION OF POLYNOMIALS 
 82. Illustrative Problem. A rectangular field is 12 rods 
 longer than it is wide ; a second rectangular field which is 4 
 rods shorter and 2 rods wider than the first has an area of 80 
 square rods less. What are dimensions of the first field ? 
 
 Solution. Let w = number of rods in the width of first field. 
 
 Then w + 12 = number of rods in the length of first field, 
 and w (to + 12) = width x length, or area of first field ; 
 also w + 2 = width of second field, 
 
 and w + 8 = length of second field. 
 
 Hence (w + 2) (w + 8) = width x length, or area of second field. 
 
 But the area of the second field is 80 square rods less than that of 
 the first. 
 
 Hence (w + 2) (w + 8) = w (w + 12) - 80. (1) 
 
 To solve this equation it is necessary to obtain the product 
 of the two binominals w -+- 2 and w + 8 without first combining 
 the terms of each binominal. In order to determine how these 
 
3-5 
 
 3-8 
 
 5-5 
 5 
 
 5-8 
 8 
 
 MULTIPLICATION OF POLYNOMIALS 83 
 
 are to be multiplied, let us consider two binomials in each of 
 which the terms can be combined if desired. 
 
 E.g. Consider the product of the binomials 5+3 and 5 + 8. This 
 may be represented by the area of a 
 rectangle 8 feet wide and 13 feet 
 long. That is, (5 + 3) (5 + 8) = 8 . 13 
 = 104 square feet. But such a 
 rectangle may be divided into four 
 rectangles, as in the figure. 
 
 Hence the area may be expressed 
 as the sum of four areas. Thus, 
 (5 + 3)(5 + 8) = 5 • 5 + 5 • 8 + 3 • 5 + 3 • 8 = 104, as before. 
 
 83. The product 5 • 5 is abbreviated to 5 2 , the small figure 
 indicating that 5 is to be used as a factor twice. It is read 
 the square of 5, or 5 squared, since it represents the area of a 
 square whose sides are each 5. 
 
 The second method here used for multiplying (5 +-3)(5 +8) 
 is applicable to any similar case, and does not depend upon the 
 possibility of first combining the terms of the binomials. 
 
 Applying this method to the binomials in equation (1) of the 
 problem in § 82, we have 
 
 w 2 + 8 to + 2 w + 16 = to 2 + 12 w - 80. (2) 
 
 Subtracting to 2 from both sides and applying Principle I, 
 
 10 w + 16 = 12 w - 80. (3) 
 
 Subtracting 10 to from both sides and adding 80 to both sides, 
 
 96 = 2 w. (4) 
 
 Hence by D, 48 = to, the width of the field ; 
 and 60 = w + 12, the length of the field. 
 
 Check by substituting to = 48 in equation (1), and also by showing 
 that the numbers 48 and 60 satisfy the conditions stated in the 
 problem. 
 
84 
 
 INVOLVED NUMBER EXPRESSIONS 
 
 84. In a manner similar to that just illustrated we may mul- 
 tiply two trinomials. 
 
 E.g. The product of a + b + c and m 4- n + r, in which the letters 
 represent any positive numbers, may represent the area of a rectangle, 
 divided into small rectangles as follows : 
 
 a 
 
 am 
 
 an 
 
 ar 
 
 b 
 
 bm 
 
 bn 
 
 br 
 
 c 
 
 cm 
 
 en 
 
 cr 
 
 Hence, the product is : 
 (a + b + c) (m 4- n + ?•) = am + bm + cm + an + bn + en + ar + br + cr, 
 in which each term of one trinomial is multiplied by every term of 
 the other, and the products are added. 
 
 Evidently the same process is applicable to the product of 
 two such polynomials each containing any number of terms. 
 
 EXERCISES AND PROBLEMS 
 
 Find each of the following products in two ways 
 
 1. (3 + 7 + 10)(2 + 6). 
 
 2. (5 + ll)(13 + 10 + 5). 
 
 3. (6-f-ll)(6 + 7). 
 
 4. (15 + 8)(15 + 4). 
 
 5. (7 + 13)(a + 6). 
 
 6. (ra + ft)(ll + 5+4). 
 
 7. 42 -36 = (40 + 2)(30 + 6). 
 
 8. 28- 73 = (20 + 8) (70 + 3). 
 
 Find as many as possible of the following indicated products 
 in two ways : 
 
 9. (a + b)(c + d). 
 
 10. (a; + 4)(x + 3). 
 
 11. ( x + y + z)(a + b + c). 
 
 12. (ll + 13)(r + f). 
 
 13. (5 + z)(5 + 7). 
 
 14. (3 + 8)(2 + 4 + 6). 
 
 15. (aj + 7)(3a? + 4). 
 
 16. (a + 4)(3a + l). 
 
 17. (3 + x)(2-\-5x). 
 
 18. (a + 6)(3a + 7&). 
 
 19. (x + y)(2x + 3y). 
 
 20. (7x + 4x)(x + 8). 
 
MULTIPLICATION OF POLYNOMIALS 85 
 
 21. A rectangle is 7 feet longer than it is wide. If its 
 length is increased by 3 feet and its width increased by 
 
 2 feet, its area is increased by 60 square feet. What are its 
 dimensions ? 
 
 22. A field is 10 rods longer than it is wide. If its length 
 is increased by 10 rods and its width increased by 5 rods, the 
 area is increased by 630 square rods. What are the dimen- 
 sions of the field ? 
 
 23. A farmer has a plan for a granary which is to be 
 12 feet longer than wide. He finds that if the length is 
 increased 8 feet and the width increased 2 feet, the floor 
 space will be increased by 160 square feet. What are the 
 dimensions ? 
 
 24. If the length of a rectangular flower bed is increased 
 
 3 feet and its width increased 1 foot, its area will be increased 
 by 19 square feet. What are its present dimensions, if its 
 length is 4 feet greater than its width ? 
 
 85. Polynomials with Negative Terms. The polynomials mul- 
 tiplied in the foregoing exercises contain positive terms only. 
 The same process is applicable to polynomials containing nega- 
 tive terms, as is seen in the following examples : 
 
 Ex. 1. Find the product of (7 - 4) and (3 + 5). This 
 product, written out term by term, would give 
 
 (7 + -4)(3 + 5) = 7- 3 + 7- 5 + -4- 3 + "4. 5 
 = 21 + 35-12-20 = 24. 
 Also (7-4)(3 + 5) = 3-8 = 24. 
 
 Ex. 2. Multiply 7 - 4 and 8 - 3. 
 
 (7 _ 4)(8 _ 3) = (7 + "4)(8 + ~3) = 7 • 8 + 7 • "3 + "4 • 8 + "4 . "3 
 
 = 56 - 21 - 32 + 12 = 15. 
 Also (7-4)(8-3)=3-5 = 15. 
 
86 INVOLVED NUMBER EXPRESSIONS 
 
 EXERCISES 
 
 Find each of the following products in two ways, as in the 
 above examples : 
 
 1. (ll-7)(6 + 5). 5. (2-3)(46-76). 
 
 2. (22 - 13) (3 + 7). 6. (3 + aj)(7*-3aj). 
 
 3. (8-5)(7-3). 7. (8-3)(8-2). 
 
 4. (17-9)(29-4). 8. (9-13)(9-17). 
 
 Perform the following indicated operations : 
 9. (a-6)(c + d). 13. (a — 6)(7a + 3 6). 
 
 10. (a — 6)(c — d). 14. (5 — y)(5 x + 3 y). 
 
 11. (a - 4) (a — 5). 15. (2 a — 3 b + c)(w + n). 
 
 12. (a + 6 — c)(m — n). 16. (y — t)(7v — 5t). 
 
 86. The preceding exercises illustrate the following principle : 
 
 Principle XIII. The product of two polynomials is found 
 by multiplying each term of one by every term of the other, 
 and adding these products. 
 
 In case some of the partial products are negative, these are 
 combined by Principle IX. 
 
 If there are similar terms in either polynomial, these should 
 usually be added first, thus putting each polynomial in as sim- 
 ple form as possible. 
 
 E.g. (3x + 2 - 2x)(4 x + 3 - 3x) = (x + 2)(x + 3) 
 
 = a; 2 + 2z + 3x + 6=a; 2 + 5ar + 6. 
 
 EXERCISES AND PROBLEMS 
 
 Perform the following indicated operations : 
 
 1. (a-7)(3a;-4 + 8). 
 
 2. (l-2a? + x-3){2x + ±a + lx). 
 
 3. (4a — x — 3a)(2x + 4:a + 7x). 
 
MULTIPLICATION OF POLYNOMIALS 87 
 
 4. (5x + 3y-4:X-2y)(6y + 3x-2y + y). 
 
 5. (13a-6-12o)(2&-3a). 
 
 6. (xy-5xy + 4:y)(Sy-3-7y). 
 
 7. (llao + 3a)(2&-3& + 5). 
 
 8. (6-4aj + 3*)(7aj + y — 3« + l). 
 
 9. (13a — 12x-y + 3)(5x-3y + xy + 5). 
 
 10. (37 — 13w + a)(a — w + 8). 
 
 11. (x-2+y)(4:y-3x). 
 
 12. (ll&-a-10&)(6a-3&-2a). 
 
 13. (7 + y-x)(2y + x-l). 
 
 14. (5as + 3y-l)(a;-2). 
 
 15. (—8a- 1 + 7 a) (5a -8 — 3a). 
 
 Solve the following equations, in each case verifying the 
 solution by substituting in the given equation the result found 
 for the unknown number. 
 
 16. (a? + 2) (a; + 3) = (x - 3) (x + 10) + 10. 
 
 17. (5a;-4)(6-aj)-97 = (a;-l)(6-5a;). 
 
 18. (3 n -1) (18 - ft) = (ft + 6) (16 - 3 n). 
 
 19. (7-a)(9a-8) = 31 + (36-9a)(a + 2). 
 
 20. (4 a + 4) (a - 3) = (4 a + 1) (a + 7) - 13 a + 221. 
 
 21. (n + 6)(3 n - 4) - 14 = (« + 8) (3 n - 3). 
 
 22. (8 n + 6) (10 - w) + 150 = (1 - w) (8 n + 3). 
 
 23. (a-l)(13-6a) = (6a-3)(8-a)-21. 
 
 24. (7 a; - 13)(6 - a;) - (a; + 4)(3- 7 a;) = 70. 
 
 25. A club makes an equal assessment on its members each 
 year to raise a certain fixed sum. One year each member 
 pays a number of dollars equal to the number of members of 
 the club less 175. The following year, when the club has 50 
 more members, each member pays $ 5 less than the preceding 
 year. What was the membership of the club the first year 
 and how much did each pay ? 
 
88 INVOLVED NUMBER EXPRESSIONS 
 
 26. There are two numbers whose difference is 6 and whose 
 product is 180 greater than the square of the smaller. What 
 are the numbers ? 
 
 27. There are four consecutive even integers such that the 
 product of the first and second is 40 less than the product of 
 the third and fourth. What are the numbers ? 
 
 28. There are four consecutive integers such that the prod- 
 uct of the first and third is 223 less than the product of the 
 second and fourth. What are the numbers ? 
 
 29. There are four numbers such that the second is 5 
 greater than the first, the third 5 greater than the second, and 
 the fourth 5 greater than the third. The product of the first 
 and second is 250 less than the product of the third and 
 fourth. What are the numbers ? 
 
 30. Prove that for any four consecutive integers the prod- 
 uct of the first and fourth is 2 less than the product of the 
 second and third. 
 
 SQUARES OF BINOMIALS 
 
 87. Just as x 2 is written instead of x • x, so (a + 6) 2 is written 
 instead of (a + b) (a + 6). The square of a binomial is found 
 by multiplying the binomial by itself as in § 86. 
 
 E.g. (a + b) 2 = (a + b)(a + b) = a 2 + ab + ab + b\ 
 
 Hence (a + 6) 2 = a 2 + 2 ab + b 2 . 
 
 This product is illustrated in the accompanying figure, and 
 is evidently a special case of the type exhibited in the figure, 
 page 83. 
 
 Translated into words this identity is : 
 The square of the sum of any two numbers is 
 equal to the square of the first plus twice the 
 product of the two numbers plus the square of 
 the second. 
 
 ba 
 
 b 3 
 
 a n - 
 
 ab 
 
SQUARES OF BINOMIALS 89 
 
 88. Similarly we obtain the square of the difference of two 
 numbers. («_*)» s «»-*«* + *• 
 
 Translate this identity into words. 
 
 While these squares are ordinary products of binomials and can 
 always be obtained according to Principle XIII, they are of special 
 importance and should be studied until they can be reproduced from 
 memory at any time. 
 
 EXERCISES AND PROBLEMS 
 
 Perform the following indicated operations : 
 
 1. (a + 6) 2 . 4. (a-2) 2 . 7. (4 + 9) 2 . 
 
 2. (17 -3) 2 . 5. (21 -bf. 8. (c-f) 2 . 
 
 3. (6 + a) 2 . 6. (z-7) 2 . 9. (x-\)\ 
 
 10. (r — s) 2 — (r-f s) 2 +(r — s)(r + s). 
 
 Check each of the above by substituting special values for 
 the letters and combining the terms of the binomial before 
 squaring. 
 
 11. Find the square of 42 by writing it as a binomial, 40 + 2. 
 
 12. Square the following numbers by writing each as a 
 binomial sum : 51, 53, 93, 91, 102, 202, 301. 
 
 13. Find the square of 29 by writing it as a binomial, 30 — 1. 
 
 14. Square the following numbers by first writing each as a 
 binomial difference : 28, 38, 89, 77, 99, 198, 499, 998, 999. 
 
 Solve the following equations, verifying each solution : 
 
 15. (a + 4) 2 + (a - 1) (2 a + 5) = (a + 4) (3 a + 2) . 
 
 16. (a - 1) (3 a - 1) - (a + 1) 2 = 2 a 2 - 18. 
 
 17. (6 - a) 2 -f (a - 3)(2 a - 5) = (3 a + 1) (a - 3) + 84. 
 
 18. (7 a -18)(a -f 4) - (a - 1) 2 = 6 (a + 1) 2 - 79. 
 
90 INVOLVED NUMBER EXPRESSIONS 
 
 19. (2&-30)(&-l)-5& 2 = 6 6-3(7> + 5) 2 + 65. 
 
 20. (5-&)(6& + 5)+4(6-3) 2 = 20-2(7> + l) 2 + 3 + 16&. 
 
 21. (5 - c) 2 + (7 - c) 2 + (9 - c) 2 =(c - 1)(3 c-58) - 93. 
 
 22. (5 c - 3) (2 + c) - 4 (c - l) 2 = (c + 1) 2 + 54. 
 
 23. (8 - 4 c) (5 - c) = (c + 1) 2 + (c + 3) (3 c - 8) + 218. 
 
 24. (4y-9)(y-5)-5(2/-4) 2 =(8-2/)(4 + 2/)-82. 
 
 25. (y + 6)"-3(y-l)»+(4-y)(5--2y) = 25y + 9. 
 
 26. (y-l)2 + 4(i/ + l) 2 + (l-2/)(5y + 6) = 15y-29. 
 
 27. a(a; + 3) + (a; + l)(a; + 2) = 2a;(a:-f-5)+2. 
 
 28. aj 2 =(*-3)(o: + 6)-12. 
 
 29. (5 + 5x)(3-x)+2(x + l) 2 +3(x + l)(x-7) = 17(x+l). 
 
 30. (8 + 3 *)(4 - x) + (a? - l)(aj - 2) + 2 (as + 5) 2 = 105. 
 
 31. There is a square field such that if its dimensions are 
 increased by 5 rods its area is increased 625 square rods. How 
 large is the field ? 
 
 Suggestion : If a side of the original field is w, then its area is to 2 , 
 and the area of the enlarged field is (w + 5) 2 . 
 
 32. A rectangle is 9 feet longer than it is wide. A square 
 whose side is 3 feet longer than the width of the rectangle is 
 equal to the rectangle in area. What are the dimensions of 
 the rectangle? 
 
 33. A boy has a certain number of pennies which he 
 attempts to arrange in a solid square. With a certain num- 
 ber on each side of the square he has 10 left over. Making 
 each side of the square 1 larger, he lacks 7 of completing it. 
 How many pennies has he ? 
 
 34. A room is 7 feet longer than it is wide. A square room 
 whose side is 3 feet greater than the width of the first room 
 is equal to it in area. What are the dimensions of the first 
 room ? 
 
SQUARES OF BINOMIALS 91 
 
 35. Find two consecutive integers whose squares differ 
 by 51 
 
 36. Find two consecutive integers whose squares differ 
 by 97. 
 
 37. Find two consecutive integers whose squares differ by a. 
 Show from the form of the equation obtained that a must be 
 an odd integer. 
 
 38. There are four consecutive integers such that the sum 
 of the squares of the last two exceeds the sum of the squares 
 of the first two by 20. What are the numbers ? 
 
 39. Two square pieces of land require together 360 rods of 
 fence ? If the difference in the area of the pieces is 900 square 
 rods, how large is each piece ? (Hint : x 2 — (90— a;) 2 = 900.) 
 
 40. There is a square such that if one side is increased by 
 12 feet and the other side decreased by 8 feet the resulting 
 rectangle will have the same area as the square. Find the 
 side of the square. 
 
 41. A regiment was drawn up in a solid square. After 50 
 men had been removed the officer attempted to draw up the 
 square by putting one man less on each side, when he found 
 he had 9 men left over. How many men in the regiment ? 
 
 42. There is a rectangle whose length exceeds its width 
 by 11 rods. A square whose side is 5 rods greater than the 
 width of the rectangle is equal to it in area. What are the 
 dimensions of the rectangle ? 
 
 89. Thus far the processes of algebra have all been based 
 upon thirteen fundamental principles, together with certain 
 obvious form changes indicated in § 37. These latter are of 
 general use in elementary arithmetic and need no special em- 
 phasis here. The principles, however, for the most part refer 
 to methods not common in arithmetic. These should now be 
 carefully reviewed and a list of them made for convenient 
 reference. 
 
92 INVOLVED NUMBER EXPRESSIONS 
 
 REVIEW QUESTIONS 
 
 1. How would 3 • 5 and 7 • 5 be added in arithmetic ? Why 
 cannot 3n and In be added in the same manner? State in 
 full the principle by which 3n and In are added. In this 
 example what number is represented by n ? Test the identity 
 3n + 7w=10nby substituting any convenient value for n. 
 
 2. How is 5 • 9 subtracted from 11 • 9 in arithmetic ? In 
 what different manner may this operation be performed? 
 Why is it sometimes necessary to perform subtraction in the 
 second way ? State in full the principle by which 12 x is sub- 
 tracted from 31 x. In the identity 31 x — 12 x = 19 x, what 
 number is represented by #? Test the equality by substituting 
 any convenient number for x. 
 
 3. How is the product 2-3-5 multiplied by 4 in arithmetic ? 
 In what different way may this multiplication be performed ? 
 Why should it ever be performed in the second way ? State 
 in full the principle by which 2 ax is multiplied by 3. 
 
 4. How is 11 + 3 multiplied by 4 in arithmetic ? In what 
 different way may this operation be performed? Why is it 
 sometimes necessary to multiply in the second way ? State in 
 full the principle by which a + 8 is multiplied by 7. 
 
 What principles are used in performing the following indi- 
 cated operations ? 
 
 ax + bx + cx, aby+cby, c(4a + 2&), 
 
 3y-by + cy, 41 a -32 b- 17 a + SOb. 
 
 5(16a-36), -3(6x-7y), a (11 -6). 
 
 5. Divide 2-4-6-20 by 2 without first performing the 
 multiplication indicated in 2-4-6-20. Do this in several 
 ways and show that all the quotients obtained are equal. 
 State in full the principle used. 
 
 What principles are used in the following operations ? 
 3 • 5 ab = 15 ab, c • 5 b = 5 (be), 20 ab + 4 = 5 ab, 
 
 16xy + x = l&y, 3* + 15* = 18<, 78^-41 A = 37 ft. 
 
REVIEW QUESTIONS 93 
 
 6. How is 12 + 18 divided by 6 in arithmetic ? In what 
 different way may this division be performed ? Why is it 
 sometimes necessary to perform division in the second way ? 
 State in full the principle used in performing the operation 
 (6 x + 9 y) -5- 3. What principle is used in performing each of 
 the following indicated operations ? 
 
 5(a + x), 3y — 4:y, (24 x + 9 y) -r- 3, 5x+ax. 
 
 7. Define equality ; equation ; identity. State in detail 
 how the equation and the identity differ. Give an example 
 of each. 
 
 8. In what ways may an equation be changed into another 
 equation such that any number which satisfies either also 
 satisfies the other ? 
 
 Describe some of the operations which change the form of 
 the members of an equation, but not their value. 
 State Principle VIII in full. 
 
 9. Name several pairs of opposite qualities all of which 
 are conveniently described by the words " positive " and " nega- 
 tive." What symbols are used to replace these words when 
 applied to numbers ? 
 
 10. When loss is added to profit, is the profit increased or 
 decreased ? What algebraic symbols are used to distinguish 
 the numbers representing profit and loss ? 
 
 11. Give an illustration by means of the number scale to 
 show that a number may represent either a change of position 
 in one direction or the other, or & fixed position with respect to 
 the zero point. 
 
 12. Why do we call positive and negative numbers signed 
 numbers ? What is meant by the absolute value of a number ? 
 
 13. State Principle IX in full. 
 
 14. By means of the number scale describe the "counting" 
 method of adding signed numbers 
 
94 INVOLVED NUMBER EXPRESSIONS 
 
 15. Make a list of pairs of opposite qualities to which posi- 
 tive and negative numbers apply. State in each case what is 
 represented by positive and what by negative numbers. 
 
 16. How is the correctness of subtraction tested in arith- 
 metic ? Is the same test applicable to subtraction in 
 algebra ? 
 
 17. Explain subtraction by counting on the number scale. 
 
 18. How do negative numbers make subtraction possible in 
 cases where it is impossible in arithmetic ? 
 
 19. What is a convenient rule for subtracting signed num- 
 bers ? State Principle X. 
 
 20. Give examples of equations which could not be solved 
 without negative numbers and show that such equations can 
 be solved by means of negative numbers. 
 
 21. Give an example in which positive and negative num- 
 bers are multiplied. State Principle XI. 
 
 22. Define division. How do we obtain the law of signs in 
 division ? State Principle XII. What is the test of the cor- 
 rectness of division ? 
 
 23. What may be the significance of a negative number 
 when obtained as a result of solving a problem ? 
 
 24. Explain how one set of signs + and — can be used to 
 indicate both quality and operation. 
 
 Show that a -f- ~b = a — b and a — + b = a — b. 
 
 25. What is a polynomial ? A term ? How are poly- 
 nomials classified? What are similar terms? By what prin- 
 ciple are similar terms added ? By what principle are they 
 subtracted ? In adding or subtracting polynomials, how may 
 the terms be arranged for convenience ? What is the prin- 
 ciple for removing a parenthesis when preceded by the sign + ? 
 By the sign — ? State Principle VII in full. 
 
REVIEW EXERCISES 95 
 
 26. Make a diagram to show how to multiply (7 + 4) by 
 (11 + 8) without first uniting the terms of the binomials. 
 Multiply (a + 6) by (c + cT) in the same manner. Multiply 
 (12 — 3) by (9 — 7) in two ways and compare results. State 
 the principle by which two polynomials are multiplied. 
 
 27. Explain why x 2 is called the square of x or x squared. 
 ■ State in words what is the square of the binomial (x + a) ; of 
 
 the binomial (x — a). 
 
 28. Show by an example how negative numbers may be used 
 in solving a problem, even though the answer to the problem 
 is positive and the statement of the problem does not involve 
 negative numbers. 
 
 REVIEW EXERCISES 
 
 In the following exercises perform the indicated operations, 
 remove all parentheses, solve all equations, verify the results, 
 and in each case state the principles used. 
 
 1. Add 3x + 4y — 3z, 5x — 2y — z, and 3y — 5 a; + 7 z. 
 
 2. From 15 a + 4 6 — 13 be subtract 3 a — 8 b + 2 be. 
 
 3. Subtract 7 x — 5y — 7 a from 6x + 5y + 3a. 
 
 4. (3x-4:y-z)(x + y + z). 5. (6 -5)(2 a 2 b -3 ab- b). 
 
 6. Add 11 axy + 13 x — 14 y, 2 y — 4 x, and 3 y + x — 8 axy. 
 
 7. (5 x - 3 b) + (2 x + b) - (4 x - 2 b - x + 5 6). 
 
 8. Add 19 b + 3 c, 2 b - 7 c, 2 c - 14 b, and c + 8 6. 
 
 9. -(a — 3 6— c) — (2 c— a- 5 6) + (a- c + 6). 
 10. Subtract 2 a; -f- 4 y + z from 13 a — 3y — 5z + 8. 
 
 "• 3 + 6 + 12 + T = " 
 
 12. 5(^-7) -3(14-z) +60= l-10ar. 
 
 13. 13 (!-»)- 6 (2 a; -5) =80 + 12 a:. 
 
96 INVOLVED NUMBER EXPRESSIONS 
 
 14 x-3 s-4 a?-8 a?-6 = 18 
 
 7 5 2^6 
 
 15. Add 7aj-3y-4, 5a; + 2y-f-5, and 3y-Sx-6. 
 
 16. Add 13 a + 4 b - 9 c, 2 c - 8 b - 16 a, and 8 a - 5 6 - 8 c. 
 
 17. (13-4-2)(5*-3 + 4aj). 
 
 18. 5(x-y) 2 -5(y-xf + (x-y)(x + y). 
 
 19. Square 73, 79, 92, 98, 1003, and 995 as binomials. 
 
 20. From 17 6 -4 a -2 c - 19 subtract 8 c-5 a — 8 6 + 4. 
 2i. 3 -(3 -2 + 6 + 8 -3) + 8- (9 -3 + 8). 
 
 22. 3(4-a)-2(5-6a;) = 8<c + 4. 
 
 Aft 0* - />• /y» iM 
 
 23. _4--4-___4-_ = — 2. 
 4 8 16 2 32 
 
 24 y-3 y + 9 = y-ll 2 
 * 4 10 4 
 
 25. ^i + ^±J + ^+8 = 2y-20. 
 
 3 3 3 * 
 
 26. 5a;-(3a;-2 + 2i/4-a;) + 13y-(6-3a; + 4). 
 
 27. From 3 — 4a — 5c + 8ar* subtract 2a? 2 — 2a — 4c + 8. 
 
 28. 12 + (2o-3c-46)-(36-c-a-8). 
 
 2 9. y + y±^+y±5 = 25. 
 
 30 . r-fciJ!+fc=»+«hI8.iBL 
 
 3 5 5 2 
 
 31. Add y - 20, 4 y + 6, 2 y + 4 x - 13, and 2 a - 8 y - 40. 
 
 32. (x - 1)(2 * - 2) + (a? - 5) 2 = (3 - a) (24 - 3 a) - 7. 
 
 33. Subtract 16 — a; + 2z — 4y from 3x — 5z — 8y. 
 
 34. 19 + (2 a; - 7) - (31 - 4 a? - 8 - 2 a?) = 5 x + 7. 
 
 35. (4 aft — 6ac — 5 ad)(b — c + a"). 
 
 36. (17 x + 8)(* - 1) + 8 = (2 - x)(6 - 17 aj) + 19. 
 
REVIEW EXERCISES 97 
 
 37. 5 - (a + b - c - d + 8) + (3 + a + c - d) - 5. 
 
 38. h- — ! = a — ^o. 
 
 10 10 10 
 
 39. Add 6 a + 9, 8 a - 13, 46 a - 8, and 6 - 54 a. 
 
 40. (a-2)(6a-4)+2(a-l) 2 =(6-a)(30-8a) + 4. 
 
 41. From 6 (a+ 2) + 3(c + 4) - 2 (6 - d) subtract 2 (a + 2) 
 -2(c + 4)+3(6-d). 
 
 a , a + 7 a — 3 _ a + 227 1 
 42 ' 3 + _ 4 3~~^ 1 ' 
 
 43. (a-2-3c-8 + 2 6)(6-a-c-& + 8). 
 
 44. ^ll + ^±l + lip3 =2 
 
 2 2 12 
 
 45. a 2 & - (3 6 - 8 a 2 - 7) + 3 a& 2 - (4 ab 2 + 8 - 2 a 2 ). 
 
 46. ^+l + ^Z^ + ^lI = 2a-26. 
 
 4 4 4 
 
 47. Add 12 a 2 6 2 c + 8 ax, 6 ax -8 a 2 b 2 c, and 2 ax + 3 a 2 6 2 c. 
 
 48. Add 5 a# 2 + 3 a?y -\- <ixy, 2 x 2 y — 6 xy 2 — 3 ay, and 4 xy. 
 
 49. Add 6ab — 3c— 2 a, 2c — 4 a& — 5 a, 5c— a + a&, and 
 3 + 5a-2c-3a6. 
 
 CA fl + l.« + 3.»-l w + 13 ,n-2 
 
 5U. — _ 1 . 1 . — — _ [_ 
 
 3 4 4 3 3 
 
 51. (n-4)(6-3n)-(6-w) 2 -10 = -4n(n-4). 
 
 52. From 35 ab - 8 x — 9 z + 13 subtract 16a6 - 4z + 5 x + 8. 
 
 53. (n + 2) 2 + (n-l) 2 +(n + l) 2 = 3n(?i + 2) + 60w + 130. 
 
 54. Subtract 5a — 8x— 6y from 13 x + 14 y — 15 z — 4 a. 
 
 55. From 9y — 4a: — 6 z — 3 & subtract 8— 9 y — 3# — 2 z. 
 
 56. 2 <e + 4 - 6 (5 x- 8 - 7 a) +2 - 4 a = 6 (2 -3 a;) -42. 
 
 57. _ (7 + 4a;-8-2a;)+4-2a; = 6a; + 25. 
 
98 INVOLVED NUMBER EXPRESSIONS 
 
 58. 16 + 5x-(8x + 9-4x + 17)=8x-3. 
 
 59. 6x-3-(4x+8-9x)-(5x-2)=x + ll. 
 
 60. (a — 1 + 6 — c — cf)(4a + 5& + 3c — 2d). 
 
 61. (4ax — 3a?/-f 5az — 8)(x + ?/ — z + 2). 
 
 62. (3ac-2ad + 4ed)(2 + 3 + 4 + 5). 
 
 63. 7-(3a-2&-4a)+6 + 2a-(3&-2a-a). 
 
 64. 8x + (5y-8) + (2y-l)-(13y + 8x-17). 
 
 65. 16 ax + 4 - (8 - 8 ax - a) - (12 ax - 13 - ax). 
 
 66. Add 15 ax 2 + 3 6c 2 , 2 6c 2 - 7 ax 2 , and 5 + 2 ox 2 - 5 6c 2 . 
 
 67. Addl6-7a6-2a 2 + 5a&, 4a 2 -2a6, and5a&-8. 
 
 68. Add 51 x 2 y - 35 + 12 a 2 , 41 - 17 a 2 - 57 x*y, and 3x*y. 
 
 69. Add 35 b 2 - 13 c 2 , 8 c 2 - 3 b 2 c 2 , and 6 & 2 - 8 c 2 - 9 6 2 c 2 . 
 
 70. Add 19-2x + 3x 2 6 + 26, 4 x 2 & + x + 5 b + 8, and 4x. 
 
 71. Subtract 2 a - 6 tfb-3x+2l from 19 - 2 * + 3 x 2 b - 7 a. 
 
 72. From 6 a - 45 -f 8 b - 3 c + 82 c& subtract 7 & -f 18 + 6 c. 
 
 73. (17 + 2a-3&-4&c)(2-a + &-c). 
 
 74. (13c-4d + 8e-3)(c-d). 
 
 75. (4xy-2y-3x-2)(y-x + y + 5). 
 
 76. (9 ax — 3 op - 5 a — 2 x + 4) (5 — x). 
 
 77. From 41 a 2 + 7 a& - 5 c 2 - 9 ab subtract 8 ab + 7 c 2 + 50 a 2 . 
 
 78. From 15 + 2x — 9xy — 3y subtract 5 xy — 4 a; — 2 ?/. 
 
 79. Subtract 12 + 8 x — 4 a —6 c — 18 abc from 5 x— 2 a + 3 c 
 
 80. 2-(71z + 42y-15x-64)-(5-91x-2y-13xy). 
 
 81. (31-2* + 3y-5)(aj + y). 
 
 82. 8x + (13 + 18x-6)-(5-6x) = 16x+10. 
 
REVIEW EXERCISES 99 
 
 83. (9z-3)(4-aO+(z-3) 2 = -8 (z + 2) 2 +94. 
 
 84. (x + l) 2 + (a; + 2) 2 + (a; + 3) 2 =(3a;-l)(a;+12)-43. 
 
 85. (2 x + 5) (a; -7)-(x- 1) 2 = (* + l)(x + 2) - 28. 
 
 86. 3 (o-xf- (2x - l)(a? - !)=(*- 7) (a; + 10) + 17 * + 50. 
 
 87. 5 - (7 - 4 x + 2 ^ - 4 b) - (8 a? - 6 y - 9 + 2 6) + 8 a. 
 
 88 . 8a>-(-3-2-4-7) + 5a; + (2 + 6 + 4)-(-3a; + 2). 
 
 89. 5y + 2a?-(6-4aj-5a;)-3y-(4a;-2y)-(-7y+8). 
 
 90. 35 y - (41 x - 16 - 12 y) + 5 a; + ( - 6 + 46 y - 18 a?) . 
 
 91. (.T-l) 2 -(a?-8)(2a;-l)=-a; 2 + 98. 
 
 92. (32 + a)(4a-l) + (5-a;) 2 + (a;-l) 2 = 6(a: + l) 2 + 194. 
 
 93. (2a?-7)(5-aj)-(2-5a;)(l - a;) = - a; (7 a? - 34) - 17. 
 
 94. (7 + x)(x-4) + (l-x) 2 =-23 + 2x 2 . 
 
 95. (12-4a)(2-a;)-4(l + :z) 2 = 5a; + 119. 
 
 96. (x - 17)(59 - 2 x) - (1 - a;) 2 = (6 - 3a;) (* - 2) + 384. 
 
 97. (3 x- 2) + (x - 1) 2 + (x - 2) 2 = 2 (a; - l)(x - 2) + 5. 
 
 98. (6-3 a>)(2 + x) + 16 (x - 1) 2 = 13 (x + 4) 2 + 364. 
 
 .. x+8 x- 9 . x- 17 4a-7 , 2a; + 6 . 5-31a; 
 
 12 6 2 3 12 
 
 100. 3j^_3jE±3 a^ = a L± _5 20. 
 
 6 3 2 6 3 
 
CHAPTER IV 
 THE SOLUTION OF PROBLEMS 
 
 90. Some of the advantages of algebra over arithmetic in 
 solving problems have been pointed out in the preceding chap- 
 ters. For instance, brevity and simplicity of statement secured 
 through the use of letters to represent numbers ; the transla- 
 tion of the words and sentences of problems into number ex- 
 pressions and equations ; and the clear and logical solution of 
 the equation, step by step. 
 
 91. A still greater advantage is set forth in the present 
 chapter ; namely, the opportunity offered by the symbols and 
 processes of algebra to summarize a whole class of problems 
 under one solution, called the formula, which is thereafter used 
 to solve all problems of the class. 
 
 PROBLEMS INVOLVING INTEREST 
 
 92. A class of problems already within the pupil's experi- 
 ence will illustrate this point. The different cases of percentage 
 or interest have been studied in arithmetic, and a large number 
 of isolated problems have been solved according to the rules. 
 In this instance, therefore, we proceed at once to summarize 
 all of this work in a few short statements. 
 
 T,etp = any principal, i.e. a number of dollars at interest. 
 i = the interest, i.e. the number of dollars accrued. 
 r = the rate, to be expressed in hundredths. 
 t = the time, to be expressed in years and fractions of a 
 year. 
 
 100 
 
PROBLEMS ON INTEREST 101 
 
 Then the rule of arithmetic for finding the interest when the 
 principal, rate, and time are given is 
 
 interest = principal x rate x time, 
 
 i.e. i=prt. (1) 
 
 If in this equation i = 150, r = .05, t = 6, find p. If i= 190.5, /> = 635, 
 r = .03, find t. If i = 665, p = 1000, t = 17, find r. Substitute other 
 values for any three of these letters and find the value of the remain- 
 ing letter. 
 
 Solve equation (1) for t in terms of i, p, and r ; also for p in 
 terms of i, r, and t, and for r in terms of p, t, and i. 
 
 It follows that if any three of the four numbers, p, r, i, and 
 t, are given, the remaining one may be found. 
 
 Let the student state four rules of interest by translating 
 these formulas into words. 
 
 Note the simplicity of these equations compared with the corre- 
 sponding rules in arithmetic. 
 
 Solve each of the following problems by substituting in the 
 proper formula : 
 
 1. What is the simple interest at 5 % on $400 for 5 years 
 and 9 months ? 
 
 Solution. i=p-r-t = 400 • T ^ • 5| = 4 • 5 • \ s = 115. 
 
 2. In what time will the simple interest on $750 at 3 % 
 
 amount to $225? Substitute in the formula t = — 
 
 pr 
 
 3. What is the semi-yearly income from an endowment of 
 $2,700,000, the rate being 4| of per annum ? (Here t = \.) 
 
 4. A father invested $1500 at 5^%, the simple interest on 
 which was to go to his eldest son on his 21st birthday. The 
 young man received $1240. How old was the son when the 
 investment was made ? 
 
102 THE SOLUTION OF PROBLEMS 
 
 5. What is the amount of money invested if it yields 
 $ 787.50 interest per annum, the rate being 5\%? (Here t = 1.) 
 
 6. A certain investment yields $8160 in 8 years. What is 
 the principal, if the rate of interest is 4 °fo ? 
 
 7. A $45,800 investment yielded $13,396.50 interest (sim- 
 ple) in 6\- years. What was the rate of interest ? 
 
 8. The endowment of a small college is $ 750,000, the yearly 
 income from which is $45,000. What is the average rate at 
 which the endowment is invested ? 
 
 9. A capitalist has investments amounting to $ 360,000, the 
 total income from which amounts to $ 1800 per month. What 
 is the average rate at which the money is invested ? (t = ^.) 
 
 10. If $ 700 is invested at 5±- % simple interest, what is the 
 amount at the end of 5 years 6 months ? This problem calls 
 for the amount, which is the sum of principal and interest. 
 
 If a = amount, then a= p +i = p + prt. 
 
 11. Solve the equation a—p+prt for p in terms of a, r, 
 and t, and translate the result into words. 
 
 12. Solve a = p +prt for r in terms of a, p, and t, and trans- 
 late the result into words. 
 
 13. Solve a =p +prt for t in terms of a, p, and r, and trans- 
 late the result into words. 
 
 14. Seven years ago I invested a certain sum of money at 6 % 
 simple interest. The amount at present is $5680. How much 
 did I invest ? (Use the formula obtained in Problem 11.) 
 
 15. Six years ago A invested $470 at simple interest. The 
 amount at present is $ 611. What is the rate per cent ? 
 
 16. Some years ago I invested $500 at 7 % simple interest, 
 which now amounts to $815. How many years ago was the 
 investment made ? 
 
 17. A merchant bought goods for $ 250, and some months 
 later sold the goods for $ 300, making a profit of 1 % per 
 month. How many months between the purchase and the sale ? 
 
PROBLEMS ON INTEREST 103 
 
 18. A real estate dealer sold a house and lot for $ 7500, 
 for which he received a commission of 4%. What was the 
 profit ? 
 
 In this case the element of time does not enter. The word 
 commission is then used for interest and the principal is called 
 the base. Letting c = commission, b = base, and r = rate, we 
 
 have 
 
 c = br. 
 
 Applying this formula, c=br = 7500 • y^ = 75 • 4 = 300. 
 
 19. Solve the equation c = br for b in terms of c and r, and 
 translate the result into words. 
 
 20. Solve c = br for r in terms of c and b, and translate the 
 result into words. 
 
 21. A merchant's commission for selling a carload of peaches 
 was $ 18.75. What was the rate of commission if the peaches 
 brought $ 375 ? (Use the formula found in 19.) 
 
 22. A broker received $25.50 commission for selling $850 
 worth of bonds. What was his rate of commission ? 
 
 23. An agent received 3 % commission for buying and 3| % 
 for selling some property. He paid $ 5750 for it and sold it 
 for $ 7200. What was his total commission ? 
 
 24. How much must I remit to my broker in order that he 
 may buy $ 600 worth of bonds for me and reserve 5 °J for his 
 commission ? 
 
 I must send him both base and commission. Calling this the 
 amount and representing it by a, we have 
 
 a = b + c = b + br = b (1 + r). 
 Hence a = 600 + 600 . fa = 600 + 6 • 5 = 630. 
 
 25. Solve the equation a=b + br for b and translate the 
 result into words. 
 
 26. Solve a = b + br for r and translate the result into 
 words. 
 
104 THE SOLUTION OF PROBLEMS 
 
 27. A merchant received $ 918 with which to buy corn after 
 deducting his commission of 2 °J on the price of corn. How 
 much was his commission and how much was used to buy 
 corn ? 
 
 Here a = 980. Find b by the formula under 25, which gives the 
 sum paid for corn. 
 
 28. A broker received $790, of which he invested $750 
 in stocks, reserving the balance as his commission. Find 
 the rate of his commission by means of the formula obtained 
 in 26. 
 
 29. An agent received $ 945 with which to buy lumber after 
 deducting his commission of 5 % on the cost of the lumber. 
 How much was his commission ? (First find the base by sub- 
 stituting in the formula of 25.) 
 
 30. A dealer sold berries for $ 18.95, and after deducting a 
 commission of 2 % sent the balance to the truck gardener. 
 How much did he remit ? 
 
 The sum he sent was the difference between the base and the com- 
 mission ; calling this d, we have 
 
 d=b-c = b-br = b(l-r). 
 
 Hence in this case d = 18.95 (1 — T fo) = 18.57. 
 
 31. Solve the equation d = 6(1 — r) for b in terms of d and r 
 and translate the result into words. 
 
 32. Solve the equation d = b — br for r in terms of b and d 
 and translate the result into words. 
 
 33. After deducting a commission of 3 % for selling bonds, 
 a broker forwarded $ 824.50. What was the selling price 
 of the bonds ? (Solve by means of the formula under 
 Problem 31.) 
 
 34. A broker sold stocks for $ 1728 and remitted $ 1693.44 
 to his principal. What was the rate of his commission ? 
 (Solve by means of the formula under Problem 32.) 
 
PROBLEMS ON AREAS 105 
 
 35. In how many years will $200 double itself at 5 % ? 
 In this case i =p. Hence, using formula (1), page 101, we have 
 
 200 = 200 x r § 5 x t. 
 Hence t = 20. 
 
 36. In how many years will any sum, p, double itself at any 
 rate, r? 
 
 Here p = prt. Hence, solving, t = 1 + r. 
 
 37. In what time will a sum of money double itself at 6% ? 
 7%? 4i%? 3|%? 
 
 38. Collect all the formulas worked out in this set of 
 problems. Translate each into words. Which were used 
 as the original ones from which the others were deduced ? 
 How many and which ones are needed in order to derive 
 all the others ? Why are such formulas better than rules 
 expressed in words ? 
 
 PROBLEMS INVOLVING AREAS 
 
 93. Another class of problems already well known to the 
 pupil in arithmetic concerns the areas of rectangles and tri- 
 angles. 
 
 If in a rectangle we let the number of units of length be de- 
 noted by I, the number of units of width by w, and the number 
 of square units in the area by a, then area = length x width; 
 
 i.e. a = Iw. (1) 
 
 If in equation (1) a = 144, I = 16, find w. If a = 1116, w = 31, find I. 
 Substitute other values for any two of these letters and find the value 
 of the remaining one. 
 
 Solve this equation for to in terms of I and a, and also for I in 
 terms of w and a. 
 
106 THE SOLUTION OF PROBLEMS 
 
 Also if b is the base of a triangle, h its altitude (height), and 
 a its area, then area = \ (base x altitude) ; 
 
 i.e. . .=**. (2) 
 
 Substitute particular values for any two of these letters and find 
 the value of the remaining one. 
 
 Solve (2) for h in terms of a and b, and also for b in terms 
 of a and h. 
 
 Let the student translate each of these equations into words. 
 Use these formulas in the solution of the following problems : 
 
 1. How many tiles each 3 by 4 inches are needed to cover 
 a rectangular floor 18 by 22 feet ? Use formula (1). 
 
 2. How long a space 25 feet in width can be covered by 
 340 square feet of roofing ? 
 
 3. The base and altitude of a triangle are 8 and 6 respect- 
 ively. Find its area. 
 
 4. The base of a triangle is 12 and its area 72. Find its 
 altitude. 
 
 5. The altitude of a triangle is 16 and its area 144. Find 
 its base. 
 
 6. A rectangle is 5 feet longer than it is wide. If it were 
 3 feet wider and 2 feet shorter, it would contain 15 square feet 
 more. Find the dimensions of the rectangle. 
 
 Let w equal the width; then construct number expressions for the 
 length, width, and area under the supposed conditions. 
 
 7. A rectangle is 6 feet longer and 4 feet narrower than a 
 square of equal area. Find the side of the square and the 
 sides of the rectangle. 
 
 8. The base of a triangle is 8 inches greater than its alti- 
 tude. If the base is increased by 4 inches and the altitude 
 decreased by 2 inches, the area remains unchanged. Find the 
 base and altitude of the triangle. 
 
PROBLEMS ON AREAS 107 
 
 9. A rectangle is 14 inches longer than it is wide. If the 
 width is increased by 5 inches and the length decreased by 4 
 inches, the area is increased by 70 square inches. Find the 
 dimensions of the rectangle. 
 
 10. A rectangle is 15 rods longer and 10 rods narrower than an 
 equivalent square. What are the dimensions of the rectangle? 
 
 11. The altitude of a triangle is 16 inches less than the base. 
 If the altitude is increased by 3 inches and the base by 2 inches, 
 the area is increased by 52 square inches. Find the base and 
 altitude of the triangle. 
 
 12. The width of a rectangular field is 20 rods less than its 
 length. If each side is decreased by 10 rods, the area is de- 
 creased by 900 square rods. What are the dimensions of the 
 field ? 
 
 13. A picture is 4 inches longer than it is wide. Another 
 picture, which is 12 inches longer and 6 inches narrower, con- 
 tains the same number of square inches. Find the dimensions 
 of the pictures. 
 
 14. A picture, not including the frame, is 8 inches longer 
 than it is wide. The area of the frame, 
 which is 2 inches wide, is 176 square inches. 
 Find the dimensions of the picture. 
 
 15. A picture, including the frame, is 
 10 inches longer than it is wide. The area 
 of the frame, which is 3 inches wide, is 
 192 square inches. What are the dimensions of the picture ? 
 
 16. The base of a triangle is 11 inches greater than its alti- 
 tude. If the altitude and base are both decreased 7 inches, the 
 area is decreased 119 square inches. Find the base and alti- 
 tude of the triangle. 
 
 17. The base of a triangle is 3 inches less than its altitude. 
 If the altitude and base are both increased by 5 inches, the 
 area is increased by 155 square inches. Find the base and alti- 
 tude of the triangle. 
 
 — * — ' — ■ — — - — —. 
 
 ■!.■ 
 
 x + 8 
 
 I Vol/,',/!,,/, ■/,,„,/,/,/,/, w,i„/>r/,W, 
 
 '■■..,■:,-) _^__/ Uiu. 
 
108 THE SOLUTION OF PROBLEMS 
 
 18. A commander in attempting to draw up his men in the 
 form of a solid square finds that there are 80 men more than 
 enough to complete the square. If he places 2 more men on 
 each side of the square he needs 84 more men to complete 
 it. How many men in his command ? 
 
 PROBLEMS INVOLVING VOLUMES 
 
 94. Still another class of problems for which the funda- 
 mental formulas are already well known concerns the volumes 
 of rectangular solids and of pyramids. 
 
 If the number of units of length, width, and height of a 
 rectangular solid be denoted by I, w, and h respectively, and the 
 number of units of volume by v, then 
 
 volume = length x width x height ; 
 
 i.e. v = lwh. (1) 
 
 If w is the width, I the length of the rectangular base of a 
 pyramid, h its altitude, and v its volume, then 
 
 volume = ^ (area of base x altitude) ; 
 
 Iwh /ox 
 
 i.e. " = ~3~' () 
 
 In equations (1) and (2) substitute particular values for any three 
 of the letters and find the value of the remaining one. 
 
 Use these formulas in solving the following problems : 
 
 1. Solve equations (1) and (2) for I, w, and h respectively 
 and translate each equation into words. 
 
 2. How many cubic feet of earth are removed in digging 
 a cellar 18 feet long, 12 feet wide, and 9 feet deep ? Solve by 
 substituting in formula (1). 
 
 3. If a cut in an embankment is 500 yards long and 4 yards 
 deep, how wide is it if 18,760 cubic yards are removed ? 
 
PBOBLEMS ON VOLUMES 109 
 
 4. How deep is a rectangular cistern which holds 500 cubic 
 feet of water if it is 6 feet wide and 8 feet long ? 
 
 5. The base of a pyramid is 16 inches long and 12 inches 
 wide. Its altitude is 30 inches. Find its volume. 
 
 6. A pyramid whose volume is 72 cubic feet has a base 
 whose area is 24 square feet. Find the altitude. 
 
 7. A pyramid whose volume is 91 cubic inches has an 
 altitude of 21 inches. Find the area of its base. 
 
 8. A rectangular room which is 10 feet high is 4 feet 
 longer than it is wide. If it were 5 feet longer and 2 feet 
 wider, it would contain 950 cubic feet more than it does. 
 Find its length and width. 
 
 9. A city building 50 feet high extends back 25 feet 
 more than its frontage. If the building were 8 feet wider 
 and extended 10 feet farther back, its capacity would be in- 
 creased 59,000 cubic feet. What are the ground dimensions 
 of the building ? 
 
 10. A pyramid whose altitude is 12 inches has a rectangu- 
 lar base 5 inches longer than it is wide. If the length of the 
 base is decreased by 1 inch and 
 
 the width increased by 2 inches, /\K 
 
 the volume of the pyramid is in- //' (, \ 
 
 creased by 72 cubic inches. Find / 1 I \ \ 
 
 the dimensions of the base of the / / I » \ 
 
 pyramid. / / I \ \ 
 
 11. The base of a rectangular / I j \ \ 
 pyramid whose altitude is 15 \* / *^~ s \ 
 
 inches is 12 inches longer than \/ x+5 ^-\ 
 
 it is wide. If the length and 
 
 width of the base are both decreased by 3 inches, the volume 
 is decreased by 675 cubic inches. Find the dimensions of the 
 base of the pyramid. 
 
110 THE SOLUTION OF PROBLEMS 
 
 12. The base of a rectangular pyramid is 15 inches wide. 
 The altitude of the pyramid is 4 inches less than the length 
 of the base. If the altitude is increased by 6 inches and the 
 length of the base by 3 inches the volume is increased by 1155 
 cubic inches. Find the altitude of the pyramid and the 
 length of its base. 
 
 13. The altitude of a pyramid is 7 inches less than the 
 length of the base. The width of the base is 13 inches. If 
 the altitude and the length of the base are both decreased by 
 3 inches, the volume is decreased by 364 cubic inches. Find 
 the altitude of the pyramid and the length of its base. 
 
 14. The width of the base of a pyramid is 2 inches greater 
 than its altitude. The length of the base is 34 inches. If 
 the altitude and the width of the base are both increased by 
 6 inches, the volume is increased by 2992 cubic inches. Find 
 the altitude and the width of the base. 
 
 PROBLEMS INVOLVING SIMPLE NUMBER RELATIONS 
 
 95. In the problems thus far considered in this chapter 
 the formulas are already known from previous study or ex- 
 perience. Algebra affords a means of deriving such formulas 
 in many cases where they are not already known. 
 
