f : 'S UC-NRLF $B 3Db MbM ^tyu'^s r]v> Digitized by tine Internet Arciiive in 2008 witii funding from IVIierosoft Corporation http://www.arcliive.org/details/firstyearalgebraOOwellricli FIRST YEAR ALGEBRA BY WEBSTER WELLS, S.B. AUTHOR OF A SERIES OF TEXTS ON MATHEMATICS AND WALTER W. HART, A.B. ASSISTANT PROFESSOR OF MATHEMATICS, UNIVERSITY OF WISCONSIN COURSE FOR THE TRAINING OF TEACHERS D. C. HEATH & COMPANY BOSTON NEW YORK CHICAGO /^ '%^i Copyright, 191 2, By D. C. Heath & Co. EDUCAnON DfclFK, IE 2 .0-. PREFACE Ix this text the authors have endeavored to present a course in algebra for the first year of high school which shall be simple, comprehensible to the students, and of high educa- tional and mathematical value. They have made the solution of equations and problems the core of the course ; they have emphasized the essentials, avoiding little-used complexities of algebra; they have taught new ideas inductively ; they have emphasized thoughtful rather than mechanical solutions of exercises ; they have tried to make the course maintain and increase the student's efficiency in arithmetic ; they have tried to make the course interesting by including varied problem material and historical notes, and valuable by including practical applications. The essential features of the course have been tried out in the classroom by many teachers. The text contains sufficient material to meet the needs of schools whose pupils have studied algebra before entering the high school ; the topics have been arranged, however, so that a class may easily cover the essentials of the course in one school year. Attention is directed to the following devices that have been employed to attain the desired ends : Each topic that is taken up is used in the solution of equa- tions. (See §§9, 10, 12, 41, 51, 60, 107, etc.) This makes the study of the various topics purposeful, allows for good grada- tion in the book as a whole, and emphasizes the equation. Problems are introduced at short intervals. Informational, geometric, and physics problems in reasonable number are used. New types of problems are introduced gradually, ap- pearing first in classified lists, are taught with extreme care, and are used thereafter in miscellaneous lists. Experimental verification is suggested for some of the facts from geometry and physics that are used. (See Exercises 7, 25, 28, 29, 38, 39, 49, 106; §§ 13, 142, 143, 190, etc.) 54f;no iv PREFACE The abstract drill examples are simple, very carefully graded, and numerous. More difficult examples are provided in an appendix. Factoring is simplified by omitting until a later chapter, XVI, the more general forms of the simple types that alone are needed in the solution of equations. If desired. Chapter XVI may be studied immediately after Chapter VIII. The chapter on square root and radicals contains only such topics as are required in the solution of quadratic equations. All of the subject of imaginaries that is necessary is given in the chapter on quadratics. Eadicals and imaginaries, as such, are beyond the proper scope of the first-year course, and will be provided for in a later text. Only a brief chapter on ex- ponents is included, and that is at the end of the book. To encourage thoughtful solution of exercises, mechanical processes like " transposition " and " clearing of fractions " are not introduced until the student is familiar with the principles underlying these processes. Special attention is given to inductive development in Chap- ters II, IV, and VIII. Checking is emphasized in a reasonable manner throughout the book. To maintain and increase efficiency in arithmetic, fractional and decimal coefficients and some large numbers are intro- duced — in a sane wa}^ it is believed. Students are encouraged to express roots of equations in approximate decimal form, especially in Chapter XV. Some short cuts in arithmetic are included in Chapter VIII. The formula is introduced as being essentially practical. (See §§ 17 and 146.) For other applications, see §§ 44, 84, 143, 150, 190. The data for the informational problems is, in the main, of permanent rather than temporary interest, and of general rather than local interest. Graphical representation is introduced in a chapter preceding simultaneous equations. The statistical data for graphical representation contains only two or three significant figures. CONTENTS I. ir. 111. IV VI. vu. Literal Numbers . . . . Positive and Negative Numbers Addition of Positive and Negative Numbers Multiplication of Positive and Negative Numbers Addition and Subtraction of Algebraic Expres SIONS ........ Addition of Monomials . . Addition of Polynomials .... Subtraction Subtraction of Positive and Negative Numbers Subtraction of Polynomials . . . , Parentheses . Removal of Parentheses Introduction of Parentheses Multiplication Multiplication of Monomials Multiplication of Polynomials by Monomials Multiplication of a Polynomial by a Polynomial Division Division of a Monomial by a Monomial Division of Polynomials by Monomials Division of Polynomials by Polynomials Simple Equations .... Properties of Equations VIII. Special Products and Factoring Quadratic Equations by Factoring VI CONTENTS PAGE IX. Highest Common Factor and Lowest Common Multiple 154 Highest Common Factor 154 Lowest Common Multiple 157 X. Fractions lilSO Reduction of Fractions 160 Addition and Subtraction of Fractions . . . 169 Multiplication of Fractions. 175 Division of Fractions 178 Complex Fractions 182 XL Simple Equations {Continued) 185 Fractional Equations 185 Solution of Literal Equations 200 XII. Graphical RepiJesentation 206 XIH. Simultaneous Linear Equations .... 221 Literal Simultaneous Equations .... 239 Equations containing Three Variables . . . 341 XIV. Square Root and Quadratic Surds . . . 244 Quadratic Surds 252 XV. Quadratic Equations 254 Complete Quadratic Equations 258 Quadratic Equations containing Two Unknowns . 275 Imaginary Roots in a Quadratic Equation . . 278 XVI. Special Products and Factoring (Advanced Topics) . •. 283 Factoring Polynomials 286 XVII. Ratio, Proportion, and Variation . . • 290 Properties of a Proportion . . . . . 293 Variation 299 XVIII. General Powers and Roots 304 FIRST YEAR ALGEBRA ALGEBRA INTRODUCTION Algebra is like arithmetic in some respects. Arithmetic consists of the study of addition, subtraction, multiplication, and division of some kinds of numbers, and of the apx)lication of this knowledge to some of the common problems of daily life and of business. Algebra continues this study of numbers. In arithmetic, numbers are represented by the digits 1, 2, 3, etc. ; sometimes also, they are represented by letters, as, for example, in interest problems, where tlie principal is repre- sented by P, the rate per cent by i2, and the interest by /. These letters make it possible to abbreviate rules ; thus, the rule " the interest for one year equals the principal multiplied by the rate per cent," may be expressed by the letters as follows : 100 In algebra, letters are regularly employed to represent num- bers. Some new kinds of numbers and many new mathemati- cal ideas are studied, and, as in arithmetic, some of the uses of this knowledge are illustrated. Algebra has a very long history. A little was known about it centuries before the Christian Era. The oldest mathemati- cal book which we have, written by an Egyptian named Ahmes, contains some problems similar to those found in our algebras. Ahmes lived before 1700 b.c. Knowledge of algebra grew very slowly indeed for many centuries; in fact it was not until the sixteenth century that algebra assumed the form which it has to-day, and since then many discoveries and im- 1 2 " ' ' ' ' ' ■ ALGEBRA provements in it have been made. Many of the wisest mathe- maticians of former days contributed to this growth. Thanks to their combined achievements and ingenuity, it is now possi- ble for any boy or girl in the first year of high school to get a much broader view of the elementary part of the subject than many of these men had. Scattered through the text, will be found historical notes calling attention to some of the epoch-making innovations in the development of algebra, together with the name and time of the man making the step forward. I. LITERAL NUMBER 1. In arithmetic, numbers are represented by the digits: 1^ 2, 3, 4, 5, 6, 7, 8, 9, and 0, and combinations of them. In Algebra, numbers are also represented by letters. Numbers represented by letters are called Literal Numbers. The follow- ing examples illustrate the use of letters as numbers. Example 1. If a boy saves 5 cents per day, how much does he save : (a) in 3 days ? (5) in 5 days ? (c) in any number of days ? This last result may be expressed by saying, "as many cents as are obtained by finding the product of the number of days and 5." In algebra, it may be expressed thus : Let n = the number of days. Then, 5 x w = the number of cents saved. So, if w is 6, 5 X n is 5 X 6 or 30 ; . if w is 8, 5 X 71 is 5 X 8 or 40. Example 2. How many inches are there : (a) in 9 feet ? (?>) in any number of feet ? Let X = the number of feet. Then, 12 r = the number of inches in x feet. 12 X is read " twelve ic." LITERAL NUMBER 3 2. Sign of Multiplication. The symbol, x, is used to in- dicate multiplication in algebra as well as in arithmetic ; it is read "a'mes" or ^^ multiplied by" A dot, • , placed above the line, is also used as a sign of multiplication, and gener- ally even the dot is omitted, so that 12 x m may be written 12 • m or 12 771. a x b may be written a • b or abj and is read " a b." Historical Notk. — The symbol, x, was first used by an English- man, Oughtred, about l()3l. The symbol, • , was introduced by Leibnitz in 1693. Multiplication was indicated as early as the thirteenth century, in Hindu and Italian books, by simply writin<. 5. 3x|m. 7. 8xJ«. 2. 7x106. 4. 9x172. 6. 5xfA:. 8. 12x|a;. 9. If one number is represented by 2 6, what will represent a number 3 times as large ? one third as lai'ge ? 10. If John is 4 times as old as James, and if James is 2?/ years of age, how old is John? Find both ages if y is. 3. 11. If the volume of a sphere is 163 1 cubic inches, what is the volume of a sphere 3 times as large? ^ 12. If the interest on a sum of a money is 25 r dollars for one year, what is the interest for 4 yeai'S ? 3 years? 6 years ? 13. There are three numbers of which the first is 4 times the second, and the third is 3 times the first. Kepresent the second number by s, and find the others. Find their values wben the number s is 5. 14. There are three numbers of which the second is 8 times the first, and the third is 4 times the second. Let/ represent the first, and then represent the others. 15. The value of A's property is 5 times that of B's, and the value of C's property is 4 times that of A's. Represent the number of dollars B possesses by 6, and then represent the number of dollars owned by A and C. 6 ALGEBRA 6. An Important Division Law. Since 2 xoa = 6a, then 6 a -=- 2 = 3 a. Similarly, 40 ic -^ 5 = 8 a;, since 5 • 8 a.- = 40 a;. Rule. — To divide the product of an arithmetical number and a literal number by an arithmetical number : l.« Find the quotient of the arithmetical numbers. 2. Multiply the quotient of step i by the literal number, EXERCISE 3 1. Divide each of the following numbers by 5 : (a) 25 t. (p) 30 a;. (c) 45 rs. (d) 75 y. 2. Divide each of the following numbers by 3 : (a) 6r. (b) 30 c. (c) 42 d. (d) 54 e. 3. Divide each of the numbers in Example 2 by 2. 4. What part of 36 ly is : (a) Siv? (b) Aw? (c) 6w? (d) Iw? 5. What part of 44 a^ is : (a) lloj? ' (6) 4a;? (c) 22a;? (d) Ix? Find the following quotients : 6. 39y-r-3. ' 9. 49 6-5-49. 12. 63 s --63. 7. 96/H-12. 10. 120^-^120. 13. 72t^72. 8. 81 a;-- 9. 11. 45 r^ 9. 14. 25 a;-- 25. Historical Note. —The symbol, -4-, was introduced by John Pell who lived during the seventeenth century. 7. Use of Literal Numbers in Solving Problems. Literal num- bers aid in solving certain kinds of problems. Example. How long will it take a bricklayer to lay 38,500 bricks if he can lay 3500 in one day ? LITERAL NUMBER 7 Arithmetical Solution Since he can lay 3500 bricks in one day, then in the unknown number of days he can lay 3500 times that number of bricks. Since this must be 38,500, according to the statement of the problem, then the number of days must be j^qj^ of 38,500 or H. Algebraic Solution Let n = the unknown number of days. Then, 3500 n = the number of bricks laid in these days, and 38,500 = the number of bricks to be laid. So, 3500 n = 38,500. Since one n is ^^^^^ ^^ ^^^ "' divide these two equal numbers by 3500. Then, w = 11. Test: 11 is correct, for 3500 x 11 = 38,500. 8. The mathematical statement 3500 n = 38,500 is called an Equation. The literal number in the equation is called the Unknown Number. An Equation expresses the equality of two numbers. The numbers on the right of the equality sign form the Right Member of the equation, and the ones on the left, the Left Member. An equation implies a question : " for what value of the un- known number is the equality true ? " For example, in the equation of § 7, n can have only one value, — the one found, 11 ; thus, n cannot be 10, for 3500 x 10 is 35,000, and not 38,500. Finding the value of the unknown is called Solving the Equation. Historical Note. — The equation is implied in Ahmes' book. To in- dicate the unknown number, he used a word hau corresponding to our word heap, Diophaiitus, a Greek mathematician of the fourth century, used for the unknown the last letter, s, of the word for number ; Vieta, a French mathematician of the sixteenth century, used the vowels, A, E^ /, O, [/"and Y; Harriot, an English mathematician of about the same S ALGEBRA time, also used the vowels but wrote them with small letters ; Descartes, a French mathematician of the same period, used the last letters of the alphabet, x, ?/, and z. 9. In solving the equation in § 7, two equal numbers were divided by the same number. It is clear that if equal numbers are divided by equal numbers, the quotients are equal. This fact is used in algebra in the following form : Rule. — Both itiembers of an equation may be divided by the same number without destroying the equality. Example. Solve the equation : 36 k = 468. Solution: 1. Since k is ^^ of 36 k, divide both members of the equa- tion by 36. 2. k = -S%^ (Rule § 9) 8. = 13. EXERCISE 4 Solve the following equations : 1. 7p=238. 6. 27?/ = 351. 2. 8n = 608. 7. 2 a; = 161. 8. 9(c = 423. 8. 5v = 21S. 4. 6A= 312. 9. 8 m = 1864. 5. 16a; = 240. 10. 10 ?t' = 2345. t'he arithmetical solution of the following examples is easy. Their algebraic solution leads to the simplest form of equation. Give the alge- braic solution. 11. What number multiplied by 13 equals 221 ? i^. Th^ product of a certain number and 17 equals 403; IftM ^he hUmber. IS. A farm consisting of 43 acres is offered for sale at the ^rice $ 3655. What is the average price per acre ? 14. tVhat number multiplied by 3.7 equals 8.51 ? LITERAL NUMBt:R % (15 J If the total expense for a picnic for a party of 18 boys anoTgirls is $ 6.94, how much must each one oontribute ? 16, A man is compelled to make a journey of 126 miles in his automobile over a poor road in 7 hours ; how many miks mu§t he average per hour ? UJ The fastest train on the Pennsylvania Railroad between »^t. Louis and New York makes the trip in 24 hours ; if the distance is 1052.4 miles, what is the average rate per hour? 10. A second rule used in solving equations is : Rule. — Both members of an equation may be multiplied by the same number without destroying the equality. This fact may "be illustrated by the scales. Sup- ^ pose that the sugar S balances the weight 11^; if tlie weight is doubled, then the weight of sugar j * " _^„_ J*"" "! must also be doubled in order to keep the balance. Example 1. The circumference of one of the large redwood trees of California is 70 feet. Find its diameter. (The circum- ference of a circle is twenty-two sevenths of its diajueter.) Solution : 1. Let d = the number of feet in the diameter. 99 * 2. Then, ^tZ = 70, the number of feet in the circutnferenOfe. 7 8. Multiply both members of the equation by 7. Then, J.^d = 7.70, (§10) or 22 d = 490. 4. Divide both members of the equation by 22. Then, d = ^ = 22t3^, or 22.2+ feet. Check : Does y of 22^^ = 70 ? 2 35 |x22A^f'x*^=70. 10 ALGEBRA Example 2. Solve the equation ^x= 142. 2. Multiply both members of the equation by 8. 1 . Then, ^ . 5 a- ^ g • 142, (Rule, § 10) or 5 a; = 1136. 3. Divide both members of the equation of step 2 by 5. Then, ^ x = ^W = 227.2. (Rule, § 9) K 28.4 Check : Does - x ^*i^= 6 x 28.4 = 142.0 ? Yes. EXERCISE 5 Solve the following equations and problems : 1. 8a=280. 6. fa; = 81. 2. 15^ = 345. 7. f2/ = 188. 3. 27 c = 1242. 8. f2! = 96. 4. 76 m = 1444. G)-^^-t = A29. 5. 27.5 X = 277.75. (w) | r = 200. 11. Three tenths of tl^e cost of a certain automobile is $210. Find the cost of the automobile. 12. The selling price of a certain book is f of its cost. Find its cost if it sells for $1.50. 13. Five eighths of a certain number is 95. Find the number. 14. Thirteen ninths of a certain number is 143. Find the number. 15. Two fifths of the area of Lake Michigan is 9200 square miles. Find the area of Lake Michigan. 16. Three eighths of the cost of the Suez Canal was $37,500,000. Find the cost of the canal. LITERAL NUMBER 11 17. Seven twenty-fifths of the distance from New York to San Francisco is 910 miles. Find the distance from New York to San Francisco. 18. Many metal articles, like a brass candlestick, are made by pouring melted metal into a mold. The piece taken from the mold is called a casting. In making a brass casting, .^1^ of the metal is lost in the melting. How much brass must be melted to make a casting which will weigh 72 pounds ? (Find the second decimal.) 19. Cottonseed meal is used as a fertilizer on farms. It contains about 7% of nitrogen, a necessary plant food. How many pounds Of cottonseed meal must a farmer purchase who wishes to distribute 15 pounds of nitrogen over an acre of ground ? 20. Tobacco stems also are used as a fertilizer. They con- tain about 8 % of potash, another necessary plant food. How many pounds of tobacco stems must a farmer purchase who wishes to obtain 12 pounds of potash? 11. Addition and Subtraction of Numbers having a Common Factor. A number which is a factor of two or more numbers is called a Common Factor of these numbers. Thus, 3 is a common factor of 6 and 9. rt is a common factor of 4 a and 7 a. 6 is a common factor of 3 x 5 and 2 x 6. A short method of adding numbers which have a common factor is illustrated in the following examples. Historical Note. — The symbol, +, was first used in print by a Ger- man mathematician, Widmann, in 1480. The origin of the symbol is much in doubt. Italian writers of this period used the symbol p, the first letter of the Latin word. plus. One explanation given for the sign, 4-, is that it comes from an inverted i, f. The Latin word, e<, means and^ and in place of it this inverted t was often used. It is easy to see how the symbol 4- may have been derived from the symbol, ?. ISi ALCxEBRA EXERCISE 6 1. 3 times 7 plus 2 times 7 is 5 times 7. 3x7 + 2x7 - 5x7 ■ for 21 + 14 = 35. 2. (12 X 9) + (8 X 9)=20 x9= ? 3. (8 X 4) + (7 x4) = (?)x 4 = ? 4. (5x 7) + (6x7) + (9 x 7)=(?)x7= ? 5. 6 times ?^ -f 4 times n = (?) times n ? 6. 6.a;-h4.a;=(?)a;? 9. 12 i; -f- 6 v + 3 -v = ? 7. 7a-f3a=^? 10. 2r4-3r-f-5r + 10r= ? 8. 11 2/4-82/ -f52/ = ? 11. (5 X 4)-(2 X 4)=3 X 4 = 12, for 20 - 8 = 12. 13,. (10 x.7)-(4 X 7) = (?)x 7 = ? 13. (12.x 8)-(5 X 8) = (?) x 8 = ? 14. 9a>-^5x=:^{?)x? 18. 13 t - 5 t + 9 t - 4:t=r 15. 16&-55 = ? 19. 12A + eA-2A-^5A=? 16. 20 2/ -1^2/ = ^ 20. 8r + 13r-llr-f 5r=? 17. 4 m + 6 m — 2 m = ? 21^ One number is four times another. Eepresent the smaller by s. Then represent the larger and find their sum. 22. One number is | as great as *an other. Let b equal the? larger. Eepresent the smaller and find their sum. 23. One number is 5 times as large as another. Let s equal tHe smaller* Represent the larger and then find their difference. 24. The base of a rectangle is three times the altitude. Eepresent the altitude by a ; then rcjpresewt the base. Find also the perimeter. (ThSi perimeter is the sum of the lengths of the sides.) JJ5. What is the perimeter of a triangle if one of its sides is 2 ^ inches, if the second side is three times as long as the first and the third side is 2i times as long as the first ? LITERAL NUMBER 1$ 26. How many incbes in n feet ? in n yaids ? in n yards -f /^ feet -I- n inches ? 27. How many cents in x nickels ? in ic dimes ? in a? dimes + X nickels -f a; cents ? Simplify the following : 28. 2c+ic. 30. 5a;-f-fa'. 32. 2h + /ABb. 34. ^a^ + J^'. 29. .Sc-h|c. 31. 3cH-.2c. 33. 4mf.l5m. 36. §^-f-|y. 12. A4ditioo of Literal Numbers used in ^.quations. Example 1. The sum of two niimbeiv 's yi. The greater number is 12 times the smaller. Fiud the u.>:nbers. Solution : 1. I^t s = the smaller number. 2. Then, 12 s = the laj-ger number. 3. Then, s + 12 s = 01 , since tlie gum of the joumhexs is 91. 4. Adding, 13 s = 91. 5. Dividing, s = 7. Chkck : If the smaller number is 7, the larger must be 84 and thair sum is Ul . EXERCISE 7 Solve and check the following equations : 1. 3a + 4a=:42. 7. 3 a; + 11 a; 4- 12 a; = 130. 2. 4m + 5m = 108. 8. 15^ + 8^-3^ = 20. 3. 76-6 = 66. 9. 7y-5r + 6r=*4. 4. 3x' + 7a;=120. 10. 18z/; - 7w + 9 w = 65. 6. lly-2y = Sl. 11. 22a; + 13a; - 6a; = 116. 6. 6z + 52 = 99. 12. 16t/-3y + 4y = 102. 13. The greater of two numbers is four times the smaller. The sum of the numbers is 60.. Find the numbers. 14. If five times a certain number is increased by three times the same number, the result is 168. Find the number. 15 Divide $56 between A and B so 'that. A shall receive seven times as. much as B. 14 ALGEBRA 16. A, B, and C together have $ 96. B has twice as much as C, and A has as much as B and C together. How much has each? 17. A man had $4195. After spending a certain sum, he found that he had left four times as much as he had spent. How much did he spend ? 18. The sum of three numbers is 120. The second is five times the first, and the third is nine times the first. Find the numbers. ^\9J The sum of thrr e numbers is 360. The second is four- teen times the first, a id the .third is the sum of the other two. Find the numbers. 20. Three m*".j: are asked to contribute to a fund. The first agrees to give twice as much as the second, and the third to give twice as much as the first. How much must each con- tribute to make a total of $ 525 ? 21. The perimeter of the triangle ABC is^40 inches. Find the lengths of its sides. 22. The perimeter of a rectangle is 132 inches. The base is double the altitude. Find the dimensions. of the rectangle. 23. The length of the fence about a rectangular field is 320 rods. If the long dimension is three times the short dimension, find the length of each. 24. The perimeter of the quadrilateral ABCD is 220 inches. The side CD is twice as long as the side AB, the side AD is three times as long; the side BC equals the sum of the sides AD and CD. Find the length of each b" side. 25. The shortest distance by railroad from New York to Chicago is 10 times the distance from. New York to Philadel- phia. The sum of the two distances is 990 miles. Find the distance from New York to Chicago and to Philadelphia. 2a LITERAL NUMBER 15 PROBLEMS ABOUT ANGLES 13. When two lines meet they form an Angle (Z). The angle ABC is a Right Angle. Angles are measured by a unit called a Degree (°). c A right angle contains 90°. Two angles whose sum is a right angle are Comple- mentary Angles ; each of the angles is called the c Complement of the other. The angles AOB and BOC are complementary ; hence a -f- 6 = 90. EXERCISE 8 1. How many degrees are there in one half of a right angle ? one third ? 2. What is the complement of 30° ? 40° ? 70° ? a°? x°? 3. Are angles of 25° and 55° complementary ? Why ? 4. If the angles 3x and 7x are complementary, what is their sum ? Fqrm an equation and determine x. What are the angles ? 5. Determine the angles 5 a and 4 a if they are comple- mentary. 6. What angle is double its complement? (Let c equal the number of degrees in the complement; form an equation.) 7. What angle is three times its com- plement ? 8. What angle is five times its com- plement ? A Straight Angle equals two right a angles. • What kind of angles are the angles AOC '^ ^ ^ and BOC? How many degrees in their sum ? How many degrees in Z AOB ? How many degrees in a straight angle ? z:^ 16 ALGEBRA 9. Find each angle in the adjoining figure, 10. There are three angles whose sum is 180°. The second is double the first, and the third is the sura of the other two. Draw a figure to illustrate this problem. Find the angles. Two angles whose sum is a straight angle are called Supple- mentary Angles ; each of th^i is called the Supplement of the other. 11. What is the supplement of 50° ? 90°? 100°? x°? 2a°? 12. The angles 5 x and 7 x are supplementary How many degrees are there in each? 13. Find the angle which is four times its supplement. 14. Find the angle which is five times its supplement. = \^ $ B ^^^ ^ Fig. 1 Fig. 2 Fig. 3 The sum of all the angles around a point is 4 right angles or 360^. Thus, a-f?> + c4-d-}-e=5 360. i. Find each of the angles in Figure 3. 16/ There are four angles whose sum is the total angle around a point. The first angle contains a°] each of the others is double the preceding. Draw a figure to illustrate this a em. How many degrees are there in each angle ? There are four angles whose sun^is the total angle around a point. The second angle is one half of the first; the third angle is three halves of the first ; and the fourth angle is four times the third. Find the angles. LITERAL NUMBER 17 If the angles of triangle ABC are torn off and placed side by side, their sum is found to be 180°. Draw a triangle with a ruler, and me if you find this to be true. From this we conclude that the sum of the angles of a triangle is 180°. \JJB/'' The second angle of a triangle is double the first, and the third angle is six times the first. How / a^b/cS many degrees are there in each ? 19/ Find the angles of a triangle when two of th^e angles ai*e equal and the third is equal to the sum of the other two. 20. Find the angles of a ti-iangle if the first is 4 times the second, and the third is 7 times the second. DEFlNITIONa 14. An Algebraic Expression, or simply an Expression, is a number expressed in algebraic symbols; as, 2, ab, 2x-3zy, -- s The Numerical Value of an expression is found by substitut- ing particular values for the literal numbers, and performing 'the indicated operations. ab indicates that a is to be multiplied by b. - indicates that r is to be divided by s. s 2x — Syz indicates that 3 times the product of y and z is to be sub- tracted from 2 times x. 15. If the same number is used as a factor one or more times to form a product, the resnlt is called a Power of the number. The number itself is called the BAse. An integer written at the right of and above the base, to 18 ALGEBRA indicate the number of times the base is used as a factor, is called an Exponent. Thus, a^, read " a square " or " a second power,'' means a x a ; a^, read " a cube " or "a third power,"" means a x a x a ; a*, read ^^ a fourth''' or " a fourth power" means a x a x a x a. If no exponent is written, the exponent 1 is understood. Historical Note. — Mathematicians sought suitable symbols for "the powers of a number for a long time. At first words were used for them. Our "a square " and "a cube" owe their introduction to Greek mathe- maticians who called the second power by a word which means the square, and the third power by one which means the cube. Herigone, a French mathematician of the early part of the seventeenth century, wrote a2,aS, «4, etc., and finally Descartes, in 1637, introduced the present symbols. The word, power, comes from a Latin word, potentia, which corresponds to the Greek word used for the second power. The word, exponent, was introduced by Stifel about 1553. 16. The Fundamental Operations are addition, subtraction, multiplication, and division. Indicated operations are to be performed in the following order : first, all multiplications and divisions in their order from left to right; then all additions and subtractions from left to right. Example. Find the numerical value of the expression, 4ab + —-d% b when a = 4, 6 = 3, c = 5, d = 2. Solution r Substituting, 4a6+ — -d3 = 4.4.3 + — -28 = 4.4-3 + — -2.2-2 = 48 + 10 - 8 = 50. EXERCISE 9 Find the numerical value of the following expressions when a — 2, b = 5, c = 3, cZ = 4, m = 4, n = 3. 1. 3a + 56. 3. 5m4-3n. 5. b^ - d\ 2. 4c -2d. 4. a'^-^c^ 6. a^ + n^ 7. 2c' -3b. 8. 2d_^3c^ m n 9. 3ab-\-2c-id, 10. 5b-{-6m-n\ 11. b d m LITERAL NUMBER 19 12. 2a2_3a4-l . X^ m' — mn -\- n\ 14) a3 + 3a26 + 3a62. (Tfi) 2aH-4c-3d 16. c«H-6«. Write in symbols the following and find their value: 17. The sum of a and b ; c and d ; m and n. 18. The difference between a and 6 ; c and d. (The difference between a and 6 is 6 — a.) 19. The product of a and b ; c and d ; m and w. 20. The quotient of a and b ; c and d ; m and n. gl^a increased by 2 6 ; c increased by 3 d. (2^ The square of m increased, by the square of n, (23^ The cube of b decreased by the cube of m. 24. 10 more than 3 a ; 5 less than 4 6. 25 3 more than the quotient of m divided by d 26. 4 less than the product of a and c. 27. The sum of the squares of a and b. FORMULAE 17. When a rule of computation is expressed by means of algebraic symbols, the result is called a Formula. Example 1. Find the formula for the area of a rectangle. Solution : The area equals the product of the base and altitude. Let a = the number of units in the altitude. Let b = the number of units in the base. Let^ = the number of square units in the area; then, A = ab. 20 ALGEBRA The following examples show how to use a formula. Example 2. Find the -area of a rectangle whose altitude is 8 inches and wkose base is 15 inches. Solution : 1. Use the formula A = ab. 2. Substitute 8 for a, and 15 for 6. Then, ^ = 8 x 15 or 120. Example 3. Find the altitude of a rectangle whose base is 75 'feet and whose area is 675 square feet. Solution : 1. Substitute in the formula A = ab. A~m^; b ~ 75. 2. Then, 675 = a x 75, or 675 = 75 a. 3. Divide both members by 75 : "9 = a. Example 4. Find the base of a rectangle whose altitude is 11 inches and whose area is 385 square inches. Solution : 1. Substitute in the formula A — ab. 2. ^ =385; a = 11. Then, 385 = 11 x 6. 3. Divide both members by 11 : W^ = b, or 6 = 35. Rule. — To solve a problem by a formula : 1. For the known letters in the formula substitute their values. 2. Perform all of the indicated operations. 3. If an equation is formed, solve for the unknown letter, if pos- sible. EXERCISE 10 1. The figure XFZTTis a Parallelogram (O). Find a formula for determining its area. (a) Let b and a equal the units in the base and the altitude, and A the square units in the area. /~~| 7 (6) How does XYZW compare in area with yZ_I_!_fL4 FGHK? ^ (c) What is FGHK? What is its base? ahi- p tude ? area ? « W (d) What then is the area of XYZW? G ^ LITERAL NUMBER 21 (e) Make a rule for finding the area of a parallelogram. (/) Expressed as a fortuuia, thie rule is A=ab (g) Find A when a = 12 and b = 20. (A) Find A when a = 15 and 6 = 26. (0 Find a when ^1 = 500 and b =40. ( j) Find a when ^ = 600 and 6 = 26. I (A;) Find b when ^ = 750 and a = 16. ^JFind b when ^ = 9G0 and a = 32. *'' 2. The figure XFZ is a Triangle (A XFZ). Find a formula for determiiiiiig its area. (a) Let 6, a, and A represent the units in the base, altitude, and area respectively. >< (6) What is the figure XiZlV? what is its /la'^^,.^ base ? altitude ? area ? '•^ b ^ (c) What part of O JT TZ W is A A' 1'^ ? /T!\l^ "7^ ^ (d) What then is the area of A XYZ ? Y^-i ^^^ (e) Make a rule for finding the area of a triangle. (/) Expressed as a formula, this rule is, A =^ab. 2 (g) Find A when « = 10 and b = 17. §Find A when = 9, c = 7. uSl) Find c when F= 240, a=(i, and 6 = 5. 22 ALGEBRA 4. The figure XYZW is a Pyramid. Its volume equals one third of the product of its base and altitude. (a) Express this rule as a formula, letting F equal the number of units in the volume. (6) Find Fwhen a = 1%, h = 15. (c) Find h when V= 160, a = 24. {d) Find a when V~ 900, h = 30. 5. The formula for the circumference of a circle is C=2 7ri?, where (7= the number of units in the circum- ference, where i? = the number of units in the radius, where 7r = 3.1416 (tt is read "pi"), (a) Express this rule in words. (§^Find, by the formula, C when B is 10 inches, r^c) ,find, by the formula, B when C is 628.32 inches 6. The formula for the area of a circle is : A =^ Trie. (a) Express this rule in words. (6) Find, by the formula, A when B is 10 inches. fc) Find, by the formula, A when ^ is 5 feet. ( .7. The numbers s, v, t, and g are connected by the formula: find s when v 50, t s=vt-\-^^gf; 3, ^ = 32.16. wv 8. From the formula £* = —-, find J5; when w=:75,v = 50, g = 32.16. (Carry the result to one decimal place.) 9. From the formula F=| ttI^^ find Fwhen E is 3. 10. From the forniula aS = 4 ttM^, find S when E = 5. II. POSITIVE AND NEGATIVE NUMBERS 18. The first numbers studied in arithmetic are the integers, such as, 1, 4, 15, etc.; the next are the common fractions, such as, ^, J, ]f, I, etc. and the decimals, such as, 2.03, 4.6, etc. The literal numbers in the first chapter represented only these same arithmetical numbers. One of the most distinctive things about algebra is its use of certain other numbers. 19. Opposite Quantities. Suppose that the temperature at a certain hour of the day was 73°, and that there was a change of 5°. To determine the new temperature, it would be neces- sary to know whether the change was a rise of 5 ° or the oppo- site, a fall of 5 °. Suppose that a person was known to weigh 85 pounds, and that during a certain time there was a change in his weight of 5 pounds. To determine his new weight, it is necessary to know whether the change was an increase of 5 pounds or the opposite, a decrease of 5 pounds. These are two illustrations of opposite quantities. Many concrete quantities exist in two such opposite states. EXERCISE 11 Tell the opposite of each of the following : 1. Sailing 35 miles north. 5. Moving 5 steps forward. 2. Sailing 25 miles east. 6. Depositing $15 in a bank. 3. Receiving $30. 7. Rise of 10° in temperature. 4. Gaining $5. 8 Walking 5 rods to the right. 9. Increasing weight by 5 pounds. 10. Adding 7. 93 24 ALGKBRA 11. What is the total result if any one of the changes in Examples 1 to 10 is followed by a second change of opposite kind and of like amount ? 12. What is the total result of two transactions, one giving a gain of % 50, and one a loss of $ 25 ? What single change will produce the same result as the two changes indicated in the following examples ? 13. If a ship sails first 6° north and then 2° south ? 14. If a ship sails first 8° east and then 10° west ? 15. If a boy, becoming ill, loses 10 pounds, and then gains 8 pounds ? 16. If the temperature first rises 12°, and then falls 15° ? 17. If a man first deposits $ 100 in a bank and then with- draws $ 125 from his account ? 18. A vessel sails from the equator due north 28°, and then due south 57°. W^hat is her latitude at the end of the voyage ? 20. In Example 10, Exercise 11, the opposite of adding 7 is subtracting 7. Addition is always indicated by the sign +, and subtraction by the sign — . These same signs are used to distinguish between opposite quantities. A quantity pre- ceded by the + sign is called a Positive Quantity, and one preceded by the — sign is called a Negative Quantity. Quan- tities preceded by the signs plus and minus are called Signed Quantities. EXERCISE 12 The following are naturally considered positive quantities. What are the corresponding negative quantities ? 1. Sailing east. 5. Eising temperature. 2. Sailing north. 6. Forward. 3. Right direction. 7. Upward. 4. Increasing. 8. Deposits. POSITIVE AND NEGATIVE NUMBERS 25 9. Assets. 10. Prolits. 11. Above zero. 12. Time a.d. fS c t > 100 - 90 - ec - : 70 - 60 50 ~ 40 - 30 20 10- :j 10 20 :_ - ao i - 40 < ^ " U 13. At 7 A.M. the temperature is — 13° ; at noon it is 8° warmer, and at 6 p.m. it is 5° colder than at noon. Required the temperature at noon and at 6 P.M. 14. At 7 A.M. the temperature is +0°; at noon it is 14° colder, and at 6 p.m. it is 2° colder than at noon. Required the temperature at noon and at 6 P.M. 15. At 7 A.M. the temperature is — 7°, and at noon -t- 9°. How many degrees warmer is it at noon than at 7 a.m.? 16. The temperature at 6 a.m. is — 14°; during the morning it grows warmer at the rate of 3° per hour. Required the temperature at 9 a.m. and at 10 A.M. 17. The positive quantities in this set indicate rise in tem- perature. What single change will produce the same result ? (a) -f 12° and + 10°. Solution : 1. A 12° rise followed by a 10' rise gives a total of 22° rise. 2. This may also be expressed thus : (+12°) + (+ 10°) = +22°. (b) + 9° and - 5°. (c) - 4° and - 5°. (d) + 7° and - 9°. 18. In this set the positive quantities refer to gains in finan- cial transactions. What is the equivalent single change? (a) +$15, -f !i>25, -$30. Solution : 1. A gain of $15 followed by a gain of §25 gives a gain of $40 ; followed by a loss of $30 gives a total result of $ 10 gain. 2. Expressed in symbols thus : (+.S15) + (+326) + (-$.30) = (+$40) + (-$30) = + $10. (b) 4- $20, + $30, - $50. (c) - $45, + $50, - $10. (d) + $35, - $20, - $25. (e) -h $14, - $20, + $19. 26 ALGEBRA In the following problems, select and mark the positive and negative quantities. Find the total result as in the preceding problems and express it as a signed number. 19. A man's income during the year is $ 1500, and his ex- penses are $1300. Find the result at the end of the year. 20. A man's monthly account book shows the items : salary $150, rent $40, food $50, insurance $25, interest on savings $ 15. Find the result at the end of the month. 21. Positive and Negative Numbers. The preceding exercises show that positive and negative quantities exist. In dealing with these quantities, positive and negative numbers are necessary. Starting with an arithmetical number like 3, a new number called Negative 3 is made ; 3 is then called Positive 3. These two numbers are opposites and have the power of destroying each other when added, just as do opposite quantities which are equal in amount. Positive 3 is written, -+- 3 ; negative 3 is written, — 3. The arithmetical number 3 is called the Absolute Value of + 3 and - 3. The negative number must always have its sign written before it ; the positive number is often written without its sign. Historical Note. — Hindu mathematicians, who knew about positive and negative numbers long before European mathematicians, in referring to them, used words which correspond to our words for assets and debits. They also were acquainted with the illustration of such numbers by means of the opposite directions on a straight line. To indicate that a number was a negative number, they placed a dot over it. European mathemati- cians did not arrive at an equal understanding of positive and negative numbers until the sixteenth century. EXERCISE 13 1. Kead the following numbers, write each in symbols, and tell the absolute value of each : POSITIVE AND NEGATIVE NUMBERS 27 (a) positive nine ; (b) negative seven ; (c) positive four ; (d) negative three fourths ; (e) negative five sixths ; (/) negative three and three tenths. 2. Kead the following numbers and tell the absolute value of each : (a) +6; (6) -2; (c) -5; (d) -1; W +i; (./•) -2.98; (y) +3.41; (7i) -45.087; (0 - 102.34, 22. Addition of Positive and Negative Numbers. The rules for addition of signed numbers are suggested by the following problems : 1. Find the sum of -f 5 and -|- 3. Just as $5 gain plus $3 gain gives $8 gain, similarly (+5)+(+3) = + 8. 2. Find the sum of — 5 and — 3. Just as §5 loss plus $3 loss gives $8 loss, similarly (_5) + (-3) = -8. 3. Find the sum of -h 5 and — 3. Just as §5 gain plus §3 loss gives .$2 gain, similarly (+6) + (-8) = +2. 4. Find the sum of — 5 and + 3. Just as §5 loss plus $3 gain gives $ 2 loss, similarly (-6) + (+3) = -2. Rule. — 1. To add two positive numbers, add their absolute values (§ 21), and prefix the plus sign to the result. 2. To add two negative numbers, add their absolute values and prefix the minus sign to the result. 3. To add a positive and a negative number, find the difference of their absolute values, and prefix to the result the sign of the number having the greater absolute value. ,28 ALGEBRA EXERCISE 14 1. Find the sum of -f 10 and — 3. Solution : Use Rule 3. Subtract 3 from 10 ; prefix + sign. (+10) + (-3) = +7. 2. Find the sum of - 12 and + 6. Solution : Use Rule 3. Subtract 6 from 12 ; prefix — sign. (-12) + (+6)=-6. 3. Find the sum of — 9 and — 5. Solution : Use Rule 2. Add 9 and 5 ; prefix — sign. (_9) + (-5) = -14. 4. To each of the numbers : + 6, +9, +14, +5, +18, +3, (a) add -4; (b) add -12; (c) add- 15. 5. To each of the numbers : -5, -15, -2, -10, -16, -20, (a) add + 6 ; (b) add + 15 ; (c) add + 12 ; (d) add -5; (e) add -6; (/) add -7. 6. To each of the numbers : _36, +48, -17, -25, +24, +29, (a) add + 9 ; (b) add - 8 ; (c) add - 7. Find the sum : 7. 8. 9. 10. 11. + 124 - 15.2 -1.35 + 2.10 -10.3 -36 + 9.1 - 1.63 -1.43 + 3.3 12. + 9 Here 9 must be taken from 17. It is necessary to be- — 17 come expert in subtracting the upper number from the - _ g lower. 13. 14. 15. 16. 17. 18. -11 + 17 -2S + 72 + 65 -59 + 43 -42 + 44 -109 -247 + 78 POSITIVE ; AND NEGATIVE NUMBERS 29 19. + Hint: There are two ' ways of doing this example. Either -2 add in oi •der fron I bottom to toi ), or from top to b ottom ; + 4 or, add first all of the positive nuinbei-s and idl of the nega- — 5 tive numbers separately, and then combine the results. 20. 21. 22. 23. 24. 25. f-4 -7 + 15 - 5 + 16 -10 -6 -5 -12 -14 -11 + 13 -2 4-9 + 6 + 11 — 7 -14 f-5 -1 + 6 + 3 + 22 + 7 -3 + 8 - 9 - 8 - 9 - 3 Find the • sum: 26. -i +i- 28. -h -h 30. +2^, -If. 27. -i 9 29. -h + f 31. -3i, + 2|. 32. Demosthenes was born in the year — 385, and died at the age of 63. What was the year of his death? 33. Pythagoras was born in the year — 580, and lived to the age of 79 years. What was the year of his death ? 