NRLF 311 IN MEMORIAM FLOR1AN CAJOR1 SIR ISAAC NEWTON (1642-1727) PRACTICAL ALGEBRA FIRST YEAR COURSE BY JOS. V. COLLINS PROFESSOR OF MATHEMATICS, STATE NORMAL SCHOOL STEVENS POINT, WISCONSIN NEW YORK .;. CINCINNATI : CHICAGO AMERICAN BOOK COMPANY COPYRIGHT, 1908, 1910, BY AMERICAN BOOK COMPANY. ENTERED AT STATIONERS' HALL, LONDON COLLINS'S FIRST YEAR ALGEBRA. W. P. 2 CAJORf PREFACE THIS book has been prepared to meet the demand for a first year course by those high schools which make the study of the more advanced parts of algebra an elective. As its title indicates, the book aims to increase the practical value of elementary alge- bra, both in the ordinary and in the cultural or educational sense. It is written as well for those who do not go to colleges as for those who do. A wave of interest in mathematical teaching has swept over the United States during the last six or seven years, probably re- ceiving an impetus from the Perry movement in England. The many periodicals and pamphlets published by educators and edu- cational organizations interested in the reform movement have been carefully studied by the author, and suggestions from them have been freely used. The plan hitherto followed largely in America has been to keep arithmetic, algebra, and geometry separate. In foreign countries, on the other hand, the aim has been to make each of these 1 tranches aid as much as possible in the study of the others. Per- haps the three most prominent features in the movement for reform in mathematical teaching have been, first, the attempt to make the work more practical ; second, the better correlation of the three branches, arithmetic, algebra, and geometry, with one another and with other branches, especially with science ; and third, the bringing down into the elementary course of the sim- plified elements of the higher mathematics. All of these features the author has sought to embody in his treatment, and a consider- able number of new practical problems have been introduced to apply these new ideas. 4 PREFACE In developing a new topic the common plan is to follow the order, (1) general theory ; (2) several examples ; (3) exercises. In this book the following plan is pursued: (1) one example; (2) exercises, with suggestions, when needed ; (3) the making of a rule by the student based on his knowledge gained in solving the exercises ; (4) more exercises, following the rule ; (5) miscel- laneous exercises after sets of similar exercises. Evidently this plan follows the usual scientific order of induction, generalization, deduction. Or, stated pedagogically, each topic is presented fol- lowing the order: presentation, comparison, generalization, appli- cation. In the making of the rules some training is acquired in an important kind of composition in which conciseness, accuracy, and completeness are essential. Thus, the student gains strength both in mathematics and in English. A feature to which attention is directed is the proving and checking which runs all through the treatment, and which obvi- ates the necessity of an answer book for the student. Just as the bookkeeper has his double-entry, and the skilled calculator his checks, which he invariably uses, so the student should have his tests to apply to his answers. Mathematics is the one subject in which checking is feasible. Doing it systematically may pro- foundly affect the pupil's mental habits. The free introduction of what have been called informational problems, if it cannot be dignified with the name correlation, at least makes the algebra circle of knowledge touch other subjects in the curriculum and the world outside. These problems are even better adapted for algebra than for arithmetic, since simple relations of the numbers are introduced in the statements. Such problems are not hard to solve, and they give training in dealing with large numbers. In his treatment of graphs, the author believes that the simple presentation and specific directions will clear the subject of its usual difficulties for young learners. Other features of the book are: great clearness in presentation; the avoidance of difficult and complicated exercises; the abridg- PREFACE 5 ment of the number of cases or kinds of problems that have to be learned; the constant use of the axioms in the solution of equa- tions; the presentation of factoring, and the deferring of the harder cases to the latter part of the course ; as well as the gen- eral conservatism in the arrangement of topics. Attention is directed to the Suggestions to Teachers on the next page. The author is indebted to Professor H. E. Cobb of Lewis Insti- tute, Chicago, for reading the chapter on practical applications and offering suggestions for its revision; to Miss Theresa Moran of the Stevens Point, Wis., High School for reading and criticiz- ing the whole manuscript ; and to Professor Robert E. Moritz, head of the department of mathematics of the University of Washington, Seattle, as well as to Mr. S. B. Todd of the Ameri- can Book Company, for valuable suggestions. In addition the following have read the proofs and have offered suggestions : Pro- fessor B. F. Finkel of Drury College, Springfield, Mo., editor of the American Mathematical Monthly; Professor W. H. Williams of the State Normal School, Platteville, Wis.; Miss Margaret French of the South Division High