 1. The sum of two numbers is 35 and their difference is 5. 
 AY hat are the numbers ? 
 
 Let g = the greater number, then 35 — g = the lesser. 
 
 2. The sum of two numbers is 48 and their difference is 
 24. What are the numbers ? 
 
 3. The sum of two numbers is 41^ and their difference is 
 23^. What are the numbers ? 
 
 4. The sum of two numbers is 8590 and their difference 
 is 3480. What are the numbers ? 
 
PROBLEMS ON NUMBER RELATIONS 111 
 
 The four problems just preceding are similar in character. 
 A simple rule can be found for solving all problems of this 
 kind. Consider problem 1. 
 
 We have g — (35 — g) = 5. (1) 
 
 By VII, 2-35 + g = 5. (2) 
 
 By I, A, 2 g = 35 + 5. (3) 
 
 By D, VI, *=? + f' (4) 
 
 By F, g = 20, the greater number, (5) 
 
 and I = 35 — g = 15, the lesser number. (6) 
 
 The results in the final form tell nothing more than the 
 
 answers to this particular problem ; but the value of g in the 
 
 35 5 
 form g = — 4- -, if examined closely, tells much more. Stated 
 
 in full it means : 
 
 the sum of the numbers . their difference 
 the greater = h ^ 
 
 Examine now the solution of the other three problems to see 
 whether this same expression will give the greater number 
 in each case. 
 
 The great advantage in not adding 35 and 5 before dividing 
 
 35 5 
 
 by 2, is that the expression 1-- preserves the original num- 
 bers as given in the problem, so that we see how each enters 
 into the result. 
 
 If the sum of the two numbers is called s and their differ- 
 ence d, then we are compelled to keep these letters separate 
 to the end of the solution. 
 
 Thus, the lesser is 1 = s — g, 
 
 and g-(s-g)=d. 
 
 By VII, g - s + g = d . 
 
 By I, A, 2g = s + d. 
 
112 THE SOLUTION OF PROBLEMS 
 
 s d 
 Hence by D, VI, the greater is g = - + - , and the lesser is 
 
 l = S -g = S-(j + f) 
 
 s d _ 2 s _ s_d_ s _d 
 2~2~T 2 2~2 2* 
 
 Hence J =| + g, am/ / = | - J- 
 
 These results put into words are as follows : 
 
 77*e greater of any two numbers is half their sum plus half 
 their difference, and the lesser is half their sum minus half their 
 difference. 
 
 Any problem of this kind is solved by substituting in this 
 formula the particular values given to s and d and simplifying 
 the results. 
 
 E.g. In problem 4, page 110, s = 8590 and d = 3480. Hence the 
 greater number is §590 + §M0 = 4295 + 1740 = 6035, and the lesser 
 
 number is — - — = 4295 - 1740 = 2555. 
 2 2 
 
 5. Find by this formula two numbers whose sum is 17,540 
 and whose difference is 11,240. 
 
 6. Find two numbers whose sum is 40 and whose difference 
 is 52. Solve also without the formula. 
 
 Evidently one of these must be a negative number in order to 
 make the difference more than the sum, but the formula applies even 
 in such cases. 
 
 7. Find two numbers whose sum is 38 and whose difference 
 is 50. Solve also without the formula. 
 
 8. The sum of two numbers is 48, and one is 3 times the 
 other. Find the numbers. 
 
 9. The sum of two numbers is 108, and one is 6 times the 
 other. Find the numbers. 
 
PROBLEMS ON NUMBER RELATIONS 113 
 
 10. The sum of two numbers is s, and one is k times the 
 other. Find the numbers. 
 
 Let 
 
 n = one of the numbers, 
 
 Then 
 
 s — n — the other number, 
 
 and 
 
 kn = s — n. 
 
 By ,4, 
 
 kn 4- n = s. 
 
 By I, 
 
 (1 + 1) n = s. 
 
 By A 
 
 n — , one of the nu 
 
 L l 1 
 
 k + 1 
 
 and kn =k , the other number. 
 
 * + l 
 
 Problems 8 and 9 may be solved by substitution in this 
 formula. State this formula in words. 
 
 11. The sum of two numbers is 195, and one is 14 times the 
 other. Find the numbers. Solve also without the formula. 
 
 12. The sum of two numbers is 75, and one is f of the other. 
 Find the numbers. Solve also without the formula. 
 
 13. The sum of two numbers is — 52, and one is 12 times the 
 other. Find the numbers. Solve also without the formula. 
 
 14. If 9 be added to a number and the sum multiplied by 4, 
 the product equals 7 times the number. What is the number ? 
 
 15. If 21 be added to a number and the sum multiplied by 5, 
 the product equals 12 times the number. What is the number ? 
 
 16. If a be added to a number and the sum be multiplied by 
 b, the product is c times the number. What is the number ? 
 
 If n = the number. Show that 
 
 ab 
 c-b' 
 
 17. If 24 be added to a number and the sum multiplied by 3, 
 the product is 9 times the number. Find the number. Solve 
 by use of the formula of 16, and also without it. 
 
114 THE SOLUTION OF PROBLEMS 
 
 18. If 3 be added to a number and the sum multiplied by 
 16, the product is 10 times the number. Find the number by- 
 use of the formula, and also without it. 
 
 19. The sum of three numbers is 108. The second is 16 
 greater than the first, and the third 25 greater than the sec- 
 ond. What are the numbers ? 
 
 20. The sum of three numbers is 98. The second is 7 
 greater than the first, and the third is 9 greater than the sec- 
 ond. What are the numbers ? 
 
 21. The sum of three numbers is s. The second is a greater 
 than the first, and the third is b greater than the second. 
 What are the numbers? 
 
 . n l 
 
 If n = the first number, show that n= - — — - — -• 
 
 3 
 
 Translate the result of problem 21 into words. It should be real- 
 ized that if in problems 19 and 20 the numbers had been kept iincora- 
 bined, as they were necessarily in 21, those results would translate 
 into words, exactly as in 21. 
 
 Use the formula derived in 21 to solve the following, and 
 also solve each one without the formula. 
 
 22. The sum of three numbers is 198. The second is 28 
 larger than the first, and the third is 25 larger than the second. 
 What are the numbers ? 
 
 23. The sum of three numbers is 31. The second is 3 more 
 than the first, and the third 2 less than the second. What are 
 the numbers ? 
 
 Since the third number is 2 less than the second, this means that 
 b of the formula is negative ; i.e. b = — 2. 
 
 24. The sum of three numbers is 91. The second is 11 less 
 than the first, and the third is 12 more than the second. What 
 are the numbers ? 
 
 25. The sum of three numbers is 69. The second is 9 less 
 than the first, and the third is 6 less than the second. What 
 are the numbers ? (How does the formula apply in this case ?) 
 
PROBLEMS INVOLVING MOTION 115 
 
 26. Divide the number 248 into two parts, such that 7 times 
 the first is 42 more than twice the second. 
 
 27. Divide the number 645 into two parts, such that 13 times 
 the first part is 20 more than 6 times the other. 
 
 28. Divide the number a into two parts, such that b times 
 the first part is c more than d times the second part. 
 
 If x is the first part, show that x = fl • 
 
 b+d 
 
 29. Translate the formula in 28 into words. 
 
 30. Divide the number 1240 into two parts such that 5 times 
 the first is 200 more than the second. Solve by use of the 
 formula, and also without it. 
 
 PROBLEMS INVOLVING MOTION 
 
 96. Problems like those in the preceding class are useful 
 chiefly in cultivating skill in deducing formulas, and so mak- 
 ing rules. Those, however, in this and most of the following 
 classes are extremely important in themselves, because of the 
 simple laws of nature which they exemplify, and of which they 
 afford a wide range of application. 
 
 97. In scientific language the distance passed over by a 
 moving body is called the space, and the number of units of 
 space traversed is represented by s. The rate of motion, that 
 is the number of units of space traversed in the unit of time, is 
 called the velocity, and is represented by v. The number of 
 units of time occupied is represented by t. 
 
 E.g. At a certain temperature sound travels 1080 feet per second. 
 Hence, in 5 seconds it will travel 5 • 1080 = 5400 feet. In this case, 
 s = 5400 feet, v = 1080 feet per second, t = 5 seconds. 
 
 We then have the formula 
 
 * = Kf (1) 
 
116 THE SOLUTION OF PROBLEMS 
 
 Solve the equation s = vt for t in terms of s and v, and for v 
 in terms of s and t. 
 
 Translate each of these formulas into words. 
 
 In equation (1) give particular values to any two of the letters and 
 find the value of the remaining one. 
 
 It is to be understood in all problems here considered that the veloc- 
 ity remains the same throughout the period of motion ; e.g. sound 
 travels just as far in any one second as in any other second of its 
 passage. 
 
 1. If sound travels 1080 feet per second, how far does it 
 travel in 6 seconds ? 
 
 2. If a glacier moves 450 feet per year, how far does it 
 move in 7 years? 
 
 3. If a transcontinental train averages 35 miles per hour, 
 how far does it travel in 2£ days ? (Given v = 35, t = 2\ • 24, 
 to find s.) 
 
 4. A hound runs 23 yards per second and a hare 21 yards 
 per second. If the hound starts 79 yards behind the hare, 
 how long will it require to overtake the hare ? 
 
 If t is the number of seconds required, then by formula (1) during 
 this time the hound runs 23 1 yards and the hare runs 21 1 yards. Since 
 the hound must run 79 yards farther than the hare, we have : 
 23 t = 21 1 + 79. 
 
 5. An ocean liner making 21 knots an hour leaves port when 
 a freight boat making 8 knots an hour is already 1240 knots 
 out. In how long a time will the liner overtake the freight ? 
 
 6. A motor boat starts 7| miles behind a sailboat and runs 
 11 miles per hour while the sailboat makes 6^ miles per hour. 
 How long will it require the motor boat to overtake the sail- 
 boat? 
 
 7. A freight train running 25 miles an hour is 200 miles 
 ahead of an express train running 45 miles an hour. How long 
 before the express will overtake the freight ? 
 
PROBLEMS INVOLVING MOTION 117 
 
 8. A bicyclist averaging 12 miles an hour is 52 miles ahead 
 of an automobile running 20 miles an hour. How soon will the 
 automobile overtake him ? 
 
 9. A and B run a mile race. A runs 18 feet per second 
 and B Yl\ feet per second. B has a start of 30 yards. In 
 how many seconds will A overtake B ? Which will win the 
 race ? 
 
 10. Two objects, A and B, move in the same direction, A at 
 v x * feet per second and B at v 2 feet per second. If A has n feet 
 the start, in how many seconds will B overtake him ? 
 
 If t is the number of seconds, then during this time A moves v x t 
 feet and B moves v<£ feet. Since B must move n feet farther than A, 
 
 wehave v 2 t=v 1 t + n. (2) 
 
 The solution of (2) for t gives the time sought. 
 
 It is of the utmost importance that formulas (1) and (2) be 
 clearly understood since they are fundamental in every motion 
 problem in this chapter. 
 
 11. A fleet, making 11 knots per hour, is 1240 knots from 
 port when a cruiser, making 19 knots per hour, starts out to 
 overtake it. How long will it require ? 
 
 12. In how many minutes does the minute hand of a clock 
 gain 15 minute spaces on the hour hand ? 
 
 Using one minute space for the unit of distance and 1 minute as 
 the unit of time, the rates are 1 and T J 2 respectively, since the hour hand 
 goes fa of a minute space in 1 minute. Letting t be the number of 
 minutes required, we have, just as in problem 10, 
 1 • t = fa t + 15. 
 
 13. In how many minutes after 4 o'clock 
 will the hour and minute hands be together ? 
 (Here the minute hand must gain 20 minute 
 spaces.) 
 
 * Vi and v 2 , read v one and v two, are used instead of two different letters to 
 represent the velocities of the first and second respectively. 
 
118 THE SOLUTION OF PROBLEMS 
 
 14. At what time between 5 and 6 o'clock is the minute 
 hand 15 minute spaces behind the hour hand ? At what time 
 is it 15 minute spaces ahead ? 
 
 Since, at 4 o'clock, it is 25 minute spaces behind the hour hand, in 
 the first case it must gain 25 — 15 = 10 minute spaces, and in the 
 second case it must gain 25 + 15 = 40 minute spaces. Make a dia- 
 gram as in the preceding problem to show both cases. 
 
 15. At what time between 9 and 10 o'clock is the minute 
 hand of a clock 30 minute spaces behind the hour hand ? At 
 what time are they together ? 
 
 In each case, starting at 9 o'clock, how much has the minute hand 
 to gain ? 
 
 16. A fast freight leaves Chicago for New York at 8.30 a.m. 
 averaging 32 miles per hour. At 2.30 p.m. a limited express 
 leaves Chicago over the same road, averaging 55 miles per 
 hour. In how many hours will the express overtake the 
 freight? 
 
 If the express requires t hours to overtake the freight, the latter 
 had been on the way t + 6 hours. Then the distance covered by the 
 express is 55 /, and the distance covered by the freight is 32 (t + 6). 
 As these must be equal, we have 55 1 = 32 (t + 6). 
 
 17. In a century bicycle race one rider averages 19^ miles 
 per hour, while another, starting 40 minutes later, averages 
 22^ miles per hour. In how long a time will the latter over- 
 take the former ? 
 
 18. A sparrow flies 135 feet per second and a hawk 
 149 feet per second. The hawk in pursuing the sparrow 
 passes a certain point 7 seconds after the sparrow. In how 
 many seconds from this time does the hawk overtake the 
 sparrow ? 
 
 19. A courier starts from a certain point traveling v x miles 
 per hour, and a hours later a second courier starts, going at 
 
PROBLEMS INVOLVING MOTION 119 
 
 the rate of v 2 miles per hour. In how long a time will the 
 second overtake the first, supposing v 2 greater than Vi ? 
 
 If the second courier requires t hours to overtake the first the latter 
 had been on the way t + a hours. Thus the distance covered by the 
 second courier is v%t and by the first Vi(t + a). As these numbers are 
 equal we have rrf«r,(f-M) (3) 
 
 20. In an automobile race A drives his machine at an 
 average rate of 53 miles per hour, while B, who starts \ hour 
 later, averages 57 miles per hour. How long does it require B 
 to overtake A ? Use formula (3). Solve also by finding how 
 far A has gone when B starts and then use formula (2). 
 
 21. A freight steamer leaves New York for Liverpool aver- 
 aging 10^ knots per hour, and is followed 4 days later by an 
 ocean greyhound averaging 20^ knots per hour. In how long 
 a time will the latter overtake the former ? 
 
 22. One athlete makes a lap on an oval track in 26 seconds, 
 another in 28 seconds. If they start together in the same 
 direction, in how many seconds will the first gain one lap on 
 the other ? Two laps ? 
 
 Let one lap be the unit of distance. Since the first covers one lap 
 in 26 seconds his rate per second is fa. Likewise the rate of the other 
 is fa. If t is the required number of seconds the distance covered by 
 the first is ^g t and by the second fa t. If the first goes one lap farther 
 than the second the equation is -fat = fat + 1; if two laps farther it 
 is fa t = fa t + 2. 
 
 23. Two automobiles are racing on a circular track. One 
 makes the circuit in 31 minutes and the other in 38£ minutes. 
 In what time will the faster machine 
 
 gain 1 lap on the slower ? y^ ,> v. 
 
 24. The planet Mercury makes a cir- / jr >v \ 
 cuit around the sun in 3 months and / / sun^.^A.lvenua 
 Venus in 1\ months. Starting in con- \ V / / 
 junction, as in the figure, how long be- \ v ^ , ^ S / 
 fore they will again be in this position ? ^s. ^ ^ y^ 
 
120 THE SOLUTION OF PROBLEMS 
 
 25. Saturn goes around the sun in 29 years and Jupiter in 
 12 years. Starting in conjunction, how soon will they be in 
 conjunction again ? 
 
 26. Uranus makes the circuit of its orbit in 84 years and 
 Neptune in 164 years. If they start in conjunction, how long 
 before they will be in conjunction again? 
 
 27. The hour hand of a watch makes one revolution in 12 
 hours, and the minute hand in one hour. How long is it from the 
 time when the hands are together until the}' are again together ? 
 
 28. One object makes a complete circuit in a units of time 
 and another in b units (of the same kind). In how many units 
 of time will one overtake the other, supposing b to be greater 
 than a ? 
 
 29. At what times between 12 o'clock and 6 o'clock are the 
 hands of a watch together ? (Find the time required to gain 
 one circuit, two circuits, etc.) 
 
 PROBLEMS INVOLVING THE LEVER 
 
 98. Two boys, A and B, play at teeter. They find that the 
 
 teeter board will balance when equal products are obtained by 
 
 multiplying the weight of each by his distance from the point 
 
 of support. 
 
 Thus, if B weigh? 
 
 U-! -^ _ _ A ____ __ y wih) 80 pounds and is t 
 
 • « £5 " ~IJ5 ' feet from the poinfc 
 
 of support, then A, who weighs 100 lbs., must be 4 feet from this 
 point, since 80 x 5 = 100 x 4. 
 
 The teeter board is a certain kind of lever; the point of 
 support is called the fulcrum. 
 
 In each of the following problems make a diagram similar 
 to the above figure : 
 
 1. A and B weigh 90 and 105 lbs. respectively. If A is 
 seated 7 feet from the fulcrum, how far is B from this point ? 
 
PROBLEMS INVOLVING THE LEVER 121 
 
 2. Using the same weights as in the preceding problem, if 
 B is 6^ feet from the fulcrum, how far is A from that point ? 
 
 3. A and B are 5 and 7 feet respectively from the fulcrum. 
 If B weighs 75 pounds, how much does A weigh ? 
 
 4. A and B weigh 100 and 110 pounds respectively. A 
 places a stone on the board with him so that they balance 
 when B is 6 feet from the fulcrum and A 5^ feet from this 
 point. How heavy is the stone ? 
 
 5. If the distances from the boys to the fulcrum are respec- 
 tively d v and d 2 , and their weights w^ and w 2 , then 
 
 d l w l = d 2 w 2 . (1) 
 
 If any three of these four numbers are given, the fourth may be 
 found by means of this equation. Solve d x iv x = d 2 w 2 for each of the 
 four numbers involved in terms of the other three. Problems 1 to 4 
 can be solved by substitution in the formulas thus obtained. 
 
 6. A and B are seated at the opposite ends of a 13-foot teeter 
 board. Using the. weights of problem 1, where must the ful- 
 crum be located so that they shall balance ? 
 
 If the fulcrum is the distance d from A then it is (13 — d) from B. 
 Hence 90tf = 105 (13 -d). 
 
 7. A and B together weigh 212^ pounds. They balance 
 when A is 6 feet, and B 6f feet, from the fulcrum. Find the 
 weight of each. 
 
 8. A, who weighs 75 pounds, sits 7 feet from the fulcrum, and 
 B, who weighs 105 pounds, sits on the other side. At what dis- 
 tance from the fulcrum should B sit in order to make a balance ? 
 
 9. If in the preceding problem C, weighing 48 pounds, sits on 
 the side with A and 4 feet from the fulcrum, where must D, 
 who weighs 64 pounds, sit so as to maintain a balance? 
 
 From problem 8, when the teeter balanced it was found that A 'a 
 weight acted like a lever with a downward force of 7 • 75 pounds and 
 B's on the other side with a force of 5 • 105 pounds. 
 
 7 • 75 = 5 • 105. 
 
122 THE SOLUTION OF PROBLEMS 
 
 Then in problem 9, C and D must sit so as to keep the board in 
 balance, that is, so as to add the same downward force to both sides. 
 Hence, as 4-48 is added on the side of A, 3 • 64 must be added on 
 the side of B, since 4 • 48 = 3 • 04. Therefore, adding these two 
 equations, we still have the balance, namely, 
 
 7 • 75 + 4 • 48 = 5 • 105 + 3 • 64. 
 
 B (105 lbs.) D(Ulbs.) (7(18 lbs) (75 lbs.) A . 
 
 L | 3-feer A J-Tevr ' ] 
 
 6 feet. ^ m 7 feet 
 
 The weight of the boy multiplied by his distance from the 
 fulcrum is called his leverage. The sum of the leverages on 
 the two sides must be the same. Hence, if two boys, weighing 
 respectively w x and w 2 pounds, are sitting at distances d x and 
 d 2 on one side, and two boys, weighing w 3 and w t pounds, sitting 
 at distances d 3 , d t on the other side, then 
 
 w x d x + wJ, = Ms + »A- (2) 
 
 10. If two boys weighing 75 and 90 pounds sit at distances 
 of 3 and 5 feet respectively on one side and one weighing 82 
 pounds sits at 3 feet on the other side, where should a boy 
 weighing 100 pounds sit in order to make the board balance ? 
 
 11. A beam carries a weight of 240 pounds 1\ feet from the 
 fulcrum and a weight of 265 pounds at the opposite end which 
 is 10 feet from the fulcrum. On which side and how far from 
 the fulcrum should a weight of 170 pounds be placed so as to 
 make the beam balance ? 
 
 PROBLEMS INVOLVING DENSITIES 
 99. If a cubic foot of a certain kind of rock weighs 2.5 times 
 as much as a cubic foot of fresh water, the density of this rock 
 is said to be 2.5. If a cubic foot of oak weighs .85 times as 
 much as a cubic foot of fresh water, the density of the oak is 
 .85. The density of water is taken as the standard with which 
 the densities of other substances are compared. 
 
PROBLEMS INVOLVING DENSITIES 123 
 
 Thus, when we say that the density of a certain kind of iron is 
 7.25, we mean that any given volume of the iron weighs 7.25 times as 
 much as a like volume of fresh water ; and when we say that the 
 density of cork is .24, we mean that a given volume of cork weighs 
 .24 times as much as a like volume of fresh water. 
 
 A cubic centimeter of distilled water at the freezing point 
 which weighs one gram is used as a standard of comparison. 
 We therefore say that the density of any substance is equal to 
 the number of grams which a cubic centimeter of it weighs. 
 
 E.g. if a cubic centimeter (ccm.) of a certain kind of marble 
 weighs 2.5 grams, then it weighs 2.5 times as much as the same vol- 
 ume of water, and hence its density is 2.5. 
 
 The weight of an object in grams is, therefore, the product 
 of its volume in cubic centimeters multiplied by its density. 
 
 E.g. if the density of cork is .24, this means that a cubic centi- 
 meter of cork weighs .24 grams. Any volume of cork, say 10 ccm., 
 weighs 10 x .24 = 2.4 grams. Hence, if we represent the weight of an 
 object in grams by w, its volume in cubic centimeters by v, and its 
 density by d, we have the relation, 
 
 w = vd. (1) 
 
 1. What is the weight of an object whose density is 4.3 and 
 whose volume is 250 ccm. ? (Here v = 250, d =4.3. Find w.) 
 
 2. What is the density of an object whose weight is 23.5 
 and whose volume is 17 ccm. ? 
 
 3. What is the volume of an object whose weight is 24 
 grams and whose density is .65 ? 
 
 4. If 500 ccm. of alcohol, density .79, is mixed with 300 
 ccm. of distilled water, what is the density of the mixture ? 
 
 The volume of the mixture is the sum of the volumes, and the 
 weight of the mixture is the sum of the weights, of the water and 
 alcohol. Hence, from formula (1) : 
 
 500 x .79 + 300 x 1 = d x 800. To find d. 
 
124 THE SOLUTION OF PROBLEMS 
 
 5. If 1200 ccm. of cork, density .24, are combined with 64 
 ccm. of steel, density 7.8, what is the average density of the 
 combined mass ? Will it float or sink ? (A substance sinks 
 if its density is greater than that of water.) 
 
 In solving problems of this kind find the expressions representing 
 weight, density, and volume, and substitute in the equation w = dv. 
 
 6. How many cubic centimeters of cork, density .24, must 
 be combined with 75 ccm. of steel, density 7.8, in order that 
 the average density shall be equal to that of water, i.e. so that 
 the combined mass will just float ? 
 
 Let v = volume of cork to be used. Then the total volume is 
 75 + v, the total weight is 75 x 7.8 + .24 v, and the density is 1. 
 Hence, 75 x 7.8 + .24 v = 1 • (75 + r). Solve this equation for v. 
 
 7. Brass is an alloy of copper and zinc. How many cubic 
 centimeters of zinc, density 6.86, must be combined with 100 
 ccm. of copper, density 8.83, to form brass whose density is 
 8.31? 
 
 8. Coinage silver is an alloy of copper and silver. How 
 many ccm. of copper, density 8.83, must be added to 10 ccm. of 
 silver, density 10.57, to form coinage silver, whose density is 
 10.38 ? 
 
 9. The density of pure gold is 19.36 and of nickel 8.57. 
 How many ccm. of nickel must be mixed with 10 ccm. of pure 
 gold to form 14 karat gold whose density is 14.88. 
 
 10. How much mercury, density 13.6, must be added to 20 
 ccm. of gold, density 19.36, so that the density of the compound 
 shall be 16.9 ? 
 
 11. What is the average density of 40 ccm. of water, den- 
 sity 1, and 180 ccm. of alcohol, density .79 ? 
 
 12. How many cubic centimeters of water must be mixed 
 with 350 ccm. of alcohol, so that the density of the mixture 
 shall be .97? 
 
PROBLEMS ON MOMENTUM 125 
 
 13. The density of copper is 8.83. 500 ccm. of copper is 
 mixed with 700 ccm. of lead, whose density is 11.35. What is 
 the density of the combined mass ? 
 
 14. "When 960 ccm. of iron, density 7.3, is fastened to 8400 
 ccm. of white pine, the combination just floats, i.e. has a density 
 of 1. What is the density of white pine ? 
 
 PROBLEMS ON MOMENTUM 
 
 100. The force with which a moving body strikes another 
 depends both upon its weight and upon its rate of motion. 
 The product of the weight and velocity of a moving body is 
 called its momentum. The weight of a body is also commonly 
 called its mass. 
 
 What is the momentum of a body whose weight is 10 pounds and 
 which moves 15 feet per second ? 
 
 What is the velocity of a body whose weight is 50 and whose 
 momentum is 350 pounds ? 
 
 What is the weight of a body whose momentum is 500 and whose 
 velocity is 25 feet per second ? 
 
 A bullet weighing \ of a pound, moving 2250 feet per second, 
 has a momentum equal to that of a stone weighing 50 pounds, which 
 is hurled at the rate of 9 feet per second, since \ • 2250 = 50 • 9. 
 
 By careful experiment it has been found that when a mov- 
 ing body strikes a body at rest but free to move, the two will 
 move on with a combined momentum equal to the momen- 
 tum of the first body before the impact. 
 
 Thus, if a freight car, weighing 25 tons and moving at the rate of 
 12 miles per hour, strikes a standing car weighing 15 tons, the two will 
 move on with the original momentum of 12 • 25. But as the combined 
 weight is now 25 + 15, the rate of motion has been decreased to 7\ 
 miles per hour, since 12 • 25 = 1\ (25 + 15). In this case the automatic 
 coupler connects the cars and they move on together. Even if the 
 
126 THE SOLUTION OF PROBLEMS 
 
 two bodies after impact do not cling together, as two croquet balls, 
 still the momentum of the one plus that of the other equals the 
 original momentum. 
 
 E.g. if a croquet ball weighing 8 ounces and moving 20 feet per 
 second, strikes another weighing 7 ounces and starts it off at the 
 rate of 18 feet per second, then if the diminished velocity of the 
 first ball is called v, we have 
 
 8-20 = 7- 18 + 8 v, 
 and solving, v — 4.25. 
 
 This indicates that the first ball is nearly stopped, which coincides 
 with common observation. 
 
 1. In a switch yard a car weighing 40 tons and moving 8 
 miles per hour strikes a standing car weighing 24 tons. What 
 is the velocity of the two after impact ? 
 
 2. A billiard ball weighing 6 ounces and moving 16 feet 
 per second strikes another ball which it sends off at the rate of 
 10 feet per second. The rate of the first ball is reduced to 9 
 feet per second by the impact. What is the weight of the 
 second ball ? 
 
 Since the momentum before impact equals the sum of the mo- 
 mentums after impact, we have 6 • 16 = 9 • 6 + 10 w, w being the 
 unknown weight of the second ball. 
 
 3. A bowler uses a 16-ounce ball to take down the last pin. 
 The ball sends the pin off at a velocity of 6 feet per second, 
 the weight of the pin being 48 ounces, while the velocity of 
 the ball is reduced to 4 feet per second. With what velocity 
 did the ball strike the pin ? 
 
 Since the momentum before impact equals the sum of the mo- 
 mentums after impact, we have 16 1> = 6 -48 + 4- 16, v being the 
 unknown velocity of the ball before impact. 
 
 4. In each of these problems we have considered the weight 
 of two bodies, which we may call w x and w 2 . If we call v x the 
 velocity with which the first strikes the second, v 2 the velocity 
 
PROBLEMS ON MOMENTUM 127 
 
 imparted to the second, and v/ the resulting decreased velocity 
 of the first,* we have 
 
 ■W = Wjft' + w 2 v 2 (1) 
 
 Translate this equation into words. It contains 5 different num- 
 bers, toi, w 2 , vi, vi', and v 2 . If any four of these are given, the fifth 
 may be found by solving this equation for that one in terms of the 
 other four. 
 
 5. Solve the equation (1) for v x in terms of w 1} w^ v/, and 
 v 2 . Translate into words. 
 
 6. Solve the equation (1) above for Wy in terms of v i} v x ', w 2> 
 and v 2 . Translate into words. 
 
 7. Solve the equation (1) above for v x ' in terms of w v iv 2 , 
 v v and Vjj. Translate into words. 
 
 8. Solve equation (1) for w 2 in terms of w v v 1} V> v 2? an( i 
 translate the result into words. 
 
 9. Solve equation (1) for v 2 in terms of w 1} w 2 , v lt v^, and 
 translate the result into words. 
 
 10. In the result of the last exercise substitute w 1 = 50, 
 w 2 = 40, v x = 10, Vy = 2, and find the value of v 2 . Make a 
 problem to fit this case. 
 
 11. In the result of problem 8 substitute w x = 1000, 
 V!=75, Vi = 25, v 2 = 50, and find the value of w 2 . Make a 
 problem to fit this case. 
 
 12. In the result of problem 6 substitute v x = 60, V = 10, 
 w 2 = 80, v 2 = 25, and find the value of w x . Make a problem 
 to fit this case. 
 
 13. In the result of problem 7 substitute v\ = 250, w 2 = 125, 
 Vy = 50, and v 2 = 50, and find the value of v x \ Make a prob- 
 lem to fit this case. 
 
 * v-l is read v one prime and is used rather than a new letter for the 
 purpose of recalling that this is another rate of the first body. 
 
128 
 
 THE SOLUTION OF PROBLEMS 
 
 PROBLEMS ON THERMOMETER READINGS 
 
 101. There are two kinds of thermometers in use in this 
 country, called the Fahrenheit and Centigrade, the former for 
 common purposes, and the latter for scientific 
 records and investigations. Hence it frequently 
 becomes necessary to translate readings from 
 one kind to the other. 
 
 The freezing and boiling points are two 
 fixed temperatures by means of which the com- 
 putations are made. On the Centigrade these 
 are marked 0° and 100° respectively and on the 
 Fahrenheit they are marked 32° and 212° respec- 
 tively. See the cut. Hence between the two 
 fixed points there are 100 degrees Centigrade 
 and 180 degrees Fahrenheit. 
 
 That is, 100 degree spaces on the Centigrade cor- 
 respond to 180 degree spaces on the Fahrenheit. 
 
 Hence 1° Centigrade corresponds to |° Fahrenheit, 
 or 1° Fahrenheit corresponds to |° Centigrade. 
 
 All problems comparing the two thermometers are solved by 
 reference to these fundamental relations. 
 
 1. If the temperature falls 15 degrees Centigrade, how 
 many degrees Fahrenheit does it fall ? 
 
 2. If the temperature rises 18 degrees Fahrenheit, how 
 many degrees Centigrade does it rise ? 
 
 3. Translate + 25° Centigrade into Fahrenheit reading. 
 
 25° Centigrade equals | • 25° = 45° Fahrenheit. 
 45° above the freezing point = 45° + 32° above 0° Fahrenheit. 
 Hence, calling the Fahrenheit reading F, we have F = 32 + f • 25. 
 
 4. Translate + 14° Centigrade into Fahrenheit reading. 
 Eeasoning as before, F = 32 + § • 14. 
 
PROBLEMS ON THERMOMETER READINGS 129 
 
 From the two preceding problems we have the formula 
 
 F = 32 + |C. (1) 
 
 Translate this into words, understanding that F and C stand 
 for readings on the respective thermometers. 
 
 5. Solve the above equation for C in terms of F and find 
 
 C = f(F-32). (2) 
 
 Translate this into words. Verify the solution of each of 
 the following by referring to the cut on page 128. 
 
 6. Translate -f 41° Fahrenheit into Centigrade reading. 
 Substitute in the proper formula. 
 
 7. Translate + 98° Fahrenheit, blood heat, into Centigrade 
 reading. 
 
 8. Translate — 20° Fahrenheit into Centigrade reading. 
 
 9. Translate — 40° Centigrade, the freezing point of mer- 
 cury, into Fahrenheit reading. 
 
 10. Translate 0° Fahrenheit into Centigrade by use of the 
 formula. Also 0° Centigrade into Fahrenheit. 
 
 11. Translate -f- 212° Fahrenheit into Centigrade, and also 
 + 100° Centigrade into Fahrenheit. 
 
 12. What is the temperature Centigrade when the sum of 
 the Centigrade and Fahrenheit readings is 102° ? 
 
 13. What is the temperature Fahrenheit when the sum of 
 the Centigrade and Fahrenheit readings is zero ? 
 
 14. What is the temperature Centigrade when the sum of 
 the Centigrade and Fahrenheit readings is 140°? 
 
 15. What is the temperature in each reading when the 
 Fahrenheit is 50° higher than the Centigrade ? 
 
 16. The Fahrenheit reading at the temperature of liquid 
 air is 128 degrees lower than the Centigrade reading. Find 
 both the Centigrade and the Fahrenheit reading at this 
 temperature. 
 
130 THE SOLUTION OF PROBLEMS 
 
 PROBLEMS ON THE ARRANGEMENT AND VALUE OF DIGITS 
 
 102. If we speak of the number whose 3 digits, in order 
 from left to right, are 5, 3, and 8, we mean 538 = 500 -f 30 + 8. 
 Likewise, the number whose three digits are h, t, and u is writ- 
 ten 100 h + 10 t + u. 
 
 Hence, when letters stand for the digits of numbers written 
 in the decimal notation, care must be taken to multiply 
 each letter by 10, 100, 1000, etc., according to the position it 
 occupies. 
 
 Illustrative Problem. A number is composed of two digits 
 whose sum is 6. If the order of the digits is reversed, we 
 obtain a number which is 18 greater than the first number. 
 What is the number ? 
 
 Solution. Let x = the digit in tens' place. 
 
 Then 6 — x = the digit in units' place. 
 
 Hence, the number is 10 x + 6 — x. Reversing the order of the 
 digits, we have as the new number 10(6 — x) + x. 
 
 Hence 10(6 - x) + x = 18 + 10 x + 6 - x. 
 
 1. A number is composed of two digits, the digit in units' 
 place being 2 greater than the digit in tens' place. If 4 is 
 added to the number, it is then equal to 5 times the sum of 
 the digits. What is the number? 
 
 2. A number is composed of two digits, the digit in tens' 
 place being 3 greater than the digit in units' place. The num- 
 ber is one more than 8 times the sum of the digits. What 
 is the number ? 
 
 3. A number is composed of two digits whose sum is 9. 
 If the order of the digits is reversed, we obtain a number 
 which is equal to 12 times the remainder when the units' digit 
 is taken from the tens' digit. What is the number ? 
 
REVIEW QUESTIONS 131 
 
 4. A number is composed of two digits whose difference 
 is 4. If the order of the digits is reversed, we obtain a num- 
 ber which is 3 less than 4 times the sum of the digits. What 
 is the number ? 
 
 5. The digit in tens' place is 1 more than twice the digit 
 in units' place. If 36 is subtracted from the number, the order 
 of the digits will be reversed. What is the number? 
 
 6. The digit in units' place is 2 less than twice the digit in 
 tens' place. If the order of the digits is reversed, the number 
 is unchanged. What is the number ? 
 
 7. The digit in tens' place is 12 less than 5 times the digit 
 in units' place. If the order of the digits is reversed, the num- 
 ber is equal to 4 times the sum of the digits. What is the 
 number ? 
 
 8. A number is composed of three digits. The digit in 
 units' place is 3 greater than the digit in tens' place, which in 
 turn is 2 greater than the digit in hundreds' place. The num- 
 ber is equal to 96 plus 4 times the sum of the digits. What is 
 the number ? 
 
 REVIEW QUESTIONS 
 
 1. In order that a problem may be solved by means of a 
 formula, how many of the letters in the formula must be given 
 by the problem ? Illustrate this by the formula i =prt. 
 
 2. State the formulas involving areas and volumes which 
 have been used in this chapter. Solve each formula for each 
 of its letters in terms of all the others. 
 
 3. The area of a circle is found by squaring its radius and 
 multiplying by 3.1416. State this rule as a formula, using the 
 Greek letter tt for 3.1416. 
 
 4. The volume of a circular column is found by multi- 
 plying the area of its base by its height. State this rule as 
 a formula, using r for the radius of the base and h for the 
 height. 
 
132 THE SOLUTION OF PROBLEMS 
 
 5. State formulas for finding two numbers when their sum 
 and difference are given. 
 
 Find a number such that if 1G be subtracted from it, 5 times 
 this difference equals the number. 
 
 Construct other problems like this and then make a formula 
 for the solution of all such problems. (See page 110.) 
 
 6. State three important formulas used in the solution of 
 motion problems. Translate each into words. Solve each for- 
 mula for each of its letters. 
 
 Of the motion problems, 4 to 29, which ones involve formula 
 (1) ? Which formula (2) and which formula (3) ? 
 
 7. State the formulas used in this chapter in working 
 problems on the lever. Solve each formula for each of 
 its letters. 
 
 Using a teeter board 12 feet long with the fulcrum in the 
 middle, how may three boys weighing 50, 75, and 100 lbs. re- 
 spectively be seated on it so as to make the board balance ? Is 
 there more than one solution ? Make a diagram for each of 
 your results. 
 
 8. What is meant by the density of a substance ? What is 
 used as a standard (unit) of density with which the densities of 
 other substances are compared ? 
 
 What is the relation between the weight of a substance, its 
 volume and its density ? 
 
 9. Define momentum. When a moving car strikes a stand- 
 ing car, free to move, what can you say of the momentum of 
 the two after impact? 
 
 10. Describe the Fahrenheit and Centigrade thermometers. 
 State the formulas for the reading of each thermometer in terms 
 of the other. 
 
 11. Write 347 as a trinomial. Write the number whose 
 three digits in order are a, b, c ; also the number whose three 
 digits in order are c, b, a. 
 
REVIEW PROBLEMS 133 
 
 REVIEW PROBLEMS 
 
 103. Most of the following problems can be solved by sub- 
 stitution in some of the formulas developed in this chapter. 
 Solve as many as possible in this way. 
 
 1. One boy runs around a circular track in 26 seconds, and 
 another in 30 seconds. In how many seconds will they again 
 be together, if they start at the same time and place and run 
 in the same direction ? 
 
 2. Divide the number 144 into two parts, so that ^ of the 
 greater is 9 more than \ of the smaller. 
 
 3. The sum of two numbers is 2890. Seven times one 
 is 266 less than 5 times the other. What are the numbers ? 
 
 4. What is the simple interest on $ 400 at 6| % for 7 years 
 and 9 months ? 
 
 5. The number of telegraph messages sent in the United 
 States in 1905 was 5 million less than three times as great as 
 in 1880, and 69 million less than twice that in 1900 ; while in 
 1900 it was 16 million more than twice as great as in 1880. 
 Find the number of messages in each of these years. 
 
 6. The same number is added to each of the numbers 8, 9, 
 10, 12. What is this number if the product of the first and last 
 sums is equal to the product of the second and third sums ? 
 
 7. Find the time between 4 and 5 o'clock when the hands 
 of the clock are 30 minute spaces apart. 
 
 8. A man buys a house for $6500. His yearly tax on the 
 property is $57. The coal costs $60 per year, repairs $50, 
 and janitor service $108. To what monthly rental are his 
 expenses equivalent if money is worth 5 % ? 
 
 9. A boatman rowing down a river makes 23 miles in 3 
 hours and returns at the rate of 3^ miles per hour. How swift 
 does the river flow ? 
 
134 THE SOLUTION OF PROBLEMS 
 
 10. A bird flying with the wind goes 65 miles per hour, and 
 flying against a wind twice as strong it goes 20 miles per hour. 
 What is the rate of the wind in each case ? 
 
 11. A steamer going with the tide makes 19 miles per hour, 
 and going against a current \ as strong it makes 13 miles per 
 hour. What is the speed of the steamer in still water ? 
 
 12. A boatman trying to row up a river drifts back at the 
 rate of 1^ miles per hour, while he can row down the river at 
 the rate of 12 miles per hour. What is the rate of the current? 
 
 13. A boatman rowing with the tide makes n miles per 
 hour ; rowing against a tide k times as strong, he makes m miles 
 per hour. At what rate does he row, and what is the velocity 
 of the stream ? State the result as a formula. 
 
 14. A beam is 16 feet long. • At what point must it be sup- 
 ported if it is to carry, when balanced, 460 pounds at one end 
 and 690 at the other, its weight being disregarded ? 
 
 15. Two numbers differ by 2 and the difference of their 
 squares is 100. What are the numbers ? 
 
 16. A beam is 12 feet long. It carries a 40-pound weight 
 at one end, a 60-pound weight 3 feet from this end, and a 
 70-pound weight at the other end. Where is the fulcrum if 
 the beam is balanced ? 
 
 17. There is a number composed of 2 digits whose tens' digit 
 is 2 less than its units' digit. The number is 1 less than 5 times 
 the sum of its digits. What is the number ? 
 
 18. A farm containing 240 acres can be rented at $3 per 
 acre. The renter finds that if he borrows money at 5 °J to 
 buy the farm he will save $125 per year. Find its value. 
 
 19. Two trains start from the same point at the same time 
 and in the same direction, one making 25 miles per hour and the 
 other 42 miles per hour. When will they be 238 miles apart ? 
 
REVIEW PROBLEMS 135 
 
 20. The number of national banks in the United States on 
 March 1, 1906, was 1322 less than twice as many as on 
 March 1, 1900, and the number in 1903 was 1009 greater than 
 in 1900. If the number in 1906 be subtracted from 3 times 
 the number in 1903 the remainder is 7736. Find the number 
 in each of the three years mentioned. 
 
 21. The earth and Mars were in conjunction July 12, 1907. 
 When are they next in conjunction if the earth's period is 
 365 days and that of Mars 687 days ? (See figure, page 119.) 
 
 22. $ 7500 is invested at 4 % simple interest. Seven years 
 and three months later the amount is used to build a house. 
 What is the cost of the house if $ 735 has to be added to com- 
 plete it ? 
 
 23. There is a number composed of two digits whose sum 
 is 12. If the order of the digits is reversed, the number is 
 increased by 18. Find the number. 
 
 24. A slow steamer sails from New York to Liverpool, mak- 
 ing 9 knots per hour. A swift liner follows 62 hours later, 
 making 20| knots per hour. In how many hours will the latter 
 overtake the former? 
 
 25. A man invested a certain sum of money at 5 % simple 
 interest. The amount 3| years later was $ 950. What was the 
 investment ? 
 
 26. The capital stock of the Bank of France is 35.6 million 
 dollars less than that of the Bank of England and 6.3 million 
 greater than that of the Imperial Bank of Germany. The 
 combined capital stock of the three banks is 134.9 million 
 dollars. Find the capital stock of each. 
 
 27. The capital stock of the Bank of Italy is 27.7 million 
 dollars less than twice that of the Imperial Bank of Kussia, 
 and the capital of the Bank of Austria-Hungary is 13.6 million 
 greater than that of the Bank of Russia. Their combined 
 capital is 99.1 million dollars. Find the capital of each bank. 
 
136 THE SOLUTION OF PROBLEMS 
 
 28. In a bicycle race A starts 32 minutes ahead of B. 
 B rides at the rate of 20^ miles per hour, while A rides 18| 
 miles per hour. How many miles from the starting point 
 does B overtake A? 
 
 29. A squadron of warships sails 13 knots per hour. A 
 torpedo boat making 27f knots leaves port 19 hours later to 
 overtake the squadron. In how many hours after leaving port 
 will the torpedo boat overtake it ? 
 
 30. A man takes out a life insurance policy for which he 
 pays in a single payment. Thirteen years later he dies and the 
 company pays $ 12,600 to his estate. It was found that his 
 investment yielded 2 °J simple interest. How much did he 
 pay for the policy ? 
 
 31. A merchant bought goods for $600 and some months 
 later sold them for $648, making a profit of 2% per month. 
 How many months elapsed between the purchase and the 
 sale? 
 
 32. In a building there are at work 18 carpenters, 7 plumbers, 
 13 plasterers, and 6 hod carriers. Each plasterer gets $ 1.90 
 per day more than the hod carriers, the carpenters get 35 
 cents per day more than the plasterers, and the plumbers 
 50 cents per day more than the carpenters. If one day's 
 wages of all the men amount to $183.45, how much does each 
 get per day ? 
 
 33. A train running 46 miles per hour leaves Chicago for 
 New York at 7 a.m. Another train running 56 miles per hour 
 leaves at 9.30 a.m. Find when the trains will be 15 miles 
 apart. (Two answers.) 
 
 34. Divide the number 280 into two parts so that \ of one 
 part is 48 less than \ of the other. 
 
 35. A merchant bought goods and sold them 5 months later 
 for $ 2687.50, making a gain on his investment of 1\ % per 
 month. How much did he pay for the goods ? 
 
REVIEW PROBLEMS 137 
 
 36. A bicyclist starts out riding 12 miles per hour, and is 
 followed 40 minutes later by another riding 16 miles per hour. 
 Find when they will be 5 miles apart. (Two answers.) 
 
 37. A father invested $ 1000 at 6^ % interest. When the 
 principal and simple interest amounted to $ 2235, it was given 
 to the son for the expenses of his college education. How 
 long had the money been invested ? 
 
 38. Two trains start west at the same time, one from New 
 York and the other from Philadelphia. If the New York train 
 runs 55 miles per hour and the Philadelphia train 47 miles per 
 hour, how long before they are 15 miles apart, the distance 
 from New York to Philadelphia being 90 miles? 
 
 39. A man bought a tract of coal land and sold it a month 
 later for $ 93,840. If his gain was at the rate of 24 % per 
 annum, what did he pay for the land ? 
 
 40. There is a rectangle whose length is 60 feet more, 
 and whose width is 20 feet less, than the side of a square 
 of equal area. Find the dimensions of the square and the 
 rectangle. 
 
 41. A number is composed of two digits whose sum is 14. 
 If the digits are interchanged, the number is decreased by 18. 
 What is the number ? 
 
 42. What time between 7 and 8 o'clock are the hour and the 
 minute hands of a clock together ? 
 
 43. Change 104 degrees Fahrenheit (fever heat) to Centi- 
 grade reading. 
 
 44. A steamer leaves Liverpool for New York Saturday, 
 9 a.m., averaging 18 knots per hour. Seven hours later an- 
 other steamer leaves New York for Liverpool, making 20£ 
 knots per hour. What time (Liverpool time) will the steamers 
 meet if the trans-Atlantic distance by their course is 2940 
 knots ? 
 
CHAPTER V 
 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 
 104. Graphic Representation of Statistics. 
 
 A graphic representation of the temperatures recorded by the 
 U. S. Weather Bureau at Chicago on December 6 and 7, 1906, 
 is shown on the next page. The readings were as follows : 
 
 3 P.M. 
 
 29° 
 
 9 P.M. 
 
 21° 
 
 3 A.M. 
 
 12° 
 
 9 A.M. 12° 
 
 4 P.M. 
 
 29° 
 
 10 P.M. 
 
 20° 
 
 4 A.M. 
 
 11° 
 
 10 A.M. 13° 
 
 5 P.M. 
 
 28° 
 
 11 P.M. 
 
 17° 
 
 5 A.M. 
 
 10° 
 
 11 A.M. 16° 
 
 6 P.M. 
 
 26° 
 
 12 m't. 
 
 16° 
 
 6 A.M. 
 
 10° 
 
 12 Noon 17° 
 
 7 P.M. 
 
 24° 
 
 1 A.M. 
 
 14° 
 
 7 A.M. 
 
 10° 
 
 1 P.M. 18° 
 
 8 P.M. 
 
 22° 
 
 2 A.M. 
 
 12° 
 
 8 A.M. 
 
 10° 
 
 2 p.m. 20° 
 
 In the graph each heavy dot represents the temperature at a cer- 
 tain hour. The distance of a dot to the right of the heavy vertical 
 line indicates the hour of the day counted from noon December 6th, 
 and its distance above the heavy horizontal line indicates the ther- 
 mometer reading at that hour. The lines joining these dots complete 
 the picture representing the gradual changes of temperature from 
 hour to hour. 
 
 Graphs of this kind are used in commercial houses to represent 
 variations of sales, fluctuations of prices, etc. They are used by 
 architects and engineers to show the comparative strength of mate- 
 rials, stresses to which they are subjected, and to exhibit multitudes 
 of other data. They are used by the historian to represent changes 
 in population, fluctuations in mineral productions, etc. In algebra 
 they are used in solving many practical problems and in helping to 
 understand many difficult processes. In the succeeding exercises the 
 
 cross-ruled paper is essential. 
 
 138 
 
GRAPHING OF STATISTICS 
 
 139 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 + 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 + 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 + 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 a, 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 3 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 p 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ± 
 
 Ifi 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 2 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 a 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 + 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 s 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 f 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 IS 
 
 M. 
 
 
 I : 
 
 ..M 
 
 
 
 
 
 
 
 
 12 
 
 
 3 A 
 
 ,H 
 
 
 
 
 
 
 
 
 1 
 
 ; 
 
 P.M. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 u Timi 
 
 Line 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 EXERCISES 
 
 Make a graphic representation of each of the following 
 tables of data:* 
 
 1. The population of the United States as given by the 
 census reports from 1790 to 1900: 
 
 1790 . . 3.9 (million) 1830 
 
 1800 . . 4.3 1840 
 
 1810 . . 7.2 1850 
 
 1820 . . 9.6 1860 
 
 12.9 
 
 1870 . 
 
 . 38.6 
 
 17.1 
 
 1880 . 
 
 . 50.2 
 
 23.2 
 
 1890 . 
 
 . 62.6 
 
 31.4 
 
 1900 . 
 
 . 76.3 
 
 *In each case the number to be represented by one space on the cross-ruled 
 paper should be chosen so as to make the graph go conveniently on a sheet. 
 Thus in (1) let one small horizontal space represent two years and one vertical 
 
140 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 
 2. The population of New York city since 1800 : 
 
 1790 . . 33 (thousand) 1830 . . 202 1870 . . 942 
 
 1800 . . 60 1840 . . 312 1880 . . 1206 
 
 1810 . . 96 1850 . . 515 1890 . . 1530 
 
 1820 . . 123 1860 . . 813 1900 . . 1850 
 
 3. The population of Chicago since 1850 : 
 
 1850 . 
 
 . 30 (thousand) 
 
 1870 . 
 
 . 306 
 
 1890 
 
 . . 1100 
 
 1860 . 
 
 . 109 
 
 
 1880 . 
 
 503 
 
 1900 
 
 . . 1698 
 
 4. The world's 
 
 yearly production of gold since 1872 : 
 
 1872 . 
 
 99.6 
 
 (million) 
 
 1883 . 
 
 102.4 
 
 1894 
 
 . . 181.5 
 
 1873 . 
 
 96.2 
 
 
 1884 . 
 
 101.7 
 
 1895 
 
 . . 194.0 
 
 1874 . 
 