34. At the beginning of the month the number of pupils in a schoolroom is 53; 3 pupils enter during the month and 5 leave school. How many pupils are there in the room at the end of the month? 35. A principal of a school finds that his six algebra classes have the enrollment indicated in line 2. He decides to make the changes indicated in line 3. Classes 12 3 Members 29 15 22 Changes — 8 +6 — 1 (a) Tell what each of the changes in line 3 means. (b) Find the result in the class membership. (c) How can he check his work from line 3 ? 4 5 6 28 18 14 -7 + 3 + 7 30 ALGEBRA 23. From the examples in Exercise 14 it is clear that adding sL negative number has the same effect as subtracting the posi- tive number of equal absolute value. Thus, (+12) + (-,5)=12-5 = 7. ( _ 10) + (_ 5) = - 10 - 5 = _ 15. 24. It is convenient to picture the positive numbers thus : +1 +2 +3 -^4 -t-5 where -f 1 is placed at any distance from the point A, -{- 2 twice as far, etc. In this sense, the positive ntimbers form a scale extending to the right. Any number precedes all larger numbers on this scale ; thus, 3 precedes 4, 5, 6, etc. Since (— 3)-i-(+3)= 0, it is natural to think of — 3 as being 3 less than zero ; and, similarly, of — 4 as being 4 less than zero. It is natural to think of the negative numbers as arranged on the left. -5 -4 -3 -2 -1 -fl 4-2 -1-3 -f-4 -4-5 < • • • ■ • • • • • • • >- A Thus, the positive and negative numbers together form a complete scale extending in both directions from zero. Starting on the left, any negative number precedes the posi- tive numbers and may he thought of as being less than the posi- tive yiumhers. MULTIPLICATION OF POSITIVE AND NEGATIVE NUMBERS 25. The terms Multiplier, Multiplicand, and Product have the same meaning in algebra as in arithmetic. The rules for multiplication of signed numbers are sug- gested by the following problems. In these problems, read the sign, x , " multiplied by." POSITIVE AND NEGATIVE NUMBERS 31 1. Find the product of +4 and +3. Since positive numbers are like arithmetical numbers, (+4) X (+3; = + 12. 2. Find the product of — 4 and 4- 3. In arithmetic, to multiply by 3 means to add the multiplicand three times. If this is done in this problem, (- 4) X (+ 3) = (- 4) + (- 4) + (- 4) = - 12. 3. Find the product of -f- 4 and — 3. In arithmetic, 4x3 = 3x4, since each equals 12. The order of the factors may be changed. If it is assumed that the same law holds in algebra, (-f 4) X (-3) should equal (— 3) x (+ 4) or, — 12, by problem 2. Then, (+4) x (-3) = -12. To multiply a number by a negative number seems to be accom- plished by multiplying it by the absolute value of the multiplier and changing the sign of the product. 4. Find the product of — 4 and — 3. The multiplier is a negative number. If the suggestion from problem 3 is followed, the result in this problem may be obtained by multiplying (—4) by 3 and changing the sign of the product. (—4) X 3 = — 12. Changing the sign of — 12 gives + 12. Therefore, (_ 4) x (- 3) = + 12. Gathering together the results of problems 1, 2, 3, and 4 : 1. (+4)x(+3)=+12. 3. (+4)x(-3) = -12. 2. (-4) X (4-3) = -12. 4. (-4) X (-3) = 4-12. Rule. — To multiply one signed number by another : 1. Find the product of their absolute values. (See all four above.) 2. Make the product positive if the multiplicand and multiplier have like signs. (See 1 and 4 above.) 3. Make the product negative if the multiplicand and multiplier have unlike signs. (See 2 and 3 above.) S2 ALGEBRA EXERCISE 15 1. Multiply the numbers : + 5, +6, +8, +7, +9, +12, (a) by +4; (h) by -3j (c) -8j (d) by -9, 2. Multiply the numbers : -6, -10, -8, -7, -15, -12, (a) by +4; (b) by -9; (c) by -6; (fZ) by +7. 3. Multiply the numbers : + 11, -9, +14, +12, -25, -15, (a) by -6; (?>) by +7; (c) by -8.; (d) by -10. Find the products of the following factors : 4. -9,-11. 8. -13, +8. 12. +1,--^. 5. -7, +12. 9. -9, +12. 13. -h+h 6. +9,-20. 10. -11,-20. 14. -|, -f 7. +6, -13. 11. +9, -16. 15- Find the product of - 2, + 3, - 4. (_2).(+3) = -6; (_6).(-4) = +24. 16. - 1, + 3, - 2. 20. -6,-3,-4. 17. +3,-4,-5. 21. -7, +2, -5. 18. -6,-2,-5. 22. - 9, +5,-4. 19. +5,-8,-3. 23. + 10, - 7, +6. 24. If the number of negative factors is even, what is the sign of the product ? 25. If the number of negative factors is odd, what is the sign of the product ? 26. Powers of Positive and Negative Numbers. (+2)2 = (+2)(+2) = + 4. (§15) (+2)^=(+2)(+2)(+2) = + 8. (+2y = (+2)(+2)(+2)(+2)= + 16. It is clear that every power of a positive number is positive. POSITIVE AND NEGATIVE NUMBERS 33 (-2/=(-2)(-2) = + 4. (-2)' = (-2)(-2)(-2) = -8. (-2)^ = (.-2)(-2)(-2)(-2) = + 16. (-2)==(-2)(-2)(-2X-2)(-2)=-32. It is clear that every even power of a negative number is positive and every odd power is negative. EXERCISE 16 Find the values of the following powers : 1. (-3)2. 5. (-5)3. 9. (-3)^ 2. (-3)'. 6. •(-2)«. 10. (-5)2. 3. (+Af. 7. (-4)\ 11. (-0)3. 4. (-1)^ 8. (+1)'. 12. (-1)'. 13. Find the value of ax^ -\-bx-^c when a = 1, 6 = 2, c = — 3, x = — 2. ax« + 6x + c = 1 . (- 2)2 + (-f 2)(- 2) + (- 3) = +4-4-3=-3. In the following examples, let a = + 1, 6 = — 2, c = + 3, and x = — 2. Find the values of the expressions : 14. Sab. 18. bc-\-ax. 22. ab'^-\-bc\ 15. a52. 19. a26 + a62. 23. ^^.^a.-^. 16. 5 6x2. 20. bx^-cx^ 24. a* - a;*. 17. — 11 6ca;. 21. a:f^—ba^. 25. a^** — a'c. III. ADDITION AND SUBTRACTION OF ALGEBRAIC EXPRESSIONS DEFINITIONS 27. A Monomial or Term is an expression whose parts are not separated by the signs + or — . Thus, 2ic2, — 3a6, and -f- 5 are the terms of the expression 2 a;^ — 3 a6 + 5. In an expression, a term whose sign is plus may be called a positive term, and one whose sign is minus, a negative term, as the terms 2 a-'-, and — 3 ah, respectively, although the algebraic value of the term depends upon the values of its literal factors. 28. If two or more numbers are multiplied together, each of them, or the product of any number of them, is called a Factor Of the product. Thus, a, h, c, ah, ac, and 6c are factors of the product ahc, 29. Any factor of the product is called the Coefficient of the product of the remaining factors. Thus, in 2 ah, 2 is the coefficient of ah, 2 a ot h, a of 2 6, etc. 30. If one factor of a product is expressed in numerals and the other factor is expressed in letters, the former is called the Numerical Coefficient of the latter. Thus, in 2 ah, 2 is the numerical coefficient of ah. If no numerical coefficient is expressed, the coefficient 1 is understood ; thus, a is the same as 1 a. 31. By § 25, (— 3) X a = — 3 a ; that is, — 3 a is the product of —3 and a. Then — 3 is the numerical coefficient of a in -3 a. Thus, in a negative term, the numerical coefficient includes the sign. 34 ADDITION AND SUBTRACTION 85 32. Terms wliich are alike in their literal parts are called Like or Similar Terms ; as, 2 xhj and —Ix^y. Terms which are not alike in their literal parts are called Unlike or Dissimilar Terms ; as, 2 ar^ and 7 xif. Sometimes unlike terms may be like with respect to one or more letters. Thus, 2 axy and 3 hxy are like with respect to xy. EXERCISE 17 1. Tell all of the factors of the monomial 4- 6 xy. 2. What is the numerical coefficient of the monomial —&xy? 3. In 2, what is the coefficient of y ? of a; ? of xy ? oi 6 x? 4. Select the sets of like terms : (a) -\-2xy^, -3xy-. (b) + 7 ?w, -6n. (c) 4- 5 abcj — 6 ab'^c. {d) —Sab, +9 ab. 33. The result obtained by adding two or more numbers is called the Sum. ADDITION OF MONOMIALS 34. Addition of Like Terms. In paragraph 11, it was found that 6 ?i + 4 7i = 10 n. G n and 4 n are like terms since they have the common factor n (§ 11). The coefficients of ii are 6 and 4, and their sum is the coefficient, 10, of the result. Rule. — To add two or more like terms : 1. Multiply their common factor by the sum of its coefficients. Example 1. Find the sum of 5 a:^ and 3 x^. Solution: 1. The common factor in hx^ and Sx^isx^, The coeffi- cients are 5 and 3 ; their sum is 8. 2. Hence, 6 a-2 + 3 j2 = g x^. Check : Let re =- 2. Sx*^ = 5 • (- 2)2 = 5 • 4 = 20, 3 x^ = 3 . 4 = 12, and 20 + 12 = 32. Also, 8 x^ = 8 • 4 = 32. Example 2. Find the sum of — 5 xhj and + 3 x^y. Solution : - 5 x^y + 3 x^y = [(- 5) + (+ 3)] x^y =_ 2 x^ 36 ALGEBRA Check: Let z = l and y = 1. Then, — 5x^ =— b ■ 1 •!=— 5, + 3a;2y = + 3. 1 . l=+3, and (_ 5) + (+ 3) = - 2. Also, -2x'h/ = C-2).l.l=-2. Example 3. Find the sum of 16 x 19 and 14 x 19. Solution : 1. The common factor is 19; its coefficients are 16 and 14. 2. 16 X 19 + 14 X 19 = (16 + 14) x 19 = 30 x 19 = 570. EXERCISE 18 Find tlie sum in each of the following : 1. 5 A and - 12 A. 8. - 17 a^ and - 15 x". 2. lip and —6 p. 9. xyz and — 9 xyz. 3. - 7 m and - 8 m. 10. 8 xY and - 29 a^/. 4. —4n and— 9 72. 11. — &cand-f6 6c. 5. + 15 J5: and - 11 jE;. 12. 12 x 16 and 8 x 16. 6. -5ab and + 13 ab. 13. 21 x 17 and 9 x 17. 7. - 13 r^s and + 36 A 14. 13 x 23 and 7 x 23. 15. 16. 17. 18. 19. -f-7a 4-9a 4- 15a^ - 4:xy -2xf -9xf 4- 5m — 15 m 20. -7a 4-9a + 8a -3a Hint. Add first the positive terms, getting + 17 a ; then the negative, getting — 10 a ; then add these results, getting + 7 a. All should be done mentally. + 7a 21. 22. 23. 24. • 25. 15 m -2m — 5m + 3m -ISx" + 2x' - Ix" + 3ar' + 8a5 -7a5 -5o1j + 3 ab 16 xyz -4 xyz — 6 xyz + 3 xyz 21 ab'C - 6abh - 17 ab'c + ab^c ADDITION AND SUBTRACTION 91 26. Find the sum of 9 a'', — 7 a^, and +8 a^ Solution : 9 a^ + (- 7 a^) -\-S a^ = 9 a^ -7 a^ -\-Sa^ = 10 a^. Note. This illustrates another arrangement of an addition example. The terms are first connected by the + sign. Then +(—7 a^) is written as — 7 a^, since adding a negative number is the same as subtracting the positive num- ber of equal absolute value. {§ 21.) Add, as in Example 26, the following: 27. Tab, — 3 aft, and + 9 a6. 29. 122/, —7 y, +9 y, and — 8 y. 28. 9?-^, -Gr', +3?'»,andl0r». 30. Wz, -^Oz, -11 2;, and -8z. 31. Find the sum of 12 a, —5x, — 3 y^, —5 a, and 8 x. Solution : 12 a + (- 5 x) -^ (- S y^) + (- 5 a) -\-(S x) = V2a'-6x-3y^-6a-\-Sx. = 7 a + 3 X - 3 y2. In this Example, 12 a and —5a give 7a; — 5 x and -|- 8 x give + 3 x ; there is no term to combine with — 3 y^. Only the like terms may be combined. Add, as in Example 31, the following : 32. Sab, — 9 cd, — ab, and + 4 cd. 33. Gx^, -10z-,2y'',4:z% -9y^,and -Sx". 34. 12r2, -3r, -8s, -5r, 4-3.9, -Tr", and lis. 35. 10 c, - 4 d, - 3 A;, + 9 c, - 4 k, + 5d, - 6 A;, and - 2 c. 36. 11 m% - 8 ?i-, + 6 Z, - 5 71*, 4- 3 m^, and -81, Simplify the following : 37. 5a-8a + 10a-12a. 38. -3x'-\-6x'-\-x^-nx'-\-a^. 39. 4:7^8-7 7^8-\-or^s -6 )^s. 40. — 8 m^n — 3 m^n -}- 2 m^n — 9 m'n. 41. 6c + 11 c?-4ci-5c-2c. 42. 2a?-3x-^x-5x^ + Sx. 88 ALGEBRA 43. 9'i'^-\-2r''-3i^-4:r\ 47. ^xy - j\ xy ~^xy. 44.. |a + ia-ia. 48. ^f^^^f-y^-^xf. 45. — \ mn + 1 mn + ^ mn, 49. 2.5 ajy — .3 a^y 4- 1.2 a;^/- 46. -2,v + irV- irV. 50. - 3.05 a + 4.4 a - 1.3 a. DEFINITIONS 35. A Polynomial is an algebraic expression consisting of two or more terms : as, a -f 6, or 2 a^^ — 3 a;?/ + 5 y"^. A Binomial is a polynomial of two terms : as, a + h. A Trinomial is a polynomial of three terms. ADDITION or POLYNOMIALS 36. Addition of polynomials is similar to addition of de- nominate numbers in arithmetic. Example 1. Find the sum of 3 yd. 2 ft. and 6 in. and 5 yd. 1 ft. and 4 in. Solution : 3 yd. + 2 ft. + 6 in. 5 yd. + 1 ft. 4- 4 in. 8 yd. + 3 ft. + 10 in. Example 2. Find the sum of 2 a+3 6+5 c and 4 a + 6+2c. Solution : 2a + 36 + 5c 4a+ & + 2c «a + 46 + 7c Rule. — To add two polynomials : 1. Rewrite the polynomials, if necessary, so that like terms are in the same vertical column. 2. Add the columns of like terms. 3. Write the results of step 2 with their proper signs for the sum. Example 1. Find the sum of 6 a-7 a;^, 3 a;^ _ 2 « + 3 2/^ and 2x^— a — mn. Solution : 6 a — 7 a;2 -2a + 3a:2 + 3?/2 — a + 2 a;2 _ inn 3a — 2a;2 ^hy^ — mn ADDITION AND SUBTRACTION 89 Check": Letting a = 2, oc = 1, y = 1, m = 2, n = 1 : 6a-7a;2=12-7 = 5;~2a + 3a;2 + 3y2 = _44.3^_3 = _^2; -« + 2x'2-m7i=-2 + 2-2=-2; and the sura of these values of the polynomials is 5 + 2 — 2 or 5. The value of the sum of the polynomials, 3 a - 2 x2 + 3 y2 _ ,n?i is C - 2 + 3 - 2 or 5. Since the two values are equal, the solution is probably correct. This is called checking by substitution. Another method of checking is to add carefully in the opposite direc- tion, as in arithmetic. Example 2. Find the sum of x^-f-\-Sxf -3 oc^y, 2a^-\- 3xtf-{- if, and 2 a;^?/ — 4 xif^. Check f or x = 1, ?/ = 1. Solution : 1. a:« - 3 ar^y + 3 a;?/^ - y* = 1 - 3 + 3 - 1 = 2 x8 + 3ici/'-2 + y" = 2 + 3 + 1 = + 2 ofiy - 4 xy'^ = 2-4 =-2 Sofi- x'^+2xy^ +4 Also, 3 a^ — a;2y + 2 xy2 _ 3 _ 1 ^ 2 = + 4. The solution is correct. 37. In Example 2 of this last paragraph, notice that the three polynomials have been rearranged. The term containing the highest power of x is placed first; then the term having the next lower power of x, and so on. The term — f, not con- taining any x, is thought of as containing the lowest power of X in this expression. The polynomials are arranged in Descend- ing Powers of x. These polynomials are arranged also in Ascending Powers of y. The exponents of y in the terms increase from left to right. In all examples in addition, subtraction, multiplication, and division of polynomials, it is advisable to arrange the polynomi- als, if necessary, according to ascending or descending powers of one of the letters when writing them preparatory to solving the example. When the polynomials are so arranged, there is less likelihood of making some error in the solution, compari- son and checking of results is facilitated, and the final solution has a more workmanlike and finished appearance. 40 ALGEBRA EXERCISE 19 Find the sums in the followinnr : 1. 2 1. 3. 7A-3B - Qf + 5n^ — 16 am ■+■ 14 &» -9A-i-2B + 12t'- 13 7l« 6 am — 5bn -\-SA- B - ^e + i^n^ 9 a?>i — 5 6n 4. 5. a^-2ah + W 7^ - 6rs^ a^ -f 4 a& - 2 b' + 6r's -10 -2a^-2ah -11^^ — 5 r-s 4- 15 rs^ 6. 7. 3a?-^x'y + 2xy' -4.f — 7 m^ 4- 5 mn^ ■^Qo^y -h^f - 6 mhi - 13 71171^ -\- 2 71^ -5a^ 7 xy^ + 4m-/i+ o7)in^ Note. In Examples 5 and 7, blank spaces are left when the first poly- nomial is written for powers which are not present in the polynomial. 8. 3a — 46 + 6c and 5a-]-7b — Sc. 9. 4A:-7 + 3m, 5A: + 2-4m, and3A;-2. 10. 2 a + 8 6 - 5 c, - 3 6 - 3 c -f 7 d, and + 6 c - 4 a + 2 cZ. 11. 12r + 6s-9^, 8r-9s + ll^, and 15r-}-7s-6t 12. x^ -\- 2xy + y% x^ - 2 xy -{- y^, and -2 x^ - 2 y^ -^ lOxy. 13. 4 m^ — 4 mn + 7i^, m^ + 4 mn -f 4 n^, and — 5 m^ + 5 n^ 14. a^ -f 3 a26 -f 3 a&^+fts and a« - 3 a^fe + 3 aZ^^ _ ft^. 15. 3a^-^9xy-'i7f, -2xy~5x--10y',and7 7/-6xy-\-Sx'. 16. 15 a^ - 6 + 4 a 4- 16 a^ and - 2 - 13 a + 8 a^ - 3 a^ 17. a^ + a^?/^+a^2/^+a;^3/4-ic?/'*+^ and x^—x*y-{-xy*-\-x^y^—y^. 18. 12 a^- 4 a + 3, -8a;4-6-f 15 a;3-f 2 a^, and -16 + 6a.-2 - 8 a.-3 - 11 a;. ADDITION AND SUBTRACTION 41 19. 9 A^ 4- 3AB' - B", - 2 A'B -f- 5 AB' + 7 2^\ and -6A^ 4-13 ^2^. 20. ia + lb — Ic and J- « — ^ ^ + iV c. '21. I ?/i — I n 4- ^j — i *^ + A ^* — i ^> ^^i + 7 ah'^ + 8 h^ Change signs mentally; add. Remainder: b a^ + 7 aV) - 3 ah'^ — 8 f/^ Check : The sum of the subtrahend and remainder is the minuend. EXERCISE 22 Subtract ; check each in some way : 1. 3. 5. 12a'-9a- 7 17 m — 12/1 -f6p 2ab-\- 5bc-Sac Sa^ + Qa-^IS 20m-lGn-5p — a6H-ll 6c — 4ac M.,.-^^ ^::0 2. 4. 6. 5a-Sx-{-2y 3a + 7 6 — c ar^ + 13a;-ll -3a-^7x-3y 2a-h46--c -Sa^-h 6x- 5 46 7. lOr^ -f6rs ALGEBRA 9. o'^Sa'b +26^ a^^Sa'b-{-ab'- b^ or' 11. x2 + 2a;y + 2/' 8. 5x'-6x 10. x' +2/^ 12. 4- xhj -{-xy^ + f -x^y-^-xy^'-f 13. From 5a — 364-4c subtract 5a + 3& — 4c. 14. From — 2 m^ — 4 mn + 9 w^ subtract Sm^ — 7 mil -\- 14 n^ 15. From ab-^bc + ac subtract ab — bc-\- ac. 16. From 4£c3-9a;2-lla;+18 subtract 3x^-8 aj2_;1^7^_j_ 25. 17. From —3y-\-Sx — 4:Z subtract — z -^ 11 x — 6 y. 18. From 8^ + 2^-7 C subtract 8^- 25 + 7 O. 19. Subtract 2a;3- 7- 4a;- 6 a;2 from 5r^-12 + 9a;'-2 a;. 20. From x^ — 2xy-{-y^ subtract x^—2xy — y^. 21. Take76-9c-2(ifrom6a-5& + 12c. 22. Take 12a5 + 4a-9 from 3a'^ + 8a2-6. 23. Subtract 1 + a^ — a — a^ from 3 a — 3 a'*+ 1 — a^ 24. From 10 a;^ _ 21 a.-^ _ 11 a; take - 15 ar^- 20 a; + 12. 25. From 17 a« - 12 ab'' + 5 6^ take 8 a^ _ 3 a^^ + 13 b\ 26. Take 6c-5d-96-4a from -10&-2c+3a-9d. 27. Subtract 4-3a;-x2 + 8a;3 + 10a;* from 9-7a; + 6a;2 -12 a;« + 5a;^ 28. From 7a-ll£i3-8 + 6a'' subtract 16a2-9 + 2rt^ + 15a - 10 a\ 29. From ar' + 3 xSj - ^y^ ■^5x^f — 4:xy* subtract 8 x*y — 7 x^y^ - 6 xY-^ 11 xy'-y\ 30. Subtract 7^<,2_5_20w3 + 13ii from -9 -Un^ + lGn + Sn^, ADDITION AND SUBTRACTION 47 31. Subtract - x^ + '6 x^y -3x1/ -\- f from ar' - 2 a^// - 2 x^- + f- 32. Subtract - 2 x" - I'd -{■ ^1 a? from x + l^x"- 18. 33. Subtract .V tn — \n + ^p from f 7rt + J ?i + f\i>. 34. Subtract fa— iV^ + A^^ ^^'^"^ 2a — ^6 — fc. 35. Subtract - | v + i w — f « from fV ■" -f | «^ — A ^*- * 36. By how much does 81 ^- + 4a2 — 36a6 exceed — 30a6 37. By how much does — 5c + 12a — 8 6 exceed 7 a— 9 c— 6? 38. By how much does exceed — 3a + 26 — c? 39. By how much does 1 exceed — 6a — 46 + 6? 40. From a^ — 2 a6 H- 6- subtract the sum of — a^ + 2 a6 — 6^ ami -2a2 + 262. Hint : The last two expressions are both to be subtracted ; they are there- fore subtrahends and should have their signs changed. Write down the min- uend, and, below it, the two subtrahends with their signs changed ; then add, thus doing the whole example at one operation. 41. From the sum of 3a2-2a6 + 6" and ha^ -^ab + Qh'^ take 6 a^ - 4 a6 - 3 6*-. 42. From 9 ar* — 8 ic -f- ar' take the sum of 5 — ar^ -|- a; and 6 a^ -7a;-4. 43. From the sum of a;-fy — 7 2 and -f-4a; — 9y take the sum of 9 X — 2y-\-z and — 5a; + 6?/ — 7z. 44. From the sum of 2a-f36 — 4(Z and 26 — 4c + 3d take the sum of4a — 46 — 3c-f-2d and 3 a + 2 c. 45. Subtract .5 aj + .25 y — 1.2 2 from 3 a; — 1.75 y -f .8 2. 41. Addition and Subtraction used in Equations. There are two more important rules used in solving equations. They will be illustrated by the scales. Review §§9 and 10. 48 ALGEBRA Suppose that the sugar S exactly balances the weight W on the scales in the figure. If a 3-lb. weight be added to the right scalepan, and 3 lb. of sugar to the left scalepan, then the scales will r^ still balance. , A '^^ ^— 1 Similarly, if anv number of pounds of weight be removed from the right scalepan and an equal weight of sugar from the left scalepan, then the scales will still balance. These facts illustrate the rules : Rule. — 1. The same number may be added to both members of an equation without destroying the equality. 2. The same number may be subtracted from both members of an equation without destroying the equality. Example 1. Solve the equation a.- — 3 = 7. Solution : 1. ic — 3 is 3 less than x\ if 3 is added to aj ~ 3, the sum is therefore x. Add 3 to both members of the equation in order to keep them equal. 2. Adding 3 to both members, x — 3+3 = 7+3, 3. or X = 10. Check : Does 10 - 3 = 7 ? Yes. Example 2. Solve the equation 18 a; — 5 = 3 a? + 55. Solution : 1. 18 x — 5 = 3 a; + 55. 2. Adding 5 to both members of the equation, 18 X = 3 X + 60. 3. Subtracting 3 x from both members of the equation, 15 X = 60. 4. Dividing both members of the equation by 15, X = 4. Check : Substitute 4 in equation 1 ; does 18 x4 — 5 = 3x4 + 56? does 72- 5 = 12 + 55? Yes. 42. In order to abbreviate the explanation of the solutions of equations, symbols A, S, M, and D will be used. Thus: A3 will mean "add 3 to both members of the equation." S2„ will mean " subtract 2 7i from both members of the equation." ADDITION AND SUBTRACTION 49 M_3 will mean " multiply both members of the equation by - 3." Dj will mean " divide bot^i members of the equation by 7." These symbols will be used in the text from now on. Pupils will find them helpful when solving equations. EXERCISE 23 Tell what the following symbols mean: 1. A^. 3. M,,. 5. D.4. 7. A^. 9. Sz. 2. S_s. 4. A_8,. 6. M_i. 8. D34. 10. M.,. 11. Solve the equation 24 — 11 7?i = 6 — 8 m. Solution : 1. 24-llm = 0-8m. 2. A8«: 24 - 3 m = 6. (Rule 1, ,§41) 3. S84: -3m=-18. (Rule 2, ,§41) 4. M.i: + 3 wi = + 18. (Rule, §10) 5. Ds: m = 6. (Rule, § 9) Check : Substitute 6 in the given equation. Does 24-11 x6 = 6-8x6? does 24 - 66 = 6 - 48 ? does -42 = -42? Yes. Note 1. In step 2, " Agm " means " add 8 m to both members of the pre- vious equation" ; the result will be equation 2. In step 4, "M-i" means "multiply both members of the previous equation by —1"; the result will be equation 4. Note 2. Whenever the coefficient of the unknown is negative, as in step 3, multiply both members by — 1 so as to make it positive. Solve the following equations and test the result : 12. a;-f3 = 12. 19. Sk = S-\-k. 13. ?/-f7 = 15. 20. 6a; = ic-f-45. 14. m-2 = 9. 21. 7 y = ^y-\~ 12. 15. 2m+5 = ll. 22. 6 2 = 63-3 2. 16. 3aH-7=:19. 23. 10r=80-6r. 17. 5c-2 = 23. 24. 2s = 99-9s. 18. 4^-7 = 21. 25. 3a = 120-7a. 50 ALGEBRA 26. 15a; = 45-15a; 34. - 2 x -\- 5 = -15. . 27. 4r + 5 = 6 + 3r. 35. -5y + 9 = ~21. 28. 3a; + 9 = 37-a;. 36. -11^ + 10 = 4-82. 29. 5^-8 = 28 + 2^ 37. 16r-l = 4r+5. 30. Gw-ll = 88 — 3t^. 38. 17t-7 = l-7t. 31. 15A;-13 = 9A; + 17. 39. llm + 6 = - 9?/i + 18. 32. 46 + 15 = 35-6. 40. 13^;- 9 = - 2^> + 36. 33. 3c-6 = c + 14. 41. 19r + ll=7r + 13. 43. In order to solve problems, it is necessary to translate the statements which give the conditions of the problem into algebraic symbols. Example. One number exceeds another number by 18. The product of the smaller number and 3 equals the larger number. Find the numbers. Solution : 1. Let s = the smaller number. 2. Then s + 18 = the larger number, 3. and '6 s = the product of the smaller and 3. 4. .*. 3 s = s + 18, since the product must equal the larger number. 5. S,: 2s = 18. 6. Dg: s = 9. Check : The smaher is 9 ; the larger 27 ; and 3 x 9 = 27. Note. The symbol, .'., means " therefore." Rule. — To solve a problem by means of an equation : 1. Represent one of the unknown numbers by some letter. 2. Represent the other unknown numbers by means of this same letter, using relations given in the problem. 3. From the conditions of the problem form an equation between the numbers; solve the equation. 4. Check the result by comparison with the statements of the problem. ADDITION AND SUBTRACTION 61 EXERCISE 24 1. What number increased by 11 e(|uals 19 ? 2. There are two numbers of which the larger is 5 times the smaller. The difference between the numbers is 24. Find the numbers. 3. One number exceeds another by 54. The larger number is 7 times the smaller. Find the numbers. 4. Five times a certain number exceeds 8 by 37. Find the .number. 6. If 12 times a certain number is diminished by 8, the result is the same as if 4 times the number is increased by 5. Find the number. 6. 15 exceeds twice a certain number by the same amount that 3 times the number exceeds 10. Find the number. 7. The age of John is double that of his brother James. What are their ages if equal results are obtained by subtracting 5 years from John's age, and adding 10 to James' age. 8. The age of A is twice that of B, and the age of C equals the sum of the ages of A and B. The sum of the ages of A and C exceeds the age of B by 40 years. Find their ages. 9. One angle is four times as large as a second angle ; if their sum is increased by 5°, the result is one straight angle. (See § 13.) Find the angles. 10. A farmer wishes to inclose a rectangular field for a past- ure, making it 15 rods wide. He wants to make it as long as possible, using 172 rods of wire fencing, which he has on hand. How long can he make it ? 11. If six times the area of Lake Superior, the largest fresh- water lake, be decreased by 12,000 square miles, the result equals the area of the Caspian Sea, the largest salt-water lake, 180,000 square miles. Find the area of Lake Superior. 62 ALGEBRA 12. If twice the height of Mt. McKinley, the highest moun- tain in North America, be decreased by 11,798 feet, the result equals the height of Mt. Everest, the highest mountain in Asia, 29,002 feet. Find the height of Mt. McKinley. 13. The longest river in the world is the combined Mis- sissippi-Missouri and the next longest is the Nile. Twice the length of theNile diminished by 2800 miles equals the length of the Mississippi ; the length of the Mississippi is 700 miles more than that of the Nile. Find the length of each river. 14. The highest velocity of wind recorded in the United States up to January 1, 1910 exceeded 25 times the lowest average velocity at any point in the United States by 2 miles per hour j the highest velocity exceeds the lowest by 98 miles. Find the highest and lowest velocity. . 15. In the United States, the lowest average annual precip- itation, rain and snow, is at Yuma, Arizona ; and the highest is at Mobile, Alabama. The precipitation in inches at Mobile is 20 times that at Yuma ; the sum of the two precipitations is 65.1 inches. Find the precipitation at each place. 44. Percentage and Interest Problems. Many of the problems involving percentage and interest may be expressed and solved by algebraic methods. EXERCISE 25 Percentage Problems 1. What does 4 % mean ? 5%? r%? m%? 2. What is 4 % of 500 ? 6 % of 250 ? 3. Express decimally 4 % of p ; 6 % of 6 ; 15 % of c. 4. If the cost of an article is c dollars, and the rate of gain is 25%, what is the gain? what is the selling price? (c-f-.25c=?.) 5. Find the cost of an article sold for $ 165 if the gain is 10%. (c + . 10 c = 165.) ADDITION AND SUBTRACTION 58 6. A grocer wishes to make 25 % on some canned goods. At what price must he buy them so as to be able to sell the goods at $1.25 per dozen ? 7. A man wishes to sell hats at $3.50 each. At what price must he buy them so as to make 12 % upon the cost? 8. A real estate agent knows that he can sell a certain lot for S3270. At what price must he buy the lot from the present owner in order that he may make a profit of 9 % ? Interest Problems 9. What is the simple interest on $200 at 6 % for 1 year? for 2 years ? for t years ? 10. What is the simple interest on R dollars at 6 % for 1 year ? for 2 years ? for t years ? 11. What is the simple interest on $200 at r % for 1 year ? for 4 years ? for t years ? 12. What is the simple interest on P dollars at r % for 1 year*? for 3 years ? for t years ? 13. If / represents the number of dollars interest on P dollars invested at r % for t years, it may be expressed by the Prt formula: / = . Express this formula in words. 100 ^ Solve the following problems by substituting in this formula : 14. If a man receives $ 1150 income from $ 3500 which has been invested for 4 years and 6 months, what rate of interest has he received ? (Since I = P . -^ ■ t.) ^ 100 ^ Solution : 1. Let r = the rate per cent. 2. P = 3500; 7=1150; t = ^. 3. 35 , .-.1150 = -3606-. ^ .44. (S 100- ^ .-. 1150 = 1 . 35 . r. 11.50 =q^r. 4.M2: 2300 = 315 r. 6. Dsis : r = 7.3+ 0/0. 64 ALGEBRA 15. What principal must be invested at 4 % to yield an income of $ 1500 per year ? 16. For what length of time must $4000 be invested at 5 % simple interest to yield $ 750 interest ? 17. What rate of simple interest has been earned on an investment of $ 2500, if the income is $ 1000 in 10 years ? A7nou7it in Interest Problems 18. The sum of the principal and interest is the amount. Indicate the amount at the end of one year if P dollars are invested at 4 %. (P + .04 P = 1.04 P.) Similarly at 6 % ; at 7 %. 19. What is the amount at the end of Wo years if P dollars are invested at 4 % ? for 3 years at 5 % ? 20. What sum of money will amount to $ 3500 if invested at 5 % simple interest for 5 years ? 21. How long will it take $ 1500 to amount to $ 2000 if invested at 5 % ? (2000 = 1500 + 1500 • yf + ic). 2. ^x-\-{z-2y). 6. 6r-[-3a-f&]. 3. 5a-(26-c). 6. 8s + {-3m-i{. 58 ALGEBRA 7. 9k-[p-2q']. 9. l-lm-n\. 8. 2a-\-{3b-c). 10. S-[-2a + b]. In the following examples, since neither symbol of grouping incloses the other, remove both at once. Tell what each example means; thus in 11: "'subtract the number (c — d) from the number (2 a — h)P 11. {2a-h)-{c — d). 16. {x-y)-{z- id). 12. (a-h)-{c + d). ^ 17. [m + 7.] + [i^-r7]. 13. [r + s]-fp-'y]. 18. \m-\-n\-\p -q\. 14. \r-s\ — \t-\-v\. 19. {-a + h)-{-{-c+d). 15. {x-y)-{z-\-io). 20. {-a + h)-{-c-d). In the following examples, combine terms after removing the parentheses : 21. 3 c- (5 c -6). 26. (3 6- ll)-(4 6 + 5). 22. (2«-6)-(5a--6). 27. {^x-\-y)-{-2x-]-y). 23. 2a-(3a-6) + 26. 28. bx''-{-2x' + x) + Q. 24. a2_^a-(a2-a). 29. (2 r- s) -{3r-2 s). 25. (5a;-6)-(2a;-4). 30. (3i9-2g) -f (-3p-2g). 31. 9m — (4m + 6n)4- (3m — m). 32. 8r + (5r-[2s + q). 33. Sx-{2y-\-\px-y']). 34. 5a-(4a-J3a-lS). 35. 66-J76-(9& + 4)-7|. 36. {lt-r)-l^t-{10r-{-t)-Sr]. 37. 3a;-(5a;-[7a;-h9a-4]-3a). 38. 5a6-[(3a6-10)-(4a6 + 7)]. 39. 7c-(5c + [12c-S6c + 2j]). 40. m — [(6 m — 7 7i) — w] — [3 m 4- 4 71 — (2 m — 3 n)]. PARENTHESES 69 INTRODUCTION OF PARENTHESES 50. Sometimes it is necessary to introduce parentheses into an expression. Development. 1. What is the rule for removing parenthe- ses preceded by a phis sign ? 2. Wliat, then, should be done with the signs of terms which are placed ivithin parentheses preceded by a plus sign ? 3. Inclose the last two terms of a-\-h-\- c — d in parentheses preceded by a plus sign. 4. What is the rule for removing parentheses preceded by a minus sign ? 5. What, then, should be done with the signs of terms which are placed vithhr parentheses preceded by a minus sign ? G. Inclose the last two terms of a + 6 -f c — d in parentheses preceded by a minus sign. Rule. — 1. To inclose terms in parentheses preceded by a plus sign, rewrite the terms without changing their signs. 2. To inclose terms in parentheses preceded by a minus sign, re- write the terms, changing the signs from + to - , and from - to 4- . " *t ^ \ Example. Inclose the last three terms of ^ + s — ?-f vMn ])arentheses preceded by a minus sign. Solution: 1. r-]-s— t-{-v = r — (— s-{-t — v). (Rule 2). Check : If the parentheses are removed, the result is the original expression. EXERCISE 27 Inclose the last three terms of the following expressions in parentheses preceded by a plus sign : 1. m-^n — c + d. 6. a- + 6^ — 2 6c -f cl 2. a^h-r- s. 7. a' -h'^-\-2hc- c". S. x-i-y + z-w. 8. a^- 4 62 + 12 6 -9. 4. r — s'-i-t + x. 9. a^ — 2 2/2 + 2/^ + z\ 5. p—q — a + b. 10. ii* — 5 ?t'^ —8n^-\-0 n + 7. 60 ALGEBRA 11-20. Inclose the last two terms of the foregoing expres- sions in parentheses preceded by a minus sign. 20-30. Inclose the last three terms in parentheses preceded by a minus sign. EXERCISE 28 Indicate and simplify, where possible, the following : 1. The sum of a and 5 ; a and b. 2. The sum of x and (a? + 3) ; x and (x — 5). 3. The difference between a and 5 ; a and b. 4. The difference between (a — 2) and 3 a; (a + 7) and 3 a. 5. The sum of (3 a -f- 5) and (2 a — 6) ; also their differ- ence. 6. The amount by which 15 is greater than 12; greater than a ; greater than (a -\- 2). 7. The amount by which 5 x is greater than 10; greater than 2 x ; greater than (x — 4). 8. The amount by which 20 exceeds 15; exceeds t; ex- ceeds (^ — 3). 9. The amount by which Sp exceeds 5; exceeds (p -{-1); exceeds (p — 3). 10. The amount by which (2 a + 3) exceeds (a — 5). 11. The smaller part of 15, if I is the larger part. 12. The smaller part of x, if 7 is the larger part. 13. The larger part of w, if 3 is the smaller part. 14. The smaller part of 2 a, if (a — 3) is the larger part. 15. The larger part of (3 c — 2), if (c + 1) is the smaller part. 16. The sum of two numbers is 50 ; the smaller is p. Eepre- sent the larger number. PARENTHESES 61 17. The sum of two numbers is 37; the larger is I. Repre- sent the smaller. 18. The difference between two numbers is 5 ; the smaller is s. Represent the larger. 19. The difference between two numbers is x-, the smaller is 3. Represent the larger. 20. The difference between two numbers is 10 ; the larger is n. Represent the smaller. 21. The integer which is consecutive to the integer a. Hint : 6 is consecutive to 5 ; 7 is consecutive to 6. 22. The second integer consecutive to a. 23. The sum of a and the consecutive integer. 24. The sum of n and the two consecutive integers. 25. If a is an odd number, what is the consecutive odd number ? Hint : 5 is consecutive to 3 j 11 to 9 ; 17 to 15. 51. Parentheses are used in equations and problems. Example. The sum of two numbers is 88. The larger number exceeds the smaller by 36. Find the numbers. Solution : 1. Let I = the larger number. (§47) 2. Then (88 — Z) = the smaller number. 3. Then i _ (88 - = 86. 4. Removing ( ), Z _ 88 + Z = 36. 2 i - 88 = 36. 5. Agg:* 2 I = 124. 6. D2: I = 62, the larger number. 88 - 62 = 26, the smaller number. Check: 62 - 26 = 36. * See § 42 for the symbol Age- 62 ALGEBRA EXERCISE 29 Solve the following equations 1. 2r-(r4-6)=ll. 2. (3a; + 4)-(2a;+9) = 15. 3. 12A-\-(2A-S)=(5A-(-7A-15). 4. (4i_5)._(2^4.5) = (2^-9)-(4«-7). 5. (- 10 a; + 12) - (6 a^ - 5) = (13 - 8 x) - (15 + 9 x). 6. (3.T + 5)-(2a;-7) = (4a; + 9)- (2a;-ll). 7. (6y/-8)-(9 2/+4) = (9-132/)-ll2/. 8. (13 -4m)- (3 m -h 9) = (2 7ii - 8) - (6 + 7 m). 9. (8^-ll)-(7-50 = 12-(13 + 40. 10. (17 ;s - 1) - (9 ^ - 10) = 15 -(13 2 + 6) - 22. 11. The sum of two numbers is 30. The greater number exceeds the smaller number by 4. Find the numbers. 12. The sum of the ages of A and B is 115 years. A is 13 years younger than B. What are their ages ? 13. Divide the number 123 into two parts such that the greater exceeds the smaller by 67. 14. The sum of the ages of A and B is 102 years. A is 26 years older than B. Find their ages. 15. Divide $93 between A and B so that A may receive $23 less than B. 16. The Library of Congress at Washington consists of three main stories, whose total height is 64 feet. The height of the second story exceeds that of the first by 7 feet, and is 8 feet less than that of the third story. Find the height of each of the three stories. 17. The total length of the Upper Steel Arch Bridge at Niagara Falls is 1240 feet. Its main span, the longest of the kind in the world, exceeds twice the remaining part of the bridge by 40 feet. Find the length of its main span. parenthesp:s 63 18. The first appropriation for a library for Congress was made in 1800. The present Library of Congress was completed at a total expense of over $ 6,000,000. The cost of the gold leaf on the dome of the present building increased by $1200 equals the appropriation made in 1800 ; and the sum of that first appropriation and the cost of the gold leaf is $ 8800. Find each of these amounts. 19. Seven major planets besides the earth revolve around the sun. The number of planets which are farther from the sun than the earth exceeds by one twice the number which are nearer to the sun than the earth. Find the number of planets nearer to and the number farther from the sun than the earth. 20. Tlie sum of two consecutive integers is 35. Find them. (See Example 21, Exercise 28.) 21. The sum of three consecutive integers is 108. Find the numbers. 22. The sum of four consecutive integers is 218. Find them. 23. The sum of two consecutive odd integers is 196. Find them. 24. Find the integer which is such that when increase.d by the first consecutive integer, and their sum decreased by the second consecutive integer, the result is 75. 25. There are four angles which make up the total angular magnitude around a point. (See § 13, Example 15). The second angle is 3 times the first ; the second exceeds the third by 10°, and exceeds the fourth by 50°. Find the angles. Illustrate with a figure. 26. An angle exceeds its supplement by 30°. Find the angle. (See Example 11, §13). 27. The complement of a certain angle exceeds the angle itself by 20°. Find the angle. (See § 13.) 28. The sum of the supplement and complement of a certain angle is 120°. Find the angle. 64 ALGEBRA 29. The fertility of farm land is maintained by adding fer- tilizers which contain certain plant foods, such as nitrogen, potash, and phosphoric acid. In 100 pounds of a good corn fertilizer, the amount of phosphoric acid should exceed the amount of nitrogen by 10 pounds, and the amount of potash should equal in weight the sum of the other two. Find the number of pounds of each plant food in the mixture. 30. Texas is the largest state in the Union, and Rhode Island is the smallest. The area of Texas exceeds 213 times the area of Khode Island by 72 square miles. The sum of their areas is 267,144 square miles. Find the area of each state. Historical Notk. — In Examples 5 and 8, negative roots are found. The mathematicians of the 16th century were slow to admit that such roots had any meaning. Cardan called them numerce fictce. Even such men as Vieta and Harriot, who contributed so much to the growth of algebra, admitted only positive solutions. Girard (1590-1632) and Des- cartes (1673) were especially instrumental in establishing the fact that negative roots should have the same meaning as positive roots. V. MULTIPLICATION 52. The Law of Signs for Multiplication of positive and nega- tive numbers (§ 25), may be written: Rule. — 1. The product of two numbers having like signs is posi- tive. 2. The product of two numbers having unlike signs is negative. Historical Note. — Until the time of Michael Stifel (looS), little progress was made toward finding the rule for multiplying powers of the same base, owing to the cumbersome notation that had been used pre- viously to denote the various powers. Some men even made a table cor- responding to our multiplication table, giving the products of some of the powers. Stifel introduced a better notation, used the word exponent, and gave the rule which we now use. 53. The Law of Exponents for Multiplication. Development. 1. Review the definitions of exponent, base, and power of a number, giv^en in paragraph 15. 2. What does a^ mean? ^^'^ r^? e^? 3. Write in exponent form : (a) b'b'b'b; (b) m-m-m-, (c) y • y - y • y - y; (d) r • r . • . r (if there are 10 factors) ; (e) a; • a; • • • a; (if there are 8 factors). 4. Find the product of a^ and a*. a^ = a- a • a; a* = a ■ a - a • a. .'. a^ • a* = (a • a ' a) X (a • a - a • a) =: a a • a ' a • a ' a ' a = d^ . 5. Find as in 4, the following products and write down the results as in part (a) : (a) a^'a* = a!. (c) r" -1''' = '^ ' (6) a^ . if* = ? ((T) m^ • m-' = ? 66 66 ALGEBRA 6. Observing the results in 5, see if you can get the follow- ing products mentally; check as in 4: (a) m^ ' m^ = ? (c) x^ - a^ = ? (b) m^ . m^ = ? (d) f • 2/' = ? , 7. The facts observed lead to the Rule. — To find the exponent of any number in a product, add the exponents of that number in the multiplicand and multiplier. EXERCISE 30 Find the indicated products 1. X- . x\ 4. t" ' t\ 1. m^^ • m\ 10. 2^ • 2\ 2. a' . a\ 5. f • t\ 8. s^' • s\ 11. 3^ • 3^ 3. iiv" . ml 6. f ' t\ 9. if . y\ 12. 5^ . 51 54. The Commutative Law of Multiplication. Development. 1. 3x4 and 4x3 each equal 12. 2. Notice, in step 1, that the same factors^ occur in each product, but that the order is changed. Does changing the order of the factors change the value of the product ? 3. Comjjare 3x4x5 and 3 x 5 x 4, by finding their values. 4. Write these same factors in some other order, and com- pare the product with the products obtained in step 3. 5. Arrange the factors 2, 4, and 5 in three different ways and compare the products. Rule.— The factors of a product may be arranged in any order without changing the value of the product. EXERCISE 31 Arrange each of the following sets of factors in two ways and find the products. 1. 3, 7. 4. 5, 2, 4. 2. 2, 5, 6. 5. 6, 5, 3. 3. 3, 4, 6. 6. 6, 4, 2. MULTIPLICATION 67 MULTIPLICATION OF MONOMIALS 55. Development. 1. Find the product of 7 a and —2 b. Solution:!. - 2 6 = (- 2)6. (See §31) 2. Then 7 «(- 2 6)= 7 • a • (- 2) . 6 Then, since the order of the factors may be changed, (§ 64) 3. 7a(-2 6) = 7. (-2).rt.6 = - 14 ab. (§ 6) 2. Find the product of —ox and — 6 x^y. Solution: 1. (- 5x)(- 6x^y)= (- 6) ■ x - (- 6) . x'^ -y. = (-">)-<-«)-^-^'^-y = + 30 x% 3. Find similarly the product of 3 mn and 2j\ 4. Find similarly the product of + 2 a^b and + 5 ab. 6. Give at sight, if possible, the following : (a) 2 r.s *5t. (c) 7 a • 3 a". (b) 3 xy ' 2z. (d) 4 m • 5 mx. Rule. — To find the product of two monomials : 1. Find the product of the numerical coefficients, using the Law of Signs for Multiplication. 2. Multiply this product by the literal factors, giving to each an exponent equal to its exponent in the multiplicand plus its exponent in the multiplier. Example 1. Multiply 2 a' by 9 a*b\ Solution : (2 a5)(9 a*b^) = + (2x9). a^^+*) ■ b^ = 18 a«&2. ^ Example 2. Multiply — 5 x^yz!^ by -f 6 xy^w. Solution : ( - 5 ofiijz^) ( + 6 xy^w) = -(5x6). x(8+i) • y(i+2) . z^ . w = - 30 x^y^z^w. ^8 ALGEBRA EXERCISE 32 Find the following indicated products: 1. af • x^ H. a^yz . xy^z. 2. a^ . a^ ' a\ 12. i^sH . rsE 3. x^ -oy^ • x^ 13. a6V . aftV. 4. y^-y^-y^. 14. rh • rs'e, 5. wi^ . 771 • m^. 15. 2^ • 2^. 6. a}- • a^ . a^ 16. 4^ • 4^. 7. E^'E^'E, 17. S^'.S*. 8. a'b^-ab. ^ 18. 10^ . 101 9. it-y . xy\ 19. 2>r2^ • r^. 10. a^x • aa^. 20. mx^ • mna^. Multiply : ' 21. 7 a"^ by 3 a^. 31. - 6 a^b^ by - 6V. 22. — 9 m^ by 8 m\ 32. + 9 ao^ by — 9 a/. 23. — 5 a6 by 2 al 33. 8 a;V by - 8 2/V. 24. - 3 rs2 by - 9 vs. 34. - 7 AB^ by + 12 ^5^(7. 25. 5 xyz by — 11 aj^/s;. 35. 12 a-6c by 6 bcd^ 26. - 11 a;/ by 9 a;^?/. 36. 13 a^ by - 7 a^ft^^. 27. - 6 d'b by - 4 ab\ 37. - 14 m'n by - 6 mn^. 28. - 6 aj2?/3 by + 12 xy\ 38. - 16 a^bc* by 4- 5 a^6«c. 29. - 9 mhi^ by 7 mVl 39. - 3 a^f^ by - 17 x^fz. 30. - 12 rst by - 9 s2. 40.-4 m/i by + 17 m^n. Find the product of : 41. 3 a^, 5 a^, and 6 a^. 42. -4.x^, -9y% and2;s^ 43. a^y^, — yh, and xz^. 44. - 12 a'b'', - b'c^, and - 8 cW 45. a^j —3 a, — 5 a% and — 4 a^. MULTIPLICATION 46. — 7 m^w, + 8 mi-^, and — 2 nh\ 47. -3 and -f-^^" i5 48. + 5 and 'Zm-" 49. 4- 4 and - 3n 2 50. -f 12 and 5r 3 ' 51. - 15 and Os 5 ' 52. - 10 and _ 11 63. — 9 and -j- 4 J) 54. +14 and -— . 7 55. 18 and 9 ' 56. + 20 and - j\ x. 57. ^-ai^and -^a'b. 58 — f mn and — | m^. 59. + .5 rs^ and - .3 j-^s. 60. - .12 f and - .7 ^. 56. Numbers, and relations between them, may be represented by geometrical figures. Example 1. The product 4x5 may be repre- sented by the rectangle in Fig. 1. Note that the ^ area of the rectangle is 20 square units. Example 2. The product 4 (3 + 7) may be represented by the Fig. 2. = 4 + Fio. 2. 4(3 + 7)=4x3+4x7. 4x10= .12 + 28. Example 3. The product a{b-{-c) may be represented by the Fig. 3. ab ab b +• Fi<;. 3. a (6 -f c) = a6 -h rtc. 70 ALGEBRA EXERCISE 33 1. Draw a figure representing 5(3 + 4), as in Example 2. 2. Draw a figure representing 4(2 + 3 + 5), as in Example 2. 