School, Milwaukee ; and Miss Mary C. Nye of the Blaine High School, Superior, Wis. STATE NORMAL SCHOOL, Stevens Point, Wis. SUGGESTIONS TO TEACHERS 1. A Shorter Course Outlined. The number of problems in this algebra is exceptionally large. On this account some teachers may find it desirable not to require their classes to solve them all. To meet this condition, an abridgment of the regular drill work is provided by placing a star (*) before the first example of sets of problems that may be omitted without sacrifice of a sufficient amount of illustration of essential principles. Like- wise a star is used to mark some of the less important topics that may be omitted. By the omission of the starred topics anu sets of problems, a briefbut practical course will be provided for one year's work. 2. Verifying, Checking, and Proving. It should be noted that the plan of proving results followed has the effect of greatly increasing the number of exercises. The verification of answers to all equations and equation problems is advised ; but checking, using figures for letters, can be easily overdone ; teachers should exercise judgment here, lest they retard the student's progress. Proving in division and factoring is urged whenever there is danger of error. This course keeps up a valuable review of multiplication and addition. 3. Drill in Applying the Axioms. It is not necessary to em- phasize to experienced teachers the importance of persistent drill in applying the axioms in the solution of equations. Inexperi- enced teachers, however, very often follow the custom of nearly all authors of not referring again to this reasoning after it has been presented once, and in so doing they make a serious mistake. This book constantly asks for the reasons underlying operations in equations. 4. Mental and Written Work. To be well balanced. Many problems throughout can be solved mentally, especially in reviews. TABLE OF CONTENTS (See Index at the end of the book for special topics) CHAPTER PAGE I. THE ALGEBRAIC NOTATION ....... 9 Algebraic Numbers presented in the Arabic Notation . . 12 Principles and Definitions ....... 24 Addition and Subtraction ....... 30 Symbols of Aggregation 40 Multiplication 45 Division 52 Integral Equations 9, 37, 61 Historical Note .64 II. THEOREMS, FACTORING, LOWEST COMMON MULTIPLE . . 66 Theorems in Multiplication and Division .... 66 Factoring 74 Quadratic Equations by Factoring .... .90 Lowest Common Multiple 92 III. FRACTIONS 96 Highest Common Factor . .98 IV. SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY . . 122 V. SIMULTANEOUS EQUATIONS. ' Two UNKNOWN QUANTITIES. . 154 Simultaneous Equations. Three Unknowns . . 168 VI. GRAPHS AND GRAPHICAL SOLUTION OF SIMPLE EQUATIONS . 172 VII. MISCELLANEOUS TOPICS .... 185 Simple Quadratic Equations ... 186 Literal Equations ...... Generalization ....... . 193 Indeterminate Equations ..... . 193 Negative Solutions 193 Supplementary Factoring 194 Historical and Biographical Notes .... . 199 VIIL INVOLUTION AND EVOLUTION 202 7 8 TABLE OF CONTENTS OHAPTEB PAGE IX. APPLICATIONS OF ALGEBRA 220 Applications in Arithmetic 220 Applications in Geometry 230 Applications with Squared Paper 236 Applications in Physics 240 Formulas 242 Manual Training and Domestic Science .... 246 X. FRACTIONAL EXPONENT QUANTITIES AND RADICALS . . 250 Radical Expressions 257 Imaginaries . 270 Equations containing Radicals 270 Historical Notes 273 XI. QUADRATIC EQUATIONS 276 Simultaneous Quadratics 293 Algebra used in Trigonometry 298 INDEX ... ........ 299 PRACTICAL ELEMENTARY ALGEBRA CHAPTER I THE ALGEBRAIC NOTATION 1. Algebra is a study of numbers. It differs from arithmetic in one way by using letters as well as figures to denote its numbers. The student should distinguish clearly between a number itself and the mark that stands for it. One definition often given is that a number is a ratio, or a quotient of one magnitude divided by another of the same kind. For example, the ratio of the length of a room to a unit length (foot, or meter) is a number. To denote this number we can use a word, or figures, or a letter. The fol- lowing examples will show how letters can stand for numbers. 1. There are three times as many girls as boys in a school con- taining 88 pupils. How many are boys ? ARITHMETICAL SOLUTION A certain number = the number of boys in the school ; then 3 times that number = the number of girls in the school ; and 4 times that number the number of boys and girls together. Hence, 4 times that number = 88, and that number = \ of 88, or 22. Thus, the number of boys in the school is 22. In the algebraic solution of this problem, which follows, instead of writing " a certain number," or "that number " each time, we abbreviate by writing x for this phrase. ALGEBRAIC SOLUTION Let x = the number of boys in the school ; then 3 x = the number of girls (3 x means 3 times #), and 4 x = the number of boys and girls together. 88 = the number of boys and girls together. Hence, 4 x = 88 ; then x = of 88 = 22, the number of boys. 9 JO THE -ALGEBRAIC NOTATION PROOF. If there are 22 boys in the school, and three times as many girls as boys, then there are 66 girls. Altogether there are the sum of 22 and 66, or 88 pupils, which is the number the problem says there are. Hence, the answer is right. 