 90.7 
 
 
 1885 . 
 
 108.4 
 
 1896 
 
 . . 202.3 
 
 1875 . 
 
 97.5 
 
 
 1886 . 
 
 106.0 
 
 1897 
 
 . . 236.1 
 
 1876 . 
 
 103.7 
 
 
 1887 . 
 
 105.8 
 
 1898 
 
 . . 286.9 
 
 1877 . 
 
 114.0 
 
 
 1888 . 
 
 110.2 
 
 1899 
 
 . . 306.7 
 
 1878 . 
 
 119.0 
 
 
 1889 . 
 
 123.5 
 
 1900 
 
 . . 254.6 
 
 1879 . 
 
 109.0 
 
 
 1890 . 
 
 118.9 
 
 1901 
 
 . . 262.5 
 
 1880 . 
 
 106.5 
 
 
 1891 . 
 
 130.7 
 
 1902 
 
 . 296.0 
 
 1881 . 
 
 103.0 
 
 
 1892 . 
 
 146.3 
 
 1903 
 
 . 325.5 
 
 1882 . 
 
 102.0 
 
 
 1893 . 
 
 157.2 
 
 1904 
 
 . 346.8 
 
 5. Thei 
 
 world's 
 
 yearly productic 
 
 n of silver 
 
 since 
 
 1872: 
 
 1872 . 
 
 65.0 
 
 (million) 
 
 1883 . . 
 
 115.3 
 
 1894 
 
 . 214.5 
 
 1873 . 
 
 81.8 
 
 
 1884 . 
 
 105.5 
 
 1895 
 
 . 208.0 
 
 1874 . 
 
 71.5 
 
 
 1885 . 
 
 118.5 
 
 1896 
 
 . 203.0 
 
 1875 . 
 
 80.5 
 
 
 1886 . 
 
 120.6 
 
 1897 
 
 . 207.0 
 
 1876 . 
 
 87.6 
 
 
 1887 . 
 
 124.3 
 
 1898 
 
 . 218.6 
 
 1877 . 
 
 81.0 
 
 
 1888 . 
 
 140.7 
 
 1899 
 
 . 217.6 
 
 1878 . 
 
 95.0 
 
 
 1889 . 
 
 155.4 
 
 1900 
 
 . 224.0 
 
 1879 . 
 
 96.0 
 
 
 1890 . 
 
 163.0 
 
 1901 
 
 . 223.7 
 
 1880 . 
 
 96.7 
 
 
 1891 . 
 
 177.0 
 
 1902 
 
 . 208.6 
 
 1881 . 
 
 102.0 
 
 
 1892 . 
 
 197.7 
 
 1903 
 
 . 220.4 
 
 1882 . 
 
 111.8 
 
 
 1893 . 
 
 209.1 
 
 1904 
 
 . 217.7 
 
 space a million of population ; and in (3) let one horizontal space represent 
 one year and one large vertical space one hundred thousand of population. 
 
GRAPHING OF STATISTICS 
 
 141 
 
 6. Temperature, at Port Conger, New York, and Singapore. 
 The average temperature for each half month is given. 
 
 
 Poet 
 
 New 
 
 Singa- 
 
 
 
 Port 
 
 New 
 
 Singa- 
 
 Month 
 
 Conger 
 
 YOKK 
 
 pore 
 
 Month 
 
 Conger 
 
 York 
 
 pore 
 
 Jan. 1-15 
 
 -35° 
 
 30° 
 
 80° 
 
 July 
 
 1-15 
 
 + 32° 
 
 66° 
 
 86° 
 
 16-31 
 
 -40° 
 
 28° 
 
 82° 
 
 
 16-31 
 
 + 37° 
 
 68° 
 
 86° 
 
 Feb. 1-15 
 
 -45° 
 
 32° 
 
 84° 
 
 Aug. 
 
 1-15 
 
 + 37° 
 
 68° 
 
 87° 
 
 16-28 
 
 -40° 
 
 35° 
 
 84° 
 
 
 16-31 
 
 + 34° 
 
 66° 
 
 86° 
 
 Mar. 1-15 
 
 -35° 
 
 38° 
 
 85° 
 
 Sept. 
 
 1-15 
 
 + 27° 
 
 62° 
 
 85° 
 
 16-31 
 
 -30° 
 
 40° 
 
 85° 
 
 
 16-30 
 
 + 20° 
 
 60° 
 
 85° 
 
 April 1-15 
 
 -20° 
 
 45° 
 
 85° 
 
 Oct. 
 
 1-15 
 
 + 8° 
 
 55° 
 
 85° 
 
 16-30 
 
 -10° 
 
 48° 
 
 85° 
 
 
 16-31 
 
 - 2° 
 
 50° 
 
 84° 
 
 May 1-15 
 
 0° 
 
 50° 
 
 85° 
 
 Nov. 
 
 1-15 
 
 -15° 
 
 48° 
 
 83° 
 
 16-31 
 
 + 8° 
 
 56° 
 
 86° 
 
 
 16-30 
 
 -20° 
 
 45° 
 
 82° 
 
 June 1-15 
 
 + 16° 
 
 60° 
 
 86° 
 
 Dec. 
 
 1-15 
 
 -28° 
 
 42° 
 
 82° 
 
 16-30 
 
 + 20° 
 
 65° 
 
 86° 
 
 
 16-31 
 
 -32° 
 
 40° 
 
 81° 
 
 When the temperature is below zero, the distance is measured 
 downward from the heavy horizontal line, as in the figure, page 149. 
 
 7. Average monthly rainfall at San Francisco, Valparaiso, 
 Chili, and Quebec : 
 
 
 (a) San Francisco 
 
 (b) Valparaiso 
 
 (c) Qttebec 
 
 January . 
 February . 
 March . . 
 
 . 5.5 (inches) 
 . 4.5 
 . 3.5 
 
 0.2 
 0.3 
 
 1 
 
 (inches) 
 
 3.2 (inches) 
 
 6.4 
 
 4.4 
 
 April . . 
 May . . 
 June . . 
 
 . 2.5 
 . 1.5 
 . 0.5 
 
 2 
 3 
 4.2 
 
 
 6.6 
 5.1 
 2.5 
 
 July . . 
 Aiagust . . 
 September 
 October 
 
 . 0.2 
 . 0.2 
 . 0.2 
 . 1.5 
 
 2.9 
 1.8 
 1.8 
 0.5 
 
 
 1.4 
 
 1.8 
 1.8 
 0.5 
 
 November . 
 
 . 3.5 
 
 0.4 
 
 
 0.4 
 
 December . 
 
 . 5.5 
 
 0.2 
 
 
 0.2 
 
 Use 10 small vertical spaces for 1 inch of rainfall and five horizon- 
 tal spaces for 1 month. 
 
142 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 
 8. After making the graphs in exercise 6, each on a separate 
 sheet, put all three on the same sheet, using a different colored 
 ink or pencil for each. In this way the relative average tem- 
 peratures are simultaneously pictured. 
 
 9. After graphing (a), (6), and (c) of example 7, each on a 
 separate sheet, combine all three, using different colors, as 
 in 8. 
 
 10. Observe the weather reports in a daily paper and make 
 a graph representing the hourly change of temperature for 
 twenty-four hours. 
 
 GRAPHIC REPRESENTATION OF MOTION 
 
 105. A useful picture of the distance traversed by a moving 
 body can be made by a graph similar to the preceding. 
 
 E.g. suppose a man is walking 3 miles per hour. We mark units 
 of time from the starting point to the right along the horizontal ref- 
 erence line, and indicate miles traveled by the number of units meas- 
 ured vertically upward from this line. (See the figure on the oppo- 
 site page.) 
 
 In the figure each horizontal space represents 1 hour, and each ver- 
 tical space 3 miles. Then in 1 hour he goes 3 miles; in 5 hours, 
 15 miles ; in 10 hours, 30 miles ; etc. The dots representing the 
 distances are found to lie on a straight line. 
 
 The graph shows at a glance the answers to such questions as : 
 How many miles does he travel in 4 hours? in 13 hours? How long 
 does it take him to go 18 miles? 23 miles? 
 
 Again, suppose 24 hours later a second man starts out on a bicycle 
 to overtake the first man, and travels 9 miles an hour. The line 
 drawn from the 24-hour point shows the distance the wheelman trav- 
 els in any number of hours counting from his time of starting. The 
 points marked in this line show how far he has gone in 1, 2, 3, 4, 5, 6 
 hours, etc., namely, 9, 18, 27, 36, 45, 54, etc. 
 
 The point where these two lines intersect shows in how many 
 hours after starting the pedestrian is overtaken and also how far he 
 has gone. 
 
GRAPHIC SOLUTION OF PROBLEMS 
 
 143 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 i-<o ^ r <? ■ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 i£ jZ J 
 
 -,** 
 
 .? j 
 
 
 
 
 
 
 
 
 In like manner solve the following by means of graphs, and 
 in each case suggest other questions which may be answered 
 from the graph : 
 
 1. In a century bicycle race A averages 17 miles per hour; 
 B, who starts 20 minutes later, averages 19 miles per hour. In 
 how many hours will B overtake A ? Who will win the race 
 and where will the loser be when the winner finishes ? 
 
 2. A starts for a town 12 miles distant, walking 3 miles per 
 hour, li hours later B starts for the same place, driving 1\ 
 miles per hour. When does B overtake A ? Where is A when 
 B reaches town ? 
 
144 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 
 3. In a mile race A runs 6 yards per second and B 5 yards 
 per second. B has a start of 250 yards. Who will win 
 the race ? How far in the lead is the winner at the finish ? 
 
 Let 1 vertical small space represent 20 yards, and 1 horizontal space 
 represent 5 seconds. 
 
 106. Illustrative Problem. A man rides a bicycle into the 
 country at the rate of 8 miles per hour. After riding a certain 
 distance the wheel breaks down and he walks back at the rate 
 of 3 miles per hour. How far does he go if he reaches home 
 11 hours after starting ? 
 
 \J 
 
 
 
 
 
 
 
 
 
 .•■>a &._ _ >__ __ 3^ K _ _ _ __ _ 
 
 3- G -f -Sw 
 
 
 
 
 
 :**:£ — zztzz *% ~ ~ 
 
 
 
 \ j f **». 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 In this graph each large horizontal space represents 1 hour, and 
 each small vertical space represents 1 mile. The problem is solved 
 as follows : 
 
 (1) Construct the line representing the outward journey at the 
 rate of 8 miles per hour, extending this line indefinitely. 
 
 (2) Beginning at the point corresponding to 11 hours, find the 
 points representing his position at each pi-eceeding hour. The line 
 connecting these points represents the homeward journey at the rate 
 of 3 miles an hour. Extend this line until it meets the first line. 
 The point where the lines meet represents 3 hours and 24 miles, 
 which is the answer required in the problem. 
 
GRAPHIC SOLUTION OF PROBLEMS 145 
 
 PROBLEMS 
 
 Solve the following problems by means of graphs. In each 
 case prepare a list of questions which may be answered from 
 the graph. 
 
 1. A man rows 18 miles per hour down a river and 2 miles 
 per hour returning. How far down the river can he go if he 
 wishes to return in 10 hours ? 
 
 2. A man goes from Chicago to Milwaukee on a train run- 
 ning 42^ miles per hour, and returns immediately on a steamer 
 going 17 miles per hour. Find the distance, if the round trip 
 requires 7 hours. 
 
 3. A pleasure trip from New York to Atlanta by steamer 
 and return by rail occupied 77 hours. Find the distance, if 
 the rate going was 16 miles per hour and returning 40 miles 
 per hour. 
 
 Let one small horizontal space represent one hour and one small 
 vertical space 16 miles. 
 
 4. A invests $1000 at 5% and B invests $5000 at 4%. 
 In how many years will the amount (principal and interest) of 
 A's investment equal the interest on B's investment? 
 
 Let one large horizontal space represent one year and one small 
 vertical space $50. Then the line representing ^4's amount starts at 
 the point marked $1000, and rises one small vertical space each year. 
 The line representing B'a interest starts at the zero point and rises 
 four small vertical spaces each year. 
 
 5. In how many years will the interest on $ 6000 equal the 
 amount on $ 2000 if both are invested at 5 % ? 
 
 6. A invests $500 at 6% and B invests $ 1000 at 5%. In 
 how many years will A's interest differ by $300 from .B's ? 
 
 Let one small vertical space represent $20. In this case both lines 
 start from the zero point. Find the point on one line which is three 
 large spaces vertically above the corresponding point on the other 
 line. 
 
146 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 
 7. Construct a graph representing the relation between the 
 Fahrenheit and Centigrade thermometer readings. 
 
 Let the hue at the bottom of the sheet and the vertical line four 
 large spaces from the left margin be the reference lines. Let one 
 small horizontal space represent a degree C, and one smal] vertical 
 space a degree F. 
 
 From F = 32 + | C (page 129), we find that if C = 0° F = 32°, if 
 C = 10° F=50°, if C = 20° F = 68°, if C = 30° F = 86°, etc. Mark 
 the point representing each pair of readings, and draw the straight 
 line connecting these points. This is the required graph. 
 
 8. From this graph read the answers to the following 
 questions : 
 
 Find C when F= 41°, F = 59°, F= 79°, F = 14°. 
 Find F when C= 35°, C = 40°, C= - 5°, C= - 15°. 
 
 From the graph it is possible to find any Fahrenheit reading when 
 the corresponding Centigrade is given, and also to find any Centigrade 
 reading when the corresponding Fahrenheit is given. Thus the graph 
 shows to the eye all the information contained in the equation F = 32 
 + 1 C. In what follows we shall consider in detail how to represent 
 equations in this manner. 
 
 GRAPHIC REPRESENTATION OF EQUATIONS 
 107. In all the graphs thus far constructed two lines at right 
 angles to each other have been used as reference lines. These 
 lines are called axes. The location of a point in the plane of 
 such a pair of axes is completely described by giving its distance 
 and direction from each of the axes. The direction to the 
 right of the vertical axis is denoted by a positive sign, and to 
 the left, by a negative sign j while direction upward from the 
 horizontal axis is positive, and downward, negative. 
 
 The horizontal line is usually called the jr-axis and the ver- 
 tical line the /-axis. The perpendicular distance of any point 
 P from the ?/-axis is called the abscissa of the point, and its dis- 
 tance from the x-axis is called its ordinate. The abscissa and 
 ordinate of a point are together called its coordinates. 
 
GRAPHS REPRESENTING EQUATIONS 
 
 147 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 + 
 
 . 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 > 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 =r> 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 <W 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 7J 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 P 
 
 ( 
 
 8^ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 M 
 
 + 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 < 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 •f 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 _H 
 
 i 
 
 
 
 
 _ 
 
 3 
 
 
 
 
 _ 
 
 ) 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 * 
 
 l 
 
 
 
 
 
 
 , 
 
 
 
 
 + :; 
 
 
 
 
 t 
 
 i 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 I 
 
 C 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Zx 
 
 
 
 )f 
 
 X 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 j 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 — 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 , 
 
 -• 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 I 
 
 - 
 
 
 J 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 I 
 
 
 
 *; 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 E.g. the abscissa of point P in the above figure is 3 and its 
 ordinate 2, or we may say the coordinates of P are 3 and 2, and indi- 
 cate it thus : P : (3, 2), writing the abscissa first. In like manner for 
 the other points we write Q: (- 1, 3), R: (- 2, 0), S : (- 3, - 4), 
 and T : (2, — 3). (For convenience upper signs are used in the 
 figure.) 
 
 We see that in this manner every point in the plane corre- 
 sponds to a pair of numbers and that every pair of numbers cor- 
 responds to a point. This scheme of locating points by two 
 reference lines is already familiar to the pupil in geography, 
 where cities are located by latitude and longitude, that is, by 
 degrees north or south of the equator and east or west of the 
 meridian of Greenwich. 
 
148 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 
 EXERCISES 
 
 1. With any convenient scale, locate the following points : 
 (2,6), (-3,5), (0,1), (1,0), (0,0), (0,-1), (0,-5), 
 (-5,0), (2i, 5ft, (-4,-8), (3,-10), (-10,3). 
 
 2. Calling north + , south — , east + , west — , so that 
 (— 5°, 8°) means the point on a map whose longitude is 5° 
 west and whose latitude is 8° north, verify the following on a 
 map: New York (- 73° 57', 40° 48'), Chicago (-87° 35', 41° 
 51'), Peking (116° 30', 39° 50'), Sydney, New South Wales, 
 (151° 15', -33° 51'), Santiago, Chili, (- 70° 39', - 33° 25'). 
 
 3. On a map of South America give approximately the lo- 
 cation of the following cities, using the notation of the pre- 
 ceding exercise : Caracas, Rio de Janeiro, Bogota, Valparaiso, 
 Lima, and Panama. 
 
 4. On a map of Africa locate in similar manner the cities or 
 islands situated at the following points: (— 5° 40', — 16°), 
 (- 14° 30', - 8°), (30° 18', 30° 1'), (- 5° 40', 36°), (40° 40', 
 - 15°), (18° 20', - 33° 55'). 
 
 5. Locate the following series of points and then see if a 
 straight line can be drawn through them : (0, 0), (1, 1), (2, 2), 
 (3, 3), (4, 4), (- 1, - 1), (- 2, - 2), (- 3, - 3). Name still 
 other points lying on the same line. 
 
 6. Locate the following and connect them by a line: (1,0), 
 (1, 2), (1, 3), (1, 4), (1, 5), (1, - 2), (1, - 3), (1, - 4), (1, - 5). 
 Name other points in this line. 
 
 7. Draw the line every one of whose points has its hori- 
 zontal distance — 2, also the line every one of whose points has 
 its vertical distance + 3. 
 
 8. Locate the following points and see if a straight line can 
 be passed through them : (1, 0), (0, 1), (2, - 1), (3, - 2), 
 (4-3), (-1+2), (-2, 3), (-3, 4), (-4, 5), & *), & f), (f, *). 
 Can you name other points on this line ? 
 
GRAPHS REPRESENTING EQUATIONS 149 
 
 Temperature chart for Port Conger. (See page 141.) 
 
150 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 
 108. In the preceding exercises, in certain cases, a series 
 of points has been found to lie on a straight line as in 
 examples 6, 7, and 8. Evidently this could not happen 
 unless the points were located according to some definite 
 scheme or law. 
 
 Illustrative Problem. Locate a series of points whose co- 
 ordinates are values of x and y which satisfy : 3 x + 4 y = 12. 
 
 We see that x = 0, y = 3, also x = 4, y = are pairs of such values. 
 Evidently as many pairs of values as we please may be found by 
 giving any value to x and then solving the equation to find the corre- 
 sponding value of y. A table may thus be constructed as follows : 
 
 Let x = 0, 4, 8, 12, -4,-8, etc. 
 
 Then y - 3, 0,-3,-6, 6, 9, etc. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 -i 
 
 n 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 -t- 
 
 - 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ' 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 4 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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 These pairs of values for x and y correspond to the points as 
 plotted in the figure, and they are found to lie on a straight 
 line. This line is called the graph of the equation. 
 
GRAPHS REPRESENTING EQUATIONS 151 
 
 Let the student find other pairs of numbers which satisfy this 
 equation and see if the corresponding points lie on this line. Also 
 find the numbers which correspond to any chosen point on this line 
 and see whether they satisfy the equation. 
 
 109. Since an equation like 3 x -f 4y = 12 is satisfied by in- 
 definitely many pairs of values of x and y, it is common to call 
 the unknowns in such an equation variables. Indeed if a point 
 be thought of as moving along the graph of this equation the 
 values of x and y corresponding to the moving point continually 
 vary, but always so that 3 x + 4y = 12. 
 
 110. Definitions. An equation is said to be of the first degree 
 in x and y if it contains each of these letters in such a way that 
 neither x nor y is multiplied by itself or by the other. 
 
 E.g. 13 x — 5 y = 14 is of the first degree, while 2 xy — x = 5 and 3 x 
 — 5 y 2 = 13 are not of the first degree in x and y. 
 
 Every equation of the first degree in two variables has for 
 its graph a straight line ; hence such an equation is commonly 
 called a linear equation. 
 
 111. To graph an equation of the first degree it is only 
 necessary to find two points on the graph and draw a straight 
 line through them. 
 
 E.g. In graphing the equation x — y = 5, we choose x = and find 
 y = — 5, and choose y = and find x = 5 and plot the points (0, — 5) 
 and (5, 0). The line through these points is the one required. 
 
 EXERCISES 
 
 Construct the graph for each of the following equations : 
 
 1. 3x + 2y = l. 5. 5-2y=12. 9. 3a-4?/=-7. 
 
 2. 5x-3y=-3. 6. 3x + 5y= — 15. 10. 3x-4y=-12. 
 
 3. 7x + 10y=2. 7. 2x-y = 0. 11. 7y = 9»— 63. 
 
 4. x + 2y = 0. 8. 3x-4y = 7. 12. x = 5y + 3. 
 
152 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 
 112. Illustrative Problem. Graph on the same axes the two 
 equations x + y = 4 and y — x = 2. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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 Solution. The two graphs are found to intersect in the point (1, 3). 
 Since the point lies on both lines, its coordinates should satisfy both 
 equations, as indeed they do. Since these lines have only one point 
 in common, there is no other pair of numbers which, when substituted 
 for the variables x and y, can satisfy both equations. 
 
 Hence x = 1, y = 3, which is written (1, 3), is called the solu- 
 tion of this pair of equations. 
 
 113. Definition. These two equations are called independ- 
 ent because their graphs are distinct. They are called simulta- 
 neous because there is at least one pair of values of x and y 
 which satisfy both. 
 
ELIMINATION BY SUBSTITUTION 153 
 
 114. Since two straight lines intersect in but one point, it 
 follows that two linear equations which are independent and 
 simultaneous have one and only one solution. 
 
 EXERCISES 
 
 Graph the following and thus find the solution of each pair 
 of equations : 
 
 L f2x-3y = 25, 6 ty + 3x = 7, 
 
 Lie -f- 3/ = 5. \2y-\-x= —6. 
 
 2 |5* + 6y = 7, ? \x-2y = 2, 
 
 \2x-y = -A. ' V2x-y=-2. 
 
 g f5*+3y»l, 8 f5x-7y = 21, 
 
 i2x + y = — 4. (x — £y=— 1. 
 
 4 J6a; + 8y = 16, Q \5x + 2y = 8, 
 \2x-3y = ll. ' (2x-3y=-12. 
 
 5 \'3x-4:y = l, 1Q tx + 3y=-6, 
 \2x-7y = 5. ' \2x-±y=-12. 
 
 SOLUTION OF SIMULTANEOUS EQUATIONS BY SUBSTITUTION 
 
 115. A pair of linear equations may be solved without con- 
 structing their graph. 
 
 Illustrative Problem. The sum of 'two numbers is 35 and 
 their difference is 5. What are the numbers ? 
 
 Solution. Let x = the greater number and y the smaller. 
 
 Then fz + y = 35, (1) 
 
 and 1 x — y = 5. (2) 
 
 From (1) by 5, y = 35 - x. • (3) 
 
 Substituting 35 — x for y in (2), x — (35 — x) = 5. (4) 
 
 Solving, x = 20. (5) 
 
 Substituting x = 20 in (1), y = 15. (6) 
 
154 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 
 116. In most problems solved heretofore there have been two 
 or more unknown numbers. In forming the equations the 
 object has been to express all but one of the unknowns in 
 terms of that one. In the case of two unknowns this is now 
 done more systematically as follows : 
 
 (a) State two equations involving the two unknowns, as (1) 
 and (2) above. 
 
 (6) Solve one of these equations for one unknown in terms 
 of the other as in (3) above. 
 
 (c) Substitute in the other equation the value of this un- 
 known as thus expressed, obtaining as in (4) one equation in 
 one unknown. 
 
 This equation (4) is the same as equation (1) on page 111, 
 where this problem was solved by means of one unknown. 
 The solution given here amounts to a formal tabulation of the 
 steps there taken in obtaining the equation, g—(35—g) = 5. 
 
 Since equation (4) contains but one of the unknowns, the 
 other is said to be eliminated. 
 
 The process here used is called elimination by substitution. 
 
 Illustrative Problem. Solve the equations : 
 
 
 
 \2x + 3y = 13. 
 \5x-6y=-8. 
 
 (1) 
 
 
 
 (2) 
 
 From (1) by 
 
 S and D, 
 
 13-2* 
 
 V= 3 * 
 
 (3) 
 
 Substituting 
 
 in (2) 
 
 5x 6 Q3-2*) = 8 . 
 3 
 
 (4) 
 
 ByF,V, 
 
 
 5 x- 2(13 - 2x)=-8. 
 
 (5) 
 
 By IV, VII, 
 
 
 5x-26+ ix = -k 
 
 (6) 
 
 By I, A, 
 
 
 9 x = 18. 
 
 (7) 
 
 By A 
 
 
 x = 2. 
 
 (8) 
 
 From (3), 
 
 
 13 - 2 • 2 - 
 
 y = ., = a - 
 
 (9) 
 
 Verify this by drawing the graphs and also by substituting these 
 values of x and y in (1) and (2). 
 
ELIMINATION BY SUBSTITUTION 
 
 155 
 
 EXERCISES 
 
 Solve the following pairs of linear equations by eliminating 
 one of the variables by substitution, and check by substituting 
 the results in the original equations. 
 
 i. 
 
 2. 
 
 4. 
 
 5. 
 
 7. 
 
 8. 
 
 9. 
 
 10. 
 
 11. 
 
 x — y = 10. 
 
 x-y = -3, 
 a; + 4 y = 12. 
 
 2x + 3y = 5, 
 7x-5y = 33. 
 
 3 a; - 4 y = 8, 
 
 4 a; + 3 y = - 6. 
 
 2 x - 4 y = 8, 
 
 3 a? + 2 y = 4. 
 
 a; + 2y = 4, 
 
 2 a; + y = - 1. 
 
 3 x — 4 y = 8, 
 2aj + 3y = ll. 
 
 5 a; + 9 y = 19, 
 
 3 x — y = 5. 
 
 4 y - 2 x = 3, 
 
 2 y + 5 a; = 6. 
 
 3 x - 7 y = - 11, 
 2 a; + y = 4. 
 
 5<c — 3y = 4— 2a;+7y, 
 
 5 y + a; = 7. 
 
 12. 
 
 13. 
 
 14. 
 
 15. 
 
 16. 
 
 17. 
 
 18. 
 
 19. 
 
 20. 
 
 21. 
 
 22. 
 
 |3y + 5a; = 12 + 2a;, 
 tl7a;-y = 4y-20. 
 
 f6y-a; = 7 + 4y, 
 \5x + 8y = l. 
 
 (6 + x + y = 2x-l, 
 l3y + z = 6y + 9. 
 
 x-y = 37, 
 
 + 3y = 314 + 13y. 
 
 2z-3y = y + 6, 
 a? + 2y = 4y + 3. 
 
 y + 5a; = 2a; + 5, 
 y - 3 x = 19. 
 
 \2x 
 
 l2y- 
 
 J5a; + 3y = 0, 
 I2aj + y = l. 
 
 2x + 3y=6x-l, 
 
 |2a; + 3y = 
 I3a;-2v = 
 
 r 
 
 y = 3. 
 5 x - 3 y = 0, 
 
 2a; + 2-6y = 2-a;. 
 
 |6a;+2y=23, 
 Il0aj-5v=21. 
 
 { 
 
 ,y: 
 
 3a-7y=15, 
 5aj + 4y = 11. 
 
 Solve 21 and 22 by means of graphs and also by elimination 
 by substitution. Notice that the latter method gives a more 
 accurate solution than can be obtained from the graph. 
 
156 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 
 SOLUTION OF SIMULTANEOUS EQUATIONS BY ADDITION OR 
 SUBTRACTION 
 
 117. Illustrative Examples. Solve 
 
 <x + 2y = 7, (1) 
 
 (Sx-2y = 5. (2) 
 
 Adding the members of these equations, + 2 y and — 2 y cancel. 
 
 Hence, 4x=12, (3) 
 
 x = 3. (4) 
 
 Substituting in (1), 3 + 2 y = 7, (5) 
 
 y=2. (6) 
 Verify this by drawing the graphs, and also by substituting x = 3, 
 y = 2, in (1) and (2). 
 
 If one of the variables does not already have the same co- 
 efficient in both equations, the solution may be obtained as 
 follows : 
 
 Solve the equations 
 
 [7a?-f3y = 4-y + 4a>, (1) 
 
 L3cc — 2/ = 4?/ — 2 — x. (2) 
 
 From (1) by A, S, 3x + iy = 4. (3) 
 
 From (2) by A, S, 4x-5y = -2. (4) 
 
 From (3) by M, 12 x + 16 y = 16. (5) 
 
 From (4) by M, 12x-15y = -Q. (6) 
 
 Subtracting (6) from (5), 31 y = 22. (7) 
 
 From (7) by D, y = f { . (8) 
 
 Substituting in (3), x = \\. (9) 
 Hence, the solution is Jf, f|, 
 
 118. The process used in the solution just given is called 
 elimination by addition or subtraction. 
 
 This method is usually simpler than elimination by substitu- 
 tion, since the latter frequently involves fractions. 
 
ELIMINATION BY ADDITION OR SUBTRACTION 157 
 EXERCISES 
 
 Solve the following pairs of equations by addition or sub- 
 traction. Substitute the results in the given equations in each 
 case to test the accuracy of the solution. 
 
 L j2a; + 3?/ = 22, 6 i5x+10y=-7, 
 
 \x — y = l. \2x+5y = — 2. 
 
 2 \5x-2y = 2l, ?> l5x + 3y=-2, 
 \x — y = 6. \3x + 2y= — 1. 
 
 3 iGx+30 = 8y, 8 |3a + 7& = 7, 
 l3y + 17=2-3a;. l5a + 3& = 29. 
 
 4 1 8 x — 4 y = 12 x, 9 fr = 3s-19, 
 l4a + 2?/ = 3 + 4y. U = 3r-23. 
 
 5. 
 
 tx + 6y = 2x-16, 1Q J2p = 5g-16, 
 
 13 a -2?/ = 24. l7g= — 3i>-|-5. 
 
 Solve the following by either process of elimination : 
 
 lt j7m = 2n-3, 6 |15 & = 10-20Z, 
 
 ll97i = 6m + 89. l25A:-30Z = 80. 
 
 2. 
 
 19n = 6m + 89 
 
 |6c + 15d=-6, ? 
 
 l21d-8c=-74. 1 
 
 6c + 15d=-6, 7 \28x+Uy = 23, 
 
 14 # — 14 y = 1. 
 
 5. 
 
 I2x-3y = 4:, 8 i5x + 2y = x + 18, 
 12?/ — 3aj= —21. l2a + 3i/ = 3:B + 27. 
 
 (u + v = 27, 9 J7?/ — # = a; — 17, 
 4v = 19-|m. 12 y + 3 a; = 38. 
 
 |7a = l + 10y, 1Q |6a; + 2 2 /=-2, 
 ll6?/=10a-l. \x-4y = -35. 
 
158 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 (3x-y = 2x-l, 16 J' 
 
 11. 
 
 3x-y = 2x-l, 16 |7o5 — 42/ = 3, 
 
 + y = 14. I5x + 8y = 6. 
 
 12. 
 
 13. 
 
 |2ar + 32/ = 5, 1? (12y - 10a = - 6, 
 
 l7a-2y = 74. l5.v-9a; = l 
 
 { 
 
 14 f6*/ + 2z = ll, 19 { 7x + 4:y = 3, 
 
 lB f4y+9x = -5, 2Q f34z + 702/ = 4, 
 
 3?/ + 12 a = 18. 12 a; + 3?/ = 25. 
 
 f 34 x + 70 ?/ = 
 ls + y = -5. l5a-8y = -36. 
 
 119. The equations thus far given have for the most part 
 been written in a standard form, ax + by = c, in which all the 
 terms containing x are collected, likewise those containing y, 
 and those which contain neither variable. When the equations 
 are not given in this form, they should be so reduced, as in the 
 second solution on page 156, before applying any method of 
 elimination, and also before solving by means of graphs. 
 
 EXERCISES 
 
 After reducing each of the following pairs of equations to 
 the standard form, solve by graphing, or by means of either 
 process of elimination, as seems best available. 
 
 1. 
 
 ix-U = 7y, 3 jr + l=-4s, 
 
 Uy + l = x. l2s = 13-5r. 
 
 f _ l-8n 
 tl6x-3y = 7x, 4. m 5 ' 
 
 Uy = 7cc + 5. l3ra + 5n = l. 
 
ELIMINATION BY ADDITION OR SUBTRACTION 159 
 
 s rm — n = 16, 
 l3» = 8-27i. 
 
 Ta?- 15 
 
 = 2/, 
 
 6. 3 
 
 1.2 a: — y = 3 
 
 fa; — 3 
 
 = -2, 
 
 7. 5y 
 
 la; + 7 y = 6. 
 
 8 |3^-5=-y, 
 
 \8y + 76 = 5x. 
 
 fa + 45 = 14, 
 l3a-6 = 14. 
 
 10. 2 ^ 2 ' 
 
 [2x-y = 16. 
 
 11. < 
 
 5 + ~3~- 8 ' 
 2x_-y_3y-x_ 
 2 4 *' 
 
 7m + 8 7« — 1 
 
 12. 
 
 2 m— 4 rt — 1 x 
 
 2 + ~3~~~*' 
 
 ( x - 3 | y +_ 
 
 13. < 
 
 a + 7 2y-4__ 5 
 
 14. 
 
 12 7 
 
 f8a-3 55-2 
 
 9 + 3 
 2a+7 35+10 
 
 13, 
 
 10 
 
 = -34. 
 
 15. 
 
 Ty-4 2a?-3 
 
 5 "*" 2 ~ ' 
 6»-3 2y+l 
 5^5 
 
 16. - 
 
 3y + 7 5a:-7 
 
 2 3 
 
 2a;-4 2y-l 
 
 10, 
 
 3 4 
 
 f 5 + 3p 5g-2 
 
 17. J 7 4 
 Up + 8? = 108. 
 
 (3x-2y = ±, 
 
 18. J 2te — 1 7y-4 
 I 5 3 
 
 -2i- 
 
 -2, 
 
 = -19. 
 
 f 5 x + 7 y = 89$, 
 19. \2x-4 6y-l 
 
 [ 3 + ~5~ ~ 16 *- 
 
 20. 
 
 f32a;-9 2/ = 299, 
 
 2jc-5 _ 3j/-jL 
 7 2 
 
 = -16. 
 
160 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 
 PROBLEMS 
 
 Solve the following problems, using two variables in each 
 case. 
 
 1. A rectangular field is 32 rods longer than it is wide. 
 The length of the fence around it is 308 rods. Find the di- 
 mensions of the field. 
 
 2. Find two numbers such that 7 times the first plus four 
 times the second equals 37 ; while 3 times the first plus 9 
 times the second equals 45. 
 
 3. A certain sum of money was invested at 5% interest 
 and another sum at 6%, the two investments yielding $980 
 per annum. If the first sum had been invested at 6% and 
 the second at 5%, the annual income would be $1000. Find 
 each sum invested. 
 
 4. The combined weight of 3 cubic centimeters of platinum 
 and 50 cubic centimeters of poplar is 84 grams, and the weight 
 of 1 cubic centimeter of platinum and 150 cubic centimeters of 
 poplar is 80 grams. Find the weight of 1 cubic centimeter 
 of each. 
 
 5. The combined distance from the sun to Jupiter and 
 from the sun to Saturn is 1369 million miles. Saturn is 403 
 million miles farther from the sun than Jupiter. Find the 
 distance from the sun to each planet. 
 
 6. The sum of the distances from the sun to the fixed 
 stars Altair and Capella is 45.3 light-years. Twice the dis- 
 tance of Altair plus 3 times that of Capella is 119.6 light- 
 years. Find the distance from each star to the sun. 
 
 7. Find two numbers such that 7 times the first plus 9 
 times the second equals 116, and 8 times the first minus 4 
 times the second equals 4. 
 
 8. The sum of two numbers is 108. 8 times one of the 
 numbers is 9 greater than the other number. Find the 
 numbers. . 
 
PROBLEMS IN TWO VARIABLES 161 
 
 9. Two investments of $24,000 and $16,000 respectively 
 yield a combined income of $840. The rate of interest on 
 the larger investment is 1 % greater than that on the other. 
 Find the two rates of interest. 
 
 10. A father is twice as old as his son. Twenty years 
 ago the father was six times as old as his son. How old is 
 each now? 
 
 11. If the length of a rectangle is increased by 3 feet and 
 its width decreased by 1 foot, its area is increased by 3 square 
 feet. If the length is increased by 4 feet and the width de- 
 creased by 2 feet, the area is decreased by 3 square feet. 
 What are the dimensions of the rectangle ? 
 
 Let I = the original length and w the width, 
 
 then , (I + 3) (w - 1) = ho + 3, (1) 
 
 and (Z + 4)(w-2) = Zw-3. (2) 
 
 From (1) by XIII, Iw + 3 w - I - 3 = Iw + 3. (3) 
 
 From (2) by XIII, Iw + 4 w - 2 I - 8 = Iw - 3. (4) 
 
 From (3) and (4) 3w- 1 = 6, (5) 
 
 and 4 w - 2 I = 5. (6) 
 
 (5) and (6) may now be solved in the ordinary manner. 
 
 12. The greatest distance from the earth to Venus is 160 
 million miles and the shortest distance is 26 million miles. 
 How far from the sun are Venus and 
 
 the earth, assuming that they move 
 around the sun in concentric circles 
 with the sun at the center ? 
 
 13. Two weights, 35 and 40 pounds 
 respectively, balance when resting 
 on a beam at certain unknown 
 distances from the fulcrum. If 
 15 pounds is added to the 35-lb. 
 
 weight, the 40-lb. weight must be moved 2 feet farther 
 from the fulcrum in order to maintain the balance. What 
 
162 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 
 was the original distance from the fulcrum to each of the 
 weights ? 
 
 If the distances from the fulcrum to the weights are di and di re- 
 spectively, then by the formula, w\d x = w 2 d2, given on page 121, 
 we have 35 d x = 40 d 2 and 50 di = 40 (d% + 2). 
 
 14. A steamer on the Mississippi makes 6 miles per hour 
 going against the current and 19^ miles per hour going with 
 the current. What is the rate of the current and at what 
 rate can the steamer go in still water ? 
 
 15. A man starts at 7 a.m. for a walk in the country. At 
 10 a.m. another man starts on horseback to overtake the pedes- 
 trian, which he does at 1 p.m. If the rate of the horseman had 
 been two miles per hour less, he would have overtaken the 
 pedestrian at 4 p.m. At what rate does each travel ? 
 
 16. A camping party sends a messenger with mail to the 
 nearest post office at 5 a.m. At 8 a.m. another messenger is 
 sent out to overtake the first, which he does in 2\ hours. If 
 the second messenger travels 5 miles per hour faster than the 
 first, what is the rate of each ? 
 
 17. There are two numbers such that 3 times the greater is 18 
 times their difference, and 4 times the smaller is 4 less than 
 twice the sum of the two. What are the numbers ? 
 
 18. Roy is three times as old as Fred was 8 years ago. Five 
 years from now Roy will be 16 years less than twice as old 
 as Fred. How old is each now ? 
 
 19. A picture is 3 inches longer than it is wide. The frame 4 
 inches wide has an area of 360 square inches. What are the 
 dimensions of the picture ? 
 
 20. The difference between 2 sides of a rectangular wheat 
 field is 30 rods. A farmer cuts a strip 5 rods wide around the 
 field, and finds the area of this strip to be 1\ acres. What 
 are the dimensions of the field ? 
 
PROBLEMS IN TWO VARIABLES 163 
 
 21. The sum of the length and width of a certain field is 
 260 rods. If 20 rods are added to the length and 10 rods 
 to the width, the area will be increased by 3800 square rods. 
 What are the dimensions of the field ? 
 
 22. In a number consisting of two digits the sum of the 
 digits is 12. If the order of the digits is reversed the number 
 is decreased by 36. What is the number ? 
 
 23. A bird attempting to fly against the wind is blown 
 backward at the rate of 1\ miles per hour. Flying with a 
 wind \ as strong, the bird makes 48 miles an hour. Find the 
 rate of the wind and the rate at which the bird can fly in 
 calm weather. 
 
 24. There is a number whose two digits differ by 2. If the 
 digit in units' place is multiplied by 3 and the digit in tens' 
 place is multiplied by 2, the number is increased by 44. Find 
 the number, the tens' digit being the larger. 
 
 25. In a number consisting of two digits one digit is equal 
 to twice their difference. If the order of the digits is reversed, 
 the number is increased by 18. Find the number, the units' 
 digit being the larger. 
 
 26. If the length of a rectangle is doubled and 8 inches added 
 to the width, the area of the resulting rectangle is 180 square 
 inches greater than twice the original area. If the length and 
 width of the rectangle differ by 10, what are its dimensions ? 
 
 27. The Centigrade reading at the boiling point of alcohol 
 is 96° lower than the Fahrenheit reading. Find both the 
 Centigrade and the Fahrenheit reading at this temperature. 
 
 Use C and F as the unknowns. Then one of the equations is the 
 formula connecting Fahrenheit and Centigrade readings obtained on 
 page 129, and the other is C + 96 = F. 
 
 28. The Centigrade reading at the boiling point of mercury 
 is 312° lower than the Fahrenheit reading. Find both the 
 Fahrenheit and the Centigrade reading at this temperature. 
 
164 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 
 29. There is a number consisting of three digits, those in 
 tens' and units' places being the same. The digit in hundreds' 
 place is 4 times that in units' place. If the order of the digits 
 is reversed, the number is decreased by 594. What is the 
 number ? 
 
 30. A man rowing against a tidal current drifts back 2\ 
 miles per hour. Rowing with this current, he can make 121 
 miles per hour. How fast does he row in still water and how 
 swift is the current ? 
 
 31. Flying against a wind a bird makes 28 miles per hour, 
 and flying with a wind whose velocity is 2| times as great, the 
 bird makes 46 miles per hour. What is the velocity of the 
 wind and at what rate does the bird fly in calm weather ? 
 
 32. A freight train leaves Chicago for St. Paul at 11 a.m. 
 At 3 and 5 p.m. respectively of the same day two passenger 
 trains leave Chicago over the same road. The first overtakes 
 the freight at 7 p.m. the same day, and the other, which runs 
 10 miles per hour slower, at 3 a.m. the next day. What is 
 the speed of each? 
 
 33. Two boys, A and B, having a 30-lb. weight and a teeter 
 board, proceed to determine their respective weights as fol- 
 lows : They find that they balance when B is 6 feet and A 5 
 feet from the fulcrum. If B places the 30-lb. weight on the 
 board beside him, they balance when B is 4 and A 5 feet from 
 the fulcrum. How heavy is each boy ? 
 
 34. C is 6£ feet from the point of support and balances D, 
 who is at an unknown distance from this point. O places a 
 33-lb. weight beside himself on the board and when 4| feet from 
 the fulcrum, balances D, who remains at the same point as be- 
 fore. C's weight is 84 pounds. What is D's weight and how 
 far is he from the fulcrum ? 
 
 35. E weighs 95 pounds and F 110 pounds. They balance 
 at certain unknown distances from the fulcrum. E then takes 
 
PROBLEMS IN TWO VARIABLES 165 
 
 a 30-lb. weight on the board, which compels F to move 3 feet 
 farther from the fulcrum. How far from the fulcrum was each 
 of the boys at first ? 
 
 36. A fast freight leaves New York for Chicago at 8 a.m. 
 At 4 p.m. the same day an express train leaves New York for 
 Chicago and passes the freight 12 hours later. Another ex- 
 press leaving New York at 6 p.m. of the same day overtakes 
 the freight 10 hours after starting. Find the rate of each 
 train if the second express goes 8 miles per hour faster than 
 the first. 
 
 37. The Centigrade reading at the melting point of silver 
 is 796° lower than the Fahrenheit reading. Find both Centi- 
 grade and Fahrenheit readings at this temperature. 
 
 38. The Fahrenheit reading at the melting point of gold is 
 992° higher than the Centigrade reading. Find both Centi- 
 grade and Fahrenheit readings at this temperature. 
 
 39. $10,000 and $8000 are invested at different rates of 
 interest, yielding together an annual income of $ 820. If the 
 first investment were $ 12,000 and the second $ 6000, the 
 yearly income would be $ 840. Find the rates of interest. 
 
 40. In a switch yard a car weighing 50 tons and going at a 
 certain rate strikes a standing car, whereupon both cars move 
 off at the rate of 4 miles per hour. If the second car, moving 
 at the same rate as the first before impact, were to strike the 
 first car when standing still, they would move off at the rate 
 of 2 miles per hour. How fast did the first car move before 
 impact and what is the weight of the second car ? (Set up the 
 equations by means of the formula iv^x = w-px + w 2 v 2 , ob- 
 tained on page 127.) 
 
 41. 200 ccm. of white oak is fastened to 25 ccm. of steel, 
 making a combination whose average density is 1.56. If 
 250 ccm. of oak is fastened to 20 ccm. of steel, the average 
 density of the combination is 1.3. Find the density of white 
 oak and also of steel. 
 
166 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 
 SIMULTANEOUS EQUATIONS IN THREE VARIABLES 
 
 120. Illustrative Problem. Three men were discussing their 
 ages and found that the sum of their ages was 90 years. If 
 the age of the first were doubled and that of the second trebled, 
 the aggregate of the three ages would then be 170. If the ages 
 of the second and third were each doubled, the sum of the three 
 would be 160. Find the age of each ? 
 
 Solution. Let x, y, and z represent the number of years in their 
 ages in the order named. 
 
 Then, x + y + z = 90, (1) 
 
 2x+S y+z= 170, (2) 
 
 and x+2y+2z= 160. (3) 
 
 Since by supposition x represents the same number in all three 
 equations, and likewise y and z, if we subtract (1) from (2), we obtain 
 a new equation from which z is eliminated. 
 
 I.e. x + 2y=80. (4) 
 
 Again, multiplying (2) by 2 and subtracting (3), 
 
 3 x + 4 y = 180. (5) 
 
 (4) and (5) are two equations in the two variables x and y. Solving 
 
 these by eliminating y, we find x = 20. (6) 
 
 Substituting x = 20 in (4), y = 30. (7) 
 
 Substituting x and y in (1), z = 40. (8) 
 
 Check by substituting the values of x, y, and z in all three given 
 equations and also by showing that they satisfy the conditions of the 
 problem. 
 
 The values of x, y, and z as thus found constitute the solution 
 of the given system of equations. 
 
 Evidently x could have been eliminated first, using (1), (2) and 
 (1), (3), giving a new set of two equations in y and z. Let the 
 student find the solution in this manner. 
 
 Also find the solution by first eliminating y, using (1), (2) and (2), 
 (3), getting two equations in x and z, from which the values of x and 
 z can be found. 
 
EQUATIONS IN THREE VARIABLES 
 
 167 
 
 121. Definition. An equation is said to be of the first degree 
 in three variables if no one of the variables is multiplied by 
 itself or by one of the others (§ 110). 
 
 The fact that the solutions are found to be the same no 
 matter in what order the equations are combined, indicates 
 that a system of three independent and simidtaneous eqtiations of 
 the first degree in three variables has one and only one solution. 
 
 As in the case of two equations, each should be first reduced 
 to a standard form in which all the terms containing a given 
 variable are collected and united and all fractions removed by M, 
 Principle VIII. 
 
 EXERCISES 
 
 Solve the following systems of equations, and check the 
 results by substituting the values found for each variable in 
 the given equations : 
 
 2 x — y + z = 18, 
 x-2y + 3z = 10, 
 
 3 x + y - 4 z = 20. 
 
 5x — 3y -\-z = 15, 
 ix + 3y— z — 3, 
 2x — y + z = 8. 
 
 4:x + 2y + z = 13, 
 x—y + z = 4:, 
 x + 2y — z — \. 
 
 6»+4y — 4z = — 4, 
 4a-2y + 8z = 0, 
 x + y+z = 4:. 
 
 (x + 2y + 3z = 5, 
 5. <4:X — 3y — z = 5, 
 
 [x + y + z = 2. 
 
 9. 
 
 10. 
 
 \2x-8y-£3z = 2, 
 |X-4y + 5z = l, 
 [3x-10y-z = 5. 
 
 x + y + z = l, 
 x + 3y + 2z = 8, 
 2x + 8y-3z=:15. 
 
 \2x — 3y+z = 5, 
 3a + 2y-z = 5, 
 {x + y + z = 3. 
 
 \x+ y + z == 6, 
 3 x _ 2 y - z = 13, 
 [2x-y + 3z = 26. 
 
 lx + y + z = 6, 
 Ax — y — z= — 1, 
 [2x + y-3z=-G. 
 
1G8 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 
 PROBLEMS INVOLVING THREE VARIABLES 
 
 122. Illustrative Problem. A broker invested a total of 
 $15,000 in the street railway bonds of three cities, the first 
 investment yielding 3 %, the second 3| %, and the third 4 %, 
 thus securing an income of $535 per year. If the second 
 investment was one-half the sum of the other two, what was 
 the amount of each ? 
 
 Solution. Suppose x dollars were invested at 3 %, y dollars at 3 \ %, 
 and z dollars at 4 %. 
 
 Then, x + y + z = 15000, 
 
 (1) 
 
 .03 x + .035 y + .04 z = 535, 
 
 (2) 
 
 and x + z = 2 y. 
 
 (3) 
 
 From (3), ar-2^ + z = 0. 
 
 (4) 
 
 Subtract (4) from (1), 3 y = 15000, 
 
 (5) 
 
 and y = 5000. 
 
 (6) 
 
 From (1), by M, .035 x + .035 y + .035 z = 525. 
 
 (7) 
 
 Subtract (7) from (2), - .005 x + .005 z = 10. 
 
 (8) 
 
 Divide (8) by .005, -x + z = 2000. 
 
 (9) 
 
 Substitute (6) in (4), x + z = 10000. 
 
 (10) 
 
 Add (9) and (10), 2 z = 12000. 
 
 (11) 
 
 z = 6000. 
 
 (12) 
 
 Substitute (5) and (12) in (1), x = 4000. 
 
 (13) 
 
 Hence, $4000, $5000, and $6000 were the sums invested. 
 
 Solve the following problems, using three unknowns : 
 
 1. The sum of three angles A, B, and O of a triangle is 180 
 degrees. | of A+\ of B+\ of C is 48 degrees, while I of A+ \ 
 of B-\-\ of C is 30 degrees. How many degrees in each angle ? 
 
 2. The combined weight of 1 cubic foot each of compact 
 limestone, granite, and marble is 535 pounds. 1 cubic foot of 
 limestone, 2 of granite, and 3 of marble weigh together 1041 
 pounds, while 1 cubic foot of limestone and 1 of granite together 
 weigh 195 pounds more than 1 cubic foot of marble. Find the 
 weight per cubic foot of each kind of stone. 
 
PROBLEMS IN THREE VARIABLES 169 
 
 3. A number is composed of 3 digits whose sum is 7. If 
 the digits in tens' and hundreds' places are interchanged, 
 the number is increased by 180 ; and if the order of the digits 
 is reversed, the number is decreased by 99. What is the 
 number ? 
 
 4. The sum of the angles A, B, and C of a triangle is 180 
 degrees. If B is subtracted from \ of A the remainder is 
 \ of C, and when C is subtracted from twice A the remainder 
 is 4 times B. How many degrees in each angle ? 
 
 5. The sum of the three sides a, b, c of a certain triangle is 
 35, and twice a is 5 less than the sum of b and c, and twice c 
 is 4 more than the sum of a and b. What is the length of 
 each side ? 
 
 6. The combined number of students at Harvard, Yale, 
 and Columbia during the year 1905-1906 was 13,390. The 
 number at Harvard minus the number at Columbia plus twice 
 the number at Yale was 6893, and the number at Columbia 
 plus 4 times the number at Yale minus twice the number 
 at Harvard was 7258. What was the number at each 
 university ? 
 