3. Draw a figure representing a(6 + c + cZ), as in Example 3. MULTIPLICATION OF POLYNOMIALS BY MONOMIALS 57. Development. 1. From the geometrical illustration in § 56, it is clear that : ^ (a) 5(7 + 3)=(5x7) + (5x3) = 35 + 15 = 50. (P) 6(4 + 5) = (6x4)+(6x5) = 24 + 30 = 54. (c) a(b-{-c-\- d) = ah -\- ac-\- ad. In every case, each term of the polynomial is multiplied by the monomial. Rule. — To multiply a polynomial by a monomial : 1. Multiply each term of the polynomial by the monomial. 2. Unite the results with the proper signs. Example. Multiply 3 a' -2 ab -\-b' hy -3 ah. Solution : (3 a2 _ 2 ah + 6-^) x (- 3 «6) = - 9 a^ft + g a%'^ - 3 ah^. Check : This result is true for any values of a and h. Let a = 1, and & = 1. 3a2_2rt& + /;2^3_2 + i =2, -3a6=-3, and 2 • (- 3) = -6 ; also, _9a36+6a26-^-3a&3 = _94.6_3=-12 + 6 = ~6. EXERCISE 34 Multiply : 1. 4 a — 9 by 5 a. 6. r- — 2 7's + s^ by — i^s^. 2. m^ — mn + v? by mn. 7. G .t^ — 5 x^ — 7 ic^ b}^ — 7 a?. 3. 3ar^ + iB-5 by -9a;-. 8. - 3 c^ - d^ -\- 6 cd hy 4. c\l\ 4. 8 0^2/ — 5 ^if by — 3 xif. 9. — 3x-y-\-i(?—3 xy- by —x-y. 5. 2a3~6a2-7by-7al 10. a^ -;x^ ^xhy -o?. MULTIPLICATION 71 11. 5 m^ — 6mn — 4 n^ 13. 3c^-5xy-{-y^ 3m^ -2ay 12. 6p--5pq + 9q^ 14. a^ - 3 a^b -{- S ab^ - b^ 4:2Jq —ab Perform the following indicated multiplications: 15. 7x'(x-5). 19. -3ab'(a'-2ab + b'). 16. -6ab'(10a^-7b'). 20. S x" - (6a^-5 x-12). 17. x^'(a^-Ax^y''-\-i/). 21. -^a^b^ - (3a^-2ab-4:b^. 18. — rs ' (r^ —rs + .s^). 22. — 5 ?n* • (8 m* — m^ — 3). 23. Simplify the expression : 3(2 x — 6)— 2(x + 6). This means that (2 x — 5) is to be multiplied by 3 ; that (a; + 6) is to be multiplied by 2 ; and that the second result is to be subtracted from the first. Solution: 3(2 a; - 5)- 2(a; + 0) = (G x - 15)-(2x+12). = 6x-15-2x- 12. = 4 a; -27. In the following examples, firsl^ tell what each means, as in Example 23; then, simplify. 24. 5(6a + 36)+4(5a-2 6). 27. 3(2 m + 8)-2(6 -5 m). 25. 2a{3x-y)-3a(2x-\-y). 28. 3 c(2 m - 4)-6c(2 m + 4). 26. x(x — y)—y(x-^y). 29. r()*^ — s)-f-s(r — s^). Multiply and then simplify the following: 3, ,2(^_2^ + ^). 3. 15(^-^1). 34. i(2a^-Ax-\-6). 40. -ix2(3a;^-24a:2^15)^ 35. 6(^x2-^a:2/ + i2/0- ^^- imw(i m^--^ m?i+i «^. 72 ALGEBRA MULTIPLICATION OF A POLYNOMIAL BY A POLYNOMIAL 58. A number may be multiplied by (2 + 3) by multiplying first by 2 and then by 3, and adding the products. Thus, (2 + 3) X 6 = (2 X 6) + (3 X 6) = 12 + 18 = 30, for (5) x 6 = 30. Similarly, (2 + 3) • (a + &) = 2(a + &) + 3(a + b) = 2a + 26 + 3a + 36 = 5a+5 6, for (5) . (a + &) = 5 a + 6 6. The multiplier in each case consists of the sum of 2 and 3 ; the multiplicand is multiplied separately by 2 and by 3, and the products are added. This illustrates the Rule. — To multiply one polynomial by another : 1. Multiply the multiplicand by each term of the multiplier. 2. Add the partial products. Example 1. Multiply 3a — 46 by 2a — 5 5. Solution : In accordance with the rule, multiply 3a— 46 by 2a and then by — 5 6, and add the partial products. A convenient arrangement is suggested by the arrangement of multiplication problems in arithmetic. 43 3a_ 4 6 12 2a- 56 86 = 2 x 43 6a^- Sab =2a(Sa-ib) 430 = 10x43 - 15 gb 4- 20 62 = - 5 6(3 g - 4 6) 516 = sum 6 g2 _ 23 g6 + 20 6^ = sum Note : In arithmetic, the multiplication proceeds from right to left ; in algebra, the multiplication proceeds from left to right. Example 2. Multiply a^ — 8 a;^ — 2 a^a; by 2 ic + a. Solution : It is convenient to arrange the multiplicand and multiplier in the same order of powers of some letter (§ 37) and to write the partial products in the same order. Leave spaces for any powers which may not be present in the multiplicand. Arranging the expressions according to the descending powers of g, we ^*^®' g3 - 2 a'^x - 8 a:8 a +2x a* - 2 a% - 8 ax» + 2 g3a; - 4 a%2 - 16x* a* - 4 g2a;2 _ g gx^ - 16 x* MULTIPLICATION 73 Check : This result should be true for all values of a and x. Let a = 1 and x = 1. Then, a8-2a2x-8x3 = l-2-8=-9, a + 2x=l-l-2 = + 3, and (_9)x(+3) = -27; also, a*-4a2x2-8ax3-16x* = l-4-8-16=-27. EXERCISE 35 Multiply : 1. a;4-3byaj + 5. 11. 2 a + & by a -f 6. 2. r-7byr-4. 12. Sc -2dhj 2 c +d. 3. 2s -5 by 8-3. 13. 5 r + 6 s by 3r - 2s. 4. 3m + 2bym-4. 14. 5x-2yhy Sx- 4.y. 5. 4«-9by< + 3. 15. 6m-3i:>by4m + 5p. 6. 3x-f 7by 2a;-f3. 16. 7 i/ - 9 z by 6 ^ + Sz. 7. 2m-|-5by 5m-l. 17. 11 a+ 5 d by 6 a- 4 d. 8. 6i)-3by2i9 + 7. 18. 12p + 7 g by 8p - 7 ^. 9. 5y-lhy(jy-S. 19. 2x' -fhy x'' -Sy\ 10. 7z4-10by42-5. 20. 9 lo' -7 v hj 11 w' + Sv. 21. m^ — m — 3bym + 3. 22. 2 a^ -h 7 a — 9 by 5 a — 1. 23. x'-2xy+Sy^hyx-Sy. 24. ar^-a;?/ + 2/2by X4-2/- 25. «'- + 4 a;?/ -h 16 2/2 by a; - 4 y. 26. m^ + mn 4- n^ by m^ — mn + n\ 27. rt3^^_2-2a2by a2 + 2a-3. 28. 3 -I- a-' - 7 a - 4 a^ by 2 a 4- 1. 29. 9a;4-2a^-5by44-3«'^-7a:. 30. 6n-8 + 4n2by -4 + 2n2-3n. 31. 9r^-5y' + eryhySry+4:y'-^7r'. 32. 3a'-5ab-Sb'hy4:a'-9ab-7b', 41. ia-i6byia-f-.i&. 42. ^m — ^7ihj ^m — ^ n. 43. \x-\yhj \x + ^y. 44. 2a-\hhj a+\h. 45. lx-\yhjix-^y. 74 ALGEBRA 33. a — h + chja—h— Co 34. r + s -|- ^ by r — s — ^. 35. 2 n^ + m^ + 3 mn by 2 7i2 _ 3 m^ 4. rn\ 36. a^ + 3 ah"- - 3 a'b - b^ by a' + '^^ - 2 a6. 37. m*-3m'^+9m2-27m4-81 by m + a 38. Aa-\-6b+10chy2a-3b-\-5c. 39. ic^ + 4 a?^ + 8 aj + 2 ar"' + 16 by x — 2. 40. a- + 6^ + c^ +.a6 — 6c + ac by a — 6 — c. 46. (3x-5y. 47. (2m-3w)3. 48. (4.r + 5)l 49. (2 a; -3)*. 50. (3 a -4. by. Find the product of the following: 51. a -\- 3, a — 4, and a + 2. 52. m + 4, 2 m — 3, and m — 5. 53. a; + ?/, flj- — a;?/ + .!/^ ^^^ x^ — y^. 54. m + 71, ?>i^ + 71^, and m — n. 55. a;^ -f x?/ + 2/^ ^ — xy -\- jf, and a^ — 3/^ PARENTHESES IN MULTIPLICATION 59. Example. Simplify (a - 2 a:)=^ — 2 (3 a + a;) (a — x). Solution: To simplify this expression, first multiply {a — 2x) by itself (§ 15) ; second, find the product of 2, 3a + x and a — X] third, subtract the second result from the first. 1. a-2ic 2. 3a + « g — 2a; a — x a^-2ax - 2 ax + 4 x^ a2 — 4 ax + 4 x2 6 a2 - 4 ax - 2 x2 3a2+ ax -3ax- X2 3 a2 - 2 ax - X2 2 MULTIPLICATION 75 3. Then (a -2a;)2-2(3a + a;)(a -x) 4. ={a^-Aax-\-ix^)-(Qa^-4ax^2x^) 5. =rt2_4aa;4.4a:2-6a2 + 4ax + 2x2 (§47) 0. =-5a2 + 6a;2. Answer. Note. Be careful to place the results of steps 1 and 2 in parentheses as iu step 4. EXERCISE 36 Tell, as in the above solution, what must be done to simplify the following ; then simplify : 1. (3a-j-8)(a-6) + (2a-h7)(4a-9). 2. (2m H- 7) (3 m - 5)- (2 m - 5) (3 m + 7). 3. (a-2x){a-{-3x)-^(a-\-2x)(a-Sx). 4. (2a-36)2-4(a-6)(a + 56). 5. 2(/iH-3)(/i-2)-(/i + 5)(/i-6). 6. 5(a;-4)(a; + l)-3(a;-3)(x-h2). 7. 2(3a; + 2)(4a;-3)-(3a;-2)(4a; + 3). 8. 3(3a4-5)(2a-8)-2(4a-7)(a + G). 9. 4(3«-2)(a; + 6)-5(2a;-7)(a; + 2). 10. (a-\-b)(a''-\-b')-(a-b)(a'-b'). USE OF MULTIPLICATION IN EQUATIONS 60. Example 1. Seven times the complement of a certain angle exceeds twice its supplement by 20°. Find the angle. SoLUTioy : 1. Let a = the number of degrees in the angle. 2. Then, 90 — a = the number of degrees in the complement, 3. and, 180 — a = the number of degrees in the supplement. 4. Hence, 7(90 - a) = 2(180 - a) + 20. 6. Multiplying, 630 - 7 a = 360 - 2 a + 20. 6. Combining, 630 - 7 a = 380 - 2 a. 7. Seao : - 7 a = - 250 - 2 a. 8. A2«: -6a =-250. 9. M_i : 5 a = 250. 10. Dg: a = 50. Check : The angle is one of 50*^. The complement contains 40° and the supplement, 130°. Does 7 x 40° = 2 x 130° + 20° ? Yes. 76 algp:bra Example 2. Solve the equation, (2a + 5)(3a-7)-(2a-5)(3a + 7)=:4. Solution: 1. (2a + 5)(3a - 7)-(2a - 5)(3a + 7) = 4. 2. Multiplying, (Ga^ + a - 35)-(6a2 - a - 35) = 4. 3. Kemoving ( ), Ga^ + a - 35 - 6«2 _(. « ^. 35 = 4. 4. Combining terms, 2 a = 4. 5. D2: a = 2. Check: Does (2 • 2 + 5)(3 • 2 - 7)-(2 -2 - 5)(3 • 2 + 7)= 4? Does (9).(-l)-(-l)(+13) = 4? Does _9-(-13) = 4? Does - 9 + 13 = 4 ? Yes. Note : lu Exercises like Example 2, be careful to put the products obtaiued in step 1, in parentheses as in step 2. EXERCISE 37 Solve the following equations : 1. 2(7^-3) = 20. 2. 4(2/-5)-h7 = 15. 3. 3(2a^-4) + 2(aj-5) = 5(x + l). 4. 6(2-3a;)-h3=3(4x-5). 5. 12-5(3a-2) = 2(a-6). 6. 2(v + 9) + 3('y-4) = 16. 7. 4(^-3) + 3(2^-f 5)=33(4-i). 8. 10-5(3Z-4) = 6(3-2Z). 9. 7^-6(2^-5)=6(6-^). 10. 37i-2(2w-7) = 3(n-2). 11. 3(4m-5)-4(m-6) = 3(m + 17)-7. 12. (x-5)(x-\-6)-(x-\-S)(x-^)=0. 13. (2/-7)(2/ + 2)-(2/-9)(^ + 3) = 0. 14. (2r 4- 3) (3r - 5)- 6(r- 4) (r- 3) + 5 = 0. 15. 3(2s-4)(s + 7)-2(3s-2)(s + 5) = -3(s-2). MULTIPLICATION 77 16. The" sum of two numbers is 75. The larger exceeds the smaller by 11. Find the numbers. 17. The sum of two numbers is 100. If four times the greater be diminished by 22, the result is 5 times the smaller. Find the two numbers. 18. The distance from New York to Paris exceeds the dis- tance from New York to London by 280 miles. Four times the distance to London exceeds three times the distance to Paris by 2900 miles. Find the two distances. 19. One number exceeds another number by 7. If 6 times the smaller is diminished by 5 times the larger, the remainder- is 5. Find the numbers. 20. Separate 60 into two parts such that 4 times the smaller shall exceed 2 times the larger by 30. 21. The sum of two numbers is 80. If twice the greater be decreased by 12, the result exceeds 4 times the smaller by 4. Find the numbers. 22. The Library of Congress stands upon a rectangular base whose perimeter is 1620 feet. The length exceeds the width by 130 feet. Find the dimensions of the building. 23. There are two consecutive numbers such that the sum of twice the smaller and three times the larger is 78. What are the numbers ? 24. There are two consecutive integers whose product ex- ceeds the square of the larger by 20. What are they ? 25. Th& total population of Chicago, Philadelphia and Greater New York (1910 Census), was 8,501,174. The popu- lation of Chicago exceeded the population of Philadelphia by 636,275 ; the population of New York exceeded twice the pop- ulation of Chicago by 396,317. Find the population of each of the cities. 78 ALGEBRA EXERCISE 38 Algebraic Expression 1. A is now 15 years of age. Express his age : (a) 5 years ago ; (6) m years ago; (c) y years ago; (d) 8 years from now ; (e) m years from now. 2. B is 6 years of age. Express his age : (a) 4 years from now ; (6) ?7i years from now ; (c) 6 years ago ; (d) x years ago ; (e) ^ years ago. 3. A is now x years of age. B's present age exceeds the age of A by 5 years. (a) Express B's present age ; (6) the sum of their ages. (c) Express the age of each 10 years ago. (d) Express the age of each 10 years from now. 4. A is now a years of age; B is twice as old. (a) Express B's present age. (h) Express the age of each 3 years ago. (c) Express the age of each 7 years from now. (d) Express the fact that B's age 5 years ago was 3 times A's age at that time. 5. Express the value of : (a) (15 — x) pounds of tea at 40 ^ per pound ; (6) X pounds of tea at 60 ^ per pound ; (c) the entire amount of tea. 6. Express the value in cents of : (a) X nickels ; (h) 2x dimes ; (c) 3 a; dollars ; (d) 4 x quarters ; (e) all of the coins. 7. Express the value in cents of : (a) d dimes; (h) (15 — d) quarters ; (c) (15 4- 3 c?) half dollars ; (c^) all of the coins. MULTIPLICATION • 79 8. Express in inches : (a) m feet ; (6) 3 m yards. (c) the combined length of m feet and 3 m yards. 9. Express in pints : (a) 3 X pints plus 2x quarts plus 5 x gallons. (6) 2 c pints plus (3 c — 2) quarts plus (5 — 3 c) gallons. EXERCISE 39 1. The sum of the ages of A and B is 50 years ; in 5 years A will be 5 times as old as B. Find their ages. Solution : 1. Let a = the number of years in A's age now. 2. Then (50 — a) = the number of years in B's age noio. 3. Then a 4- 5 = the number of years in A's age in 5 yr. 4. and (50 — a + 5) or 55 — « = the number of years in B's age in 5 yr. 5. .-. (a + 5) =5(65-rt). Complete the solution. Note. Represent with care the present ages of both persons ; also their ages at the other time mentioned ; then form the equation. 2. A father is now 9 times as old as his son. In 9 years he will be only 3 times as old as his son. What are their present ages ? 3. The difference between the present ages of a father and son is 25 years. In 10 years the father will be twice as old as his son. .What are their present ages ? 4. A is 5 times as old as B. In 9 years he will be only 3 times as old as B. What are their ages ? 5. B is twice as old as A. 35 years ago he was 7 times as old as A. What are their present ages ? 6. A is 68 years of age, and B is 11. In how many years will A be 4 times as old as B ? Hint. Let N equal the number of years. Find the age of each in N" years, and then form the equation. 80 ALGEBRA 7. A is 25 years of age and B is 65. How many years ago was B 6 times as old as A ? 8. A grocer has two grades of tea, a 60j^ grade and a 90^ grade. He wishes to make a mixture which he can sell for 80/ per pound. How many pounds of each must he use in a mixture of 120 pounds? Solution : 1. Let n = the number of pounds of 60 ^ tea used. 2. .-. (120 — n) = the number of pounds of 90^ tea used. 3. .'. 60n = the value of the 60 ^ tea in cents. 4. and 90(120 — w) = the value of the 90 ^ tea in cents. 5. . •. 60 w 4- 90(120 — n) = the value of the mixture in cents. 6. But 120 X 80 = the value of the mixture in cents. 7. .-. 60 w + 90(120 - n)= 9600. 8. From the equation n = 40, 120 — w = 80. Check : 40 pounds of tea at 60/ are valued at $24. 80 pounds of tea at 90 fi- are valued at f 72. Total value of the mixture is !$96. Also, 120 pounds at 80 ^ are valued at $ 96. 9. A grocer has tea worth 70/ per pound and other tea worth 40/ per pound. How many pounds of each must he take to form a mixture of 50 pounds which he may sell at 49/ per pound ? 10. A grocer has coffee which he sells at 36/ per pound, and other coffee which he sells at 20 / per pound. How many pounds of each must he take to make a mixture of 100 pounds which he may sell at 25/ per pound ? 11. A seedsman wishes to make a mixture of grass seed consisting of clover seed and blue grass seed. He sells his clover seed at 40/ per pound, and his blue grass seed at 22/ per pound. How many pounds of each must he take to make a mixture of 200 pounds which he may sell for 25/ per pound ? 12. A sum of money amounting to $2.80 consists of dimes and quarters. The number of dimes exceeds the number of quarters by 7. Find the number of each kind of coin. MULTIPLICATION 81 Solution : 1. Let q = the number of quarters. 2. Then g -j- 7 = tlie number of dimes. 3. .*. 25 g = the number of cents in the quarters, and \(i(q -f 7) = the number of cents in the dimes. 4. .-. 25^4-10(^ + 7) =280. Complete the solution and check it. 13. A man has two kinds of money, dimes and fifty -cent pieces. If he is oifered $4.00 for 20 coins, how many of each kind must he give? 14. A sum of money amounting to S2.20 consists of five- cent pieces and quarters. There are in all 16 coins. . How many are there of each kind? 15. A sum of money amounting to $24.90 consists of $2 bills, fifty-cent pieces and dimes. There are 5 more fifty-cent pieces than $2 bills, and 3 times as many dimes as' $2 bills. How many are there of each denomination ? 61. Equations having Fractional Coefficients. Example 1. If the sum of a certain number and one half of itself be diminished by three iifths of the number, the re- mainder is 9. Find the number. Solution : 1. Let x — the number. (§10) (§57) (§9) 10 + 5-6 = 9? Yes. Note. In order to eliminate the denominators, multiply the equation by the Lowest Common Multiple of the denomiuators. Then -!-¥-• 2. The denominators must be eliminated. Mio: ^ rr 2 lOaJ + ie-- ~-^ ' — = 90. 3. 4. 10 a: + 5 a; - 6 a; = 90. 5. Combining, 9x = 90. 6. D9: a; = 10. 5 2 Check: Does 10 + ^-3.^=9? (§10) 82 ALGEBRA Example 2. Solve the equation ^-4^ = — ^ . 6 3 5 4 Solution : 1. The L. C. M. of 6, 3, 5, and 4 is 60. 60(Lp-|) = 60(?^i^-l). 3. 70 m - 100 = 30 m - 15. 4. Aioo : Ss&m' 70 m - 36 m = 100 - 15. (§ 41) 5. .-. 34m = 85; w = ff = I =2.5. ' Check : This solution may be checked by substitution or by going over the solution again. If the latter method is used, great care must be taken, as it is easy to overlook an error. EXERCISE 40 Solve the following equations : , X ' X 5 .- 12. .• ''-^ 3 14. 4d-^' = '' + ?3, 5 2 20 15 li_^' = ?_?i ■ 9 6 3 2' 2 3 6 m 7n _ 8 2. 3" 5 "15* 3. -i-r 4. ¥=i«^ 5. 5a 3a 11 3 4 6 ^ 2r 3 ' 5r 6 = 6. m- 3 ?/i _ q '"5~'--' m 2 c 5 3c ■ 2c 2 3 4 3 * 2 5 2 3 7 "3 — — ■[ 5 ~ ■ 4 i ic_4 X _3 14~T'"2 : ?/ _5 8'~2 7. ?i = l4-l. 17. 7 3^ 8. — = 1+ — . 18. -^--_ — - 9. ^ = i^-?. 19. 10. ^ + il = ^. 20. 22 9 52 5 3 8 6 4* 3.V _^_ .5?/ 1 2 3 4 8* 4^ -1 = 7^ 16 9 9 ' 3 ' 7?/i 4 m 11 2m 2 ~~3 6 5 MULTIPLICATION 83 EXERCISE 41 1. One fifth of a certain number exceeds one eighth of the same number by 3. Find the number. 2. The sum of three numbers is 65. The second is one half of the first, and the third is two thirds of the first. Find the numbers. 3. What number increased by one half of itself equals the sum of two thirds of itself and 25 ? 4. What number exceeds the sum of its third, sixth, and fourteenth parts by 18? 5. What number is such that if four sevenths of it be sub- tracted from itself, the result equals the excess of three fourths of the number over 18? 6. What number is such that if two thirds of it be in- creased by 100, the result equals four fifths of it ? 7. Seven eighths of a certain number is as much less than 21 as three tenths of it exceeds 2|. W^hat is the number ? 8. The difference between the third and fifteenth parts of a certain number is 28. Find the number. 9. In a triangle commonly used by draughtsmen, the second angle is two thirds of the first, and the third angle is one half of the second. Find the angles of the triangle. (§ 13) 10. In another triangle used by draughtsmen, there are two equal angles, each of which is one half of the third angle. Find the angles of this triangle. 11. There are three consecutive numbers such that the sum of the second and third exceeds three halves of the first by 9. Find the three numbers. 12. A man has $4.35 in dollars, dimes, and cents. He has one fourth as many dollars as dimes, and five times as many cents as dollars. How many coins of each kind does he have ? 13. The Treasury at Washington is one of the most impos- ing of the Rational buildings. Its perimeter is 1400 feet. Its 84 ALGEBKA width exceeds one half of its length by 25 feet. Find its dimensions. 14. The greatest depth of Lake Superior is one half that of Lake Michigan ; the greatest depth of Lake Huron exceeds one sixth that of Lake Michigan by 700 feet. Tlie depth of Lake Huron exceeds that of Lake Superior by 100 feet. Find the depth of each. 15. Probably the largest room in the world under one roof is the passenger concourse of the Union Station in Washing- ton, D.C. Its perimeter is 1780 feet. One fifth of its length exceeds its width by 22 feet. Find its dimensions. 16. Ten times the population of the United States in 1820, in millions, exceeded the population in 1910 by 3.8 millions ; the population in 1910 exceeded 7 times the population in 1820 by 25 millions. Find the population in both years. 17. Plants feed upon certain plant foods present in the soil, such as potash, nitrogen, and phosphoric acid. A fair crop of potatoes will remove from an acre of ground about 99 pounds of these three foods. The amount of potash removed is 5 times, and the amount of nitrogen 2J times that of phosphoric acid. Find the number of pounds of each removed. 18. The length of the foundation of the Capitol in Wash- ington exceeds twice the width by 51^ feet. The perimeter of the foundation is 2202| feet. Find the dimensions of the foundations of the Capitol. 19. The average wholesale value of oak lumber in 1899 was $13.78 per thousand feet. This exceeded one half of the wholesale value in 1909 by ,$3.53. W^hat was the wholesale value per thousand in 1909 ? 20. The distance from San Francisco to London via New York is 6990 miles. The part of the journey by rail is 50 miles less than |^ of the part by water. Find the part of the journey on land and the part on water. VI. DIVISION 62. Division is the process of finding one of two numbers when their product and the other number are given. To divide 15 by 3 means to find the number by which 3 must be multi- plied to give the product 15. The Dividend is the product of the numbers ; it is the num- ber divided. The Divisor is the other given number ; it is the number by which the dividend is divided. The Quotient is the required number. 63. It is clear that a -^ a = 1 ; for d x 1 = a. 64. It is agreed that the product \)f zero and any number is zero. This makes division by zero impossible. Thus, if we try to find the quotient of 6 -v- 0, and let q equal the quo- tient, we should have the relation 6 = . g. But ■ q =0 and not 6, so there cannot be any ordinary number to use as q. Hence, there is no number to represent the quotient of 6 -f- 0. 65. Division is indicated by writing a fraction whose numera- tor is the dividend and whose denominator is the divisor. Thus, the quotient of 15 h- 5 is written ^. The quotient of 7 a&c -r- 3 xy is written ^^-^ • Sxy Note. The line, — , was used to indicate division long before the sym- bol, -r. 66. Division of a Product by a Number. Example. Divide 6 x 8 by 2. Solution:!. 6x8--2=^-^. 2 85 86 ALGEBRA 2. If 6 is divided by 2, ^Ji3 =12^ = 24. 2 )2 4 3. If 8 is divided by 2, 1x8^6^^24. 2 ^ 4. If both 6 and 8 are divided by 2, 3 4 6x8 ^ ^x^ ^-^o 2 ^ 5. Since we know that ^ = 48 h- 2 or 24, it is clear that the results 2 obtained in steps 2 and 3 are correct, but that the result in step 4 is in- correct. Hence, Rule. — To divide the product of two or more numbers by a num- ber, divide any one of the factors by the number, but divide only one of them by it. EXERCISE 42 Find each of the following indicated divisions in two ways: 6 1. 9x12 2 18x24^ 3 28x56 67. The Law of Signs for Division. Since (-h 2) x (+ 3) = + 6, then (+ 6) -f- (-f 2) = + 3. Since (- 2)x (+3) = - 6, then (_ 6) -- (-2) = + 3. Since (-f 2) x (- 3) = - 6, then (- 6) Since (- 2) x (-3)=-f 6, then (+6) (+2)=-3. (-2) = -3. If the signs of the dividend, the divisor, and the quotient in each of the previous statements are examined, the following rules become clear : Rule. — 1. The quotient of two numbers having like signs is positive. 2. The quotient of two numbers having unlike signs is negative. DIVISION 87 EXERCISE 43 1. Divide each of the following numbers by -t-3 : -hl2; +15; +27; -18; -36; -42; -57. 2. Divide each of the following numbers by + 2 : _18; +48; +72; -24; -96; +54; -108. 3. Divide each of the numbers in Example 2 by : (a) -3; (6) +6; (c) -2; (d) -12. 4. Divide each of the numbers in Example 1 by — 3. 5. Divide each of the numbers in Example 1 by — 2. 68. The Law of Exponents for Division. Development. 1. Review the definitions of exponent, base, and power of a number in § 15. 2. Divide a^ by a^. Ill Solution : \ = ^-^ ^^— ^ = 1 . a • a = a^. a° jar.jgr.iir 1 1 1 Therefore a^ -^ a^ = a^. Check : a^ . a^ = a^. Each a in the denominator is divided into one of the a's in the numerator. The quotient in each ease is 1, since a -^ a = 1. 3. Find as in step 2 the following quotients and write the results as in part a : (a) a^-7-a^ = a^. (c) m^ h- m^ = ? 4. Examine carefully the exponents in the dividend, the divisor, and the quotient. In the following problems, try to give the results immediately without going through the solution as in step 2. Test by multiplication. (a) p'^p'=? (c) b'^-^b^=? (6) a«-=-a' = ? (d) c^^c^=? 88 ALGEBRA 5. Divide a'^b^ by a^b^. Ill 11 SoLUTiox: a^b^^a^b^= ^^ = ^'^'^' <^ ' ^ '^ ■ f> ■ b - b • b ^^^, 11111 Rule. — The exponent of any number in the quotient is equal to its exponent in the dividend minus its exponent in the divisor. Historical Note. — This rule was known to Stifel (see note § 53). DIVISION OF MONOMIALS BY MONOMIALS 69. Example 1. Divide - 14 a^b^ by + 7 aK Solution : Use the Law of Signs, § 67, and the Law of Exponents, ~ ^^ ^^^^ Z3 - 2 a(3-2)6--2 = _ 2 a62. + 7 a^ Check: {+ 1 a^) {- 2 ab'^) = - U a^b'^ . Example 2. Divide 54 a^b-& by - 9 a'bh\ Solution : ^^ ^'^^^^ =-Q a(7-4)5(2-2)c(3-2) = _ e a%^c = - 6 aH. - 9 a*&^c2 Check : (- 6 aH) x (- 9 a^b'^c'^) = + 54 d'bH^. Notice that by the law of exponents, 6^ _j_ ^2 _ ^2-2 _ ^o^ No meaning has been given to the zero poAver of a number. Since b^ -^ W must equal 1, we agree that W = 1. Hie zero power of any number is 1. Thus : aO = l; 5« = 1; c' = l. Rule. — To divide a monomial by a monomial : 1. Make the quotient positive, if the dividend and divisor have like signs ; make it negative, if they have unlike signs. 2. Find the quotient of the absolute values of the numerical co- efficients. 3. Multiply the quotient of step 2 by the product of the literal factors, giving each its exponent in the dividend minus its exponent in the divisor. 4. Omit any literal factor which has the same exponent in the dividend and divisor. DIVISION 89 Example. Divide - 33 a%x^y^ by + 3 aVy. Solution. ( - 33 af^bx^y*) -4- ( + 3 a^x^y ) = - 1 1 aby^. Ans. Check : These solutions may be checked by substitution, for they must be correct for all values of the literal numbers (except sometimes) . A better way is to use the rule that the divisor times the quotient equals the dividend. Here, does ( + 3 a^x^y) x ( - 1 1 aby^) = - 33 a^x^y* ? Yes. EXERCISE 44 Divide : 1. oc^hja^. 21. - 24 a*b^c by - 8 a*b\ 2. r'hyr*. 22. 2S a^y^ by -7 a^y. 3. y by J3». 23. - 33 a^x'y* by - 3 ay. 4. m* by m\ 24. 65 oi^y^^i^ by — 13 xy^. 5. a^b^ by ab. 25. 28 a^^^c® by - 14 a^bcl 6. 7^9^ by 7^8. 26. - 72 a^y by - 6 ^y. 7. x^y* by a^. 27. - 40 a^6V by - 8 6c. 8. aW' by a^6. 28. - 55 ar'/a;^ by - 11 y-z^. 9. 12 a* by 2 a». 29. - 70 aJb^c' by 14 od^c. 10. 15 a^ by 3 a^. 30. - 96 m^n* by - 12 mw*. 11. 20 ?V by 4 r2.^ — i I B D| Hint : A is at C when B starts ; B overtakes A at/>. 15. Suppose A, traveling 15 miles per hour, starts 4 hours before B. At what rate must B travel to overtake A in 10 hours ? 16. Two hours after A left, B starts after him in an auto- mobile at the rate of 27 miles an hour and overtakes him in 2i hours. At what rate was A traveling ? 17. A and B travel toward each other from points separated by 250 miles, A at a rate wljich exceeds B's rate by 8 miles an hour. If they meet in 5 hours, at what rate did each travel ? 18. Some boys who are boating on a river know that they can go with the current 6 miles per hour and can return against the current at the rate of 3 miles per hour. How far may they go if they have only 3 hours for the trip ? 19. A man has 11 hours at his disposal. How far may he go in a buggy at the rate of 10 miles an hour if he plans to return at an average rate of 7 miles per hour ? 20. An automobile is traveling at the rate of 25 miles an hour. In how many hours will a second automobile overtake the first if the second starts 2 hours later than the first, and travels at the rate of 35 miles an hour ? SIMPLE EQUATIONS 107 21. An express train whose rate is 36 utiles an hour starts 54 minutes after a slow train and overtakes it in 1 hour and 48 minutes. What is the rate of the slow train? 22. An automobile party is traveling at the rate of 20 miles per hour. At what rate must a second automobile travel in order to overtake the first if it starts 2 hours after the first and wishes to overtake it in 3 hours ? 23. Chicago, and Madison, Wisconsin are about 140 miles apart. Suppose that a train starts from each city toward the other, one at the rate of 35 miles per hour and the other at the rate of 40 miles per hour. How soon will they meet? 84. Problems about Thermometers. There are two kinds of thermometers in common use, the Fahrenheit and the Centigrade. The Fahrenheit is the, one with which most of us are familiar. The Centigrade is used by scientists throughout the world. It is necessary at times to change a temperature reading on one scale to the corresponding reading on the other scale. The temperature at which water boils and that at which it freezes are called the "boiling" and ioo° the "freezing" points. On the Fahrenheit scale these points are marked 212° and 32° ; on the Centi- grade scale, 100° and 0° respectively. The num- ber of Fahrenheit degrees between these two points is 180, and the number of Centigrade is 100. Hence, t 100 Centigrade degrees correspond to 180 Fahren- S heit degrees, or 1 Centigrade degree to f Fahrenheit degree. 17.8 Note. This does not mean that a temperature of 1° C. is the same as f° F. 1° C. is one degree above 0; the corresponding Fahrenheit reading is §° above 32 (the freez- ing point), or 33|^. Thus, a temperature of 1° C. = a tem- perature of SSI*^ F. f\ ^ II 108 ALGEBRA EXERCISE 50 1. How many Fahrenheit degrees are equal to the following number of. Centigrade degrees? (a) 15; (b) 25; (c) 50; (ri) 100. 2. Remembering that Centigrade degrees above freezing are counted from zero, and Fahrenheit from 32, what Fahrenheit temperature corresponds to the following Centigrade tempera- ture? (a) +15°C. Solution : 1. 15 Centigrade degrees = 27 Fahrenheit degrees. 2. 15° C. above freezing = 27° F. above 32 = 59° F. .'. 15° C. corresponds to 59° F. (b) +30°C.; (c) +55'^C.; (d) - 10° C. 3. Derive a formula for changing Centigrade temperature readings into Fahrenheit readings. Solution : 1. Let C° = the Centigrade reading. Let F° = the Fahrenheit reading. 2. C Centigrade degrees =(f C) Fahrenheit degrees. 3. C° counted from 0, the Centigrade freezing point, =(§ C)° counted from 32, the Fahrenheit freezing point. .-. i^ = 32 + f C. Check : Let C = 0. .-. i^ = 32 + § • = 32. Let C = 100. .-. if = 32 + 1 . 100 = 32 + 180 = 212. Since the freezing and boiling temperatures correspond, the solution is correct. 4. The formula can be used to change Fahrenheit into Centi- grade readings. Change — 13° F. to Centigrade. Solution : 1. - 13 = 32 + | C. (Substituting in the formula.) 2. .-. - 65 = 160 + 9 C. 3. .-.- 225 = 9(7, or C=- 25°; i.e. 25° below zero Centigrade. SIMPLE EQUATIONS 109 6. In Physics and Chemistry, the temperature — 273° C. is important. To what Fahrenheit temperature does this cor- respond ? (Substitute in the formula.) 6. The following substances melt at the temperatures indi- cated. To what Fahrenheit temperatures do these correspond ? ParaflBn + 55° C. Iron + 1200° C. Tin + 232° C. Mercury - 39° C. 7. Attempts have been made to get record-breaking low temperatures. The following table gives low temperatures produced, the name of the experimenter, and the date of the experiment. To what Fahrenheit temperatures do these correspond ? • Experimenter Fahrenheit Faraday Dewar Onnes 8. The temperatures at three places in the United States on a certain day were : (a) -foO°F.; (b) -f- 12° F. ; (c) - 8° F. What would these temperatures be on a Centigrade scale? 9. The following liquids boil at the temperatures indicated : Alcohol 172.4° F. Turpentine 320° F. Give the boiling temperatures on the Centigrade scale. 10. Air can be liquefied by reducing its temperature until it reaches — 182° C. To what Fahrenheit temperature does this correspond ? Datx Temperatttbe 1714 -17°C. 1823 - 102° C. 1898 -262°C. - 269° C. 1908 VIII. SPECIAL PRODUCTS AND FACTORING 85. In arithmetic, it is found necessary to memorize the multiplication table as an aid in multiplication, division, and factoring. In algebra, also, certain forms of number expres- sions occup frequently, which must be multiplied, divided, or factored by inspection. 86. To Factor an algebraic expression is to find two or more expressions which will produce the given expression when they are multiplied together. Review the definitions of factor (§ 28) and common factor (§ 11)- * 87. A number which has no factors except itself and unity is called a Prime Number ; as, 3, a, and x-\-y. A monomial is expressed in items of its prime factors thus : 12 a^hh = 2 '2'3'a'a'a'b'b'C. 88. Squaring a Monomial. Development. 1. What does x^ mean ? (xyY? (2r^sy? 2. Find each of the following squares by multiplication : (a) (2xyy; (b) (Sa^by-, (c) (- 2 .^s^. 3. Compare the exponent of each letter of the square with the exponent of that letter in the given monomial. Rule. — To square a monomial : Square its numerical coefficient, and multiply the result by each of the literal factors of the monomial, giving each letter twice its original exponent. 110 SPECIAL PRODUCTS AND FACTORING 111 89. Cubing a Monomial. Development. 1. Find each of the following cubes by multiplication : (a) {2x^yf; (b) (3 7-V)^ (c) (-2^fy. 2. Compare the exponent of each letter of the cube with the exponent of that letter in the given monomial. Rule. — To cube a monomial : Cube its numerical coefficient, making the result negative if the given monomial is negative, and multiply the result by the literal factors of the monomial, giving each letter three times its original exponent. Example 1. Find (-5 xYf. Solution : ( — 5 x^y^)^ =4-25 rr^j/^. Example 2. Find ( - 5 x^ff. Solution : ( - 5 x^)^ = - 125 ofiy^. EXERCISE 51 1. What sign does the square of any number have ? 2. What sign does the cube of a negative number have ? 3. Learn thoroughly the squares of the numbers from 1 to 20. 4. Learn thoroughly the cubes of the numbers from 1 to 6. Give the values of the fol lowing indicated powers : 5. (a^bf. 15. (-IxYf. 25. (-18)1 6. (-a'b'y. 16. (-{-Sa'bJ. 26. (-i6ty. 7. (2a^f)'. 17. (-5mhiy. 27. (-\-Gmny. 8. (-{-abcf. 18. (- 9 r^sty-. 28. a«)^. 9. (+ 2 a'bf. 19. (+4(r>d2)l 29. (imnf. 10. (-^Sxyzf- 20. (11 ab'cy. 30. (-in^f. 11. (-5 my. 21. (- 12 m'ny. 31. (+fa^6)l 12. {-2x^y)\ 22. (-\-5m*ny. 32. (-icdy- 13. (-3x*yy. 23. (-15v*zy. 33. (-i^yy- 14. i-e^fy. 24. {- 10 c'dy. 34. (-i^sy. 112 ALGEBRA 90. The Square Root of a Monomial. If an expression can be resolved into two equal factors, it is said to be a Perfect Square, and one of the factors is said to be its Square Root. Thus, 4 a%^ is equal to 2ab^ x2 ab^ ; hence it is a perfect square and 2 ab^ is its square root. Note. 4 a^b^ is also equal to (— 2 ab^) x (— 2 ab^) , so that — 2 ab^ is also a square root. In this chapter, only the positive square root will be considered. The following questions lead to the rule for extracting the square root of a perfect square monomial. Development. 1. What sign does the square of any mo- nomial have ? 2. When squaring a monomial, what do you do with the ex- ponents of the literal factors ? with the coefficient ? 3. In finding the square root, then, what should you do with the exponents of the literal factors ? with the coefficient ? 4. Find the square root of each of the following monomials, and test the result by multiplication : (a) a^; (b) Ax'y^; (c)16r2s^; (d) 25 a^yh\ Rule. --1. A perfect square monomial is positive, has a perfect square numerical coefficient, and only even numbers as exponents. 2. To find its square root : find the square root of its numerical coefficient, and multiply the result by the literal factors of the mo- nomial, giving each letter one half of its original exponent. The symbol for extracting the square root is the Radical Sign, V ; the vinculum is usually combined with it, V . Example. Find the square root of 25 m%®. Solution : \/25 m^n^ = 5 mhi^. 91. The Cube Root of a Monomial. If an expression can be resolved into three equal factors, it is said to be a Perfect Cube, and one of the factors is said to be its Cube Root. Thus, since 27 a%^ is equal to 3 a^ft • 3 a^b • 3 a^ft, it is a perfect cube, and 3 a^b is its cube root. SPECIAL PRODUCTS AND FACTORING 113 The following questions lead to the rule for extracting the cube root of a perfect cube monomial. Development. 1. What sign does the cube of a positive number have ? of a negative number ? 2. When cubing a monomial, what do you do with the ex- ponents of the literal factors ? with the coefficient ? 3. In finding the cube root, then, what should you do with the exponents of the literal factors of the monomial? with the coefficient ? 4. Find the cube root of each of the following monomials and test the result by multiplication : (a) a^ (b) a^«; (c) f; (d) mV; (e) 8 a' ; (/) 27 m« ; (g) - c« ; (h) - 64 ^f. Rule. — 1. A perfect cube monomial has a perfect cube numerical coefficient, whose sign may be + or -, and its literal factors have exponents which are exactly divisible by 3. 2. To find its cube root : find the cube root of the numerical co- efficient, making it positive or negative, according as the sign of the monomial is + or - ; and multiply the result by the literal factors of the monomial, giving each letter one third of its original exponent. The symbol for extracting the cube root is the radical sign with the Index 3, as follows : -y/ . Example. Find the cube root of — 125 aW. Solution : V - 125 a%^ = - 5 a^b^. EXERCISE 52 Find the indicated roots : 1 V4m^ 6. -^27mV. 11. ■\/-64:a^y\ 2. V9m«. 7. V36a2fe«. 12. V225 a^y . 3. V25 d'b\ 8. -^-Sa'b^ 13. V2567¥. 4. Va«6^ 9. V-125m«ni2. 14. V+216c3dV. 5. V8a36«. 10. V169mV. 15. V-27r3p. 114 ALGEBRA 16. ^ + V 14. r — 3s. 21. 2xy-{-9. SPECIAL PRODUCTS AND FACTORING 119 22. 3a^-6b. 25. 9 a^ + o r\ 28. 2xy-9z\ 23. 10 r -f 4 t-. 26. 7 — 2 al 29. 7 a6 — 5 cd. 24. lls-5<. 27. 8c + 3mV. 30. 9a*-h()b^ Note. For additional drill problems, if desired, square a binomial like 3 a: + 6, making b successively 1, -, 3, etc. up to 10. Then change 3 to 4 or any other number. Short daily drills of this sort afford good mental arithmetic. Expand the following : 31. (m-if. 34. (x + iy. 37. (r + i/. 32. (y-if. 35. (n + if. 38. (s -\- ^ tf. 33. (2-h|)^ 36. (p-iy. 39. (x-iyy. 40.~ Square 29 mentally. Solution : 292 = (30 - 1)2 = 900 - 60 + 1 = 901 - 60 = 841 . This should be done mentally. 41. Square 32. (Hint : 32 = 30 + 2). Square mentally the following numbers : 42. 21. 46. 32. 50. 29. 54. 52. 43. 22. 47. 33. 51. 28. 55. 43. 44. 23. 48. 19. 52. 39. 56. 57. 45. 31. 49. 18. 53. 38. 57. Problem. Find a rule for squaring any number ending in 5. Solution : 1. 35 = 3 x 10 + 5 ; 45 = 4 x 10 + 6 ; 56 = 5 x 10 + 5. 2. Similarly, any number ending in 5 may be represented by 10 n + 5. Thus, for 95, n is 9, since 9 x 10 + 5 = 95. 3. (10 n -\- 5)2 = 100 w2 4- 100 w + 25 = 100n(n+ l) + 26, or, n ■ (n + 1) hundreds + 25. Thus the square of 95, in which n = 9, is . 9 . (9 + 1) hundreds + 25, or 9025. 120 ALGEBRA Rule. — To square a number ending in 5, drop the 5, multiply the balance of the number by the consecutive integer, and affix 25 to the result. Example. 85^ = 7225. I.e. 8 X 9 = 72 ; affixing 25, the result is 7225. 58. Find by this rule the squares of some numbers ending in 5, such as 35, 45, 105, 115, etc. 96.. Factoring Perfect Square Trinomials. In algebra, it is necessary to be able to recognize a perfect square trinomial. Development. 1. Square the following binomials, and write the result as in part (a) r (a) (a + 6)2 = a^ + 2 a6 + h\ (c) (3 o^ + 4 yf = ? (b) (2 a + 3 6)2 = ? (d) (4 m - 5 n)' = ? 2. How many terms are there always in the square ? 3. What sign does the first term of the square have ? the third term ? 4. Notice that the first and third terms are perfect squares and that the second term is twice the product of the square roots of these two terms. 5. Are the following perfect squares ? Give the reason for your opinion. Give the square roots of the perfect squares : (a) c'-{-2cd-\-d\ (c) r^-6r + 9. (6) m^ + 2 7nn -\- n\ (d) a^ _ 10 a; - 25. Rule. — 1. A trinomial is a perfect square when two of its terms are perfect squares and positive, and when the remaining term is twice the product of the square roots of the perfect square terms. 2. To find the square root of a perfect square trinomial : extract the square roots of the two perfect square terms, and connect them by the sign of the remaining term. Example 1. Is 4 ar^ + 9 ^/'^ — 1 2 a;/ a perfect square ? Solution : 4 cc^ is a perfect square ; its square root is 2 x. 9y^ is a perfect square ; its square root is 3 y^. 12x2/2 = 2(2a;)(3j/2). SPECIAL PRODUCTS AND FACTORING 121 Hence 4 x^ + 9 y^ — 12 xxf- is a perfect square, and its square root is 2 X - 3 y2. Example 2. Is 9 a^ + 7 a6 + 4 ^- a perfect square ? Solution : 9 a^ and 4 6^ are perfect squares. Their square roots are 3 a and 2 6, respectively. 2(3 a) (2 6) = 12 ah. Since the third term is 7 aft, and not 12 a6, 9 a2 _). 7 (jft _(. 4 52 ig not a perfect square. To be a perfect square, the term 7 ah would need to be changed to -|- or — 12 ah. Then V9 a2 i 12 aF+TP = 3 a ± 2 6. EXERCISE 56 Supply the missing term so as to make perfect square tri- nomials of the following expressions, and then give the square roots : 1. ar + (?) 4- in^' H. >«'' + 4 m + (?)• " 2. a2 - (?) ^i\ 12. a^ - 6 a; 4- (?)• 3. m^ - (?) H- n2. 13. f - 12 ?/ + (?)• 4. ^6 _ (?) ^_ ^2. 14. z'^ - 10 2^ + (?). 5. a;< _ (?) 4. 9 2/2. 15. 9a2 4-6a + (?). 6. 9 ar^ + (?) + 4 /. 16. 144 A^ - (?) -f 25. 7. If) r< - (?) + 25 «2. 17. 9 62 - (?) + 36 c}. 8. 100 m2 -f (?) + 4 n". 18. 25 x'^ + (?) -f 36 f. 9. 25 c-'- (?) + 9(^2. 19. 49 r^ - (?) + 25 s2. 10. 81 c« - (?) -f 25 d\ 20. 9 a;2 _ (?) ^_ 64. In the following, determine whether the trinomials are per- fect squares ; find the factors when possible. 21. 4m2- 20 7mr 4- 25 ?i^ Solution : 1. This is a perfect square, according to the rule. (§ 96, Rule 1.) 2. Hence, 4 m^ - 20 mn'^ + 25 n* = (2 m - 6 it^)^ = (2m- 5n2)(2m-6n2). 122 ALGEBRA 22. m^ — 10 mn + 25 n\ 34. 169 m^ - 26 m^n -f- w^. 23. .T^ + 12 ^2^ + 36 /. 35. QA a'b- -{- U abed -\- c'd^. 24. mV 4- 18 maj -f 81. 36. 100 x^ - 80 a:^ + 16. 25. 64 a^ + 15 ab + 61 37. 49 ^4 _|_ 112 m-' + 64. 26. 100x'-60xhj^9i/. 38. 9 a- + 42 «6 + 49 52. 27. 49 a;-/- 70 aj?/;^ + 25 22. 39. 121 a^ft-^ + 130 a6c + 36 c^. 28. 4 a^ - 22 ax + 25 a^. 40. 64 a' + 176 «& + 121 b\ 29. 81 X- + 16 ?/ — 72 xy. 41. a^^ + a.- + J. 30. 4a2-28aa.' + 49a^. 42. y^-^-^y + i- 31. 25 x^ -I- 16 2/2 — 40 a:?/. 43. z^ + /^ «^^ — I ^w- 32. 9 mV+ 25 r^- 30 7)i?i?'2 44 ^ ^^,2 ^ 25 ^i^ _ i_o rnn. 33 4 ^4 _^ 36 a;2 - 12 a-^2 45 _4^ ^2 _^ x2_ ^y _^ g y2^ 97. Complete Factoring. When a number is to be factored, all of its prime factors should be found. The factors found first may sometimes be factored again. Thus, 48 = 8.6 = 4.2-2.3=:2.22.2.3. In algebra, this sort of factoring is frequently necessary. Example 1. Find the prime factors of 5 0771^ — 50 amn + 125 aiv^. Solution : 5 am^ — 50 amn + 125 a7i^ = 5a(m2- 10m/i + 25^2) (§94) = 5 a (m - 5 n) (m - 5 n). (§ 96) Do not fail to rewrite all factors, like the 5 a, which are not factored again. Rule. — To find the prime factors of an expression : 1. First remove any monomial factor which may be present. 2. Then factor the resulting expression, when possible, rewriting all expressions which cannot be factored. SPECIAL PRODUCTS AND FACTORING 123 EXERCISE 57 Find all of the prime factors : 1. 