 7. The total number of students at the universities of 
 Illinois, Michigan, and Wisconsin during the year 1905-1906 
 was 12,216. Twice the number at Illinois plus 3 times that at 
 Michigan plus 4 times that at Wisconsin was 36,145. If the 
 number at Michigan is subtracted from the sum of the num- 
 bers at Illinois and Wisconsin, the remainder is 3074. Find 
 the number at each university. 
 
 8. The total number attending schools in the United States 
 in 1904-1905 was 17,953,844. If the number in secondary 
 schools and colleges combined be subtracted from the number 
 in elementary schools, the remainder is 15,924,656; while if 
 twice the number in colleges be added to the number in ele- 
 mentary and secondary schools combined, the sum is 18,092,388. 
 Find the number of students of each kind. 
 
170 INTRODUCTION TO SIMULTANEOUS EQUATIONS 
 
 9. The combined foreign trade in 1905 at the three ports, 
 London, Liverpool, and New York, was 3598 million dollars. 
 If the amount at London is subtracted from the combined 
 amounts at New York and Liverpool, the remainder is 988 
 million ; and if the amount at New York is subtracted from 
 the combined amounts at London and Liverpool, the remainder 
 is 1384 million. Find the amount of foreign trade at each 
 port. 
 
 In the following three examples find the number of seconds 
 in each record and reduce the results to minutes and seconds. 
 
 10. If x is the number of seconds in the Eastern inter- 
 collegiate record for a mile run, y the number in the Western 
 intercollegiate record, and z the number in the world's record, 
 then 
 
 x + y + z = 781.15, 
 
 —x+2y+z= 519.35, 
 
 2 x — y + 1 = 514.55. 
 
 11. If a; is the number of seconds in the Eastern inter- 
 collegiate record for a half mile run, y the number in the West- 
 ern intercollegiate record, and z the number in the world's 
 record, then 
 
 2x + 3y + z = 697.7, 
 
 3x + 2y + 2z = 809.8, 
 2x-y + z = 22S.l. 
 
 12. If x — number of seconds in the world's mile trotting 
 record in 1806, y = number of seconds in the world's record in 
 1885, and z = number of seconds in the. world's record in 1903, 
 then 
 
 * + y + z = 426.25, 
 2z + 4?/ + 6z = 1584, 
 - a> + y + 2 « = 186.75. 
 
REVIEW QUESTIONS 171 
 
 REVIEW QUESTIONS 
 
 1. How may a point in a plane be located by reference to 
 two fixed lines ? What are these fixed lines called ? What 
 names are given to the distances from the point to the fixed 
 lines ? Why are negative numbers needed in order to locate 
 all points in this manner ? 
 
 2. Draw a pair of axes in a plane and locate the following 
 points: (5,0), (-2,0), (0,3), (0,-1), (0,0). 
 
 3. State a problem involving motion and solve it by means 
 of a graph. 
 
 4. How many pairs of numbers can be found which satisfy 
 the equation x — 2 y = 6 ? State five such pairs and plot the 
 corresponding points. How are these points situated with 
 respect to each other ? What can you say of all points corre- 
 sponding to pairs of numbers which satisfy this equation? 
 What is meant by the graph of an equation ? 
 
 5. How many pairs of numbers will simultaneously satisfy 
 the two equations 3x+2y=7 and x +y = 3 ? Show by means 
 of a graph that your answer is correct. 
 
 6. Describe elimination by the process of substitution ; also 
 by the process addition or subtraction. Under what conditions 
 is one or the other of these methods preferable ? 
 
 7. Why is the solution by elimination in some cases pref- 
 erable to the solution by means of the graph ? 
 
 8. Describe the solution of a system of three linear 
 equations in three unknowns. Is it immaterial which of the 
 three variables is eliminated first ? 
 
 9. Can you find a definite solution for two equations each 
 containing three unknowns ? Illustrate this by means of the 
 equations 4 a; — 3y — z = 5 and x + y + z = 2. 
 
CHAPTER VI 
 SPECIAL PRODUCTS AND FACTORS 
 
 123. Repeated Factors. Number expressions containing re- 
 peated factors have already been considered in Chapter III. 
 x-x was written x 2 and called the square of x, or x square; 
 similarly, (a + b) (a -f b) was written (a + 6) 2 and read the 
 square of the binomial a + 6 or the binomial a + b squared. 
 
 124. Definitions. Any number written over and to the right 
 of a number expression is called an index or exponent and, if 
 a positive integer, shows how many times that expression is to 
 be taken as a factor. 
 
 A product consisting entirely of equal factors is called a 
 power of the repeated factor. The repeated factor is called the 
 base of the power. 
 
 E.g. x % means x • x -x and is read the third power of x or x cube; x 6 
 means x • x • x • x • x, and is read the fifth power of x or briefly x fifth, 
 (x — y) s = (x — y)(x — y)(x — y) and is read (x —y) cubed or the 
 cube of the binomial (x — y). 
 
 Notice two important differences between an exponent and 
 a coefficient. 
 
 (1) 5a = a+a-r-a + a + a, while a 5 = a • a • a • a • a. 
 
 (2) In 5 abc the coefficient 5 applies to the product abc, 
 while in abc 5 , the exponent 5 applies only to the factor c. In 
 order to make it apply to the product it is necessary to use a 
 parenthesis, thus, (abc) 5 means the product abc taken five 
 times as a factor. 
 
 172 
 
PRODUCTS OF POWERS OF SAME BASE 
 
 173 
 
 EXERCISES 
 
 Perform the following indicated multiplications 
 
 1. 
 
 2 3 
 
 , 2 4 , 2 5 , 
 
 2 6 . 
 
 9. 
 
 10 2 , 10 3 , 10 4 , 10 5 . 
 
 17. 
 
 (x — y — 5) 2 . 
 
 2. 
 
 3 2 
 
 3 3 , 3*, 
 
 3 5 . 
 
 10. 
 
 (a + bf. 
 
 18. 
 
 (w + z — 4) 2 . 
 
 3. 
 
 4 2 
 
 4 3 , 4«, 
 
 4 5 . 
 
 11. 
 
 (c-d)\ 
 
 19. 
 
 (2—z — x + w) 2 . 
 
 4. 
 
 5 2 
 
 5 3 , 5 4 , 
 
 5 5 . 
 
 12. 
 
 98 2 = (100-2) 2 . 
 
 20. 
 
 (x — y — 5) 2 . 
 
 5. 
 
 6 2 
 
 6 3 , 6 4 . 
 
 
 13. 
 
 (a + b + cy. 
 
 21. 
 
 (a - y)\ 
 
 6. 
 
 7 2 
 
 7 s , 7 4 . 
 
 
 14. 
 
 (a + 6 - c) 2 . 
 
 22. 
 
 (x + y) 3 . 
 
 7. 
 
 8 2 
 
 8 3 , 8*. 
 
 
 15. 
 
 (3 - a) 2 . 
 
 23. 
 
 (a + b)\ 
 
 8. 
 
 9 2 
 
 9 s , 9 4 . 
 
 
 16. 
 
 (3 _ 6 _ C )2. 
 
 24. 
 
 (c - d) 4 . 
 
 
 125 
 
 . In the case 
 
 of factors expressed 
 
 in Arabic figures mul- 
 
 tiplications like the following may be carried out in either of 
 two ways. 
 
 E.g. 3 2 • 3* = 9 • 81 = 729. 
 
 or 3 2 • 3 4 = ( 3 • 3)(3 • 3 • 3 • 3) = 3 2 + 4 = 3 6 = 729. 
 
 But with literal factors the second process only is possible. 
 E.g. a 2 . a* = (a • d)(a • a • a • a) = a?+* = a 6 . 
 
 The process in which the exponents are added applies only 
 when the factors are powers of the same base. 
 
 E.g. 2 8 . 3 2 = 8 • 9 = 72 cannot be found by adding the exponents. 
 
 2 1 or 2 is called the first power of 2. 
 Thus 2 • 2 8 = 2 1 • 2 8 = 2 1 + 8 = 2 4 . 
 
 EXERCISES 
 
 In the following exercises carry out each indicated multipli- 
 cation in two ways in case this is possible : 
 
 1. 2 5 .2 6 . 3. 2-2 4 . 5. 3-3 4 . 7. 4 2 -4 3 . 
 
 2. 2 2 -2 3 . 4. 3 2 -3 3 . 6. 3 2 -3 5 . 8. 4-4 4 . 
 
174 
 
 SPECIAL PRODUCTS AND FACTORS 
 
 9. 5-5 2 . 
 
 12. 7-7 3 . 
 
 15. x 7 -x\ 
 
 18. 2 3 -2 2 .2 4 . 
 
 10. 5 2 -5 8 . 
 
 13. a 2 -a 3 . 
 
 16. * 3 -t 4 . 
 
 19. 3-3 2 -3 3 . 
 
 11. 6 2 -6 2 . 
 
 14. tf-x 2 . 
 
 17. J 2 -* 3 -* 4 . 
 
 20. 2 2 .2 3 -2 2 .2 
 
 126. Illustrative Problem. To multiply 2* by 2 n , k and n 
 being any two positive integers. 
 
 Solution. 2* means 2 • 2 . 2 • 2, etc., to k factors, 
 and 2 n means 2 • 2 • 2 • 2, etc., to n factors. 
 
 Hence 2* . 2» = (2 . 2 . 2 • . . to n factors) (2 • 2 • • • to i factors) 
 
 = 2.2.2.2... to fc + n factors in all. 
 That is, 2*.2»=2*+». 
 
 The preceding examples illustrate the following principle : 
 
 127. Principle XIV. The product of two powers of the 
 same base is found by adding the exponents of the powers 
 and making this sum the exponent of the common base. 
 
 
 
 
 EXERCISES 
 
 
 Perform the 
 
 following 
 
 indicated 
 
 multiplications by means 
 
 of Principle XIV : 
 
 
 
 
 1. 2 3 -2 7 . 
 
 
 8. 
 
 3 2« . 3 » 
 
 15. w x -w y+Sx . 
 
 2. a 3 • a 7 . 
 
 
 9. 
 
 5 2+» , 5 2-» 
 
 16. n 2 -n 3c+ih . 
 
 3. 3 4 -3 5 . 
 
 
 10. 
 
 J2x+v . J2x 
 
 ~". 17. c x • c 2_x . 
 
 4. x A • X s . 
 
 
 11. 
 
 a m • a n . . 
 
 18. ar-z 40 . 
 
 5. 3*. 3". 
 
 
 12. 
 
 /2a . fb+a 
 
 19. r 30 -?' 40 . 
 
 6. sc*.z B . 
 
 
 13. 
 
 y* a - y 3 "- 
 
 20. s^-s™. 
 
 7. 4 a • 4 J 
 
 
 14. 
 
 qAb , rtfa+ib 
 
 21. -y 2 *" 1 . <y3*+2 # 
 
 Perform the following multiplications by means of Princi- 
 ples IV, III, and XIV : 
 
 22. 2 3 (2 2 + 2 4 ). 25. aX^b-ab 3 ). 
 
 23. 2 2 (3 • 2 4 + 5 • 2 s ). 26. a^ a (4 of + 3 a 88 ). 
 
 24. 4 4 (3-4 3 -5.4 2 ). 27. ^(5^-3^). 
 
PRODUCTS OF MONOMIALS 175 
 
 Multiply the following and state all principles used : 
 
 28. a 2 m 2 - b 2 m 3 by m 4 . 37. 3 2 ■ 4 2n - 6 3 • 4 4n by 4 3 \ 
 
 29. 4.3 2 -5-7-2by 2 4 . 38. aV-a c ^bya fc . 
 
 30. 2-3-4-3 2 -i-7.3by 3. 39. 6 2 • 4 3 -7 4 • 4 2 + 4 4 by 4 3 . 
 
 31. 4z 2 -3ar 3 + 6a 4 by x 2 . 40. 12 x 2 f - 6 x?y 2 + 3 xy by x 3 . 
 
 32. ox-3^ + 2» 4 by x 4 . 41. 2 a 36 - 3 a 2c - 4 a 45 by a 2 *. 
 
 33. 3 y 2 + 4 y* - y 3 by y. 42. 6 ar 3 "- 25 + 8 x 2 "- 3 * by x 2a+2i . 
 
 34. 7x 4 -5x?-2xby &. 43. y 3m+2n - y 2n 3m+l by ^ m+3n . 
 
 35. 3 a 2 & 4 + 2 d 2 b - 4 a 2 6 2 by a 3 . 44. ar 5c - 3a + ar k+2a + a; 2a byz 4a - 2c . 
 
 36. 4 a 26 — a ;?c + a 2d by a 26 . 45. a"*-** — a bx - ,lx —a x by a 5 * +n *. 
 
 PRODUCTS OF MONOMIALS 
 128. In multiplying two numbers each of which is in the 
 factored form, if the factors are all expressed in Arabic figures, 
 the operation may be carried out in two ways. 
 
 E.g. (2 • 3) (2 • 3 • 5) = 6 • 30 = 180. 
 
 Also (2 • 3) (2 • 3 • 5) = 2 2 • 3 2 • 5 = 4 • 9 • 5 = 180. 
 
 In the second process one of the factors, 2 or 3, is multiplied in 
 
 and this product is then multiplied by the other factor, 2 being 
 
 combined with the factor 2 and 3 with 3 by Principles III and XIV. 
 
 In the case of literal factors the second process only is available. 
 
 E.g. (a 2 6 2 ) (5 a*b»c) = 5 a 4 + 2 6 2 +» c = 5a% 5 c. 
 
 EXERCISES 
 
 Perform the following indicated multiplications, each in two 
 ways where possible : 
 
 1. (5 • 7 2 )(5 3 • 7 s ). 4. (3.4 2 )(4-3 2 ). 7. 6t(S$P). 
 
 2. (3 2 -5)(3.2 3 .5 2 ). 5. 3a6(5a 2 6 2 ). 8. 7m(4mV). 
 
 3. (4 • 5 3 )(5 • 4 3 ). 6. 4 x(3 xy). 9. 6 k 2 (6 2 tf). 
 
 By definition (§ 77), expressions in the factored form, such 
 as 3 ab, 5 a 2 b 2 , 3 • 2 3 • 5 2 , etc., are monomials. 
 
176 SPECIAL PRODUCTS AND FACTORS 
 
 The preceding examples illustrate the following principle : 
 
 129. Principle XV. The product of two monomials is 
 found by multiplying either one by each factor of the 
 other in succession. 
 
 Each factor of the multiplier is associated with any desired factor 
 of the multiplicand according to Principle III, and where the bases 
 are the same the exponents are added according to Principle XIV. 
 
 If there are no factors common to multiplier and multiplicand, the 
 product can only be indicated by writing all the factors of both in 
 succession. 
 
 Multiply : exercises 
 
 1. 4- 7 -8-9 by 2 -3 -5. 5. 3 *V by 4 ay*. 
 
 2. 3 .8- 2 by 2- 5- 6. 6. 5 a 2 b 3 c by ab*c. 
 
 3. 2 xyz by 3 z?yz. 7. 2 a^&V by 5 xb*c. 
 
 4. 6 3 • 2* • x 5 by 6 • 2 3 . x. 8. 9 x° yV by 4 x 2a y 2h z c . 
 
 9. 6 x* n+l f n ~h n by 3 x l -* n y 5 ~ n ^~ n - 
 
 10. 3 n 2x+4 m y - 5 r 21 - 1 by w^m'-V*. 
 
 In each of the following exercises state which of the Prin- 
 ciples I-XV are used : 
 
 11. 3 2 -2 3 (2 4 .3 2 -4.2 2 -2.3 2 ). 
 
 12. 6.3 4 (4 2 -2 2 .3 2 +4 3 .2 4 -5.3 4 ). 
 
 13. 2x(4: + 7x i -3x 2 y), 16. 4*y(3 <m— 4 bg +2 xy). 
 
 14. 4yz(3y 2 x 3 — fx 2 -{- y*x*) . 17. 2x a (x a — xb — 3c). 
 
 15. 5 a 2 b\a s - b 3 + a 2 b 2 ). 18. 2 yaf (y - as* + 3 y). 
 
 19. 4a^ ,+ X^ n+1 -4y 2M+5 + 5a 2n y 4 ). 
 
 20. 4 a? !n+ y ,+2m (y , 2/ m — ccy + £"* + y n ). 
 
 21. (a + by(a-b). 24. (a 3 +a 2 6 + a6 2 +6 3 )(a-6). 
 
 22. (a - 6) 3 (a + 6). 25. (a + * — c) (a+ & + c). 
 
 23. (a + 6) 2 (a-&)\ 26. (3a>-2y-l)(2a> + y). 
 
MONOMIAL FACTORS 
 
 177 
 
 27. 
 
 (5a-3&)(6a 2 + 26 2 -l). 
 
 39. (i 
 
 28. 
 
 (3a 2 -26 2 -3)(4a + 3 6 3 ). 
 
 40. (i 
 
 29. 
 
 (1 + a + « 2 ) (1 - a). 
 
 41. (: 
 
 30. 
 
 (1 - a + a 2 - a 3 ) (1 + a). 
 
 42. ( 
 
 31. 
 
 (a + &)(a-6)(a 2 + 6 2 ). 
 
 43. ( 
 
 32. 
 
 (a + 6) (a 2 -aft + 6 s ). 
 
 44. (. 
 
 33. 
 
 (a - b) (a 2 + a& + b 2 ). 
 
 45. (i 
 
 34. 
 
 (x-\-y)(x — y). 
 
 46. (i 
 
 35. 
 
 (100 + 1) (100 - 1). 
 
 47. 
 
 36. 
 
 (100 + 2) (100 - 2). 
 
 48. (i 
 
 37. 
 
 (x + 3) (cc + 5). 
 
 49. (i 
 
 38. 
 
 (a? - 3) (a? - 5). 
 
 50. (< 
 
 (t + $)(t-3). 
 (x-2y)(x + 3y). 
 (x* + x + l)(x*-x + l). 
 (a; + y) (a 3 — a; 2 ?/ + xy 2 —f). 
 (x -y) (tf+xty+xf+f). 
 (x 2 - xy + y 2 ) (x 2 + an/ + /). 
 (ar' + 2 any + */ 2 ) (a; - y) 2 . 
 (tf-y^^ + xttf + y*). 
 (x 2 + y^(x*-x 2 y 2 + y*). 
 (x 2 + y 2 ) (x + y) (x - y). 
 (x—y) (xf+xy+y 2 ) (ar'+y 5 )- 
 (a n - 6") (a 2rt -f a n 6 n .+ Z> 2 "). 
 
 FACTORS OF NUMBER EXPRESSIONS 
 
 130. The factors of numbers are of great importance in 
 arithmetic. For instance, the multiplication table consists of 
 pairs of factors whose products are committed to memory for 
 constant use. Likewise in algebra the factors of certain special 
 forms of number expressions are so important that they must 
 be known at sight. 
 
 An expression containing no fractions is said to be prime if 
 it has no integral factors except itself and 1. 
 
 Thus 2, 3, x, x + 2, a 2 + b 2 , are prime expressions. 
 
 A prime expression may be factored by using fractions or 
 radicals. See p. 214. 
 
 Thus 
 
 5 = 2-2^ and 2= V2- V2- 
 
 Such factors are not included in what are here called prime 
 factors. 
 
178 SPECIAL PRODUCTS AND FACTORS 
 
 131. Monomial Factors. If the terms of a polynomial con- 
 tain a common factor, they may be combined with respect 
 to this factor according to Principles I and II. The ex- 
 pression is thus changed into a product of a monomial and 
 a polynomial. 
 
 Illustrative Examples. 
 
 1. ax + ay = a(x + y), by Principle I. 
 
 2. a 2 — ab = a(a — b), by Principle II. 
 
 3. 9.8 + 3.4.5 = 3.4(3.2+5), by Principle I. 
 
 4. 6 a 2 b - 4 ab 2 = 2 ab (3 a - 2 b), by Principle II. 
 
 5. 5 xy — 3 x 2 y + 4 x*y = xy (5 — 3 x + 4 x 2 ), by I and II. 
 
 Observe that factoring each term of a polynomial does not 
 factor the polynomial. 
 
 E.g. 330 - 210 - 60 is not factored by writing it 2 • 3 • 5 • 11 
 — 2> 3.5.7 + 2 2 .3-5; but by adding the coefficients of the common 
 factor 2 • 3 • 5, thus, 
 
 2.3.5.11 -2- 3- 5- 7 + 2 2 .3.5 = 2. 3- 5 (11 -7 + 2). 
 
 Likewise 10 a 2 bc — 15 ab 2 c + 20«6c 2 is not factored, although each 
 term is in the factored form. But if 10 a?bc — 15 ab 2 c + 20 abc 2 is 
 written in the form 5abc (2 a — 3 b + 4 c), it is then factored. 
 
 These examples illustrate the factoring of a polynomial 
 when it contains a monomial factor common to every 
 term. 
 
 When such a common factor has been found the whole ex- 
 pression is then written in the form of a product by means of 
 Principles I and II. The result may be checked by Princi- 
 ples IV and XV. If the terms of a polynomial which is to be 
 factored contain a common factor, this should always be re- 
 moved at the outset. 
 
TRINOMIAL SQUABES 179 
 
 EXERCISES 
 
 Factor the following polynomials : 
 
 1. 8-12-18 + 48. 7. 15 xy* - 20 afy + x>y\ 
 
 2. 3-4- 5-15-20 + 35. 8. 9 v s w* + 21 v 4 w 3 - 18 v 2 w 2 . 
 
 3. 3- 11- 4 -22- 2 + 44 -6. 9. 12 a*b 3 - 8 cfb 4 - 6 a 2 b 2 . 
 
 4. 3 4 -2 4 -3-2 3 -5.2 2 . 10. 32.3 4 -64.3 3 -16-2 3 .3 2 . 
 
 5. 6.5 4 .7+3-5 3 .2-5 4 .9.2 2 . 11. 27 • 2 3 + 54 • 2 4 -36 • 2 3 . 
 
 6. 13 a*b - 16 a 3 b* - 2 a 3 b\ 12. 11 aV - 44a 3 x 4 + 33ax. 
 
 13. 1.2- 3-4 + 2- 3- 4- 5- 3- 4- 5- 6+4- 5- 6. 
 
 14. 72 x 4 b 3 a - 36 x*b 2 a 2 - 48 xWa\ 
 
 15. 84, x 9 b 7 y 4 + IS x 3 b s y* + 12 a^by. 
 
 16. 19 • 3 4 • 2 4 • 5 3 - 13 • 3 4 • 2 3 . 5 2 - 29 ■ 3 2 • 2 2 . 5 4 . 
 
 17. 17 a 4 6 3 c 2 + 51 aW - 34 aW. 
 
 18. 38 a 12 6 14 c 4 - 76 a u & 12 c 3 - 76 a M & u c». 
 
 19. 4,x 2a y Sb -^Qx 3a y 2b -8x' ia y ib . 
 
 20. 3 a 4n+2 6 3n+4 + 6 a 6n+4 6 3n+3 - 12 a 5n+3 6 5n+6 . 
 
 TRINOMIAL SQUARES 
 132. In §§ 87 and 88 we found by multiplication : 
 
 {a + b) 2 = a 2 + 2ab + b\ (1) 
 
 and (a - b) 2 = a 2 - 2 ab + 6 2 . (2) 
 
 By means of these formulas we may square the sum or dif- 
 ference of any two number expressions. 
 
 E.g. (3x+2y) 2 =(3xy+2'3x.2y+(2y) 2 =9x 2 +12xy+4;y 2 . 
 Also[(5 + r)-(s-l)] 2 =(5 + r) 2 -2(5 + r)(s-l) + (s-l) 2 . 
 
 The last expression may now be reduced by performing the 
 indicated operations. 
 
180 SPECIAL PRODUCTS AND FACTORS 
 
 In this manner write the squares of the following binomials. 
 Verify the first ten results by actual multiplication. 
 
 1. 
 
 t + 8. 
 
 11. 
 
 7(a-3)-2(6 + c). 
 
 2. 
 
 r-12. 
 
 12. 
 
 ±(x-ij) + 2(z + tf). 
 
 3. 
 
 Sx — 5y. 
 
 13. 
 
 (3a-26) + 5. 
 
 4. 
 
 6a- 7. 
 
 14. 
 
 7 x — (4 r — s). 
 
 5. 
 
 m 5 + 3». 
 
 15. 
 
 7(m 2 -3)-6(m 3 + rc). 
 
 6. 
 
 Itf + Zx. 
 
 16. 
 
 3(z + y)-2(3 + x). 
 
 7. 
 
 3tf-2y\ 
 
 17. 
 
 4(ar 5 -3y) + (ar ! + 4y 2 ). 
 
 8. 
 
 8 m 2 n — 7 win 2 . 
 
 18. 
 
 2x 2 y — b (x — y). 
 
 9. 
 
 5a 3 -36 3 . 
 
 19. 
 
 3(5x 2 -z)-4:x'z 2 . 
 
 10. 
 
 4 aWy 3 — 3 a?xy*. 
 
 20. 
 
 7(r-s) + 3(rV-2rs). 
 
 133. The binomial a + 6 is one of the two equal factors of 
 a? + 2ab + b 2 , and is called the square root of this trinomial. 
 
 Likewise a — b is the square root of a 2 — 2 ab + & 2 - 
 
 In each case a is the square root of a 2 and b of ft 2 . Hence 
 2 ab is twice the product of the square roots of a 2 and b 2 . 
 
 From the squares obtained in the last article, we learn to 
 distinguish whether any given trinomial is a perfect square, 
 as in the following examples : 
 
 1. sc 2 -}- 4 # + 4 is in the form of (1), since x 2 and 4 are squares 
 each with the sign +, and 4 a; is twice the product of the square 
 roots of x 2 and 4. Hence 
 
 a? + 4« + 4 = x 2 -f 2 (2a?) + 2 2 = (a? + 2) (x + 2) = (x + 2) 2 . 
 
 2. cc 2 — 4 a; + 4 is in the form of (2), since it differs from (1) 
 only in the sign of the middle term. Thus 
 
 ^-4^ + 4 = ar ! -2(2a0+2 2 = (as - 2) (x-2)= (x - 2) 2 . 
 
 A trinomial which is the product of two equal factors is 
 called a trinomial square. 
 
TRINOMIAL SQUARES 181 
 
 EXERCISES 
 
 Determine whether the following are trinomial squares, and 
 if so indicate the two equal factors. If any trinomial is not a 
 square, make it so by modifying one of its terms. 
 
 1. x 2 + 2xy + y 2 . 10. 64 + t 2 - 16 t. 
 
 2. x 2 -2xy-\- y 2 . 11. 16 + x 2 — 8 x. 
 
 3. x 4 + 2 arty 2 + y 4 . 12.9-6?/ + y 2 . 
 
 4. a; 4 - 2 arty 2 + y 4 . 13. 25ar ! + 16?/ 2 + 40ar?/. 
 
 5. m 2 + n 2 — 2 mw. 14. 4m 2 +« 2 +2 m». 
 
 6. r 2 + s 2 + 2rs. 15. 100 + s 2 + 20s. 
 
 7. 4a^-8^ + 4?/ 2 . 16. 64 + 49 + 112. 
 
 8. a 6 + & 6 + 2 a 3 6 3 . 17. 16 a 2 + 25 6 2 - 50 a&. 
 
 9. a 8 + 6 8 -2a 4 6 4 . 18. 16 a 2 + 25 b 2 + 40 a6. 
 
 19. 81- 270& + 225 6 2 . 
 
 134. From the foregoing examples we see that a trinomial is 
 a perfect square if it contains two terms which are squares 
 each with the sign +, while the third term, whose sign is 
 either + or — , is twice the product of the square roots of the 
 other two. Then the square root of the trinomial is the sum 
 or the difference of these square roots according as the sign 
 of the third term is + or — . 
 
 Since on multiplying we find (a — 6) 2 and (b — a) 2 give the same 
 result, we may write the factors of a 2 — 2 <xb + b 2 either (a — b) (a — b) 
 or (b — a)(b — a). See page 214. 
 
 EXERCISES 
 
 Factor the following. If any one of the trinomials is found 
 not to be a square, make it so by modifying one of its terms. 
 
 1. 9 +2-3-4 + 16. 3. 9x 2 + I8xy + 9y 2 . 
 
 2. x 2 +±y 2 +4xy. 4. 4 x 2 + 4 xy + y 2 . 
 
182 SPECIAL PRODUCTS AND FACTORS 
 
 5. 4 x 2 + 8 xy + 4 y 2 . 18. 81 a 2 - 216 a + 144. 
 
 6. 25x 2 + 12xy + 4:y 2 . 19. 4 a 2 + 8 a6 2 + 46 2 . 
 
 7. 16ar 9 + 16a*/ + 4?/ 2 . 20. 9 6 4 + 18 6 2 c 4 + 9 c 8 . 
 
 8. 9 r 2 + 36 rs + 25 s 2 . 21. 4 x 2 + 4 y 2 - 8 ay. 
 
 9. 16x 6 + 8x*y + 2/ 2 . 22. 9 a 2 - 16a6 + 4 6 2 . 
 
 10. 4a 8 + 12z 4 a 3 + 9a 4 . 23. 9z 4 - 24^6 + 166 2 . 
 
 11. a 10 + 6 a 5 6 + 9 6 2 . 24. 25 + 49 z 2 - 70a. 
 
 12. (a + l) 2 + 2(a + 1)6 + 6 2 . 25. - 30 a6 2 + 9 a 2 + 25 6 4 . 
 
 13. (z + 3) 2 +4(;c + 3)2/ + 4?/ 2 . 26. 16 a 2 - 24 a6 + 9 6 2 . 
 
 14. ^ + 12 3* + 36. 27. 36 or* - 84 a: + 49. 
 
 15. a 4 + 18 a 2 + 12. 28. 25-90 + 81. 
 
 16. 121 + 4 a 8 - 44 x 4 . 29. 64 a? 2 - 32 a; + 9. 
 
 17. 16x 4 + 642/ 4 -64^ 2 . 30. (3 + a) 2 + 6 2 -26(3+a). 
 
 31. (2 - a) 2 - 2(2 - x) (x -1) +0 - l) 2 . 
 
 32. (2 +yf + 2(2 + y)(l +4) + (1 + 4) 2 . 
 
 33. (3a-26) 2 -10(3a-26) + 25. 
 
 34. (6a-6) 2 + (2a + l) 2 -2(6a-6)(2a + l). 
 
 35. 25(a + 6) 2 + 50(a + 6) (a - 6) + 25(a - 6) 2 . 
 
 36. x? + 12 x(a + 6 + c) + 36 (a + 6 + c) 2 . 
 
 37. 49 (m - 3) 4 + 36 (m + I) 6 - 84(m - 3) 2 (m + l) 3 . 
 
 38. W(x - y) 2 - 16 (a; - y) (x + y) + 4(a; +y) 2 . 
 
 39. - 30(a + 6) (a - 6) 2 +25 (a - 6) 4 + 9(a + 6) 2 . 
 
DIFFERENCE OF TWO SQUARES 
 
 183 
 
 THE DIFFERENCE OF TWO SQUARES 
 135. By multiplication, 
 
 (a + b) (a - b) = a 2 - b\ 
 
 Translate this formula into words. By means of this formula 
 the product of the sum and difference of any two number 
 expressions may be found. 
 
 E.g. (3 a + 2 b) (3 a - 2 b) = (3 a) 2 - (2 6) 2 =9 a 2 - 4 A 2 . 
 Also (a; + y - z) (x + y + z) = [(x + y) - z] [(x + y) + z] 
 
 The last expression may now be simplified by performing 
 the operations indicated. 
 
 In this manner form the following products. Verify the 
 first ten by actual multiplication. 
 
 1. (4a + 5ft)(4a-5 6). 9. 
 
 2. (24 x + 12 y) (24 x -12 y). 10. 
 
 3. (16a 2 ft 3 -3c)(16a 2 ft 3 +3c). 11. 
 
 x m + y n ) (x m — y n ). 
 
 2n+l_i_£ > 2n-l\ (tfn+l ffn-\\ 
 
 4. (5 -6Z 2 ) (5 + 6* 2 ). 12. 
 
 5. (3x- 2y) (Sx + 2y). 13. 
 
 6. (a 8 - f) (a 3 + f). 14. 
 
 7. (Sa^-S^XSa^ + Sy 4 ). 15. 
 
 8. (>+(y-«)]0-(y-«)]. 16. 
 
 -(a-6)][c+(«-6)]. 
 
 «-(# + «)][>+& + «)]. 
 a + ft + c)(a — ft — c). 
 a — 6 + c) (a— 6 — c). 
 r-y-z) (r-y + z). 
 a + ft +c) (a + 6 — c). 
 
 17. [a + ft+(c-ef)][a+&-(c — d)]. 
 
 18. + y + (w+v)][>+?/-(w + v)]. 
 
 19. [4 x - (a - 2 ft)] [4 x + (a - 2 ft)]. 
 
 20. [ a .+ 2b-(x-y*)][a+2b+(x-y*)l 
 
 21. (11 b 3 x - 3 bx>) (11 b 3 x + 3 ftz 3 ). 
 
184 SPECIAL PRODUCTS AND FACTORS 
 
 From the preceding examples we see that every binomial 
 which is the difference between two perfect squares is com- 
 posed of two binomial factors ; namely, the sum and the 
 difference of the square roots of these squares. 
 
 E.g. 16 x 2 — 9 y 2 is the difference between two squares, (4 x) 2 and 
 (3 y) 2 . Hence we have 
 
 16x 2 -9# 2 = (4x) 2 - (3y) 2 = (4x + 3*/)(4x-3?/). 
 
 EXERCISES 
 
 Factor each of the following : 
 
 1. x'-Ay 2 . 8. a 2 -l. 15. 4:-(x-2y) 2 . 
 
 2. 9a?-36y*. 9. l-9x 4 . 16. 16a-25a& 2 . 
 
 3. x*-b 2 . 10. 4 -36 a 2 . 17. 49 a -4 a?/ 2 . 
 
 4. 4ar>-9& 8 . 11. 1-64 a 8 . 18. 225-64afy 4 . 
 
 5. 16a 4 -9Z> 4 . 12. I44z 2 & 4 -1. 19. 576 a -144a?/ 2 . 
 
 6. 64 -b 2 . 13. 256 a*b 6 -c\ 20. 5 8 -3 8 . 
 
 7. 1-& 2 . 14. l-(x + y) 2 . 21. x*-8ly 2 . 
 
 136. It is important to determine whether a given expres- 
 sion can be written as the difference of two squares. 
 
 E.g. a 2 + b 2 + 2 ab - c 2 = (a + b) 2 - c 2 = (a + b + c) (a + b - c). 
 Also, c 2 - a 2 + 2 ab - b 2 = c 2 - (a 2 - 2 aft + 6 2 ) = c 2 - (a - 6)* 
 
 = (c — a + b)(c + a — b). 
 
 EXERCISES 
 
 In the following determine whether each is the difference of 
 two squares, and if so, factor it accordingly ; if not, make it 
 so by modifying one of its terms. 
 
 1. ^-(y-z) 2 . 5. 4a 2 & 2 -(a 2 + 6 2 -c 2 ) 2 . 
 
 2. (x-yf- z \ 6. a 2 -(6 2 +c 2 + 2 6c). 
 
 3. a 2 + & 2_ 2a 5_4. 7 . (2a-5) 2 -(3a + l) 2 . 
 
 4. a? + y 2 -2xy-z 2 . 8. (3 x 2 - yf - (x + y) 2 . 
 
THE SUM OF TWO CUBES 185 
 
 9. (3a-2 6) 2 -(8a + 5&) 2 . 18. 9a 2 + 6ab + b 2 -c 2 . 
 
 10. (3 m _4) 2 -(2m + 3) 2 . 19. 16aj*-a s +4a6 -46 2 . 
 
 11. (2r + s) 2 -(3r-s) 2 . 20. a 2 -2a& + 6 2 -c 2 . 
 
 12. 81-(a + 6+c) 2 . 21. c 2 -(a 2 + 2a6 + 6 2 ). 
 
 13. x * + ^ + i_4a 2 . 22. <?-(a 2 -2ab + b 2 ). 
 
 14. a 2 -0 + 2i/) 2 . 23. (a + 6) 2 -(a-6) 2 . 
 
 15. 9 s 2 -(a -ft) 8 . 24. a 2 + 4a6 + 6 2 -a^. 
 
 16. 25m 2 - (3 r + 2s) 2 . 25. a 2 + 4a& + 45 2 - x 2 . 
 
 17. 4c 2 -(4a 2 + 12a6-9 6 2 ). 26. 9a 2 + 16 b 2 - 32ab-x*. 
 
 27. a 2 + ±ab + 4:b 2 — (x 2 — 2xy + y i ). 
 
 28. (3a;-2) 2 -(4^ + 9?/ 2 -12a^). 
 
 29. a^ + 4a;y-i/ 2 -(a 2 + 2a6 + 6 2 ). 
 
 30. 16 x A y 2 - (4 a 2 + 9/ + 12 ay). 
 
 31. (a + 6) 2 -(4a 2 + 9& 2 -12 6c). 
 
 THE SUM OF TWO CUBES 
 137. By multiplication we find 
 
 (a + 6)(a 2 -a6+6 2 ) = a 3 +6 3 . 
 
 The pupil should perform the multiplication indicated in this 
 formula and carefully note how the terms cancel in the product. 
 
 By means of this formula obtain the following products : 
 
 1. ( x + y)(x 2 -xy + y 2 ). 6. (x 2 + y 2 ) (x* - x?y 2 + y*). 
 
 2. (a + 2)(a 2 -2a + 4). 7. (5 a + b) (25 a 2 - 5 ab + b 2 ). 
 
 3. (3a + 6)(9a 2 -3a&+6 2 ). 8. (1 + oz 2 )(l -5x 2 -\-25x i ). 
 
 4. (l + 4cc)(l-4 1 r+16ar J ). 9. (1 + 2xy)(l-2xy+4:X 2 y 2 ). 
 
 5. (w> 2 +3a 2 )0 4 -3a 2 w; 2 +9a 4 ). 10. (2x+3y)(4;X 2 -6xy+9y 2 ). 
 
186 SPECIAL PRODUCTS AND FACTORS 
 
 These products show that every binomial which is the sum 
 of the cubes of two numbers is the product of two factors, one 
 of which is the sum of the numbers, and the other is the sum 
 of their squares minus their product. 
 
 E.g. (1) x* + y* = (x + y) (x 2 - xy + rf). 
 
 (2) 8 a 3 + 27 b s = (2 a) 3 + (3 b) s 
 
 = (2a + 3J)(4a 2 -2a.36+ 9 b 2 ). 
 
 (3) X« + y* = ( X 2)8 + (,,2)3 _. (X 2 + ^ (x 4 _ x y + ^4). 
 
 . Notice the difference between the trinomial x 2 — xy + y 2 and the 
 trinomial square x 2 — 2 xy + y 2 . 
 
 EXERCISES 
 
 Determine whether each of the following is the sum of two 
 cubes, and if so, find the factors ; if not, make it so by modi- 
 fying one of its terms. Check the results by multiplication. 
 
 1. tf + tf. 9. 27a^ + l. 17. 125 a? + y\ 
 
 2. a 3 + 8 6 3 . 10. 8a 3 + 27 6 3 . 18. 1 + rr 8 . 
 
 3. 27a 3 + 6 3 . 11. 8a 3 + 64& 3 . 19. 64^ + 27^. 
 
 4. 8a 3 + l. 12. wV + ar>a 3 . 20. 8 3 +10 3 . 
 
 5. 1 + 64 ar 5 . 13. 1 + 8 a?b s . 21. 1 + 729 a*. 
 
 6. 2 3 + 3 3 . 14. 64 a- 5 + 343. 22. » 6 + y 12 . 
 
 7. 125 + 729. 15. 1 + a 3 . 23. a 9 + Z> 3 . 
 
 8. 1 + 125 a 6 . 16. a 3 + 9 ft 3 . 24. 27^ + 125 s 3 . 
 
 THE DIFFERENCE OF TWO CUBES 
 138. By multiplication, we find 
 
 (a - b)(a 2 + ab + b 2 ) = a 3 - b 5 . 
 
 In performing this multiplication, note carefully how the 
 terms cancel in the product. 
 
THE DIFFERENCE OF TWO CUBES 187 
 
 By means of this formula obtain the following products. 
 
 1. (x-y^tf + xy + y-). 6. (w 2 -2a 2 )(w 4 +2io 2 a 2 + 4a 4 ). 
 
 2. (&-3)<7> 2 + 3Z>+9). 7. (3a-2&)(9a 2 +6a&+4& 2 ). 
 
 3. (2a-&)(4a 2 + 2a&+6 2 ). 8. (3a 2 -l) (9x 4 + 3a 2 + l). 
 
 4. (l-4«)(l+4aj + 16aj 2 ). 9. (1 - 2 a#) (1 + 2 ajy + 4 ofy 2 ). 
 
 5. (a 2 -6 2 )(& 4 + a 2 & 2 + & 4 ). 10 - (2a?-3y) (4aj ! + 6*y + 9jr ! ). 
 
 These products show that the difference of the cubes of two 
 numbers is the product of two factors, one of which is the 
 difference of the numbers, and the other the sum of the squares 
 of the numbers plus their product. 
 
 E.g. (1) x s - y a = (x - y)(x 2 + xy + if). 
 
 (2) 8 a 3 - 64 b* = (2 a)« - (4 6) 8 
 
 = (2a -46)(4a 2 -2a-45+ 16 b 2 ). 
 
 (3) a 9 - b 9 = (a 3 ) 3 - (i 3 ) 3 = (a 3 - i 3 )(a 6 + « 3 6 3 + &«). 
 
 Notice the difference between the factor x 2 +xy-\-y 2 , and the trino- 
 mial square x 2 +2 xy+y 2 . 
 
 EXERCISES 
 
 Determine whether each of the following is the difference of 
 two cubes, and if so, find the factors ; if not, make it so by 
 changing one of its terms. Check the results by multi- 
 plication. 
 
 1. a 8 — 6 s . 8. 64 or*-?/ 3 . 
 
 2. 8a 3 -& 3 . 9. 27 -125 a 3 . 
 
 3. a 3 -8 6 3 . 10. x 6 -y 6 . 
 
 4. 8 a 3 - 8 K, 11. 1 - x 6 . 
 
 5. 3 3 -2 3 . 12. a 9 -8. 
 
 6. 1-a 10 . 13. 1-125^. 
 7.1-8 a 3 . 14. 8-27 X s . 
 
 22. Also factor 10, 11, 19, and 20 as the difference of two 
 squares. 
 
 15. 
 
 27^-64. 
 
 16. 
 
 2*s"-l. 
 
 17. 
 
 8 x 4 - f. 
 
 18. 
 
 64 a 3 - 27 b s . 
 
 19. 
 
 1-729 a". 
 
 20. 
 
 x 6 -y 12 . 
 
 21. 
 
 27 r 3 - 125 s 3 . 
 
188 SPECIAL PRODUCTS AND FACTORS 
 
 TRINOMIALS OF THE FORM x 2 +(a + 6) x+ ab. 
 139. In §§ 82 to 85 were found such products as 
 
 (1) (x + 5)(x + 2)= x 2 + Ix + 10. 
 
 (2) (x - 5)(x - 2)= x 2 - 7 x + 10. 
 
 (3) (a? + 5)(oj - 2)= a,- 2 + Bx - 10. 
 
 (4) (a; - B)(x + 2)= ar> - 3a; - 10. 
 
 In each case the binomials to be multiplied have one term x 
 in common, and the other two terms unlike. The trinomial 
 product in each case is the square of the term common to the 
 binomials, the algebraic sum of the unlike terms times the 
 common term, and the product of the unlike terms. 
 
 Thus, the coefficient of x in (1) is 5 + 2, in (2) it is (— 5) + (— 2), 
 in (3) itis(+ 5) + (- 2), and in (4) it is (-5)+(+ 2). 
 
 The third term of the product in (1) is (+ 5)(+ 2), in (2) it is 
 (- 5)(- 2), in (3) it is (+ 5)(- 2), and in (4) it is (- 5)(+ 2). 
 
 With these points clearly in mind, such products may be 
 written out at once, without the formal work of multiplication. 
 
 EXERCISES 
 
 Find the following products by inspection : 
 
 1. (a + 7) (a +9). 9. (a?- l)(a?-+ll). 
 
 2. (6-5)(6 + 7). 10. (c-3)(c + 12). 
 
 3. (x -7) (x -17). 11. (e»-l)(c»- 5). 
 
 4. (y + 8)0 - 3). 12. (x* - 5) {x* - 3). 
 
 5- (V + 7)(y - 9). 13. (a 2 - 2)(a 2 - 4). 
 
 6- (y-7)(y-l). 14. (a 3 -l)(a 3 + 4). 
 
 7. (*-3)(aj-4). 15. (2a-l)(2a-3). 
 
 8. (x-S)(x-9). 16. (2a + 3)(2a-4). 
 
THE TYPE x 2 + (a + b)x + ab 189 
 
 17. (x-5)(x + 2). 25. (3&c + 4)(3 6c-7). 
 
 18. (3 a -7) (3 a + 2). 26 . (100 + 3) (100 + 5). 
 
 19. (a 2 -8)(a 2 -4). 2? (2 a « + &)(2 a « + c). 
 
 20. (&3-3)(6 3 -4). 
 
 21. (2a? + a)(2a;+6). V ' A ' 
 
 22. (5c-4)(5c + 5). 29 ' (* + «)<*-»)• 
 
 23. O»y+0)(ajy-7). 30 - t? r- «)(»-. &)• 
 
 24. (l-3a)(l-2a). 31. (3x* + 2 2/)(3^-52/). 
 
 140. It is possible to recognize such products at sight, and 
 thus to find the factors by inspection. 
 
 Illustrative Examples. Determine whether the following tri- 
 nomials are of the kind just considered : 
 
 1. x 2 + 7 x + 12. The question is whether two numbers can- 
 be found such that their sum is + 7 and their product 12. 
 The numbers 3 and 4 answer these conditions. Hence, 
 
 x 2 + 7 x + 12 = (a; + 3) (x + 4). 
 
 2. x 2 — 5 x — 14. Since the product of the numbers sought is 
 — 14, one number must have the sign — and the other -{-; and 
 since their sum is —5, the one having the greater absolute 
 value must have the sign — . Hence, the numbers are — 7 and 
 + 2, and we have x 2 — 5x— 14=(# — 7)(x + 2). 
 
 3. x 2 -7 x + 12 = (x-3)(x -4). Since (-3)(-4)= + 12 
 and (-3) + (-4) = -7. 
 
 4. x 2 + 4:X-12 = (x + 6)(x-2). Since (+6)(-2)=-12 
 and(+6) + (-2) = + 4. 
 
 It is to be noted that it is not always possible to find inte- 
 gers to fulfill these two conditions. 
 
 E.g. given x 2 + 5x + 3. By inspection, it is easily seen that there 
 are no two integers such that their sum is + 5 and their product -f 3. 
 
190 SPECIAL PRODUCTS AND FACTORS 
 
 EXERCISES 
 
 Determine whether each of the following trinomials can be 
 factored by inspection, and if so, find the factors ; if not, mod- 
 ify one term so as to make such factoring possible. 
 
 1. ^ + 3^ + 2. 11. b 2 + 8b + 15. 21. z 4 + 18ar° + 77. 
 
 2. tf + x-6. 12. b 2 -b-56. 22. a 4 -5a 2 -104. 
 
 3. ^_a;_6. 13. b 2 + b-5S. 23. a 2 +32a + 240. 
 
 4. x 2-6 x +8. 14. c 2 -3c-15. 24. a 4 -lla 2 + 28. 
 
 5. a2 + 6a+8. 15. rf-lSx + m. 25. a 4 -lla 2 -60. 
 
 6. 3^-3^ — 8. 16. x 2 + 15a;-54. 26. a 2 -14 a — 51. 
 
 7. a,* 2 + 2 a -8. 17. 0^-14 # — 95. 27. a 2 -3a-54. 
 
 8. a 2_4 a _32. 18. y 2 + 21 y + 98. 28. » 4 -8a^-32. 
 
 9. a 2 + 4a-32. 19. y 2 -7y-98. 29. a 6 -3a 3 -154. 
 10. 6 2 + 15 6 + 56. 20. a 2 -19 a; + 78. 30. a?-10x + 25. 
 
 31. a 2 6 6 - 13 a& 3 - 30. 41. 9 a 2 + 24 a + 16. 
 
 32. x 2 - 17 xyz + 72 y 2 z 2 . 42. 81 a 2 -99 a + 30. 
 
 33. r 8 + 6 r 4 s - 91 s 2 . 43. # 2 + 26^ + 133. 
 
 34. a 4 c 4 + 9a 2 c 2 -162. 44. x 2 + 5xy-8±y\ 
 
 35. a 2 + lla-210. 45. *+3r— 154. 
 
 36. m 4 + 4mV + 4w 2 . 46. u 2 + 38 uv + 165 v 2 . 
 
 37. sV-Wst-te. 47. (a + 6) 2 -19<a + &) + 88. 
 
 38. a 2 b 2 - 27 ab + 26. 48. (z-?/) 2 -14(a;-y) + 40. 
 
 39. Z 2 + 13Z + 42. 49. (r-s) 2 -17(r-s) + 60. 
 
 40. «Y — 11 xy — 180. 50. x 2 + (a + 6) x + a&. 
 
THE TYPE atf + bx + c 191 
 
 141. Trinomials of the form ax* + bx + c. 
 
 Ex. Find the product of 2 x + 5 and 3 x + 2. 
 
 2z + 5 
 3a + 2 
 
 6^ + 15* 
 
 4z+10 
 6a 2 + 19a + 10 
 
 The products 3 a; • 2# and 2 • 5 are called end products and 
 2 • 2 a; and 5 • 3 x are called cross products. In this case the 
 cross products are similar with respect to x and are added. 
 Hence the final product is a trinomial two of whose terms are 
 the end products and the third term is the sum of the cross 
 products. 
 
 This fact enables us to write such products at once. 
 
 Kg. (3a + 4)(5a-7)=15cr-a-28. 
 
 In this case 15 a 2 is the first end product and — 28 the second, while 
 — a is the sum of the two cross products, 20 a and — 21 a. 
 
 In this manner obtain the following products : 
 
 1. (2a + 3)(a + 3). 11. (3<-5)(* + 4). 
 
 2. (4o-l)(3a + 2). 12. (5x-y)(2x + 3y). 
 
 3. (2 a + 5) (a -7). ' 13. (3x-2y)(x + 3y). 
 
 4. (7r + 8)(3r-6). 14. (4a-3y)(a+jf). 
 
 5. (2 x + 8) (9 x -4). 15. (3r-2s)(2r + s). 
 
 6. (3 m — l)(4m + 3). 16. (5m-n)(2m + »). 
 
 7. (5s-7)(2s-4). 17. (5a + 3x)(3a-±x). 
 
 8. (2aj-l)(7* + 4). 18. (4a-56)(a + 3&). 
 
 9. (4n-9)(5w-7). 19. (3a + 5 6)(a-6). 
 10. (&y-l)(5y + ll). 20. (3 c- 7 d)(2 c + 3d). 
 
192 SPECIAL PRODUCTS AND FACTORS 
 
 142. Trinomials in the form of the above products may some- 
 times be factored by inspection. 
 
 Ex. 1. Factor : 5 x? + 16 x + 3. 
 
 If this is the product of two binomials they must be such that the 
 end products are 5x 2 and 3 and the sum of the cross products 16 x. 
 
 One pair of binomials having the required end products is 5x + 3 
 and x + 1, another is 5 x — 1 and x — 3, and still another 5 x + 1 and 
 ar + 3. 
 
 The sum of the cross products in the first pair is 8 x, in the second 
 pair — 16x, and in the third pair 16 x. Since the sum of the cross 
 products in the last pair is the one required, the factors are 5x+l 
 and x + 3. 
 
 Ex. 2. Factor: 6 a 2 - 5 a -4. 
 
 In this case as in the one preceding there are several pairs of bi- 
 nomials whose end products are 6 a 2 and —4, such as 2 a — 2 and 3 a + 2, 
 6 a— 1 and a + 4, etc. By trial we find that among these 3 a — 4 and 
 2 a + 1 is the only pair the sum of whose cross products is — 5 a. 
 Hence 6 a 2 - 5 a - 4 = (3 a - 4)(2 a + 1). 
 
 Obviously not every trinomial of this form can be factored in this 
 manner. Thus, for example, in 6 a 2 + 5 a + 4, no pair of binomials 
 whose end products are 6 a 2 and 4 has the sum of its cross products 5 a. 
 
 In this manner factor the following : 
 
 1. 3^+5^ + 2. 9. 5r*+18r-& 
 
 2. 9a 2 + 9a + 2. 10. 14 a -2 - 39 a + 10. 
 
 3. 2** +11* +12. 11. 5x 2 + 26a-24. 
 
 4. 9z 2 + 36z + 32. 12. 2a 2 -5a + 2. 
 
 5. 2 a* -a; -28. 13. 2rft 2 -m-3. 
 
 6. 12.s 2 + lls + 2. 14. 7c 2 -3c-4. 
 
 7. 6t 2 + 7t-3. 15. 5a; 4 + 9^-18. 
 
 8 . 6^-3-2. 16. 7a 4 + 123a 2 -54. 
 
FACTORS FOUND BY GROUPING 193 
 
 17. 6c 2 -19c + 15. 24. 6-5a;-4iB 2 . 
 
 18. 3a 2 -21a + 30. 25. 3h 2 -13h + U. 
 
 19. 6^ + 4^-2. 26. lor 2 -r-2. 
 
 20. 20a 2 -a-99. 27. 2t 2 + llt + 5. 
 
 21. 12c 2 + 25c + 12. 28. 10-5X-15X 2 . 
 
 22. 8 + 6a-5a 2 . 29. 5 x 2 - 33 x + 18. 
 
 23. 15-507-10^. 30. 20-9a-20ar ! . 
 
 FACTORS FOUND BY GROUPING 
 143. Besides the methods of factoring which have been 
 applied to the types of expressions thus far considered, there 
 are various other processes which will be considered in the 
 Advanced Course. One other method of general application 
 will suffice here. 
 
 Ex. 1. Find the factors of ax + ay -f- bx + by. 
 
 By Principle I, the first two terms may be added and also 
 the last two. 
 
 Thus, ax + ay + bx + by = a(x + y) + b(x + y). 
 
 The given expression has thus been changed into the sum of 
 two compound terms which have a common factor, (x + y), and 
 may be added with respect to this factor by Principle I. 
 
 Thus, a{x + y) + b(x + y) = (a + b)(x+y). 
 
 Hence, ax -{-ay + bx-\-by = (a-\-b)(x-\-y). 
 
 Ex. 2. Factor ax — ay — bx + by. 
 
 Combining the first two terms with respect to a and the 
 second two with respect to — b, we have, 
 
 ax — ay — bx + by = a(x — y) — 6 (x—y). 
 
194 SPECIAL PRODUCTS AND FACTORS 
 
 Again combining with respect to the factor x — y, 
 ax—ay — bx-\-by=(a — b)(x —y). 
 
 The success of this method depends upon the possibility of 
 so grouping and combining the terms as to reveal a common 
 compound term, § 77. The order of the terms in the given 
 polynomial may be changed as found desirable. 
 
 EXERCISES 
 
 Factor the following expressions by means of Principles I 
 and II : 
 
 1. aft 2 + ac 2 — db 2 — dc 2 . 11. 2 n 2 — en + 2nd — cd. 
 
 2. 6 ms— 15 nt + 9 ns — 10 mt. 12. 5 ax — 15 ay — 3 bx + 9 by. 
 
 3. 8 ax — 10 ay + 4 bx — 5 by. 13. 3xa — 12 xc — a + 4 c. 
 
 4. 2 a 2 + 3 a& — 14 an — 21 wfc. 14. 3 xy — 4 wm -f- 6 my — 2 xn. 
 
 5. ac + 6c + ad + 6d*. 15. 1 mn + 1 mr — 2n 2 — 2nr. 
 
 6. ax 2 — ftcc 2 — ay 2 + 6y 2 . 16. a — 1 + a 3 — a 2 . 
 
 7. 8 ac - 20 ad- 6 6c +15 bd. 17. 3s + 2 + 6s 4 +4S 3 . 
 
 8. 2ax — Q>bx + Zby—ay. 18. as 2 — 3 bst — ast + 3 bt 2 . 
 
 9. 5 + 4 a — 15 c — 12 ac. 19. 3 m?i + 6m 2 — 2 am — an. 
 10. 156-6-206c + 8c. 20. 2ar + 2as + 26r + 26s. 
 
 MISCELLANEOUS EXERCISES 
 
 Classify the following expressions according to the fore- 
 going types for factoring, and indicate the factors : 
 
 l.<o? + 5z+6. 5. 4:X 2 + 9y 2 -12xy. 
 
 2. 1 — X s . 6. 5 ar> + 4aa; + 7 xy. 
 
 3. a? + 11 ai + 30. 7. 2w 2 — 6wc-3wy + 9cy. 
 
 4. 4ar J + 9y 2 + 12:ry. 8. 4x- 2 -# 2 . 
 
MISCELLANEOUS FACTORING 195 
 
 9. a 3 + 6 3 . 35. 3(a+l) 3 +4(a+l) 2 +a + l. 
 
 10. 9x 2 + y i + Qxy 2 . 36. (x + a) 2 - (x— a) 2 . 
 
 11. 2y 2 a s + 4:ya 2 -8ya. 37. 15 m 2 + 224 m - 15. 
 
 12. ^ + 7^+6. 38. 3.^4- 27 a +42. 
 
 13. 9 0^ + 36^ + 36 a*/ 2 . 39. cc 4 + 49 a 2 + 14 aa 2 . 
 
 14. 9y-9z-2xy + 2xz. 40. 27 a 6 -aV. 
 
 15. a'-l. 41. S^ + aV. 
 
 16. a 4 + 6 4 + 2a 2 6 2 . 42. 8M-406e + 3cd-15ce. 
 
 17. a 4 -25. 43. a 2 -11 a + 30. 
 
 18. 27 a 3 -125. 44. 0-+2) 2 -4(a-2) 2 . 
 
 19. 4a 2 + 4a6 + a& 2 . 45. x* + 9y 2 -6yx?. 
 
 20. 4 a 2 + 9 a 4 - 12 ax 2 . 46. 4a 2 -7ca 2 - 4d 2 + 7cd* 2 . 
 
 21. 1 + ar 5 . 47. a 2 + 15a -16. 
 
 22. 2x 2 + 5a + 3. 48. 18-27c + 166-24c6. 
 
 23. 36 + 4 a 6 +24 or 5 . 49. 4-(a 2 + & 2 -2a&) 2 . 
 
 24. 0_1)2_0 + 1) 2 . 50. 10r + 36s-66r-5s. 
 
 25. 8 + 64 a 6 . 51. 25 + 64^ + 80a 3 . 
 
 26. a,c — ax — 4&c + 4&c. 52. 1000 — X s . 
 
 27. 27 -216 a 3 . 53. 10 3 + ar*. 
 
 28. 3 3 + 6 3 a 3 . 54. 8a 3 + a 3 & 3 + & 2 a 2 . 
 
 29. 25(x + l) 2 -4. 55. 100 -49a 4 . 
 
 30. 5cx-10c + 4:dx — 8d. 56. 100 + 625 + 500. 
 
 31. 4t{x+2) 2 +y 2 + ±{x + 2)y. 57. a 2 -17a + 72. 
 
 32. ra + 2 r/t - 5 sa- 10 sA. 58. a 2 + 17 a + 72. 
 
 33. -2a 2 6 + a 4 + 6 2 . 59. a 2 +166 2 -8a6. 
 
 34. 2/ia — hb+ 6 a- 36. 60. x s — y 9 . 
 
196 SPECIAL PRODUCTS AND FACTORS 
 
 61. 24 a 2 + 37 a -72. 87. 24 a 4 c 4 + a 6 + 144 c 8 a 2 . 
 
 62. (8* + 15s 2 — 100. 88. 9s 2 + 4?/ 4 -12s?/ 2 -16. 
 
 63. 9« + 8 3 . 89. 81 + 100s 8 - 180 s 4 . 
 
 64. 9s 4 + 16?/ 2 + 24sy 90. a 4 + 27a 2 + 180. 
 
 65. 1-1000. 91. a 4 + 3a 2 -180. 
 
 66. 16a 2 6 2 +24a& + 366 3 . 92. a 4 -3 a 2 -180. 
 
 67. 64 + 8. 93. 144-(a 4 + fc 2 -2a 2 6). 
 
 68. 16a 2 6 2 + 9aV + 24a 2 6c. 94. 81 a 2 6 4 + 49 c 2 - 126 ab 2 c. 
 
 69. a 2 + 46 2 + 4a&-4s 2 . 95. 12 s 2 - 23 st + 10 t 2 . 
 
 70. a s b 6 + c?. 96. 36 s 4 + 12 s 2 ?/ 4 + ?/ 8 . 
 
 71. 5^ + 10s 3 ?/ 2 + 30 or 5 ?/ 4 . 97. 16s 6 + 9y 4 + 24sy~49. 
 
 72. 16aV + 4c 2 s 2 +16ac 2 s. 98. y 2 + 35y + 300. 
 
 73. a 6 y 3 -z\ 99. 5 ?/ 2 -80?/ + 300. 
 
 74. s 4 - 7s 2 - 120. 100. 100 - (16s 2 + ?/ 6 - 8s?/ 3 ). 
 
 75. 9a 4 & 2 — 12a 3 & + 4a 2 . 101. ac—bc + ad—bd. 
 
 76. 8ao + 27a& 7 . 102. 6 rd -15 re +22 cd-55ce. 
 
 77. s 4 + 4s 2 + 4 — s 6 . 103. z 3 + ya — ifz 3 — ay*. 
 
 78. l-125a 3 6 6 . 104. s 4 + 2s 2 + l-s 2 . 
 
 79. 16 + 16a6+4a 2 6 2 . 105. 60 s 2 + 7 s?/ - ?/ 2 . 
 
 80. 64a 3 + 8a 2 & 3 . 106. s 2 - 20 s?/ + 75 ?/ 2 . 
 
 81. 36a 2 6 4 + o 2 6 4 + 12 ab*c. 107. s 2 - 17s -60. 
 
 82. 4a 2 + 9& 4 + 12a6 2 -16a 4 . 108. 65^ + 8 r-1. 
 
 83. s 6 + 17s 3 + 30. 109. a 2 -13a -140. 
 
 84. 25-(a 4 -2a 2 Z> 3 + & fi ). 110. 39s 4 -16s 2 + l. 
 
 85. -112a 2 c 3 + 49a 4 + 64c 6 . 111. 625- (31 -4a 2 ) 2 . 
 
 86 . a 2_ a _380. 112. 36a 2 -29a& + 5 6 2 . 
 
EQUATIONS SOLVED BY FACTORING 197 
 
 113. c 4 - 31c 2 + 220. 115. 26 + 39w-22m-33mw. 
 
 114. ac + d 5 a— b 4 c-b*d 5 . 116. 12 a 2 + 11 a - 56. 
 
 117. a 2 + 4a&+46 2 -(a 2 -4a& + 4& 2 ). 
 
 118. (3a;-l) 2 -(a,' 2 + 4^-4^) 2 . 
 
 119. (x + 3 y y+(x-2 y y +2(x + 3y)(x-2y). 
 
 120. 16 (a + b) 2 - 8 (a -6) (a + 6) + (a - b)\ 
 
 121. 256a,- 2 - (49 x 2 + 4?/ 4 - 28a?/ 2 ). 
 
 122. (2a -a) 2 +100 (a -3a) 2 +20 (2a - a) (a - 3a). 
 
 123. -48(a-6)(a+6) + 36(a-&) 2 + 16(a+6) 2 . 
 
 EQUATIONS SOLVED BY FACTORING 
 
 144. Illustrative Problem. There are two consecutive num- 
 bers the sum of whose squares is 61. What are the numbers ? 
 
 Solution. Let x = one of the numbers, then x + 1 is the other. 
 Hence, x 2 + (x + l) 2 = 61 (1) 
 
 By F, x 2 + x* + 2x + 1 = 61 (2) 
 
 By F, I, S, 2x 2 + 2x =60 (3) 
 
 By D, x 2 + x = 30 (4) 
 
 These equations differ from any which we have studied here- 
 tofore in that they contain the squares of the unknown num- 
 ber, which cannot be removed by addition or subtraction. 
 
 It is evident on inspection that a = 5 satisfies equation (4). 
 
 That is, 5 2 + 5 = 30. 
 
 Hence 5 is one of the numbers sought, and 5 + 1 is the other, and 
 these numbers satisfy the conditions of the problem, 
 
 Since 5 2 + 6 2 = 25 + 36 = 61. 
 
 145. Definition. Equations which involve the second but 
 no higher degree of the unknown number are called quadratic 
 equations. 
 
198 SPECIAL PRODUCTS AND FACTORS 
 
 One method of solving quadratic equations is now to be 
 considered. 
 
 By S, equation (4) above may be written 
 
 x 2 + x - 30 = 0. (5) 
 
 Factoring the left member, 
 
 (x + 6) (x - 5) = 0. (6) 
 
 Equation (6) is satisfied by any value of x which, substituted in 
 the left member, reduces it to zero ; and, since the product of two fac- 
 tors is zero if either factor is zero, we seek values of x which make 
 x — 5 = and also x + 6 = 0. Hence the equation is satisfied by 
 x = 5 and also by x = — 6. Thus, 
 
 (5 + 6) (5 - 5) = 11 • ss 0, 
 and also (- 6 + 6) (- 6 - 5) = . - 11 =0. 
 
 Therefore — 6 and — 6 + 1 = — 5 are two numbers which meet 
 the condition of the problem. 
 
 That is, (- 6) 2 + ( - 5) 2 = 36 + 25 = 61. 
 
 Hence this problem has two solutions, namely, the numbers 5 and 6 
 and the numbers — 6 and — 5. 
 
 Illustrative Problem. A rectangular flower bed 10 feet long 
 and 6 feet wide is surrounded by a gravel walk whose area is 
 192 square feet. How wide is the walk ? 
 
 Solution. Let the width of the walk be x feet, then 
 
 2 • 10 x is the area of the sides, 
 2 • 6 x is the area of the ends, 
 4 x 2 is the area of the corners. 
 
 Hence the total area is 
 
 4x 2 + 2-10x + 2.6x = 192. (1) 
 
 By F, 4x 2 +32x = 192. (2) 
 
 ByZ>, x 2 + 8x=48. (3) 
 
EQUATIONS SOLVED BY FACTORING 199 
 
 By S, x 2 + 8 x - 48 = 0. (4) 
 
 Factoring, (x + 12) (x - 4) = 0. (5) 
 
 But (5) is satisfied if x + 12 = 0, and also if x — 4 = 0. (6) 
 
 Hence x = — 12 and also x = 4. (7) 
 
 Check by substituting in equation (3) : 
 
 ( - 12) 2 + 8 (- 12) = 144 - 96 a 48. 
 Also 4 2 + 8 • 4 = 16 + 32 = 48. 
 
 Hence the quadratic equation to which this problem gives rise has 
 two solutions, but since the width of a walk cannot be a negative 
 number, only the number 4 satisfies the conditions of the problem. 
 
 Ex. 1. Solve the quadratic equation 
 
 5a 2 + 30z + 3 = 3-5». (1) 
 
 By A, S, 5x 2 + 35x = 3-3 = 0. (2) 
 
 Factoring, 5 x (x + 7) = 0. (3) 
 
 But (3) is satisfied if 5 x = 0, and also if x + 7 = 0. (4) 
 
 Hence x = 0, and also x = — 7. (5) 
 Check. Substitute x = and also x = — 7 in (1). 
 
 Ex. 2. Solve the quadratic equation 
 
 6x 2 + llx = 10. (1) 
 
 By S, 6x 2 +llx-10 = 0. (2) 
 
 Factoring, (3 x - 2) (2x + 5) = 0. (3) 
 
 But (3) is satisfied if 3 x — 2 = and also if 2 x + 5 = 0. 
 Hence x = § and x = — |. 
 
 Check by substituting each of these values in (1). 
 
200 SPECIAL PRODUCTS AND FACTORS 
 
 Ex. 3. Solve the quadratic equation 
 
 3z 2 -5a;-7 = 2a: 2 -a;-ll. (1) 
 
 By .4andS, x 2 -4x + 4 = 0. (2) 
 
 Factoring, (x - 2) (x - 2) = 0. (3) 
 
 It follows from the first factor and also from the second that (3) is 
 satisfied if x = 2. 
 
 In this case the two solutions turn out to be identical, while in 
 Example 1 one solution was zero and the other was — 7. 
 
 If we count each of these results as two solutions, then for every 
 quadratic equation thus far solved we have found two values of the 
 unknown number. 
 
 146. The method of solution above explained consists of 
 three steps : 
 
 (1) Transform the equation so that all terms are collected 
 in one member, with similar terms united, leaving the other 
 member zero. This can always be done by Principle VIII. 
 It is convenient to make the right member zero. 
 
 (2) Factor the expression on the left. 
 
 (3) Find the value of x which makes each of these factors 
 zero. This is readily done by setting each factor equal to 
 zero and solving it for the unknown. 
 
 EXERCISES 
 
 Find two solutions for each of the following quadratic 
 equations : 
 
 1. ar , -3» + 2 = 0. 6. a 2 + 3a = 10 a + 18. 
 
 2. x , + 7« = 30. 7. a 2 + 10a=-24-4a. 
 
 3. a 2 -lla=-30. 8. 2z 2 -6:r=-40+12a;. 
 
 4. a 2 + 13 a = 30. 9. 3x + x 2 = 20^-72. 
 
 5. a 2 + 10a + 8 = -3a-34. 10. 17a + 30= -^-40. 
 
EQUATIONS SOLVED BY FACTORING 201 
 
 11. 7aj» + 2a; = 30a>-21. 21. x 2 + 12 a + 6 = 5a-4. 
 
 12. lla + 3x 2 = 20. 22. 2a 2 -7a=60 + 7tf. 
 
 13. l§-5x+x 2 = -2x*-20x-2. 23. 60 a; + 4 ^ + 144 = 8 a. 
 
 14. or* -16 = 0. 24. 18 se = 63 - x 2 . 
 
 15. ^-1=0. 25. 24x 2 = 12a + 12. 
 
 16. x 2 — x = 0. 26. 2 a; = 63— x 2 . 
 
 17. x 2 + a- = 0. 27. 22 x + x 2 = 363. 
 
 18. 4 a 2 = 25. 28. 3a^ + 7a; = 6. 
 
 19. ar J + 4a; + 4 = 0. 29. 2^ = 2- 3». 
 
 20. ar J + 8a; + 16 = 0. 30. x-2 = -3x 2 . 
 
 147. It is sometimes possible to solve other equations than 
 quadratics by the above process. 
 
 Ex. 1. Solve the equation : 
 
 a 8 + 30 x = 11 x 2 . (1) 
 
 By S, x»-llx 2 +30x = 0. (2) 
 
 By §131, x(x 2 -llx + 30) = 0. (3) 
 
 B\ § 139, x (x - 5) (x - 6) = 0. (4) 
 
 (4) is satisfied if x = 0, if x — 5 = 0, and if x — 6 = 0. 
 Hence the solutions are x = 0, x = 5, x = 6. 
 
 Ex. 2. a>(» + l)(a;-2)(a;+3) = 0. 
 
 Any value of x which makes one of these factors zero re- 
 duces the product to zero and hence satisfies the equation. 
 Hence the solutions of the equation are found from 
 
 x = 0, a; + 1=0, x — 2 = 0, and # + 3 = 0. 
 
 That is x = 0, x= — 1, x = 2, x= — 3 are the values of x 
 which satisfy the equation. 
 
 Notice that this process is applicable only when one member 
 of the equation is zero and the other member is factored. 
 
202 SPECIAL PRODUCTS AND FACTORS 
 
 EXERCISES 
 
 Solve the following equations by factoring : 
 
 1. x 3 — x 2 = 6x. 8. x 2 — ax — bx + ab = 0. 
 
 2. 5a; = 4ar ! + ar J . 9. 4 ( x _ 2) 2 _ (a? + 3) 2 = 0. 
 
 3. X s — 25 a = 0. 10. ar'-aic + fo; — ab = 0. 
 
 4. X s — 3x2=— 2 x. 11. x 2 + ax — bx — ab = 0. 
 
 5. 3 ^ = 15 or* + 42 a. 12. 9(> + 2) 2 -4 (z-3) 2 = 0. 
 
 6. 5 a 3 + 315 a = 80 a 2 . 13. ^-aj-3^ + 3 = 0. 
 
 7. a* + ax + &a; + a& = 0. 14. or 5 -4a;- 8 a 2 + 32 = 0. 
 
 15. 5 ^ + 120 a 2 = 119 a -2a^ + 8a^. 
 
 16. (a?+6aj-16)(a>*-7a; + 12) = 0. 
 
 17. (aj + 7)(a 2 -T7a; + 72) = 0. 
 
 18. (a?-16)(« 8 + llaj 8 +30aj) = 0. 
 
 148. Illustrative Problem. The paving of a square court 
 costs 40^ per square yard and the fence around it costs $ 1.50 
 per linear yard. If the total cost of the pavement and the 
 fence is $ 100, what is the size of the court ? 
 
 Solution. Let x — the length of one side in yards. 
 Then 40 x* = cost in cents of paving the court, 
 
 150 • 4 x = 600 x = cost of the fence in cents. 
 
 40 x 2 + 600 x = 10000. (1) 
 
 By A x 2 + 15 a; = 250. (2) 
 
 By S, x* + 15s - 250 = 0. (3) 
 
 Factoring, (x - 10) (x + 25) = 0. (4) 
 
 Whence, x = 10, and also x = — 25. (5) 
 
 It is clear that the length of a side of the court cannot be — 25 
 yards. Hence 10 is the only one of these two results which has a 
 meaning in this problem. 
 
 It happens frequently when a quadratic equation is used to 
 solve a problem that one of the two numbers which satisfy this 
 equation will not satisfy the conditions of the problem. 
 
EQUATIONS SOLVED BY FACTORING 
 
 203 
 
 PROBLEMS 
 
 In each of the following problems find all the solutions pos- 
 sible for the equations and then determine whether or not each 
 solution has a reasonable interpretation in the problem. 
 
 1. The dimensions of a picture inside the frame are 12 by 16 
 inches. What is the width of the frame if its area is 288 
 square inches ? 
 
 2. There are two consecutive integers such that the sum of 
 their squares is 3961. What are the numbers ? 
 
 3. An open box is made from a square piece of tin by cutting 
 out a 5-inch square from each corner and turning up the sides. 
 How large is the original square if the box 
 contains 180 cubic inches ? 
 
 If x — length of a side of the tin, then the volume of 
 the box is : 5(x - 10) (x - 10) = 180. (See the figure.) 
 
 4. A rectangular piece of tin is 4 inches 
 longer than it is wide. An open box contain- 
 ing 840 cubic inches is made by cutting a 6-inch square from 
 each corner and turning up the ends and sides. What are 
 the dimensions of the box ? 
 
 5. A farmer has a rectangular wheat field 160 rods long by 
 80 rods wide. In cutting the grain, he cuts a strip of equal 
 
 width around the field. 
 How many acres has 
 he cut when the 
 width of the strip is 
 8 rods ? 
 
 6. How wide is the 
 strip around the field 
 of problem 5, if it con- 
 tains 27^ acres ? 
 7. In the northwest a farmer using a steam plow starts plowing 
 around a rectangular field 640 by 320 rods. If the strip plowed 
 the first day lacks 16 square rods of being 24 acres, how wide is it ? 
 
 160 
 
 80 
 
 X 
 1 
 
 8 
 
 X 
 
 *(l60-2a:) 
 160-2* 
 
 \x ** 
 
 X 
 
 H 
 M 
 
 8 
 
 160-2* 
 
 f 
 
 8 
 
 
 
 
 .so 
 
 160 
 
204 
 
 SPECIAL PRODUCTS AND FACTORS 
 
 8. A rectangular piece of ground 840 by 640 feet is divided 
 into 4 city blocks by two streets 60 feet wide running through 
 
 840 feet it at right angles. How many square 
 
 feet are contained in the streets ? 
 
 9. A farmer lays out two roads 
 through the middle of his farm, one 
 running lengthwise of the farm and the 
 other crosswise. How wide are the 
 
 roads if the farm is 320 by 240 rods, and the area occupied 
 
 by the roads is 1671 square rods ? 
 
 QUADRATIC AND LINEAR EQUATIONS 
 
 149. When two simultaneous equations are given, one quad- 
 ratic and one linear, they may be solved by the process of sub- 
 stitution, which was used (§ 116) in the case of two linear 
 equations. 
 
 Illustrative Example. Solve the equations : 
 r x 1 — y' 1 = — 16. 
 I a? — 3 2/ = — 12. 
 
 From (2) by S, x = 3 y - 12. 
 
 Substituting (3) in (1), (3 y - 12) 2 - y 2 = - 16. 
 From (4) by F, 9 y 2 - 72 y + 144 -y 2 =- 16. 
 
 From (5) by F, A, 8 y 2 - T2 y + 160 = 0. 
 
 By A y 2 - 9 y + 20 = 0. 
 
 Factoring, (y — 5)(y — 4) = 0. 
 
 Hence, y = 5, and y = 4. 
 
 Substitute y = 5 in (2) and find x = 3. 
 
 Substitute y = 4 in (2) and find x = 0. 
 
 Therefore (1) and (2) are satisfied by the two pairs of values, 
 
 x = 3, y — 5, and x — 0, y — 4. 
 
 Check by substituting these pairs of values in (1) and (2). 
 
 (1) 
 (2) 
 (3) 
 (4) 
 (5) 
 (6) 
 (7) 
 (8) 
 (9) 
 
QUADRATIC AND LINEAR EQUATIONS 205 
 
 EXERCISES 
 
 In the manner just illustrated solve the following : 
 \x+2y = 8, \x-y=-l, 
 
 2. 
 
 3. 
 
 5«« + 12?/ 2 = 128. Ua; 2 + 3i/ 2 = 147. 
 
 1^ + ^ = 1. ' Ix 2 — 52/ 2 = 4. 
 
 f 2 a? — y = 6, 13 fas-yssi, 
 
 U^ + 5?/ 2 = 36. ' 1 3 as 2 — 2 2/ 2 = — 5. 
 
 fa; + 3y = 6, J5z-7y=-28, 
 
 la 2 + 3/ =12. ' 115 a 2 + 49 1/ 2 = 784. 
 
 a; _2 2/= -2, j6z-7*/ = 18, 
 
 x 2 -6?/ 2 = 10. " 1 36 ^-7^ = 324. 
 
 .8* -16?/ = -120, 16 Ja-9y = 2 
 
 7. 
 
 \x-\ 
 7 x* + 2 y 2 = 585. ' Ick 2 - 45^ = 4. 
 
 j7a + 9 ? / = 88, f« + y=8, 
 
 17^4-9^ = 736. ' ll3» 2 + 32/ 2 = 160. 
 
 8 ix-y = 6, \2x-5y=-16, 
 
 U 2 -7?/ 2 =36. ' 14^ + 15/ =256. 
 
 |3a;4-22/ = 7, J7a + 4y = 7, 
 
 l3^ + 8y 2 =35. ' l49a 2 -8?/ 2 = 49. 
 
 f»-8y«-ll, fa»-3y^-12, 
 
 * 1 3 cc 2 - 16/ = 11. ' lx 2 -y 2 = -16. 
 
 150. Illustrative Problem. The fence around a rectangular 
 field is 280 rods long. What are the dimensions of the field, 
 if its area is 30 acres ? 
 
 Solution. Let w = the width of the field in rods, 
 
 and I sr length of field in rods. (1) 
 
 Then 2 1 + 2 w = 280, 
 
 and lw — 4800 (one acre = 160 square rods). 
 
 (2) 
 
206 SPECIAL PRODUCTS AND FACTORS 
 
 From (1), I + w = 140, and w = 140 - I. 
 
 Substituting this value of w in (2), 
 
 Z(140 - 0= 4800, 
 or, Z 2 -U01 + 4800 = 0. 
 
 Then, (I - 60) (/ - 80) = 0. 
 
 Whence 60 and 80 both satisfy the equation. 
 If I = 60, then from equation (1) w = 80, and if I = 80, then w = 60. 
 
 We group these pairs of numbers as follows : 
 
 | / = 60 >and M = 80, 
 I w = 80, I w = 60. 
 
 Substituting these pairs of values in both (1) and (2), we have 
 
 | 2 • 60 + 2 • 80 = 280, j 2 • 80 + 2 • 60 = 280, 
 
 I 60 • 80 = 4800, an 1 80 • 60 = 4800. 
 
 Hence, we obtain two pairs of numbers which satisfy both of these 
 equations. The solution I = 60 and w = 80 is applicable to this prob- 
 lem only by calling the greater side the width instead of the length. 
 AVe have seen before that solving a quadratic often results in one 
 solution which is without meaning in the problem that gives rise to 
 the equation. 
 
 In any case each solution should be carefully examined to ascertain 
 whether under any reasonable interpretation they are both applicable 
 to the problem. 
 
 151. If squares are constructed on the two sides, and also on 
 the hypotenuse of a right-angled triangle, then the sum of the 
 squares on the sides is equal to the square on the hypotenuse. 
 This is proved in geometry, but may be verified by counting 
 squares in the accompanying figure. This proposition was first 
 discovered by the great philosopher and mathematician Py- 
 thagoras, who lived about 550 b.c. Hence it is called the 
 Pythagorean proposition. 
 
 We now proceed to solve some problems by this proposition. 
 
QUADRATIC AND LINEAR EQUATIONS 
 
 207 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ^ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 s 
 
 t 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 11 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 rf 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 6 
 
 s 
 
 1- 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 *k 
 
 f<- 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 s 
 
 1- 
 
 ft 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 i 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 PROBLEMS 
 
 1. The sum of the sides about the right angle of a right tri- 
 angle is 35 inches, and the hypotenuse is 25 inches. Find the 
 sides of the triangle. 
 
 Solution. Let a = the length of one side in inches, 
 
 and b = the length of the other. 
 
 Then a + b = 35, (1) 
 
 and a 2 + b 2 = 25 2 = 625 (Pythagorean proposition). (2) 
 
208 SPECIAL PRODUCTS AND FACTORS 
 
 From (1), a = 35 - ft. 
 
 Substituting in (2), (35 - ft) 2 + ft 2 = 625, 
 
 or 1225 - 70 b + ft 2 + ft 2 = 625, 
 
 2 ft 2 - 70 ft + 600 = 0, 
 ft 2 -35ft + 300 = 0, 
 (ft- 20) (ft- 15)= 0. 
 
 Whence ft = 20, and ft = 15. 
 
 From (1), if ft = 20, a = 15, and if ft = 15, a = 20; that is, the 
 sides of the triangle are 15 and 20. 
 
 2. The difference between the two sides of a right triangle 
 is 2 feet, and the length of the hypotenuse is 10 feet. Find 
 the two sides. 
 
 3. The sum of the length and width of a rectangle is 
 17 rods, and the diagonal is 13 rods. Find the dimensions of 
 the rectangle. 
 
 4. A room is 3 feet longer than it is wide, and the 
 length of the diagonal is 15 feet. Find the dimensions of 
 the room. 
 
 5. The length of the molding around a rectangular room 
 is 46 feet, and the diagonal of the room is 17 feet. Find its 
 dimensions. 
 
 6. The longest rod that can be placed flat on the bottom 
 of a certain trunk is 45 inches. The trunk is 9 inches 
 longer than it is wide. What are the dimensions of the 
 bottom ? 
 
 7. The floor space of a rectangular room is 180 square feet, 
 and the length of the molding around the room is 56 feet. 
 What are the dimensions of the room ? 
 
 8. A rectangular field is 20 rods longer than it is wide, and 
 its area is 2400 square rods. What are its dimensions ? 
 
 9. A ceiling requires 24 square yards of paper, and the 
 border is 20 yards long. What are the dimensions of the 
 ceiling ? 
 
QUADRATIC AND LINEAR EQUATIONS 209 
 
 10. The area of a certain triangle is 18 square inches, and 
 the sum of the base and altitude is 12. Find the base and 
 altitude. 
 
 11. The altitude of a certain triangle is 7 inches less than 
 the base, and the area is 130 inches. Find the base and 
 altitude. 
 
 12. The sum of two numbers is 17, and the sum of their 
 squares is 145. Find the numbers. 
 
 13. The difference of two numbers is 8, and the sum of their 
 squares is 274. Find the numbers. 
 
 14. The difference of two numbers is 13, and the difference 
 of their squares is 481. Find the numbers. 
 
 15. The sum of two numbers is 40, and the difference of 
 their squares is 320. Find the numbers. 
 
 16. The sum of two numbers is 45, and their product is 
 450. Find the numbers. 
 
 17. The difference of two numbers is 32, and their product 
 is 833. What are the numbers ? 
 
CHAPTER VII 
 
 QUOTIENTS AND SQUARE ROOTS 
 
 QUOTIENT OF TWO POWERS OF THE SAME BASE 
 
 152. Illustrative Problem. To divide x 6 by x\ 
 
 Since by § 66 the quotient times the divisor equals the dividend, 
 we seek an expression which multiplied by x* equals x 6 . 
 
 Since by Principle XIV two powers of the same base are multiplied 
 by adding their exponents, the expression sought must be that power 
 of x whose exponent added to 4 equals 6. Hence the exponent of the 
 quotient is 6 - 4 = 2. That is, x* ■*■ a; 4 = x 6 - 4 = x 2 . 
 
 EXERCISES 
 
 Perform the following indicated divisions : 
 
 1. 2 4 -=-2 2 . 8. 5 13 -=-5 12 . 15. x* + x*. 
 
 2. 2 3 --2 2 . 9. T- 4 --! 22 . 16. t u + t*. 
 
 3. 2 4 -=-2. 10. 8 3 -=-8. 17. m 3 -*-ra. 
 
 4. 3 3 -=-3 2 . 11. 6 4 -=-6 2 . 18. n 6 -=-n 2 . 
 
 5. 3 4 h-3. 12. a s + a 2 . 19. (20) 4 h-(20). 
 
 6. 3 4 -5-3 2 . 13. a*--- a 3 . 20. (101) 14 --(101) 13 . 
 
 7. 9 u -*-9*. 14. m 4 -t-m*. 21. 41 7 -;-41 6 . 
 
 153. The process of division by subtracting exponents 
 leads in certain cases to strange results. 
 
 Thus, according to this process, x 4 + x 4 = a; 4-4 = x°, which is as yet 
 without meaning, since an exponent has been defined only when it 
 is a positive integer. It cannot indicate, as in the case of a positive 
 integral exponent, how many times the base is used as a factor. We 
 know, however, that x 4 + x 4 — 1, since any number divided by itself 
 
 210 
 
QUOTIENTS OF POWERS OF THE SAME BASE 211 
 
 equals unity. Hence if we use the symbol x° it must be interpreted to 
 mean 1, no matter what number x represents. It is sometimes con- 
 venient in algebraic work to use it in this way. 
 
 Again by this process x 2 + x* = x 2-4 = x -2 , which is as yet without 
 meaning, since negative exponents have not been defined. Cases of 
 this kind are considered in the Advanced Coui-se. 
 
 The preceding exercises illustrate the following principle : 
 
 154. Principle XVI. The quotient of two powers of the 
 same base is a power of that base whose exponent is the 
 exponent of the dividend minus that of tJie divisor. 
 
 For the present only those cases are considered in which the expo- 
 nent of the dividend is greater than or equal to that of the divisor. 
 
 Notice that Principle XVI does not apply to powers of different 
 bases. 
 
 E.g. 3 7 + 2 4 does not equal any integral base to the power, 7 — 4. 
 This division can be performed only by first multiplying out both 
 dividend and divisor. 
 
 EXERCISES 
 
 Perform the following indicated divisions by means of Prin- 
 ciple XVI : 
 
 1. 2 7 -r-2 3 . 6. x in + x 2n . 11. a?* +h -+- x° +b . 
 
 2. a 7 + a s . 7. 3 2 "- 1 -f- 3 a " 2 . 12. ic^ + w'. 
 
 3. 3 4 -=-3 2 . 8. 5"+ 5 --5" +2 . 13. (17) 14 h- (17) 13 . 
 
 4. x* + x*. 9. x a+i -^x a+2 . 14. 4 3 -i-4. 
 
 5. 3 3 "-*-3 2n . 10. t 4a + t a . 15. (12) 4 --(12) 3 . 
 
 In the following use Principles V, VI, and XVI : 
 
 16. (2 4 + 2 3 ) -=- 2 3 . 21. (a 2 ra 4 -& 2 m 3 )-r-m 3 . 
 
 17. (3 • 2 4 + 5 • 2 3 ) -=- 2 2 . 22. (4 • 3 2 - 3 3 • 5 • 7) -f- 3 2 . 
 
 18. (3.4 3 -5-4 4 )-=-4 2 . 23. (2 3 • 3 + 2 4 • 3 3 - 2 3 ) -2 3 . 
 
 19. (a 3 b - a 4 6 2 ) -*- a 2 . 24. (12afy- llaV+5a; 4 ) + X 2 . 
 
 20. (4 ^ + 3 a 4 )-*- a? 2 . 25. (x 3m+4 +x 2m+3 -5x m+2 )-i-x m+ \ 
 
212 QUOTIENTS AND SQUARE BOOTS 
 
 DIVISION OF MONOMIALS 
 
 155. In finding the quotient of two numbers each in the 
 factored form, if the factors are represented by Arabic figures, 
 the operation may be carried out in either of two ways. 
 
 E.g. 2 8 • 3» . 5 - 2 • 3 = 1080 - 12 = 90. 
 
 Also 23 • 3 3 • 5 - 2 2 • 3 = 2 • 3 2 • 5 = 90. 
 
 In the second process we divide by one of the factors, 2 2 or 3, 
 and divide this result by the other. 2 3 • 3 3 • 5 divided by 2 2 
 gives, according to Principle V, 2 • 3 3 • 5, and this result di- 
 vided by 3 gives by the same principle 2 • 3 2 • 5. In practice 
 such operations may readily be performed simultaneously. 
 
 In the case of literal factors the second process only is 
 available. 
 
 E.g. 5 aWc + a 2 6 2 = 5 a 4 - 2 6 8 ~ 2 c = 5 a'Wc = 5 a 2 bc. 
 
 EXERCISES 
 
 Perform the following indicated divisions in two ways when 
 possible : 
 
 1. 5 8 -7 9 -h5-7 2 . 5. 15a s b* + 5ab. 
 
 2. 2 3 -3 3 -5 3 -^2.3 2 -5. 6. 12 x*y ~ 4 x. 
 
 3. 44.5<h_4.5 3 . 7. 18 ^--3 J 2 . 
 
 4. 3 4 -5 2 -5-3-5. 8. 28m 3 n-=-7m. 
 
 The preceding examples illustrate the following principle : 
 
 156. Principle XVII. TJie quotient of two monomials is 
 found by dividing the dividend by each factor of the divi- 
 sor in succession. 
 
 Each factor of the divisor is associated with any desired 
 factor of the dividend according to Principle V, and when the 
 bases are the same the exponents are subtracted according to 
 Principle XVI. 
 
DIVISION OF MONOMIALS 213 
 
 157. If there are factors in the divisor not found in the 
 dividend, this process terminates before the operation is com- 
 pleted. The remaining steps of the division must then be indi- 
 cated, which is usually done in the form of a fraction. 
 
 E.g. x 2 -r-3 x s y = — — — = — . 
 
 3 x 1 • xy 3 xy 
 
 Again, 15 a% 2 c + 3 a 2 bx 2 y = 15 ^ >% = £«*? . 
 
 3 a-bx 2 y x' 2 y 
 
 By this process all factors common to dividend and divisor 
 have been canceled. Notice that in the first example the fac- 
 tor 1 remains when x 2 of the dividend is divided by x 2 of the 
 divisor. 
 
 _L. . , EXERCISES 
 
 Divide : 
 
 1. 4 • 7 • 9 by 2 . 3. 6. 5 a 4 b n c by ab*<*. 
 
 2. 12 • 8 • 20 by 2 • 4 • 5. 7. 10 aWV by 2xb 4 c. 
 
 3. 6 x?y 2 z by 2 xyz. 8. 36 x 4 f by 6 xSf . 
 
 4. 6 4 • 3 4 • x 3 by 6 2 • 5 2 • x 2 . 9. 35 a*- y+i by 5 x a ~hf+\ 
 
 5. 12 x 12 y 13 by 4 xtfz. 10. 2 m 20 * 4 ™ 3 — 2 by m a+2 n a - 2 . 
 
 In each of the following exercises state which of the Prin- 
 ciples I-XVII are used : 
 
 Divide : 
 
 11. 2 3 .3 2 -2 4 .3 3 by 2 3 . 3 2 . 
 
 12. 5-2 7 .3 8 + 7-2 5 .3 4 by 2 s • 3 4 . 
 
 13. 4 aV- 3 x?y 2 by x 2 y 2 . 
 
 14. 18 xy - 12 x A f + 6 x 2 y 2 by 6 afy 2 . 
 
 15. 49 a 4 -f 21 a 3 - 7 a by la. 
 
 16. 12 ax 4 }/ 3 — 16 a 2 ary -f 8 a 3 xy by 4 axy. 
 
 17. 2 ar 3 * + 4 x 4 " — 8 x 2 " by 2 a 8 . 
 
 18. 6 x 2 " f J -f 12 ar 3n+] - 10 iC +1 by 2 a n+1 . 
 
 19. 4 a; 13 - 6 « n 6 — 10 x 4 c by 2 x 4 . 
 
 20. 10 a 3 6 2 - a 2 b 3 + 15 u 4 6 4 by 5 a 2 b 2 . 
 
214 QUOTIENTS AND SQUARE ROOTS 
 
 SQUARE ROOTS OF MONOMIALS 
 
 158. Definition. The radical sign, Vj indicates that we are to 
 find one of the two equal factors of the number expression 
 which follows it, and the vinculum is attached to it, V , to 
 show how far its effect is to extend. 
 
 E.g. V9 is read the square root of 9. 
 Similarly, Va 2 -f 2 ab + b 2 is read the square root of a 2 + 2 ab + b 2 . 
 
 The square root of any number is at once evident if we can 
 resolve it into two equal groups of factors. 
 
 E-g. 
 
 V576 = V2- 2- 2- 2- 2 -2. 3. 3= V(2 3 . 3)(2 8 . 3) = V24 . 24 = 24. 
 
 It should be noted that every square has two square roots. 
 E.g. V9 = - 3 as well as + 3, since (- 3) 2 = 9 and 3 2 = 9. 
 
 In obtaining a square root the two results should always 
 be indicated. This is usually done by attaching the double 
 sign ± to the square root, e.g. V9 = ± 3. 
 
 
 
 EXERCISES 
 
 
 
 Find by i 
 
 inspection the following 
 
 ; square roots : 
 
 
 
 i. VI. 
 
 8. 
 
 VI21. 
 
 15. V324. 
 
 22. 
 
 V2 4 ". 
 
 2. V9. 
 
 9. 
 
 VI69. 
 
 16. V289. 
 
 23. 
 
 V5 1 " 2 . 
 
 3. VI6. 
 
 10. 
 
 V225. 
 
 17. V625. 
 
 24. 
 
 V7 1 ". 
 
 4. V25. 
 
 11. 
 12. 
 
 V196. 
 
 V256. 
 
 18. V900. 
 
 25. 
 26. 
 
 Va 12 . 
 
 5. V36. 
 
 19. Vioooo. 
 
 V3 1 " 4 . 
 
 6. V49. 
 
 13. 
 
 V576. 
 
 20. Va 4 . 
 
 27. 
 
 Va 14 . 
 
 7. V8T. 
 
 14. 
 
 VIoo. 
 
 21. Vi?. 
 
 28. 
 
 V3^. 
 
SQUARE BOOTS OF MONOMIALS 215 
 
 159. The square root of the product of several factors, each 
 of which is a square, may be found in two ways if the factors 
 are expressed in Arabic figures. 
 
 E.g. V4 • 16 • 25 = V1600 = V40 . 40 = ± 40, 
 
 or V4 • 16 • 25 = V2' 2 . 4 2 • 5 2 = ± 2 • 4 • 5 = ± 40. 
 
 But with literal factors, the second process only is available. 
 
 E.g. Vl6 a*b 4 c* = V4 2 a%*c 2 = ± 4 ab 2 c. 
 
 EXERCISES 
 
 Find the following indicated square roots, keeping the re- 
 sults in the factored form : 
 
 1. V2 2 -3 2 . 7. V3 12 -5 14 . 13. V9zy 2 . 
 
 2. V81 • 121. 8. V2 22 • 3 12 . 14. V121 a¥. 
 
 3. V49-25-169. 9. Vl6aW. 15. V7 4 a 4 6 2 . 
 
 4. V8 2 .5 2 -3 2 . 10. V64aV. 16. V625 x*y\ 
 
 5. V5 4 .3 2 -4 4 . 11. V4 4 a 2 6 4 . 17. Vl225a 2r . 
 
 6. V25 • 36. 12. ^W¥y\ 18. V36 r & 4 ™c 2 ". 
 
 Notice that the square root of a sum is not obtained by taking 
 the square roots of the terms separately. Thus, V9 + 16 is 
 not equal to V9 + Vl6. 
 
 The preceding exercises illustrate the following principle : 
 
 160. Principle XVIII. The square root of a product is 
 obtained by finding the square root of each factor sepa- 
 rately and then taking the product of these roots. 
 
 In order that a factor may be a perfect square it must be 
 a power whose exponent is even. Its square root is then a 
 
216 QUOTIENTS AND SQUARE ROOTS 
 
 power of the same base whose exponent is equal to one-half 
 the given exponent. 
 
 Thus, Vx^= Vx 3 • x s = x s . The exponent 3 of the root is one-half 
 the exponent 6 of the power. Hence to find the square root of a mo- 
 nomial we divide the exponent of each factor by 2. 
 
 EXERCISES 
 
 Find the following square roots : 
 
 1. V4a%«. 6. VlO 4 a 4 b\ 11. V81 x*y 8 c w . 
 
 2. V3VV 4 . 7. Vo^m". 12. V729 a«y™z H . 
 
 3. V5- • 3- f M . 8. V5 4 ^3 8 ^7 2 . 13. V64 • 625 a 2 b\ 
 
 4. V121 xy\ 9. V3 14 - 7 12 a 4 . 14. V216 arV". 
 
 5. V576 a 2 b*. 10. V2^W 2 . 15. V3 2 * • 5 2 ". 
 
 DIVISION BY A POLYNOMIAL 
 161. The simplest case of division by a polynomial is that 
 in which the dividend can be resolved into two factors, one 
 being the given polynomial divisor and the other a monomial. 
 
 E.g. To divide 4 x 3 + 4 x 2 y by x + y, factor the dividend and we 
 have 4 x 2 (x+y) h- (x + y) = 4 x 2 . 
 
 In case the dividend cannot be factored in this manner, 
 then, if the division is possible, the quotient must be a polyno- 
 mial. The process of finding the quotient under such circum- 
 stances is best shown by studying a particular case. 
 
 Illustrative Example 1. Consider the product 
 
 (x 2 + 2xy + y 2 )(x+y) = x 2 (x + y) + 2 xy (x +y) + y 2 (x + y). 
 
 The products, x 2 (x + y), 2xy(x + y), and y 2 (x + y) are called partial 
 products, and their sum, x 3 + 3 x 2 y + 3 xy 2 + y 3 , the complete product. 
 
 In dividing x 3 + 3 x 2 y + 3 xy 2 + y 3 by x + y the quotient must be 
 such a polynomial that when its terms are multiplied by x -f y the 
 
DIVISION BY A POLYNOMIAL 217 
 
 results are these partial products, which in the solution are called 1st, 
 2d, and 3d products. 
 
 The work may be arranged as follows : 
 
 Dividend or product : x 3 + 3 x 2 y + 3 xy 2 + y 3 
 1st product, x 2 (x+y) : x 3 + x-y 
 
 x + y, divisor. 
 
 x 2 + 2 xy + y 2 , 
 
 Dividend minus 1st product: 2 x 2 y + 3 xy 2 -f y 3 [quotient. 
 
 2d product, 2xy(x + y): 2 x' 2 y + 2 xy 2 
 
 Dividend minus 1st and 2d products : xy 2 + y 3 
 
 3d product, y\x+y): xy 2 + y 3 
 
 Dividend minus 1st, 2d, and 3d products : 
 
 Explanation. Since the dividend or product contains the 
 term x 3 , and since one of the factors, the divisor, contains the 
 term x, the other factor, the quotient, must contain the term 
 x 2 . Multiplying this term of the quotient by the divisor, we 
 obtain the first partial product, x 3 + x*y. 
 
 Subtracting the first partial product from the whole product 
 X s + 3 x 2 y + 3 xy 2 + y 3 , the remainder is 2 x 2 y + 3xy 2 + y 3 , which 
 is the product of the divisor and that part of the quotient 
 which has not yet been found. Since this product contains the 
 term 2 x?y and the divisor contains the term x, the quotient 
 must contain the term 2 xy. The product of 2 xy and x + y is 
 the second partial product. 
 
 Subtracting this second partial product from 2x?y + 3 xy 2 
 -t-y 3 , we have xy^ + y 3 still remaining after the first and second 
 partial products have been subtracted from the whole product. 
 This remainder is the product of the divisor and the part of the 
 quotient not yet found. Since the product contains the term 
 xy 2 and the divisor contains the term x, the quotient must contain 
 the term y 2 ; hence, the third partial product is xy 2 + y 3 . 
 
 Subtracting the third partial product the remainder is zero. 
 Hence the sum of the three partial products thus obtained is 
 equal to the whole product, and it follows that x 2 + 2 xy -f- y 2 is 
 the required quotient. 
 
218 QUOTIENTS AND SQUARE ROOTS 
 
 162. Problems in division may be checked by substituting 
 any convenient values for the letters. For example, in this 
 case, x = l, y = l, reduces the dividend to 8, the divisor to 2, 
 and the quotient to 4, which verifies the correctness of the 
 result. 
 
 Since division by zero is impossible (see Advanced Course), 
 care must be taken not to select such values for the letters 
 as will reduce the divisor to zero. 
 
 Illustrative Example 2. Divide 2x i -\-x z — 7 a? + 5 x — 1 by 
 
 Solution. [divisor. 
 
 x 2 + 2x-l, 
 
 Dividend or product: 2 x*+x i — 7 x 2 +5 x— 
 
 1st product, 2x 2 (x 2 +2a;-l): 2x*+ix 8 -2x 2 12 x 2 - 3 x + 1, 
 
 Dividend minus 1st product : — 3 x 8 — 5 x 2 + 5 x— 1 [quotient. 
 
 2d product, -3 x(x 2 + 2x-l): -3x 8 -6x 2 +3x 
 Dividend minus 1st and 2d products : +i 2 +2i-1 
 
 3d product, 1 • (x 2 + 2 x- 1): x 2 +2x-l 
 
 Dividend minus 1st, 2d, and 3d products : 
 
 Check. Substitute x = 2 in dividend, divisor, and quotient. 
 
 Illustrative Example 3. Divide 20 a? - 8 + 18 a* + 22 a - 19 a 3 
 by 2u 2 -3a + 4. 
 
 Solution. Arranging dividend and divisor according to the de- 
 scending powers of a, we have Tdivisor. 
 Dividend or product: 18a 4 -19a 8 + 20a 2 +22q-8[ 2 a 2 -3 a + 4, 
 
 1st product : 18a 4 -27a 8 +36a 2 9a 2 +4a-2, 
 
 Dividend minus 1st product : 8 a 3 — 16 « 2 + 22 a— 8 [quotient. 
 