3mx'-^24:7nx-\-ASm. 6. 20 a^ - 20 a^bd" -t 5 b'c*. 2. lSt~12at + 2aH. 7. 30 a^b - 120 ab-^ 120b. 3. 7 m-n^i- 70 mn'-{-175n\ 8. 3aa^ -Saxy -\-3ay'. 4. 5 x^yh-\- 70 xyz-^ 215 z. 9. 75cx' + IS cy^ -120 cxy. 6. 11 mx'y^-^ 22 7nxy-j-U 711. 10. 5 7^s - 10 rst -h 20 st\ CASE III Type Form : (a -{-b){a — b)=a^- Ir. 98. The Product of the Sum and the Difference of Two Numbers. Development. 1. Find by multiplication the following products, and write the results as in part (a) : (a) (a;-f-3)(.r-3) = a^-9. (c) (A; + 10) (A; - 10) = ? (6) (m + 7)(m-7)=? {d) (r + 9)(r-9) = ? 2. Observe the results in step 1 ; try to find the following products mentally. Check the results by multiplication. (a) (a + 6)(a-6) = ? (c) (d + 4)(d-4) = ? (6) (c-h8)(c-8) = ? id) (2/ + 5)(y-5) = ? 3. Write the sum of x and y ; write the^ir difference. Find the product of the results. 4. Prove by multiplication the following fact : (^a + h){a-b)=a'-b\ Rule. — To find the product of the sum and the diiference of two numbers : 1. Square each of the numbers. 2. Subtract the second square from the first. Example 1. Find (5 a^ -\- m) (5 a^ — m). Solution : (6 a^ + in) (5 a^ - m) = (6 a^)^ - {my = 25 a* - wi^. Example 2. Find mentally the product of 24 and 16. Solution : 24 x 16 = (20 + 4) (20 - 4) = 400 - 16 = 384. 124 ALGEBRA EXERCISE 58 Find by inspection the products : 1. (a + 2) (a -2). 11. (10 0^2/ -11) (10 0^2/ + 11). 2. (r-3)(r + 3). 12. (13m2-12)(13m2 + 12). 3. (s^ + 4)(s2-4). 13. (a'-h')(a'-\-b'). 4. (x^-^5y)(x^-5y). 14. (^~f)(^-^f). 5. (3 m + 4 7i) (3 7n — 4 w). 15. (aj«-8/)(a^ + 8/). 6. (m^-l)(m^ + l). 16. (7a:/^-10)(7a;?/2^ + 10). 7. (z-3d)(z-{-3a). 17. (|^-*)(l^ + i)- 8. (Sk'-9t){SJc' + 9t). 18. (im + |)(|m-|). 9. (Sab-7c){3ab-\-7c). 19. (I^^-A)(^'^+A). 10. (4r-5s2)(4r + 5s2). 20. an^-i)(in' + i). 21. (9 + 5) (9 -5). 25. 32 28. 29. 55-65. 22. (25 + 2) (25 -2). 26. 53 47. 30. 33-37. 23. 22 . 18. 27- 62 58. 31. 41-49. 24. 23 . 17. 28. 98 .102. 32. 22-28. 33. Find the cost of 18 dozen of eggs at 22^ per dozen. 34. Find the cost of 16 yards of gingham at 24^ per yard. 35. Find the cost of 45 yards of scrim at 55 j^ per yard. 99. Factoring the Difference of Two Perfect Squares. Development. 1. What is the product of (a + 2) and (a — 2) ? What, then, are the factors of a^ — 4? 2. Find the factors of : (a) a2-9; (c) k^-P; (b) m'-W; {d) 9r2-4s2. 3. Write the square of r; of t; the difference of these squares. Factor the result. 4. Similarly, the factors a^ — b^ are : (a'-b') = (a + b){a-by SPECIAL PRODUCTS AND FACTORING 125 Rule. — To factor the difference of two squares : 1. Find the square root of the two perfect square terms. 2. One factor is the sum of the results ; the other factor is the difference of the results. Example 1. Find the factors of 25 r* - 16 tf^. Solution : 25 r* - 16 <« = (5 r^)* - (4 t^y^ = (5r2 4-4«8)(5r-2_4«8). Example 2. Find mentally the value of 13^ — 71 Solution ; 132 - 72 = (13 + 7) (13 - 7) = 20 X 6 = 120. This example shows how the above rule can be used to simplify arith- metical work. EXERCISE 59 Factor the following token possible : 1. a* - b\ 12. 36 - 49 y'. 23. 100a262c2-l. 2. (f-9. 13. 1 - 36 a262. 24. c*-|dl 3. 16 -dl 14. o^-i. 25. 81a;i0-196y«r 4. a^-1. 15. 100 2V -49. 26. 256n^-m2. 5. l-y\ 16. /'-im2. 27. 225-64^2. ^ 6. x^-4.y\ 17. ia^-i^<^- 28. 85a^-y2 7. 8. 9 -ml 18. 19. 64m^-81n«. 169 a2- 196. 29- -^-i- a^ d^ 9. 16^^-25/. 20. 25 a? - 1. 30. ?5-^. 10. 81r2-16 2^ 21. 144 x^ - 121 2/2. r2 36 11. 25a«-8l2«. 22. ^^>^-A- Find mentally the following : 31. 162-91 36. 352 -152. 32. 232-72. 37. 272 - 132. 33. 242 - 62. 38. 262 -42. 34. 332-172. 39. 952 -52. 35. 242-162. 40. 752 -252. 126 ALGEBRA Find the prime factors of the following : 41. 5 x^ - 5 2/2. (See § 94.) 46. 36 xv^ - xiA 42. mH — 2ot. 47. ttR- — tts\ 43. Zai"" -12 as\ 48. \ irdr - \ -rrm'. 44. 4c-c?— 9d^ 49. m^ — n\ 45. 32?/ — 2 05^. 50. o^-y^. Find the following quotients : 51. ix'-y'-)-^{x-y). 55. (4 s^ - ^^^ -- (2 s - ^). 52. {a''-w})-^{a + m). 56. (9x-2-16 ?/2)--(3x-|-4 ?/). 53. (r2-9)-(r-3). 57. (2om'-Un'')^{^m + 4.n). 54. (^2 _ 25) - (? + 5). 58. (169 - 100 a^)- (13 -10 a). Tell what binomial will divide each of the following ; give the quotient : 59. 9cc2-4 2/*. 62. 144/ -121. 60. m^- 16 7*2. 63. 256 c* -400. ea. 25ci«-36 6^ 64. 100r2-36i^ EXERCISE 60 Review Expand the following : 1. (x'-y-y. 4. 2pg(^''-s')(r^ + ^). 2. 2a(x-\-yXx-y). 5. 3c(2a-6)l 3. 3m{c(^-y%a^-\-y^. 6. 5ab{a^-by. Factor completely the following : 7. r*-2rV + s^ 10. 3 ax^ - 6 axY + S ay*. 8. mC* — 2 mc^ 4- m. 11.5 mx- — 10 mxy -\- 5 my^. 9. 5tx^-5ty\ 12. 2 r^m* - 4 ?^m V -f 2 ?''V. SPECIAL PRODUCTS AND FACTORING 127 CASE IV Type Form : (x + a){x ^ h) = x- -{- (a -\- h)x + ab. 100. The Product of Two Binomials having a Common Term. Development. 1. Find by actual multiplication the fol- lowing products, and write the results as in part (a) : (a) (a;-|-2)(a; + 3)=ir2 4- 5x-f 6. (6) (x 4- 5)(a: -f 3) = ? {d) (m - 7)(m - 2) = ? (c) (a-\-b){a-^Q>) = ? (e) (s-5)(s-8)=? 2. Observe carefully the results in 1. Try to find the following products mentally; check the results by multipli- cation : (a) (6 + 4)(6 + 2). (c) (x-3)(a;-7). (6) (cH-4)(c-f3). (cZ) (y-4)(2/-5). Rule. — To obtain the product of two binomials having a common term: 1. Square the common term. 2. Multiply the common term by the algebraic sum of the second terms of the binomials. 3. Find the product of the second terms. 4. Add the results. Example 1. Find the product of a; — 8 and a; -|- 5. Solution: (x - 8)(x + 5) = a;2+(- 8 + 6)ic+(-8)(+ 5) = a;2 - 3 a; - 40. Example 2. Find {ah + 2){ab - 11). Solution : {ah + 2) {ah - 11) = a^"^ - 9 a6 - 22. Note. Here, one glances at + 2 and — 11, notes that their sum is — 9 and that their product is — 22, and writes the result as above. Example 3. Find mentally the product of 23 and 24. Solution : 1. 23 x 24 = (20 + 3) (20 + 4) = 400 + 7 • 20 + 12 = 400 + 140 + 12 = 662. 58 ALGEBRA EXERCISE 61 Find the following products : 1. (a; + 2)(a^ + 3). 26. (rs-St)(rs — St). 2. (x-\-2)(x + A). 27. (a + 3&)(a + 136). 3. (a + 3) (a + 5). 28. (c - 4. d)(c -12 d). 4. (a+4)(a + 6). 29. (x^-Sy){x^-10y). 5. (m + 5)(m + 9). 30. (x'-^5f)(a^-2f). 6. (b-S)(b-7). 31. (p^ + 11 q)(p^^ 15 q) 7. (5_4)(&-8). 32. (r«-12)(r^ + 20). 8. (c_5)(c-7). 33. (s3-15)(s3 + 8). 9. (,,_6)(r-8). 34. (t^ - 16 W)(f- 4: w). 10. (r-9)(r-10). 35. (a^-lSy)(x^-^5y), 11. (r + 2)(r-8). 36. (a-\-lTt')(a-3t'). 12. (m + 3)(m-10). 37. (x-9a){x + 15a). 13. (n + 5)(n- 11), 38. (x-hllb)(x~Sb). 14. {s + 6)(s- 12), 39. (f - 13 z)(y' -7 z). 15. (t + m-11). 40. (t -i- 15 r)(t- 10 r). 16. (x-2){x-^10). 41. (a'-Ub)(a'-10b) 17. (y + ll){y-3). 42. (c-22x)(c + 20x). 18. {z^l0){z-5). 43. (x + 19y)(x-5y). 19. (w-4:)(w + 9). 44. (a -33 b)(a + 6b). 20. («-l)(a+8). 45. (c' + 25 d)(c'- 10 d). 21. (s2_6)(s2 + 10). 46. {x-35)(x + 20). 22. (^_12)(^ + 9). 47. (m-i)(m-5). 23. (xy-10){xy + 15). 48. (.+!)(. + 1). 24. (r-5s){r+Ss). 49. (2/ + i)(2/ + 3). 25. (x-2y)(x + Sy). 50. (^ + i)(^ + i). 51. Find mentally the product of 62 x 68. Solution : 62 x 68 = (60 + 2) (60 + 8) = 3600 + 600 + 16 = ? SPECIAL PRODUCTS AND FACTORING 129 Find mentally the following arithmetical product^: 62. 22 X 28. 55. 24 x 26. 58. 52 x 54. 53. 33 X 37. 56. 32 x 38. 59. 23 x 25. 64. 34x36. 57. 44x46. 60. 33x34. Perform mentally the multiplications in the following problems : 61. Find the area of a rectangle whose base is 26 inches and altitude is 24 inches. 62. Find the cost of 12 yards of lawn at 18 ^ per yard. 63. Find the cost of 22 yards of scrim at 25 ^ per yard. 64. Find the cost of 53 bushels of corn at 55 ^ per bushel. 65. Find the cost of an 11-pound roast at 18 ^ per pound. 101. Factoring Trinomials of the Form :^ -\-px-\-q. A trinomial of the form y?-\-i)x + q can be factored if it is the product of two binomials having a common term (§ 100). Development. 1. {x -\- 5)(if — 3) = s? -\-2x— 15. In obtaining this product, the algebraic sum of + 5 and — 3 is taken for the coefficient of oc, and the product of + 5 and — 3 is taken for the third term. To factor x^ + 2 x — 15, it is necessary to find two numbers whose product is — 16, and whose algebraic sum is + 2. 2. Factor «2 _|_ 7 ^ ^ 12. Two numbers whose product is + 12, are -j- 3 and + 4, and their sum is + 7. Try as the factors {x + 3) and (x -f- 4). Check : Does (x + 3) (x + 4) = x2 + 7 x + 12 ? Yes. Rule. — To factor a trinomial of the form x- -\- px -\- q: 1. Find two numbers whose algebraic product is g and whose al- gebraic sum is p. 2. One factor is jt + one number ; the other factor is x 4 the other number. 130 ALGEBRA Example 1. Factor x^-26x- 192. Solution : It is necessary to find two numbers whose sum is — 26 and whose product is — 192 ; the number of greater absolute value must be negative. If necessary write all possible pairs of factors of — 192, one of which is + and the other — . We have : (4-l)x(-192); sum = -191. (+4)x(-48); sum =-44. (4-2)x(- 96); sum = - 94. (+ 6) x (- 32); sum = - 26. (+8)x(- 64); sum = - 61. .-. + 6 and — 32 are the numbers required. .-. a:2 - 26 X - 192 = (a: + 6) (x - 32). EXERCISE 62 Factor : 1. or + 14 a; + 45. Solution : 1. Find two numbers whose product is 45 and whose sum is +14. 2. Factors of 45 are, 1 and 45 ; 3 and 15 ; 5 and 9. 3. 5 + 9 = 14, hence the factors are (x + 5)(a; + 9). Check : (x + 5) (x + 9) = x^ + 14 x + 45. Note, This solution should all be done mentally ; decide upon a pair of factors of 45 and immediately determine their sum. 2. x'-^-Bx-^-G, 5. a2 + lla + 28. 3. a^ + 8x + 15. 6. m2 + 9m + 20. 4. ,^ + 10r + 21. 7. f + 9t+lS. 8. y_i2p + 32. Solution : 1. Since 32 is positive, the second terms of the factors must have the same sign ; since the sum of the second terms is — 12, the sfecond terms are negative. 2. (-8)x(-4)= +32; and (-8) + (-4)=-12. Therefore the factors are (p — 8) (p — 4). Check : (j? - 8) (j9 - 4) = p2 _ i2p + 32. 9. aj2-7a?+12. 12. iv^ - 10 w -\- 24. 10. 2/2- 9?/ + 14. 13. a^- 11 a + 30. 11. z'-llz + 24.. 14. c2-12c + 35. SPECIAL PRODUCTS AND FACTORING 131 15. x2_|_6.T_16. Solution : Since — 16 is negative, the second terms of the factors must have unlike signs ; since the sum of the second terms is + 6, the term of greater absolute value must be positive. 2. Factors of - 16 of this sort are : (+16, - 1) and (+8, — 2). 3. (_|_ 8) 4- ( - 2) = + 6, therefore the factors are (x + 8) and (x — 2). Check: (x + 8)(x - 2) = x2 + 6 x - 16. (By § 100.) 16. a^ + 3a;-40. 19. z--\-4:Z-32. 17. a:^ -}- 2 a; - 24. 20. w^-\-4:W-60. 18. y--\-4:y'-21. 21. a^ + Sa-Sl. 22. ?7i2 - 4 m — 21. SoLDTiON : 1. The factors of — 21 must have unlike signs ; the one of greater absolute value is negative, since the sum is — 4. 2. Such factors of - 21 are : - 21 and + 1 ; - 7 and + 3, 3. ( — 7) + ( -t- 3) = — 4 ; therefore the factors are (m — 7) and (m + 3). Check : (m - 7) (m + 3) = m^ - 4 ?n - 21. (See § 100.) 23. x^-2x- 35. 36. a^-13xw-\-22 w\ 24. n^-7n- 18. " 37. m^ - 7 m - 44. 25. a^-8a-33. 38. r^ + 2 r^s - 48 s^. 26. 7^ - 4 r - 45. 39. s^ - 6 st - 55 t'\ 27. m* - 11 m2 4- 30. 40. w^ - 18 tv^ + 72. 28. a^ - 15 a + 54. 41. / -f 4 ?/ - 96. 29. r^-\-r-2. 42. a^-70—3a. (Rearrange it.) 30. X* + 6 3^-^ 8. 43. 6^ - 24 - 10 b. 31. c^ -^ 15 cd-h 36 d\ 44. c^ + 20 c -f- 84. 32. x^ -12xy-h 32 y\ 45. z^ - 36 4- 5 z. 33. z^ -{-2z~ 63. 46. a' + 25 a^ - 150. 34. c^d'^ -cd-6, 47. - 19 m^ + m^ -f 84. 35. 2/- + 10 7/2 + 9 z\ 48. a^^ -h 50 - 27 x". 132 ALGEBRA 49. t^-st-20s\ ' 55. a^ft^ - 16 a6c + 28 cl 50. a" -\-l ah -m h\ 56. x^ - 21 xyz + 108 yh\ 51. ^2 - 20 aa; + 99 a''*. 57. a" -- 21 ah'' + 110 h\ 52. r^ 4- rs - 72 si 58. c^d^ - 10 cd - 96. 53. x^ + ^Oy'-l^xy\ 59. aV - 26 ac + 160. 54. a2 + 16a + 15. 60. r^ - 7 rs - 120 s^. Find all of the prime factors : 61. 3 ah^ _ 15 a& + 18 a. Solution : ^ ah'^ - lb ah + 1% a - ^ a {h'^ - bh ■\- Q) 62. 5f + S5t-i- 60. 66. ahn^ - 7 abn -}- 6 ah. 63. mr2 + 2 mr - 15 m. 67. 11 a^ - 11 a; - 66. 6^. Sca^ + 6cx-9 c. 68. 8 aj^ + 48 a; - 56. 65. 7 yh ^ 21 yz - 126 z. 69. m^ri^ -\-2m^n- 63 ml CASE V Ti/P^ i^brm .• (ax + 6) (cjf + d). 102. The Product of Two Binomials of the Form (ax + b). Development. 1. Find by actual multiplication the fol- lowing products, and write the results as in part (a) : (a) (2 a! + 3) (3 a; 4- 1) = 6 a^ + 11 a: + 3. (h) (3m + 2)(2m + 3)=? (c) (2y -j-A)(3y -2)=? 2. Examine carefully the results in 1, then try to get the first and third terms of the following products mentally ; pos- sibly you can get the second term also. Check by actual multiplication. (a) (2 r 4- 4)(3 r + 1). (c) (3 s + 5)(2s + 1). (h) (2r + 5)(r + 2). . (d) (4.x + 2y)(2x + ^y).- SPECIAL PRODUCTS AND FACTORING 138 3. Below is given the product of (5 x — 4 y) and (2 x -|- 3 y). 5x — 4:y 2x -\-3y 10 ar*- Sxy -\-15xy-12f 10ar^+ lxy-12f ... (5a;_4^) {2x -h ^y) = 10x^ + {-^ + 15)xy -12^. Rule. — 1. The first term of the product is the product of the first terms of the binomials, {bx 2x = l0x^.) 2. The third term of the product is the product of the second terms of the binomials, {—^y) ( + 3 1/) = - 12 y-. 3. The second term of the product is the algebraic sum of the " cross products." (Notice the position of these cross products on the left of the equation : —^y-2x and b xZy). Example 1. Find the product (5 r — 6 s) (2 r+ 3 s). Solution : In all of these examples, the only difficulty is that of find- ing the " middle term." 1. 5r-38 = 15rs, -6s.2r =- 12rs, and (15rg) + (-12rs) = -|- 3 rs. 2. .-. (5 r - 6 s) (2 r + 3 s) = 10 r2 + 3 rs - 18 s\ Example 2. Find the product (9 a; -f 4 2/) (3 a; — 6 ,y). Solution : 9 • - 6 =- 54, 4 • 3 = 12, and (-54) + (12) =-42. .-. (9x -f 4 y) (3 X - 6 y) = 27 a;2 _ 42 xy - 24 y^. In this example, the coefficient of the middle term was found first and then xy was affixed. In practice all of this should be done mentally as in the following example. Example 3. Find the product (7 m — 4n)(8mH-5n). Solution : (7 ?n - 4 n) (8 w -f 5 w) = 56 m^ + 3 mn - 20 n^. [Middle terra : 7 • 6 = 35, - 4 • 8 = - 32, 35 +(- 32) =3.] Pupils should try to acquire such skill that they can find the correct products mentally. This is the manner in which experts do it. 134 ALGEBRA EXERCISE 63 1. (2x + 2) 26. (llm2-4)(5m2 + 6). 62. (Sm^- 9 M)(4m2 4-5 n>mq SPECIAL PRODUCTS AND FACTORING 135 53. (7ar^ + 3/)(5ar^-2 2r). 57. (lom-2n)(6m -h 5n). 54. (llc2-5cZ)(4c2-5d). 58. (6m + |)(4m + i). 55. (15p'-^2q)(92^-2q). 59. (^ x-^y){9 x- iy). 56. C20a-7b)(Sa-hSb}. 60. (12a + f 6)(10 a-| 6). 103. Factoring Trinomials of the Form mjr + nx + />. The product of two binomials like (2 a; -|- 3) and (3x+5) is a bi- nomial of the form mx^ + nx -|- p. This means that there is a term containing the second power of x, one containing the first power (as a rule), and one free from x. The following discussion shows how to factor trinomials of the form m3i?-\-nx-^p, when they are factorable. Development 1. Find the products : (a) (2a;-f 6)(3a; + 2). (6) (3 a! -5) (4 a: + 7). 2. How do you obtain the " middle term " of the product ? 3. Factor 12 a^ + 23 a; + 5. Solution : 1, The first terms of the binomials might be 2 a- and 6 x, for their product is 12 x'^. Place them in parentheses thus : (2x ) (6 z ). 2. The second ternis of the binomials are both positive since 5 and 23 X are positive. Place the factors -|- 5 and + 1 in the parentheses and note the middle term which results. (a) (2 a; + 6)(6 a; + 1) ; middle term, + 32 a;. Incorrect. (6) (2 X + 1)(6 X + 6) ; middle term, + 16 x. Incorrect. 3. Step 2 shows that the factors 2 x and 6 x for 12 a;^ are incorrect. Try 3 X and 4 X for 12 x2, thus : (3 X )(4x ). (a) (3x -f l)(4x + 6) ; middle term, + 19x. Incorrect. (6) (3 X + 6)(4 X + 1) ; middle term, + 23 x. Correct. Check : (3 x + 5)(4 x + 1) = 12 x^ + 23 x + 5. NoTK. This may seem a long process at first, but practice soon develops such skill that most of the trial of factors can be done mentally. 186 ALGEBRA EXERCISE 64 Factor : 1. Sx^-i-5x + 2. 5. 6a;2 + 7a; + 2. 2. Sa^^7a + 2. 6. 12f-^13ti-S. 3. 5m2 + 7?ii + 2. 7. 7a2 + l0a + 3. 4. 7/ + 9:f/ + 2. 8. 6^- + 13« + 6. 9. 12i^2~17w + 6. Solution : 1. Since 6 is positive, its factors must have like signs ; and, since —17 is negative, the cross products must both be negative. Therefore the factors of 6 must both be negative. 2. To get 12 W52, use {2w )(()W ). To get + 6, try - 2 and - 3. (a) (2 10 — 3) (6 10 — 2) ; middle term, — 22 w. Incorrect. (6) ' (2io — 2)(6w — Z); middle term, — 18 w. Incorrect. 3. To get 12 io2, use (3 wj ) (4 w ). To get + 6, try — 2 and - 3. (a) (3 tc - 2) (4 «? - 3) ; middle term, — 17 to. Correct. Check: Does (3io - 2)(4 wj - 3)= 12 m?2 _ 17 ^^ + G ? Yes. Note. In step 2, the factors 3 and 2 were used in both ways, as in (a) and (&) ; usually this is a wise plan, although in this case an explanation could be given to show that it was unnecessary. 10. 2w^ — llw-\'5. 14. 2w;2_9it' + 4. 11. 2w'~-7iv-{-3. 15. 6m2-ll7/i + 3. 12. Sf/-19b-^6. 16. 8«2_22a + 15. 13. 6m^-7m + 2. 17. 5c'' -23c -\- 12. 1 8. Factor 15 x- 4- 14 a? - 8. Solution: 1. The factors of —8 must have unlike signs. Arrange the signs so that the cross product of greater absolute value is positive. 2. For 15a:2, try (3x )(5x ). For 8, try 2 and 4. (a) (3 X — 2) (5 a- + 4) ; middle term, +2x. Incorrect. The sign of 4 is made +, because 4 • 3 is greater than 2 • 5. (6) (3 X + 4) (5 X - 2) ; middle term, + 14 x. Correct. The sign of 4 is made +, because 4 • 5 is greater than 3 • 2. Check: Does (3x + 4)(5x - 2)= 15x2 + 14x - 8 ? Yes. SPECIAL PRODUCTS AND FACTORING 137 19. Factor 24 m^ -m- 10. SoLDTioN : 1. The factors of —10 must have unlike signs. The cross product of greater absolute value must be negative. 2. For 24 m"^, try (0 7rt )(4 m ). For 10, try 5 and 2. (a) (6 m 4- 2) (4 ?>i — 6) ; middle term, — 22 m. Incorrect. (6) (6 m — 6) (4 »n + 2) ; middle term, — 8 m. Incorrect. 3. For 24 m^, try (3 m ) (8 w ). (a) (3 m — 6) (8 m 4- 2) ; middle term, —34 m. Incorrect. (6) (3 m — 2) (8 m + 5) ; middle term, — m. Correct. Check : Does (3 m- 2) (8 ?» + 5) = 24 m^ _ m - 10 ? Yes. Note. In this last step, for example, after placing 2 and 5 in the parenthe- ses, we see that 2 x 8 is greater than 3x5; therefore make 2 negative and 5, positive. 20. 2a^-3a;-5. 37. 14 mV - 31 mn - 10. 21. 3»i* + 4m-7. 38. 4:i^s^ + irs-15. 22. 5t^-2t'-7. 39. 12 ic^ 4- 17 a; + 6'. 23. 7?'2-|-4r-ll. 40. 24m«-18m-15. 24. 6s2-7s-5. 41. 12 ^^ _^ 13 ^a; - 35 a:^ 25. 1282-f5s-3. 42. 2w'-3wr-20r'. 26. 2x2 + a;-15. 43. 15y' ■\- 19 yz-\- 6 z'. 27. 9r2-6r-8. 44. lOn^ -^9nt-9t\ 28. 15c2-4c-3. 45. 9/r + 3 ftc-SGc^. 29. 21A' + 2A-S. 46. 15a2_26a6 -21 61 30. 5/ + 16y + 3. 47. 7ar*-26ir.y-8?/. 31. 18«2_3^_10. 48. 15a2-|-29a?>4-12 62. 32. 4:x'-24:X + S5. 49. 21 a;^ _ 29 xt/ - 10 y*. 33. 10ar'-13a;-30. 50. G r" - 2rj rs -\- 25 s". 34. 322_|.22z4-7. 51. 4c2-8cd-21d2. 35. 18m2-f-17m-15. 52. 9d*-6d2~35. 36. 6x2 + 31 a; + 35. 63. 8i)« + 18p»-35. 188 ALGEBRA 54. 10 m* -\- 19 m^n — 15 nK 57. 35 df -\- at — 12 a. 55. Uv'-\-29v-15. 58. 24.x^7^-7xy7^-6y^r^. 56. 15k'^x — 16kx — 15x. 59. 55 m V+2 mnar*— 21/1 V. (Remove the monomial first.) 60. 24 C^d^ -\-2cd^ — 15 d^. EXERCISE 65 Perform the various steps of the following indicated multi- plications mentally : 1. 3a(2a + 5)(a-4).. Solution : 3 a(2a + 5) (a - 4) = 3 a(2 a^ - 3 a - 20) = 6 a^ - 9 a2 - 60 o. 2. 2m(3m-l)(2m + 6). 5. 5(2r-5s)l 3. 4.ab(5x-2){5x-{-2). 6. -3(6 a- 5)(2 a + 3). 4. 7(3cH-5d)(3c-4d). 7. -2(7 r-4 s)(2r4-3 s) Solve the following equations, performing mentally all of the steps of the solution : 8. 2(3x-2)(x-^4:)-(6x-^5)(x-S) = 65. Solution : 1. 2(3 x - 2) (x + 4) - (6 x + 5) (x - 3) = 65. 2. 2(3x2 + 10x-8) - (6x2-13x-15) =65. 3. Complete this example by removing parentheses, combining terms, etc. 9. (5 a; - 4) (3 a; -h 4) - 3 (5 aj + 6) (a; - 7) = 10 (9 a; + 15). 10. (4 r - 5)(3 r + 7) - 2 (r + 1)(6 r - 7) = 3 (2 r + 3). Find all of the prime factors in the following ; remember to remove the monomial factor first : 11. 24 m^a H- 18 mna — 15 A. 12. lSbc'-2bx'. 13. 12xHj-Sx'y^-4.x^f. 14. 15 c^d - We'd- 25 cd., 15. IS Ic'r"-- 60 Mr' -{- 501^7^. 16. 4.S cd' + 120 cde + 75 ceK SPECIAL PRODUCTS AND FACTORING 139 17. 39 m-nar -f- 20 mhixy - 4 mhiy\ 18. 3x*-17x'-mx'. 19. ■ixy-^2x^f-7a^y\ 20. 4:5 r*s - SO r^s'. CASE VI Type Form : a^-b^ = (a- b)(a- + a6 + ft^). 104. Factoring the Difference of Two Cubes. Development. 1. Divide a^ — 27 by a; — 3 ; by » -}- 3. 2. Divide x^ — y^hj x — y; hy x-\-y. 3. Write the cube of a; the cube of 6; the difference of these cubes. 4. Examine the results in 1 and 2 ; by what do you think you can divide a^— b^ and get an exact quotient? 5. Prove by division the following fact : Hence, «» - 6« = (a - b)(a' -\-ab-\- b^. Rule. — 1. The difference of the cubes of two numbers may be divided exactly by the difference of the numbers. 2. The quotient is the square of the first number, plus the product of the two numbers, plus the square of the second number. Example 1. Find a divisor of 8 a^ — 27 ; find the quotient. Solution : 1. 8 x8 - 27 = (2 x)8 _ 38. 2. .'. It is the difference of the cubes of 2 a: and 3, and can be divided exactly by (2x -3). 3. .-. The quotient is (2 xy + (2 x)(3) + (3)2, or 4 x2 + 6 :k + 9. Check by substitution or by multiplication. Example 2. Factor 8 a^ — 64 i/^. Solution : 1. 8 x^ - 64 2/« = (2 x^)^ - (4 yY 2. =(2x2-4y){(2x2)2+(2x2)(4y) + (4 2/)2} 3. =(2x2-4«/){4x4 + 8x2?/ + 16 2/2}. NoTK. The middle term is not twice the product of the first and second numbers ; try not to confuse this type with the one in § 96. 140 ALGEBRA EXERCISE 66 Factor : 1. a'-S. 2. 63_(34. 3. c^ — 125. 4. c'-SdK 5. m^ — 64 n^. 6. r«-8. 7. s«-125. 8. a^2_g4^s^ 9. Sb^-a^ 10. 64m«-l. 11. 8a;«-125?/3. 12. 3ar^-24?/«. (Remove monomial first.) 13. AOa'-Bb^ 14. 24aj<'-81?/l 15. 4.a — 500ax^ 16. 320m3-5w«. 17. a^-i- 18. /-^. 19. z^- 125 What is the quotient of 20. (a^-a3)^(a;-a)? 21. (7^-x')-^(r-x^)? 24. (S-x^)-^(2-x')? 25. (mi^-27)-^K-3)? 22. (27a.-«//-l)--(3ir2/-l)? 26. (27 w'-6iv^)-r-(3w'~Av)? 23. (64a«-6«)-(4a2-?>2)? 27. (1 - 125 m«) - (1 - 5 m^)? CASE VII Type Form : (a^ + b') = (a + &) (a' - fl& + &")• 105. Factoring the Sum of Two Cubes. Development. 1. Divide (a^ + 27) by (x + 3) ; hj x — 3. 2. Write the cube of m; of 7i; the sum of these results. Write the sum of m and n. 3. Divide the sum of the cubes of m and n, by the sum of m and n; also by their difference. 4. Prove by division the following fact : (a" + b^) -^ (a -{-b) = a^ - ab + b\ Hence (a^ + 6») = (a + ?>)(a- - tt6 + Z>')- SPECIAL PRODUCTS AND FACTORING 141 Rule 1. — The sum of the cubes of two numbers may be divided exactly by the sum of the two numbers. 2. The quotient is the square of the first number, minus the prod- uct of the two numbers, plus the square of the second number. Example 1. Find an exact divisor of x*^ -}- 8 ; find the quotient. SoLL'TioN : a* -f 8 = (x:^y + (2)8. Since it is the sum of the cubes of two numbers, it may be divided ex- actly by the sum of the two immbers (jc^ -j- 2). (Rule 1.) The quotient is {x^y - (a;2) . (2) + (2)2 or x* - 2 x + 4. (Rule 2.) Note. The middle term is not twice the product of the two nambers. Example 2. Find the prime factors of 3 mx^ + 81 mi^. Solution : 3 wx^ + 81 ?«?/« = 3 m{x^ + 27 f). ' =3m[(x)3+(3y2)3-| = 3 m[x + 8 2/2] [;r2 - (3 y^) • x + (3 y2)a] = 3 w(x 4- 8 y2)(:i--i - 3 xy'^ + 9 y^). EXERCISE 67 Change the sign in each example of Exercise 66 from minus to plus. This will give 27 examples. 106. Summary. In this chapter, seven special forms of algebraic expressions have been considered. These type forms are collected here for reference : I. a{x + y + z-\-. . .) =ax+ay-\-az + . . . . II. {a±bf =a2±2a&+&2^ III. {a-\-h){a-h) =d'-tr. IV. (x + a){x + b) =x'-^(a-{-b)x-^ab. V. {ax-^ b) {ex -\-d) = acxr + {ad -\-bc)x-\- bd. VI. (fl-6)(a-4-a& + 6') = ff'-fr'- VII. (a + b){(^ -ab^b") =a?-\-b\ Familiarity with these forms makes it possible : (a) to perform many multiplications in algebra and arith- metic mentally ; 142 ALGEBRA (6) to perform many divisions mentally ; (c) to factor many algebraic expressions. These are the more important forms. Others are discussed in a later chapter. (XVI.) Rule. — To perform an indicated multiplication of two or more factors by means of the type forms : 1. First find the product of all binomial and poljmomial factors. 2. Then multiply by the monomial factor. Example. Find the product Sx(2 x-\-y)(2x — y)(4:X^ -\-y^. Solution : S x(2x -h y)(2x - y)(4x -{- y^) = 3 X . (16 a;4 _ y^) = 48 x^ -Sxy*. Rule. — To find the prime factors of a given expression : 1. First remove any monomial factor of the expression. 2. Factor the resulting polynomial factor or factors by the proper methods, until all of the prime factors have been found. Example. Factor 3 ax^ — 21 aoif — 24 a. Solution : 3 ax^ — 21 ax^. — 24: a = 3 a(x^ - 7 x3 - 8) = 3a(x3-8)(x3 + l) = 3a(x- 2)(x2 + 2 X + 4)(x -f l)(x2 - x + 1). EXERCISE 68 MISCELLANEOUS EXAMPLES Expand the following expressions : 1. (3x-\-2y)(4:X-3y). 6. (a + 6)(a- 6)(a2 + 6=^. 2. (3^x-6y)(j\x-^6y). 7. (9x-2y)(x-\-y). 3. (a' + a^ ■\- l)(a2 - 1). 8. 3 x(a -b)(a + b). 4. (2 a -by. 9. (5a-4b^(5a-6b'). 5. (7 a^ + b-){S a2 - 8 b^. . 10. 3 m(m* - 4 mhi^ + 2 7i% SPECIAL PRODUCTS AND FACTORING 141 11. 7/i(n«-Hl)(n3-l). 21. 3y(2 - y)(5 +3,v). 12. 3a(a-|-l)(a-l). 22. 7(3 m + l)(m- 6). 13. {a? - 12 y){a^ -f 9 y). 23. {x^ +f){x'- y"). 14. {x-\-y){x'-xy+y-){x^-f). 24. (a^y - 4)(iC2/+ 16). 15. (a;-f?/)^ 25. 2{x-4.yy. 16. (m - \){m - \). 26. (a; + V){x - l)(ar^ - 3). 17. 2<3a;-hl)(a;-4). 27. (a^ +i)(a^- i a; + J). 18. (B'-AACXS^-eAC). 28. (Ia^-f)(iar»4-J). 19. a(a^ + 3)(a3-3). 29. (2?i-^)2. 20. 5(a-6)(2a+3 6). 30. (.^ - f)(a; + f). Find the prime factors of the following : 31. 5 r^- 40a; + 80. 47. 125a^-Sy^. 32. aa^ + 6aa:^ + 9a. 48. Sa^-Sa-S. 33. a;22,2 _ 5 a^a^y + 6 a^ftl 49. 169 a^^ 78 a& + 9 6^. 34. 3x^-60 a^- 288 y. 50. 30 m^ - 47 m - 5. 35. c2 4- 44 c -45. 51. 24: x^m + SI fm. 36. c3-64d^. 52. a«-f 14a« + 49. 37. 3 a:^ ^ 24 f. 53. ar^^/ - 2.16 y*, 38. m*+23w='n + 132?il 54. 3^^ + 33 A; +72. 39. a^-18a2 + 77. 55. 9x^-4:Xy-lSf. 40. x^-3a^y-10Sy\ 56. 33^-3/. 41. 4 ar^- 28 a;?/ + 49 2/1 57. a* -256. 42. 169mV-raV. 58. 3 a^- 108 a- 43. a2 + 17a-38. 59. c'cP + 9 cd-52. 44. 3c2 + 132c-135. 60. 5x^^-5 7/^. 45. x^-y\ 61. 25r2 + 60rs + 36s*. 46. 2d2_26d-136. 62. ar^ + Ja; + ^. [4: ALGEBRA 63. 196f-^\%a\ 67. xh^-20xz-69. 64. - 25 a; + «2 -f- 100. 68. 18 - 19 c+c2. 65. 300 m* - 243 ar^. 69. b - K) + 15 b\ 66. y + 12 y _ 108. 70. — 50a + a2 + 49. 107. The skill acquired in this chapter in performing some multiplications mentally is of use in solving certain equations. Example. Solve the equation : 6(4.-xy-5{2x-\-7){x-2) = 22-{2x + Sy. Solution : Perform the operations mentally. 1. 6(4~a:)2_5(2a; + 7)(x-2)=22 _(2 x + 3)2. 2. . •. 6(16 - 8 X + x2) - 5(2 x-^ + 3 a; - 14) r^ 22 - (4 ic2 + 12 a; + 9). 3. .-. 96 - 48 X + 6 x2 - 10 x^ - 15 x + 70 = 22 - 4 x^ - 12 x - 9. 4. .-. 166-63x-4x2=13- 12x-4x2. 5. .-. - 63 x + 12x3= 13 -166. 6. .-. -51x=- 153. 7. .-. x = 3. 1 13 1 81 Check : Does 6(|^-=^)2_ 6(6,.f-^ (lU^ z= 22 - 66-r^)2 ? Does 6 - 65 = 22 - 81 ? Does - 59 = - 59 ? Yes. EXERCISE 69 Solve the following examples performing all of the work mentally : 1. (5ic + 7)(3x-8) = (5a; + 4)(3a;-5). 2. (4m-7)«=(2m-5)(8m + 3)-2. 3. (5-3 r)(3 + 4 7-) - (7 + 3 ?-)(l - 4 r)= - 28. 4. (l_3^)V-(p + 5)2=40>+l)(2i7-3). 5. 2t{t-\-7)-(t-5y = (t + 3){t-ll). 6. (3a;+5)(2a;-3)-6(a;-2)(ic + 13)=ll. 7. (7a-2)(3a + 5)-(4a + 3)2 = 5a(a + 2)+l. SPECIAL PRODUCTS AND FACTORING 146 8. (5 n - 6) (5 71 + 6) - 5(2 n - 3)- = 5(n -f 12) (n - 1). 9. 3(4a_5)2_(6a- 1)- = 7 + (2a - 9)(6 a- 11). 10. (2x-7y-(ox-2y+3x{7x-{-5) = -i. 11. Find two numbers whose difference is 6, and the differ- ence of whose squares is 120. 12. Find two numbers whose sum is 13, and the difference of whose squares is 65. 13. Divide the number 20 into 2 parts such that the square of one exceeds the square of the otlier by 40. 14. Find four consecutive numbers such that the product of the first and third shall be less than the product of the second and fourth by 9. 15. Find two consecutive numbers such that the difference of their squares, plus 5 times the greater number, exceeds 4 times the less number by 27. 16. Find two consecutive numbers such that the sum of their squares exceeds twice the square of the smaller by 261. 17. One man travels a certain distance in as many hours as he travels miles per hour ; another man travels the same dis- tance in two hours less time by going three miles per hour faster. What was the rate and time of the first man ? EXERCISE 70 Problems about Area 1. Express the area of each of the following figures, assum- ing for each an altitude of 20 inches and a base of 30 inches : (a) A rectangle (§ 17). (b) A triangle (§ 17). (c) A parallelogram (§17). 2. Express the same areas if the altitude is 2 a; and the base (x— 6). 146 ALGEBRA 3. Express the same areas if the base is y and the altitude exceeds the base by 4. 4. The side of a square is s inches. (a) Express the area of the square. (b) Express the dimensions of a rectangle if its base is 4 inches more and if its altitude is 3 inches less than the side of the square. , (c) Express the area of the rectangle. {d) Form an equation expressing the fact that the area of the rectangle exceeds the area of the square by 50 square inches. 5. The base of a rectangle exceeds twice its altitude by 5 inches ; the base of a triangle exceeds the base of the rec- tangle by 4 inches, and its altitude exceeds the altitude of the rectangle by 3 inches. Let a represent the altitude of the rectangle. (a) Express the dimensions and area of the rectangle. (6) Express the dimensions and area of the triangle. (c) Form an equation expressing the fact that the area of the rectangle exceeds the area of the triangle by 25 square inches. Equations 6. The base of -a certain rectangle exceeds its altitude by 8 inches. If the base and altitude are both decreased by 2 inches, the old area exceeds the new by 36 square inches. Find the dimensions of the rectangle. 7. The base of a rectangle is 9 feet more and the altitude is 8 feet less than the side of a square. The area of the rec- tangle exceeds the area of the square by 15 square feet. Find the dimensions of the rectangle. 8. A man planned a house whose length exceeded its width by 10 feet. He found that it would be too expensive to build the house as planned, so he decided to decrease both SPECIAL PRODUCTS AND FACTORING 147 dimensions 5 feet. He found that this made a difference of 425 square feet in the area covered by the house. What were the original and the new dimensions ? 9. The main shaft of Washington's Monument is square at the bottom and top. The side of the lower square exceeds the side of the upper square by 21 feet. The area of the lower square exceeds the area of the upper square by 1869 square feet. Find the dimensions of the two squares. 10. A man planned to set out an apple orchard with two more trees to each row than he had rows, but found that that plan left one tree over. He found that if he decreased the number of rows by 3, and increased the number of trees per row by 5, he used all of his trees. How many trees had he ? QUADRATIC EQUATIONS SOLVED BY FACTORING 108. Not all equations are simple or first degree equations. (§77-) Example. Find the number whose square exceeds the num- ber itself by 6. Solution : 1. Let n = the number. 2. Then n^ = the square of the number, 3. and .-. n^ = n-\-6. 4. .-. n2 - n - 6 = 0. 5. Factoring : (n - 3) (n + 2) = 0, 6. If (w — 3) = 0, then • (n + 2) would also equal zero. n - 3 = 0, if n = 3. 7. If (n + 2) = 0, then (n — 3) • would also equal zero. n 4- 2 = 0, if n = — 2. 8. These values of n, 4-3 and — 2, should satisfy equation 4. Check : If n = 3, does (3)2 -3-6 = 9-3-6 = 0? Yes. If 71 =-2, does (-2)2- (-2) -6 = 4 + 2-6 = 0? Yes. 9. Moreover, both of these numbers satisfy the conditions of the problem : 32 is 9 ; 9 exceeds 3 by 6. (- 2)2 is 4 ; 4 exceeds - 2 by 6. 148 ALGEBRA 109. An equation like n^ — n — 6 = is called a Quadratic Equation or an Equation of the Second Degree. Other examples are : 4 x^ — 9 = 0, and Sy^-^ = -y-\-9. Notice that the equation has only one unknown ; that this unknown does not appear in the denominator of any fraction ; that it does appear with exponent 2 ; that it may or may not appear with exponent 1. Every quadratic equation has two roots, just as the equation in § 108. 110. Solution of Equations by Factoring depends upon the following numerical fact. If one of the factors of a product is zero, the value of the product is also zero. Thus, 3x0 = 0;(-6)x0 = 0;2x0x(-3)=0x(-3)=0. Example 1. Solve the equation 4 ic^ — 9 = 0. Solution : 1. Factor : (2 x - 3) (2 x + 3) = 0. 2. If 2 X - 3 = 0, then (2 x - 3) (2 x + 3) = 0. 2x-3 = 0, if2x = 3orxi= + |. 3. If 2 X + 3 = 0, then (2 x - 3) (2 x + 3) = 0. 2x + 3 = 0, if 2x=-3, or x=-|. 4. The roots of the equation are + f and — f . 5. Check : Does 4(f )-2 - 9 = 0? ^ 9 Does ^.-- 9 = 0? ie. 9-9 = 0? Yes. Does 4(-|)'2- 9 = 0? 1 9 Does < . ::_ 9 = ? i.e. 9-9 = 0? Yes. Rule. — To solve an equation by factoring : 1. Transpose all terms to the left member. 2. Factor the left member completely. 3. Set each factor equal to zero, and solve the resulting equations. 4. The roots obtained in step 3, are the roots of the given equation. Check by substitution in the given equation. SPECIAL PRODUCTS AND FACTORING 149 , Example 2. Solve the equation -— — — = —. o J o Solution: 1. Mg:* 2wi2-3m = 36. 2. Sss : 2 1»2 _ 3 w - 85 = 0. 3. Factor: (2 w + 7)(»» - 5) = 0. 4. 2 m + 7 = if w = — J. m — 5 = if 7rt = + 6. 5. The roots of the equation are + 5 and — J. Check by substitution. EXERCISE 71 Solve by factoring the following quadratic equations. 1. ar'-12.T-}-32 = 0. 11. x'-5x = 0. 2. y'-6y = 55. 12. Sc^-2c = 0. 3. n^ = 6S-2n. 13. 8a--5a-3 = 0. 4. m2-18m = -72. 14. 4p^-^Sp = 21. 5. ar'-21a; + 110 = 0. 15. 24. i^ -\- 2 r = 35. 6. c2 = 2-c. 16. 4ar^ = 8a;-3. 7. (p-7d-^G=0. 17. Sw^ = -iiv-l. 8. ^2-21 = 4^. 18. 3?^^2^i-§r=l. 9. 9m2-4 = 0. 19. 2'--^ 2 + 1 = 0. 10. 36x^-25 = 0. 20. x^ + jx + l=0. 111. A Literal Equation is one in which some or all of the known quantities are represented by letters ; as, 2x-j-a = hx'-10. Example 1. Find two numbers whose difference is a, and whose product is 6 times the square of a. Solution: 1. Let a; = the larger number. 2. Then x -- a = the smaller number. * For the symbol M^ see ^ 42. 150 ALGEBRA 3. .-. x{x-a) = Qa\ 4. .♦. a;2-aa;-6a2=0. 6. .-. (ic-3a)(ic + 2a)=0. .6. .*. x = 3 a or a; = — 2 a, the larger number. Check : If x = ^a, then 3 a — a = 2 a, the smaller number ; and 3 a • 2 a does equal 6 a^. If ic = — 2 a, then — 2a — a=— 3 a, the smaller number ; and — 2a- — 3 a does equal + 6 a^. Each solution is correct, no matter what a may be. Thus, if a is 5, 3 a is 15, 2 a is 10, and 15 x 10 is 150 ; also Q.a^ = 150. Example 2. Solve the equation ^-^ = 5c2. Solution : 1. ?! _ §_^ = 5 c2. 2 2 2. Ma: x'^ -Zcx- 10c2 = 0. 3. Factoring : (a; - 5 c) (x + 2 c) = 0. 4. If a; — 5 c = 0, then a; = 5 c. 5. If X + 2 c = 0, then a; = - 2 c. Check by substitution. EXERCISE 72 Solve the following equations for x : 1. Q^-\-2ax-35a' = 0.\ 9. 9x^ = 4:Xv-hlSv^ 2. 16x2-96^ = 0. 10. 7x^ + 41xk = 6k\ 3. x'-\-23mx-^130m' = 0. 11. x^ 3 a" ^ ax — - = 0. 2 2 /»»2 4. 25a^ = 9c2. 5. 2x'-^7ax-\-Sa' = 0. 12. - = 5p'-2px. 4 6. 7. 8. 7a^-10bx-]-Sb' = 0. 6x'-\-7cx = 5c^ 13. 14. X , n 9 ex SPECIAL PRODUCTS AND FACTORING 151 15. ar — o = ^ ^'' ^^- ~ ~ io 7xt ' 15 ^6* 2 5ra; 6 11 r* 3 6x2 4xs 3^ :0. 5 5 2 16. bar = m\ 19. 15 3 ^5 112. AVheii solving a problem whose solution leads to a quadratic equation, two sets of results are obtained. In some cases, both sets satisfy the conditions of the problem; in other cases, only one set satisfies the given conditions. Example 1. Find three consecutive odd numbers such that when the product of the first and third is increased by twice the square of the second, the sum equals 23. Solution : 1. Let x = the smallest number. 2. .*. X + 2 and a; + 4 are the other two numbers. (For example, if 3 is the smallest, 3 + 2 or 5 and 3 + 4 or 7 are the other two.) 3. .-. a-(a; + 4)+2(x + 2)2 = 23. .-. a;2 + 4 X + 2(a;2 + 4 a; + 4) = 2.S. .-. 3 x2 + 12 X + 8 = 23. .-. 3 x2 + 12 X - 15 = 0, or x2 + 4 X - 6 = 0. .-. (x + 6)(x- 1)=0. x = — 6orx= + l. 4. When x= - 6, x + 2 =— 3, and x + 4 = — 1. When x= + l, a- + 2 = 3, and x + 4 = 6. The solutions are — 5, — 3, and — 1 ; and 1, 3, and 6. Check: (-5) . (-1) + 2(-3)2 = 5 + 18=23. 1 • 6+2 • 32 =5+18=23. In this case, both solutions are satisfactory. Example 2. Determine the base and altitude of a triangle when the area is 110 square inches and the base exceeds the altitude by 9 inches. 158 ALGEBRA Solution : 1. Let a = the number of inches in the altitude, 2. .*. a + 9 = the number of inches in the base. 3. ... ei±±3 = the area. 2 . a(a + 9) ^ 120 2 .-. a2 + 9 05 _ 220 = 0. T+g" .-. (a + 20)(a- ll)z:z0. .-. a = 11, or - 20. 4. When a = 11, the base is + 20 and the area is (11 x 20) -j- 2 or 110. This satisfies the conditions of the problem. When a = — 20, the result can have meaning only if we have triangles with negative altitude. In such cases we agree to take only the positive solution. EXERCISE 73 1. Twice the square of a certain number equals the excess of 15 over the number. Find the number. 2. Find three consecutive numbers whose sum is equal to the product of the first two. 3. Divide 18 into two parts so that the sum of the squares of the parts shall be 170. 4. Find two numbers whose sum is 7 and the sum of whose cubes is 133. 5. Determine the base and altitude of a triangle such that the area shall be 15 square feet and the altitude shall be 7 feet less than the base. 6. Central Park in New York covers an area of about 800 acres. Its length exceeds its width by 2 miles. Find the dimensions of the park. (A square mile contains 640 acres.) 7. A merchant sold goods for $ 18.75 and lost as many per cent as the goods cost dollars. What was the cost ? SPECIAL PRODUCTS AND FACTORING 153 8. The length of a certain rectangular farm is three times its width. If its length should be increased by 20 rods, and its width by 8 rods, its area would be trebled. Of how many square rods does the farm consist ? 9. The standard size city lot in 'parts of Chicago is five% times as long as it is wide. The lots in parts of Indianapolis are 10 feet wider and 5 feet longer than those in Chicago. Three times the area of a Chicago lot exceeds twice the area of an Indianapolis lot by 275 square feet. Find the dimensions of the lots in both cities. 10. An architect who has made plans for a house with a base 30 x 42 feet finds that he must reduce the size. By what equal amount must he reduce the two dimensions of the house in order to make the area of the new base 925 square feet ? IX. HIGHEST COMMON FACTOR AND LOWEST COMMON MULTIPLE 113. A monomial is said to be Rational and Integral when it is either an arithmetical number, or a single literal number with unity for its exponent, or the product of two or more such numbers. Thus, 3, a, 2 a^bc^ are rational and integral. 114. The Degree of a rational and integral monomial is the sum of the exponents of its literal factors. Thus, a^bc^ is of the eighth degree. 116. A polynomial is said to be rational and integral when each term is rational and integral ; as, 2 a^6 — 3 c + d^. The degree of a rational and integral polynomial is the degree of its term of highest degree. Thus, 2a^b — Sc-\- d^ is of the third degree. 116. Recall the definition of prime factor of an expression (§ 87), and common factor (§ 11) of two or more expressions. Thus the prime factors of : (a) 6 m2(x2 - 1) are 2, 3, w, m, (x - 1), and (x + 1). (b) 9 m\x^ - 1) are 3, 3, m, w, w, (x - 1), and (x^ -\- x + 1). Common factors of (a) and (6) are 3, m, m, and (a; — 1). EXERCISE 74 Select the common factors in the following sets of ex- pressions: ^^ 5.mn'(x + l). 3 • 5 • m^n(x — l). 2. 2'3'af-y'(x + 4:)(x-S). 2'2'3'x''y\xi-A). 164 HIGHEST COMMON FACTOR 166 3. 7 '^•7^'S^(x-\-3)(x-2)(x-l). 3-7 '5'7^^s{x- 3){x - 2)(x - 1). 4. 6.7. m^'n^x - a){^ + "O- 6.7.7 .mV(ar + a)(ic' + a')- 5. 3.5.7.2.11. 2 . 2 . 3 . 5 . 13. 6. 3.7.5.2.2.2. 2.2.7.11. 117. The Highest Common Factor (H.C.F.) of two or more rational and integral expressions is the expression of highest degree (§ 114), with greatest numerical coefficient, which will divide each of them without a remainder. Example. Find the H. C. F. of 6 mH\x - l){x H- 2) and 15 m^n (x + l){x + 2). Solution : 1. The greatest integer which will divide both expressions is 5. 2. The highest power of m which will divide both is w»*. The highest power of n which will divide both is n. The highest power of {x + 2) which will divide both is (x + 2). Neither (jc — 1) nor {x -\- \) will divide both expressions. 3. . •. The H. C. F. is 5 m^n{x + 2). Check : 5 m^r\:\x - l)(a; + 2)-t- 6 mH{x + 2) = n{x - 1). 15m%(x + l)(x + 2) -f-5m3n(x + 2) = 3m(3-+ 1). Rule. — To find the H. C. F. of two or more expressions : 1. Find all of the prime factors of each expression. 2. Select the factors common to all of the given expressions, and give each the lowest exponent it has in any of the expressions. 3. Form the product of the common factors selected in step 2. Example. Find the H.C.F. of 68 (m + ^0^ (w - n)* and 85 {m -t- iif (m — n). Solution : 1. 68(w + n)2(w - n)* = 2 • 2 • 17 • (m + ny(m - n)*. 86(7n + w)3(m - n) = 5 • 17 • (in + n)8(m - n). 2. . •. The H. C. F. = 17(m + ny\m - n). 166 ALGEBRA EXERCISE 75 Find the highest common factor of : 1. 16 and 56. 6. 14 x^y, 21 xy\ and 35 a^y\ 2. 64 a and 96 b. 7. 15 m^< 45 mV, and 25 miiK 3. 72 a;2 and 27 xy. 8. 12 a^, 18 x^^/', and 24 a^y. 4. 5 a^62 and 2 a^ft^ ^ 16mV,56mV,and88mV 5. 20 a^^ and 15 xy': 10. 18 j-^s, 27 r^^, and 45 rsl 11. (a + 6)(a-&) and2a(a + 6). - o 12. 3ir(a;4-2/)(aJ-2/) and 2(x + y)(x-y)(x'-{-y^. "'■'' * 13. (aj + iy(x - 3) and (a? + l)2(a; + 2). 14. (r -\- sf (r - sf and (r + s)^ (r - s)l 15. 3(a;-2 2/)(a;-h2^) and6(a;-2 2/)'. 16. 2 ««a; H- 4 a^ar^^- 2 aa:^ and 3 a^a; + 3 aa;*. Solution : 1. 2 a% + ia'^x^ + 2ax^ = 2 ax{a^ + 2ax + x^) = 2 ax(a + x){a + x). 3 a^a: + 3 ax* = 3 ax(a^ + a;^) = 3 ax(a + a;)(a2 _ ax + a;2). 2. The H. C. F. = a . X • (a + ic) = ax(a + a;). 17. a2-62anda2_2a6 + 62. 18. a;-+ 2a;-24, scr- 14a; + 40, and a;2-8a; + 16. 19. 2r2_7r-h6and6r2_llr4-3. 20. a:»-27andar^-lla^ + 24. 21. m^ — 8 m^ and m^ + 2m + 4. 22. 6 a^b'-~ 15 a^b^sind 12 a'b-\- 21 a^bK 23. 3a3 4-192and a2-7a-44. 24. 3 x^ -16xy-\- 5 y^ and x"- -{-lOxy- 75 y^ 25. 