 2d product : 8a 8 -12a 2 + 16a " 
 
 Dividend minus 1st and 2d products : — 4a 2 + 6 a — 8 
 3d product : - 4a 2 + 6a-8 
 
 Dividend minus all products : 
 
 Check. Substitute a = 1 in dividend, divisor, and quotient. 
 
DIVISION BY A POLYNOMIAL 219 
 
 163. From a consideration of the preceding examples the 
 process of dividing by a polynomial is described as follows : 
 
 1. Arrange the terms of dividend and divisor according to 
 descending (or ascending) powers of some common letter. 
 
 2. Divide the first term of the dividend by the first term 
 of the divisor. This quotient is the first term of the quotient. 
 
 3. Multiply the first term of the quotient by the divisor 
 and subtract the product from the dividend. 
 
 4. Divide the first term of this remainder by the first term 
 of the divisor, obtaining the second term of the quotient. 
 Multiply the divisor by the second term of the quotient and 
 subtract, obtaining a second remainder. 
 
 5. Continue in this manner until the last remainder is zero, or 
 until a remainder is found whose first term does not contain 
 as a factor the first term of the divisor. In case no remainder 
 is zero, the division is not exact. 
 
 EXERCISES 
 
 Check the result in each case, being careful to substitute 
 such numbers for the letters as do not make the divisor zero. 
 
 Divide the following : 
 
 1. a 2 -f- 2 ab + b 2 by a + b. 
 
 2. a 2 -2 ab + b 2 by a-b. 
 
 3. a 3 - 3 a 2 b + 3 ab 2 -W by a- b. 
 
 4. 2 a^ -f 2 x*y — 4: x 2 — x — 4= xy — y by x -\- y. 
 
 5. x 3 -f- %y 2 — x?y — y 3 by x — y. 
 
 6. tf + lrf + x — 6bycc + 3. 
 
 7. a^ + 4 xP + x — 6 by x — 1. 
 
 8. :B 4 -6ar , + 2ar ! -3:K + 6by x — 1. 
 
220 QUOTIENTS AND SQUARE ROOTS 
 
 9. x 3 + 3 x?y -f- 3 xy 2 + y 3 by x 2 + 2 an/ + y 2 . 
 
 10. a? — 8 3? + 75 by a? — 5. 
 
 11. 2 a? + 19 a 2 b + 9 ab 2 by 2a+b. 
 
 12. a; 4 — 4 ar 3 ?/ + 6 a% 2 — 4 a$ 3 + ?/ 4 by x — y. 
 
 13. x 4 + 4 arfy -f 6 ar 2 / + 4a# J -j-?/ 4 byar ! + 2ar?/ + 2/ 2 . 
 
 14. a; 4 -f- a^y + a;?/ 3 + y* by a; + ?/. 
 
 15. x* + a% 2 + y* by x* — xy + 1/ 2 . 
 
 16. x* — y 4 by x — y. 
 
 17. a 3 + 6 3 + c 3 — 3 a&c by a + 6 + c. 
 
 18. 2a 4 +ll3 3 -26a- 2 + 16a;-3by a^+7»-3. 
 
 19. x 5 + 5 afy + 10 x'y 2 + 10 a,- 2 ?/ 3 + 5 xy*+ y 5 by x 2 + 2 xy + y 2 . 
 
 20. a,- 5 - x* - 27 ar 3 + 10 x 2 - 30 x - 200 by a; 2 - 4 x - 10. 
 
 21. 3a^ — 4a# + 8a;z — 4y 2 + 8y« — 3z 2 by x — 2y + 3z. 
 
 22. 9rV-4r¥ + 4rs£ 2 -s¥by3rs-2rt + s*. 
 
 23. 9 a 2 b 2 + 16 x 2 - 4 a 2 - 36 Vx 2 by 3 a& + 6 bx - 2a - 4 a;. 
 
 24. a^ + a? 2 ?/ + a^z — a;?/2 — ?/ 2 z — yz 2 by x 2 — ?/z. 
 
 25. a 5 + a 4 & + a 3 - a 3 6 2 - 2 a& 2 + & 3 by a 2 + ab- b 2 . 
 
 26. a 3 -{- 2Z 3 — 2 s + 3 xyz by a; + y — z. 
 
 27. a 3 + & 3 + 3 ab - 1 by a + b — 1. 
 
 28. 6 x 5 *- 11 a; 4 * + 23 0^ + 13^ -3 a;* + 2 by 3o* + 2. 
 
 29. a 3 * - 3 a 2k b k + 3 a*6 2 * - ft 3 * by a* - 6*. 
 
 30. 32 s ia - 9 s^P + 12 sH 2h - 18 s a P - 17 £ 46 by -«•-*». 
 
 164. Since the dividend is the product of the divisor and quo- 
 tient, it follows that if one factor of an expression is given, the 
 other factor may be found by division. 
 
SQUARE ROOTS OF POLYNOMIALS 221 
 
 EXERCISES 
 
 Obtain the factors of each of the following products, one 
 factor being given in each case : 
 
 Product Given factor 
 
 1. x^ — y 8 . x — y. 
 
 2. ar 3 + y*. x + y. 
 
 3. a 5 — b 5 . a — b. 
 
 4. a 5 + b 5 . a + b. 
 
 5. a 6 + & 6 . a? + b 2 . 
 
 6. a 9 + 6 9 . a 3 + 6 3 . 
 
 7. r 3 — s 3 . r 2 + rs -{- s?. 
 
 8. r 4 -s 4 . r 3 + r 2 s + rs 2 +s 3 . 
 
 9. r 5 4- s 5 . r 4 — ^s -f rV — rs 3 -|- s 4 . 
 
 10. a 3 -12a 2 + 27a + 40. a-5. 
 
 11. ar*- 5 a; 4 ?/ + 11 arty 2 a 2 - 3 xy + 2 y 2 . 
 — 14 a: 2 ?/ 3 — 51 xy* + 54 f/ 5 . 
 
 12. a; 4 + xPy 2 + y*. aP — xy + y 2 . 
 
 13. a 3 + 5a 2 -2a-24. a 2 + 7a + 12. 
 
 14. a 5 -5a 4 & + 10a 3 6 2 a 2 -2a6 + 6 2 . 
 
 - 10 a 2 b 3 + 5ab*- b s . 
 
 15. x 5 — 5 x?y 2 — 5 x 2 ^ 3 + y 5 . x 2 — 3 xy + y 2 . 
 
 SQUARE ROOTS OF POLYNOMIALS 
 
 165. In §§ 87, 88, we found certain trinomials which were 
 perfect squares, namely, 
 
 a 2 + 2a& + 6 2 = (a + &) 2 , (1) 
 
 a 2 -2ab + b 2 =(a-b) 2 . (2) 
 
 Hence we know the square roots of all trinomials which are 
 in either of these forms. These trinomial squares may be used 
 
222 QUOTIENTS AND SQUARE BOOTS 
 
 to discover a process for finding the square root of any poly- 
 nomial which is a perfect square. 
 
 Finding a square root may be regarded as a process of 
 division in which divisor and quotient are equal and both are 
 to be found simultaneously. 
 
 Illustrative Example. Find the square root of 4ar J +12sc?/+9?/ 2 . 
 
 Considering the formula (1) we are to pass from the square 
 a 2 + 2 ab + b 2 to the square root a + b, and for this purpose 
 we write a 2 + 2 ab -f b 2 in the form a 2 + 6 (2 a + b) and arrange 
 the work as follows : 
 
 Square or product, 4 x 2 + 12 xy + 9 y 2 \ 2 x -f 3 y, square root. 
 4 x 2 1st par'l product. 
 
 1st par'l divisor, 4 x 
 1st compl. divisor, 4 a: + Sy 
 
 \2 xy + 9 y% square minus 1st par'l prod. 
 12 xy + 9 y 2 , 2d par'l product. 
 
 
 
 Supposing that 4 x 2 is the a 2 of the formula, a is then 2 a, 
 which is the first term of the root. Squaring 2 x gives 4 x 2 , the 
 first partial product. Subtracting 4 cc 2 from the total product 
 leaves 12 #?/ + 9 y 2 , which is the b (2 a + b) of the formula. 
 
 Since b is not yet known, we cannot find completely either 
 of the factors of b (2 a + b) ; but since a has been found, we can 
 get the first term of the factor 2 a -f- b, viz. 2aor2-2sc = 4a;, 
 which is the first partial divisor. Dividing 12 xy by 4 a; we 
 have 3y, which is the b of the formula. Then 2a + b = 4:X + 3y 
 the first complete divisor. 
 
 To obtain the second partial product, b (2 a + b) or 12 xy + 9y 2 , 
 we multiply 4 x + 3 ?/ by 3 y. On subtracting, the remainder is 
 zero and the process ends, whence the required root is 2 x + 3 y. 
 
 It should be clearly understood that the sum of the first and 
 second partial products is a square, viz., the square of 2 x + 3 y, 
 because it has been constructed just as a? + b(2 a + b) was 
 formed from a + &. 
 
SQUARE ROOTS OF POLYNOMIALS 223 
 
 EXERCISES 
 
 Find in the manner just described the square roots of the 
 following trinomials : 
 
 1. 36<c 2 -84a^ + 49?/ 2 . 4. 81 m 4 + 144 m?n 2 + 64 n*. 
 
 2. 16 a 2 - 40 ab + 25 b 2 . 5. 49 s 6 - 84 s 3 + 36. 
 
 3. 121 t 2 + 264 ta + 144 a 2 . 6. 1 - 20 a 4 + 100 a?. 
 
 166. The process just applied to trinomials is applicable to poly- 
 nomial squares having a larger number of terms, by regarding the 
 a of the formula at each step as representing the part of the root 
 already found and b as the term of the root about to be found. 
 
 Illustrative Example. Find the square root of x 2 + 2 xy + y 2 
 + 6 sc+ 6 y + 9. The work may be arranged as follows : 
 
 Square or product, x 2 + 2 xy + y 2 +Gx + Qy + 9\x + y + 3, square root. 
 x 2 1st par'l product. 
 
 2xy + y 2 +6x+Qy + 9, 1st remainder. 
 
 2xy + y 2 , 2d par'l product. 
 
 6 x + 6 y + 9, 2d remainder. 
 
 6 x+ 6 y+ 9, 3d par'l product. 
 
 1st par'l div'r, 2x 
 
 1st compl. div'r, 2x + y 
 2d par'l div'r, 2 x+ 2 y 
 
 2d compl. div'r, 2x + 2y + 3 
 
 ~ 0" 
 
 After the first two terms of the root have been found, namely, 
 x and y, then we consider x + y as the a of the formula and 
 call it a'. The new b, which we call b', is then found to be 3. 
 Subtracting the first and second partial products is the same 
 as subtracting (x-\-y) 2 , that is, the square of a'. Hence, the 
 second partial divisor, which is twice a', is 2 (x-\-y). 
 
 In case there are four terms in the root, then the sum of the first 
 three, when found as above, is regarded as the new a, called a". 
 The remaining term of the root is the new b, and is called b". 
 
 The formula (a — b) 2 = a 2 — 2 ab + 6 2 is not needed, since this may 
 be written (a + (- ft)) 2 = a 2 + 2 a (- b) + (- by, which is in the 
 form of (a + b) 2 = a 2 + 2 ab + 6 2 . 
 
224 QUOTIENTS AND SQUARE ROOTS 
 
 EXERCISES 
 
 Find the square roots of the following polynomials: 
 
 1. x 2 + y 2 + z 2 -2xy + 2xz-2yz. 
 
 2. 9ar J + 4?/ 2 + z 2 — 12xy + 6xz — Ayz. 
 
 3. a?-8ab + 16b 2 -2ac + c 2 + 8bc. 
 
 4. a 4 + 4a 3 + 6a 2 + 4a + l. 
 
 5. c 4_4c 3 + 6c 2 -4c + l. 
 
 6. ?/ 4 - 4 ?/ 2 - 8 a-?/ 2 + 16 a + 163? +4. 
 
 7. x 4 — 2 a 8 + 3 a 2 — 4* + 4. 
 
 8. a 2 + a 2 6 2 - 2 a 2 6 + 2 ode - 2 a& 2 c + & 2 c 2 . 
 
 9. a 4 + 4a 3 6 + 6a 2 & 2 + 4a& 3 + & 4 . 
 
 10. a; 4 — 4 a??/ + 6 x^y 2 — 4 xy 3 -f 2/ 4 . 
 
 11. a; 6 - 4 as" + 4 x 4 + 6 a; 3 - 12 cc 2 + 9. 
 
 12. a 4 + 53 a 2 + 14 a 3 + 28 a + 4. 
 
 13. a- 2 + 16afy 2 + 289 + 8a: 2 ?/ + 34a: + 136a:?/. 
 
 14. 9 a 4 + 4 a 2 + 256 -12 a 3 -96 a 2 + 64 a. 
 
 15. a 6 + 6a 5 + 15a 4 + 20a 3 + 15a 2 + 6a + l. 
 
 16. a 6 -6a 5 + 15a 4 -20a 3 + 15a 2 -6a + l. 
 
 17. 16x 6 +4,y 2 -[-l-16x i y + 8x i -4:y. 
 
 18. 25 + 49 a 2 + 4 a 4 -70 a -20 a 2 + 28 or 5 . 
 
 19. 64 x 6 + 192 x 5 + 240 a 4 + 160 ar 5 + 60 x 2 + 12 x + 1. 
 
 20. 4a 6 -12a,- 5 + 13a; 4 -14a* ! + 13av ! -4a; + 4. 
 
 21. « 4 J 4 -2aV + a 2 -2fl 2 6+2a6+5 2 . 
 
 22. 16a 6 + 24a 5 + 25a 4 + 20a 3 + 10a 2 + 4a + l. 
 
 23. x H f + 2 ary + 3 a; 4 ?/ 4 + 4 ary + 3 tfy 2 -\-2xy-\-l. 
 
 24. l + 2a; + 3ar 2 + 4ar 3 + 5ar 4 + 4ar 5 + 3a; 6 + 2a; 7 + ar 8 . 
 
 25. 9 a 2 - Gab + 30 ac + 6 ad + 6 2 - 10 6c -2 M + 25C 2 
 
 + 10cd + d 2 . 
 
SQUARE ROOTS OF ARABIC NUMBERS 225 
 
 SQUARE ROOTS OF NUMBERS EXPRESSED IN ARABIC FIGURES 
 
 167. The square root of a number expressed in Arabic 
 figures may be found by the process just used for polynomials. 
 
 Illustrative Example. Find the square root of 5329. 
 
 In order to decide how many digits there are in the root we 
 observe that 10 2 = 1000 and 100 2 = 10000 ; hence the root lies 
 between 10 and 100, i.e. it contains two digits. Since 80 2 = 
 6400 and 70 2 = 4900, we see that 7 is the largest number pos- 
 sible in tens' place of the root. The work is arranged as 
 follows : 
 
 The given square, 
 
 5329 |70 + 3, square root. 
 
 a 2 = 70 2 
 
 4900, 1st partial product. 
 
 2 a = 2 x 70 = 140 
 
 ft = 3 
 
 2 a + b = 143 
 
 429, 1st remainder. 
 
 429 = 6 (2 a + b). 
 
 
 
 Having decided as above that the a of the formula is 7 tens, 
 we square this and subtract, obtaining 429 as the remaining 
 part of the power. 
 
 The first partial divisor, 2 a = 140, is divided into 429 giving 
 a quotient 3, which is the b of the formula. Hence the first 
 complete divisor, 2 a + b, is 143, and the second partial prod- 
 uct, b (2 a + b), is 429. Since the remainder is zero, the pro- 
 cess is exact and 73 is the square root sought. 
 
 EXERCISES 
 
 Find the square roots of the following : 
 
 1. 3249. 5. 6241. 9. 1849. 
 
 2. 8836. 6. 7056. 10. 7225. 
 
 3. 7569. 7. 9409 11. 3025. 
 
 4. 8281. 8. 9801. 12. 9216. 
 
226 QUOTIENTS AND SQUARE ROOTS 
 
 168. The square of any integer from 1 to 9 contains one or 
 two digits ; the square of any integer from 10 to 99 contains 
 three or four digits ; the square of any integer from 100 to 999 
 contains five or six digits ; etc. 
 
 Hence it is evident that, if the digits of a number which is 
 a perfect square be separated into groups of two each, count- 
 ing from units' place toward the left, the number of groups 
 thus formed is the same as the number of digits in the square 
 root. 
 
 Illustrative Example. Find the square root of 120,409. 
 
 Since this number is divided into three groups, the first digit is in 
 hundreds' place. The work is arranged as follows : 
 
 Square, 12 04 09 1300 + 40 + 7 = 347, square root. 
 
 a 2 = 300 2 9 00 00, 1st partial product. 
 
 2a = 2 x 300=600 
 
 b= 40 
 
 2 a + b = 640 
 
 2 a' = 2 x 340 = 680 
 
 3 04 09, 1st remainder. 
 2 56 00 = b (2 a + b). 
 
 V= 7 
 
 2 a' + V = 687 
 
 48 09, 2d remainder. 
 48 09 = V (2 a' +b'). 
 
 
 
 The first partial divisor, 2 x 300, is completed by adding the second 
 term of the root, 40, that is, 2 a + b — 600 + 40. At first glance it 
 might be supposed that the second digit is 5 instead of 4, since 600 is 
 contained in 30,409 50 times, but account must be taken of the addi- 
 tion to be made to the partial divisor, and when this is done the 
 quotient is 40, not 50. 
 
 In the second partial divisor, 2 a' stands for 2 times (300 + 40) = 680, 
 and V stands for the third digit of the root. 
 
 In case a square contains an odd number of digits the last 
 group at the left will have one instead of two digits. 
 
 E.g. 3 47 21 has three places in its square root, of which the first, 
 hundreds' digit, is the largest square in 3, namely, 1. 
 
SQUARE BOOTS OF ARABIC NUMBEBS 
 
 227 
 
 EXERCISES 
 
 Find the square root of each of the following : 
 
 1. 294,849. 5. 3481. 9. 100,489. 13. 35,721. 
 
 2. 37,636. 6. 7569. 10. 26,569. 14. 16,641. 
 
 3. 872,356. 7. 1849. 11. 874,225. 15. 32,761. 
 
 4. 599,076. 8. 73,441. 12. 170,569. 16. 223,729. 
 
 169. Since the square of any decimal fraction has twice as 
 many places as the given decimal, it is evident that the square 
 root of a decimal fraction contains one decimal place for every 
 two in the square. 
 
 E.g. (.15) 2 = .0225 ; (.012) 2 = .000144. 
 
 Hence, for the purpose of determining the decimal places in 
 the root, the decimal part of a square should be divided into 
 groups of two digits each, counting from the decimal point 
 toward the right. 
 
 Illustrative Example. Find the square root of 4.6225. Ac- 
 cording to §§ 168, 169 the root contains one digit in the integral 
 part and two in the decimal part. The work is as follows : 
 
 a 2 = 2 2 
 2a = 2 x 2 
 b = 
 2a + b = 
 
 4.62 25 1 2 + .1 + .05 = 2.15 
 4 
 
 4.0 
 4.1 
 
 2 a' = 2 x 2.1 = 4.2 
 b'= _ J; 05 
 
 2 a' + b'= 4.25 
 
 .62 25, first remainder. 
 
 .41 = b (2 a + b). 
 
 .21 25, second remainder. 
 .21 25 = b' (2 a' + b'). 
 
 
 
 This process is also applicable for the purpose of approxi- 
 mating the square root of a number which is not a perfect 
 square 
 
228 
 
 QUOTIENTS AND SQUARE ROOTS 
 
 Illustrative Example. Find the approximate square root of 
 582 to three decimal places. The solution below shows all the 
 steps of the work. 
 
 5 82 1 20 + 4 + .1 + .02 + .004 = 24.124 
 4 00 
 
 a 2 = 20 2 
 
 
 2 a = 2 x 20 
 
 = 40. 
 
 b = 
 2a + b 
 
 ' 4. 
 = 44. 
 
 2 a' = 2 x 24 
 
 = 48. 
 
 b' = 
 
 .1 
 
 2 a' + V = 
 
 48.1 
 
 2 a" = 2 x 24.1 
 
 = 48.2 
 
 b" = 
 
 .02 
 
 2 a" + b" = 
 
 48.22 
 
 182, 
 176 
 
 6.00, 
 
 first remainder. 
 
 = b (2 a + b). 
 
 second remainder. 
 
 4.81 =b' (2 a' + 6'). 
 1.1900, third remainder. 
 
 .9644 = b" (2 a" + b"). 
 
 2 a"' = 2 x 24.12 = 48.24 
 
 V" = .004 
 
 2 a'" + V" = 48.244 
 
 .225600, fourth remainder. 
 
 .192976 = V" (2 a'" + 6'"). 
 
 .032624 
 
 The decimal points are handled exactly as in division of deci- 
 mals in arithmetic, the chief care being needed in forming the 
 divisors. 
 
 170. Evidently the process in this example may be carried 
 on indefinitely. 24.124 is an approximation to the square root 
 of 582. In fact, the square of 24.124 differs from 582 by the 
 small fraction .032624. 24.12 is the nearest approximation 
 using two decimal places. If the third figure were 5 or any 
 digit greater than 5, then 24.13 would be the nearest approxi- 
 mation using two decimal places. Hence three places must be 
 found in order to be sure of the nearest approximation to two 
 places. 
 
SIMPLIFYING RADICALS 229 
 
 
 EXERCISES 
 
 
 
 Find the square 
 
 roots of the folio 
 
 wing, 
 
 correct to two deci 
 
 mal places : 
 
 
 
 
 1. 387. 
 
 7. 2. 
 
 
 13. .02. 
 
 2. 5276. 
 
 8. 3. 
 
 
 14. .003. 
 
 3. 2.92. 
 
 9. 5. 
 
 
 15. .5. 
 
 4. 27.29. 
 
 10. 7. 
 
 
 16. .005. 
 
 5. 51. 
 
 11. 8. 
 
 
 17. .307. 
 
 6. 3.824. 
 
 12. 11. 
 
 
 18. 200.002. 
 
 SQUARE ROOTS OF FRACTIONS EXPRESSED IN ARABIC FIGURES 
 
 171. Since in arithmetic the product of two fractions is 
 found by multiplying their numerators and their denominators, 
 a fraction is squared by squaring its numerator and its denom- 
 inator separately. 
 
 Hence, to extract the square root of a fraction, we find the 
 square root of its numerator and its denominator separately. 
 
 E.g. Vif = |, since f x | = if. 
 
 However, in approximating the square root of a fraction whose 
 denominator is not a perfect square, if the final result is to be 
 reduced to a decimal, the direct use of this method is cumber- 
 some since it usually involves the approximation of two square 
 roots and always necessitates dividing by a long decimal fraction. 
 
 E.g. Vf = a/2 ■+■ V3 = 1.4142 + 1.7321, in which we are now 
 obliged to divide by 1.7321. 
 
 This difficulty may be avoided in either of two ways : 
 
 1. The fraction may be reduced to a decimal before the 
 root is approximated. 
 
 E.g. Vy = V.666 • • • = .8165. 
 
230 QUOTIENTS AND SQUARE BOOTS 
 
 2. The denominator of the fraction may be made a perfect 
 square before approximating the root. 
 
 y *3 *9 3 3 
 
 By either of these two processes only one square root is 
 approximated, and there is only a short division by 3 instead 
 of a long division by 1.7321. 
 
 It is clear that any fraction can be changed into an equal fraction 
 whose denominator is a perfect square by multiplying numerator and 
 denominator by the proper number. 
 
 E-ff- J = h I = M> j = il> e tc. If a fraction is given in the form 
 
 — , it may be written Vf = vf = |V3. In like manner, 
 V3 
 
 J__ = 7 . _J_ = 7 JT =7 JH = 7 ^ 
 
 vn vn v n X i- J i ii 
 
 EXERCISES 
 
 Find approximately correct to two decimal places the follow- 
 ing square roots. 
 
 In the first ten obtain the results in three different ways : 
 (a) Find the root of each numerator and denominator separately ; 
 (6) reduce each fraction to a decimal ; (c) reduce each fraction 
 so as to make its denominator a perfect square. 
 
 •In the remaining exercises use methods (6) and (c) only. 
 In each case compare the results obtained. 
 
 5. 
 
 v|. 
 
 6. VJ. 
 
 ii. V3?. 
 
 16. 
 
 Vf. 
 
 Vf. 
 
 Vf 
 
 7. V*f. 
 
 8. vjf 
 
 12. V^. 
 
 13. V|. 
 
 17. 
 
 1 
 
 V5* 
 
 
 9. V|. 
 
 io. Vf 
 
 14. Vf. 
 
 15. Vf 
 
 18. 
 
 5 
 V13 
 
 19. 
 
 3. V*. 8. V*f. 13. Vf. V5 20. 
 
 21. 
 
 3 
 
 V7* 
 
 7 
 y/Tf' 
 
 Vf 
 
SIMPLIFYING RADICALS 231 
 
 172. Principle XVIII may be used to advantage in approxi- 
 mating the square roots of certain integral numbers. 
 
 E.g. suppose V2 has been computed, and VS is desired. It is 
 unnecessary to compute the V8 directly, for by XVIII, 
 
 VS= vTT2 = y/i- V2 = 2V2. 
 
 This sort of simplification is possible whenever the number 
 under the radical sign can be resolved into two factors, one of 
 which is a perfect square. 
 
 E.g. suppose the V5 to have been computed, then 
 
 VT25 a V25V5 = V25 • V5 = 5 V5. 
 
 In like manner, Va 5 6 8 may be written 
 
 Va 4 6 2 • ab = Va*P ■ Vab = a?bVab. 
 
 Definition. A radical expression is said to be simplified 
 when the number under the radical sign is in the integral form 
 and contains no factor which is a perfect square. 
 
 E.g. the simplified forms of Vl^5> VaW VT — ~> are respectively, 
 5 V5, a% Vab, i V5, i >/3. 
 
 EXERCISES 
 
 Given V2 = 1.4142, V3 = 1.7321, V5 = 2.2361, compute 
 the following, correct to three places of decimals, without 
 further extraction of roots : 
 
 1. 
 
 V80. 
 
 6. 
 
 V2-3. 
 
 11. 
 
 V27 + V|. 
 
 2. 
 
 vj. 
 
 7. 
 
 V72. 
 
 12. 
 
 V45 + VJ. 
 
 3. 
 
 VJ- 
 
 8. 
 
 V98. 
 
 13. 
 
 V50-VJ + V8. 
 
 4. 
 
 V48. 
 
 9. 
 
 V363. 
 
 14. 
 
 V48+V75-V3. 
 
 5. 
 
 V75. 
 
 10. 
 
 VI25. 
 
 15. 
 
 V32+V72-V18. 
 
232 QUOTIENTS AND SQUARE ROOTS 
 
 Simplify the following : 
 
 16. V32a 2 6. 19. VioVy^. 22. V500 x 7 a s b. 
 
 17. VSlxW 2 . 20. V63 b(fd\ 23. -yJ'Stf + Kxy + dy 2 . 
 
 18. VoO aW. 21. V900 o6V. 24. V8« 2 -12y 2 . 
 
 25. V32a 2 -64a6 + 32 6 2 . 26. VF25 x 2 + 250 ajy + 125 y\ 
 
 27. Find approximately to four decimal places the sides of 
 a square whose area is 120. 
 
 28. Approximate to four decimals the side of a square having 
 an area equal to that of a rectangle whose sides are 15 and 20. 
 
 29. How many rods of fence are required to fence a square 
 piece of land containing 50 acres, each acre containing 160 
 square rods? 
 
 30. A square checkerboard has an area of 324 square inches. 
 What are its dimensions ? 
 
 173. In adding or subtracting expressions containing radi- 
 cals it is always best to first reduce each radical expression to 
 its simplest form, since this often gives opportunity to com- 
 bine terms which are similar with respect to some radical 
 expression. 
 
 Ex.1. V32+ V72- Vl8 = 4V2 + 6V2-3V2 = 7V2by 
 Principles XIII, I, and II. 
 
 Ex.2. VJ+Vi2-V| = |V3 + 2v / 3-|V3 
 
 = a + 2-i)V3=l|V3. 
 
 EXERCISES 
 
 Simplify each of the following as far as possible without 
 approximating roots. 
 
 1. V27 + 2V48-3V75. 
 
 2. V20 + V125- VT80. 
 
 3. 3V432-4V3 + V147. 
 
APPLICATIONS OF SQUARE BOOT 233 
 
 4. 3V2450-25V2 + 4V13122 
 
 5. 3y 2 ^x J z + 2^x J z?-yz i yJ- 2 
 
 6. V4 x*y + V2o a-?/ 3 — x -\/xy 
 
 7. Vaa- 2 — bar + V4 cwV — 4 
 
 8. 4A/|-fV^-3V27. 
 
 9. 2 V| + V60 + Vf. 
 
 10. 5 V3- 2 V48 + 7 VIM. 
 
 11. Va 3 - a 2 b - Vai* - b 3 - V(a + 6)(a 2 - 6 2 ). 
 
 12. A/a + 3 V2~cT- 2 V3a + V4~a- V8a+ Vl2a. 
 
 13. Va- 3 -\-2xry + xy 2 — Va? — 2 x 2 y + a;?/ 2 — V4 xy\ 
 
 14. Vr - s + Vl6 r - 16 s + Vrt 2 - st 2 - V9(r— *). 
 
 15. V(m — w) 2 a + V(m + n) 2 a — -\/am 2 + Va (1 — ra) 2 — 
 
 16. V32 x 2 y* + V162 arY - V512 arfy 4 + V1250 tfy*. 
 
 APPLICATIONS OF SQUARE ROOT 
 
 174. Some of the most interesting and useful applications of 
 the square root process are concerned with the sides and areas 
 of triangles. 
 
 The fact that the sum of the squares on the two sides of a 
 right triangle equals the square on the hypotenuse was used 
 in Chapter VI. (Pythagorean Proposition, page 206.) 
 
 If a and b are the lengths of the sides, and c the length of 
 the hypotenuse, all measured in the same unit, this propo- 
 sitionsays: <? = a 2 + b\ (1) 
 
 Hence, by S, a 2 = <? - b\ (2) 
 
 and b 2 = c 2 - a 2 . (3) 
 
234 QUOTIENTS AND SQUARE EOOTS 
 
 Taking the square root of both sides in each of these 
 
 equations, 
 
 c=Va 2 + 6 2 . (4) 
 
 a = Vc 2 ^^6 2 . (5) 
 
 b = V^^a 2 . (6) 
 
 The negative square root is omitted here, as a negative 
 length cannot apply to the side of a triangle. By these formu- 
 las, if any two sides of a right triangle are given, the other 
 may be found. 
 
 E.g. if a = 4, b = 3, then, by (4), 
 
 c = V4 2 + 3 2 = Vl6 + 9 = V25 = 5. 
 
 If c = 5, b = 3, then, by (5), 
 
 a = V5 2 - 3 2 = V25 - 9 = Vl6 = 4. 
 
 If c = 5, a = 4, then, by (6), 
 
 b = Vo 2 -4 2 = V25 - 16 = \/9 = 3. 
 
 Illustrative Problem. If the two sides of a right triangle are 
 8 and 12, find the hypotenuse correct to two decimal places. 
 
 Solution. We have c - Va 2 + b 2 = V64 + 144 = V208, 
 
 V208 = VI6TT3 = VIS • V13 = 4VI3 = 4(3.605)= 14.420. 
 
 PROBLEMS 
 
 In solving the following problems, simplify each expression 
 under the radical sign before extracting the root. Find all 
 results correct to two decimal places. In each case construct 
 a figure. 
 
 1. The sides about the right angle of a right triangle are 
 each 15 inches. Find the hypotenuse. 
 
 2. The hypotenuse of a right triangle is 9 inches and one 
 of the sides is 6 inches. Find the other side. 
 
APPLICATIONS OF SQUARE ROOT 235 
 
 3. The hypotenuse of a right triangle is 25 feet and one 
 of the sides is 15 feet. Find the other side. 
 
 4. The hypotenuse of a right triangle is 7 rods and one of 
 the sides is 5 rods. Find the other side. 
 
 5. The hypotenuse of a right triangle is 12 inches and the 
 two sides are equal. Find their length. 
 
 Let s equal the length of one of the equal sides. 
 
 Then s 2 +s 2 =144. 
 
 2 s 2 = 144. 
 
 s 2 = 72. 
 
 s = V72 = 6V2= 6 x 1.414 = 8.484. 
 
 6. The hypotenuse of a right triangle is 30 feet and the 
 sides are equal. Find their length. 
 
 7. The hypotenuse of a right triangle is h and the sides 
 are equal. Find their length. Solve 5 and 6 by means of 
 the formula here obtained. 
 
 8. The diagonal of a square is 8 feet. Find its area. 
 
 9. The diagonal of a square is d. Find an expression in 
 terms of d representing its area. 
 
 10. The side of an equilateral triangle is 6 inches. Find 
 the altitude. 
 
 A line drawn from a vertex of an equi- 
 lateral triangle perpendicular to the base meets / 
 the base at its middle point. Hence this / 
 problem becomes : the hypotenuse of a right / 
 triangle is 6 and one side is 3. Find the / 
 remaining side. 
 
 11. The side of an equilateral triangle is 10. Find the alti- 
 tude. 
 
236 QUOTIENTS AND SQUARE BOOTS 
 
 12. The side of an equilateral triangle is s. Find the 
 altitude. 
 
 This is equivalent to finding a side of a right triangle whose hypote- 
 nuse is s, the other side being -. Let a equal altitude. 
 
 Then • = V"-(§)'-V^ 
 
 ; V 4 -^=V¥ ! =V^ 
 
 -V 
 
 j-VS=fv^. 
 
 This formula gives the altitude of any equilateral triangle in 
 terms of the side. By means of this formula solve 11 and 12. 
 It is interesting to notice that the square root of 3 is the only 
 root required in finding the altitude of any equilateral triangle 
 whatever. 
 
 13. Find the altitude of an equilateral triangle whose side 
 is 41. Substitute in the formula under 12. 
 
 14. Find the area of an equilateral triangle whose side is 5. 
 
 Since the area of a triangle is \ the product of the base and alti- 
 tude, we first find the altitude by means of the formula under 12, and 
 then multiply by J the base. 
 
 15. Find the area of the equilateral triangle whose side is s. 
 
 s 2 — 
 Show the result to be — V3. 
 4 
 
 16. If the area of an equilateral triangle is 16 square 
 inches, find the length of the side. 
 
 Let s equal the length of the side. Then bv the formula derived 
 
 s 2 
 
 under 15, 16 = — V3. 
 
 Hence (§§ 171, 172), s 2 = -^4 = — V3 = 21.33 x 1.732. 
 V3 3 
 
APPLICATIONS OF SQUABE BOOT 237 
 
 17. The area of an equilateral triangle is 50 square inches. 
 Find its side and altitude. 
 
 18. The area of an equilateral triangle is a square inches. 
 Find the side. 
 
 Solve the equation a = — V3 f or s, and simplify the expression, 
 
 finding s 2 = -J , and s = V ~V~ = - V3aV3. 
 
 V3 6 
 
 19. The area of an equilateral triangle is 240 square inches. 
 Find its side. (Substitute in the formula obtained under 18). 
 
 20. Find the area of a regular hexagon whose 
 side is 7. 
 
 A regular hexagon is composed of 6 equal equi- 
 lateral triangles, whose sides are each equal to the 
 side of the hexagon (see figure). Hence this prob- 
 lem may be solved by finding the area of an equi- 
 lateral triangle whose side is 7, and multiplying the result by 6. 
 
 21. Find the area of a regular hexagon whose side is s. 
 (Solve 20 by substituting in the formula obtained here.) 
 
 22. The area of a regular hexagon is 108 square inches. 
 Find its side. 
 
 If the area of the hexagon is 108 square inches, the area of one of 
 the equilateral triangles is 18 square inches. Hence this problem can 
 be solved like 18. 
 
 23. The area of a regular hexagon is a square inches. Find 
 its side. (Solve 22 by substituting in the formula obtained 
 here.) 
 
 24. Find the radius of a circle whose area is 9 square inches. 
 
 The area of a circle is found by squaring the radius and multiply- 
 ing by 3.1416. The number 3.1416 is approximately the quotient 
 obtained by dividing the length of the circumference by the diameter 
 of the circle. This quotient is represented by the Greek letter tt 
 
238 
 
 QUOTIENTS AND SQUARE BOOTS 
 
 (pronounced pi). In this chapter we use 3^ as an approximation to ir. 
 This differs from the real value of it by less than .0013, and hence is 
 accurate enough for most purposes. If a represents the area of a 
 circle, the above rule may be written 
 
 a 
 
 irr'. 
 
 Hence if a = 9, 
 and 
 
 9 = 9^ _ 63_ _ 2 g63 
 
 j 3f 22 ' 
 
 25. Find the radius of a circle whose area is 68 square feet. 
 
 26. Find the radius of a circle whose area is a square feet. 
 
 We have 
 
 Hence 
 
 a = irr i , or r z = 
 
 1 7T \ 7H IT 
 
 In problems stated in terms of letters, the results, of course, cannot 
 be reduced to a decimal. In such formulas it is best not to replace 
 the letter ir by any of its approximations. 
 
 27. Find the sum of the areas of a circle 
 of radius 6 and the square circumscribed 
 about the circle. 
 
 The area of the circle is 6 2 7r = 36 ir, and the 
 area of the square is 4 • 6 2 = 4 • 36 ; i.e. the 
 square contains 4 squares whose sides are 6. 
 The sum of the areas is 
 
 4 • 36 + 36 7T = (4 + tt) 36 = (4 + 3f) 36. 
 
 28. Find an expression for the sum of the areas of a circle 
 of radius r and the circumscribed square. (Solve 27 by sub- 
 stituting in the formula here obtained.) 
 
 29. If the sum of the areas of a circle and the circumscribed 
 square is 64, find the radius of the circle. 
 
 By the formula obtained under 28, 
 
 64= (4 + 7r)r 2 = ^r 2 . 
 
 Hence, 
 
 r 2 = 64^ = 8.96, 
 r = V8.96 = 2.99. 
 
APPLICATIONS OF SQUARE ROOT 
 
 239 
 
 30. If the sum of the areas of a circle and the circumscribed 
 square is 640 square feet, find the radius of the circle. 
 
 31. The sum of the areas of a circle and the circum- 
 scribed square is a. Find an expression representing the 
 radius of the circle. (Replace w by 3| before simplifying.) 
 
 32. If the radius of a circle is 12, find 
 the difference between the area of a circle 
 and the circumscribed square. 
 
 33. If the radius of a circle is r, find the 
 difference between the area of the circle and 
 the circumscribed square. (Solve 32 by 
 substituting in the formula obtained here.) 
 
 34. If the radius of a circle is 16, find the 
 
 area of the inscribed square. (This is the same problem as finding 
 the area of a square whose diagonal is 32. See problems 8 and 9.) 
 
 35. If the radius of a circle is r, find an expression repre- 
 senting the area of the inscribed square. 
 
 (This is problem 9, the hypotenuse being 
 2r.) 
 
 36. If the radius of a circle is 12, find 
 the difference between the area of the circle 
 and the area of the inscribed square. 
 
 37. If the radius of a circle is r, find 
 an expression representing the differences 
 
 between the areas of the circle and the inscribed square. 
 
 38. The radius of a circle is 10. Find the area of an 
 inscribed hexagon. See note, problem 20. 
 
 39. The radius of a circle is 6. Find 
 the difference between the areas of the 
 circle and the inscribed hexagon. 
 
 40. Find an expression representing the 
 difference between the areas of a circle with 
 radius r and the inscribed regular hexagon. 
 
240 QUOTIENTS AND SQUARE ROOTS 
 
 SOLUTION OF QUADRATIC EQUATIONS BY MEANS OF SQUARE ROOT 
 
 175. Illustrative Problem. A rectangular field is 18 rods 
 longer than it is wide and its area is 50 acres. What are its 
 dimensions ? 
 
 Let x = the width of the field. 
 
 Then, x + 18 = the length of the field. 
 
 And x (x + 18) = the number of square rods in the field. 
 
 Hence, rr 2 + 18 x = 8000, (1) 
 
 or x 2 + 18 x- 8000 = 0. (2) 
 
 We are not able to factor the left member of the equation 
 by any method thus far known, and hence we cannot solve the 
 equation as in § 145. We therefore proceed to study a general 
 method of solving quadratic equations. 
 
 Consider the equation in the form (1) above. We seek a 
 number k 2 such that x 2 + 18 x + k 2 shall be a perfect square. 
 By § 134 the middle term of a trinomial square is twice the 
 product of the square roots of the two square terms. Hence, 
 18 x=2 kx, that is, k must equal 9. 
 
 Hence, adding 9 2 (=k 2 ) to each member of (1), 
 
 a^+18.T + 9 2 = 8000 + 9 2 , (3) 
 
 or, x 2 + 18 x + 81 = 8081. (4) 
 
 Since the left member is now a perfect square, we may 
 extract the square root of both sides, approximating the root 
 on the right. 
 
 Hence, x + 9 = ± V8081 = ± 89.89, 
 
 giving x = - 9 + 89.89 = 80. 89, 
 
 and also, x = - 9 - 89.89 = - 98.98. 
 
 In this case the negative result is not applicable to the 
 problem. Hence the width of the field is 80.89 rods, which 
 is correct to two decimal places. 
 
QUADRATIC EQUATIONS 241 
 
 Illustrative Example. Solve the equation : 
 
 x 2 - 12 x + 42 = 56. (1) 
 
 By S, x 2 -12x = 14. (2) 
 
 By A , x 2 - 12 x + k 2 = 14 + F. (3) 
 
 By § 175, - 12z = 2 fee or *= - 6. 
 
 Hence, x 2 - 12 x +(- 6) 2 = 14 + 36 = 50. (4) 
 
 Taking square roots, x — 6 = ± V50 = ± 5 V2. (5) 
 
 By ,4, * as 6± 7.071. (6) 
 
 Hence x = 6 + 7.071 = 13.071, 
 
 and also x = 6 - 7.071 = - 1.071. 
 
 176. This process is called solving the quadratic equation by 
 completing the square, since in each case a number is added to 
 both sides which makes the left member a trinomial square. 
 
 Since the process always involves extracting the square root 
 in order to find the value of the unknown, the two solutions 
 of a quadratic equation are commonly called the roots of the 
 equation. By analogy the solutions of any equation are some- 
 times called its roots. 
 
 EXERCISES 
 
 In solving the following quadratic equations the result may 
 in each case be reduced so that the number remaining under 
 the radical sign shall be 2, 3, or 5. (§ 172.) Use these 
 square roots only. 
 
 1. x 2 -4a; = 8. 9. a 2 - 4a; = 16. 
 
 2. JB»=3-6aJ. 10. 2x =14 + 4 x. 
 
 3. 4 a; = 16 -a 8 . 11. 24 = 3 x 2 + 12 x. 
 
 4. a; 2 + 6 a; = 9. 12. 69 - 18 x = 3 a; 2 . 
 
 5. a; 2 + 6a; = ll. 13. 84 + 24 x = 12 x>. 
 
 6. af-^x = 12. 14. 25-3^ = 5*;. 
 
 7. a?-8a;=-14. 15. x 2 + ix = 2. 
 
 8. aj 8 = 2a;+l. 16. a^-fa; = - 2 ^. 
 
242 QUOTIENTS AND SQUARE ROOTS 
 
 177. In case the coefficient of x 2 is not unity, both members 
 may be divided by this coefficient. 
 
 Example. Solve 3 x 2 + 8 x = 4. (1) 
 
 By A i»+J* = f (2) 
 
 By A, *P + |* +*» = ! + *■. (3) 
 
 By § 175, §x = 2£xor&=f. 
 
 Hence, **+ t* + (*)? = ♦ + V = ¥• (4) 
 
 Taking square roots, * + f = ± V-^- (5) 
 
 Hence, x= -|±|V7^ (6) 
 
 and the two roots are x = 0.43, 
 
 and x = — 3.10. 
 
 The preceding example may also be solved as follows : 
 
 Multiplying each member of (1) by 4 • 3 = 12, 
 then, 36 x 2 + 96 x = 48. 
 
 By A, 36 x 2 + 96 x + t* = 48 + k 2 , 
 
 where - 12 kx = 96 x or & = 8. 
 
 Hence, 36 x a + 96 x + (8) 2 = 48 + 64 = 112, 
 and 6 x + 8 = ± VTl2 = ± 4 VT. 
 
 Therefore x = - f ± ] VT. 
 
 178. The advantage of this solution is that fractions are 
 avoided until the last step, and the value of k is found to be 
 the same as the coefficient of x in the given equation. This 
 may always be accomplished by multiplying the members of 
 the given equation by four times the coefficient ofx 2 . 
 
 In the solution of the following equations only the square 
 roots V2, V3, V5, V6, V7, need be used. In all cases where 
 the roots are integers or exact fractions the solution may also 
 be obtained by factoring as in § 145. 
 
 1. 2x* + Zx=2. 4. 6x + l = -Sx 2 . 7. 2x 2 -3x = U. 
 
 2. Sx 2 + 5x = 2. 5. 2ar 2 = 5z + 3. 8. 3^ = 9 + 2^. 
 
 3. 3z = 9-2ar ! . 6. ±x = 2x 2 -l. 9. 4x 2 = 2x + l- 
 
QUADRATIC EQUATIONS 243 
 
 10. §x-l = 3x 2 . 23. 2z-l = -4 a 2 . 36. 6^-12^ = 2. 
 
 11. 2^ + 4^ = 23. 24. 5x»+16:r=-2. 37. 3x 2 + 2x = 5. 
 
 12. 3^-7 = 40;. 25. 4a* + l = 8ar. 38. 2 + 3x = 2x>. 
 
 13. 2x*-5 = 3x. 26. 2^-3^ = 20. 39. 8« + l = -4ar J . 
 
 14. 4ar> = 6a;-l. 27. 2ar J -3 = -5z. 40. 8 + 4a; = 3a; 2 . 
 
 15. 2x = l-5x 2 . 28. 3ar + 4a; = 8. 41. 10 + 4^ = 5^. 
 
 16. 3*-20«-2a£ 29. 10-4:r = 5a;*. 42. 2 + 5^ = 3^. 
 
 17. 2x + 3x 2 = 9. 30. l+4x 2 = -6x. 43. 3x-l± = 2x 2 . 
 
 18. 4a^-l = 3ic. 31. 5-3a = 2z 2 . 44. 3x 2 -2x = 5. 
 
 19. 4z = 7-2;r ! . 32. 7 + 4x = 2x i . 45. 2a^+4*»l. 
 
 20. 2x+l = bx 2 . 33. 6 a 2 + 12 a = 2. 46. 3x-l = -4or s . 
 
 21. 3x 2 + 4x=7. 34. 6x 2 -12x=-2. 47. 23 + 4 a = 2 a 2 . 
 
 22. 3z + 9 = 2x 2 . 35. &x 2 +12x=-2. 48. Zx-l = 2x 2 . 
 
 179. Solution by Formula. Solve the equation 
 
 ax 2 + bx + c = 0. (1) 
 
 By S, M, 
 
 4 a 2 x 2 + 4 aftx = — 4 ac. 
 
 (2) 
 
 By ,4, 
 
 4 a 2 x 2 + 4 aftx + £ 2 = - 4 ac + k 2 , 
 
 (3) 
 
 where 
 
 4 a£x B 4 aix or k = b. 
 
 
 Hence 
 
 4 a 2 x 2 + 4 a&x + b 2 = b 2 - 4 ac. 
 
 .(4) 
 
 Taking square roots, 2 ax + b = ± \/& 2 — 4 crc. 
 
 (8) 
 
 By 5, A 
 
 _ - 6 ± Vb 2 - 4 ac 
 
 (6) 
 
 Calling the two values of x in the result x x and x 2 we have, 
 
 _-6+V6 2 -4ac _ -6-V6 2 -4g C 
 ri ~~ ~2«r~ ~ ; Jr2 ~~ ~2^~ "7 
 
244 QUOTIENTS AND SQUARE ROOTS 
 
 Any quadratic equation may be reduced to the form of (1) 
 by simplifying and collecting the coefficients of x 2 and x. Hence 
 any quadratic equation may be solved by substituting in the 
 formulas just obtained. 
 
 Ex. 1. Solve x 2 -4x + l = 0. 
 
 In this case a = 1, b = —4, c = 1. 
 
 Hence 
 
 , _-(-4)±V(-4)2_4.1.1 
 
 2-1 
 x 1 = 2 + VS 
 
 From which 
 
 and 
 
 x 2 = 2 - V3. 
 
 Ex. 2. Solve 
 
 3a: 2 + 16 x - 12 = 0. 
 
 Here 
 
 a = 3, b = 16, c = - 12. 
 
 Here 
 
 _ _ 16 ± V(16) 2 -4-3(- 12) 
 
 2-3 
 
 From which 
 
 Zj = §, and x 2 — — 6. 
 
 180. A quadratic equation may be proposed for solution 
 which has no roots expressible in terms of the numbers of 
 arithmetic or algebra so far as yet studied. 
 
 Example. Solve x 2 + 4 x = - 8. (1) 
 
 By .4, a: 2 + 4x + 4=-4. (2) 
 
 Taking square roots, x + 2 = ± V— 4. (3) 
 
 V— 4 is unknown to us as a number symbol, since there is no num- 
 ber thus far considered whose square equals — 4. (See Principle XI.) 
 Such symbols are defined and used in the Advanced Course, and are 
 called imaginary numbers. For us any quadratic equation which gives 
 rise to such a solution is to be interpreted as stating some impossible 
 condition. 
 
 EXERCISES 
 
 1. 7-3^ = 5 x 2 . 4. 12 -51 a = 36 + 6^. 
 
 2. 51^-33 = 3^. 5. 5aj + a? + 8 = 0. 
 
 3. Ux + 8-x*=52-3x 2 . 6. 5x*-31x=-§. 
 
QUADRATIC EQUATIONS 245 
 
 7. x 2 -8 + 3a;=-15a;. 10. 37 - 4 x 2 - 12x = 79 -5a* 
 
 8. llar 9 -49a: + 57 = 0. 11. 10 a? + 41 + 7 a: = 44. 
 
 9. 3a^ + 18-16x=5. 12. 45 + 3^-85-2^ = 0. 
 
 MISCELLANEOUS QUADRATICS 
 
 Solve as many as possible of the following equations by fac- 
 toring. When this is not convenient, use the formula of § 179, 
 or complete the square independently in each case. Show 
 which equations state impossible conditions. Approximate all 
 square roots to two decimal places. 
 
 1. x 2 + 11 x = 210. 21. 2a: 2 + 3a;-3 = 12a;+2. 
 
 2. 5^-3^ = 4. 22. 3^-7^ = 10. 
 
 3. 7x + 3 x 2 -18 = 0. 23. 17a + 31 + 2ar° = 0. 
 
 4. 2= 5 a; + 7 a* 24. 18 - 41 a = 3 + a* 
 
 5. 6z-lla: 2 = -7. 25. 10 x + 25 = 5 - 2 x -x 2 . 
 
 6. -51+42a;-3ar ! = 0. 26. 3a-59 + ar> = 0. 
 
 7. 3ar ! + 3a: = 2x- + 4. 27. 5ar+7z-6 = 0. 
 
 8. 13-8a + 3ar = 0. 28. x>+A2=lx. 
 
 9. 2^ + 11 a=32 a; -a?- 27. 29. 8 a; -5 a? 2 = 2. 
 
 10. 176 + 3a;-a; 2 = 2x. 30. 5^ + 3^-22 = 0. 
 
 11. x 2 + 6x-54 = 0. 31. 50 +20 ^ + ^ = 5 x. 
 
 12. 5 x 2 + 9 x + 12 = 4 x 2 + a;. 32. x 2 + x + 4 = 0. 
 
 13. 2a*-4x-25 = 0. 33. 20 05+2 3? + 42=33 a> + a* 
 
 14. 7.^ + 11 z = 6. 34. 17x-3x- 2 =-6. 
 
 15. 2z 2 -ll£ + 5 = 0. 35. 8a? + 5a? 2 =-2. 
 
 16. 2ar ! -lla; = 6. 36. 10 + 15 x + a; 2 = 26 x. 
 
 17. 25 x - 95 = x 2 . 37. 3 x 2 - 2x -7 = 0. 
 
 18. 11 x 2 - 42z = 2. 38. 5x 2 -9aj-18 = 0. 
 
 19. a; 2 - 8 a;- 4 = a?-22. 39. 7 a; -7 a; 2 + 24 = 0. 
 
 20. 8a; 2 + 5a; = -8. 40. 31 +2 a? + a; 2 = 0. 
 
 41. 7z 2 + 7a:-5a; 2 + 20 = ar 2 -2a; + 2. 
 
246 QUOTIENTS AND SQUARE BOOTS 
 
 181. The solution of two equations in two variables, one of 
 which is linear and the other quadratic, can be reduced to the 
 solution of a quadratic equation in one variable. (See § 149.) 
 
 Example. Solve f x + y = 3, (1) 
 
 (3x 2 -y 2 = U. (2) 
 
 From (1), y = 3-x. (3) 
 
 Substituting in (2) and reducing, 
 
 2x 2 +6x-23 = 0. (4) 
 
 Substituting in formula § 179, * = - 6 ± V36 - 4 -2(- 23)^ _ 
 
 4 
 Hence x l = 2.21 and x 2 = — 5.21. 
 
 Substituting these values of x in (1) we have as the approximate 
 roots, 
 
 = 2.21 1 
 = 0.79 J 
 
 yi = 0.79 J am S 2 = 8.21 | 
 
 y x and y 2 are here used to designate the values of y which correspond 
 to x x and x 2 respectively. 
 
 In this manner solve the following equations simultaneously, 
 finding in each case two pairs of roots. In the case of roots 
 which are neither integers nor exact fractions, find the approxi- 
 mate results to two places of decimals. 
 
 f*-y=l. K_2/! = 3 
 
 ' lar , + w 8 = 13. 
 
 # 2 = 13. 6 - 9 4 
 
 [x — y = 4. 
 x + y=9. J 
 
 ^ + 2/ 2 = 41. [x-y = l. 
 
 3 - W = 42. l» 16 
 
 . j3x-y = 5. o \2x+y = 5. 
 
 fa: + 4y = 26. f*-y»a 
 
 ' laj , -«' = ll. ' 1^-3^ = 13. 
 
 !: 
 
QUADRATIC EQUATIONS 247 
 
 10. J*- 3 ^ 1 - 16. J* + 2/= 9 - 
 
 f3a-4</ = l. ly-2x = 5. 
 
 * 1 ic 2 — y 2 = 24. ' l?/ 2 -3a;y = 16. 
 
 fz + 2/ = 4. f22/-3a,- = 0. 
 
 ' l2^-3^4-2/ 2 = 8. ' l?/ 2 + a 2 = 52. 
 
 13 J*-*-*- 19 \y-2x = 5. 
 
 \±x t + 2xy-y 2 = 19. ' lar J + y 2 = 40. 
 
 f5* + y = 12. 2Q f*-4 y = 12. 
 
 12^-3^ + ^ = 0. I3a 2 + 2a:y-6y = 44. 
 
 |x-2 2 / = 3. Jy-3a = f. 
 
 ' l2y 2 -a 2 = 4. " 12^-3^ = 4. 
 
 PROBLEMS 
 
 In each problem find the two roots of the quadratic equation 
 and determine whether both are applicable to the problem : 
 
 1. The area of a window is 2016 square inches and the length 
 of the frame is 180 inches. Find the dimensions of the window. 
 
 2. The area of a rectangular city block, including the side- 
 walk, is 19,200 square yards. The length of the sidewalk when 
 measured on the side next the street is 560 yards. Find the 
 dimensions of the block. 
 
 3. A farmer starts to plow around a rectangular field which 
 contains 48 acres. The length of the first furrow is 376 rods. 
 Find the dimensions of the field. 
 
 4. A rectangular blackboard contains 38 square feet and 
 the length of the molding is 27 feet. Find the dimensions of 
 the board. 
 
 5. A park is 120 rods long and 80 rods wide. It is decided 
 to double the area of the park, still keeping it rectangular, by 
 adding strips of equal width to one end and one side. Find 
 the width of the strips. 
 
248 QUOTIENTS AND SQUARE ROOTS 
 
 6. A fancy quilt is 72 inches long and 56 inches wide. 
 It is decided to increase its area 10 square feet by adding a 
 border. Find the width of the border. 
 
 7. A city block is 400 by 480 feet when measured to the outer 
 edge of the sidewalk. At 4 cents per square foot it costs $ 416.64 
 to lay a sidewalk around the block. Find the width of the walk. 
 
 8. A farmer starts cutting grain around a field 120 rods 
 long and 70 rods wide. How wide a strip must he cut to 
 make 12 acres ? 
 
 9. The sides of a right triangle are 6 and 8 inches respec- 
 tively. How much must be added to each side so as to increase 
 the hypotenuse 10 inches ? 
 
 10. A rectangular lot is 16 by 12 rods. How wide a strip 
 must be added to one end and one side to obtain a rectangular 
 lot whose diagonal is 1 rod greater ? 
 
 11. A picture is 15 inches by 20 inches. How wide a frame 
 must be added to increase the diagonal 3 inches ? 
 
 12. An athletic field is 800 feet long and 600 feet wide. 
 The field is to be extended by the same amount in length and 
 width so that the longest possible straight course (the diagonal) 
 shall be increased by 100 feet. What will be the dimensions 
 of the new field ? 
 
 13. A starts north from a certain place going 4 miles per 
 hour and B starts east from the same place at the same time 
 going 3 miles per hour. In how many hours will they be 
 16 miles apart, the earth's surface being considered as a plane? 
 
 Let t equal the required number of hours. 
 Then (4 2 + (3/) 2 = 16 2 = 256. 
 16 P + 9 fi = 256, 
 25 fi = 256. 
 5t= ±16. 
 t = ± 3£. 
 
 The solution t = — 3£ is not applicable to this 
 problem. 
 
QUADRATIC EQUATIONS 
 
 249 
 
 14. In the preceding problem if A goes 5 miles per hour and 
 B 4 miles per hour, in how many hours will they be 24 miles 
 apart ? 
 
 15. A rectangle is 12 inches wide and 16 inches long. How 
 much must be added to the length to increase the diagonal 
 4 inches ? 
 
 Let x = number of inches to be added to the length. The diagonal 
 of the original rectangle is Vl2 2 + 16 2 = 20. Hence the diagonal of 
 the required rectangle is 24. 
 
 Then 12 2 + (16 + x) 2 = 24 2 , 
 
 or x 2 + 32 x - 176 = 0. 
 
 Solving, x, = - 16 + 12 V3 = 4.78, 
 
 and x 2 = - 16-12 V3 = -36.78. 
 
 The negative solution obtained here may be taken to mean that if 
 the rectangle is extended in the opposite direction from the fixed 
 corner, we shall get a rectangle which has the required diagonal. See 
 the figure. 
 
 16. How much must the width of the rectangle in problem 
 15 be extended so as to increase the diagonal by 4 ? 
 
 17. A trunk 30 inches long is just large enough to permit an 
 umbrella 36 inches long to lie diagonally on the bottom. How 
 much must the length of the trunk be increased if it is to 
 accommodate a gun 4 inches longer than the umbrella? 
 
 18. A rectangle is 21 inches long and 20 inches wide. The 
 length of the rectangle is decreased twice as much as the 
 width, thereby decreasing the length of the diagonal 4 inches. 
 Find the dimensions of the new rectangle. 
 
250 QUOTIENTS AND SQUARE ROOTS 
 
 19. In a rectangular table cover 24 by 30 inches there are two 
 strips of drawn work of equal width running at right angles 
 through the center of the piece. What is the width of these 
 strips if the drawn work covers one-tenth of the whole piece ? 
 
 20. A certain university campus is 100 rods long and 80 
 rods wide. There are two driveways running through the 
 center of the campus at right angles to each other and parallel 
 to the sides. What is the width of these driveways if their 
 combined area is 356 square rods ? 
 
 21. A farm is 320 rods long and 280 rods wide. There is a 
 road 2 rods wide running around the boundary of the farm and 
 lying entirely within it. There is also a road 2 rods wide 
 running across the farm parallel to the ends. What is the area 
 of the farm exclusive of the roads ? 
 
 22. A rectangular park is 480 rods long and 360 rods wide. 
 A walk is laid out completely around the park and a drive 
 through the length of the park parallel to the sides. What is 
 the width of the walk if the drive is three times as wide as 
 the walk and the combined area of the walk and the drive is 
 3110 square rods ? 
 
 23. The sum of the sides of a right triangle is 18 and the 
 length of the hypotenuse is 16. Find the length of each side. 
 
 24. The length of a fence around a rectangular athletic field 
 is 1400 feet, and the longest straight track possible on the field 
 is 500 feet. Find the dimensions of the field. 
 
 Using 100 feet for the unit of measure the equations are 
 
 ix + y = 7, 
 \x* + y* = 25. 
 
 25. The difference between the sides of a right triangle is 8 
 and the hypotenuse is 42. Find the lengths of the sides. 
 
 26. A. room is 5 feet longer than it is wide and the distance 
 between two opposite corners is 25 feet. Find the length and 
 width of the room. 
 
QUADRATIC EQUATIONS 251 
 
 27. One side of a right triangle is 8 feet, and the hypotenuse 
 is 2 feet more than twice the other side. Find the length of its 
 hypotenuse and of the remaining side. 
 
 28. A vacant corner lot has a 50-foot frontage on one street. 
 What is the frontage on the other street if the distance between 
 opposite corners along the diagonal is 110 feet less than twice 
 this frontage. 
 
 In an old Chinese arithmetic said to have been written about 2600 
 B.C., we find the following two problems in each of which the square 
 of the unknown occurs but cancels. 
 
 29. In the middle of a square pond whose sides are 10 feet 
 there grows a reed which reaches 1 foot above the water. 
 When the reed is bent over to the side of the pond, it just 
 reaches the top of the water. How deep is the water ? 
 
 30. A bamboo reed 10 feet high is broken off so that the top 
 reaches the ground just three feet from its base. How far 
 from the ground is the reed broken off ? 
 
 31. The sum of the squares of two consecutive integers is 
 13,945. Find the numbers. 
 
 32. The product of two consecutive integers is 4422. Find 
 the numbers. 
 
 33. A square piece of tin is made into an open box, contain- 
 ing 864 cubic inches, by cutting out a 6-inch square from each 
 corner of the tin and then turning up the sides. Find the 
 dimensions of the original piece of tin. 
 
 34. A rectangular piece of tin is 8 inches longer than it is 
 wide. By cutting out a 7-inch square from each corner and 
 turning up the sides, an open box containing 1260 cubic inches 
 is formed. Find the dimensions of the original piece of tin. 
 
 35. By cutting out a square 8 inches on a side from each 
 corner of a sheet of metal and turning up the sides, we obtain 
 an open box such that the area of the sides and ends is 4 times 
 the area of the bottom. Find the dimensions of the original 
 sheet if it is twice as long as it is wide. 
 
252 
 
 QUOTIENTS AND SQUARE BOOTS 
 
 36. An open box whose bottom is a square has a lateral area 
 which is 400 square inches more than the area of the bottom. 
 Find the other dimensions of the box if it is 10 inches high. 
 (By lateral area is meant the sum of the areas of the four 
 sides.) 
 
 37. A box whose bottom is 4 times as long as it is wide 
 has a lateral area 600 square inches less than 4 times the area 
 of the bottom. Find the dimensions of the bottom if the box 
 is 6 inches high. 
 
 38. A train approaching Chicago from the south at the 
 rate of 50 miles per hour is 75 miles away when a train 
 starts west from Chicago at the rate of 25 miles per hour. 
 How long after the second train starts will they be 50 miles 
 apart measured diagonally across the country ? 
 
 If t is the number of hours required, then (75 - 50 1) 2 + (25 <) 2 = (50) 2 . 
 This may be written : (25) 2 • (3 -2 2 + (25) 2 • P = 2 2 - (25) 2 . 
 
 Hence dividing both members by (25) 2 , we have (3 — 2 t) 2 + t 2 = 4. 
 
 39. An automobile run- 
 ning northward at the rate 
 of 15 miles per hour is 20 
 miles south of the inter- 
 section with an east and 
 west road. At the same 
 time another automobile 
 running westward on the 
 cross-road at the rate of 20 
 miles per hour is 15 miles 
 east of the crossing. How 
 far apart (diagonally) will 
 they be 15 minutes later ? 
 One hour later ? 
 
 40. Under the condi- 
 tions of problem 89 how 
 long after the time speci- 
 
 4/ C 
 
 / CO 
 
 / <± 
 
 / 15-3 n 
 
 15 
 
 
 s J 
 
 15-r 2 / 
 
QUADRATIC EQUATIONS 253 
 
 fied will the automobiles be 10 miles apart? Is there more 
 than one such position? 
 
 41. What are the rates of motion of the automobiles in 
 problem 39 if one hour later they are 5 miles apart and 3 
 hours later they are 35 miles apart? (See the figure.) 
 
 If the rates of the automobiles are r x and r 2 , then after 1 hour we 
 
 have (20-r 1 ) 2 + (15-r 2 ) 2 = 5 2 , (1) 
 
 and after 3 hours we have (20 - 3 r,) 2 + (15 - 3 r 2 ) 2 = 35 2 . (2) 
 
 Simplifying (1) and (2), r* + r 2 2 - 40 r x - 30 r 2 + 600 = 0, (3) 
 
 and 9 r t 2 + 9 r 2 2 - 120 r x - 90 r 2 -600 = 0. (4) 
 
 Multiplying (3) by 9, subtracting from (4), and simplifying, 
 
 4 r t + 3 r 2 = 100. (5) 
 
 The solution of this problem may now be completed by solving (5) 
 and (1) simultaneously. 
 
 Note. In equation (2), 20 — 3 r± and 15 — 3 r 2 are both negative 
 numbers. This means that in this problem the distances north and 
 west of the crossing are negative, while those south and east are 
 positive. 
 
 REVIEW QUESTIONS 
 
 1. Define exponent. Explain the difference between an 
 exponent and a coefficient. 
 
 2. Explain why and under what circumstances exponents 
 are added in multiplication. 
 
 Show that (a 2 ) 3 = a 8 , (a 3 ) 4 = a 12 . 
 
 3. Explain the method of multiplying two monomials. 
 How is Principle III involved in Principle XV ? 
 
 4. What is meant by factoring ? Is the following expres- 
 sion factored ? x (a + b) + y (a + b). Why ? 
 
 5. What are the characteristics of a trinomial square ? 
 Are the following trinomials squares ? If not, change one 
 
 term in each so as to make it a square, x 2 + xy + y 2 ; x* + x*y 2 
 + y *. a 2 -2ab-b 2 ; 4a 2 + 4a6 + 46 2 . 
 
254 QUOTIENTS AND SQUARE ROOTS 
 
 6. What are the factors of the difference of two squares ? 
 Factor x 6 — ?/ 6 as the difference of two squares. 
 
 7. What are factors of the difference of two cubes ? 
 Factor x 6 — y 6 as the difference of two cubes. 
 
 8. What are the factors of the sum of two cubes ? 
 Factor x 6 + y 6 as the sum of two cubes. 
 
 9. Explain how to factor a trinomial by inspecting the 
 end products and cross-products of two binomials. 
 
 By this method factor, 
 
 3^-7 a-10; a 2 -9a: + 18; 3^ + 5^-12. 
 
 10. By means of the following examples explain the pro- 
 cess of factoring by grouping. 
 
 x 2 + ax + bx + ab ; x 3 — x — 3 x 2 + 3. 
 
 11. How may a quadratic equation be solved by factoring ? 
 Why is it essential that one member of the equation be zero 
 
 when the other member is factored ? 
 
 12. Explain the method of solving the quadratic equation 
 by completing the square. 
 
 13. How many roots has a quadratic equation ? When does 
 the solution of a quadratic equation indicate that the condi- 
 tions of a problem are impossible ? 
 
 14. Explain why and under what circumstances exponents 
 are subtracted in division. 
 
 15. Explain the method of finding the quotient of two mo- 
 nomials. Show how Principle V is involved in Principle XVII. 
 
 16. State Principle XVIII. Given V7 = 2.645, find V28 by 
 means of this principle. 
 
 17. Explain how a number in Arabic figures is divided into 
 groups for the purpose of finding its square root. 
 
 18. Show how the value of the following may be approxi- 
 mated by finding only one square root. 
 
 5 V20 + 2 V45 - 3 V80 + 2 VJ. 
 
CHAPTER VIII 
 LITERAL FRACTIONS 
 
 COMMON FACTORS 
 
 182. If a number is a factor of each of two or more numbers, 
 it is said to be a common factor of these numbers. 
 
 Thus, 8 is a common factor of 16 and 48, and 12 is a common 
 factor of 12, 36, and 48. 
 
 If each of a given set of numbers is factored into prime factors, 
 any common factor which they may have is at once apparent. 
 
 Illustrative Example. Find the common factors of 72, 96, 
 120, and 288. 
 
 Factoring, we have 
 
 72 = 2 
 
 . o 
 
 . 2 
 
 3 
 
 • 3 = 2 s • 3 2 . 
 
 
 96 = 2 
 
 ■ 2. 
 
 , 2 . 
 
 ■ 2 
 
 • 2 • 3 = 26 . 3. 
 
 
 120 = 2 
 
 • 2 
 
 • 2 
 
 • 2 
 
 • 3 . 5 = 2 4 • 3 
 
 •5. 
 
 288 = 2 
 
 ■ 2 
 
 • 2 
 
 • 2 
 
 • 2 • 3 • 3 = 26 
 
 8*. 
 
 The common prime factors are 2, 2, 2, and 3. The other 
 common factors are the various combinations of these, namely, 
 2.2 = 4,2.2.2 = 8,2.3 = 6,4.3 = 12, and 8-3 = 24. 24 
 is, therefore, the greatest common factor of these numbers. 
 
 To find common factors of literal expressions we proceed 
 in the same manner. 
 
 Illustrative Example. Find the common factors of x + y, 
 x 2 — y 2 , and x 2 + 2 xy + y 2 . 
 
 Factoring, we have x + y = x + y. 
 
 x 2 -y 2 = (x + y)(x - y). 
 x 2 + 2 xy + y 2 = (x + y) {x + y) . 
 
 Hence, x -f y is the only common factor of these expressions. 
 
 255 
 
256 LITERAL FRACTIONS 
 
 Illustrative Example. Find the common factors of 
 10 x 2 + 20 xy + 10 y 2 , 5(x + y)(x> - y 2 ), and 15{x + y)(x* + f). 
 
 Factoring, 10 x 2 + 20 xy + 10# 2 = 2 • 5(x +y ) (x + y~). 
 
 5(x + #)(x 2 - # 2 ) = 5(x + y)(* + #)(x - y). 
 15(x + ^)(x 8 + y s ) = 3 • 5 (x + y)(x + ?/)(x 2 - xy + # 2 ). 
 
 The common prime factors are 5, x + y, and a; + ?/. The 
 other factors, obtained by combining these, are 5(cc + y), (x + y) 2 , 
 and 5(x + y) 2 . The last factor, 5(sc 2 + 2a$ + 2/ 2 ), is called the 
 highest common factor. 
 
 The name highest instead of greatest is used in algebra referring 
 to the degree of the factor. Thus x 2 is of higher degree than x, al- 
 though if x = \, x 2 is not greater than x. 
 
 183. Definition. In general that common factor which con- 
 tains the greatest number of prime factors is the highest common 
 factor. This is usually abbreviated to H. C. F. (See § 130.) 
 
 EXERCISES 
 
 Find the H. C. F. of the following sets of expressions : 
 
 1. x-y, x?-y 2 , x 2 — 2xy + tf. 
 
 2. x 2 + 2x + l,3x + 6x? + 3x i . 
 
 3. x 2 + 4 x + 4, x 2 - 6 x — 16. 
 
 4. x 2 -8x + 16,x 2 + 10x-56. 
 
 5. a s -b 3 , a 2 -2ab + b 2 . 
 
 6. xP+y 3 , x 2 — y 2 , x 2 + 2xy-\-y 2 . 
 
 7. x 2 — 7 x + 12, ax - 3 a — bx + 3 b. 
 
 8. a 2 -13a + 42, a 3 -216, a 2 -a-30. 
 
 9. 27 + f,y 2 + 9y + 18, f-9. 
 
 10. b 2 + 7b-S0, 6 2 +H6-42, 6 2 -&-6. 
 
COMMON MULTIPLES 257 
 
 11. a 3 + 2a 2 + a, a 2 + a, a 3 + 5 a 2 + 4 a. 
 
 12. X s + y 3 , X s + afy + #?/ 2 + S/ 3 . 
 
 13. a 4 + 3 ar 5 + 2 a 2 , & + x 2 , x* + 7 a?+ 6 x 2 . 
 
 14. ar 2 — 11 x -f 30, xz - 5 2 + x 2 — 5 x. 
 
 15. ra 3 — w 3 , 2 xfm 2 4- 2 tfmn + 2 a^w 2 . 
 . 16. x>-l,a?- 1,3^-13 x + 12. 
 
 17. 1 - 64 a,- 3 , 1 - 16 ar 2 , 5 - 2 2 - 20 x + 8 a*. 
 
 18. 1 + 125 a 3 , 1 + 10 a + 25 a 2 , 1-25 a 2 . 
 
 19. ac — ax + 3 be — 3bx, a 3 +27& 3 . 
 
 20. 5 c — 2, 5 ac + 20 c — 2 a - 8. 
 
 21. 4x* -x 2 ,2x i + x i -x 2 ,2x 4 -3x i + x 2 . 
 
 22. 3a 3 -3a, 3a 3 -6 a 2 +-3a, 6a 3 + 12a 2 -15a. 
 
 23. 6 a; — 10 xy + 4 xy 2 , 18 a; — 8 xy 2 , 54 a; — 16 xy 3 . 
 
 24. 3x 5 + 9x 4 -3x 3 ,5x 2 y 2 -{- 15xy 2 -5y% 7 ax 2 + 21ax-7a. 
 
 25. 18 3^-57 3^ + 30 a:, 9 a 3 - 15 a: 2 + 6 a;, 18 ar» - 39 x 2 + 18 a;. 
 
 COMMON MULTIPLES 
 184. A number is said to be a multiple of any of its factors. 
 In particular any number is a multiple of itself and of one. 
 
 Thus, 18 is a multiple of 1, 2, 3, 6, 9, and 18, but not of 12. 3 a 2 x 2 
 is a multiple of 3, 3 x, 3 x 2 , etc. 
 
 Since a multiple of a number is divisible by that number, it 
 must contain as a factor every factor of that number. 
 
 E.g. 108 is a multiple of 54 and contains as factors all the factors 
 of 54, namely 3, 3, 3, and 2, and also 2, 6, 9, 18, and 54. 
 
 Definition. A number is a common multiple of two or more 
 numbers if it is a multiple of each of the numbers. 
 
 Thus, 18 is a common multiple of 6, 9 and 18. Evidently 3 • 18, 
 4 • 18, 5 • 18, etc. are also common multiples of 6, 9 and 18. Of all 
 these common multiples 18 is the smallest and is called the least or 
 lowest common multiple. 
 
258 LITERAL FRACTIONS 
 
 185. The process of finding the least common multiple of a 
 set of numbers in Arabic figures is shown as follows : 
 
 Illustrative Example. Find the least common multiple of 
 16, 24, and 98. 
 
 Finding the prime factors of each number, 
 16 = 2 • 2 • 2 • 2 = 2* 
 24 = 2- 2. 2- 3 = 2 8 - 3 
 98 = 2 • 7 • 7 = 2 • 7 2 
 
 Any multiple of 16 contains all its prime factors, namely, 
 2, 2, 2, and 2. Any common multiple of 16 and 24 contains in 
 addition to the prime factors of 16 any prime factors of 24 not 
 in 16, namely 3. Any common multiple of 16, 24, and 98 contains 
 in addition to 2, 2, 2, 2 and 3 those prime factors of 98 not in 16 
 or 24, namely 7 and 7. Hence 2- 2- 2- 2- 3-7-7 = 1176 is 
 a common multiple of 16, 24, and 98. Moreover, it is the least 
 common multiple because no unnecessary factor has been included. 
 
 We proceed with a set of literal expressions in the same 
 manner as above. 
 
 Illustrative Example. Find the L. C. M. of 
 
 ar 2 — 2Z 2 ; x 2 + 2 xy + y 2 ; x?-2xy + y\ 
 
 Factoring, x 2 — y* = (x — y) (x + y). (1) 
 
 x* + 2xy + y*= (x + y)(x + y). (2) 
 
 x*-2xy + y* = (x-y)(x-y). (3) 
 
 In order that an expression may be a multiple of (1) it must con- 
 tain the factors x — y and x + y. To be a common multiple of (1) 
 and (2) it must contain a second factor x + y, giving (x — y), (x+y), 
 (x + y). To be a common multiple of (1), (2) and (3) it must 
 contain a second factor x — y, giving (x — y), (x + y) , (x + y), (x — y). 
 The product thus found contains the fewest prime factors possible 
 in order to be a common multiple of (1), (2), and (3). Hence 
 (x — y)(x + y) (x + y)(x — y) = (x — y) 2 (x + y)' 2 is called the lowest 
 common multiple of (1), (2), and (3), since it is the common 
 multiple of lowest degree. 
 
COMMON MULTIPLES 259 
 
 In general, the process may be described as follows : to 
 obtain the lowest common multiple of a set of expressions, 
 factor each expression into prime factors; use all factors of 
 the first expression together with those factors of the second 
 which are not in the first, those of the third which are not in 
 the first and second, etc. It is evident that in this manner we 
 obtain a product which is a common multiple of the given 
 expressions, but such that if any one of these factors is omitted, 
 it will cease to be a multiple of some one of the expressions ; 
 that is, it will no longer be a common multiple of them all. 
 
 Thus, if in the example above either of the factors x — y is omitted, 
 the product will no longer be a multiple of x 2 — 2 xy + y 2 . 
 
 186. Definition. We now define the lowest common multiple 
 
 of a set of expressions as that common multiple which contains 
 the smallest number of prime factors. The lowest common 
 multiple is usually abbreviated to L. C. M. 
 
 EXERCISES 
 
 Find the L. C. M. of the following expressions : 
 
 1. 2-3.4; 3-7-8; 2 s - 3 -4. 9. 25^-1, 125^-1. 
 
 2. 5«Y, lOafy, 25ofy. 10. 2^-7 a+6, 4^-11 a>+6. 
 
 3. 2 ab, 6 a 2 , 4 b 2 c. 11. x^ — y^jX — y^ + xy-^y 2 . 
 
 4. x 2 — y 2 , x 2 — 2xy + y 2 . 12. x 3 — y 3 , X s + y 3 , x 2 — y 2 . 
 
 5. x-y^ + y^tf — y 2 . 13. 5 ^+7^-6, x 2 — 15a;— 34. 
 
 6. 4 — x 2 , 2 — x, 2 + x. 14. cc 3 + 2/ 3 , x 2 — y 2 , (x— y) 2 . 
 
 7. a 2 +2a&+& 2 ,a 2 -2ao + o 2 . 15.' 3a6c,a 2 -4ac+4c2, a-2c. 
 
 8. x2+ 3 x + 2, x 2 - 4,^-1. 16. x 2 -l, x + 1, a 2 + 8a + 7. 
 
 17. 4 x 3 y - 44 x 2 y + 120 xy, 3 a?x 2 - 22 a?x + 35 a 3 . 
 
 18. x 2 + 2xy+y 2 , 2aa^— 10ax + 12a. 
 
260 LITERAL FRACTIONS 
 
 19. 3 6a^-216a; + 36 6, ar*- 5a; + 4. 
 
 20. 5a 2 6 2 -5aV,6 2 + 26c + 6 + c + c 2 . 
 
 21. 15 (fax 2 + 16 <?ax + (?a, 2 cax 2 + 10 cax + 8 ca. 
 
 22. ac — 4 a + be — 4 6, aa: -f ac + 6a; + 6c. 
 
 23. a^ + 5a; + 4, a; 2 + 7a;-f 12,a; 2 + 5a; + 6. 
 
 24. x 2 + 2x+l,x 2 -2x + l,x 2 -l. 
 
 REDUCTION OF FRACTIONS TO LOWEST TERMS 
 187. In arithmetic the denominator of a fraction is usually- 
 regarded as indicating the number of equal parts into which a 
 unit is divided, while the numerator designates a certain num- 
 ber of these parts. 
 
 Thus, | means 2 of the 3 equal parts of a unit. 
 
 5 
 However, a fraction such as — cannot be regarded in this 
 
 5 
 way, since a unit cannot be divided into 3^ equal parts. — 
 
 5 
 indicates that 5 is to be divided by 3£. I.e. — = 5 -*- 3\> 
 
 In algebra any fraction is usually regarded as an indicated 
 division in which the numerator is the dividend and the de- 
 nominator is the divisor. 
 
 Thus, - is understood to mean a ■+■ 6. 
 6 
 
 The numerator and denominator are together called the terms 
 of the fraction. 
 
 In case the numerator and denominator have common factors, 
 these may be removed, without changing the value of the frac- 
 tion, by means of Principle XVII, as applied in § 157. 
 
 Thus, — - — '——— = — ■ — — , where the common factor 3 is can- 
 , ' 3 • 7 • 11 7 • 11 ' 
 celed. 
 
REDUCTION TO LOWEST TERMS 261 
 
 Similarly — - — ■ — - = - — -, the factors 2 3 3, and 4 being can- 
 
 2 4 • 3 • 4 2 2 • 4 
 
 celed; and <*" 7 * + 12 > = (»-3)(«-*) = fez^. 
 ' (^-Sa-t-e) (a? -2) (a? -3) (a;-2) 
 
 If the terms of a fraction have no common factor, the frac- 
 tion is said to be in its lowest terms. Evidently the process of 
 canceling common factors in the numerator and denominator 
 may always be continued until the fraction is reduced to its 
 lowest terms. 
 
 EXERCISES 
 
 Reduce the following fractions to lowest terms : 
 
 t 3 . 9 2 . 2 G aW x*-f 
 
 ' 2 4 .5 3 -9 4 ' ' a 2 6 2 ' ' 2x 2 -3xy + y 2 ' 
 
 4 a W x* + 2xy + y 2 64 -ft 3 
 
 8aW* " tf-tf ' ' 16-86 + & 2 ' 
 
 . afyV a a: 2 + 7a;-30 _ ar 3 + 27z 3 
 
 a^V a? 2 — 7 a; + 12 xy — 5 x+ 3 yz — 15 z 
 
 10 1-216C 3 2 7 • 3 s . 5 4 - 2 6 . 3 4 • 5 7 
 
 a;-4y-6ca; + 24c2/' ' 2 4 -3 2 • 5 5 -2 5 • 3 2 - 5 2 ' 
 
 ,j 14 6z — 2 6a; + ax — 7 az 5 x?y* — 12a*Y + 7 a 3 ?/ 2 
 
 ar ! -492 2 ' 6 afy 2 + 3 x?y 2 
 
 .„ 3a 2 -29a + 56 nQ 5c + 106- 6c-26 2 
 
 63 -9 a- 7m + ma 8c 3 + 64 6 3 
 
 13 a ( x - Vf 19 4 a; 4 -28 3^ + 48 0^ 
 (a?-tf)(x-y) ' 2x*-$a? + 6x 2 ' 
 
 14 a; 3 +27 20 9a 2 6 4 + 18a 2 6 3 c+9a 2 6 2 c 2 
 4a 2 + 24a; + 36' " 3a6 3 -3a6c 2 
 
 15 a 2 -3a-36 + a6 7 a?y 2 - 133 xy + 126 a; 
 
 (a 2 -b 2 )(a-3) ' 15xy 2 -36xy + 21x ' 
 
262 LITERAL FRACTIONS 
 
 22 20x i + 20x 2 y + 5xy 2 (a?-l)(s-2)(a?-3)(s-4) 
 
 . Wtf-Wafy 2 ' (a>-l)(a>-3)(a-3)(a;-4)' 
 
 23 3a^-3a5 2 c 2 (s 8 -y g )(as» + 2sy + y 8 ) 
 27 a 3 6 2 + 27 a 3 6c * (a 2 - 2 ^ + ?/ 2 ) (a? + y) ' 
 
 24 4a 3 -42a 2 + 20a (s»-l)(a3 + l)(3s? + 3) 
 
 2a 4 6 6 -20a 3 & 6 3(» 4 -l) 
 
 28 ( 3 a * ~ 3 a 2 b 2 ) (a 2 + 13 a + 42) 
 
 (a 2 + 3a + 2)(a 2 + 5a-6) ' 
 
 29 2 3 • 4 3 • 5 4 (x 2 - b 2 ) (x 2 + 19 x + 90) 
 
 ' 2 2 .4-5 3 (« 2 + 9a;-10)(ar i + 10a; + 9)' 
 
 REDUCTION OP FRACTIONS TO A COMMON DENOMINATOR 
 188. Since we have just seen that a common factor may be 
 removed from the numerator and denominator of a fraction 
 without changing the value of the fraction, it is evident that a 
 factor may be introduced into both terms without changing 
 the value of the fraction. 
 
 Thus since — ^— is reduced to - by removing the factor 3 from both 
 3 • 5 5 
 
 4 3-4 
 
 numerator and denominator, so - is changed to by introducing 
 
 o o • o 
 
 the factor 3 in both the terms. Likewise any other factor may be 
 
 introduced. E.g. * = ^ , ^ = («-»)(« + ») = «^J* # ' 
 y 5 5-7 a + b (a + b)(a + b) (a + 6) a 
 
 In this manner any fraction may be changed into an equal 
 fraction whose denominator is any given multiple of the de- 
 nominator of the given fraction. 
 
 Thus, f can be changed into a fraction whose denominator is 
 
 q o _ -i o 
 
 72 (a multiple of 4) by multiplying both terms by 18, i.e. '-. — j— ttt • 
 
REDUCTION TO COMMON DENOMINATOR 263 
 
 and can be changed into a fraction whose denominator is 
 
 x- y 
 
 x 2 — y 2 by multiplying both of its terms by x + y, i.e., 
 
 Sa -2b = (Sa-2b)(x + y) 
 x — y x 2 —y 2 
 
 Any two or more fractions may therefore be changed into 
 respectively equal fractions which shall have a common de- 
 nominator, namely, a common multiple of the denominators of 
 the given fractions. 
 
 Illustrative Example. Keduce ~ , . „ , to frac- 
 
 as +1 x — 1 x l — 1 
 
 tions having a common denominator. 
 
 The L.C.M. of the denominators is (x — l)(x + 1). Multiply the 
 numerator and denominator of each fraction by an expression which 
 will make the denominator of each new fraction (x — l)(x + 1). 
 
 x-1 __ (x- l)(x- 1) _ x 2 - 2 x + 1 
 x + 1 (x + l)(x-l) (z + l)(x-l)' 
 
 X+1 = (X+1)( X +1) = (X + 1)2 . 
 
 x -l ( x -l)(x + l) (x + l)(x-l) ' 
 
 2x + 3 _ 2x + 3 
 x 2 -l (x + l)(x-l)" 
 
 It is best to indicate the multiplication in the common denomina- 
 tor, since this makes it more easily apparent by what expression the 
 numerator and denominator of a fraction must be multiplied in 
 order to reduce it to a fraction with the required denominator. 
 
 It should be noticed that there are three signs in connection 
 with a fraction, the sign of the fraction itself, the sign of the 
 numerator, and the sign of the denominator. It follows from 
 the law of signs in division (Principle XII) that any two of 
 these signs may be changed simultaneously without changing 
 the value of the fraction. 
 
 -p a _ — a — a _ _ a 
 
 ' 9 ' b~ -b~ ' b -b 
 
 Thus, 
 
264 LITERAL FRACTIONS 
 
 This is useful in cases like the following: 
 
 x + 1 x 1 
 
 Reduce , — — -, and to fractions having a com- 
 
 1 — x x'- — 1 x-\-l 
 
 mon denominator. 
 
 x + 1 _ —x-l _ (x + l)(—x — l) _ —x* — 2x — l 
 1-x x-1 ' ' (x + l)(x — l) '' x*—l 
 
 X X -, 1 X — 1 X— 1 
 
 , and 
 
 tf-\ tf-l' x + \ (aj + l)(a?-l) x*-l 
 
 EXERCISES 
 
 Reduce each of the following sets of fractions to equivalent 
 fractions having a common denominator. 
 
 1 ^+3 4 4 2 g - 1 a? + l _ 
 
 x—y' x* — 2xy + y' 2 3 — x' x + l' x — 3 
 
 . a — 1 a+1 c a + & a— 6 a 
 
 a 2 -& 2 ' a 2 + 2a6+6 2 ' b-a (a + &) a 2 -& 2 
 
 3-a a;+4 .1 1 
 
 ^-9^+20' 7<e*-26a;--8 ' a 2 — 3<e — 4' a; 2 + 3a; + 2' 
 a+1 a— 1 
 
 7. 
 
 8. 
 
 a 2 -2a& + & 2 ' a 2 + 2a&+6 2 
 1 1 1 
 
 a 3_&3> 6 _ a ' a 2 +a& + & 2* 
 
 a & 
 
 10. 
 
 5a 2 -4a-12' a 2 + 4a-12' a-2 
 a; — 2 a + 2 cc-1 
 
 ^-Sa-G' rf + lZx-lOtf x* + VZx + 18 
 
 In each of the following exercises reduce the given fractions 
 to a common denominator and check by substituting a con- 
 venient number for each letter, taking care that no denomina- 
 tor becomes zero : 
 
REDUCTION TO COMMON DENOMINATOR 265 
 
 11 mr d 16 *{T-t) W(T-Q) 
 m _l' i + n ' ' w(Q-T)' ~ w (Q-t) 
 
 12 Rr 1 17 P d(V-w) 
 (B + r)(m-l)' B + r ' W(p-2t+l)' W-w 
 
 CS Br + Sr + S B t . V V 1 
 
 B + BS' BSs ' ' V-v' V+v' V'-v 1 
 
 . . a b , « BA 1 1 
 
 14. , • 19. 
 
 15. 
 
 n—an—b a — A'B-\-r'A — a 
 
 W(T-Q) V 9ft B r 1 Bx 
 
 w((J — t) "jo x y x — y x-\-y 
 
 189. Since any number may be written as a fraction with 
 the denominator 1, the above process may be used to reduce 
 an integral expression to the form of a fraction having any 
 desired denominator. 
 
 Thus, 3=^5; x-y= ( x -M xi -V ,ete. 
 5 x 2 — 1 
 
 It is sometimes convenient to reduce expressions, some of 
 which are not fractions, to the form of fractions having a com- 
 mon denominator. 
 
 Illustrative Example. Reduce 5 x, ~ ' , —r^, to 
 
 ar — 1 x — 1 
 
 fractions having a common denominator. The lowest common 
 
 denominator is ar 2 — 1. 
 
 „,. 5 x (x 2 — 1) 5 x 8 — 5 x 
 
 Thus, 5x=- — \ — , - — • — T~> 
 
 x 2 — 1 x 2 — 1 
 
 5x — 1 _ 5x — 1 
 
 x 2 - 1 " x 2 - 1 ' 
 
 2 x — y _ (2 x — ?/)(x + 1) _ 2x 2 + 2x — yx — y 
 x-1 " (x-l)(x+l) ~ x 2 -l 
 
266 LITERAL FRACTIONS 
 
 EXERCISES 
 
 Reduce the following mixed expressions to fractions having 
 a common denominator : 
 
 - 5x— 3 oil . 2 «+ 1 
 
 1. 5, a>-y,. • 3. 1 + a + a 2 , — ^rr- 
 
 ar + 2 #?/ + ?/ J a — 1 
 
 2. 3a ~ c , !•=•, 2c + 2> 4 *+«*+*•, *±*. 
 
 x—y x+y x — y 
 
 5. rf — xy + y 2 , x* — y 2 , 
 
 7. 3 a — 2& — c, 
 
 8. x*-l, a?-l, 
 
 9. x 2 + 2z?/ + 2/ 2 , 
 
 10. x + y, x-y, - -*-, : 
 
 x? + y 2 x-y 
 
 In each of the following reduce the expressions to the 
 form of fractions having a common denominator and check 
 the results by substituting convenient numbers for the 
 letters : 
 
 „. ** ,. „. SA.,B-r. 13. r, «±- ( - 
 
 a — A a — A 2 
 
 14 . TP(g-Q sT} 8(t-q) m u H} hd, 
 
 w 2 Z> 
 
 # > 
 
 x + y 
 
 X 
 
 y . 
 
 x—y 
 
 x + y 
 
 5 
 a — &' 
 
 2 
 b-c 
 
 x + 1 
 x-1 
 
 
 1 
 
 x + y' 
 
 1 
 1-x 
 
 x-y 
 
 x + 1 
 
ADDITION AND SUBTRACTION 267 
 
 ADDITION AND SUBTRACTION OF FRACTIONS 
 
 190. Fractions having a common denominator may be added 
 or subtracted, exactly as in arithmetic, by adding or subtracting 
 the numerators and dividing the result by the common 
 denominator. 
 
 For we have, by Principle VI, — - — = — + — = 4 + 5. Like- 
 
 o | e o k 
 
 wise — ■ — = t + t , the division being indicated in this case. Hence, 
 4 4 4 ° 
 
 3 5 3 4-5 
 reading this identity from right to left, we have ■? + r = — -z — . 
 
 ti • 6 4 _6-4 2 T 
 
 Likewise — = . In general, 
 
 x — y x — y x — y x — y 
 
 a b _a + b , a b_a — b 
 c c~ c ' c c~ c 
 
 If fractions which are to be added or subtracted do not have 
 a common denominator, they must be reduced to this form 
 
 Example. Add ^ and a2 + 2ah + ** . 
 
 a + b a 2 - 2 ab + b 2 
 
 Reducing the fractions to the common denominator 
 
 (a-b)(a - b)(a + V), 
 we have 
 
 a-b __ (a-b)(a-b)(a-b) _ a 8 - 3a 2 & + 3 ab 2 - b* } 
 a + b (a + b)(a - 6)(a -b) (a + b) (a-b) (a-b) 
 
 &nd a 2 +2ab + b* _ (a + b) (a 2 + 2 ab + b 2 ) _ a 8 + 3 a 2 b + 3 ab 2 + b s 
 a 2 -2ab + b 2 (a + b)(a - b)(a - b) ~ (a+b)(a - b)(a- b)' 
 
 Adding the numerators, we have 2 a 8 + 6 ab 2 ; whence the sum of 
 
 the fractions is 
 
 2 a 8 + 6 ab 2 
 
 (a + b)(a- b)(a-b)' 
 
268 LITERAL FRACTIONS 
 
 EXERCISES 
 
 Perform the following additions and subtractions : 
 
 1. 3 | 2 I a + b 8 2x * y y 
 
 4 7 3 a— b x? — y 2 x — y x + y 
 
 x-y x-y a + 1 q-1 
 
 (x + y) 2 tf — y 2 tf + a + l a 2 — a + 1 
 
 3 ^-9^ + 18 | x 1Q _&_, & b 2 
 
 or 2 -13 a; + 36 4 -a l_&'i + &2 1 _ &2 
 
 4 2 I CT I & 11 a^ + 1 ■ a + 1 , 3a+2 
 
 3 a + 6 a-& * x-2 x + 2 x 2 — ! 
 
 3 5 2 .. »-l aj+1 , ar>-5 
 
 2 3 -3 2 2 2 -3 4 2 4 -3 3 a + 1 a?-l a 2 -l 
 
 6 a 2 — 9 & 2 a 2 — 6 q& y 2 y y 
 
 a 2 +6& + 9& 2 a 2 -96 2 ' ' y 2 -l y + 1 l- y ' 
 
 7. ^±1 + ^^-2. 14. i-I L. + ^l. 
 
 »— y » + y « y x—y x+y 
 
 15 I ^ . 2 a 2 — a 
 2 6_ a " l "&2_ a 2 + & + a - 
 
 16 ^ + 4^ 
 
 a^ + y 3 » + y tf — xy + y 
 
 W. .- JL -. + : * * 
 
 1 — ar* 1 — a; 1 + aj + a^ 
 
 18. a ~ 3 CT ~ 1 
 
 q 2 -3a + 2 a 2_5 a + 6 o 2 -4a + 3 
 
ADDITION AND SUBTRACTION 269 
 
 19- . * .. + 2 
 
 20. 
 
 a^-5a;-14 tc-7 a,- 2 -9 a; + 14 
 
 a & 
 
 ac + ad—bc — bd a? — 2 a& + & 2 
 
 ,!. » +.,39 x. 3 
 
 a-5 a 2 + 3a-40 a + 8 
 
 22. ; " 1 " „-- ^,+ 4 
 
 2/ 2 + 82/+16 y(y + 4) y 2 (y+4) 
 
 23. = — + 
 
 a* + 4:X-60 v*-±x-12 
 
 n . a—c , c — a 2 
 
 24. -= „ + 
 
 25. 
 
 a 2 — c 2 a 2 + 2 ac + c 2 a — c 
 
 _9 8 
 
 ! + 7a;-18 ^ + 6^-16' 
 
 26. a + 2 | «~ 4 « + 2 
 
 a 2 -a-6 a 2 -7a + 12 a 2 -2a-8 
 
 27 . *^_ * + 1 
 
 ^-11*5 + 30 ^-36 a^-25 
 
 (z-l)(a + 2) + (a; + 2)(a-3) + (a;-l)(3-z)' 
 
 29 1 1 , 1 
 
 («-&)(&-c) (&-a)(c-d) (b-c)(c-d) 
 
 30 4 a-1 a 2 -38a-3 
 
 'a-3 a 2 + 3a + 9 a 3 -27 
 
270 LITERAL FRACTIONS 
 
 MULTIPLICATION AND DIVISION OF FRACTIONS 
 
 191. The product of a fraction and an integer. In arithmetic 
 the product of a fraction and an integer is obtained by multi- 
 plying the numerator of the fraction by the integer. 
 
 Thus, 2 x 4 = — and 7 x - = — • 
 3 3 5 5 
 
 Since in algebra a fraction is an indicated quotient, and since 
 multiplying the dividend multiplies the quotient, it follows that 
 in algebra also the product of a fraction and an integral expres- 
 sion is obtained by multiplying the numerator by the integral 
 expression. 
 
 That is, in general, a • - = = — 
 
 c c c 
 
 It is best to factor completely the expressions to be multi- 
 plied and keep them in the factored form until all possible 
 cancellations have been made. 
 
 EXERCISES 
 
 Find the following indicated products and reduce the frac- 
 tions to the simplest form. 
 
 1. (l-o) x 1 + a + a \ 6 . <V+9a;+18)x X ~ 
 
 a — 1 x 2 —2x — 15 
 
 2. (a 3 -?/ 3 ) X < ^ty. 7 . (1 _^) x - 1 ~ X • 
 
 x-y v J 1+x + a? 
 
 3. (a»-2«a + a ! )x^ 8. (27 a 3 - 1) x — ~^~. r- 
 
 x — a 9or+3a + l 
 
 5 - Jfra x(o! - 5a+4 >- 10 - *-****t& . 
 
MULTIPLICATION AND DIVISION 271 
 
 192. In arithmetic a fraction is divided by an integer by 
 multiplying its denominator or dividing its numerator by the 
 integer. 
 
 — ***-& ¥^fH- 
 
 Since multiplying the divisor or dividing the dividend divides 
 the quotient, it follows that an algebraic fraction is divided by 
 an integral expression by dividing the numerator or multiply- 
 ing the denominator by the integral expression. 
 
 That is, in general, - ■+■ c = — = — j- — 
 b be b 
 
 EXERCISES 
 
 Find the following indicated quotients and reduce the frac- 
 tions to their lowest terms : 
 
 , x 3 — w 3 / o , . 2 \ n # 2 4-4# + 4 , j, .. 
 
 1. ?--i- (ar + xy + y 2 ). 3. „„ — — — 5-(ar — 4). 
 
 z + y v * * } x 2 -2x-\-l v ; 
 
 2. t±t+(x>-y*). 4 . ^^^(1 + 3^+9^. 
 
 a — ?/ 1 — 9ar* 
 
 2^-35 ^_^_ 4a ._ 5)> 
 
 a? + 10 x + 21 
 
 6. ^7 ^ X + , 3 r 9 -*- (s 2 - x - 156). 
 
 or — 8 a; -f- 15 
 
 7. ^ + 6c-ad-M c _ d 
 
 xy — 4:X—3y + 12 x ' 
 
 x 2 + ax + bx + a6 
 a 2 -fax — 3#— 3 6 
 
 (cc 2 + aa? — 5 x — 5 a). 
 
 9. : -s- (3 r — xr + 3 s — xs). 
 
 mx — m—nx + n 
 
 10 - ^~Q 3; ~99 ^( a:2 + 9a; + 2 )- 
 
 ar — 9 x — 22 
 
272 LITERAL FRACTIONS 
 
 TO MULTIPLY A FRACTION BY A FRACTION 
 
 193. In arithmetic, to multiply a number by the quotient of 
 two numbers is the same as to multiply by the dividend and 
 then divide the product by the divisor. 
 
 E.g. 10 • — = 10 • 6 = 60; or, 10 • — = (10 • 18) + 3 =60. 
 2 5_/2.*\ . „_»«6 - 2-5 
 
 Likewise, | • - = U -5J--7 
 
 3 3-7 
 
 Since in algebra a fraction is an indicated quotient, to mul- 
 tiply a number by an algebraic fraction we multiply by the 
 numerator and divide the product by the denominator. 
 
 rru fl o fa \ . ac . . ac 
 
 Thus - r^-fr-v* *" w 
 
 Hence, the product of two algebraic fractions is a fraction 
 whose numerator is the product of the given numerators and 
 whose denominator is the product of the given denominators. 
 
 Illustrative Example. Multiply ^^ by £=§*±f 
 and reduce the resulting fraction to its lowest terms. 
 
 x 2 - 1 ^-3x+2 ^ (g-l)(ic + l)(x-2)(x- 1) 
 
 x 2 -7x + 10 x 2 + 2x+l (x - 2)(x - 5)(x + l)(* + 1) 
 
 _ (x- l)(x - 1) _ x 2 - 2 x + 1 
 (x-5)(x+l) x 2 -4x-5* 
 
 It is desirable to resolve each numerator and denominator 
 into prime factors, and then cancel all common factors before 
 performing any multiplication. 
 
MULTIPLICATION AND DIVISION 273 
 
 EXERCISES 
 
 Find the following indicated products and reduce each frac- 
 tion to its lowest terms : 
 
 1 3 arfy 2 6 az 3 2 . 4 3 10-2 
 
 2yz 2 9x*' 5 2 . 2 4 3 4 
 
 5o(q-6) 9(a + &) 2 6 3 2 . 2 3 5 4 . 7 2 6 • 3 2 
 
 3 c (a + 6) 15 (a 2 - 6 2 ) ' 5 2 3 4 • 2 3 5 4 • 7 3 
 
 12<*b 35 (c 2 +c5 + ft 2 ) . 7 fl 2 -^ 2^ + 43; + 2 
 ' 5(c 3 -6 3 ) 14c*&» ' ' x 2 -! 3a? + 6x 
 
 4 y» + 3y + 2 y»-7y + 12 g a 2 -10q+16 a + 3 
 ' y«_5 y + 6 f + Sy + 7 ' a 2 +6a + 9 a 2 -4' 
 
 9# (x + yf-z 2 x a; x (a; - y) 2 - z\ 
 
 x 2 + xy — xz (x — zf—y* xy — y 2 — yz 
 
 10 a 2 + 7a + 12 3a 3 + 27a 2 + 42a 
 a 3 + 5a 2 + 6a 6a 2 + 66a + 168 ' 
 
 3<(a - 5)(a + 2) 2 2 (a - 3)(a - 5) 
 ' 2 3 (a-3)(a-2) 3 3 (a-5) 2 
 
 12. 
 