27 a» + 8 5^ 9 a- - 4 6^ and 9 a^ + 12 a& 4- 4 ^^l LOWEST COMMON MULTIPLE 157 LOWEST COMMON MULTIPLE 118. A Multiple of a number is any number whieh contaitis the given number as an exact divisor. Thus : (a) Some multiples of 3 are (3, 9, 12, and 30. (6) Some multiples of (x + y) are 2(x + y) and (x + y)(x — y). 119. One number may be a multiple of two or more different numbers. Thus : (a) 24 is a multiple of 2, 3, 4, 6, 8, and of 12. (6) 5 a^^bc is a multiple of 5, a=^, 6, c, and of a. (c) 3)^' 18. 2 a(m 4- a) and 6 a\m + a)(m — a). 19. (a; + 3)(a;-2) and (a;-2)(a;-3). LOWEST COMMON MULTIPLE 159 20. 3(r -f s)(r - s) and 2(r -f s)(r - t). 21. (a-4)(a-3) and (a-4)((X-5). 22. (1 - xy, (1 - x)(l 4- x), and (1 - xf. 23. (2-3 a;)2, 3(2 - 3 x)(2 -f- 3 x), and (2 -f- 3 xf. 24. (2 - a;)(3 - x), (3 - a;)(4 - x), and (4 - x)(2 - x). 25. 4 ic^ — 4 wi^, 6 a; + 6 m, and 3 ar* — 3 7/i\ Solution : 1. 4x^ — im^ = i(x^ — m^) = 4(x - m) (x-\-m). 6x + 6m = 6(a: + m). 3 x8 - 3 w8 = 3(a^-»/i») =3(r- w)(3:2 + mx + m^. 2. The L. C. M. = 12(a; - m) (x + w) (a:2 + mx + m^). 3. L. C. M. -?- 4(x - m) (X + m) = 3(x2 -f wix + m2) . L. C. M. ^ 6(x + m) = 2(x - rn) (x2 + »nx + m*). L. C. M. -^ 3(x - w)(x2 + mx + m^) = 4(x + m). 26. r2-16and7-2 4-llr-}-28. 27. a^ + 2 ttic 4- a^ and a^ + xl 28. Hx^y — y £^nd 22 ar^z — 9 a;2 — z. 29. 62_i264.35, 62-}- 2 6 - 63, and 6^-3 6 - 108. 30. 4a:2_25and4a:2_20a;-|-25. 31. 3 m^- 6 m - 72, 4 m^ + 8 m - 192, and 2 m'- 24 m +72. 32. 9n2-27n + 8and3n2-2n-16. 33. l-|-27a:»andl-5a;-24ar^. 34. ar^ + a; - 42 and x2 _ a; _ ^q 35. a*' — a and a^ — 9 a* — 10 a. X. FRACTIONS 122. The quotient of a divided by b is written -• The b expression - is called a Fraction ; the dividend, a, is called the Numerator, and the divisor, b, is called the Denominator. The numerator and denominator are called the Terms of the frac- tion. - is read " a divided by bJ' The denominator, b, must never be zero (§ 64). REDUCTION OF FRACTIONS TO LOWEST TERMS 123. A fraction is said to be in its Lowest Terms when its numerator and denominator have no common factor except . unity. Thus : (a) - , -, ^-i-^ are in their lowest terms. 3 b X — y (.b) i~ , — T" ) ; , ^1 are not in their lowest terms. 124. To reduce a fraction to lowest terms, a principle, easily illustrated by arithmetical fractions, is used. If, in the fraction y , both terms be divided by 4, a new fractionf is obtained. But, since -2/ = 3, and f = 3, therefore ^^^ = |. The value of the fraction is not changed ; its form is changed. Rule. — If the numerator and denominator of a fraction are both divided by the same number, the value of the fraction is not changed. Example 1. Reduce to lowest terms 40 a^b'c Solution: 1. gjoB^ ^2^ ■ 3 . gB . 52 . c 40 a^b-^c 23 . 5 . a^ . 62 . c 2. Divide both numerator and denominator successively by their com- mon factors, 23, a^, b^, and c. 160 FRACTIONS 24 a^b^c I a H^ 1 3a AOa^b-'c 1 1 1 1 5 161 Then, Note 1. Dividing by all of the common factors is equivalent to dividing both numerator and denominator by their highest common factor. Note 2. Tlie process of removing a common factor from numerator and denominator of a fraction is called cancellation. It depends upon the rule in §m. Note 3. It is wise to write the quotients 1, a, etc., as they are obtained. Example 2. Reduce to lowest terms -I- o4 6 9a2-f 24a6-}-166« 1 Solution- 27 a^ + 64 6« _ QloM-^) (0 a'^ - 12 q5 -f 16 ft^) ■ 9a2 + 24a6 + 16 6-^ C2^z-h) 1 ^ 9a2-12a& + 16 62 (3 a + 4 &) Here, the numerator and denominator are both divided by (3 a + 4 6). Check : These examples may be checked by substitution. It is im- portant to remember that the original fraction and the simplified result are equal for all values of the literal numbers except such as make the denominator zero. (§ 122.) Rule. — To reduce a fraction to lowest terms : 1. Express numerator and denominator in terms of their prime factors. 2. Divide both numerator and denominator by all of their common factors, i.e. by their H. C. F. 125. Errors in Reducing Fractions. One common error occurs in reducing fractions such as 3win2(x + 2/) ^^ifitC0 {^[.^Y^' Pupils sometimes think that the result is 0, because all factors have been cancelled. If, as suggested, the quotients are indicated, this danger will be avoided ; thus, 1 11 1 3mn2(x + j/) '^jntn^ {xA^) 1 1111 162 ALGEBRA Another, and more common, error is illustrated in the fol- lowing faulty solution : 1 1 2a + 6 aiJk-r5 2 + 1 ah ;ib \ 11 = 3. Testitfora = 2, 6 = 2; ?A±^ = idL2 = 6 ^ 1 a6 4 4 2 The error is in dividing one part of the numerator by a and another part of it by h. Neither a nor 6 is a factor of the nu- merator. They cannot be canceled in this problem. The frac- tion is already in its lowest terms. EXERCISE 77 Reduce the following fractions to their lowest terms : . ^ 12 5^. ^^r'sf 15' * Sxf ' eSr^s^^^- 36* ' 42mV' * 40xy*z' ^ 3ab'c(x-^y) ' ^^ 2af-2y\ 2abc{x-\-y) ' 5a^ — 5y^ 8 12(c^-3)(a + 3) ^^ Sam' -3 an'' 4(a-f3) * * 3m2 + 6mnH-3w2* . 18(2r-gy(r-f^) ^g «' + ^'^ * 9(2r-s)(r + s)2 ' a^c - 2 a6c - 3 fe^c ^^^ a^ + 7a-flO 17, 8a^-125 (x2-^4a — 5 ' 2 ax' -\- ax — 15 a m^ + m — 56 -g ma;^ — ma; — 12 m ■ m^ - m - 42 * ' 3 ic'^ + 13 a; + 12 ' a^_9a;4-18 ^^ 169-^ + 47-^ + 1 . 3a^^-3a;-36* ' 64r«-l ,, a2-lla6 + 2862 ^^ a3-8a2 + 12a ^_14a6 + 4962 Sa^-- 60a + 180 FRACTIONS 168 21. 22. 23. ^-f aa^ — 2 aacy -\- ay^ a^-64 a% — 16 m («^-4)(a-l) 24. 25. 4 7n^ -f 16 m?? 4- 15 n^ 6 m^ — mn — 15 n^ 15 a;^y + 10 x^y^ 6a:«2/' + 4ar^/ 26. ^^-^y^ . 126. Signs in Fractions. A fraction is an indicated quotient. Its sign is governed by the law of signs for division (§ 67). Thus: From 1 and 2, it is clear that •4-4 From 1 and 3, it is clear that 4-12 ^ + 4 12 + 4 -12 = -3. = -h3. + 12 -4' -12 + 4'J since +3 = — (- 3).. Rule. — If the sign of one term of a fraction is changed, the sign of the whole fraction must be changed. From 1 and 4, it is clear that + 12^-12 + 4 -4 , since both equal + 3. Rule. — If the signs of both terms of a fraction are changed, the sign of the fraction must not be changed. Changing the signs of an expression is accomplished by multiplying it by — 1. Example 1. Reduce to lowest terms : 12 + 2x-2ar»* 164 ALGEBRA Solution : Multiply the denominator of the given fraction by — 1, thus changing its signs ; also, change the sign of the fraction. 1 x2-9 ^ x2-9 ^ (aL^^3)(x + 3) ^ a; -f 3 12 + 2 X - 2 x-2 2 x2 - 2 a: - 12 20-^') (x + 2) 2{x + 2) ' 1 a:2-9 1-9 -8 2 Check. Let x — \, Then also 12 + 2x - 2x2 12 + 2-2 12 3' X + 3 1+3 4 2 2(x + 2) 2(1 + 2) 6 3 The value of x selected must not make any denominator zero (§ 64). Example 2. Reduce to lowest terms: (9 - m^) Solution : Multiply the numerator by — 1, thus changing its signs ; also change the sign of the fraction. ^ _ 9-^2 ^ w2 - 9 w2 - 7 m + 12 m2 — 7 m + 12 1 2 ^ ( m^-^)(m + 3) ^ (m + 3) (m - 4) (W2^-^) (m - 4) ' 1 EXERCISE 78 Reduce to lowest terms : 42/2-0^ 6. 27-^ ' 2x'-7xy-]-6y^ e-9 Sx-371 7. a?-l^ "' f^xy-^x" 64 -ar^ 3 «' - ^^ 8. at^ - as" 24 - 2 a - tt^ 2 as^ + 4 ars — 6 ar^ ^ 3 7?r - 3 71^ 9. 3c«_ 6c2rZ + 3cc?2 n^ + 4 mri — 5 wi^ bd^-bc" a^2 ^ a; - 6 10. 18 ma?2 _ 8 ^^2 24 W2/3 - 81 nx^ FRACTIONS 165 TO REDUCE A FRACTION TO AN INTEGRAL OR MIXED EXPRESSION 127. An Integral Expression is an expression which has no fractional literal part ; as, a^ — 2 ah + fr, or | ah-. An integral expression may be considered a fraction whose denominator is 1 ; thus, a + 6 is the same as —^ — 128. A Mixed Expression is an expression which has both integral and fractional literal parts ; as, a + - or x + ^ • C 7J-Z Rule. — A fraction may be changed to an integral or mixed expres- sion by performing the indicated division. Example 1. Reduce — ~ " ■ to a mixed expression. 3 X Solution : Using the method of short division (§ 70), 6^+16x-2^6^ 150. -j^, _2_. Ans. 3x 3x 3x 8« 3x Example 2. Reduce ~ — -— ^ ^ ~ - to a mixed ex- 4 af' + 3 pression. Solution : Using the method of long division (§71), 3 a: -2 4t«a + 3|l2a:*t- -8x--J-h4x- -5 12a:8 -l-9a; .8a:2-5a-- -6 ■8aa -6 -5ar-f 1 12x8 _8x2 + 4a;-5 4a:'-i + 3 = 3a;-2- = 3 a; - 2 + 4a;2-f 3 -6r r + 1 + 3 The first term of the numerator of the fraction in the resalt is negative. Change the signs in the numerator by multiplying the numei-ator by — 1, and also change the sign of the fraction. See § 126. 166 ALGEBRA EXERCISE 79 Reduce the following to mixed expressions : 1. W- 2. «. 3. ^K 4. ^\K 5 12a^-16ft + 7 ^ 15j9^H-12iJ-4 4a ' * 3i) * 15m3-6m2 + 3m-8 „ 30a;« + Sa^- 15a^- 7 o. ■ a? + 2f x-y a« + 8 6« 13. x-y 14 a^4-^-^y+6y^ a-2h ' x-2y 16 3a« + 8a^-7 4a-l * a2-2a-3 TO REDUCE FRACTIONS TO THEIR LOWEST COMMON DENOMINATOR 129. In arithmetic, fractions having a common denominator may be added or subtracted without difficulty. Thus : Fractions which do not have the same denominator must be changed to equal fractions having a common denominator. (a) -^ + I = A + T% - H, for f = j%. In algebra, also, fractions which do not have a common de- nominator must be changed to a common denominator before they can be added or subtracted. 130. To change fractions to a common denominator, a prin- ciple, easily illustrated arithmetically, is used. If both terms of the fraction | are multiplied by 4, the result is ^. I = ^ since both equal 3. The value of the fraction is not changed ; its form only is changed. FRACTIONS 167 Rule. — If the numerator and denominator of a fraction are both multiplied by the same number, the value of the fraction is not changed. 131. Two fractions which have the same value but different form are called Equivalent Fractions. Thus, - and — are equivalent fractions ; they differ in form, but b mb have the same value, since the second is obtained by multiplying both terms of the first by m. 132. The Lowest Common Denominator (L. C. D.) of two or more fractions is the lowest common multiple of their denom- inators. Example 1. Reduce to their lowest common denominator i!?^ and ^^. 2 ab' 3 a'b Solution : 1. The L. C. M. of 2 ah^ and 3 a'6 is 6 aW. (§ 121) 2. To change the denominator 2 ab^ into a^ft^, we must multiply 2 ab^ by 3 a^, that is by 6 a^b'^ -r- 2 ab'^. In order not to change the value of the fraction, the numerator 3 mx must also be multiplied by 3 a^. Then, Sn^^Sa^.Zmx^^Qa^mx^ . 3. For the second fraction, 6 a^b^ -4- 3 a^ft = 2 6. Multiply both nu- merator and denominator by 2 b. Then, ^^ = UilAm. = lOfcnj^ . (§ 130) Rule. — To reduce fractions to their lowest common denominator: 1. Find the prime factors of the denominators. 2. Find the L. C. M. of the given denominators ; this is the L. C D. 3. For each fraction, divide the L. C. D. by the given denominator ; multiply both numerator and denominator by the quotient. Example 2. Reduce to their lowest common denominator : 4« and 3a a'-4 a'-Ba + 6 168 ALGEBRA Solution : 1. 2. 4a 4a a2-.4 (a-2)(a + 2) 3a 3a a^ - 5 a + 6 (a - 2) (a - 3) 3. TheL. CD. is(a+2)(a-2)(a-3). • 4. L. C. D. -^ (a - 2)(a + 2) = a - 3. 4a 4 a(a — 3) " (a-2)(a + 2) ~ (rt-2)(a + 2)(a-3)' 5. L. C. D. - (a - 2) (a - 3) = (a -f 2). . 3 a ^ 3aCa-t-2) " (a-2)(a-3) (« - 2)(a + 2)(a - 3)' Check : The final fractions in steps 4 and 5 may be changed into the original fractions by cancellation. EXERCISE 80 Reduce the following to equivalent fractions having their lowest common denominator : 1- i; I; |. 9- 2- f; tVj tu- 12 6 3. 4. 3a 55 T' 6 3 ' 5 ' 2 * 12. 5_mw, 7 mp ^ 3_np 5 ' 13. 14. 6 > 4 ' 2. a 3 5 • 2x a . 3.V bz ah c , A. xi xz 2/2 15. 8. — ; -; — . 16. 2 mhi 3mn^^ 5 7nhi^ i^s ^ ^ 2m^n 3mn'' 5 77iV 2a-5 c. 4a-f36 ^ab ' 12 ac ZX-A: z 6x —5z 4.x^ ' 3 x'z 2ab b a'-b'' a-{-b A a' 2 4a2_s 1' 6a'-9a 2x . 3?/ x^2y' ' ^'-2?/ ^x 5 5 a; -10' 2 a; -4 FRACTIONS 169 17. ^ ; ^ . 19. «+^ - ^-^ af-Qx-i-S' a^-16 a«-2a6' aft-f-fc^ 18. -AZL; _-^ 20 ^ ^ ^ a H- 5 a + 3 21. 22. 23. 24. 25. a2_a-6' a^ + Ta-l-lO a + 36 a — 36 a^ _ 7 a/^ _^ 12 6^' a- - a6 - 12 6^ x^—2xy-\-if^ {a-{-b)(x — y) x-\-y ^ xy a-j-2 a— 3 a 4- 1 a2-|_2a-3' a2_3^^2' a^~^a-6 ADDITION AND SUBTRACTION OF FRACTIONS 133. Example 1. Perform the indicated addition : 4aH-3 l-6b' 4 a^b 6 ab^ Solution : 1. The fractions cannot be added because they do not have a common denominator. By the methods of § 132, change the two fractions to equivalent fractions having their lowest common denominator. 2. The L. C. D. = 12 a^b^. Multiply the terms of the first fraction by 12 a^b^ H- 4 a^b, or 3 b'^ ; and the terms of the second fraction by 12 a^b^ ^ 6 ab^, or 2 a. Then, 3 4 g -t- 3 1 - 6 ?)2 ^ 3 feg X (4 g + 3) 2 a x (1 - 6 S^) 4 a-^b 6 g68 3 6^ x 4 g^6 2 a x (6 g&8) ^ (12 afo2 4.9^2) (2a-12a62) 12 g268 "^ 12 g268 _ (12 ab^ + 9 &2) + (2 g - 12 ab^) 12 g263 _ 12 aft^ + 9 62 + 2 g - 12 ab^ 12 0*63 _ 9 62 + 2 g 12aa68 * 170 ALGEBRA Check : Let a = I ; h = 1. 4^4-3 ^7 l-6fe2 ^-5 ^^^7 (-5)^11. 4a-^6 4' 6a&3 6 ' 4 6 12* also, 9b^±2a^9_±2^n^ 12rt253 12 12 Rule. — To add or subtract fractions : 1. Reduce them, if necessary, to equivalent fractions having their lowest common denominator. 2. For the numerator of the result, combine the numerators of the resulting fractions, in parentheses, preceding each by the sign of its fraction. 3. For the denominator of the result, write the L. C. D. 4. Simplify the numerator by removing parentheses and combin- ing like terms. 5. Reduce the result to lowest terms. Example 2. Simplify ^^^^ii' - I^^^^j . 6 14 Solution: L 5a^ - 4 y _ 7 x - 2 y ^ 7(5x- 4 y) _ 3(7 a: - 2 y) 6 14 42 42 2 ^ 7(5a;-4y) -Snx-2y) 42 ^ 42 4 . ^ 14 a; - 22 y 42 5. Check ; Let x = 1 ; y = 1. 20x-ny) ^ 7x-n 42 21 5x-4y ^ 5-4 ^ 1 lx-2y ^b^ ^^^1 5^-8^-4. 6 6 6' 14 14' 6 14 42 21 ' also, 1 x-\\y _ 7-11 ^ ^ ^ r^^^ solution is correct. ' 21 21 21 Note. In the first step of the solution, the numerator and denominator of the first fraction are multiplied by 7, and of the second fraction, by 3. FRACTIONS 171 EXERCISE 81 Perform the indicated additions and subtractions : 9 27 3 3 6^15 2 ±4-?_l 9 ^-l£^lA ' 15 5 20 '9 3 "^27 3. A + ?_i. • 10. ^+^_^. 16 8 4 6 10 15 I4.I-A. 11 6(1-5 3a + 7 • 14 "^21 84* * 8 "^ 12 ' ^ 5a , 3a -„ 2a-8^3a + 5 6. ^^ -f^. 12. -^_+--^^. - 6m 3 m -„ 3 7^1 + 4 2m-|-5 *• T'u:' ^^' ~i2 ^9 „ 15a; 7a;, 3a; .. 4a-9 3a-8 7 u - — . 14. 8 4 2 9 12 ,, 5a-f 1 , 26 + 3 7c-4 ''• "6~ + -8 12-' 3a + 4 4a-3 . 5a + 2 16. _ __4.___. 2a-36 3a + 6 4a-56 9 18 27 * 6m + l 5m — 2 8m — 3 7m + 4 ^®' "~3 6~~"^~^ i2~* ,- 4r-3s.6r + 5s 5r+2s 3r-10s 10 15 20 «^ 3t-2x 7t-Sx ^ 9t-{-4x lOt-^lx iiU. ^ 1 — — ' 172 ALGEBRA 21. ?-A. 26. 22. — - + -^. 27. 3_ 5 a 2a 2 Sx 42/ 2 7' S t .7 X z a b c_ 2 m^ mw 5 n^ 5a4-l . 3a 6a 2a 23. i_l+^. 28. lQ^ + 3y 3a; + 5y 2 ic^i/ iC2/^ 24. - + ---. 29. ^:zl + fc2_?_j_5--3^ a»/ 2yz 3zx 25. - + -_-. 30. 2a-6^26-c_^2c- 31. a6 6c ca 1 Solution: 1 2. cc^ -f- a? x^ — X 1 1 1 a:2 + a; x^-x x(x + 1) a;(x - 1) (0^-1) (a^ + 1) ic(a;+ l)(x- 1) a:(x + l)(a; - 1) 3 ^ (a;-l)-(x + l) a;(x+l)(x-l) ^ x—\—x—\ x{x-^ l){x— 1) -2 ^ +2 Notice that in line 2, the L. CD. is a;(x + l)(x— 1) ; that the numerator and denominator of the first fraction are multiplied hy (x — 1) , and of the second fraction by (a: + 1) ; that parentheses are used in lines 2 and 3 ; that in line 5, the signs of both numerator and denominator are changed. Check by substitution. 32. Simplify « + ^ ^"^ 4 a^ - 9 6=^ (2 a + 3 b)' Solution: 1. « + ^ ^"^ 2. = a 4-h a (2a-3&)(2a + 36) (2 a + 36)(2 a + 3 6) 3 ^ (q + 6)C2a + 3&) C«-6)(2a-3 6) (2a-36)(2a + 36)(2a + 3&) (2a-3 6)(2a+36)(2a+3 6) FRACTIONS 178 . ^ r2 gg 4- 6 orft -f 3 ^2)-(2 a^-^ab + S &«) (2 a - 3 6) (2 a + 3 6) (2 a + 3 6) 2 a2 + 5 a6 + 3 62 _ 2 a2 + 5 aft -3 62 o. 6. (2 a - 3 6) (2 o + 3 6)(2 a + 3 6) 10rt6 (2a-3 6)(2a +3 6)(2a + 3 6) Check : Let a = 1, 6 = 1. « + ^ =^ = -^-, '-' ^-0^0,and-^-0^-g; 4a2-962 -5 6' (2 « + 3 6)^ 25 ' 5 6 also 10 «& 10 ^ 10 ^ 10^ 2 (2a-3 6)(2a + 36)2 (2-3)(5)2 -25 26 5 Notice that the indicated products in step 3 in tlie numerator are found and inclosed in parentheses in step 4. All of this solution should be done mentally. 33. ^ 1 ■ 41. 2-0? 2-\-x ^ ' rn — l m-\-l 2-\-x 2 — X 34 -^^4--^. 42 4p^ + l 2;)-! • r+a r-3 ■ 4:^^-1 2p + l 35. -i ?-. 43. -1 i2«-6)_^ 5m-2 2m-h3 2a + 6 Sa^ + ft^ 3 6 5 6 36. 3a-4 5a+6 37. ^y ^^ • * 3y — X 2a;— 3y 38 ^ I y - • 2x-\-2y 3x-3y 1 2 39. 40. 3a-.7 6a4-15 __3 4 4i)-6 15;)- 12* 2 49. 44. x-^S ar'^-27 x-S x'~21 45. x-\.2 x-2 16 x-2 x+2 x'-4. 46. 11 2 a a^h a-h a^-h^ 47. 1 Sx ax a — x a^ — x' a*— x^ 48. m-i-n m — n 4 mn m — u m +71 m^ — n^ 5 ic2_5a;+6 x2_,.2a._i5 174 ALGEBRA KA q+l (^ — 4 , a 4- 3 oO- -s t: o \ —7: + a2_a-6 a2-4a-f-3 a2H-a-2 3a-|-2 , a + 3 a-2 62. Ba^-a-l 3a2+7a + 2 2a2 + 3a-2 r a r^ — 6 ar + 9 a^ r^ + 4 ar — 21 a^ g3 a; + 2 2(a^-l) a;-3 a;24-4a; + 3 ar^ + a;-6 a;2-a;-2* -, 1 a , a^ — 4 a + l a^-a + l a^ + l 55. Simplify -^ + ^^-±4- ^ ^ a-b b'-a' Solution : 1. Notice that b^ — a^ is not in the same order as a ~h. Change the signs of the denominator and also the sign of the fraction (§ 126). 2 3 2h+a_ 3 a4-26 a-h b'^~ aP- a-b a^ - b^ 9 3 a + 2b a-b (a-6)(a + 6) 4. _3(a + 6)- (a + 2 6) 5. _3a + 3?)-a -26 6. {a — b)ia + b) 2rt + & a2 - 62 Check by substitution. 56. -1^ + -^. 59. 57. — ^t 1 60. -^+ ^ y^ — xy x^ — xy a + 1 1 — a tt^ — 1 x-\-4: 2 X — 5 r r _ r^ 3a;-6~8-4a;* ' r + 2"~2-r 7^-4' FRACTIONS 62. a b 2 b' -a' 67. 5a-b rt -f 6 a-b b'^ ' 1 5a+b 63. 2x ?^x 68. 3a + l ^-".?- 2a — 3 Q-x-x' x'-^^x -10 64 ^'-i-y 69. . «' - b' a' + ab + b^ 65. '-'*'-^- 70. 3^ + 4 «™' + f 3 wi — 4 66. X+7J 71. r' rs + s' ^^. 176 MULTIPLICATION OF FRACTIONS 134. In arithmetic, the product of two fractions is the product of their numerators divided by the product of their denominators. Thus, f x ^^ = ^^. In algebra, the same rule is followed. Example 1. Multiply — by — • 2 X 15 a^ ' Solution: 1. ^-« . i^ = ?Oax^ 2x15 a2 30 a'^x 2. Reduce to lowest terms : 2 X 20aa^_^a^_2x 30a^x~mrth7~ Sa' 3 a It is customary to cancel the common factors in step 1 as in the following example. Example 2. Sin.plity ?^±2a^ . a^. a^ — 16 a^ — 1 Solution : 1. q-^ + 2 g - 3 a^-4a a'i - 16 ' a2 - 1 1 1 1 1 _ a(a + 3) ^ gg + 3a (a + 4)(a + l) a2 + 5a + 4' 176 ALGEBRA Check : Let a = 2. a2 + 2a-3_4-f4-3 a^-4a also, 4-16 _4 4 -8_- - 1 a2 4-3a 5 -12 -5 12 -4 3 ' 12 3 36 5 a2_ 1 4 + 6 10 5' a2 + 5 a + 4 4+10 + 4 18 9 Notice that the factors (a — 1) of the first numerator and of the second denominator are each divided by (a— 1), or, are cancelled ; similarly the factors (a — 4) of the first denominator and the second numerator. Rule. — To find the product of two or more fractions : 1. Find all of the prime factors of the numerators and denomina- tors. 2. Divide out (cancel) factors common to a numerator and a denom- inator. 3. Multiply the remaining factors of the numerators for the nu- merator of the product, and of the denominators for the denominator of the product. EXERCISE 82 Find the following indicated products : , 3 21 10 . 5a'b Jy'G ^c'a A. T~r • 777: * ^:r ' o. 3 21 10 14 20 9 5 6 12 18 10 7 ^' ^ ' ~ ' ^' 7. 6 a^h . ^xy^ 15 m^y 2 ah 3 am" 25 W 20 U' 3 aW _ B a 5. — - • 9. ^a 66 8c 4 6 10 c 12 a 10. ^^'-^ m^ — 16 m ^^ a2 + 2a-35 6a3 ' {a 2a(a-3) + 7)(a-3) 2oc2 3 6a^ 5 b'c 4a« 9 6^ 15 6^ 21 10 a'' 27 7ri'y ' 2,0 n T.xy 20 m^x 14 t-y ISn' x' -a' a^ ' 2x+2a x^-\-2 ax +- Sx 4m2- -1 m' -f 4m 2m + l FRACTIONS 17T 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. p2 4- 4 p - 45 ^ Spr-Sr" 4: pr — 20 r p2-81 a*-Sa^ + 12a^ a" - 36 a.'^ + S o^-\- 2x^-^x xif-,f x"- xy-2 if a^ + :ry x^- 2xy-\-y' 2 a^ + 5 « 2 + 2 2 a^ - 3 a 2 a^ + a - 6 6 a^ + 3 a'^ 6 r^ - r8 - 2 s2 8 /-^^ - 6 rs H- s^ 12 7^ + 5 rs - 2 s2 4r^- s^ 5m4-10 3m — 9 8m2- -2 8 m - 4 Vdm^-h 3 m^ - 12 x^ — y^ 3? -\-xy - -2.V^ 5 a; + 5 y 2 a^ + 2 2/^ a^ + 3 a.-?/ + 2 2/^ x'-2xy -\-y' 8 a?^ - 27 ^ 4 ar^ + 12 a; + 9 ^ 4a;-6 4a^-9 * 8a^ + 12a;4-18 ' 6a;2_}_53._6* m"* — 1 4 m^ — 3 »i 4 m^ + 3 w 16 m^ - 9 n2 2 m^ -f 2 2a^-a 4a2 + 4a + l 4a2-l 1-ar^ 5a ar^-4 m — 1 10 g^ - 10 g 4g2_2g-2 3 a) -6 2aj-4 ar^_a;-2 ar^ + a; - 2 aP ^ 5x 6a-\-6x g2 + 2 ga; + ar* 3 x- 3 a 5 a^ + 5 aa; + 5 x^ m^ -f ^^^ . n — m [(m -f ^Q^ — 'mn'] rn? — n^ n -\- m [(m — nif + mw] 178 ALGEBRA DIVISION OF FRACTIONS 135. In arithmetic, to divide a fraction by another fraction, we multiply the dividend by the inverted divisor. Thus : (a) 1.^3_^^_3^ ^"^^ io-5-;0''3"2 2 ^ ^ ^ ^ 3 5 3 Z; 3 In algebra, it is advisable to factor the expressions first, and then use the same process as in arithmetic. Example. Divide ^-11^ by ^-^^ + 1 . Solution:!, o.^^ ^ ?iziM±i 2 a;(x-l) . (x^l)(x-l) (x + l)(ar.2-a; + l) ■ a;(x2-a;+l) 1 1 3 x(o^^^) x(x2.^='-x-^fT:) (X + 1)(^!^=^TT) (^^-l)(iK - 1) 1 1 x^ x^ (x + l)(a;-l) x^-l Check : Let x = 2. x2 - a; ^ 2 x2 - 2 X + 1 ^ 1 ^^^^ 2 ^1^1. a^s+l Q' ofi^x^ + x 6' 9 6 3' also, ^ — = - • ' a:2 - 1 3 Rule. — To divide one fraction by another : 1. Factor the numerators and denominators of the fractions. 2. Invert the divisor fraction. 3. Multiply the dividend fraction by the inverted divisor. FRACTIONS 179 EXERCISE 83 Perform the following indicated divisions : 5. 7. 5 . 10 18 • 9 ' 3 21c2 . 3c ' 32d ' Sd^ 7a . 5a 15b 66 15ar^ . Sx" 3a . 2? 10 aj2_6a;-|-8 * ar^ -xll2 '''• 4r-'-25?r^ . ,_2_ rc-5^^ 11. 32 m*' 15 n^ 35 w8 _ II* tf U/ « -1-tf a/ u (^ .(/ ^4 at) /^ 91.9 2. ^z;^ -^ rrr • 4. — — ^ H- t— „ • 6. —-7-^ -^ 6 a26l OX*jf (c-rf)- ' (c + d) ' 16r2-9y2 ' 49-2 -4-3n; ""' r^ + rs-j-s^ * 4r-f 45* e'-e . (e-iy ^^ (t-^2sy . <8-K2.s^ 13. 14. 15. a^m H- 10 am H- 21 y« a^m^ — 9 m^ tv^ — H , iv' + 2ic-{-A w^-\-7w-^10 ' w^-\-2w a^-5ab-Ub^ a^-3a6-286' a^-f 5a6-2462 * a2-8a64-156«' m^ — y^ . ^^ 4- ^^/ H- ^?/ mY — y* ' mif-irf Perform the indicated multiplications and divisions ^^ x' + 7xy + 10tf ^ x + y 18. 19. 20. x^ -\- Q xy -\- 5 y^ op^ -\- 4 xy -\- 4: y^ x-\-2y a:'-b^ ^ ab-2b\ (a- bf a2-3tt64-26=* * 62 + 06 'a{a-b)' r2-|-4rs-|-4.s2 * 4r^ -^ ^^ hr"^ ^lOrs am* — an* _^ m^ + n^ bm^ -\- bmn -\- bn^ bm^ — bn^ m^ — n- aw? + 2 amn + ai^ 180 ALGEBRA 136. Sometimes there are mixed expressions in an algebra problem. These mixed expressions should always be reduced to fractional form as in arithmetic. Example 1. Simplify 12f -^ 3\. Solution : 123 -.31=^-. 1^ = ^x1 = ^. * * 8 4 8 13 26 Example 2. Simplify (5 a' -19:. 4:X^ ^ ) \ a-2xj Solution : In the first parentheses, the fraction is to be subtracted from 5, and in the second, the fraction is to be subtracted from 3. (Use rule, § 133.) ■ V a2-4xV ■ V a -2x1 / 5(a2-4x2)-((f2_i9a;2) \ ^ / 3(a - 2a;)-(a - 5x) \ \ a^-4x^ ) ' \ a- 2x J / 5a2-20a;^-a-2 + 19x^ \ _^ /S a-6x- a-]-6x \ [ a^-^x^ j ' \ a-2x J W-Ax-^l ■ [a- 2x {a^^^){a + 2x) (2,0--^ a + 2x Check : Let a = 1 ; x = 1. g_a^-]9^^5_l-a9^5 18 a--2-4 x2 1-4 = 6-6: 3- ^~^^ = 3-J^^ = 3- — =3-4=-l, also, a-2x 1-2 2aH-x ^ 2 + 1 ^3^.^ a-{-2x 1 + 2 3 and (-l)H-(-l)=l; Note. In such examples, first perform carefully the indicated additions and subtractions loithin the parentheses and, afterwards, the multiplications and divisions. FRACTIONS 181 EXERCISIi: 84 Perform the indicated operations : 1. (3+t).(3 + A).. 7. (2-|)(J^,). 2. (5-i).(2-T%). fa' a\^f2a' a\ V yJ vf-^J 9, ... (.,._^),(._.-?). / 2m'^ + 7m-15 \ / 2 m'^ - 19 m + 42\ . A ■ [2m'-3m-Uj'\ 8m-12 y\ V^-jr -ua ary \ a; - vj \v xj 20 f ^^^'^ ^ ' r a;'^-3a; + 9 x' + x-6 ]_ ■ \x' + x-12j' [x'-^-'^x-S' 3a;-9 J* 3m+6 182 ALGEBRA COMPLEX FRACTIONS 137. A Complex Fraction is a fraction having one or more fractions in either or both of its terms ; as, (^^ «_\ d \a — b a + bj 138. A complex fraction is simply a case of division of fractions. The problems are very similar to those in the last paragraph. The numerator and denominator should be sim- plified separately and then the division performed. Example 1. Simplify d Solution : This means divide a by the result obtained by subtracting - from 6. d ad a _ a _ I ^ \ _c~ / bd-c \~^' \bd-c) d \ d J bd Check : Let a=:l, 6 = 2, c = l,d = l. g _ 1 _ 1 ^1 d 1 also, «^ ^ -^ = 1; bd-c 2 / a a \ Example 2. Simplify ^ — 7 r This means : find the difference of the fractions in the numerator, and the sum of the fractions in the denominator ; divide the first result by the second. The work may be arranged as follows : FRACTIONS 183 Solution : 1. — ; 4. = a — h a -\- b 2 _ a(a-\-b)— a(a-b) , -bCa + h)+ a(a- b) Ca-6)(a + &) (a-6)(a + 6) o _ a^-^ab — a^-^ ab ^ ab + b'^ -h a^ - ab ia-b)ia + b) ' (a-6)(a+6) 2 a6 . ga + ?>^ (a-?v)(a + ?>) ■ ia-b)(a + b) 5 _ 2 aft ^ (a-b)(a +b) _ 2ab (a - 6)(a + 6) a^ + b'^ a^ + b'^ ' Check : Let a = 2 ; 6 = 1. a a 2_2 4 a - & a + 6 1 3_3_4 3_4 J) ■ a 1,2 6 3 5 5 :-b a+b 13 3 * a2 + 62 4 ^. 1 5 EXERCISE 85 «^^ e. ^. 10. -^. 2. i^t_SL. 7. '' ar'-v' 1-i ?• 6 11- 8. 3 4 1+^ 6-* 2 771 12. a- 1 '4a P- 1 1 - 1 1> 1 - X 1 — ^ m or 184 ALGEBRA 13. ^^. 16. ^. 17. -^-JL 5 d a; a + 2 6 14. ^-^^ „^". 16. ^. 18. ^ " • 2/ a? X y X ,2^ n m — n 9 x^ — y^ n m-\-n ' -. 2(x+2 y) m 0?- m — n -^ 6 + 3a.' +2/ 17 a; + 22 a^-3i»-4 20. : 23. 5 2+ ^ a X — 4: a — x 21. " + f . . 24. \+ "^ ■ ■1,0 H g^ — go? a — 6 l-\-ax 2a— h . 2a+h 25. a4-3& CT-3& 2«-6 2ct + 6' a — 36 a + 3 h XL SIMPLE EQUATIONS— (Conf/ni/eJ) FRACTIONAL EQUATIONS 139. If the unknown number, or numbers, of an equation do not appear in the denominator of a fraction, the equation is called an Integral Equation ; as, 3 a; — 5 = 2 a; -f- 7. 140. If the unknown number, or numbers, of an equation do appear in the denominator of a fraction, the equation is called 3 2 a Fractional Equation ; as, 5 = - + 7. X X 141. A fractional equation may be changed into an integral equation by Clearing of Fractions. Example. Solve the equation — — ^- = 4h — -^ — ^ 4 5 10 Solution : 1. The lowest common multiple of 4, 5, and 10 is 20. Mul- tiply both members of the equation by 20. 2. /p . ('^^-^)-/p . (i^^^ = 20 . 4 + ;|. ^I^±il. 3. .-. 6(3x- l)-4(4x-6) =80 + 2(7ar + 6). 4. .-. loa;-5 - 16j; + 20 = 80 + 14a: + 10. 6. .-. 15 - a; = 90 + 14 x. 6. /. - 15 a; = 76. 7. D_i5: x=-5. Check: Does 3(- 5)- 1 _4(- 5W 5 ^ 7(- 5)+ 5 ^ 4 5 10 Does ml^ _ ^J^ = 4 4- ZL?P ? Does -4 + 5 = 4- 3? Yes. 4 5 10 18ft 186 ALGEBRA EXERCISE 86 Solve the following equations : 1 a^4-5 a^-fl _3 ^ ?r -f- 12 _ ?r — 9 2 4 10 2. -^"^ + 2 = •^^"^^ - 8. "'^ "^ "" — ""^ ~ '"^ = — 1. 11 + m, 10 — m _ ^ 9 2 7?l + ll 10-??i 6 3 ^-1, f-M4 6 3 3 9 4. l(a + 5)-4 = ^^^^. 10. 1+1?_5(._3)=4. 3^ ■ 4 4 7^^ 5. K^ + 2)-i(r-2)=2. 11. ^^{Uu+l)-\(i^-^) = l. c-2 c-4^2 J2 3(.T-1) ■ 5a?H-7 ^17/ 4 6 3* * lo" 3 6* 13. il±i_l(2Z>+3)=^i^. 2 o 4 14. ^-|(4n4-9)=l(5n + 8). 15. ^^-^(8^ + 3) = l(4^-3). 7a; — 8 7 x-\-6 _ x — r) 4a;-f9 14 4 a; " 2 7 a; * Solution: 1. Multiply both membei's by 28a:, the L. C. M. of the denominators. 8. .-. 2 a-C7 X - 8) - 7(7 x + 6) = 14 x(x - 5)- 4(4 x + 9). 4. /. 14 a;2 _ 16 x - 40 x - 42 = 14 x^ - 70 x - 16 x - 36. 6. ... _65x-42 =-86x-36. Q. .•..+ 21x=4 6. 7 . /,. _ 6 _ 2 Check the solution either by substitution or by going over the steps carefully. SIMPLE EQUATIONS 187 It is essential that roots of fractional equations be checked. For reasons that are given in a later course in algebra, whenever the apparent root makes the denominator of any fraction zero, that number is not a root at all. n. 6_U5. 22. a a o 18. A_A= 7. 23_ 6_ a 1 5 a 3* 5 4:m 2 3 m 7 48* 3 4a; + ^ = 15 X h X - ■2 4 2. 5 X X 12 X 2x-\-S 5x + 3 = 0. 5^4-4 ll«-2_3 2i ■ 6^ -•"• 3m -5 4 9 7^1 - 7 2 12 3 m 19. ^ + ^ = 151. 24. 30_1^9+ 7 ^0_ 7 21 a; 3x 5x-4. 10a; + 9 ^51 5 10 6 a;' r_3 1 3r-7 20. r=-_2. 25. 21. ±r_!l^^_^+3=0. 26. 2 r 3 2 r 2 5 2 27. Solve the equation = 0. 1 a;-2a; + 2a;2_4 Solution : 1. The L. C. M. of (x-2), (x+2), and (x^ - 4) is (a;2-4). Multiplying both members of the equation by x^ — 4, 3. .-. 2{x + 2) - 5(x - 2) - 2 = 0. 4. .-. 2 ic + 4 - 5 a; + 10 - 2 = 0. 5. .-. -3a; + 12 = 0. G. .♦. - 3 X = - 12, or, a; = 4. 28. -12. 3. 32. l^ = -i2_ x-2 1-x d-x 29. .^0_ = _9_. 33. 2x _4x+5 3 771 — 3 m — 5 3x— 4 6a;— 1 3a;— 4 30. i±5 + _l_ = 5. 34. Igj^'-Sa^-S^g x-3 a;-3 3x» + 6a; + 4 31. 5^ + 1^3^ X _ 35 6» + 5 ^ 2 _ 3a: 2a;-3 2a:-3 2ar'-2a! x'-l af^-l 188 ALGEBRA 3m 2m 2m^-15 36. 37. 38. 39. 2m + 3 2m-3 4m^ 9 2^1 3^-5 t-2 t-3 3 4 1 ,._2 2r-l r + 4 2(y-7) y-2 ?/ + 3^q 40 a + 2 2a-3_ 26 - a 41. 42. 43. a — 4 a + 3 a^ — a — 12 3w 4 ^1 2m-6 5m- 15 2* 3a 7 ^3 2a-5 3a + 12* 2:24-1 2^-1 9^4-17 44. 45. 22-16 2:^4-12 z'-2z-^S 2t + 7 3t — 5 ^ 17t + 7 6f-4 9^4-6 9«2_4' 3a;-2 ^ 36-4a; 24-3a; a;4-3- x' — 9 S — x ,' o 1 ^v 4.- 6x4-1 2a;-4 2aj-l 46. Solve the equation — — ^ -— = — ^ 15 7 a; -16 5 Solution: 1. Clear of fractions only partially at first by multiplying by the L. C. M. of 15 and 5. Mis: 6a^4-l- ^y^~^^ = 6x-3. 1 X — lb QQ /v. go 2. Transposing and uniting terms, 4 = - —- • 7 x — 16 3. M(7 x-16) : 28 a: - 64 = 30 a; - 60. 4. Completing the solution, x = ~2. Check it by substitution. SIMPLE EQUATIONS 189 47. 48. 49. 2a-l a+2 6a-5 2 2a + 5 6 y+ ^y 3^33/- ' _ 9 ?/ - -4: 9 -2 2a; 4-7 5a;-4 JB-f 6 14 3«-hl 7 ot + 1 4t-\-7 3f-2 5 5« + 8 3 EXERCISE 87 1. Divide 56 into two parts such that five eighths of the greater shall exceed seven twelfths of the less by 6. 2. If the base of a certain rectangle be increased by 2 feet, the altitude is equal to one third of tlie result. The perimeter of the rectangle is 36 feet. Find the base, altitude, and area of the rectangle. 3. A has $52 and B has $38. After giving B a certain sum, A has left only three sevenths as much money as B then has. AVhat sum was given to B ? 4. Divide 45 into two parts such that the sum of four ninths of the greater and two thirds of the smaller shall equal 24. 5. The denominator of a certain fraction exceeds the numerator by 27 ; if 9 be subtracted from both terms of the fraction, the value of the fraction becomes \. Find the fraction. 6. A's age is three eighths of B's, and 8 years ago it was two sevenths of B's age. Find their ages at present. 7. Washington was admitted to the Union 18 years before Oklahoma, and may therefore be said to be 18 years older than Oklahoma as a state. One fourth of Washington's age in 1911 exceeded Oklahoma's age by 1^ years. Find the year when each was admitted.' 190 ALGEBRA 8. If a certain number be diminished by 23, one fourth of the result is as much less than 37 as the number is greater than 56. Find the number. 9, If the number of states admitted to the Union since its formation by the 13 original states is diminished by 9, the quotient obtained by dividing that number by 2 equals the original number of states. Find the present number of states (1912). 10. The numerator of a certain fraction is 6 less than the denominator; if the denominator is increased by 1 and the resulting fraction be multiplied by 3, the product equals |. Find the fraction. 11. Find the angle such that 3 times its complement in- creased by two thirds of its supplement equals 137°. 12. I buy some bulbs from a seed store for $3, paying 75 p per dozen for one variety, and 50^ per dozen for- another variety. The number of the first variety purchased exceeds the number of the second variety by 18. Find the number of each variety purchased. 13. If a railroad train consists of a certain type of passen- ger engine, one parlor car, and five sleeping cars, its value is $ 129,200. The value of each sleeping car exceeds the value of the engine by $300; the value of the parlor car is five sixths of the remainder when the cost of the engine is dimin- ished by $ 100. Find the value of the engine, the parlor car, and of a sleeping car. 14. A man has $3000 invested, part at 5 % and part at 6 %. His total income per year is $157. How much has he in- vested at each rate? 15. In 1912, the "age" of Maine was 4y2-j- times that of Wyoming ; in 1920, it will be 3i times it. Find when each state was admitted to the Union. SIMPLE EQUATIONS 191 16. The denominator of a certain fraction is 7 less than the numerator ; if 5 be added to the numerator, the value of the fraction becomes f . Find the fraction. 17. The income at 5% on one sum of money exceeds by $ 35.50 the income at 4 % on a sum which is $ 350 Jess than the first. Find the two sums invested. 18. The denominator of a certain fraction exceeds the numerator by 5 ; if the denominator be decreased by 20, then the resulting fraction, increased by 1, is equal to twice the original fraction. Find the fraction. 19. The supplement of a certain angle divided by its com- plement gives as quotient 2^. Find the angle. 20. If twice a certain number be diminished by 5 and the result be divided by the number, the quotient exceeds 1 by a* fraction whose numerator is 7 more than the number and whose denominator is 3 less than the number. Find the number. 142. Work Problems. If a man can do a piece of work in 8 days, then in one day he can do one eighth of it, and in three days he can do thi'ee eighths of it. If a man can do a piece of work in x days, then in one day 1 15 he can do - part of it, and in 5 days he can do 5 x - or - X XX part of it. EXERCISE 88 1. If a man can plow a field in 15 days, what part can he plow in one day? in 4 days? in x days? 2. If a machine can do a piece of work in x days, how much can it do in one day ? in 7 days ? 3. A can do a piece of work in 5 days, and B the same work in 8 days. (a) How much can A do in one day? in x days? (6) How much can B do in one day ? in a; days ? 192 ALGEBRA (c) How much can they do together in one day ? in a; days ? (d) How much can A do in 2 days ? (e) How much can B do in 3 days ? (/) How much can they do together if A works 2 days and B 3 days? Equations 4. A can do a piece of work in 10 hours and B can do it in 5 hours ; how long will it take them to do it together? Solution : 1. Let x = the number of hours it will take them to do it together. 2. A does 4t in 1 hour ; .-. he will do — in aj hours. B does ^ in 1 hour ; .-. he will do - in x hours. 5 .-. they will do ( — + - ) in a; hours. VlO 5y 3. They complete the task in x hours. Represent the whole task by ^ of itself or by 1. 4. .-. -^ + ^ = 1. 10 5 5. Mio : a: + 2 a; = 10. 6. 3 a; = 10, a: = 3^ hours, or 3 hours and 20 minutes. 5. A painter can paint a house alone in 5 days, and an apprentice can do it alone in 15 days. In how many days can they do it if they work together? 6. A man can plow a certain field in 6 days; his son can plow it in 9 days. How long will it take them to plow the field if they work together ? 7. In a newspaper office there is one machine which can print the morning issue of the paper in 2 hours, and another which can do it in 3 hours. In how many hours can they turn out the edition, if they are run together? 8. A can do a piece of work in 15 hours, and B can do it in 18 hours. If A works for 7 hours, how many hours must B work to complete the task? SIMPLE EQUATIONS 193 9. A can do a piece of work in 15 hours, while B can do it in 25 hours. After A has worked a certain time, B com- pletes the work. If B works 9 hours longer than A, how long did A work? 10. A can do a piece of work in 18 days. If he and B can do three fifths of it in 6 days, how long will it take B alone to do the work ? 143, Problems about the Lever. A teeter board is one form of lever. The point which supports the lever is called the ful- crum; the parts of the lever to the right and left of the fulcrum are called the lever arms. If the weights L and R just balance, it is well known that, if R moves farther to the right, while L is stationary, then the right side goes down ; and if R moves toward the fulcrum, then h ^ the right side goes up. Thus, the influence of R upon the lever depends both upon the weight of R and its distance from the fulcrum. The influence of a weight upon a lever is called its leverage. It can be shown that the leverage of a weight is measured by the product of the weight and its distance from the fulcrum. Thus, if R weighs 50 pounds and is 4 feet from the fulcrum, its lever- age is 200 ; and, if L weighs 80 pounds and is 2^ feet from the fulcrum, its leverage is 80 x 2^ or 200 also. The two wiU balance. This truth may be tested in the following manner. 1. Remove a side and an end from a crayon box. Balance a stiff ruler on the edge of the box (the fulcrum). 2. Place two pennies 6 inches from the ful- crum on the left side. Find where four pen- nies must be placed on the right side of the fulcrum to balance them. (It should be 3 inches to the right.) Notice that 3x4=12 and that 6x2 = 12,- that the leverages are equal. 3. Find where 3 pennies must be placed, on- the right, to balance the 2 pennies on the left. Find the leverage of the 3 pennies and compare it with the leverages in step 2. 194 ALGEBRA 4. Keeping the two pennies C inches to the left, place two others 2 inches to the right, and 2 four inches to the right. The ruler should be in perfect balance again. The leverage of the first two pennies on the right is 2 x 2 or 4 ; of the second two, is 2 x 4 or 8 ; 4 + 8 — 12. Example. Suppose that the weights and distances in the figure are : A. 40 pounds ; distance 5 feet. B. 45 pounds ; distance 4 feet. C. 55 pounds ; distance 6 feet. D. 60 pounds ; distance 6 feet. E. 50 pounds ; distance x feet. Where must E be placed so that the lever will balance ? CAB D E Solution : 1. The leverages are : A. 6 X 40 = 200. C. 6 X 55 = 330. J?. 50 x a; = 50 x. B. 4 X 45 = 180. D. 6 x 60 = 360. A B C D U 2. . •. 200 + 180 + 330 = 360 + 50 x. .-.350 = 50 a; .-. 7 = x. Rule. — To make a simple lever balance : The sum of the leverages of all weights (forces) on one side of the fulcrum must equal the sum of the leverages of all weights (forces) on the other side of the fulcrum. EXERCISE 89 1. A boy weighing 70 pounds sits 6 feet from the fulcrum and balances a boy who is sitting 3^ feet from the fulcrum on the other side. Find the weight of the second boy. 2. A, weighing 96 pounds, sits 5^ feet to the left of the fulcrum. If B weighs 66 pounds, where must he sit on the right in order to balance A ? SIMPLE EQUATIONS 195 3. A, who weighs 92 pounds, and B, who weighs 115 pounds, wish to sit at the ends of a teeter board wliich is 9 feet long. How far from A must the fulcrum be placed so that they will balance ? 4. A boy wishes to carry two heavy packages over his shoulder by balancing them at the ends of a stiff rod which is 4 feet long. If one package weighs 20 pounds and the other 80 pounds, how far from the end upon which the 20 pound package is carried must the rod rest upon his shoulder ? 5. Three children, weighing G2, 75, and 89 pounds respec- tively, arrange themselves upon a teeter board, the first sitting 4 feet from the fulcrum, and the second 5 feet from the ful- crum on the same side. Where must the third sit in order to balance the other two ? 6. Three boys, A weighing 73 pound;, B weighing 95 pounds, and C weighing 6o pounds, sit on a teeter. A is 5 feet from the fulcrum on the left side, B is on the other side 4 feet away, C is on the left side 4 feet away. Can two other boys weighing 80 pounds and 115 pounds respectively arrange them- selves one on each side and at equal distances from the ful- crum so as to balance the teeter ? Where must the boy weighing 80 pounds sit ? 144. Additional Distance, Rate, and Time Problems. EXERCISE 90 1. What is meant by "the rate"? "the time"? "the dis- tance " ? (§83.) 2. Give a simple arithmetical problem involving time, rate, and distance. 3. What is the rule for finding the (a) distance when the rate and time are known ? (6) rate when the time and distance are known ? (c) time when the rate and distance are known? 196 ALGEBRA 4. The rate of one train exceeds that of another by 5 miles per hour. Let r represent the rate of the slow train. (a) Express the rate of the faster train. (p) Express the time required by each train in going 100 miles. 5. The rate of one train is f that of another. Let x repre- sent the rate of the faster train. (a) Express the time each requires for a trip of 50 miles. (6) Form an equation to express the fact that the time of the slow train exceeds that of the faster train by 1 hour. 6. The time required by one train in going a certain dis- tance is f that of another train. Express the time of the slow train by t. (a) Express the rate at which each train travels in going a distance of 100 miles in the time mentioned. (b) Form an equation to express the fact that the rate of the faster train exceeds the rate of the slow train by 20 miles per hour. Equations 7. The rate of an express train is three times that of a slow train. It covers 180 miles in 8 hours less time than the slow train. Find the rate of each train. 8. A messenger starts out to deliver a package to a point 24 miles distant, at the rate of 8 miles per hour. At what rate must a second messenger travel to arrive at the same time as the first messenger, if he starts 1 hour after him ? 9. The rate of a passenger train exceeds twice the rate of a freight train by 5 miles per hour. It can go 350 miles while the freight train goes 150 miles. Find the rate of each train. 10. A man just missed a train. He knew that it would stop at a station 15 miles distant from the central station and SIMPLE EQUATIONS 197 decided to try to catch it by going to the second station in an automobile. If the train runs at the rate of 20 miles per hour, at what rate must he travel in the automobile in order to arrive at the station 10 minutes ahead of the train, if it takes him 5 minutes to get the automobile and if it is 20 miles to the station by road ? River Problems On a river, the direction in which the water is flowing is called down- stream, and the opposite direction is called upstream. When going downstream, a boat is carried along by the current of the river and what- ever force is exerted within the boat ; when coming upstream, its progress is retarded by the current of the river. This is something like the effect of the wind upon a person who is walking ; when going with the wind, he is carried along by it, and when going against the wind, he is retarded by it. 11. The rate of the current of a river is 3 miles per hour: (a) at what rate will some boys go downstream if their own rowing is ' at the rate of 5 miles per hour in still water ? (h) upstream ? 12. How long will it take the boys in Example 11 : (a) to go 24 miles downstream ? (b) 24 miles upstream ? (c) for the trip down and back ? 13. (a) How long will it take the same boys to go d miles downstream ? (h) d miles upstream ? (c) Form an equation to express the fact that the time down and back is 5 hours, (d) Find d from the equation in step (c). 14. Some boys who can row 4 miles an hour in still water made a trip on a river whose current is 2 miles an hour. If it took them 8 hours for the trip, how far did they go ? 15. A party take a trip in a motor boat which runs at the rate of 15 miles an hour. They take 3 hours for the trip. What distance did they go, if the rate of the current is 3 miles an hour ? 198 ALGEBRA EXERCISE 91 Supplementary Problems 1. Three times the difference between one fourth and one tenth of a certain number exceeds five by one fifth of the number. Find the number. 