 3(g + 4) 2 Q-7) 2 
 
 4(« + 4)(as-7) 3(a> + 4)(a>-7)' 
 
 13 a 3 +5a 2 -36a (q-16)(a-3) 
 
 a 2 -7a-144 a(a-4)(a + 2)" 
 
 14 3a(a-7)(a-5) 6(a-3)(a+10) 
 7 6(a-3)(a-7) a(a-5)(a-10) 
 
 15 42(6-3)0-4) 6(6 + 7)0-5) 
 
 3(6-4)(6+7) 14(6-3)(6-6) 
 
 a(»-<*) (6»-q»)» (6 + a) 2 
 
 6(6 + a) b 2 + ba + a 2 (6 - a) 2 
 
274 LITERAL FRACTIONS 
 
 17 a 2 -4g + 3 a 2 - 9a + 20 a 2 -la 
 a 2 - 5 a + 4 a 2 - 10 a + 21 X a 2 - 5 a' 
 
 18 3 a(a - 7)(o - 5) 6(a - 3)(o + 10) 
 
 ' 7 6(a-3)(a-7) a(a-5)(a- 10)' 
 
 19 a 2 - 12a + 35 a 2 -8a + 15 
 
 ' c(a 2 - 10a + 21) a 2 + 3a-70' 
 
 20 c*-10c + 21 c^-3c-88 
 
 c 2 - 18 c + 77 c 2 -8c + 15' 
 
 ar^ar 8 - 16) y 2 - 12 y + 35 y«( g + 5) 
 
 ' y 2 (y 2 -3 y- 28) a?-9x+2Q a?(x-4) 
 
 22. 
 
 3 a 2 6 2 (c 2 -14 c+33) 2(c 2 + c - 56) c 2 -2c-15 
 
 4 a* (c 2 - 10 c + 21) a 2 (c 2 - 16 c + 55) 6 2 (c 2 + 12 c + 32) ' 
 
 St 2 -2t-l 2t 2 + 5t-S 4t 2 + 10t + 4, 
 2t 2 + t-l 3t 2 + 7t + 2 4:t 2 -2t-2' 
 
 m . 6^-7^ + 2 „6a?-5x-l 10a^ + 3a;-l 
 ' 10a^-7a; + l 6^+a-l Ztf-kx-l 
 
 45 2 -175+4 105 2 -216 + 9 36 2 -5 5 + 2 
 • 66 2 -76 + 2 56 2 -236 + 12 46 2 -56 + l" 
 
 TO DIVIDE A FRACTION BY A FRACTION 
 
 194. In arithmetic, to divide a number by the quotient of 
 two numbers is the same as to divide by the dividend and 
 multiply the result by the divisor. 
 
 E.g. 72 + — = 72 + 6 = 12 ; 
 
 if 3 
 
 or, 72 - ^ = (72 - 18) . 3 = 4 . 3 = 12. 
 
 Likewise, jUf = (§ + B ) . 7 = (^) . 7 
 
MULTIPLICATION AND DIVISION 275 
 
 Since in algebra a fraction is an indicated quotient, to divide 
 a number by an algebraic fraction we divide by the numerator 
 and multiply this result by the denominator. 
 
 y b d \b J \b-cl 
 
 ad 
 be' 
 
 Hence, in algebra, as in arithmetic, a number is divided by a 
 fraction by inverting the fraction and multiplying by the new- 
 fraction thus obtained. 
 
 Thus, *^* S5S f x ^»«f. 
 
 b d b c be 
 
 EXERCISES 
 
 Perform the following indicated divisions, and reduce the 
 resulting fractions to their lowest terms : 
 
 q3-[-53 a + b g^-6o;-16 . x*+9x+U 
 
 d 2 -9b 2 ' a + 3b' ' x 2 +4a;-21 : tf-Sx+W 
 
 x 2 + x-2 . a 3 4-2^ x 2 -! . a? + 2x-3 
 
 x>-3x ' x* + 9x-36' ' ^-4^-5 ' x>-25 
 
 a 2 -11 a -26 , a 2 - 18 a + 65 
 a 2_3a_18 * a 2 -9a + 18 ' 
 
 e x 2 +9xy + lSy 2 . x 2 + 6xy + 9y 2 
 x 2 — 9 xy + 20 y 2 xy 2 — ky* 
 
 x 2 4- mx + nx + mn a? — n 2 
 x 2 — mx — nx + mn x 2 — m? 
 
 3a 4 -9a 8 -54a» . a 3 4-8a 2 4-15a 
 9 a 3 -117 a 2 + 378 a ' 3a 2 -33a + 84 # 
 
 ft a 2 - 11 a + 30 a 2 - 3 a . a 2 -9 
 
 a 3 -6a 2 4-9a a 2 -25 ' a 2 4-2a-15 
 
276 LITERAL FRACTIONS 
 
 10. 
 
 x* + x-56 x- 2 + 4x-32 
 
 a 2-3a + 2 a 2 -8a + 15 . a 2 -9a + 14 
 a 2 -7a + 12 « 2 -6a + 5 ' a 2 -12a + 32* 
 
 12 n 2 -ll?i + 18 x n 2 -8n-9 , 6n-12 
 m 2 — 7 ?« — 18 m 2 — 5 m — 14 an + a 
 
 13 a 2 -6 2 &(a-5) . b(a + b) 
 ' ab 2 x a 2 + 2a6 + 6 2 ' a 2 -2a& + & 2 
 
 14 a + 5 a 2 -& 2 . (a- &) 2 (a + fr) 2 
 ab 3(a 2 +6 2 ) ' 3 a 3 ?/ + 3 a?/ 3 
 
 5a5 4 -5ar> ^-9^ + 8; 
 
 15. 
 
 7a 2 -56x-63 14 a; 2 + 14 a; -1260 
 
 8y 2 (,y + 4)(y + 5) , y( y + 4)(y + 8) . 
 ' 2*(y + 6)(y-7) ' 2 3 G,-7)(j/ + ll) 
 
 17 (c + 4)(c-4) (c + 2)(c-5) . ( c + 4)(c-13) 
 
 • ( C _3)( c _2) ( c -5)(c-6) ' (c-6)(c + ll) 
 
 2a 2 6(6-4)(& + 4) (& + 6)(5+8) , a(& + 8)(&-4) 
 
 • 5( & + 4)(6 + 6) (6 + 8)(6-9) ' (&-9)(6-5) 
 
 19 ^(s-?) 2 (s + 2)(a-3) , x\x-3)(x- 5) 
 ' (x + 2) 2 (a>-2)(a>-7) ' (s-5)(a>-7) 
 
 a^ 2 (c + 5)(c-4) ( C -8)(c + 9) , o6(c+-9)(c-l) 
 
 • ( c _4)(c-8) (c + 4)(c + 7) " (c + 7)(c + l) 
 
 21^ + 23a;- 20 6 x 2 - 11 »- 10 . 7a 2 + 17a;-12 
 " 10^-27^ + 5 3a»+2o;-5 "" 5x> + 9x-2 ' 
 
COMPLEX FRACTIONS 277 
 
 s* + 6a; + 9 9a^ + 9a;-10 , 3a?-7a;-20 
 ' 9^-4 7a 2 + 20a;-3 ' 35a?-12a> + l" 
 
 23 (a + 4) 2 -6 2 x 3as-4 . a b* + 4:b 2 + b s 
 12 as 3 — 16 x* ay? 4- 4 a 2 — 6X 2 jc 3 
 
 12 ya^ 2 - 18 ya; 3 6s 8 4- 7 6a; . 3&x 3 + 76a; 2 
 15a; 2 + 32aj-7 6a^4-15y ' 10a; 2 4-23a;-5' 
 
 ac + bc — ar — br 6 a 2 + 23 a 4- 7 . 3 a 2 4- a 4- & + 3 afr 
 3 a 2 — 8 a — 3 c& — r& + c 2 — re ' a& — 3 6 4- ac — 3 c 
 
 COMPLEX FRACTIONS 
 
 195. Sometimes fractions occur whose numerators or de- 
 nominators, or both, contain fractions. 
 
 1 + 1 1 + . 1 
 
 E.g. 2 and x + 1 x ~ 1 . 
 
 * 1 1 _1 1_ 
 
 a x — 1 x + 1 
 
 Such fractions are called complex fractions. A complex frac- 
 tion is said to be simplified when it is reduced to an equal 
 fraction whose numerator and denominator are in the integral 
 form. 
 
 1+1 «+l 
 
 a a a a+1 a — 1 
 
 Ex.1. 
 
 ■a 1 a _ 1 a a 
 
 a a a 
 
 __ <x + l a _ a4-l 
 a a—1 a—1 
 
 This result may also be obtained directly by multiplying both 
 
 I 1 
 
 terms of the given fraction by a, finding at once "*" . 
 
 a — 1 
 
278 LITERAL FRACTIONS 
 
 1 + x 
 
 Ex.2. 
 
 x + 1 x — 1 
 
 x — 1 x+1 
 
 X — 1 + 
 
 x + 1 
 
 X 2 - 
 
 1 
 
 x + 1 — 
 
 x + 1 
 
 X 2 - 
 
 2x 
 > 
 
 ■ 1 
 
 . x 2 -! 
 
 x 2 -! 
 
 By multiplying the terms of the given fraction by (x + V)(x — 1) 
 
 ■r 1 4- X + 1 
 
 we may also get directly — — — :! — = x. 
 
 x — j~ X — X ~r x 
 
 Such fractions seldom occur in the problems of elementary 
 algebra. A more extended discussion is given in the Advanced 
 Course. 
 
 EXERCISES 
 
 Reduce each of the following complex fractions to its 
 simplest form : 
 
 m + n + 1 a? — 8 b 3 
 
 1. 1±£. 4. 3 7. 
 
 1 m — n — 1 3a — 2b 
 
 x 
 
 i_l! i i a + ft Jf 
 
 - ■ S 2 fl. 
 
 1 + a' . a — b 1 + d 2 
 
 4- 
 
 2 Z>2 
 
 , x tf-tf H hd 
 
 3. —1. 6. 4 9. c cD 
 
 a g-f-y 1 + £ 
 
 *~2 2 c. 
 
RATIO AND PROPORTION 279 
 
 RATIO AND PROPORTION 
 
 196. Definitions. A fraction is often called a ratio. Thus - 
 may be read the ratio of a to b, and is also written a : b. 
 
 The numerator is called the antecedent of the ratio, and the 
 denominator the consequent. The antecedent and consequent 
 are called the terms of the ratio. 
 
 An equation, each of whose members is a ratio, is called 
 a proportion. 
 
 Thus, - = - is a proportion, and is also written a : b = c : d. 
 b d 
 
 It is read the ratio of a tob equals the ratio of c to d, or briefly, 
 a is to b as c is to d. 
 
 The four numbers a, b, c, and d, are said to be in proportion. 
 a and d are called the extremes of the proportion, and 6 and c 
 the means. 
 
 IMPORTANT PROPERTIES OF A PROPORTION 
 
 1. If, in the proportion - = -, both members of the equation 
 b d 
 
 be multiplied by bd, we have, ad — be. 
 
 That is : If four numbers are in proportion, the product of the 
 means equals the product of the extremes. 
 
 2. Show that if - = c , then - = -. 
 
 b d a c 
 
 Hint. Divide 1 = 1 by the members of the given equation. 
 This process is called taking the proportion by inversion. 
 
 3. Show that if -=-, then - = -• 
 
 b d c d 
 
 Hint. Multiply both members of the given equation by - . 
 
 c 
 
 This process is called taking the proportion by alternation. 
 
280 LITERAL FRACTIONS 
 
 4. Show that if a - = c -, then ?±± = c _±*. 
 
 b d b d 
 
 Hint. Add 1 to both members of the given equation. 
 
 This process is called taking the proportion by composition. 
 
 5. Show that if - = -, then *I2*»£i±i!. 
 
 b d b d 
 
 Hint. Subtract 1 from each member of the given equation. 
 This process is called taking the proportion by division. 
 
 6. Show that if ?«*, then ^±& = ^M. 
 
 b d a— b c — d 
 
 Hint. Divide the members of the equation obtained under 4 by 
 the members of the one obtained under 5. 
 
 7. Show that if g = ^ = g, then a + c + e = g. 
 
 b d f b+d+f b 
 
 Let - = - = - = k : then a = bk, c = dk. e = fk. 
 b d f ■ J 
 
 Hence, a + c + e = bk + dk +fk =(b + d +/) k, 
 a + c + e 7 , a c e 
 
 and 
 
 b+d+f b d f 
 
 That is, If several ratios are equal, the sum of the antecedents 
 is to the sum of the consequents as any antecedent is to its 
 consequent. 
 
 EXERCISES 
 
 a c 
 
 1. If ad = be, show that- = -• Hint. Divide by bd. 
 
 b d 
 
 2. If ad = be, show that - = - • 
 
 c d 
 
 3. If ad = be, show that - = - • 
 
 c a 
 
RATIO AND PROPORTION 281 
 
 4. If ad = be, show that - = - ■ 
 b a 
 
 5. If - = -, show that — - — =— — - 
 b d a c 
 
 m T£ a c i ,i , o- b c — d 
 
 6. If - = - , show that = 
 
 b d a c 
 
 m T£ Cb C 1 i/U i. <* — b C — d 
 
 7. If -=-, show that = 
 
 b d a+b c+d 
 
 8. If ^ = C -, show that 1+b^a^dL 
 
 b d c+d c—d 
 
 9. If - = -, show that — ! — = -• 
 b d c+d c 
 
 10. If - = -, show that — - — =-• 
 b d b+d b 
 
 11. If- = -, show that 
 
 a — b a 
 
 b d c—d c 
 
 12. If - = -, show that ^=^ = -- 
 b d' b-d b 
 
 13. Solve the equation - = - for each letter in terms of all 
 
 6 d 
 
 the others. If a = 3, 6 = 5, c = 8, find d. If b = 7, c = 9, 
 d = 3, find a. If c = 13, d = 2, a = 5, find 6. Ifd = 50, a = 3, 
 6= -7, find c. 
 
 14. If - = -, then x is said to be a fourth proportional to a, b, 
 , b x 
 
 and c. 
 
 Find a fourth proportional to 3, 5, and 7 ; also to 9, 5, and 1, 
 and to 3, — 2, and — 5. 
 
282 LITERAL FRACTIONS 
 
 15. If - = -, then x is called a mean proportional between 
 a and b. 
 
 Solve the equation ~ = ~ for * in terms of a and & - Snow 
 
 that there are two solutions, each of which is a mean propor- 
 tional between a and b. 
 
 Find two mean proportionals between 4 and 9 ; also between 
 5 and 125, and between — 4 and — 36. 
 
 16. Which is the greater ratio „ , — or — t— — 
 
 5 + 4d 5+5d 
 
 Hint. Reduce the fractions to a common denominator and com- 
 pare numerators, (d is a positive number.) 
 
 17. Which is the greater ratio, a "** , or a ~*~ , t ♦ 
 
 a + 86 a + 10 6 
 
 18. Which is the greater ratio, - or ~*~ , if b and c are posi- 
 
 b b + c 
 
 tive, and a less than b ? a equal to b ? a greater than b ? 
 
 19. Find two numbers in the ratio of 3 to 5 whose sum is 160. 
 
 20. Find two numbers in the ratio of 2 to 7 whose sum is 
 -108. 
 
 21. Find two numbers in the ratio of 3 to — 4 whose sum 
 is -15. 
 
 22. What number added to each of the terms of the ratio 
 ^ makes it equal to f| ? 
 
 23. What number must be added to each term of the ratio 
 Jy to make it equal to the ratio f ? 
 
 24. What number added to each of the numbers 3, 5, 7, 10, 
 will make the sums in proportion, when taken in the given 
 order? 
 
 25. Two numbers are in the ratio of 2 to 3, and the sum of 
 their squares is 325. Find the numbers. 
 
EQUATIONS INVOLVING FRACTIONS 283 
 
 EQUATIONS INVOLVING FRACTIONS 
 
 197. We have already seen, § 40, that in solving an equa- 
 tion involving fractions the first step is to multiply both 
 members of the equation by a number which will cancel all 
 the denominators. Evidently, any common multiple of the 
 denominators is such a multiplier. For the sake of sim- 
 plicity, and for reasons which are fully discussed in the 
 Advanced Course, the lowest common multiple is always used 
 for this purpose. 
 
 Illustrative Example. Solve the equation : 
 
 x .x — 6_1— 2 a; _ 55 ,^. 
 
 4 +_ 6 8 "~8" U 
 
 Solution. The L.C.M. of 4, 6, and 8 is 24. Multiplying both 
 members of the equation by 24, 
 
 24* , 24(z-6) 24(1 - 2 x) _ 24 • 55 /0 . 
 
 ~T + 6 ~ 8 ~~ 8" {2) 
 
 ByF,V, 6x + 4(:r-6)-3(l - 2x)= 3 • 55 = 165. (3) 
 
 ByF, 6z + 4x-24-3 + 6x = 165. (4) 
 
 ByF, A, 16 a; = 192. (5) 
 
 By A x = 12. (6) 
 
 As in this solution, so in general, each denominator is can- 
 celed by Principle V, after multiplying by the L.C.M. In 
 practice, however, equation (2) may be omitted, the canceling 
 being done mentally, and equation (3) may be written down 
 at once. 
 
 The dividing line of the fraction acts like a parenthesis with 
 respect to the sign preceding the fraction. 
 
284 LITERAL FRACTIONS 
 
 EXERCISES 
 
 Solve the following equations, and check each solution by 
 substituting in the original equation : 
 
 4cc — 5 2x + 5 __ x — 3 . 7a; + 5 , 
 3 5 = 2 + 8 
 
 _ x + 3 5x — 3 5z — 3_ , 
 
 2a? + 5 = 5a; + 20 7a: + 13 15fl + 3 
 3 15 2 4 
 
 3s + 10 7a: + 15 5 a:-14 = 3 a;-12 2 
 5 3 7 4 
 
 ,3a + 5,7-3a; 15 — a; , 3 — 11 x 
 
 5. X-\ -—- + — 7> = — o 1 a 
 
 4a: 2x 2x 4 
 
 . 7aj-2 3a;-24 3z + 4 5-4a? 
 
 "• — ~ 7i " 
 
 2^_5» 7_a_^ = 45 
 3 12 8 2 
 
 2_x _ Zx 5x _ 11a; _ n* 
 
 3 4 7 " 12 = 
 
 „ s-3 . 2s+3 5s-3 3s-l Q 
 9 - -2^ + "^ 6~- = ~^ 3 - 
 
 ,- 2t + l £-1 , 4*-8 3*-l,„ 
 10 - "1 8 _ + "l5~ = ^0" + 3 - 
 
 ,„ z + 3 4» + 5 . 3 a-5_5a;-7 , . 
 "' ~2 7~~ + ~4 ^2~ + 4, 
 
EQUATIONS INVOLVING FRACTIONS 285 
 
 8fc-6 13fc 21k-12 = k-2 Uk-3 d 
 ' 5 2 5 10k 5 
 
 5a; — 1 2a; + 3 5a; + l _ 13a: + 5 1 
 ' 2 3 4 11 
 
 14 o ft-15 ft-20 4fe + 2 = Q 
 2^3 11 
 
 17 + ra 2m-7 . 5m— 3 . Q 3 
 15. . _ 1 _ — __f- ii __. 
 
 4 3 2 m 
 
 3r + 4 5r + l 8r + 4 = o 6r 
 5 12 10 7 * 
 
 17 9 + 2 */ 4y + 3 _ 27 + 3y s 
 17> "^ 3» 6~ " 
 
 1Q 2 a; + 5 5 a; + 2 , 4a;-5 _ 3a;-f 4 
 
 z+3 2z-15 5z-ll 11 =Q 
 '2 « 8 2' 
 
 OA 3a5 + 20.5a;-3 4a>-l Q 
 20. __ + — — =3. 
 
 198. Illustrative Example. Solve the following equation : 
 
 2a; — 1 4 _ 3a; „ m 
 
 x-1 x + 1 a*-l ' K ' 
 
 Solution. The L. C. M. of the denominators is x 2 — 1. In multi- 
 plying both members of the equation by x 2 — 1, x — 1 is canceled in 
 the first fraction, x + 1 in the second, and x 2 — 1 in the third, giving 
 
 (2x-l)(x+ l)+4(x-l)- 3x = 2(x 2 - 1) (2) 
 
 Solving, 2x 2 + x- l + 4x-4-3x = 2x 2 - 2, (3) 
 
 2 x = 3, (4) 
 
 and x = §. (5) 
 
 Check by substituting x = § in (1). 
 
286 LITERAL FRACTIONS 
 
 EXERCISES 
 
 Solve the following equations and check each solution by 
 substituting in the original equation : 
 
 - Sx-1 4s + 3 x 2 27 1 
 
 2 3a; + 5 2x + l_ x — 1 
 
 x-9 x + 2 a?-7x-18 
 
 3 3a;-4 4cc-l ar>+44 _ Q[ 
 
 aj + 5 a?+4a? + 9aj + 20 - 
 
 4 4-s 2 3a; + 6 = 2a: + l 
 
 ic 2 — 5ic — 14 a:+2 ~ a- — 7 
 
 a;_4 3a-15 3^-114 
 
 5. 
 
 2z-10 2z-6 4a 2 -32a:+60 
 
 6 6(a;+4) 3(2a?-l) = 7 [ 
 a + 5 a + 1 2* 
 
 „ 3z-4 . 5a-7 9a^-38 
 
 7 - r + ; 
 
 a;-4 2a;-2 2^-10^ + 8 
 
 g a; + 17 2(a; + 6) = a;-l 
 a; + 5 ar + 3 ' a:-f-3 
 
 9 a; + 2 3a;-15 = 3a:-21 
 ic — 5 a; — 3 x — 3 
 
 2a-3 . 3a + l 4z + 17 
 
 10. 
 
 -4a; <c — 2 a; — 2 
 
EQUATIONS INVOLVING FRACTIONS 287 
 
 11 3x-2 = 2x i + 15x + 2$ 2x-\ 
 " 23 + 3 2^ + 53 + 3 3 + 1 ' 
 
 23-3 3-8 _ x + 2 
 
 23 + 2 53 + 2 23 + 2* 
 
 12. 
 
 20^ + 73-3 3a; + l =1 
 9a^-l 33-1 
 
 73 2 + ll3 + 4 3 + 3 ^ 73 + 11 
 6^ + 13a; + 5 23 + 1 33 + 5 
 
 33 + 1 a;-3 _ = 2ar 2 -10a; + 12 
 5o;-7 23-7 10.-^-493 + 49' 
 
 199. Sometimes it is best to add fractions before multiplying 
 by the L. C. M. and in other cases to multiply by the L. C. M. 
 of part of the denominators first, and, after simplifying, multiply 
 by the L. C. M. of the remaining denominators. 
 
 Ex. 1. Solve the equation 
 
 111 
 
 3—2 3 — 1 3 — 4 3 — 3 
 
 Adding fractions on the right and left, 
 1 1 
 
 (3-2)(3-l) (3-4)(3-3) 
 
 (1) 
 
 (2) 
 
 Multiplying by L. CM., (3-4)(3-3)=(3-2)(3-l). (3) 
 
 Hence 4 3 = 10, (4) 
 
 and 3 = 2\. (5) 
 
 Check by substituting 3 = 2£ in equation (1). 
 
288 LITERAL FRACTIONS 
 
 Ex. 2. Solve the equation : 
 
 4t-3 t-2 = 2t-2 
 16 4 5t + 2 
 
 3. 
 
 8 dx + 2 
 
 7* + 3 21t + 9 = 17<-3 2 
 ' 5 15 3* + ll 
 
 11^-15 33^ + 15 ^5^ + 5 
 10 30 v-5 ' 
 
 x — 1 x — 2 _ x — 3 a — 4 
 #— 2 3 — a; a; — 4 a; — 5 
 
 (1) 
 
 Multiplying by 16, 4 i - 3 -4 « + 8 = ^|f=^- (2) 
 
 of-)-* 
 
 Hence, ^W" < 3 > 
 
 Multiplying by 5 1 + 2, 25 £ + 10 = 32 £-32. (4) 
 
 Hence, 7* = 42, (5) 
 
 and I-.6L (6) 
 
 Check by substituting £ = 6 in equation (1). 
 
 EXERCISES 
 
 3x + 6 9a; + 3 _ x + 1 * 
 5 15 6 x -8 
 
 7a:+l 14s-22 = lla; + 5 
 ' 12 24 8 a- 28* 
 
 3z + 4 12a + l 5ic-l 
 
EQUATIONS INVOLVING FRACTIONS 289 
 
 ? 1 2 1 4 
 
 X 
 
 -1 
 
 2 
 
 X + l 
 
 a; 
 
 -2 
 
 4a: + l 
 
 X 
 
 X 
 
 -2 
 -3 
 
 a; 
 
 X 
 
 -3 
 -4 
 
 X — 
 X — 
 
 i* 
 
 x — 5 
 6 — x 
 
 X 
 
 9 
 
 -7 
 
 X 
 
 9 
 
 -2 
 
 5 
 
 X — 
 
 8 
 
 5 
 
 x + l 
 
 X 
 
 -1 
 
 + x 
 
 -2 
 
 X — 
 
 3 
 
 x — 5 
 
 9. 
 
 10. 
 
 x— 2 3— # #— 4 a— 6 
 
 SIMULTANEOUS FRACTIONAL EQUATIONS 
 200. When pairs of fractional equations are given, each 
 should be reduced to the integral form before eliminating, ex- 
 cept in special cases like those in the second following illustra- 
 tive example. 
 
 Ex. 1. Solve the equations : 
 
 x — y x + y tf — y* 
 3 2 -18 
 
 (2) 
 
 \2x-y x—3y (2x — y)(x-3y) 
 
 From (1) by M, 4 (x + y) + 6(a> - y) = 36. (3) 
 
 BjF,D, 5x-y = l8. (4) 
 
 From (2) by M, 3(x - Sy) - 2 (2 x -y) = - 18. (5) 
 
 VyF, D, x + ly = l%. (6) 
 
 From (4) by M, 35x-ly = 125. (7) 
 
 Adding (6) and (7), 36 x = 144. (8) 
 
 ByZ>, a = 4. (9) 
 
 Substitute x = 4 in (6), y « 2. (10) 
 Check by substituting x = 4, y = 2 in (1) and (2). 
 
290 
 
 LITERAL FRACTIONS 
 
 Ex. 2. Solve the equations 
 
 x y 
 
 20_21 
 Las y 
 
 = 3. 
 
 (1) 
 
 (2) 
 
 In this case it is best to solve the equations for - and - in- 
 stead of for x and y. 
 
 From (1) by M, — + 
 
 x 
 
 ?i = 14. 
 
 y 
 
 (3) 
 
 Adding (2) and (3), 
 
 21 = 17. 
 
 X 
 
 (4) 
 
 Hence by D, 
 
 1_1 
 
 x 2 
 
 (5) 
 
 Substituting - = - in (1), 
 
 1 1 
 
 y 3* 
 
 (6) 
 
 From (5) and (6) by M, 
 
 x = 2,y = 3. 
 
 (7) 
 
 EXERCISES 
 
 
 Solve the following equations : 
 
 
 1. { 
 
 2a?-l 3y-l _ 
 
 — «y 
 
 oj + 1 y+1 (aj + lXy + l)' 
 
 cc -f- 2 2x-l_ 5xy 
 
 I2y-1 y + 1 (2y-l)(y+l) 
 
 2. 4 
 
 3a5 + 2_:B + l 
 3^-5 f — 1* 
 3a;-2 3z-l 
 
 2 
 
 y + 1 y-1 (2,_l)(y + l) 
 
3. 
 
 PROBLEMS INVOLVING FRACTIONS 
 (3 2 
 
 291 
 
 5 - 6 - = 2, 
 t v 
 
 V t 
 
 f - + - = 21, 
 
 x y 
 
 I-S— la 
 
 a; 
 
 y 
 
 a o 
 3 
 
 ft 
 
 + 1 -? = 24. 
 
 la & 
 
 6. < 
 
 !l 
 
 = -4, 
 
 - 6 + Ll = 5 2. 
 
 7.4 
 
 ri2_io_ 1 
 
 a; ?/ 
 
 a 6 
 
 a 2/ 
 
 c , d 
 
 U y 
 
 h 
 1. 
 
 9. < 
 
 10.^ 
 
 r 1 . 1 ia 
 
 - + - = 16, 
 
 a; j/ 
 
 1 + 1=14, 
 
 y z 
 
 i+i=i2. 
 
 2 X 
 
 1 ■ 1 
 
 - + -=«, 
 
 x y 
 
 - + -=&, 
 y z 
 
 - + - = c 
 Iz a; 
 
 <> 
 
 In 9 first add all three equations, and from half the sum subtract 
 each equation separately. Likewise in 10. 
 
 PROBLEMS LEADING TO FRACTIONAL EQUATIONS 
 
 In solving the following problems use one or two unknowns 
 as may be found most convenient. 
 
 1. There are two numbers whose sum is 51 such that if the 
 greater is divided by their difference, the quotient is 3^. Find 
 the numbers. 
 
 2. There are two numbers whose sum is 91 such that if the 
 greater is divided by their difference, the quotient is 7. Find 
 the numbers. 
 
 3. There are two numbers whose sum is s such that if the 
 greater is divided by their difference, the quotient is q. Find 
 an expression in terms of s and q representing each number. 
 Solve 1 and 2 by substituting in the formula just obtained. 
 
 4. What number must be subtracted from each term of the 
 fraction \\ so that the result shall be equal to ^ ? 
 
292 LITERAL FRACTIONS 
 
 5. What number must be subtracted from each term of the 
 fraction f ^ so that the result shall be equal to f ? 
 
 6. What number must be subtracted from each term of the 
 
 fraction" - so that the result shall be equal to - ? Solve 4 and 
 o d 
 
 5 by substituting in the formula obtained under 6. 
 
 7. What number must be added to each term of the fraction 
 \ to obtain a fraction equal to \$ ? 
 
 8. What number must be added to each term of the fraction 
 
 - to obtain a fraction equal to — ? 
 b H d 
 
 By means of the formula thus obtained, solve problem 7, and also 
 4, 5, and 6. (See work under Problem 23, p. 114.) 
 
 9. There are two numbers whose difference is 153. If their 
 sum is divided by the smaller, the quotient is equal to f£. 
 Find the numbers. 
 
 10. There are two numbers whose difference is b. If their 
 sum is divided by the smaller, the quotient is q. Find the 
 numbers. Solve 9 by substituting in this formula. 
 
 11. Divide 548 into 2 parts, such that 7 times the first 
 shall exceed 3 times the second by 474. 
 
 12. There are two numbers whose sum is 48 such that 3 
 times the first is 8 more than 5 times the second. Find both 
 numbers. 
 
 13. There are two numbers whose sum is s such that a times 
 the first is b more than c times the second. Find both numbers. 
 
 14. What number must be subtracted from each of the num- 
 bers 12, 15, 19, and 25 in order that the remainders may form a 
 proportion when taken in the given order ? 
 
 15. What number must be added to each of the numbers 13, 
 21, 3, and 8 so that the sums shall be in proportion when taken 
 in the order given ? 
 
PROBLEMS INVOLVING FRACTIONS 293 
 
 16. What number must be added to each of the numbers' a, 
 b, c, d so that the sums shall be in proportion when taken in 
 the given order ? 
 
 17. What number must be subtracted from each of the 
 numbers a, b, c, d so that the remainders shall be in proportion 
 when taken in the given order ? 
 
 Compare the results in 16 and 17 and explain the relation between 
 them. (See remark under Problem 23, page 114.) 
 
 Solve 14 and 15 by substituting in the formulas obtained in 16 
 and 17. 
 
 18. There is a number composed of two digits whose sum is 
 11. If the number is divided by the difference between the 
 digits, the quotient is 16f. Find the number, the tens' digit 
 being the larger. 
 
 19. There is a number composed of two digits whose sum 
 is s. If the number is divided by the difference between the 
 digits, the quotient is q. Find the number, the tens' digit being 
 the larger. 
 
 20. Illustrative Problem. A can do a piece of work in 8 days, 
 B can do it in 10 days. In how many days can they do it 
 together ? 
 
 Since A can do the work in 8 days, in one day he can do \ of it, 
 and since B can do it in 10 days, in one day he can do T ^ of it. If x 
 is the number of days required when both work together, in one day 
 
 they can do - of it. Hence we have the equation, 
 
 x 
 
 1 + 1 = 1 
 8 10 a; 
 
 21. A can do a piece of work in 12 days and B can do 
 it in 9 days. How long will it take both working together 
 to do it ? 
 
 22. A pipe can fill a cistern in 11 hours and another in 13 
 hours. How long will it require both pipes to fill it ? 
 
294 LITERAL FB ACTIONS 
 
 23. A can do a piece of work in a days and B can do it in b 
 days. How long will it take both together to do it ? 
 
 24. A cistern can be filled by one pipe in 20 minutes and 
 emptied by another in 30 minutes. How long will it take to fill 
 the cistern when both are running together ? 
 
 25. A pipe can fill a cistern in 12 hours, another in 10 hours, 
 and a third can empty it in 8 hours. How long will it require 
 to fill the cistern when they are all running ? 
 
 26. A man can do a piece of work in 18 days, another in 21 
 days, a third in 24 days, and a fourth in 10 days. How long 
 will it require them when all are working together ? 
 
 27. A and B working together can do a piece of work in 12 
 days. B and C working together can do it in 13 days, and A 
 and C working together can do it in 10 days. How long will 
 it require each to do it when working alone ? 
 
 Suggestion : Let a = the fraction of the work A can do in one day, 
 b = the fraction of the work B can do in one day, and c = the fraction 
 of the work C can do in one day. 
 
 Then a + b-^, b + c = ^, c + a = ^. 
 
 28. A and B working together can do a piece of work in I 
 days. B and C can do it in m days and C and A can do it in 
 n days. How long will it require each working alone ? 
 
 29. The circumference of the rear wheel of a carriage is 1.8 
 feet more than that of the front wheel. In running one mile 
 the front wheel makes 48 revolutions more than the rear wheel. 
 Find the circumference of each wheel. 
 
 If x is the number of feet in the circumference of the front wheel, 
 
 then — — is the number of revolutions in going one mile. 
 x 
 
 30. The circumference of the rear wheel of a carriage is 1 
 foot more than that of the front wheel. In going one mile the 
 two wheels together make 920 revolutions. Find the circum- 
 ference of each. 
 
MISCELLANEOUS PROBLEMS 295 
 
 31. The distance from Chicago to Minneapolis is 420 miles. 
 By increasing the speed of a certain train 7 miles per hour the 
 running time is decreased by 2 hours. Find the speed of the 
 train. 
 
 4°0 
 If r is the original rate of the train, then — — is the running time. 
 
 r 
 
 32. The distance from New York to Buffalo is 442 miles. 
 By decreasing speed of a fast freight 8 miles per hour the 
 running time is increased 4 hours. Find the speed of the 
 freight. 
 
 33. A motor boat goes 10 miles per hour in still water. In 
 10 hours the boat goes 42 miles up a river and back again. 
 What is rate of the current ? 
 
 34. A train leaving New York over the Pennsylvania Boad 
 requires 9 hours to overtake a train leaving Philadelphia west- 
 ward at the same time. If the Philadelphia train had started 
 toward New York, they would have met in one hour. Find the 
 rate of each train, the distance from New York to Philadelphia 
 being 90 miles. 
 
 35. The average of the numbers x, 34, 0, — 58, — 19, 0, — 20, 
 and y is 12; while the average of 2x,3y, — 18, 50, and — 30 is 
 — 4. Find x and y. 
 
 36. The length of a rectangle is 8 feet greater, and its width 
 is 4 feet greater, than the side of a certain square. The sum 
 of the areas of the square and rectangle is 736 square feet. 
 Find the dimensions of each. 
 
 37. The difference between the areas of a circle and its cir- 
 cumscribed square is 12 square inches. Find the radius of 
 the circle. (See problem 33, p. 239.) 
 
 38. The difference between the areas of a circle ard its 
 inscribed square is 12 square inches. Find the radius of the 
 circle. 
 
296 LITERAL FRACTIONS 
 
 39. The difference between the areas of a circle and the 
 regular inscribed hexagon is 12 square inches. Find the radius 
 of the circle. 
 
 40. The altitude of an equilateral triangle is 6. Find its 
 side and also its area. Find the side and area, if the altitude 
 is h. 
 
 41. The radius of a circle is 3 feet. Find the area of the 
 regular circumscribed hexagon. Find the area if the radius 
 is r feet. 
 
 42. The radius of a circle is r. Find the difference between 
 the areas of the circle and the regular circumscribed hexagon. 
 
 43. The difference between the areas of a circle and the 
 regular circumscribed hexagon is 9 square inches. Find the 
 radius of the circle. 
 
 44. A circle is inscribed in a square and another circum- 
 scribed about it. The area of the ring formed by the two cir- 
 cles is 25 square inches. How long is the side of each square? 
 
 45. A square is inscribed in a circle and another circum- 
 scribed about it. The area of the strip inclosed by the two 
 squares is 25 square inches. Find the radius of the circle. 
 
 In the following five problems solve the resulting literal equations 
 for each letter in terms of the others, and for each solution state a 
 corresponding problem. 
 
 46. A hound pursuing a deer gains 400 yards in 25 minutes. 
 If the deer runs 1300 yards a minute, how fast does the hound 
 run ? If the hound gains v x yards in t minutes and the deer 
 runs v 2 yards per minute, find the speed of the hound. 
 
 47. A disabled steamer 240 knots from port is making only 
 4 knots an hour. By wireless telegraphy she signals a tug, 
 which comes out to meet her at 17 knots an hour. In how 
 long a time will they meet ? If the steamer is s knots from 
 port and making i>j knots per hour, and if the tug makes v 2 
 knots per hour, find how long before they will meet. 
 
MISCELLANEOUS PROBLEMS 297 
 
 48. A motor boat starts 7| miles behind a sailboat and runs 
 11 miles per hour while the sailboat makes 6^ miles per hour. 
 How far apart will they be after sailing 1^ hours ? If the 
 motor boat starts s miles behind the sailboat and runs v y miles 
 per hour, while the sailboat runs v 2 miles per hour, how far 
 apart will they be in t hours ? 
 
 49. An ocean liner making 21 knots an hour leaves port 
 when a freight boat making 8 knots an hour is already 1240 
 knots out. In how long a time will the two boats be 280 knots 
 apart ? Is there more than one such position ? If the liner 
 makes v^ knots per hour and the freight boat, which is s t knots 
 out, makes v 2 knots per hour, how long before they will be s 2 
 knots apart? 
 
 50. A passenger train running 45 miles per hour leaves one 
 terminal of a railroad at the same time that a freight running 
 18 miles per hour leaves the other. If the distance is 500 miles, 
 in how many hours will they meet ? If they meet in 8 hours, 
 how long is the road ? If the rates of the trains are v x and v 2 
 and the road is s miles long, find the time. 
 
 51. In going 1200 yards the rear wheel of a carriage makes 
 60 revolutions less than the front wheel. If the circumference 
 of each wheel be increased by 3 feet, the rear wheel will make 
 only 40 revolutions less than the front wheel. Find the cir- 
 cumference of each wheel. 
 
SCIENCE. 
 
 High School Physics. 
 
 By Professor Henry S. Carhart, of the University of Michigan, and 
 H. N. CHUTE, of the Ann Arbor High School. New edition, thor- 
 oughly revised. i2rao, cloth, 440 pages. Price, $ 1.25. 
 
 NO other text-book on Physics, published in this country, has 
 ever enjoyed the popularity or the success that has, from 
 the first year of its publication, attended Carhart and Chute's 
 High School Physics. Throughout the country the demand for 
 the book has been far in excess of that for any other manual 
 covering the same field, while in many states it has been used 
 more widely than all its competitors combined. The new edition 
 of the book is a distinct improvement on its predecessor. A 
 comparison of the two books will show numerous changes in 
 details and a smoothing down of the rougher spots. Physics is 
 not an easy subject ; but the authors have made an honest effort 
 to relieve the difficulties for immature students without such an 
 emasculation of the subject as to diminish its value for either 
 discipline or scientific information. 
 
 The problems have all been replaced by entirely new ones, in 
 which the purpose kept in view is the concrete illustration of 
 principles, with a minimum of arithmetical computation. Atten- 
 tion has been given also to the careful grading of the problems. 
 None of them have been inserted as puzzles to test the student's 
 intellectual skill, while, as a whole, they are distinctly easier than 
 were those of the preceding edition. 
 
 The book remains, as it was before, the most attractive manual 
 for the study of Physics that has been published. Its method 
 cannot be improved. The principles of the science are stated in 
 a simple, clear, and direct manner ; their application is illustrated 
 by apt experiments ; finally, numerous practical problems test the 
 pupil's mastery of the subject. 
 
 Its arrangement of material, its completeness along the lines of 
 the most recent developments of physical science, its excellent 
 woodcuts, its wealth of problems, and the ease and precision of 
 its style deserve and have received unstinted praise. 
 
 62 
 
SCIENCE. 
 
 First Principles of Chemistry. 
 
 By Raymond B. Brownlee, Far Rockaway High School; Robert 
 W. Fuller, Stuyvesant High School ; William J. Hancock, Eras- 
 mus Hall High School; Michael D. Sohon, Morris High School; 
 and Jesse E. Whitsit, DeWitt Clinton High School; all of New 
 York City. 
 
 THIS manual includes a treatment of the common elements 
 and their important compounds, together with a full dis- 
 cussion of the fundamental theories of chemistry. These theo- 
 ries are simply, clearly, and accurately stated without being buried 
 in a confusing mass of detail. The recent developments in chem- 
 ical theory are given due recognition. 
 
 In the development of laws and hypotheses, the historical order 
 is followed, the experimental facts in each case being described 
 before stating the discovery that resulted from them. Theoret- 
 ical topics are introduced as soon as the pupil is able to take in 
 their full significance ; this, on the ground that to delay their 
 presentation is to deprive the pupil of a very useful tool in his 
 acquisition of chemical ideas. 
 
 This book marks a step in advance by its omission of much 
 material that has found a place in other elementary manuals 
 through tradition rather than for its real value. On the other 
 hand, a number of the metallic elements, sometimes neglected, are 
 given the thorough treatment that their industrial importance 
 deserves. One of the chief aims of the descriptive matter is to 
 show the student the many points of contact between the life 
 about him and the science he is studying. 
 
 An important feature of the book is the brief summary and the 
 test exercises given at the end of each chapter. The summary 
 is a series of pithy statements emphasizing the essentials and 
 affording systematic review. 
 
 This book has been prepared by five teachers of long experi- 
 ence in both college and secondary school work. Three years of 
 careful discussion and revision by the authors have produced a 
 unified text-book especially planned for the beginner in chemistry. 
 
 56 
 
SCIENCE. 
 
 Practical Physiography. 
 
 By Dr. Harold Wellman Fairbanks, of Berkeley, California. 
 8vo, cloth, 570 pages. 403 Illustrations, Price, $1.60. 
 
 IN this volume the author has tried to work out a practical, con- 
 crete treatment of the subject of Physical Geography. The 
 book is intended as an aid to study, not as a compendium of 
 information ; consequently a description of the world as a whole 
 is omitted. Attention is devoted specifically to the region of the 
 United States, and the typical examples afforded by it are studied 
 as representatives of universal processes. 
 
 The purely descriptive method has been discarded as far as 
 practicable, the object being to lead the student to investigate 
 and find out for himself. Instead of being told everything, he is 
 asked to use his observing and reasoning powers. 
 
 No separate chapters are devoted to the relation between 
 physical nature and life, but, instead, this relation is brought 
 out in its appropriate place in connection with each topic through- 
 out the book. Such an arrangement, it is believed, will make the 
 whole matter much more vital. 
 
 Another feature which the author trusts will meet with favor 
 from the practical teacher is the distribution of the questions and 
 exercises throughout each chapter, in close connection with the 
 descriptive portions to which they refer. The placing of ques- 
 tions and exercises by themselves at the close of each chapter, as 
 is done in many text-books on the subject, puts a premium upon 
 mere memorizing of the text and the omission of all practical 
 work. It is not expected, however, that the pupils will be able 
 to answer all the questions without aid and direction from the 
 teacher. 
 
 The illustrations are a marked feature of the book. Photo- 
 graphs have been used wherever possible, as they appeal with 
 much more force to pupils of high school age than do diagrams 
 or sketches. Most of the views are from the author's own 
 
 negatives. 
 
 * 68 
 
SCIENCE. 
 
 Lessons in Elementary Botany for Secondary 
 Schools. 
 
 By Professor Thomas H. Macbride, Iowa State University. i6mo, 
 cloth, 244 pages. Price, 80 cents. 
 
 THIS is a new edition of Professor Macbride's well-known 
 text-book, enlarged and revised. This edition contains 
 ten new full-page plates, and while it calls for no material 
 but such as is easily accessible to every teacher, its method 
 is nevertheless in perfect harmony with the best instruc- 
 tion of the time. The student is not asked to study illustra- 
 tions or text ; he is sent directly to the plants themselves 
 and shown how to study these and observe for himself the vari- 
 ous problems of vegetable life. The plants passed in review are 
 those which are more or less familiar to every one, and the life 
 history of which every child should know. 
 
 Austin C. Apgar, State Normal School, Trenton, N.J. : There are many 
 points in the book which please me. Not the least of these is that the 
 author has not been misled by the craze for that " logical order " which 
 begins with protoplasm and, some time or other, if the subject be pursued 
 long enough, reaches such things as can easily be found and examined. 
 He begins with the known and gradually advances to the unknown, — 
 the only order in which successful teaching can be accomplished. 
 
 A Key to Native Plants. 
 
 To accompany Macbride's Elementary Botany. 
 
 IN response to a desire expressed by many teachers, Professor 
 Macbride has prepared a Key to the More Common Species 
 of Native and Cultivated Plants occurring in the Northern United 
 States. The material thus furnished has never before been 
 offered in such convenient form for class-room work. Its use will 
 in no way change the original plan of the Elementary Botany, 
 but it offers many advantages for additional study. 
 
 The Key will be furnished without charge to those ordering 
 the author's Botany. Copies ordered without the Botany will be 
 
 furnished at twenty-five cents each. 
 
 60 
 
MA THE AT A TICS. 
 
 A Practical Course in Bookkeeping. 
 
 By Carlos B. Ellis, of the Central High School, Springfield, Mass. 
 8vo, cloth, coo pages. Price, $0.00. 
 
 THIS book is the outgrowth of long experience in teaching 
 and in actual business. It is designed to give pupils the 
 knowledge of the principles of bookkeeping and of the customs 
 of business that every man needs, together with sufficient practice 
 in the application of these principles and customs to meet the 
 requirements of the best business offices. 
 
 A feature of the book is its logical development of principles 
 whereby the pupil makes a careful study of the essential accounts 
 instead of merely memorizing rules. The pupil who understands 
 the uses of the principal accounts and the relations which each 
 account sustains to the others, will not need to remember rules 
 for journalizing. In a word, the book is designed to teach the 
 pupil to think. 
 
 A large amount of work is provided for supplementary drill, 
 and there is abundant practice in the use of business forms and 
 of the variations in method adapted to special requirements. 
 
 Introduction to Geometry. 
 
 By William Schoch, of the Crane Manual Training High School, 
 Chicago. i2mo, cloth, 142 pages. Price, 60 cents. 
 
 THE aim of the author is to furnish a series of exercises and 
 problems that will enable the pupil to pass easily from the 
 individual notions of his old experience to a firm knowledge of 
 the elementary facts and concepts of Geometry. The pupil is 
 to convince himself of geometric truths through measuring and 
 drawing, and thus gradually to come into possession of the mate- 
 rial with which formal Geometry deals. Along with this, the 
 practical side of the subject is taken into account, the pupil's 
 knowledge being tested constantly by his power to apply it. 
 
 61 
 
MA THE MA TICS. 
 
 Plane and Spherical Trigonometry. 
 
 By President ELMER A. Lyman, Michigan State Normal College, and 
 Professor EDWIN C. GODDARD, University of Michigan. Cloth, 146 
 pages. Price, 90 cents. 
 
 THIS text-book aims to furnish sufficient material in analytical 
 trigonometry, and also in the solution of the triangle, both 
 of which have been adequately met heretofore by no single book. 
 Particular attention has also been given to the proofs of the 
 formulae for the functions of a ± y8. Nearly all other text-books 
 treat the same line as both positive and negative in the same 
 discussion, thus vitiating the proof, and in many cases proofs 
 are given for acute angles and are then supposed to be estab- 
 lished without further discussion for all angles. These diffi- 
 culties have been avoided by so stating the proofs that the 
 language applies to figures involving any angles and proving 
 the general case algebraically to avoid drawing an indefinite 
 number of such figures. 
 
 Computation Tables. 
 
 By Lyman and Goddard. Price, 50 cents. 
 
 Plane and Spherical Trigonometry, with Com- 
 putation Tables. 
 
 By Lyman and Goddard. Complete edition. Cloth, 214 pages. 
 Price, $ 1.20. 
 
 Principles of Plane Geometry. 
 
 By J. W. MacDonald, Agent of the Massachusetts Board of Educa- 
 tion. i6mo, paper, 70 pages. Price, 30 cents. 
 
 A Primary Algebra. 
 
 By J. W. MacDonald, Agent of the Massachusetts Board of Educa- 
 tion. i6mo, cloth. The Complete Edition, 218 pages (containing the 
 Teacher's Guide, the Student's Manual, and Answers to Problems). 
 Price, 75 cents. The Student's Manual, cloth, 92 pages. Price, 30 cents. 
 64 
 
MATHEMATICS. 
 
 Plane Trigonometry. 
 
 By Professor R. D. BOHANNAN, of the State University, Columbus, 
 Ohio. Cloth, 379 pages. Price, $2.50. 
 
 SOME of the features of this book are the use of triangles hav- 
 ing variety of form, but always such form as they should 
 have if they had been actually measured in the field ; the sugges- 
 tion of simple laboratory exercises ; problems from related sub- 
 jects, such as Physics, Analytical Geometry, and Surveying. 
 
 With a view to minimize the difficulties of the subject, the 
 author has ventured to present the subjects in the following 
 order: the sine, its inverse and reciprocal; the cosine, its in- 
 verse and reciprocal ; the sine and the cosine family in union ; the 
 tangent, its inverse and reciprocal ; all the functions in union. 
 
 A pamphlet containing four-place mathematical tables is fur- 
 nished free with the book. 
 
 Calculus with Applications 
 
 By Ellen Hayes, Professor of Mathematics at Wellesley College. 
 i2mo, cloth, 170 pages. Price, $1.20. 
 
 THIS book is a reading lesson in applied mathematics, intended 
 for persons who wish, without taking long courses in mathe- 
 matics, to know what the calculus is and how to use it, either as 
 applied to other sciences, or for purposes of general culture. 
 Nothing is included in the book that is not a means to this 
 end. All fancy exercises are avoided, and the problems are for 
 the most part real ones from mechanics or astronomy. 
 
 Logarithmic and Other Mathematical Tables. 
 
 By William J. Hussey, Professor of Astronomy in the Leland Stan- 
 ford Junior University, California. 8vo, cloth, 148 pages. Price, $1.00. 
 
 VARIOUS mechanical devices make this work specially easy to 
 consult ; and the large, clear, open page enables one readily 
 to find the numbers sought. It commends itself at once to the 
 eye, as a piece of careful and successful book-making. 
 
 65 
 
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 THIS BOOK ON THE DATE DUE. THE PENALTY 
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 THE UNIVERSITY OF CALIFORNIA LIBRARY 
 
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