2. A's age 11 years from now divided by his age 11 years ago is the fraction -i/-- Find his present age. 3. The width of a room is three fifths of its length; if 12 feet be added to the width and taken from the length, the room will be a square. Find its dimensions. 4. Five lines radiate from a point making angles such that the second is one half of the first, the third is twice the first, the fourth is the sum of the second and third, and the fifth is three times the third. Find the angles. (See § 13.) 5. Find the three angles of a triangle if the second angle is one half of the remainder obtained by diminishing the first angle by 1°, and if the third angle is | of the remainder ob- tained by diminishing the first angle by 7°. (See § 13.) 6. Two men, A and B, 57 miles apart, travel towards each other, B starting 20 minutes after A. A travels at the rate of 6 miles an hour and B at the rate of 5 miles an hour. How far will each have traveled when they meet ? 7. Washington's Monument, the highest piece of masonry in the world, consists of a main shaft surmounted by a pyra- mid. The height of the main shaft exceeds that of the pyramid by 445 feet ; if the height of the pyramid be decreased by 5 feet and the result be divided by the height of the main shaft, the quotient is the fraction -^. Find the height of the two parts of the monument. 8. The rate of an express train is five thirds of that of a slow train. It travels 36 miles in 32 minutes less time than the slow train. Find the rate of each train. SIMPLE EQUATIONS 199 9. If the third of three consecutive even numbers is divided by each of the first two in turn, the difference of the fractions obtained is equal to the quotient of 7 divided by three times the first number. Find the numbers. 10. A workman does one third of a piece of work in 5 days ; he and a second workman complete the task together in 4 days. How manys days would it take the second man to do the work alone ? 11. Some boys row on a river whose current is known to be 2^ miles per hour. They find that it takes them as long to go upstream 2 miles as downstream 7 miles. What is their rate of rowing ? 12. A's age exceeds twice B's age by 7 years. The sum of three fifths of B's age 2 years ago and of four sevenths of A's age 4 years from now is 26 years. Find their present ages. 13. The income on one sum of money at 4|^ % and the in- come on a sum $ 600 greater at 3^ % together amount to $ 421 per year. Find the total amount invested. Hint: H%=n = 'Eh- 14. The rate of an express train is three halves that of a slow train. It covers 270 miles in 3 hours less time than the slow train. Find the rate of the train. 15. A water reservoir can be filled by an old pump in 12 hours. After a new one is installed, it is found that the reservoir can be filled by the two pumps together in 4 hours. How long would it take the new pump alone to fill the reservoir ? 16. A man invests a sum of money in 4^% stock and a sum % 180 greater than the first in 3^ % stock. If the incomes from the two investments are equal, find the sums invested. 17. Some boys are rowing on a river whose current is 5 miles per hour in one stretch and 3 miles in another. They find when going downstream that they can go 4 miles where 200 ALGEBRA the current is rapid in the same time that they can go 3 miles where the current is slower. Find the rate at which they row in still water ? The following three equations arise in the solution of three problems in applied mathematics. Solve them. 18. S0T-^gT'=:S0(T-2)^^g(T-2y,wheveg = S2. De- termine T. 19. ^Jil:! = 13.6 + ^^:^ . Get the result to two decimals. X X 20. ^^ + 'M^ = xv. Determine a.. SOLUTION OF LITERAL EQUATIONS 145. A Literal Equation is one in which some or all of the known numbers are represented by letters. Example 1. 2x-\-a=:7a — x. The problem is to determine a number x, such that the equation is satisfied for all values of a. Solution : 1. 2 x -{- a — 1 a — x. 2. Transposing, Sx = 6 a. 8. Dg:. x = 2a. Check : Does 2(2 a) -\-a = 1a — 2a? Does 4a -\- a = 7 a — 2a? Yes. This solution means that the number x is always double the number a. Example 2. Determine the number x in the equation (p _ cxf^ (a - cxy= b(b - a). Solution : 1. (b — cx)^ — (a — cxy = b(b — a). 2. Expanding, (62 -2bcx-\- c%2)_ (^2 _2acx + d^x^) =b^- ah. 3. .-. 62 _ 2 hex + c2a;2 _ a2 + 2 acx - cH^ - b^ - ab. 4. Transposing and uniting terms, 5. Factoring, 6- I>2c(a-6): 2 acx — 2bcx = = ««- ah. 2cx{a -6) = -a(a - -6). 2cx{a -&)- a(a -h) 2c{a- -6) 2c(a -h) .*. x = a SIMPLE EQUATIONS 201 Check: Does (b - — Y - (a -^Y = b{b - a) ? Does 62 _ a6 ^. ^ _ rt-i 4. a-i _ ^ = b^-ab ? Yes. 4 4 Note. After simplifying the eiiuation until it takes the form of the • ■<}uation in step 5, divide both members of the resulting equation by the coefficient of x. EXERCISE 92 Solve the following equations for x : „, s ^^x^ — bb — x2xb 1. 3x-5a = 2(a-hx). 12. = ^ ' ax a a X 2. 5(x-3b)==3x-llb. (n'mx\ f , n«\ 3. a{3bx-2a)=b{2a-3bx). ^^' ^*'V2"^ VJ"'^!,'^'"^ 4 j 4. 5rx— 6 5 = 3(ric — s). 5. ax — ac=^bx — be. 6. ax — V? = —b{b-\-x). 15. 7. T{x — r)-= s (s -\-2 r — x). X — " 1 2x 3. X x—a 3 X- -4 5m — 271 3a; + 4 5m + 2n 2a;-f 3a ^ 3a;-f 46 8. c{x-(^ = d(d^-x). • 2x-3b Sx-4.a' 9. x-l-^-=^ = — . 17 a;4-n a;-n ^ 2(m+ny w. m^ * a; — m a;+??i x^—ni^ ,. m2 1_1 7i2 2na;-3 - 9nx-{-2 nx m n mx nx—1 3nx — \ a — x _ 10b — a 1 iq ^ ^ a^ — b^ XX. — — ~~ — — — ~~ — — — • At/. bbx 16 ax 3 a x — a x — b b(x-~b) 20 ^^ ~ ^ I hx-\-a _2 I <^^ — ^ , 6a; ttic abx 2j 10a;-3a ^ 9a;-5a» 3(a;-f-2a^ 3 a 3 a" a« ' 22 » _ a — 2 6ca; _ 5_x _ S ac — Sbx — 9 a 2 46c ~6c i26c 202 ALGEBRA 23. 24. 25. x — 2 a x-\-2 a ar' — 4a^ 3x(a — b) a — 2b_a — b a^ — 6^ x + b x—b 2 X 3a 2ar^4-aa^ 3a; — a a; + 2a 3ar^ + 5aa; — 2a^ 146. Applied Algebra. Some of the equation problems given up to this point illustrate one use of algebra; an unknown number may sometimes be found by means of an equation if numerical relations involving the number are known. In order to solve the equations that arise, skill in the funda- mental operations (§ 16), factoring, and fractions, is necessary. In § 17 and § 94, another application of algebra is illus- trated ; a formula is often used to express in convenient form a rule of computation. In deriving formulce and in using them, all of the algebra so far studied, and much more, fre- quently, is necessary. 147. Deriving and Using Formulae. Three examples of the methods of deriving formulae will be given, besides those in §17. Example 1. Derive a formula for the total area of the walls of a room in terms of the height, length, and width. Solution : 1. Let h = the height of the room in feet. Let w = the width of the room in feet. Let I = the length of the room in feet. Let S — the total area of the walls in square feet. 2. Then, hw = the area of one end wall in square feet. hw =z the area of the other end wall. hi — the area of one side wall. hi — the area of the other side wall. 3. .'. S^hvo^-hw^-hl^rhl ^Ihw^'l hi. .'. 8=2h{w + l). Thus, if the height is 9 feet, the width 14 feet, and the length 18 feet, the total area of the walls is /S = 2 X 9 X (14 + 18) = 18(32) = 676 square feet. SIMPLE EQUATIONS 203 Example 2. Derive formulie for two numbers whose sum is a and such that the larger exceeds the smaller by h. Solution : Let s = the smaller number. Then, s -\- b = the larger number. 2. .-. s + (s + 6) = a, or 2 s + 6 = a. .-. 2s = a-6, ors = ^^:i^. 2 The smaller number is ^ ~ and the larger number is ^^-^t — 25 — 7 18 Thus, if a is 25 and 6 is 7, the smaller number is or — , or 9, 2 2' and the larger is — ^ , or 16. id , 9 + 16 = 25 and 16 = 9 + 7. Example 3. Derive a formula for the rote in terms of the < I mount, the principal, and the time. Solution : 1. The fomiula for the amount (§ 44) is A = P + ^. 100 2. Mioo : 100 ^ = 100 P + Prt. 3. Sioop: 100 ^ - 100 P = Prt = {Pt)r. 4 100^-lOOP^ ^^,.^100U-P). Pt ' I't This formula enables us to find r when A,, P, and t are known. Thus, if a man receives §3500 at the end of 6 years from an investment of $ 2400, what rate of simple interest has his money earned ? Here, A = % 3600, P=$ 2400, and « = 6. . ^ ^ 1^(3500 - 2400) ^ 1100 ^ 7 g^ . -8406- X 6 144 ' ' 24 that is, the rate is 7 .6 %. In the following list of examples, a number of formulae, taken from physics, chemistry, geometry, and engineering, are given. It is impossible to show in this text how these formulae are derived, as that calls for special knowledge of these various subjects. 204 * ALGEBRA EXERCISE 93 (a) Express each of the first 6 formulae in words. (6) Solve each of them for the letters indicated. 1. ^ = ab. (a) Solve for a; (b) for b. (§ 17) 2. A=^' (a) Solve for a; (b) for 6. (§17) 3. (7 = 27rr. Solve forn (§17) 4. F= Iwh. (a) Solve for w ; (b) for h. (§ 17) 5. F= i bh. (a) Solve for 6; (6) for 7i. (§ 17 j 6. i^ = |(7+32. Solve for 0. (§84) 7. ^ = ^ ^ is a formula from geometry. (a) Find ^ when a = 15 ; ?> = 24 ; and c = 20. (6) Find c when ^ = 550 ; 6 = 30 ; and a = 22. (c) Solve the formula for a ; (c?) for 6. 8. A = F+?n. 100 (a) Solve for P. (5) for ^. 9. r= - + ^ is a formula from physics. a (a) Solve it for t. (b) for a. 10. mg — T=mfis 3. formula from physics, (a) Solve it for T; (b) for/; (c) for m. 11. u = ( — — — ]v is a formula from physics. V3/H- ml (a) Solve it for m; (&) for iJf. O = — is a formula from b — a (a) Solve it for a; (b) for b. 12. O = ^^ is a formula from physics. b — a SIMPLE EQUATIONS 205 13. s = ^^ — is a formula from physics. b — a (a) Solve it for 6; (b) for a. 14. h = k(l 4- ^Ij t) is a formula from physics, (a) Solve it for i; (b) for k. T JIf 15. I = is a formula from physics. Mt (a) Solve it for L; (b) for M. 16. w = j5 — is a formula from engineering. 1 -f — — 600 fZ^ (a) Find u^ correct to two decimals, when a = 18, 1=12, and cZ = 10. (6) Solve the formula for a. 17. p = — + d is a formula from engineering. V (a) Find p when a = .56, cZ = ^, ^ = f . (b) Solve the formula for t. E 18. O = is a formula from physics. R + - n (a) Simplify it in the right member. (b) Solve it for n ; (c) for r. 2 19. F= is a formula from physics. (a) Find F when m = 150, v = 25, f/ = 32 and r = 5. (6) Solve it for r ; (c) for m. 20. - = - + - is a formula from physics. f P -1 ^_ . _ __j_ __:(10,1). Draw the lines AB, BC, CD, and AD. What figure is formed ? Draw the diagonals of the figure, and find the co- ordinates of the point where they meet. 6. Locate the points ^ : (3, - 5) ; J5 : (4, - 2) ; C: (7, - 2) D : (S, - 5) ', E : (7, - S) : F: (4, - 8). Draw the lines join- ing the points in order. What figure is formed ? 152. The numbers which are represented graphically in mathematics are usually connected by an equation. Example. Draw the graph of the equation 2^4 Solution : 1. Solve the equation for y. M4 : 2x + y = i. S2x: y=4-2x. 2. Select any value of x and find the corresponding value of y. Thus, if X = 1, y = 4 — 2 = 2. Similarly : when X = - 5 - 8 -1 + 2 + 4 + 6 + 7 then y = + 14 + 10 + + 4 - 4 -8 -10 .3. Use the pairs of numbers so obtained as coordinates of points ; thus, locate the points (-5, + 14) ; (—3, + 10) ; (-1, + 6) ; etc. 4. Draw the line connecting the points. (Graph on page 214.) Notice that the graph seems to be a straight line. The coordinates of the point A on the graph are +3.5 and — 3. Do these satisfy the equa- tion ? Does M + :i3 = i? 2 4 Does 1.75+(- .75) = 1? Yes. 214 ALGEBRA " ■■ n -r- "" ~ "1 ~ ~ -1 ~ Y ~ — ~ - - — n "l 1—] n ^ \ .,. s ,..,. \ \ s r 1 - ^ s s i \ \ r ' b^ X X - ^ . J ■6 . -4 ". .' ?" . ' -7 k s - T Ff T jT _ 3_ —Q - ■ s A* S s s V' s s s * Y 1 Fig. 6 EXERCISE 97 1. Determine the coordinates of B, C, and D. Determine whether they satisfy the equation of the graph ? 2. Select any point not on the graph, find its coordinates, and determine whether they satisfy the equation of the graph. 3. Similarly, draw the graph of y = x-\-S. For x select the values -1, - 2, - 3, - 4, + 1, + 2; -f 3, + 4, 0. What does the graph appear to be ? 4. Select three new points which are on the resulting graph, find their coordinates, and determine whether they satisfy the equation given in Example 3. 5. Draw the graph of ^ 1. Select at least four negative and four positive values of x, and from them determine the corresponding values of y. What sort of graph do you obtain ? GRAPHICAL REPRESENTATION 215 6. Select any three points on the graph and determine whether their coordinates satisfy the equation. 153. The equations in the preceding paragraph have each had tn:o unknown numbers. The following facts are to be remem- bered from the examples of § 152 : 1. The graph of an equation of the first degree (§ 77) having two unknowns is a straight line. For this reason, first degree equations are also called linear equations. 2. The coordinates of every point on the line satisfy the equation. 3. The coordinates of every point not on the line do not satisfy the equation. 154. A Solution of an equation having two or more unknowns is a set of values of the unknowns which satisfy the equation. Be careful not to confuse the word "solution " in the sense of this para- graph with the same word when it is used to mark the process of solving an example as is done in the text. Example. Consider the equation x + y = 5. (1, 4) is a solution because 1+4 = 5. (— 8, + 18) is a solution because — 8 + 13 = 5. 155. Number of Solutions. There are an indefinitely large number of points upon a straight line. This is expressed by saying that there are an injinite number of points on a straight line. Since the coordinates of each point satisfy the equation of the line, then : there are an infinite number of solutions of a linear equation ivith two unknowns. This fact is evident also because for every value of one of the unknowns, a value of the other may be found. Example. Consider the equation 2x-\-3y=zl5, When x = 1, 2 . 1 + 3 2/ = 16 or 3 y = 13, 2/ = 4| ; then a- = 1, y = 4| is one solution. Similarly when, x=2; y = i^. x=-S; y = + 7. x = 3; y = 3. x=-5, y= + S\. 216 ALGEBRA Thus as X changes in value, in such an equation, y also changes in value, acquiring a new value, x and y are said to vary, and are called Variables. Hereafter these equations will be called equations having two variables. An equation of the first degree having two variables has an in- finite number of solutions. Such equations are called Indeterminate Equations. 166. The graph of an indeterminate linear equation with two variables is always a straight line by § 153. A straight line may be drawn with a ruler as soon as two of its points are known. This leads to the Rule. — To draw the graph of a linear equation having two var- iables : 1. Select one value for one variable and determine the correspon- ding value of the other variable ; this gives one solution. 2. Determine a second solution as in step 1. 3. Plot the two points whose coordinates are the pairs of numbers and connect them with a straight line. 4. Check the result by finding a third solution, and plotting the corresponding point. The third point should fall upon the graph ob- tained in step 3. Example. Draw the graph oiAx — Sy = 6. Solution : 1. Let x = ; then y =—2. (4.0~3y = 6; - 3 ?/ = 6 ; i/=-2.) 2. Let x = S', then y = 2. (4.8-3!/ = 6; -3?/ = 6-12; -Sy=-6; y = 2.) 3. Plot the points and draw the line. See Fig. 7. Check : Let x = Q ; then y = 6. Is this point on the line ? Note. To get the best results, it is necessary to use coordinate paper for the following exercises. Fair results may be obtained by using paper ruled by the pupil. GRAPHICAL REPRESENTATION 217 Fig. 7 EXERCISE 98 Draw the graphs of the following equations, each on a^ sepa- rate sheet of graph paper : 1. a; + 2/ = 5. 2. 2x — y = 6. 3. 2x-\-Sy = 6. 4. 3x-2y = 12, 5. 5 a? -f 4 2/ = 20. 6. Sx — 5y=15. 7. 7 a; 4- 4?/ = 2. 8. 5x — Sy = —6. 9. x-6y=-10. 10. Sx=5-7y. 157. Independent Equations and Simultaneous Equations. Example. Draw upon the same sheet the graph of 2x-y==4.. (1) 2x-\-Sy = 12. Solution : 1. For equation 1 : 2 x — y = 4. Ifx = 0, y=-4;ify = 0, 5c = -f-2. Solutions : (0, - 4) and (+ 2, 0). (2) 218 ALGEBRA 2. For equation 2 : 2x + Zy = l2. li X = 0, y = 4 ; if y = 0, x = 6. Solutions : (0, + 4) and ( + 6, 0). 3. See Fig. 8 for the graphs. ~ "" "~ ~ ~ ~ ~ " V n """ " "■ "■ "" "~ " 1 "" ( (1 ( 2) ^ J s . i ^i /n / 1 / ^01 J 1 <. / ■n 'sV' 1 / ^■f / ' f\ ^ ^s x' ~ 4i (5) : x = 2. (6) 5. Substitute this value of a; in (1) : 10 - 3 y = 19. (7) -3y = 9. (8) y = - 3. (9) 6. The common solution is x = 2, ?/ := — 3. Check : Substitute in (1) : 10 + 9 = 19. Substitute in (2) : 14 - 12 = 2. * See § 42 for the symbol J/4. Read this " multiply both members of equa- tion (1) by 4." SIMULTANEOUS LINEAR EQUATIONS 223 Notice : 1. That, in equations (1) and (2), the coeflScients of y are not the same. 2. That, by multiplications, the coefficients of y are made the same in absolute value in equations (3) and (4). 3. That, by addition, 12 y disappears in equation (5), and that 20 x and 21 a; are combined, giving 41 x. This is allowable because we assume that t and y represent the same numbers respectively in all of the equations, namely, the particular numbers which together form the common solution of equations (1) and (2). y is said to be Eliminated. Elimination means to cause to disappear ; thus y was made to disappear by adding the two equations (3) and (4). 4. That the remaining number, x, is then easily found. The solution is an example of Elimination by Addition. Example 2. Solve the equations j 1'^^ « + ^ 6 _ 1. (1) (2) Solution : 1. Mj (1) : 30 a + 16 ft = 2. (3) 2. Ma (2): 30 a - 21 6 = - 72. (4) 3. Subtract (4) from (3) : 37 6 = 74. (5) 4. D37 (5) : 6=2. (6) 6. Substitute the value of 6 in (1) : 16 a + 16 = 1. (7) .-. 15 a = -15. (8) .-. a=-l. (9) The solution is: a = — l;?) = +2. Substitute the values of a and h in equations (1) and (2). The solution is an example of Elimination by Subtraction. Rule. — To solve two simultaneous linear equations having two variables by the addition or subtraction method of elimination : 1 . Multiply, if necessary, both the first and second equations by such numbers as will make the coefficients of one of the variables of equal absolute value. 2. If the coefficients have the same sign, subtract one equation from the other ; if they have opposite signs, add the equations. 3. Solve the equation resulting from step 2 for the other variable. 4. Substitute the value of the variable found in step 3 in any equa- tion containing both variables, and solve for the remaining variable. 5. Check the solution by substituting it in both of the original equations. 224 ALGEBRA Historical Note. Little progress was made in solving linear equa- tions having more than one unknown until the latter part of the 15th cen- tury, although mathematicians before that time had considered such problems. After Stifel and Stevin had introduced somewhat simple no- tations for several unknowns and their powers, definite methods for sol- ving equations of the first degree with two unknowns were developed. Johannes Buteo, a French monk, (1492-1572), solved equations with three unknowns in a text on algebra which appeared in 1559. EXERCISE 101 Solve by the method of addition or subtraction : 1. 3. 5. 6. 7. 8. 9. 10. 11. i \Sx-\-y = ll. [5x-y = lS. [5a-hb = 7. |r- 6 s= -10. \2r-7s=-15. f5 m-l- 4 71 = 22. 13 m + n = 9. [7c-2d! = 31. 4c-3d = 27. 5a + 3b=-9. 3a_46=-17. 6x-h2y=-3. 5x — 3y= — 6. |3s-|-7^ = 4. [7 s-\-St = 26. [Sp-6q = 10. {5r-9y = l, i8r-102/=-5. f8c-h9^ = ^. 12. 13. 14. 15. 16. 17. 18. 19. 20. l8c-9^ = 77. 6x 15 ~lly=-4 x-^4.y = 53. 3x + 7y = 2. .7x + 82/=-2. 4.V 14 + 15(/ = 7. ?; 4-6^ = 9. |28 x-3Qy=:l. 114 a; + 15 2/=6. i7x + 92/ = 8. 19 a; - 8 2/ = 69. 2x -f2/ = 13. X y_U, [3 5 15 \i' ^2 = .. s 4" 1-*- fl^^ . + ^ = -1-. m .2 ^^_1 5 - 7 -Itt. ^ ■I='- 2a [ 9 «=!■ 4. SIMULTANEOUS LINEAR EQUATIONS 225 168. Elimination by Substitution is a second common method of solving simultaneous equations. Example. Solve the equations I „ ^ ~ ^ ^ ~~ ^' ^J Solution : 1. Solve (1) for x in terms of y : 7x = 15 + 9i/;x = (l^i^V (3) 2. Substitute this value of a; iu (2) : 8y-5( ^^ + ^n = -I7. (4) 3. M7(4): 66y- 5(15 + 9y) = - 119. (5) 4. Expanding: 5(3 ?/ - 75- 45y =- 119. (6) 5. Combining : 11 y - 75 =- 119. (7) 6. A75: 11 2/ =-44. 7. Dn: 2/ =-4. 8. Substituting the value of y in (3) : a.^ l5 + 9(-4) _ 15-36 ^ -21^ 3 7 77' 9. The solution is : x = — S, y = —4. Check it by substitution in equa- tions (1) and (2). Note. In step 3, when multiplying 5f ^' ' ^Y '^> ^^"^ obtains 7 . 5(15 + 0y) . ^jjg ^.g cancel, giving the result 5(15 + 9 1/) . Rule.— To solve two simultaneous linear equations having two variables by the substitution method of elimination : 1. Solve one equation for one variable in terms of the other vari- able. 2. Substitute for this variable in the other equation the value found for it in step 1. 3. Solve the equation resulting in step 2 for the second variable. 4. Substitute the value of the second variable, obtained in step 3, in any equation containing both variables and solve for the first variable. 6. Check the solution by substituting it in the original equations. 226 ALGEBRA Historical Note. The earliest use of the Substitution Method of Elimination in print, of which we have any record, is in Newton's Arith- metica Universalis, in 1707. EXERCISE 102 Solve by the substitution method : l4?/i + w = 16. ' ll2p-hl0g = -5. 2 |r-6s = 2. ^^ j^6w-9x = 19. [3s-Sr==29. ' ll5w+7a; = -41. g \5x-\-Sy = + 2. ^3 |15^-125 = -54. [10 a.' -12 2/ = +32. ' 110^+9jB = -2. |7a.--42/ = -19. ' l9Jf-6iV^=57. 16aj-3?/ = 10. ' l4p + 6^ = + 7. g |8i> + 5g = 5. |6a;-102/ = 5. I3p-2g = 29. ' ll5.v-14a; = -15. „ {5m-12n = -31. ^^ |9c + 8d5 = -6. l3m + 22w = -4. * [12c + 10d = -7. g |2a^-32/ = -14. ^g f3e + 7/=-23. l3aj + 7^ = 48. ' t5e + 4/=-23. ^ |5a.- + 92/ = 8. ^g_ J7^ + 8A: = -10. [6 2/ -9a; = -7. ill ^ + 6 A; = - 19. {7r-6f = 63. |5r-8.s = 60. I9r + 2^ = 13. ' i6r + 7s = -ll. 169. As a rule, the equations containing two variables do not occur in as simple form as those given in §§ 167, 168. Example. Solve the equations \— ^=0. (1) [x(y-2)-y(x-5)=-13. (2) SIMULTANEOUS LINEAR EQUATIONS 227 Solution : 1. Simplify equation (1) 7 3 = 0. x-\-S y + 4 .-. 7(y + 4)- -Six + 3) = = 0. 3x -7y = = 19. 2. Simplify equation (2) : x(y-2)~ -y(x -5) = :-13. 2x -6y = = +13. (8) (4) 3. The equations (1) and (2) thus become in (3) and (4) : 3x-7y = 19. (3) 2a;-5y = 13. (4) Solve these equations by either of the two methods of elimination ; the solution will be found to be 0- = 4, y =— 1. Check : Substitute the values of x and y in equations (1) and (2). Equations (3) and (4) are called the Standard Form of equations (1) and (2). Rule. — To solve complicated simultaneous linear equations hav- ing two variables : 1. Reduce the equations to the standard form by clearing of frac- tions and simplifying. 2. Solve the resulting equations by either of the two methods. 3. Check by substituting in the original equations. EXERCISE 103 Reduce to the standard form and solve : \2x 3 -'-i = - -7 2 X A 5 11 2 • \Sp + 7g = 12. 1. 2. \p±2q + ^=1. 4. 10 m 8 2/ 11. m + 3 = -17. 3 2 3 + 2^ l-j-5s 5 11 = 4-2. 228. ALGEBRA 5. 6. 7. 8. i 3 "^ 4 6 5^-8?^ = 22 9. 1 ' +..^ = 0. m - 2 2n-3 1 +,,^=0. .2 7/1 + 5 3n-7 r + ^ r — ^ 2 3 10. 3 ■ 4 c + cZ-2 ^ 1 c ~d 3 3c + d-3 re + a - 5 1- a- 6 2a-|-36 = 2d-G 11 { a? 4- 5 y 2 y + a; 13 11 3x~7j = 2. ' +^. = 0. i» — 1 2/ — 1 5 7 12. = 0. .2 x - 3 2 y + 13 |(m + l)(7 ' 12 m + 3^1 + 9 = 0, r — s 25 r + s 2/ a;_„ 3"2-^- 2x 4 + 5?/ 11 13. J (''' + ^^ ("^ + ^) ~ ^'"^ + ^) (^ - '^) =112. 14. < 15. 2 6 3 r + s — 9 s — r — 6 = 0. w — 2 _ 10 -i(? _ p-10 ^ Q 5 3 4 ~ ' 16. 17. j) + 2 2w+p ID + 13 6 32 16 x-y 2^+y ^Q 3 2 a; + 2?/_a;__ll 2 4~~T* 4a + & 6a-36 0. 5 3 8a + 5& 10 a -6 = -3. = -4. SIMULTANEOUS LINEAR EQUATIONS 229 170. Certain equations in which the variables occur in the denominators of fractious may be solved readily without clearing the equations of fractions. Example. Solve the equations ' X y Vx y (1) (2) Solution : Eliminate the term containing y : 1. Ms (1): ^_ 1^ = 40. X y (3) 2. M,(2): 24^46^_3 X y (4) 3. Add (4) and (3): ^ = 37. (6) .-. 74 = 37x or x = 2. (6) 4. Substitute 2 for X in (1): 6 -- = 8. y (7) 5. Solve (7) fory: y=-3. Check : Substitute x = 2, y = - 3 in equations (1) and (2). EXERCISE 104 Solve the following sets of equations : 1. 2. a 10 _ b :-l. 5+ a 15 _ h = + 1. 10 c 9 d :4. 8_ c l'>_ d 9 2* =-11. 21 = — . 2 fi+i= 5. iC y 1 1^ 1. — — .x y 3_ i_ 9. r « 4 + 3^ -1. Vr s 6 , 4 4 + - w v "5* 9 . 5 _ 7 + - .to i; ~10 230 7. < 8. < ALGEBRA '2 3 . A B 5 6_ 1 .A B 2* 9. ^ 5 7_29 3x y 9 ' 3 5 9 [x^Ay 8* '^-5 = 17. r t 5 6 3 .r^t 2* 10. - r 5 4 1 2 a; 3 2/ 2 2 17 13 a; 2y~72 171. In solving problems where two or more letters are used to represent unknown numbers, as many independent equations must be obtained from the conditions of the problem as there are letters used. Example 1. Four seventeenths of the greater of two num- bers exceeds the less number by 5 ; if the greater be divided by the less, the quotient is 5 and the remainder 10. Find the two numbers. Solution : 1. Let g = the greater number, and I = the less number. 2. Then, ^\g = l+^. (1) 3. When the greater is divided by the less, the quotient is 5 and the remainder 10 ; therefore, ^ = 5 Z + 10. (2) 4. Solving equations (1) and (2), g = 85, I = 15. Check : j\ of 85 = 20 ; 20 - 15 = 5, and 85 = 5 • 15 + 10. Note. In equation (2) , use is made of the fact that the dividend equals the divisor times the quotient plus the remainder. Thus, when 17 is divided by 2, 17 = 8-2+1. Example 2. If 3 be added to both numerator and denomi- nator of a fraction, its value becomes f ; and if 2 be subtracted from both numerator and denominator of the fraction its value becomes ^. Kequired the fraction. SIMULTANEOUS LINEAR EQUATIONS 231 Solution : 1. Let n = the numerator, and d = the denominator. /. - = the fraction. d 2. By the first condition : !i±l = ^. (1) ^ d + 3 3 ^ ^ By the second condition : ^ ~ = - • (2) d — 2 2 3. Solving the equations (1) and (2), « = 7, d = 12. Therefore the fraction is /tj. Check: • l+± = }^ =^- ■ J^l = l =1. 12 +J^ 15 3 12-2 10 2 Note. Check the solution by going back to the conditions of the problem. EXERCISE 105 1. Divide 59 into two parts such that two thirds of the less shall be less by 4 than four sevenths of the greater. 2. Find two numbers such that two fifths of the greater ex^ ceeds one half of the less by 2, and four thirds of the less exceeds three fourths of the greater by 1. 3. If 5 be added to the numerator of a certain fraction, the value of the fraction becomes f ; and if 6 be subtracted from its denominator, the value of the fraction becomes f . Find the fraction. 4. If 9 be added to both terms of a fraction, its value be- comes 4 ; and if 7 be subtracted from both terms of the frac- tion, its value becomes |. Find the fraction. 5. In 1910, the cost of 3 tons of anthracite coal in Phila- delphia exceeded the cost of 4 tons of bituminous coal in Bal- timore by $3.10 ; and the cost of 9 tons of the bituminous coal exceeded the cost of 5 tons of the anthracite by 90^. Find the cost of the anthracite and of the bituminous coal in 1910. 6. A's age is three fifths of B's age ; but in 16 years A's age will be five sevenths of B's age. Find their ages at present. 232 ALGEBRA 7. If twice the greater of two numbers be divided by the less, the quotient is 3 and the remainder is 7 ; if five times the less be divided by the greater, the quotient is 2 and the re- mainder is 23. Find the numbers. 8. On the tower of the City Hall of Philadelphia is a statue of William Penn. If the total height of the tower and statue be divided by the height of the statue, the quotient is 14 and the remainder is 29 ; the height of the statue exceeds J^ of the height of the tower by 7 feet. Find the height of the tower and of the statue. 9. If the numerator of a fraction be trebled, and the denom- inator be increased by 8, the value of the fraction becomes f ; and if the denominator be halved, and the numerator be de- creased by 7, the fraction becomes \. Find the fraction. 10. The City Hall of Philadelphia is said to cover a greater area than any other building in the United States. One fifth of its width exceeds one sixth of its length by 13 feet; and one ninth of its length exceeds one tenth of its width by 7 feet. Find its dimensions. 11. The perimeter of a certain isosceles triangle (see p. 102) is 140 inches. The side exceeds the base by 10 inches. Find the three sides of the triangle. 12. Three years ago A's age was -| of B's age ; but in nine years his age will be ^- of B's age. Find their present ages. 13. If the age of a university is reckoned from the date of its founding, then, in 1912, the age of Yale exceeded the age of Princeton by 46 years. Four years later, one half of the age of Princeton will exceed one third of the age of Yale by 13 years. Find when each was founded. 14. Twice the shorter side of a parallelogram exceeds the longer side by 5 inches ; one third of the sum of the shorter side and 9 exceeds one fifth of the longer side by 3. Find the sides of the parallelogram. SIMULTANEOUS LINEAR EQUATIONS 233 15. In 1912 the sum of the ages of the oldest English uni- versity, Oxford, and the oldest American university. Harvard, was 1316 years ; the age of Oxford exceeded three and three fourths times the age of Harvard by 5 years. Find when each was founded. 16. If one weight is placed 8 inches from the fulcrum of a lever, it balances another weight which is 6 inches from the fulcrum ; if the first weight be decreased by 3 pounds, and the second be increased by 4 pounds, the resulting weights will balance if placed 10 feet and 5 feet respectively from the ful- crum. Find the two weights (§ 143). 17. The sum of the reciprocals of two numbers is ^. Twice the reciprocal of the greater number exceeds the reciprocal of the less number by ^. Find the two numbers. j Hint. The reciprocal of 3 is - ; of a is - . j 18. The sum of the reciprocals of two numbers is ^. If twice the reciprocal of the less be increased by four times the reciprocal of the greater, the sum is 8. Find the two numbers. 19. A purse contained $ 6.55 in quarters and dimes ; after 6 quarters and 8 dimes had been taken out, the number of quarters equalled three times the number of dimes. How many of each kind of coin were there ? 20. In 7 years, A will be three times as old as B, and 8 years ago he was 6 times as old. What are their present ages ? 21. The length of a room exceeds its width by 4 feet. If 2 feet are added to the length, and 3 feet to' the width, the area is increased by 103 square feet. Find the dimensions. Solution: 1. Let Z = the number of feet in the length, and w = the number of feet in the width. /. he = the area. 2. l = w-\-^. (1) 234 ALGEBRA 3. l-\-2 = the new length. w -\-S = the new width. .'. (I + 2) (w + S) = the new area. 4. .-.(«+ 2) (MJ 4- 3) = ho + 103. . (2) 5. Solve the equations (1) and (2), and check. 22. If a rectangular lot were 6 feet longer and 5 feet wider than it is now, it would contain 839 square feet more ; if it were 4 feet longer, and 7 feet wider, it would contain 879 square feet more. Find its length and width. 23. If one weight, increased by 10 pounds, be placed 6 feet from the fulcrum, it will balance a second weight, placed 4i feet from the fulcrum ; if the second weight, increased by 10 pounds, be placed 3^ feet from the fulcrum,, it will balance the first weight placed 6 feet from the fulcrum. Find the two weights. 24. Will weighs 50 pounds. When Will seats himself 6 feet from the fulcrum and John seats himself 4 feet from the fulcrum on the same side, they exactly balance James, who is sitting 7\ feet from the fulcrum on the other side. When Will and John change places, they find that James must sit 7| feet from the fulcrum. How much do John and James weigh ? 25. A crew can row 10 miles down stream in 50 minutes, and 12 miles up stream in an hour and a half. Find the rate in miles an hour of the current, and of the crew in still water. (Hint. Review Biver Problems, Exercise 90, § 144 ; express 50 min- uted as f hour. ) 26. A motor boat which can run at the rate of 15 miles an hour in still water went downstream a certain distance in 4 hours. It took 6 hours for the trip back. What was the dis- tance and the rate of the current ? 27. A crew are rowing on a stream the rate of whose current is known to be 2 miles an hour; they find that it takes them SIMULTANEOUS LINEAR EQUATIONS 235 one and one third hours to go down, and four hours to come back a certain distance. Find the distance and the rate of the crew in still water. 28. An express train travels 30 miles in 27 minutes less time than a slow train. If the rate of the express train were J as great, and if the rate of the slow train were ^ as great, the express train would travel 30 miles in 54 minutes less time than the slow train. Find the rate of each train in miles an hour. 29. If a field is made 5 feet longer and 7 feet wider, its area would be increased by 830 square feet; but if its length is made 8 feet less, and its width 4 feet less, its area is dimin- ished by 700 square feet. Find its length and width. 30. The fore wheel of a carriage makes 8 revolutions more than the hind wheel in going 180 feet ; but if the circumfer- ence of the fore wheel were | as great, and of the hind wheel f as great, the fore wheel would make only 5 revolutions more than the hind wheel in going the same distance. Find the circumference of each wheel. 31. A man has $ 2500 invested from which he receives a total income of $ 135. Part of the money is invested at 6 % and part at 4^ %. How much is invested in each way ? 32. A man has altogether $5000 of savings. He has part invested in a 5 % bond, and the balance invested in a mort- gage drawing 6%. If his total income is $280, how much has he invested in each way ? 33. A man has $1200 invested at one rate of interest and $ 500 at a rate which is 1 per cent greater than the former rate. The income from the first investment exceeds the in- come from the second investment by $23. Find the rate at which each sum is invested. 34. The simple interest on $800 at 5 % for a certain num- ber of years exceeds the simple interest on $300 at 6 % for a 236 ALCxEBRA second period of years by $ 60. If the second period of years exceeds the first by 4 years, find the number of years each sum is invested. ' 35. There are two supplementary angles such that ^ of the larger exceeds -f of the smaller by 5°. Find the angles. (§ 13.) 36. One angle of a triangle is 35°. If the number of degrees in one of the remaining two angles is divided by the number in the other, the quotient is 9 and the remainder is 10. Find the three angles of the triangle. (See § 13.) 37. The fastest train on the Pennsylvania R. E,. makes the trip between Chicago and Fort Wayne, Indiana, a distance of 148 miles, in 1 hour and 20 minutes less time than one of the ordinary trains. Its rate is fthat of the ordinary train. Find the rate of each train. 38. A and B working together can do a piece of work in 6 days ; they can also complete the work if A works 10 days and B 3 days. How many days would it take each of them to do the work alone ? (See § 142.) 39. A and B working together can do a piece of work in 7^ days. If A works alone for 3 days, and B alone for 6 days, they would complete -j\ of the work. Find how long it would take each alone to do the work. 40. A man has a sum of money invested at a certain rate of interest. Another man has a sum greater by $ 3000, invested at a rate 1 % less, and his income is $45 less than that of the first. A third man has a sum less by $2000 than that of the first, invested at a rate 1 % greater, and his income is $ 40 greater than that of the first. Find the capital of each man, and the rate at which it is invested. 172. Relations between the Digits of a Number. Integral numbers are written by means of the digits. Thus, 372 is a number of three digits; 3 represents 300 units, 7 represents 70 units, and 2 represents 2 units. SIMULTANEOUS LINEAR EQUATIONS 237 Similarly, if t is the tens^ digit and u is the units' dlrjit, the number is 10 t 4- u. When the digits of a number are reversed, a new number is formed; thus, reversing 52 gives 25. Notice that 52 = 10 X 5 4- 2 and 25 = 10 X 2 -f- 5. Similarl}^, if x and y are the tens and units digits of a number, the number is 10 x-\- y, if the digits are reversed, the new tens' digit is ?/, the new units' digit is it', and the new number is 10 y + x. EXERCISE 106 1. Write the number whose units' digit is a, tens' digit 6, and hundreds' digit c. 2. Write the number represented by reversing the digits in Example 1. 3. Representing the tens' digit of a number by t, and the units' digit by u represent: (a) the sum of the digits ; (6) the number ; (c) the product of the digits; (d) the quotient when the number is divided by the sum of the digits. 4. Representing the tens' digit of a number by x, and the units' digit by y. (a) express the original number; (b) express the number obtained by reversing the digits; (c) express the quotient of the new number divided by the old number; (d) express by an equation the fact that when the original number is divided by the sum of its digits the quotient is 5 and the remainder is 3. (See note, Example 1, § 171.) 238 ALGEBRA 5. The sum of the two digits of a number of two digits is 11 ; if the digits be reversed, the quotient of the new number divided by the old is 2 and th>e. remainder is 7. Find the number. Solution : 1. Let t = the lens' digit, and u =the units' digit. .■.t + u = IL (1) 2. lot -\- u= the original number, and \0u + t = the new number. .-. 10« + «=: 2(10^4- w)+ 7. (2) 3. Solving the pair of equations, (1) and (2), gives i = 3, w = 8. .*. the number is 38. Check : The sum of the digits is 11. 83 -~ 38 gives the quotient 2 and the remainder 7. 6. The tens' digit of a number exceeds its units' digit by 4. If the digits be reversed, the new number is 6 more than one half of the old number. Find the number. 7. The sum of the two digits of a number is 9. If the digits be reversed, the quotient of the new number divided by the units' digit of the given number is 13 and the remainder is 1. Find the number. 8. The sum of the two digits of a number is 16 ; and if 18 be subtracted from the number, the remainder equals the number obtained by reversing the digits. Find the number. 9. If the digits of a number of two figures be reversed, the sum of the resulting number and twice the given number is 204 ; and if the given number is divided by the sum of its digits, the quotient is 7 and the remainder is 6. Find the number. 10. If a certain number be divided by the sum of its two digits, the quotient is 4 and the remainder is 3. If the digits be reversed, the sum of the resulting inimber and 23 is twice the given number. Find the given number. SIMULTANEOUS LINEAR EQUATIONS 239 173. Literal Simultaneous Equations. In solving literal si- multaneous equations, the addition or subtraction method of elimination is usually the best. Example. Solve for x and y the equations : ax-^by = c. (1) rx-{- sy = t. Solution : Eliminate y. (^) 1. M, (1) : sax -\- shy = sc. (3) 2. Mft (2) : brx -f- bsy = bt. (4) 3. Subtract (4) from (3) : sax — brx =sc — bt. (5) .'. {sa — br)x = (sc - -bt). -hn -br) (0) Now, going back to (1) and (2), eliminate x. 4. Mr (1): rax + rby = re. (7) 5. Ma (2) : rax + say = ta. (8) 6. Subtract (8) from (7) : (rb - sa)y = (re - -ta). ^ (rb- - ta)^ -sa) The solution is : x = ^^ ~ ^S 2/ = ^ ta ta — re sa —br rb — sa sa — br Note. In step 6, when subtracting say from rby, we notice that these are like terms since they have the cominon factor y, and then subtract by multiplying that common factor y by the difference of its coefficients, (rb — sa). Or, we may think of the difference as being rby — say, which, factored, becomes (rb — sa)y. EXERCISE 107 \Sx + 4:y = 7a. {2ax-\-.y = b. [2 x — 5y = 6b. ' lax— 2y= c. {Sx-\-ay = 5. imx-^ny=p. ' [2 x — by = 6. ' \cx-^dy = e. {4:X-\-my = n. (ax-{'by = a^ + Sab. \x-{-py = q, ' \2ax — 3 by =2a^ — 4: ab. 240 ALGEBRA 7. 8. X + ay = ab — 2 a. hx-{-ay = — ah. {ax— by = 2 ah. [2hx-\-2ay^3h^-a^. hx-\-ay = a^ + ah. x — y = a-{-h. 2cx + dy = 2c'-\-d\ 10. \x + = 3. 11. 12. 13. 9Lj^'^^a'' + h\ x y ah 2 1,2 X y rx ■}-sy = 0. sx-\-ry = s^— r^o mx — ny = 0. x-y = myi 14. x-\-ay = h. hx-\-y = a. 15. is n. Find two numbers whose sum is m and whose difference 16. Divide the number c into two parts so that when the larger is divided by the smaller the quotient is d. 17. Divide the number ?- into two parts such that when the larger is divided by the smaller the quotient is s and the re- mainder is t. 18. A and B can do a piece of work together in k days ; if A works 3 days and B works 5 days they can do -i- of it. Find how long it would take each alone to do the piece of work. 19. The sum of the digits of a number of two digits is a; the number itself equals b times the units digit. Find the digits of the number. 20. If a be added to the numerator and b be added to the de- nominator of a certain fraction, its value becomes 1 ; if h be added to the numerator and a to the denominator, its value be- comes 2. Find the fraction. 21. Find two numbers whose sum is c and such that h times the first exceeds a times the second by d. 22. Find two numbers such that the quotient of a divided by the greater exceeds by c the quotient of b divided by the SIMULTANEOUS LINEAR EQUATIONS 241 less ; and such tliat the quotient of h divided by the greater exceeds by d the quotient of a divided by the less. The following two sets of equations arise in the solution of problems in applied mathematics; find Tand/: 23. I ^^ - ^ = '^^/- 24. ) ^"^ ~^^ ^*•^• T=vf. [T-ng = nf. EQUATIONS CONTAINING THREE VARIABLES 174. In the preceding paragraphs equations having two variables have been solved; in each case two equations were given. It is interesting to study equations with more than two variables. For three variables, three equations are necessary. Example. Solve the set of equations, \2 x — y -\-z = 5. (1) \Sx + 2y-{-3z = 7. (2) l4a;_3y-5^ = -3. (3) Solution : Eliminate z by combining (1) and (2) ; the resulting equa- tion will contain only x and y. 1. Ma (1) : 6x-Sy-\-Sz = lb. (4) 2. Subtract (2) from (4) : 3 a; - 5 ?/ = 8. (5) Now. eliminate z by combining (1; and (3); the resulting equation will contain only x and y. 3. M5(l): 10 x-5?/ +52 = 25. (6) 4. Add (3) and (6) : 14 x - 8 ?/ = 22. (7) This gives the equations : 3x-6y=S. (5) Ux-Sy = 22. (7) Solve this set of equations for x and y. The solution is, x = 1, y =— \. Substitute these values of x and y in (1) and obtain z. 2 + 1 + 2; = 5. .-. z = 2. The complete solution is .r = 1, y = — 1, z = 2. Check the solution by substituting it in each of the given equations. 242 ALGEBRA EXERCISE 108 a - 2 6 + c = 0. a-b-{-2c = -ll. 2a-6 + c = -9. {Sx~2y = l. 2. 30^4-4^ = 5. [5z-\-3y = 4:. {12 m — 4:71 -{-p= 3. 3. Un — 71 — 2 p = — 1. [5m — 271 = 0. i5r-3s-4t = -l. 4. 2s + 3^ = H-9. [4r-^ = 3. pa^-f 4r+2« = -5. 5. 2x-3r-4^ = -10. [4a; + 2r + 3^ = -21. (6A-2B-3C = 3. 6. 2^ + 35 + 5(7=0. [8^-55-60=1. 7. 8. 10. 2 1 , 1 f, + - = 5. X y z 111^ - + - + - = 9. a; 1/ 2; ^-^ = -2. iC 2 11,1 1 -+- + - = ;:• X y z 3 1_1_1__5 X y z 3 1_1_1^25 y z X 3 x-\-y-\-2z = a. 2x-y + z=:b. x — 2y±z = c. 1_1_1_^ X y z ' 1-1-1 = 6. y z X 111 = c. z X y 11. The angle .^ of a triangle exceeds the angle B by 20°; and the angle C exceeds the angle A by 20°. Find the three angles of the triangle. (See § 13.) 12. The perimeter of a triangle is 175 inches. The side a is 20 inches less than twice the side 6; and the side h exceeds the side c by 5 inches. Find the sides of the triangle. 13. The sum of the sides a and 6 of a triangle is 147 inches ; the sum of the sides h and c is 135 inches ; and the perimeter is 219 inches. Find the three sides of the triangle. SIMULTANEOUS LINEAR EQUATIONS 243 14. The total area of the three largest oceans is 134 million square miles. The area of the Pacific Ocean exceeds twice the area of the Atlantic Ocean by 1 million square miles; and the area of the Atlantic Ocean exceeds the area of the Indian Ocean by 7 million square miles. Find the area of each of the oceans. 15. A recipe for a fondant for candy calls for a total of three and five sixths cups of sugar, water, and glucose ; twice the total amount of water and glucose exceeds the amount of sugar by one sixth cup ; and three times the amount of glucose and twice the amount of sugar make six cups. How many cups of each ingredient is required for the fondant ? 16. The sum of the three digits of a number is 13. If the number, decreased by 8, be divided by the sum of its units and tens' digits, the quotient is 25 ; and if 99 be added to the number, the digits will be reversed. Find the number. Solution : 1. Let x = the hundreds' digit, y = the tens' digit, and z = the units' digit. 2. Then 100 r -\- \0 y -]- z = the number, 3. and 100 2 + 10 y + x = the number with its digits reversed. 4. By the conditions of the problem, x + y + z = l'S, lOOx+lOy + g — 8 _25 y-h z and 100x-\-10y + z + 99 = l00z-\-l0y + x. 5. Solving these questions, x = 2, y = S, z =S. Therefore the number is 283. 17. The sum of the three digits of a number is 23 ; and the digit in the tens' place exceeds that in the units' place by 3. If 198 be subtracted from the number, the digits will be reversed. Find the number. 18. A and B can do a piece of work in 10 days, A and C in 12 days, and B and C in 20 days. In how many days can each of them alone do it ? XIV. SQUARE ROOT AND QUADRATIC SURDS 175. Square Root by Inspection. The square root of a per- fect square monomial (§ 90), and of a perfect square trinomial (§ 96) have been found by inspection. Eeview these paragraphs. 176. Tv/o Square Roots are obtained for each number. They are of equal absolute value, but have opposite signs ; they may , be written together, by means of the double sign, ± . Example 1. VOo^p" ^ _j_ 3 ^sjr, . since (+ 3 gi%Y = 9 a^^^ ; and (- 3 a26)2:^ 9^2452. Example 2. VOa^- 12aa: + 4x2 = ±(3 a - 2x) ; since {+(3 a- 25c)}2 = + (3 a - 2ic)2 = Qa^ - 12ax + 4ic2, and {_ (3 a _ 2 x)f = + (3 a - 2 x)2 = 9 a2 - 12 ax + 4x2. 177. The square root of a large number may sometimes be obtained by inspection by factoring the number. Example. VlTeTa* = \/4 • 441 a* =±2 -^Icfi =±^2. a^. EXERCISE 109 Find the square roots of : . ^o 4 6 ^ 16ai« ^ 169 a^y 2. 64a;V2^ ' 4a^ ^ 121 cH^ 3. 144aW. * 1^9^'* * 144^2/^' ^ 196 m« ,^ 225 s«^^ 4. 225 /m 7. ^^^. 10. ^^^. 11. When is a trinomial a perfect square? Find the value of : 12. Va^-6a262 + 9 6^ 14. Vl69 a^ - 26 ar^ + r«. 13. Vm«-10m3n + 25n2. 15. V4 aj^ - 20 a;?/^ + 25 2/\ 244 SQUARE ROOT AND QUADRATIC SURDS 245 16. V3136. 18. V4225a262. 20. V5184r«s». a^ + 2 a6 + b'^ 2a 2fl + 2a6 + 63 + 2a& + 62 17. V2916a;y. 19. V5625mV. 21. Vll025ar?/V. 178. Square Root found by Long Division. If it is not pos- sible to factor readily the number under the radical sign, the square root, if there is one, may be found by a process like long division. Example 1. Find the square roots of a- -f- 2 a6 + b^. _ a+ 6 Solution : 1. s/a^ = a. Place a in the root. 2. Square a ; subtract. 3. 2 X a = 2 a. Trial divisor. 2ab -T-2a = b. Add b to the trial divisor and to the root. Complete divisor. 4. & X (2 a + 6) ; subtract. The square roots are : +(« + &) and —(a-\-b). Explanation : 1. Find the square root of the first teim, obtaining a, the first term of the root ; place it in the root. 2. Square the first term of the root and subtract it from the given number, obtaining the first remainder, '2ab + b'\ 3. Double the first term of the root, obtaining 2 a, the trial divisor. Divide the first term of the remainder by 2 «, obtaining 6, the second term of the root. Add b to the root and to the trial divisor ; the complete divisor is 2 a + 6. 4. Multiply the complete divisor by b and subtract. Step 3 is suggested by the process of squaring a binomial. When squaring a binomial, the middle term is obtained by taking twice the product of the fii-st and second terms ; this is equivalent to taking twice the first tenn and multiplying by the second. Reversing the process, the second term, 6, will be found, if 2 ab is divided by 2 a. After a^ is sub- tracted from a^ -\-2ab + 6^ the remainder 2a6 + b'^ equals 6(2 a + b). This suggests adding b to the trial divisor and multiplying the sum by b. Example 2. Eind the square root of 20ar^-70a; + 4a;*-h49-3ar^. 246 ALGEBRA Subtract. ix^+^Ox^- 8:r-^-70x+4d 4x4 4x' +5x 20x^- 3x^-70x+49 4a:H6x 20x3 + 25x2 4x2 + 10 a: -7 -28x2-70x+4d 7). 4a;H10:« — 7 -28x2-70x+49 Solution : 1. Arrange it in the descending powers of x : 2x2 + 5x-7 2. V4 x4 = 2 x2. 3. (2x2)2=4x4. 4. 2 X (2 x2) =: 4 x2. 20 x3 -4- 4 x2 = 5 X. 5. 5x(4x2 + 5x). 6. 2x(2x2 + 5x). 7. - 28x2 --4x2=- 7- 8. _7(4x2 + 10x The square roots are Rule. — To find the square root of an algebraic expression : 1. Arrange it according to ascending or descending powers of some letter. 2. Write the positive square root of the first term of the given expres- sion as the first term of the root. Square it and subtract the result from the given expression. 3. Double the root already found, for the trial divisor. Divide the first term of the remainder by the first term of the trial divisor. Add the quotient to the root and also to the trial divisor,- obtaining the com- plete divisor. 4. Multiply the complete divisor by the new term in the square root ; subtract the product from the remainder obtained in step 2. 5. Continue in this manner : (a) double the root already found for a new trial divisor ; (6) divide the first term of the remainder by the first term of this product for the new term of the root ; (c) add the new term of the root to the trial divisor, obtaining the complete divisor ; (d) multi- ply the complete divisor by the new term of the root ; (e) subtract. EXERCISE 110 Find the square roots of the following : 2. 25m^-\-S0mn + 9n\ 5. l-6a + lla^- 6a^ + a\ 3. S6a^-12ab + b\ 6. 9x'-24.a^-\-4.x' +Wx + 4:. 7. 49a2_30a3 + 16^9a*-40a. 8. 25a''-20a^y-26x'y'-\-12xy +9y*. SQUARE ROOT AND QUADRATIC SURDS 247 9. 9a^ + l-4a34.4a*'-Ga--f 12a^ 10. 16 m* + 8 ?/iV - 23 m^x* - {jmx^ + 9 a-^. 11. l-2x-^3x^-4x^ + ^x*-2x'^-^x^'. 12. a^ - 4 x'a' + 10 a%'' + 4 ar^a^ - 20 xa' + 25 a«. 13. 9x^ + 25y^-\-16z^ + 30xy-24:Xz-40yz. 14. a2 + 62 4.c2-2a6-2ac + 2^;c. 15. 20 iib^ + 9 a* - 26 a-?>2 + 25 6* - 12 r/6. 16. 20ar*'-70a- + 4a-^ + 49-3ar^. 17. 49 m* — 14 mhi — 55m^n^ + 8 mii^ + 16 n*. 18. 7/i2 + 8m + 12- — + 4/ 19. ^ ^^ ^ +2a:^4a^ _ 16 , 8a- 13ar^ 4ar' , 4a^ 20. -7r-\-- — -T^^ r + — r' 9 6 a 6a^ (t a* 179. Square Root of an Arithmetical Number. The square root of 100 is 10; of 10,000 is 100; 6tc. Hence the square root of a number between 1 and 100 is between 1 and 10 ; the square root of a number between 100 and 10,000 is between 10 and 100 ; etc. That is, the integral part of the square root of a number of one or two figures contains one figure ; of a number of three or four figures, contains two figures; and so on. Hence, if the given number is divided into groups of two figures each, beginning with the units figure, for each group in the number there will be one figure in the square root. The groups are called Periods. Thus, 2346 becomes 23 45 ; it has two periods and its square root has tw(5 'figures, a tens' and a units' figure. 34038 becomes 3 40 38 ; it has three periods and its square root has tbi'fee figures. A number having an odd number of figures will always have only one figure in its left-hand period, as in this case. 248 ALGEBRA A decimal number is divided in the same manner, starting from the decimal point in both directions. Thus, 3257.846 becomes 32 57.84 60. The last decimal period is always completed by aniiexijig a zero. This number has two figures before the decimal point and two after it, in its square root. 180. The first figure of the square root of a number is found by inspection ; the remaining figures are found in the same manner as the square root of a polynomial. Example 1. Find the square root of 4624. There are 60 + 8 46 24 36 00 Solution, 1. Divide 4624 into periods ; this gives 46 24. in the square root a tens' and a units' figure. 2. The tens' figure must be 6 ; 7 is too large for 70^ = 4900, which is more than 4624. 3. The rest of the square root is found as follows : 3600 is the largest square less than 4000. VSOOO = 60 ; place 60 in the root. Square 60 and subtract. Double 60. Trial divisor. 120 102 -=- 12 = 8+. Place 8 in root and add to trial divisor. __8 Complete divisor 128 Multiply complete divisor by 8. The square roots are -f 68 and — 68. It is customary to abbreviate the solution by omitting the zeros as in the following example. Example 2. Eind the square root of 552.25. 10 24 10 24 Solution. The largest square less than 5 is Place 2 in the root. 2x2 = 4; annex 0. Trial divisor. 15-^4 = 3+; add 3 to the trial divisor. Complete divisor. Multiply by 3. 2 X 23 = 46 ; annex 0. Trial divisor. 230 -4- 46 = 6+ . Add 5 to the trial divLsor. Complete divisor. Multiply by 5. The square roots are + 23.5 and — 23.5. 4 ; V4 = 2. 23.5 5 52.26 4 40 162 3 43 129 460 23 26 . 5 466 23 25 SQUARE ROOT AND QUADRATIC SURDS 249 Rule. — To find the square root of an arithmetical number : 1. Separate the number into periods (§ 179). 2. Find the greatest square number in the left-hand period ; write its positive square root as the first figure of the root ; subtract the square of the first root figure from the left-hand period, and to the result annex the next period. 3. Form the trial divisor by doubling the root already found and an- nexing zero. 4. Divide the remainder by the trial divisor, omitting the last figure of each. Annex the quotient to the root already found ; add it to the trial divisor for the complete divisor. 5. Multiply the complete divisor by the root figure last obtained and subtract the product from the remainder. 6. If other periods remain, proceed as before, repeating steps 3, 4, and 5 until there is no remainder or until the desired number of decimal places has been obtained for the root. Notp: 1. It sometimes happens that, on multiplying a complete divisor by the figure of the root last obtained, the product is greater than the remainder. In such cases, the figure of the root last obtained is too great, and the next smaller integer must be substituted for it. Note 2. If any figure of the root is 0, annex to the trial divisor and annex to the remainder the next period. Example 3. Find the square root of 4944.9024. 70.32 Solution : .49 44.90 24 49 140 1 44 90 3 1403 42 09 14060 2 8124 2 140 62 2 8124 The square roots are -f 70.32 and — 70.32. The first trial divisor is 140. Since this is greater than 44, the first remainder, annex to the root, obtaining 70. The second trial divisor is 1400; (2x70=140; annex 0, 1400). Bring down the next period 90, getting for the second remainder 4490. Divide 44 by 14 gives 8+ ; annex 8 to the root and add 3 to 1400, etc. 250 ALGEBRA EXERCISE 111 Find the square roots of : 1. 1521. 5. 23409. 9. 462.25. 2. 4489. 6. 54756. 10. 9.8596. 3. 5625. ^ 7. 173889. 11. 11.9716. 4. 8836. 8. 42025. 12. 17.8929. 181. The Approximate Square Root of a number which is not a perfect square is often desired. Obtain usually the first three figures following the decimal point. Example. Find the approximate square root of 2. Solution : 1.414 2.00 00 00 20 2-t 1 100 96 28 21 4 00 2 81 2820 119 00 2^ ll 112 96 7 04 The square roots are -}- 1,414+ and — 1.414+. Note. In order to obtain the desired number of decimal places, annex zeros until there are three periods. EXERCISE 112 Find the approximate square roots of : 1. 3. 3. 6. 5. 10. 7. 13. 9. 15. 11. 19. 2. 5. 4. 7. 6. 11. 8. 14. 10. 17. 12. 21. 182. Table of Square Roots. In the remainder of the course, it will be necessary to use frequently the square roots of some numbers. Retain some of the square roots as they are SQUARE ROOT AND QUADRATIC SURDS 251 found, either in a notebook or in some other convenient place. Make a list of the numbers from 1 to 50, and write their square roots beside them, thus : NrMBER Square Root 1 1.000 2 1.414 3 1.732 After working Exercise 112, twelve of the numbers of this table may be tabulated. These roots may be used to obtain the square roots of other numbers. Example 1. Find the square root of 8. Solution : \/8=V4x2 = 2x\/2 = 2x (1.414+) = 2.828+. Example 2. Find the square root of 12. Solution : Vl2 = V4 x 3 = 2V3 = 2 x (1.732+) = 3.464+. EXERCISE 113 1. Find the following square roots to three decimals : (a) V18. (b) V20. (c) V24. (d) V27. (e) V28. 2. Comjilete your table of square roots up to 50. Get as many roots as possible by inspection (§ 175) ; get as many of the remaining roots as possible as in Example 1. Find the others by the long division method (§ 180). 183. The square root of a fraction which is not a perfect square may be found as follows : >'2 >'2x2 ^4 v5^ VO V4 2 ' ±^ = ±2_i49+^ 224+ 2 2 Rule. — To find the square root of a fraction : 1. Change the fraction into an equivalent fraction with a perfect square denominator. 1. 1- 4. i- 2. 5 5. i- 3. A- 6. f 252 ALGEBRA 2. The square root of the new fraction equals the square root of its numerator divided by the square root of its denominator. 3. If desired, express the result of step 2 in simplest decimal form, prefixing the double sign, ±- ExAMPLK. Find the approximate square root of |. Solution: 1. The smallest square number into which 8 can be changed is 16 ; multiply both terms of the fraction by 2. ^S >'2 X 8 ^lO 4 4 EXERCISE 114 Find the approximate square roots of : 7. f 10. f 13. ^, 8. |. 11. |. 14. A. 9. f 12. yV 15. 2V QUADRATIC SURDS 184. The indicated square root of a number which is not a per- fect square is called a Quadratic Surd ; as, V3, \/- , ^x \/- + 1- ^ o ' ^ y 185. Surds should be simplified as in the following examples: Thus, a quadratic surd is in its simjjlest form when the number under the radical sign is an integer which does not contain any perfect square factor. While a quadratic surd has two values, one positive and one negative, it is agreed to consider only the positive root, in order to avoid ambiguity. This root is called the principal root. 186. Addition and Subtraction of Surds. Example 1. Find the sum of V20 and V45. Solution : 1. V20 + V45 = V4T5 + V9T5 = 2 VS + SVS = 6V5. SQUARE ROOT AND QUADRATIC SURDS 253 This solution assumes that surds may be added like other numbera. The coefficients of Vo are 2 and 3 ; the sum is found by multiplying V6 by the sum of its coefficients (§ 34). The advanta is 5Vo or 5x (2.230+) or 11.180+. This same result could be obtained by adding the square roots of 20 and 45. Example 2. Simplify V| + VJ. Solution: 1. J2+ Jl = J5+^= 3v^+ V2^3v^+2V2^5V2^ A/8^\2 A/i6^>/4 4 2 4 4 4 2 5V2^ r)X(1.4U+) ^7.070+_^.^., 4 4 4 Example 3. Simplify | + V|". 2 . ./1_2 , ./3^2 V3_2+ V3 Solution: 1. ?+Ji=£_j_ /§ 3^>'3 3 \j) 3 3 >9 3 3 3 2 2-1- V3 ^2 + 1.732+ ^ 8.732+ ^.^^^^^ 3 3 3 ' * Note. The results of problems involving surds are often left in the surd form as in step 1 of Examples 2 and 3. There are advantages in finding the approximate decimal value of the result. EXERCISE 115 Simplify the following : 1. V12+V75. 11. | + V|. 2. V98-V18. 12. i-V|. 3. VSi") - V20. 13. I + V^. 4. 3V27-V48. 14. -j + VA- 5. V28+V63. 15. _3_4.V^. 6. 6V2-V50 + VI8. 16. +|+V|. 7. Vl2+Vi. 17. -i-\-^l 8- Vf+V^. 18. -f + Vf «• Vl + V^. 19. -f + VI^ 10. V| - V24. 20. f - V V-. XV. QUADRATIC EQUATIONS 187. A Quadratic Equation is an equation of the second degree (§ 161) ; it may have one or more unknowns. 188. A Pure Quadratic Equation is a quadratic equation. hav- ing only one unknown, which contains only the second power of the unknown, as, ax^ = b. Example 1. An acre of ground contains 43,560 square feet. How long must the side of a square field be in order that the area of the field shall be one acre ? Solution : 1. Let s = the number of feet in one side. 2. Then s^ = the number of square feet in the area. 3. Then s2 ^ 43,660. Extract the square root of both members of the eciuation. 4. Then s= ±208.7+. Since this is a field, only the positive root has meaning ; hence the side of the field must be 208.7+ feet. 189. A pure quadratic equation has two roots, because two square roots are obtained in extracting the square roots of the two members of the equation. Rule. — To solve a pure quadratic equation. 1. Clear the equation of fractions, transpose, and combine terms until the equation takes the form x'^ = a number. 2. Extract the square root of both members of the equation, plac- ing the double sign, ± , before the root in the right member. Note. After extracting the square roots of both members of an equa- tion like x^ = a^, we get ±x = ±a. This gives : -{- x = -\- a, -{- x= — a, — x=+a, and — x=— a. 254 QUADRATIC EQUATIONS 265 If both members of the last two equations are multiplied by — 1, the equations become + x = — a, and -{- x = -\- a. These are the first two of our four equations. Thus, it is clear that, from x'^ = a'^, we get only two equations, x = -{- a and x = — a^ or x = ±a. Example. Solve the equation — — - -\ — = t^t H • ^ 3 m 12 m Solution: 1. f^ + ^ = ^+^- 3 w 12 m 2. Mxim-. 8m2+36 = m2 + 144. 3. Simplifying : 7 m^ = 108. 4. 2>7 : m^ = 1^^. 5. V":* m=±\/^^=±Qy/} = ±^\^. 6. a/2I=4.582: m =± -• (4.582) = ± ^Ii^ = ± 3.927. 7 7 7. mi = + 3.1)27; m2 =-3.927. " wii " is read " m one." The numeral 1 is called in such cases a sub- script. *' m2 " is read " m two." These subscripts are used in this case to distinguish between the two roots of the quadratic. Check : In cases such as this, it is better to check by going over the solution a second time. Great care must be taken, however, for it is easy to overlook an error. Note. — Get the result in the radical form first; that is, m = ± ? V21 ; then it is wise, for many reasons, to get it in decimal form as finally given. EXERCISE 116 Solve the following equations : 1. 7 p^- 175 = 0. 5. 3(m-2)+2m(m-l)=m. 2. 5ar^-48 = 80-3a:2^ 6. ^t -{- 5) - t(t - 1) = 4. t. 3. 12c2-140 = 9c2-32. 7. 9a'-7 = 0. ^ 2m^_3jn^-7^11. g. lla^-3 = l. 5 3 15 * The symbol " V : " placed in the left margin will mean, " take the square root of both members of the previous equation." 256 ALGEBRA Review the statement made at the top of page 187; bear it in mind whenever solving fractional equations. 9 'j_ 11^5 3 ^^+2 5^^-3 ^ 4^^-fl 3/S2 ^S^ 6* " 5 10 25 * 13 r/+4 gr-4^10 • ^_4 ^+4 3 r*_3r2+4 7^-3 o. 14. — 10. n. 2r- 4- -3 9-f-r r 3r4-2 x-\-l x"^— x + 1 x + 1 3r*-i-2?-2-4 3?'2 + 2 15. a2-2ca^ = = 3 Ir. Solve for ic. Solution: 1. - 2 0x2 zz 3 62 _ a2. 2. 2 ca;2 = a2 _ 3 ^,2. 3. ^2_«2-3&2 2c = ± — V2 a2c _ 6 62c. 2c Solve for x : P, jl6. a2+a;2 = m. 19. 2cx^=cP, 17. a4-2a;^ = c. 20. i^ar^ = c. 18. aa^ = m. 21. 3m3^ — n=p, 190. A Right Triangle is a triangle which has a right angle for one of its angles ; as triangle ABC, in which angle jB is a ' [ right angle. The side opposite the right angle is the hypote- ^b nuse; as, side AC. The side BC is the base and AB is the altitude. In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. Thus, b' = a'-\- cl To verify this fact, draw a right triangle with BC 3 inches m^ AB 4 inches; measure AC. Substitute the lengths of the sides in the equation b^ = a^ + cl QUADRATIC EQUATIONS 257 EXERCISE 117 Carry out all results in this set to one decimal place: 1. Find the altitude of a right triangle whose base is 13 feet and whose hypotenuse is 30 feet. Solution : 1. Let x = the number of feet in the altitude. ' 2. Then x^ 4- 132 ^ 302, (^l^y ?) 3. Complete the solution. 2. Find the base of a right triangle whose hypotenuse is 45 feet and whose altitude is 25 feet. 3. If the diagonal of a rectangle is 68 inches and the base of the rectangle is three times the altitude, what are the di- mensions of the rectangle ? 4. If the altitude of a rectangle is a and the base is 4 timias the altitude, what is the length of the diagonal? 5. Solve the formula W = d? + (? : (a) for a ; (6) for c. In the isosceles triangle ABC, AD, which is perpendicular to BC, is the altitude and BC is the base. BD and i?C are equal; this may be verified by measuring them. ^ 6. If AB is 15 inches and BC is 18 inches, find ^W. 7. If the equal sides of an isosceles tri- angle are each 30 inches, and the altitude is 18 inches, find the base. 8. If the equal sides of an isosceles triangle are each 3 a inches and the base is 2 a inches, find the altitude ? 9. An equilateral triangle is one which has all of its sides equal. Find the altitude of an equilateral triangle whosQ sides are each 10 inches. 10. Find the altitude of an equilateral triangle whose sides are each s inches; (b) find the area of the triangle. 11. The area of a circle is found by the formula A=7r7'^, where r is the radius, and tt = 3|. f Find the area of a circle whose radius is 9 inches.. 258 ALGEBRA 12. What is the radius of a circle whose area is an acre ? (Express the radius in feet. See § 188.) Express the results of the following in simplest radical form). 13. Solve the formula A = ttt^ for r: (a) letting 7r=3|; (6) without substituting the value of tt. 14. The volume of a circular cylinder is given by the for- mula F= TTT^h ; where r = the radius, h the altitude. Find V when r = 5 and h = 13. 15. Find r when V= 759 and h = 14. 16. Solve the cylinder formula for r. 17. Solve the formula S = 4: irr^ for r. 18. Solve the formula V = ^ irr^h for r. 19. The distance s in feet through which an object will fall in t seconds is given by the formula s = i gt^, where g is 32. Suppose that a stone is allowed to fall from a tower; how far will it fall in : (a) 3 seconds ? (6) 5 seconds ? 20. How long will it take a ball to fall 900 feet? 21. Washington's Monument in Washington, D. C. is 555 feet high. How long will it take a ball to fall that distance ? 22. Solve the equation s = \gt^ for t. 23. Solve the formula/=^^ for V. COMPLETE QUADRATIC EQUATIONS 191. A Complete Quadratic Equation is a quadratic equation having only one unknown, which contains the first power of the unknown as well as the second power; as, 23^-3x-5=0. 192. Complete quadratic equations have been Solved by Fac- toring in § 108. QUADRATIC EQUATIONS 259 EXERCISE 118 Solve by factoring : 1. ar-5a; = 24. ;^3 J__Jt5_^_l o02 if:a '8ar^24x 2* 2. 2 ??r — m — 15 = 0. 3. 2a:2 + i3^._^20 = 0. 14. ^4-1 = -??. 4. 12/- 77/ = 10. ^ "^^ ^^ 5. 5 lu^ — 7 w = 0. 6. 9a,'2-49x = 0. (Solve for j-). 15. 1-t' 12 3-t 16. 7 s-2 s-3 s-4 17. 7 3 1 8. 8 ic^ + 14 7/ia: = — 3 m^ ' y __ 3 2^ — 4 2 9. 6 ar»- 11 a!A; = 10 Ar'. , 2 6 18. 1+—=^ ^^=0. 10. 5ar^ + 152>' = 28icp. P-3 i> - 8 a^ a; 35 lo 3 2 _5 19. 326 7^(1, + 6m — 54 12. 1_A=_^. 20. ?:ii?=^^ 1. 2 ^x" 12 a; r + 3 (r-hS)^ 193. Graphical Solution of Equations With One Variable. Many facts about equations containing one variable can be discovered by the aid of graphical rejjresentation. Example 1. Consider the equation 3 a; — 12 = 0. The expression Sx — 12 has a different value for each value of x: Thusif x = 2, 3x-12=-6; if x =- 3, 3x - 12 =- 21. The problem is to find the value of x for which the expression 3 x — 12 will equal zero. Graphical Solution : 1. Let y = 3 x — 12. 2. Find values of y for some values of x : if X = 0, ?/ =- 12. if x=-2, y =- 18. . if X = + 5, y = + 3. if X = + 6, y = + 6. 260 ALGEBRA 3. Use these pairs of numbers as coordinates of points and draw the graph. ■ 1 1 1 V ' M I/C 1 1 \l\ 1 / J- J-^m- - Z - j ■+!<) ~ r 1 / 1 f -'ib r^ - 1 ( 1 j* 1 1 * / 1 .- X ,. 1 1 1 A/ _^__^ j_ _X - ) II- 1 1 ji 5 13 1 i i 1/ 1 i 1 / t: __ _ ^ Z : ^ ^ -U .x 1 in / 1 1 n) J , Sl y'" Fig. 9 4. jBC crosses the x axis at point A. The coordinates of ^ are : x = 4, 2/ = 0. 6. Hence when x = 4, 3 x — 12 = 0. (y is the expression 3 x — 12.) .-. X = 4 is the desired solution of the equation, for we were looking for a value of x for which 3 x — 12 = 0. Rule. — To solve graphically an equation containing one variable : 1. Simplify the equation as much as possible. 2. Transpose all terms to the left member. 3. Represent by y the expression found in step 2. 4. Find for y the values which correspond to selected values of the variable in the equation. 6. Use the pairs of values obtained in step 4 as coordinates of points ; plot the points ; draw the graph, making the vertical axis the y axis. 6. The graph crosses the horizontal axis at points whose ordinates are zero, and whose abscissas are the desired roots of the equation. QUADRATIC EQUATIONS Example 2. Solve the equation ar — a; = 6. Solution : 1. x^ — ar = 6, or x"^ — x — = 0. 2. Let y = X- — X — a. 3. If a;=-4, y= (-4)2- (_ 4) -6= 10 + 4-6 = 4-14. 4. 261 Similarly \f x = + 1 + 2 + 4 + 5 - 1 -2J-3 -4 then y = -6 -6 -4 + 6 + 14 -4 + 6 + 14 ■MllUllllIM iliiii :::: " " ~ 5 : -^ - -- -> + — S / : ^— >^ _J ^ ^4^^ J LJ__^ 1 L-te 1 :: ::: ::: ::x'-:::::: :: .: Fig. 10 5. The graph crof3ses the horizontal axis at the points ^jand B. Ac- cording to the rule, the abscissaB of these points are the two roots of the equation. At^: X = - 2, y = ; i.e. x^ - x - (i = 0. I AtB: x=-\-S,y = 0; i.e. x^-x-(i = 0. Check : x = - 2 ; does (_ 2)2 _ (_ 2) - 6 = ? Yes. x = + 3; does (+3)2- (+3) -6 = 0?: Ted. 262 ALGEBRA EXERCISE 119 9olve graphically the equations : 1. ic H- 3 = 0. 4. a^ - a; = 12. 2. 2 ic = 7. 5. x^—7x-\-6 = 0. 3. x^ = 16. 6. x'-\-6x-\-5 = 0. 194- Some quadratic equations cannot be solved readily by factoring ; for example, ic^— 6cc — 2 = 0, since x^ — 6x — 2 does not have any rational factors. The graphical solution shows that this equation has two roots. Solution: 1. ic2_6a;-2 = 0. 2. Lety = a;2- -6x- -2. When X = -2 -1 + 1 + 2 + 3 + 4 + 6 + 7 + 8 + 14 then y = + 14 +-5 -2 - 7 ^10 -11 -10 -2 + 5 y: 1 'U- Jl I 4 x t i *«^n: - L jL:E it t: it 41 V J \ I j ,, _ r _ \ T t 4 A A .7 ^ ^ I- JL- X A ^ ' 4^ ^ -i -."^ -; - \ i+l i+2 +3 +' + +')/+7 "t : _L_ I it L- X i i i^ ^'^ ^^ r 7 >^ ^ !^ 7 ^^ V -^^ S ^Z^- -',° ^ - --k 1 i . , J- XL -- Fig. 11 3. The graph crosses the horizontal axis at the points A and B. The abscissa of A is about — .3 ; the abscissa of JB is about +6.3. QUADRATIC EQUATIONS 263 This indicates that the roots of this equation are approximately — .3, and + 6.3. There are two methods of solving such equations which give the roots more accurately. 195. Solution by Completing the Square. Development 1. Find: (a) (x — 4)^; (b) (a? + 5)^; (<-■) (^-r; (<*) (y+if- 2. When is a trinomial a perfect square ? (See § 96.) 3. Make a perfect square trinomial of ic^ — 10 a;. Solution : 1. ^ of 10 = 5 ; 62 = 26 ; add 25. 2. The perfect square is x'^ — 10 ic -f- 26 or {x— 5)2. 4. Make perfect square trinomials of the following : (a) ar'-12a?; (b) y^-Uy, (c) z^-20z. 5. Solve the equation a^ — 12 a; + 20 = 0. Solution : 1. S20 '■ a;2 - 12 a; = - 20. 2. Make the left member a perfect square by adding 36 ; therefore add 36 to both members : (§41). ^36 : x2 - 12 X + 36 = 36 - 20. or (x - 6)2 = 16. 3. V": x-e=±4. 4. .'. a; — 6 = + 4, or a; = 6 + 4 = 10, one root, and x — 6=— 4, orx = 6 — 4 = 2, another root. Check : x = 10 ; does (10)2 - 12(10) + 20 = ? Yes. x = 2] does (2)2 - 12(2) -f 20 = ? Yes. 6. Solve the equation a^ — Sx — 5= 0. Solution : 1. x2 — 3 x — 6 = 0. 2. As: x^-Sx=+b. 3. i(- 3) = - I ; (- f )« = + f ; add I to both members. 4. A^: x2-3x + J = 6 + | = ^. 6. V: x-^ = ±V'^- = ±iy/29. 6. ,.^ = 3 1^29=3-±^. 2 2 2 264 ALGEBRA ; 7. Radical results, x-^ = ^jty^ and X2 =^-^^. 2ii 2 8. Decimal results, xi = ^ + ^-^^^ and x^ = ^~f-^^^ . _ 8.885 ^ - 2.385 2 ~ 2 = 4.192+ =-1.192+. Check : To check the solution by substituting the roots in either their decimal or their radical form is a long process, with many opportunities for errors. Persons skillful in algebra check by going over the solution carefully. A quick check, the reason for which will be learned later in algebra, is to find the algebraic sum of the roots ; this result should equal the nega- tive of the algebraic coefficient of x in the equation in which the coefficient of x^ is 1. Here: +4.192+ The coefficient of cc^ is 1. The coefficient of x — 1.192+ is — 3. This equals the negative of the algebraic Sum. + 3 sum of the roots. If the coefficient of x^ is not 1, first imagine the equation divided by that coefficient, and then select the coefficient of x. Rule. — To solve a quadratic equation by completing the square : 1. Simplify the equation; transpose all terms containing the unknown number to the left member, and all other terms to the right member so that the equation takes the form ax^ -\-bx=c. 2. If the coefficient of 3^ is not l, divide both members of the equation by it, so that the equation takes the form Z, Find one half of the coefficient of x\ square the result; add the square to both members of the equation obtained in step 2. This makes the left member a perfect square. 4. Write the left member as the square of a binomial ; express the right member in its simplest form. 5. Take the square root of both members, writing the double sign, ± , before the square root in the right member. QUADRATIC EQUATIONS 265 6. Set the left square root equal to the + root in the right member of the equation in step 6. Solve for the unknown. This gives one root. 7. Repeat the process, using the — root in step 5. This gives the second root of the equation. 8. Express the roots first in simplest radical form, and then, if desired, in simplest decimal form. EXERCISE 120 Solve by completing the square : 10 z^-10z = 5. 11. x^-\-5x= —^. 12. s2_|.io = 7s. 13. w;2 - 3 w == - 2. 14. m2-f-m = 30. 15. 7-2-13r-|-30=0. 16. z^-6 = 5z. 17. ^2 + 9^ = 11. 9. t^-St=4:. 18. /-152/-fl6 = 0. 19. Solve the equation 7? — ^x = l, Solution : 1. i of f = \. {\Y = \. 2. Aj: a;2-|x-f i=l + i. 3. {.x-\y = {^)-_ 5. x-\=^-\V^. x-\=-\\/\0. x = i + ^ViO x = i-i\/lO _ 1 + VlO _ 1 - VlO ~ 3 3 _ 1 + 3.162 ^ 1-3.162 O Q 3 3 1. ar^_4a;-5 = 0. 2. ar^ + 8 a; - 33 = 0. 3. ic2_6a;-27 = 0. 4. x2 + 10ic4-24 = 0. 5. ar^_2a;-15 = 0. 6. 2/s_42/ = 4. 7. a2 + 6a = l. 8. m^ - 2 m = 1. 266 ALGEBRA Check : + 1.387 Coefficient of a: = — | = — .6666. - .720 = _ .667. Sum. + .667 +.667 = -(-.667). 20. x^-^x^^. 23. a^-la = ^. 21. f-iy = ^. 24. m2-fm = |. 22. z^-\-^z = l. 25. f-^t = T. 26. Solve the equation 3ic^ — 2aj — 1=0. Solution : 1. Da: 2. H-' 3. Complete the solution £ 27. 3a^ - 2 a; - 5 = 0. 28. or'-\-2r-S = 0. 29. 4.f- -8^ + 3 = 0. 30. Sx'- -4aj-7 = 0. 31. 2 a'. - 3 a - 9 = 0. 32. 5p'-^Sp = l. 33. 6w' -5w = 10. 34. 3 5 a' -2-5=0 a 38. t St 5 ^-3 4 2 39. -^ + 1 = 0. «— 4 5 42. Sw w — 1 + 2 4 1 a — 5 — 2 a — 2^ 2 1 40. -i^i^+_-:^=0. 41. 3 5* 15 c-2 c+2 8* 1 V.O 3 2 . 35.^+2-2^ = 0. 43. _^___ = 1. 36. !^_2+4 = 0. 44. ^-^ + A= 2 . 3 m 22/-7 15 2/-3 37. I + ?-5 = 0. 45. 2 ^- 1 r'r to^ — 9w-f-3w — 3 196. Solution of Literal Quadratic Equations. Example. Solve the equation ax^ — 3 6a; — c = 0. Solution : 1. aoH^ — 3 &a: — c = 0. QUADRATIC EQUATIONS 267 2. D,: x^- — x^^ = 0. a a 3. A,: a;2_§^x = ^. - . a a 4. The coefficient of x is (^-^) ; one half of it is (^r^)- The square of (^^^-^) is l^^\ • Add this to both members of equa- tion 3. Ix 3&\2^ 96^ + 4, ^ 2a) 4a2 36 6. a;-~ = ±r^V962 + 4ac, 2 a 2 a 7 ^ _ +36 j: V9 62 + 4 ac I, X — 2a Q . ^ + 8 6 + \/9 62 + 4 ac . ^ _ + 3 6 - V9 62 + 4 ac 2a 2a Check : xi + iC2 = -^^^ — - = H Since this is the negative of tho 2a a coefficient of x in step 2, the roots are correct. EXERCISE 121 Solve the following equations for x : 1. x^ + 2ax-S = 0. 8. x^-\-2'mx = 2m + l. 2. x^-\-2ax-\-b = 0. 9. x"- 4.ax = 9b^-4:a\ 3. x^-{-4tX-c = 0. 10. aa^4-2a; + l = 0. 4. xr-i-Sx+7rt = 0, 11. ra;^ + 4 ^a; — 5 = 0. 5. 2a;=^4-3a;-n=0. 12. car^ + 2 c?a; + p = 0. 6. 2x^-^4: ax -\-b = 0. 13. .-c^ + ma; + n = 0. 7. x^ + 3ax-4:t = 0. 14. aa;^ _f_ ^^ ^ ^ ^ 197. Solution of Quadratic Equations by a Formula. All quadratic equations having one unknown may be put in the form ax' + bx + c^O. 268 ALGEBRA This equation may be solved like the equations in Exercise 121. The roots will be found to be ^ 2a This result may be used as di formula for solving any quad- ratic equation of the form ax^ -\-hx-\-c. Example 1. Solve the equation 2a^ — 3aj— 5 = 0. Solution : 1. Comparing the equation with ax^ + 6a; + c = : a =2, 6 =-3, c=-5. 2. Substitute these values in the formula : -h± V&2 - 4 ac 2a 3. Then x = -(-3)± V(- 3)^-4(2)(- 5) _2(2) _ + 3 zt: V9 + 40 4 _ 3 ± \/49 ^ 3 j: 7 4 4 4. .. xi- ^ -^_2,a:2- ^ - ^ - 1. Check : a^i = f ; does 2 (|)2 _ 3(f) - 6 = ? does2.^:f - V---6 = 0? does Y-V- 5 = 0? Yes. X2=-l; does 2(- 1)2 -3(-l)- 5 = 0? does 2 + 3-5 = 0? Yes. Example 2. Solve the equation 2ic2_3a._3_.0- Solution:!. a = 2, 6 =—3, c=— 3. 2. Substituting in the formula, x — — — ^— - — — — ~: za _ 3j-v/9^-24 _ 3rb V33 _ 3 + 5.744+ ^~ 4 ~ 4 ~ 4 * 8.744+ 1QA+ ^ - 2 .744+ ^q^, .'. xi = = 2.186+ ; X2 = =— .o8d+. QUADRATIC EQUATIONS 269 Check: 2.186+ The coefficient of x is — |, when the coeffl- - .686+ cientof a;2 = 1 ; 1.5 =_(_ |)_ 1.600 (For this method of checking, see § 195.) EXERCISE 122 Solve the equations : 1. 4r2-7r + 3 = 0. 12. 3id^ -T w-^2 = 0. 2. 6<--f 13^ + 5 = 0. 13. 5r + 8c = 4. 3. 2/2 = 62/4-72. 14. 9x' + 16x + S = 0. 4. a2_7a-30 = 0. 15. 15^2-22^-5 = 0. 5. 3ar^-2a;-33 = 0. 6. 3m2-f-5m-f 1 = 0. 7. os^ = 5s — l. 8. 6w'-llw-\-2 = 0. 9. 23^-3x-l = 0. 10. a^ + a;-l = 0. 11. 2/2-4^ + 2 = 0. 198. Summary of Methods of Solving a Quadratic. Four methods of solving a quadratic equation have been given : the graphical, by factoring, by completing the square, and by the formula. The first is useful mainly as a means of illus- tration ; the third is useful mainly in solving the general quad- ratic ax^ + bx-\-c = 0, and, thus, in deriving the formula. Historical Note. Greek mathematicians as early as Euclid were able to solve certain quadratics by a geometric method, about which the student may learn when he studies plane geometry. Heron of Alexandria, about 110 B.C., proposed a problem which leads to a quadratic. His solution is not given, but his result would indicate that he probably solved the equa- tion by a rule which might be obtained from the quadratic by completing its square in a certain manner. Diophantus, 275 A.r>. , gave many problems which lead to quadratic equations. The rules by which he solved his 16. 3 2 2 17. a; — - = 0. X 3 18. ,^ 2^ = 0. 19. 2 ^ = 5. d-1 d 270 algp:bra equations appear to have been derived by completing the square. He considered three separate kinds of quadratics. He gave only one root for a quadratic, even vi^hen the equation had two roots. The Hindu mathematicians, knowing about negative numbers, con- sidered one general quadratic. Cridharra gave a rule much like our formula. The Hindus knew that a quadratic has two roots, but they usu- ally rejected any negative roots. The Arabians went back to the practice of Diophantus in considering three or more kinds of quadratics. Mohammed Ben Musa, 820 a.d,, had five kinds. He admitted two roots when both were positive. Alkarchl gave a purely algebraic solution of a quadratic by completing the square, and refers to this method as being a diophantic method. In Europe, mathematicians followed the practice of the Arabians, i^nd by the time of Widmann, 1489, had twenty four special forms of equations. These were solved by rules which were learned and used in a mechanical manner. Stifel, 1486-1567, finally brought the study of quadratics back to the point that had been reached by the Hindus one thousand years before. He gave only three normal forms for the quadratic ; he allowed double roots when they were both positive. Stevin, 1548-1620, went still farther. He gave only one normal form for the general quadratic, as do we ; he solved this in both a geometric and an algebraic manner, giving the method of completing the square. He allowed negative roots. EXERCISE 123 Miscellaneous Examples Solve the following equations by any of the preceding methods. As a rule, solve by factoring if possible ; otherwise by the formula. 1 (3 a; + 2) (2 a? -f- 3) = (2; -- 3) (2 a; - 4). 2. 9(?/-l)2-4(2/-2)2 = 44. ^ 30 30 _^ g ic-f-2 4.-x_l m m -f-1 x—1 2 X 3 _3 ^==1, 7. -1^L_ 2__^ a — 6 a — 5 ' 2a-f-5 a — 1 _5 ^ = 6 8 2-3^ 4-^^11 x-l-4 x-2 ' * 4 t-2 4 QUADRATIC EQUATIONS 271 3a — 5 2a-f 5 _-. -- 2 w -\- S ^ — ^ _2 2a — 5 3a-f5 5 w; — 4 4v-3 2v 7-r 2r + l 14 2 20. 13. x-e 3(a;-l) 3x 14 y + ^ I y-2^ 6y + 16 _ ' 2^-2 ?/ + 2 3i^ 15. ^-+J— ^ = -1 1-t'^l + t 1-t 8 3r-6 7 11 -2 r 16. 5-r 2 2(5 - 2 r) 3-2a; 2 + 3x ^1 IGx + a^ 2 + a; 2-a; ~3 aj2-4 18. .V + 1 y + 2^ 2y + 13 7J-1 y-2 y-1 19. l-f---I^_+ ?J^ = 0. 3a; + l^(3a; + l)(7a;+l) 1 15a; (3a!+l)(l-5a;)' 2(1-5 a;)(7a;+l) (3a;+l)(7a;+l) 21. Solve the equation 2 p^x^ — 3 pqx — q^ = 0. Solution . 1. a = 2p^ ; h = (- Spq) ; c = (- q"^). 2. Formula: ^ ^-b ±Vb^ - i ac ^ 2a ^ - (- 3pg) zb V(- Spq)^ - 4(2p2)(- g2) ^ + 3pq± V9pV + 8pV 2(2 p2) 4p'^ f_Spq ± Vn p'^q^ ^ Spq±pqy/17 ^ pq(S ± VlT) ^ g(3 j, VlT) 4|)2 41)8 41)8 4i) 272 algp:bra Solve the following equations for x : 22. x^-6cx-{-5c^ = 0. 27. 2 P-2 Px -{- x^ = l. 23. 3x^=:2rx-\-2r^. 28. x'-(a-l)x = a. 24. a;2 + c2 — 2 ajp = 0. 29. oi^ + ax + bx -{- ab = 0. 25. x^-xp-\-(c^-d^) = 0. 30. ax^-(a-2b)x=z + 2b. 26. lgx^-hax=s. 31. ma;2+ (2mn-3 n)aj-6 7i2=0. EXERCISE 124 Review § 112 before solving this set of examples. 1. Twice the square of a certain number equals the excess of 3 over that number. Find the number. 2. If three times the square of a certain number be increased by the number itself, the sum is 10. Find the number. 3. Find two consecutive integers whose product is 306. 4. If the product of three consecutive integers be divided by each of them in turn, the sum of the three quotients is 74. Find the integers. 5. The sum of the squares of two consecutive integers is 685. Find the integers. 6. The sum of a certain number and its reciprocal is ^. Find the number. Hint : The reciprocal of a number is obtained by dividing 1 by the number. The reciprocal of a; is - • X 7. Find the dimensions of a rectangle whose area is 357 square feet if its length exceeds its width by 4 feet. 8. The numerator of a certain fraction exceeds its denomi- nator by 3. The fraction exceeds its reciprocal by ff . Find the fraction. 9. The main waiting room of the Union Railway Station in Washington, D.C., has an area of 28.600 square feet. The length exceeds the width by 90 feet. Find the dimensions. QUADRATIC EQUATIONS 273 10. Find the base and altitude of a triangle whose area is 63 square inches, if the base exceeds twice the altitude by 4 inches. 11. Find the dimensions of a rectangle whose area equals that of a square of side 24 feet, if the difference of the base and altitude of the rectangle is 14 feet. 12. Find the dimensions of a rectangle if its area equals that of a square of side 35 feet, if the difference of the base and altitude is 24 feet. 13. Find the dimensions of a rectangle whose area is 3750 square feet, if the sum of its base and altitude is 155 feet. 14. Find the dimensions of a rectangle whose area is 1701 square feet, if the sum of its base and altitude is 90 feet. 15. Find the dimensions of a right triangle if its hypotenuse is 20 feet and the base exceeds the altitude by 4 feet. 16. Find the dimensions of a right triangle if its hypotenuse is 26 feet and the sum of whose base and altitude is 34 feet. 17. Find the sides of an isosceles triangle if the perimeter is 35 inches and if the number of inches in the base is the quotient of 75 divided by the number of inches in one of the sides of the triangle. 18. A man travelled 105 miles. If he had gone 9 miles more an hour, he would have performed the journey in 1^ hours less time. Find his rate in miles an hour. (See p. 105.) 19. If a man travels 120 miles by one train and returns on a train whose rate is 10 miles an hour more, he will require 7 hours for the trip. What is the rate of the first train ? 20. A crew can row 8 miles downstream and back again in 4| hours ; if the rate of the stream is 4 miles an hour, find the rate of the crew in still water. (See § 144.) 21. A man travels 10 miles by train. He returns by a train which runs 10 miles an hour faster than the first, accomplish- 274 ALGEBRA ing the whole journey in 50 minutes. Find the rate of the first train. 22. A tank can be filled by two pipes running together in 2 hours. The larger pipe by itself will fill it in 3 hours less time than the smaller pipe. How long will it take each pipe to fill the tank alone ? (See § 142.) 23. Some boys are canoeing on a river, in part of which the current is 5 miles an hour, and in another part 3 miles an hour. If, when going downstream, they go 4 miles where the current is rapid and 8 miles where it is less rapid in a total time of 1| hours, what is their rate of rowing in still water ? 24. I have a lawn which is 60 by 80 feet. How wide a strip must I cut around it when mowing the grass to have cut half of it ? Hint : Referring to the figure, it is clear that if w = the number of feet in the width of the border cut, then the dimensions of the uncut part of the lawn are (60 - 2 ^/?) and (80 - 2 w?). Hence, (60 - 2 lo) (80 - 2 w?) = ^ • 60 • 80. Complete the solution. 25. A farmer is plowing a field whose dimensions are 40 rods and 90 rods. How wide a border must he plow around the field in order to have completed | of his plowing ? 26. The numerator of a certain fraction is 8 less than the denominator. If the denominator and numerator each be increased by 5, the resulting fraction is twice the fraction obtained by increasing the original denominator by 1. Find the fraction. 27. An automobile made a trip of 50 miles, 10 miles within the city limits and 40 miles outside the city limits. Outside of the city, the rate was increased 15 miles an hour. If the trip took 2| hours, find the rate at which they travelled within and outside of the city limits. w 'w wlw ' 80-2W 1 J CM |S W_" Iw --w"!w- QUADRATIC EQUATIONS 2T5 28. The numerator of a certain fraction exceeds its denom- inator by 5. If the numerator be decreased by 3 and the de- nominator be increased by 4, the sum of the new fraction and the original fraction is 3. Find the original fraction. 29. A bicyclist rides a number of hours at a number of miles an hour which exceeds the number of hours by 3 ; an automobilist, starting 3 hours after him, overtakes him by going two and one half times as fast as he did. Find the rate of each. 30. The circumference of the fore wheel of a carriage is less by 4 feet than the circumference of the hind wheel. In travelling 1200 feet, the fore wheel makes 25 revolutions more than the hind wheel. Find the circumference of each wheel. 199. Quadratic Equations having Two Unknowns. Example 1. Find two integral numbers the sum of whose squares is 90, and such that the larger exceeds twice the smaller by 3. Solution : 1. Let I = the larger number, and 8 = the smaller number. 2. Then 1^ + 8^ = 90. (1) and 1-2 8 = 3. (2) 3. Eliminate I by the substitution method (§ 168). From (2) l={2s + S). Substitute in (1): .-. (2 s + 3)2 + s^ = 90. .-. 4 s2 4. 12 s + 9 4- s^ - 90 = 0, or 6 s2 + 12 s - 81 = 0. Factoring : (5 s + 27) (a - 3) = 0. .-. s = 3, or - V. Check : If s = 3, Z = 9 ; 3^ + 92 = 9 + 81 = 90. — ^ cannot be a solution, as integral numbers were desired. As in simultaneous linear equations, when there are two unknowns, two equations must be given. These two equations may both be quadratic equations, or one may be a quadratic and the other a linear equation as in the example above. In this course, only the combination of 276 ALGEBRA one quadratic equation and one linear equation will be given. The substitution method of eliminating one variable is usually the most convenient. Example 2. Solve the pair of equations ~ ' \A [xy = — 15. (2) Solution : 1. Solve (1) fora; : x = (2 — y). 2. Substitute in (2) : y(2-y) = - 15. (3) 3. .-. 2y- 2/2+ 15 = 0, or y2_2y- 15=0. (4) 4. ... (y_ 5)(y + 3)=:0. .-.?/ = + 5 and 2/ =-3. 5. Since x = 2 — y; when y = 6, a; = 2 — 5 = — 3; when y = _3, x = 2-(-3) = + 5. First solution : oj = — 3, y = -f 5. Second solution : x =+ 6, y = — 3. Check : The solutions may both be checked by substitution. EXERCISE 125 Solve the following pairs of equations : fa;-f2/ = — 3. ^ [m^-f- mn — n^ = — 19. ' [xy = — 54:. " i m — n = — 7. 2 ^a^ + 6^ = 113. g ix-y = -i. a — b = — l. ' [xy = 4:5. 5x'-3y^ = -7. fy-2x=10, y + 2x = l. i3a^-32/' = -5. 27'2_rs=6s. f?.?^^ • V + 2s = 7. 10. a 6~5* g 2y + 2x = hxy. [a + 6 = 16. 2x + 2y = 6. f 18 14 ^g 3c + 2d = -2. 11- \r-s r + s cd + 8c = 4. [r-2s = l. 12. Find two numbers whose sum is 15 and the sum of whose squares is 113. QUADRATIC EQUATIONS . 277 13. Find two numbers whose difference is 9 and the sum of whose squares is 221. 14. Find two numbers whose difference is 7, and whose sum multiplied by the greater gives 400. 15. Find two numbers whose difference is 4, and the sum of whose reciprocals is |. 16. Find the number of two digits in which the units' digit exceeds the tens' digit by 2, and such that the product of the number and its tens' digit is 105. (See § 177.) 17. The sum of the squares of the two digits of a number is 58. If 36 be subtracted from the number, the digits of the remainder are the digits of the original number in reverse order. Find the number. 18. The area of a rectangular field is 216 square rods, and its perimeter is 60 rods. Find the length and width of the field. 19. A and B working together can do a piece of work in 6 days. It takes B 5 days more than A to do the work. Find the number of days it will take each to do the work alone. 20. The difference in the rates of a passenger train and a freight train is 10 miles per hour. The passenger train requires 1 hour more for a trip of 175 miles than the freight train re- quires for a trip of 100 miles. Find the rate of each. 21. The altitude of a certain rectangle is 2 feet more than the side of a certain square ; the perimeter of the rectangle is 7 times the side of the square, and the area of the rectangle exceeds twice the area of the square by 32 feet. Find the side of the square and the base of the rectangle. 22. Find the sides of a parallelogram if the perimeter is 24 inches and the sum of the squares of the number of inches in the long and short sides is 80. 23. One of two angles exceeds the other by 5°. If each is multiplied by its supplement, the product obtained from the 278 ALGEBRA larger of the given angles exceeds the other product by the square of the smaller of the given angles. Find the angles. 24. Two angles are supplementary. The square of the number of degrees in the larger angle exceeds by 4400 the product of the number of degrees in one angle by the number in the other angle. Find the number of degrees in each angle. 25. A man has two square lots of unequal size, together containing 74 square rods. If the lots were side by side, it would require 38 rods of fence to surround them in a single inclosure of six sides. Find the length of the side of each. IMAGINARY ROOTS IN A QUADRATIC EQUATION 200. Example. Solve the equation x^ — 2 x -[- 5 = 0. Solution : 1. Use the formula method of solving the equation. a = l, 6 = - 2, c = 5. 2 ^ ^ - 5 rfc V62 - 4 gc ^ -f 2 ::b V4 - 4 • 1 • 5 2a 2 The question arises what does V — 16 mean ? Is — 4 the square root of - 16 ? No, for ( _ 4)^= + 16. Is + 4 ? No, for ( + 4)^ = -f 16. Thus, no number with which the student is familiar will produce — 16, when it is squared. 201. No number raised to an even power will produce a negative result ; hence an even root of a negative number is impossible up to this point. To avoid this difficulty, a new kind of number is introduced. An Imaginary Number is an indicated square root of a negative number; asV— 16; V — 3; V— a^. The numbers previously studied, rational and irrational, are called Real Numbers. QUADRATIC EQUATIONS 279 202. Every imaginary number can be expressed as the product of a real number and V— 1. V — 1 is indicated by /, and is called the Imaginary Unit. Thus, V- 1«= \/l(>(-l)=± 4V - 1 =±4i. V-a^= Va-^(-l)=± aV-l =±ai. v/^^5 = VS ( - 1) = ± \/5 ■ y/^^ - ± iV5. HisTOKicAL Note. The symbol i for V— 1 was introduced by Euler, one of the greatest mathematicians of the eighteenth centuiy. EXERCISE 126 Express the following in terms of i : 1. V-49. 2. V-:31>. 3. V-100. 4. V-81al 5. V-25. 6. V-B4. 7. V-144. 8. V -121 a^b^. 9. V^ 10. V-^ 11. V^^. 12. V^=^. 13. V^^. 14. V-32. 15. V^T8. 16. V^^^T27 17. V-5. 18. V-27. 19. V-12. 20. V-28. 21. Simplify V-^. ■■■H- 22. \/--. 25. a/-". 28 25. i- 8 "9* 26. i- 75^ '4 * 27. <-- 20 23. \/-|- 26. \/-'^. 29 36 24. Xl-:^' 27. V'-^- 30 :i = ± 3tV3 2 .V: 32 "25* '■V- 125 64 -V^ 63 lOO' 280 ALGEBRA 203. Addition and Subtraction of Imaginary Numbers. EXERCISE 127 1. Add V^^ and V-36. Solution : V— 4 + V— 36 = 2i-\-Gi = 8i. Note. While every imaginary number, like \/— 4, has two values, cue positive and one negative, in i^roblenis such as the one in this exercise, only the principal root, the positive one, is used, as in the case of surds (§ 185). 2. V-16-hV-25. 3. 6. V-1+V-49-V-64. 7. V--36 4-V-100-V-81. 4. V-81-V-(54. 8. V-a2_V_4«2^y_9^.^2^ 5. V-100 + V-l()9. 10. 3 + V-2^ 9. V-lGx^ 13. V^28- V-25a;2 + V-49a;2. 11. V-18-V-8. 14. V-24-V^^r54 7H-V-63. 12. V^^^ + V^45. 15. V-44-f 2V-11-V-99. 16. Simplify 4-|±V-f Solution : 1. -± V- 27 _ 5 Vir27 ^5 3v/3- VC-l) 4 "2 2 2 2 ^n 3rV| 2 2 _5±3i\/3 The numbers in Examples 1-15 are called Pure Imaginaries, The sum, or difference, of a pure imaginary and a real number, § 201, as in this exercise, is called a Complex Number. QUADRATIC EQUATIONS 1 Simplify the following : .'■ i-H- » h^- u. iW-l- "'■ io*>/ ™ " IW-fe- - il^xFl 281 204. Meaning of Imaginary Roots of a Quadratic on the Graph. Example. Consider the equation x^-^ x-{-2 = 0. Solution : 1. Solve the equation by the formula : a = l; 6 = 1; c = 2. Xi = l+iV7. X2 -l-^V-7 ^ -l-fct V7 2 2 - 1 - I V7 2 2 2. Solve the equation graphically. (Review rule § 193.) Let y = x2 + x + 2 : When X = + 1 + 2 + 3 -1 -2 -3 -4 then y = 4-2 + 4 + 8 + 14 + 2 + 4 + 8 + 14 3. The graph has the same shape as the graphs obtained when solving other quadratic equations ; but the graph does not cross the horizontal axis at all. Hence, y or x^ + x + 2 is never zero for any real value of X. This is charaQteristic of the graph of a quadratic which has imaginary roots. h 4'--3- ♦K) ♦2_+3. t .t$_ +5. 282 ALGEBRA V EXERCISE 128 Solve the following equations. Express the roots in simplest form. Draw the graphs for the first three equations. 1. a;2 + 2.T-|-3 = 0. 2. x2 + 3x-|-4 = 0. 3. 2x''-x-\-2 = 0. 4. 3m2-2m + 5 = 0. 5. 5c2-7c + 3 = 0. 6. St' = 9t--5. 7. ll?-24-6 = 15r. 9. 2 x-\-S ' 3a; 5 = 0. 10. 2a- + 3 _x X + 7 11. 30^1 2 "^a; -1 = 0. 12. 2a;2-5aa; + 7a2 = 0. 13. Ss^-4.xw-^2w^=:0. ^_3^ 1_^ ^ 5 10 2 * 14. 5x^-7 xt-\-3t^ = 0. XVI. SPECIAL PRODUCTS AND FACTORING ADVANCED TOPICS 205. In paragraph 98 is the rule : " The product of the sum and the difference of any two numbers equals the difference of their squares " ; thus, {x + y){x — y) = x^ — y^ for all numbers X and y. lix = 2a and t/ = 3 6, (2 a + 3 6)(2 a - 3 6) = 4 a2 - 9 62. If a; = 14 and y = 5, (14 + 5)(14 - 5) = 169 - 25 = 144. If a; = (a 4- b) and y = {c -\- d), then similarly [(a + &) + (c + (?)][(« + h) - (c + d)^ = (a + 6)2 _ (c + dy. Likewise, in any of the type forms studied in chapter VIII, the numbers may be general number expressions. Example 1. Multiply (a + 6 + c) by (« + 6 — c). Solution : 1. (a + 6 + c)(a + 6 — c) = {(« + 6) + c}{{a + 6) — c} = (a + 6)2 _ c2 = a2 + 2 a6 + 62 - c\ Here x = {a -\- h) and y = c. Example 2. Multiply {r -^ s -\- 1 — n)hy {r -\- s — t -\- n). Solution : 1. (r j^ s -\- 1 — n){r -{- s — t -\- n) = {(r -hs) + (t- n)}{(r + s) - (t - n)} = (r + s)2_ (t - ny = r2 + 2 rs + s2 _ ^2 4. 2 «n - n2. Here, x = (r + s) and y = (t — n). Note. In such examples, the rules for introducing parentheses (§ .50) are used. The various terms of the expressions may be rearranged, if necessary, so that one factor becomes the sum and the other the difference of the same two numbers, when the terms are grouped. 283 284 ALGEBRA EXERCISE 129 Find the following products mentally : 1. S(a-\-b)+5ll(a + b)-5l. 2. l(m-^n)-2pl{{m-\-n)-\-2pl. 3. \10-(r-^s)lllO-\-{r + s)\. 4. sSp-{c+d)l\S2i-{-(c-{-d)l. 5. \(c + 2 d) - 11 all(c + 2 d)-h Hal. 6. (a — b + c)(a — 6 — c). 7. (x-y -{-z)(x-y -z). 8. (a' + a - l)(a2 - a + 1). 9. (a^ -f a6 4- ?>')(a2 - ah + 6^). 10. (a -{-2b -3 c)(a _ 2 5 + 3 c). 11. (3a; + 42/ + 22)(3aj-42/-22;). 12. (a^ + aj- 2)(a^-a;-2). 13. (a + r — c + f^) (a-{-r-{- c — d). 14. (a — 6 + m -{- 7^) (a — b — m — n). 15. (2 £c 4- 2; — ?/ + w)(2 X — z — y — ^'). 16. J(« + ^) + 2(a - 6)} J(a + &) - 3(a - b)}. Solution : Just as (x -\-2y)(x—Zy) = x^ — xy — 6y^ so {(a + 6) +2 (a -&)}{(« + &) -3 (a -ft)} = (a + 6)2- (rt + 6)(a_6) _6((i-6)2 = (a2 + 2 a& + &2) _ (^2 _ ft-2)_ 6(a2 _ 2 a& + ft^) = a2 + 2 a& 4- &2- a2 _{_ ?^2 _ a2 _^ 12 a6 - 6 62 = Uab-6a'^-4b^. Note, ic = (a + 6) and y = (a — 6). 17. J(^^^ + *0- 4H0^ + w)- 5J. 18. ;(a;_iy) + 8n(aj-?/)-6(. 19. {Sx-(y-hz)il2x-(y-{-z)l. SPECIAL PRODUCTS AND FACTORING 285 20. \x-}-3y-^15zl\x-\-3y-10z\. 21. lr + 2s-3tl]r + 2s-^7t\. 22. \Sp-4{q-\-r)\\4.p-5((j-{-r)l. 23. \^-\-2x-hl\lx'-{-2x-5l. 24. [(a + b) - 5Y. 28. [ft -\-3b- cy. 25. [0+(m-n)]'- 29. [a - 6 + c - d]". 26. [2rt-(c + d)]2. 30. [2r + s-«H-a;]'- 27. [a + 6 + c]2. 31. [3a-6 + 2c-dJ. 206. General Problems in Factoring. Example 1. Just as scF—if = {x-{- y){x — y) so (m - nf - 25 a' = \{m - n)+ba\ \{m-n)-6a\. Example 2. Just as ar^ - 3 a; — 88 = (a; — 11) (a; + 8) %o{a-2hf-3{a-2h)-%^=\(a-2h)-ll\\{a-2h) + ^\. = (a-26-ll)(a-26 + 8). Note, x is (a — 2 6). EXERCISE 130 Factor completely the following expressions : 1. (a + 6)2_c2. 11. {x'-4.y-{x + 2)\ 2. {m-nY-x'. 12. 9(m-r2)2- 12(m -7i)+4. 3. ar^__(y_^.2;)2. 13. (a; - 2/)2 - (m - ?i)l 4. m2-(7i-i))2. 14. (a2-2a)^+2(a2-2a)+l. 5. ilx-2yy-y\ 15. (l+w')'-4«l 6. (a + 6)2-|-23(a-f6)-f 00. 16. (ar'+3 a;)^ 4-4(a;24-3a;) + 4. 7. (a;_2/)2 + 2(a;-?/)-63. 17. (9 a^ + 4/- 144 al 8. (a;4.2/)2-5(aj + 2/)-36. 18. (a2-|-7a)H20(aH7«)-96. 9. {r-\-8f+4:{r + s)t-^f. 19. (m + w)2 4-7(7?i4-rO- 144. 10. (j9-g)2-i-8(/>~9)y-20 7^. 20. (a;^ ^ ^j _ 9^2 „ 9^ 286 ALGEBRA 21. (x-\-yy-z\ 26. (x-^yy-\-(x-y)\ 22. (r-\-sy-{-8t^ 27. ^x^-ix"" -{-ly. 23. {m + 7if-(m-ny. 28. 27 m^ -(m -nf. 24. a3+(a + l)^ 29. (2 a- 6)«- (a + 2 ?))^ 25. a3-8(a + 6)^ 30. {x + Syy- {x-3 yy. FACTORING POLYNOMIALS 207. Polynomials Reducible to the Difference of Two Squares. Certain polynomials may be put into the form of the differ- ence of two squares by grouping certain terms. Example 1. Factor 2 mn -\-m^ — l-\- v?. Solution : L 2 mn + w^ — 1 + ti^ = (m^ + 2 mn + n^)— 1. = (m 4- w)2 - 1. = (w + n + l)(m + w - 1). (§206) Example 2. Factor o? - (? ^h^ ^d? -2 cd- 2 ah. Solution : 1. a^ - c^ + IP- -d:^-2cd-2ah = (a2 -2ab-\- b'^)-(c^ + '2cd + cP) = (a_6)2_(c + (Z)2 = {(«- 6) + (c 4- d)]{(a - b)-(c + d)} = (^a — b + c-\-d)(a — b — c — d). EXERCISE 131 Factor : 1. a^-2ab-\-b^-c^, 6. 2 mn-n^-^1 -m\ 2. 7rv'-^2m7i + n^-p\ 7. 9a2-24a6-f 16&2_4c2. 3. a^-x^^2xy-y\ 8. 16ar^-4?/2 4.202/:3-2522. 4. a^ — i/'^ — 2^ + 2 2/2. 9. 4: 71^ -\- m^ — x^ — 4: mn. 5. 52_44.2a6 + a2. lo. 4^2-66-9-62. 11. 10xy-9z^-^y^-^25x\ 12. a2-2a6 + 62-c24-2cd-d2. 13. a^ — b'^ + x" — y^-{^2ax-^2by. SPECIAL PRODUCTS AND FACTORING 287 14. J^ -(- m^ — if — ir — 2 mx — 2 ny. 15. 2xij-a' + x^-2ab-b--\-if. 16. 4a2 + 4rt6 + ^/-9c^ + 12c-4. 17. 16y'-3(^-Sxi/-z^ + x'-12z. 18. m^ - 9 n^ + 25 a^ - b- - 10 cwi + 6 6m. 19. 4a2-c2-12a6 + 2crf + 962_d2. 20. 9x*-4:X^-\-z^-6x'z-20x7j-2ri!/'. 208. Certain polynomials can be factoreil by (jronpiny their terms. Type Forvi : ab + ac + bd -\- cd = {a -\- d){b -\- c). Example 1. Just as ax -\-bx = (a-\- b)x (§ 34) so, a(x + 2/) + b{x + y) = (a-^b)(x-\- y) (§ 34) Example 2. Factor 6 a;^ - 15 a^ - 8 a; -f 20. Solution: 1. Qx'^-\bx^-^x-\-20={(d j?-\^x^)-{'^x~2Q) (§47) = 3x2(2x-5)-4(2x-5) (§94) = (3x2_4)(2x-5). EXERCISE 132 Factor < 1. 2a(a; + 2/)-3(.r4-2/). H. 2 + 3 a-8 a^- 12a^ 2. 5 m(r -h s) + 2 n(r 4- .*J). 12. 3 ar^ + 6 a^ + a; + 2. 3. Sp(2x — y) — r(2x-y). 13. lOma;— 15 >ia;- 2m+3n. 4. 8(f + 1«) — wi(^ 4- w). 14. a'^a; + atca; — a^by — bhy. 5. a{h-\-c) — dih + c). 15. a^bc — cw^d -\- abhl — bccE 6. a6 4- a?t + 67M + m7i. 16. 30 a'^ — 12 a^ _ 55 a + 22. 7. ax-ay + bx-by. 17. 56 - 32 a; 4- 21 a.-^ - 12 arl 8. ac — ad — 6c 4- 6d. 18. 3 aa; — a?/ 4- 9 6a; — 3 by. 9. a« + a2 4-a + l. 19. 4 ar' 4- ar^/ - 4 / - 16 a;?/. 10. 4ar' — 5»* — 4 a; — 5- 20. rt — rn — sn -\- st. 288 ALGEBRA 21. ar -[- as -^ br -{- hs — cr — C3. 22. ax-\-a}j ~ az + bx — bz-^ by. 23. am — bm — cp -\- ap — mi — bp. 24. x^ — ic^;- + a;^?/ + xy'^ + ^/"^ — yz'^. 25. 2 ax + ex -i- S by — 2 ay — 3 bx — cy. 209. In the following list of examples, the types of factor- ing stndied in this chapter will be used. Before taking up the list of examples, review, if necessary, the rules for obtain- ing the H. C.F. of .two or more expressions in chapter IX and the rules for operations with fractions in chapter X. EXERCISE 133 Find the H. C. F. and the L. CM. of the following : 1. 3 a^ - 21 a^ - a + 7, and a^ + 6 a - 91. 2. ac + ad —be — bd, and a^ — 6 ab -\- i) b"^. 3. a^-\-b^-c' + 2 ab, and a^ -b~-c^-\-2 be. 4. m^ - 4 m, m^ + 9 m^ — 22 m, 2 m"^ — 4 m^ _ 3 m^ + 6 m. 5. 3a^-a'b-\-Sab- b\ 27 a^ - b\ 9 a^ - 6 «?> + b\ 6. 16 mi'^ — 71'^, 16 711^ — 8 771^71^ + n*, 2 7nx + 2 my — 7ix — ny. 7. a^ — d~x — av? + ar^, 3 a^ — 3 a^x -f 5 aa^ — 5 a^. Reduce the following fractions to lowest terms : x^ — y^ -\- z^ -\- 2 xz ^ ax — bx — ay -f by ' d? — y^ — z^ -\-2yz a? — V^ 4m-^-10?yi^-6m+15 ^^ 2ae-2bc- ad-^bd 6m'^ + 8m=^-9m-12* * d^-^(? Simplify the following : 13. x^ -\- y''^ — z^ — 2 xy x^ — y'^ — 2 yz a^ + &-^ f. ab + c2 64.C V a'-ab + b'- SPECIAL PRODUCTS AND FACTORING 289 a^—h- — c^ — 2hc . a — h — c rt2_52_^_,.2^,c * a + h-c ar — as-{- br —bs ar + as 4- ^^f + ^^ r^-^ ' a^^2ab-^b' 16. Solve the equation : x^+ax — 3ab — 3 bx = 0. Solution : 1. a;'^ + aa: — 3 a6 — 3 hx = 0. 2. Factoring: ic(x + a)- 3 &(»; + rt) = 0. (x-3&)(x + a) = 0. 3. .-. a: = 3 Z> or a; = - a. (§ 110) Check : When x = 8 6, does (3 6)* + a(3 6) - 8 aft - 3 6(3 6) = ?, , Does 62 4. 3 a?^ _ 3 a?, _ 9 ?;2 = ? Yes. Whenx=- a, does (- a)2 + a(- a)- 3 a6 - 36(- a) =0? Does a2 - a2 - 3 a6 + 3 a6 = ? Yes. 17. x^ -\-ax — ab — bx=0. 19. ax'^ — bx— acx-\-bc = (). 18. ar^ — 2?w7i-m-— ?r = 0. 20. ax- —2 dx-Q>d-\-3 ax=0. XVII. RATIO, PROPORTION, AND VARIATION 210. The Ratio of one number to another is the quotient of the first divided by the second. Thus, the ratio of a to 6 is -; it is also written a : h. The h numerator is called the Antecedent and the denominator is called the Consequent. All ratios are fractions and are subject to the usual rules for operations Avith fractions. 211. The ratio of two concrete quantities may be found if they are of the same kind and are measured in terms of the same unit. Thus, the ratio of 3 lb. to 2 lb. is | ; and the ratio of 350 lb. to 2 tons is ?¥A or A. EXERCISE 134 Express the following ratios and simplify them : 1. 3 to 9. 3. 5.Tto2a;. 5. | to yV 7. 25 to 375. 2. 12 to 2. 4. 6anol5al 6. y\ to i 8. d'-hHo o?-h\ 9. A line 15 inches long is divided into two parts which have the ratio 2 : 3. Find the parts. Solution: 1. Let x — the short part. 2. Then 15 — a; = the long part. 3. Then -^^_ = ?. 15 -X 3 Complete the solution. 10. Divide a line 63 inches long into two parts v/hose ratio is 3 : 4. 290 RATIO, PROrORTTOX, AND VARTATTON 291 11. Divide 36 into two parts such that the ratio of the greater diminished by 4 to the less increased by 3 sliall be 3 : 2. 12. The ratio of the height of a tree to the length of its shadow on the ground is 17 : 20. Find the height of the tree if the length of the shadow is 110 feet. 13. Divide 99 into three parts which are as 2 : 3 : 4. Hint : Let the parts be 2 a;, 3 a;, and 4 x. 14. Divide a farm consisting of 720 acres into parts which are as 3 : 5. 15. Divide $ 1000 into 3 parts which are as 5 : 3 : 2. 212. A Proportion is a statement that two ratios are equal. The statement that the ratio of a to 6 is equal to the r^tio of c to d is written either — = -, or a: = c : (6. h d This proportion is read " a is to 6 as c is to d.'' Thus 3, 9, 5 and 15 form a proportion since | = ^^3^. HisTORicAF. Note. Leibnitz, 1646-1716, was instrumental in estab- lisliing the use of the form a : b = c : d. 213. The first and fourth terms of a proportion are called the Extremes, and the second and third the Means. In the proportion a: b = c: d, a and d are the extremes, and b and c are the means ; a and c are the antecedents, and b and d are the consequents. EXERCISE 135 Find the value of the literal number in the first six of the following exercises and of x in the remaining ones : 1 - = A 3 Jl=i: 5 ^ — x _^ ' 3 27' * 16 5* 3 ~2* < 16^ c "5 9 24' z 2 ?=^^. 4 ^ = ? 6 2— ? = 5 ' y iO ' 24 « "4 + ^2' 292 7. a _x b~~c 8. a X 2b~3c ALGEBRA 9. ^ = ?:. SX t 11. a — x a X b 0. ^ = A. 12. a n np nx x—m X 214. A Mean Proportional between two numbers a and b is the number x in the proportion a : x = x : b. 2 X A mean proportional between 2 and 3 is aj in : - = - • . X 3 .-. x2 = 6 ; x= ± V6. Thus, there are two mean proportionals between any numbers. Gener- ally the positive one is used. 215. The Third Proportional to two numbers a and b is the number x in the proportion a : b = b : x. 2 3 Thus, the third proportional to 2 and 3 is x in : - = - ; 3 X .*. 2 ic = 9 and x = 4.5. 216. The Fourth Proportional to three numbers a, b^ and c is the number x in the proportion a:b = c: x. 2 4 Thus, the fourth proportional to 2, 3, and 4 is the number x in : - = -< O X y .-. 2 a: = 12 and x = 6. Note. The numbers must be placed in the proportion in the order in which they are given, as in the illustrative examples. EXERCISE 136 Find the fourth proportional to : 1. 2,5, and 4. 4, 35, 20, and 14. 2. 5, 4, and 2. 5. 6 a, 2 b, and c. 3. 7, 3, and 14. 6. x, y, and xy. Find the mean proportionals between : 7. 18 and 50. 9. 2 a and a. 8. 2^ and f . 10. 12 m^w and 3 mn^. RATIO, PROPORTION, AND VARIATION 293 11. — and — -^- — 12. ar^ — v^ and — - — '^ ^ « a+4 a-f2 ^ x — y 13-16. Find the third proportional to the numbers in ex- amples 7, 8, 9, and 10. 17. Find the third proportional to a^ — 9 and a — S. 18. Find the third proportional to 10 a; and 3 y. 19. Find the fourth proportional to : 2 a^ — 2 2/^ a^ — y^ ^ax — by-\-ay — bx a-\-b ' ^^^^^' ^^ a^J^xy-^f ' 20. Find the mean proportionals between : ax — ay — bx -\- by a^ — y^ x^ + xy-\-y'^ (a — bf PROPERTIES OF PROPORTIONS 217. In a proportion, the product of the means is equal to the product of the extremes. This property of a proportion is proved as follows : If - = -, then ad = be, by clearing of fractions. b d Example. Since § = |, 2 • 9 should equal 3 • (y. Does it ? 218. If the product of two numbers is equal to the product of two other numbers, one pair may be made the means and the other the extremes of a proportion. If mn = xy, then ^ = ^ . X n Prove this by dividing both members of the given equation by nx. Prove that the following proportions also are true : (a) ^ = ^ (divide by ny). (ft) -^ = ^ . (c) - = -^ • ^ ^ y n ^ ^ ^^ ^ ^ m y ^ ^ x m ExAMPLK 1. Since 3 • 8 = 6 • 4, f should equal |. Does it ? Example 2. Write three other proportions which should be true ac- cording to the property given in this paragraph. 294 ALGEBRA ' 219. In any i^roportion, the terms are in proportion by Alter- nation ; that is, the first term is to the third as the second is to the fourth. f rx: a c ah it - = - , prove - = - . b d^ c d Suggestion. Use § 217 and then divide both members of the equation by cd. Example. Since | = ^\, then | should equal j%. Does it ? 220. In any proportioii, the terms are in proportion by Inver- sion ; that is, the second term is to the first as the foiirth is to the third. TP a c b d If - = -, prove _ = - . b d a c Suggestion. Use § 217, and then divide both members of the equation by ac. Example. Since | = j%, then | should equal \2,. Does it ? 221. Bi any proportion, the terms are in proportion by Compo- sition ; that is, the sum of the first tivo terms is to the second as the sum of the last two terms is to the fourth. yp a c a-\-b c-\- d If - = - , prove ^ = ^ - b d^ b d Suggestion. Add 1 to both members of the given equation. 24 24-6 44- 12 Example. Since - = — , then "^ should equal — ^^^^ — . Does it ? 6 12 6 ^12 222. In any proportion, the terms are in proportion by Divi- sion ; that is, the difference of the first two terms is to the second, as the difference of the last two is to the fourth. xnttc a — b c — d It - = - , prove = • b d' ^ b d Suggestion. Subtract 1 from both members of the equation. Example. Since — = ^, then ^^-=^ should equal i^^ • Does it ? 2 8 2 ^8 RATIO, PROPORTION, AND VARIATION 295 223. //* cm 1/ proportion, the terms are in proportion by Compo- sition and Division ; that /.s, the sum of the first two terms is to their dijference as the smn of the last two terms is to their difference. T* « c _„^„^ a-\-h c-\-d It - = -, prove — ——■= — ■ h d a—bc—d Proof. 1. Since ?=-, then ^^Jt^ = !^-±^. (Composition) b d b d 2. Since - = -, then ' ^Lsil = ^--A, (Division) b d b d ^ ^ 3. Divide the members of the equation in step 1 by those of the equa- tion in step 2 : , , 7 , ^ i ^ a + b . a — b _ c + d . c — d b b ~ d d ' 4. Simplifying step 3: a±b^c±d a— b c — d Example. Since ^ = ^ , then, ^^-±^ should equal i^-±-^ • Does it ? 2 8 '10-2 ^ 15-3 224. In a series of equal ratios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent. yo a c e . a -\- c -\- e -\- etc. a If - = - = -, etc., prove — — — — — — = - • b d f ^ 6 4-ci+/+etc. b Proof. 1. Let v = the common value of the equal ratios ^ , -, - , etc. b d f 2. Then since - = v, a = bv. b - = V, c = dv. d - =v, e=fv. 3. Then (a -\- c + e)= bv + dv -\-fv = v{b + d +/). 4. U(b+d+/) ' — ' ■ — = v. .•. — ■ = or - or — • b+d+f h^d+f b d f Example. Since^-=-=A J_±A±A should equal ^ . Does it? 2 6 10 2 + 6+ 10 ^ 2 Historical Note. All of these properties of a proportion were known to Euclid, 300 B.C. 296: ALGEBRA 225. There are several other properties of a proportion which follow directly from properties of an equation or of a fraction. . (a) If - = - then — = — • Raise both members to the third b d^ 6^ # power. (h^ Tf - — - thpTi '^^ — Extract the cube root of both b~d' \/?> ~ \/d members. Multiply numerator and denomi- (c) If T = -J then — - = -— • nator of the first ratio by m, and b d mb nd (a) If - = -, then — - = — -• -u rn b d nb nd equation by — of the second by n. Multiply both members of the 11 226. In the preceding paragraphs, some of the simple prop- erties of a proportion have been given. There are many others which may be derived by means of these simple properties. ^ J. a c 2 a 4- 3 6 2a- 3 6 Example. If - = -, prove !—-—=- -— • h d' ^ 2c + 3d 2c-3d Proof. 1. Since ? = ^. then |^ = |^; • (§ 225, d) b d 6b od 2. Then 2 a + 36 ^ 2 c + 3d, (By § 223) 2a-3& 2c-3d! K y ^ J 3. Then 2 a -f 3 6 ^ 2a_-3j, (By § 219) 2c+3(Z 2c-3(Z ^ ' ^ ^ EXERCISE 137 1. Write by inversion : / \ 3 15 ,,x 2 m ,. a X (") 4 = 20- ^*) 5=x- ^"^ b = V 2. Write these same three proportions by alternation. 3. Write these same three proportions by composition. 4. Write these same three proportions by division. 5. Write the proportion (c) in Example 1 : (a) by inversion and the result by composition ; RATIO, PROPORTION, AND VARIATION 297 (&) by alternation and the result by division ; (c) by composition and the result by alternation ; (d) by division and the result by inversion. 6. -r ,. m X J.^ 1. VI -\-n n If _ = - , prove that = - • n y x-^y y 7. If -=z-, prove that — ^-- = ■ • s b ^ r a 8. If _ = - , prove that = - • b d^ c~d d 9. If -=^, prove that „.= ,.• 10. If 2_^, prove that 2«-36^2c-3cJ EXERCISE 138 Proportion in Geometry 1. In a triangle in which DE is parallel to BCj m : r To test this truth : (a) measure m, n, r, and s ; (6) find the value of the ratio m : r and of n:s; (c) compare these two ratios. *This truth may be tested in any triangle. It may be expressed thus : the upper segment on one side is to the lower segment on that side as the upper segment on the other is to the lower segment on the other. 2. Write the proportion — = - by alternation. Express the resulting proportion in words as in Example 1. 3. Write the proportion of Example 1 by composition and express it in words. 4. Write the proportion of Example 1 by inVBrsion and express it in words. 6. If ^ID = 7, DB = 4, and .AE = 8, find EO. 298 ALGEBRA 6. If AB = 12, AD = 5, and AC = 14, find AE. Hint. Let AE = x, and C^ = 14 - x. 7. If AD = Z>jB, how does AE comi)are with EO? 8. If AD = 20, />B = 8, and AC = 30, find AE and ^Xl 9. If two perpendicular lines BC and DE are draAvn from one side of an angle to the other, then BC : AC = DE'.AE. Test this statement by measuring the lines in the figure and finding the value of the ratios. 10. Draw any other perpendicular as XY. Find the ratio of Xy to AY and compare the ratio with those found in Example 9. What do you conclude about all ratios obtained by dividing the length of the perpen- dicular by the distance from A to the foot of the perpendicular (like AY)? 11. Using the fact stated in Exam- ple 9, tell how to find the height of the tree in the ^gure, if the height of the rod and the lengths on the shadows of the tree and the rod are as indicated. 12. Suppose that EF and AC are perpendicular to DC in the adjoining figure. Suppose that EF = 10 feet, OF = 12 feet, OC = 150 feet, and BC = 20 feet. Determine AB. 13. Suppose that CD and AB are perpendicular to AE in the adjoining figure ; that AX = 5 feet, YB = ^ feet, AE = im feet, CE = 25 feet, and CD = 30 feet. Find XY RATIO, PROPORTION, AND VARIATION 299 VARIATION 227. Some quantities change or vary and are called Variable Quantities. Thus, the distance between a moving train and its destination varies, — ihat is, it decreases ; the age of an individual varies from moment to moment, — that is, it increases. 228. A quantity which is fixed in any given problem is called a Constant. Thus, if a workman receives a fixed sum per day, the total wages due him changes from day to day if he works and remains unpaid. His daily wage is a constant ; his total wages is a variable. 229. A change or Variation in one quantity usually produces a variation in one or more other quantities. Such variables are called Related Variables. For each value of one variable there is a corresponding value of the other variable, or variables. Thus, if the side of a square is increased, the perimeter and the area of the square are also increased. 230. Variation is the study of some of the laws connecting related variables. Instead of the quantities themselves, their measures in terms of certain units of measure are used. Thus, distance is expressed as a number of miles, rods, or other units of lengtli ; weight is expressed as a number of units of weight ; area is expressed as a number of units of area. 231. One quantity varies directly as another when the ratio of any value of the one to the corresponding value of the other is constant. Thus, the ratio of the perimeter of a square to the side of the square is always 4, becau.se the perimeter is 4 times the length of the side ; there- fore the perimeter varies directly as the side of the square. 232. The symbol, (x, is read "varies as"; thus, acchis read " a varies as 6." 800 ALGEBRA If xcjzy, then - = m, where m is a constant, expresses the relation between any two corresponding values of x and y. (See § 231.) Since - = m, then x = my. Either equation may be used to express direct variation. 233. One quantity is said to vary inversely as another when the product of any value of the one and the corresponding value of the other is constant. Thus, the time and rate of a train going a distance d are connected by the equation rt = d. If the distance remains fixed, then the time varies inversely as the rate ; for example, if the rate is doubled, the time is halved. If X varies inversely as y, then xy = m, where m is a con- stant, expresses the relation between them. If xy = m, then also a; = - . Either equation may be used to express inverse variation. 234. One quantity is said to vary jointly as two others when it varies directly as their product. If x varies jointly X as y and 2;, then — = m, where m is a constant, expresses the relation between the variables. Thus, the wages of a workman varies jointly as the amount he receives per day and the number of days he works ; for, letting W equal his total wages, w his daily pay, and n the number of days he works, then W= nw. Here m = 1. Again, the formula for the area of a triangle is A = \ah. This shows that the area of a triangle varies jointly as the base and altitude, (Here m = \.) RATIO, PROrORTION, AND VARIATION 301 235. One quantity may vary directly as a second and inversely as a third. Let x vary directly as y and inversely as z ; then my x = -^ z expresses the relation between the variables. Notice that this combines the equation for direct variation of y and inverse variation of z. 236. Variation of more complicated related variables needs to be expressed sometimes. Example 1. xcKy^ may be written x = my^. Example 2. af oc y^ may be written ar' = 7ny^. Example 3. The volume of a circular cylinder varies jointly as the altitude and as the square of the radius. This may be expressed : v oc ai-^, or v = ka'i'^. Example 4. a varies directly as q, and inversely as (P. This may be expressed : a= -—• EXERCISE 139 Express the following relations both by means of the symbol oc and by an equation: 1. The area of a rectangle varies jointly as the base and altitude. , , 2. The area of a circle varies as the square of the diameter. 3. The volume of a rectangular prism varies jointly a3 the length, width, and height. 4. The distance a body falls from a position of rest varies as the square of the number of seconds in which it falls. 5. The interest varies jointly as the principal, the rate,' and the time. 302 ALGEBRA Express the following relations by mearis of equations : 6. The rate of a train varies inversely as the time, if the distance is constant. 7. The rate of gain varies inversely as the capital invested, if the total gain is constant. 8. The weight of an object above the surface of the earth varies inversely as the square of the distance from the center of the earth. 9. The per capita cost of instruction for pupils in a school room "Varies directly as the salary of the teacher and inversely as the number of the pupils. 10. The volume of a circular cone varies jointly as the alti- tude and the square of the radius. 11. It z varies jointly as x and y, and equals f when .V = 4- and ic = f , find z when y =^ and x = ^. Solution. 1. According to the conditions z = mxy. 2. . •. - = m • - . ^ , or m = - , since s = ^, when x = | and x = L 6 ^ 5' 3' " ^ ^ 3. . •, = I xy, substituting | for m. 4. . • . ^ = f • ^ ■ ^ = J^, when x = | and y = l. Note. In such problems, first find the constant, as in step 2. 12. li y ccx and is equal to 40 when x = 5, what is its value when x = 9? 13. If yocx^ and is equal to 40 when a; = 4, what is the equation for y in terms of a;? 14. If X varies inversely as y and is equal to | when y = f? what is the value of y when a? is | ? 15. If (5 a; -f 8) oc (6 2/ — 1) and x = 6 when y = — 3, what is the value of x when y = 7? 16. The distance fallen by a body/ from a position of rest, varies as the square of the number of seconds in which the RATIO, PROPORTION, AND VARIATION 303 body falls. If it falls 256 feet in 4 seconds, how far will it fall in seconds ? 17. The interest on a sum of money varies jointly as the rate of interest and the principal. If the interest is ??875 when the rate is 5% and the principal is $3000, what is the interest when the rate is 6% and the principal is $2500? 18. The principal varies directly as the interest and inversely as the rate. If the principal, $4000, produces $250 interest at 4%, what princij)al must be invested for the same time to yield $500 at 5% ? 19. The number of tiles required to cover a given area varies inversely as the length and width of the tile. If it ta,kes 270 tiles 2 inches by 5 inches in size to coyer a certain area, how many tile 3 inches by 6 inches will be required for the same area ? 20. The number of posts required for a fence varies inversely as the distance between them. If it takes 80 posts when they are placed 12 feet apart, how many will be required when they ate placed 15 feet apart ? XVIII. GENERAL POWERS AND ROOTS 237. In the preceding chapters, only positive integers have been used as exponents. The fundamental definition, when w is a positive integer^ is : a^ = a ' a ' a '" a (m factors). (§ 15) More general powers occur in mathematics. When m and n are positive integers : I. Multiplication Law. Just as a^^ • a* = a^*, (§ 53) so a'" X a" = a'"+". II. Division Law. Just as aF -i- o^ = a^, (§ 68) soa'" -^(f=:if"-". - III. Power of a Power. Just as (a^ = a'% (§ 89) so {(f"y = V)i 64. a ' Va. 72. (7/^2'^)l 80. {mhi')K 65. .T-" . -^. 73. (m~^ . n^)^. 81. (x-WK 66. m- - Vm. 74. (x-^y. 82. («-W)+l 67. x-^-i-x-^. 75. (a-^6^)«. 83. -v^m • Vn. 68. z-*^z-\ 76. (a2&4)i. 84. v^.^. 85. (J + ah^ -\-b^)(J- •b^). 86. (4.x-^-6x-'-\-9)(2 x-^ + S)^ 87. (2 a-i - 7 - 3 a)(4 a-^ + 5). 88. (a;-^ + 2)(x-^ - 2). 92. (05- - r)(^" + r)- 89. {x^ 4- 7)(a;^ + 3). 93. ^r - 2/")^- 90. (a« + ^'O -^ (^^^ + '->^)- 94. (a;2- -f: y^)2. 91. (a-* — 1) -^ (ot"^ — 1). 95. (a^ + 4)(aj'» — 7). 96. (x"" + 3)(a;'« — 5). 97. (aj^*" -I- 2 a;"' 4- l)(a;'^ + 1). 98. (a^" + a"6" + 62'»)(a-" — a'^^" + ^'"). 99. (ar^ + 3 a:^"^ 4- 3 aj^ 4-' /) -^ (a?" + Z/)- 100. (r^ — s^*)-H(r"* — s*). MISCELLANEOUS EXERCISES A. ADDITION AND SUBTRACTION Simplify : 1 .; . ax 4- hx + ex. 3. 2 aa^ + 6?/^ -f 3 dx^ + c^/^ 2. mx + Zi!/ — nx + A;i/. 4. px^ -f ^i/^ + »'^^ + 5 2/^. 6. (a -f 6) a; H- (2 a — Z;) cc. 7. 3 (m-n)^-4(m-w)H 5(m + ?i)- 7(m+w)+ 8(m-?i)'l 8. Ada 3(r + s)+ 6(i + s)- 7 (?• + 0^ ^0' + ^•)— 3(t -f- s) + 5(r + 0, and -5(r + 6') - 2(i +- s) +4(r + ^). 9. Add 7 a^« -^ h x^' + S-oj" - 1, - 2if«« + 5 - 3 a.- + Grc^n^ and — 5 ic^" — 4 a;^ + 8 X"" — 7. 10. From 7 a;" — 3 a^"-^> + 5 a;^"-^) + 4 subtract 2 a;" + 5 -|- 11. By how much does as'* 4-3 aj^"-^^?/ + 4 ir^"-^y + 5 x^^'-^hf exceed «" — 2.a;^"-^^2/ + 5 a?^"-^)?/^ -|- 8 a;("-^>2/l 12. Subtract , 5 (a; + 2/)^ + 6(a; + 2/) 4- 9 from 13 (x 4- 2/)^ 4- 2(a;4-2/)-4: 13. Subtract — xT -\- nx^^'-'^^y 4- (n — 1) x^^^-^^y- from 0. 14. How much must be added to x'^ — x^y 4- x-y^ 4- xy^ 4- y* to give: (a) x*—y*? (b) x*-\-y*? 15. Subtr act aar 4- 2 fta-?/ + C2/^ 4- cZa? 4- 62/ 4- / from Ax^ 4- 2 ^a-?/ 4- Cy^ + Dx + Eyi-F. 810 MISCELLANEOUS EXERCISES 311 B. EVALUATION OF ALGEBRAIC EXPRESSIONS Let a = 4, 6 = 5, c = 6, p = 2, m = 2, n = 3 ; iiud the nu- merical value of the following : 1. a". 2. h"". 3. c''->rp'^. 4. 6'* — 4/>'". '263* * 4 a- 36 -f 6 c* ' \^a - 6y * 8. h^-\-c^-2hp. 9. c(a+6)«. 10. 2a + (yi-l)j9. When ^ = 28, r = 6, s = 4, and tt = 3|: 11. Find F: (a) if F=7r?-7i; (6) if V^^irr". 12. Find S: (a) if aS = 2 7rr/i ; (6) if /S = 2 tti^ + 2 ttt^. 13. Find Fif F= ^ 7r/i(r2 + s^ + rs). 14. Find a; if w = 1000, T = 35, « = 15, and s = 50, when s 15. Find F if P = 500, i> = 5, d = i and ^ = 72,000, when Ed' C. PARENTHESES Remove the parentheses and combine terms : 1. c-[2 c-(6 a- 6)- ;c- 5 a-f-2 6 -(-5a + 6a- 36) j]. 2. a - (2 a - [3 a - 54 a - 5 a - If]). 3. X -[y - Ix - z - X - y + zl + (2 X - {- X + yl);i. 4. 5a;-[2aj-(-a;- j2a;-^~^S-3a;)-3a;]. 5. a — I — a ~ [— a — ( — a — I — a — a — l\)']l. 6. 28 - 5 - 16 - (- 4 + [55 - 31 +47]) J. 7. a: - (2/ -h 2r - [a: - (_ a; - 2/) + 2]) -h fz - 2 X - 2/|. 8. 2n-[3?i-|4n-?r^^i -(-5n~9)]. 812 ALGEBRA D. MULTIPLICATION Perform the following indicated operations : 1. (x-y + zf-(x + y- z)\ 3. {a - h - c -{- d)\ 2. {2x-\- 3)2(2 X - 3)2. 4, (« + hf -{a- hf. 5- (« -y){y-^)-{^- ^) (y-^)-(x- y) (x - z). 6. {^x-^yf-n{x-y){x-5y), 7. f (« - &) («' + a'6 + «&' + &') H K + ?>')'- 2 a^ft^j. 8. {a-\-h+cy-{a + hf-G(2a + 2h + c). 9. 2(a + 2 a?) (a - 2 x) -[(a + 2 a;)^ + (a - 2 it')^]. 10. {a-{-x){o?-^)[_a^-x{a-x)']. 11. (aj»-f6)^. 13. {x^ + y^f. 12. (a** — 6") (a" + ft**). 14. (ar'^« + a^« + a;") . (a;» + 1). 15. («2'* + 3 !»-« 4- 3 a;" 4- 1) • (•'«" + 1). 16. lut + i/^2-, _ [-^(^ _ 1) + |y.(^ _ i)2-j. E. DIVISION Divide : 1. 70a-50-a^-37 a2by6a-5-a3-2a2. 2. a^-a^'^-a^/^ + ^^^by a2-2a6 + 62. 3. a^ H- 2/^ + 2;^ — 3 xyz hy x -[- y -\- z. 4. a3-86-^-27c^-18a6cby a-25-3c. 5. 8m^-14m2-18m + 21 by 4m^ + 6m-7. 6. (2m2-m-l)(3m2 + m-2) by (2 m + 1)(3 m- 2). 7. I a^-l a^b + ^ab'-ib^hy ^a-ib. 8. (x-yy-{-lhy(x-y)-\-l. 9. 6(m + 7i)2-(m + n)-15by 3(m + w)-5. 10. ix'-lx^ + ^x-^hj^x^-^-^x-^. 11. (ar^'" + i»2m _|. ^>j _^ ^« ;^2. (a^4.5a;«4-6)-^(af-^-2). 13. (f-^Sy^--^Sy--^l)-h(y^'{-l). 14. (a:^"* — y^) ~r {xT — y™). 15. a;*" — -^^ (pf* -|- ^'''). MISCELLANEOUS EXERCISES 313 F. FRACTIONS Simplify the following by performing the indicated operar tions : 1. 12' V ^ 2a^6 ; 3a; + 2 14. 5x-l 6x2_3,_i2 10x2_i9^,^g m^ + 4 mn -|- 4 n^ * \m^ — 4 w^ * rn^ + mn — Qn^j 16 ar^-2a?.V + 4y^4a:^-9/ . 2a;-3y 2a; + 32/ ' x'^^f ''a^-4:y^' 17. X— m X ■\- m. — n" a; 4- n a; — m (oj — m) (a; -f n) 314 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 1/1^1 a\x — a x-\-2aJ x^-\-ax—2d ALGEBRA 3 4m-3 + 2m-5 6m2 + 13m-5 12 m^ + Sm- 3 2 x2-3 ' ^-\^(. I 1_>) x + 2 x-lj \x'-3x-\-2 x-2) 4- (a — 6) (a — c) (6 — c) (6 — a) ?— 1 + 1 (x-y)(y-^) (y-z)(^-^) (z-^)(^-y) g + c (g + &) . (&4-C) (a-5)(6-c) , (6-c)(c-a) (c-a)(a-6)' 1 1 a(a — b)(a — c) b(b—c)(b — a) c(c — a)(c — b) 1 1 2ft (a - &) a + b (5^ - ft^ * a-1 g + lj * |a-l"'"a4-lj ' J(a + l)2_2J (l ^^— Vl- \ 1*^ — rs 4- sy \ x'^-2 jfy^ 4- y^ ^ ( x^ - ;V^ ^ (a^ + y)^ - 2 a;y ' a;?/ «y 1 +^+ 2 1-x l-\-x 1 + a^ First, add the first two frac- tions, then add to the result the third fraction. 1 + ^ 2 a ^aW a-b a-\-b g2-f 62 a' + b' 1 1 + 1 1 -• Combine the first two , ' . fractions ; then the last t-1 t + 1 t-2 t-\-Z ^^^,^^ ^^^ ^^^ ^jjggg results. _1 1 d d c-d c + d c2-fd2 c^_(p' MISCELLANEOUS EXERCISES 315 33. _!_ + -J_ + -2y_ + _4/_. y — x y-\-x f-^^ .V*H-a;* 34. -i L_+_25__„4a^. a-h a + ^> a' + ft' «* + ^' 112 2 ^-1 «+l «+2 i-2 G. FRACTIONAL EQUATIONS 7a:^ + 10a;-24 5 ^„ {x^\f (x' + l) ■ 6 ^ .2(2a! + l) 4 (a?H-l)(a;-h3) ^a;-6 ^ 3a?^ f 5a;-4 ^ 3a; + 5 (a;-f5)(a; + 7) aV+2' ' 4a;2_3^_|.2 4a;-3 ^ a; — 2 « — 3 a? — 5 u? — 6 « — 3 a; — 4 a; — 6 x — 7 First combine the fractions in the first member ; then the fractions in the second member. a; — 4 x — h x — \ x — 2 x—5 x—6 x—2 x—S - m-\-2 m + 6 _ m -\-5 m + 3 ' m + 3 m-\-l ?/i -f 6 m + 4 Make this example like Example 6 by transposing. 1 a 3a-4 9. a + 1 a^-a + l a^ + l r + 2 2(r-l) ^ r2^-4r + 3 7^-\-r-Q r* 316 ALGEBRA H. LITERAL LINEAR EQUATIONS 2. 2(x-b)(2a-Sb-Sx)-(2a-Sx)(b + 2x). 3. (x + ay-^(b-j-cy=(x-ay+{b-cy. 0. 5. {x-2a-^3by ax bx _ x-{-b x + a 2(a + b) ^ x + b X x — b 1 -(x-2a)(x-\-Sb)-6ab = 0. a + b. 7. b b_ C C— ft _A x—c x — a a—x a 9. 10. 11. 12. + x-\-a 1 8 (2^-3m)^ ^ a;-39?i (2x-3ny x-Sn' 2x—a—b (x — a) (x — b) x(x — a — b) x + Aa + b 4:X-\-a-\-2b x-\- a-\-b x-\-a— b 5. T-\-agp(h-cQ) Nn r Mm p — r agph. Find r. Find r. 1. I. SIMULTANEOUS LINEAR EQUATIONS ±i+s=2x-y^^. 7 4 Sx- x-hy 2y-S = 2y-4.. ^^-^^ + 3 = 0. 7 3. 4. 1. S-^_2x±3^y±S 5 4 4 -5 11 3 5 m + 6 11 7* 10 7 m 55 r ~5" 21 12 11. 25 37. 6. 2(3a + 6)_i(2a + 6 + l)=iV- 3a-f(4a + & + |) = 56. f3s-i(2s + « + 6) = 5^. 20-i(2i-s) = -8. MISCELLANEOUS EXERCISES 317 8. 9. 10. X 4- ay 6-2 = a. hx -\- ay h a. -^-^ = -1. m n X y _. 3m 6n 2 3* 3a^36 + b. x-y = 2{a^ -b^ 2x-b 3x -y. a 2x-b a + 26 a-2y 13. y * 22 2/ 14. x y 1+1=1, y z a z X 15. a:-|-?/ + « = — 8. y-^z-\-u = h. ^Z-\-ll,-\-X — — 10. 16. 2*_2a; = -13. i»-3?/ = 13. 2/ — 4 z = 5. z-5w=23. a 6 12a;-4?/ + 2; = 3. 11. {x — y — 2z=^ — l. 5x-2y = 0. 3x — y — z = 7. 12. r«^w'Ni4^*i5lv '-."S... JW 54i.no UNIVERSITY OF CALIFORNIA LIBRARY