2. There are 7 sheep for every lamb in a pasture, and 96 sheep and lauibs altogether. How many lambs are there ? Work by both arithmetic and algebra, following the models on p. 9, and prove answer. 3. There are three packages, the second of which weighs twice as much as the first, and the third three times as much as the first. When put together on the scales they balance a 24 Ib. weight. What is the weight of each package ? SOLUTION. Let x = the number of pounds in weight of first package ; then 2 x the number of pounds in weight of second package, and 3 x = the number of pounds in weight of third package. Then x + 2 x + 3 x = 24, the sum of the weights of all, or, 6 x 24, by adding x, 2 #, and 3 x ; then x = of 24 = 4, number of pounds in 1st package, 2 x = 2 x 4 = 8, number of pounds in 2d package, 3 x = 3 x 4 = 12, number of pounds in 3d package. PROOF. 4 + 8 + 12 = 24 ; also 8=2x4, and 12 = 3 x 4. 4. Three persons together weigh 350 Ib. The second weighs twice as much as the first and the third 4 times as much as the first. What is the weight of each ? Prove answer. 5. Two men, each working for $ 1 a day, received $ 132 for their labor. A worked 11 days every time B worked 1 day. How many days did B work and how many did A work ? Prove answer. SUGGESTIONS. Let x = the number of days B worked. Then how many dollars did B receive? Ans. $x. If B worked x days and A worked 11 tinies as many clays, how many days did A work ? Ans. 11 x. Then how many dollars did A receive ? Now how many dollars does the problem say x + 1 1 x equals ? 6. A horse and saddle cost $ 150. If the horse costs 5 times as much as the saddle, what is the cost of each ? Prove answer. THE ALGEBRAIC NOTATION 11 7. Of 240 stamps collected by a boy and his sister, the boy col- lected 4 times as many as his sister. How many did each collect ? Prove. 8. Twice a certain number added to 5 times the number equals 98. What is the number ? Prove answer. 9. A farmer raised a certain number of bushels of wheat, 4 times as much corn, and 3 times as much oats. If there were in all 4000 bu. of grain, how many bushels of each kind did he raise? 10. A manufacturer who has 720 men working for him wishes to put a certain number in one building, 3 times as many in a second, and twice as many in a third building as he puts in the second. How many men must he put in each ? 11. The area of the land surface of the earth in round num- bers is 51,200,000 square miles. The area of the rest of the earth's land surface is 7 times that of North America. What is the area of North America in square miles ? Prove answer. 12. The total population of the Philippines in round numbers is 7,634,000. Of this number there are 10 times as many civilized people as there are uncivilized people. How many are there of each ? Prove answer. 13. In 1900 when the population of the United States in even thousands was 76,303,000, there were 3 times as many people living in small cities (under 30,000) and in the country as in large cities. How many lived in large cities ? Prove answer. 14. The Panama Canal will be 46 miles long. Of this distance the lower land parts on the Atlantic and Pacific sides will together be 9 times the length of the Culebra Cut, or hill part. How many miles long will the Culebra Cut be ? Prove answer. 15. The value of all other property in the United States in 1900 was about 5.8 times the value of manufactured articles, and the total wealth was estimated to be 88.5 billions of dollars. What was the number of billions of dollars in the value of the manufac- tured articles ? Prove answer. 12 THE ALGEBRAIC NOTATION r 2. The value of a letter, as we see from the preceding article, does not change throughout the solution of any problem. But the same letter, as x, may have different values in different prob- lems. This is not the case with figures. Thus, 4 always stands for four. 3. Symbols of Omission or Continuation. Dots or periods are commonly used in algebra to indicate that figures or letters not written are to be understood. They are read " and so on to" or (t and so on." Thus, 1, 2, 3, 4, ... 20 is read "one, two, three, four, and so on to twenty " ; while 1, 2, 3, 4 ; is read " one, two, three, four, and so on indefinitely." 4. Algebra and arithmetic are unlike, as explained in 1, in that algebra uses letters as well as figures to denote numbers. They are unlike also in another important respect, namely, arith- metic has but one series of numbers, while algebra has two series, extending in opposite directions from zero. Arithmetic series, 0, 1, 2, 3, 4, 5, 6, Algebra series, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, The height of a tree, the age of a man, the number of silver dollars a man has in his pocket, etc., are all arithmetical numbers because their lowest value is 0. The number of dollars a man has to his credit on the books of a bank, a temperature, a date A.D. or B.C., etc., are all algebraic numbers, because such num- bers can lie on either side of zero. There are vastly more con- crete numbers that are arithmetical than are algebraical. Examine the room you are in for examples of the two kinds of numbers. 5. Algebra may be described as that branch of mathematics which uses letters as well as figures in the study of numbers, and separates its numbers into two opposite kinds. 6. Addition of Algebraic Numbers. Addition is literally " put- ting together." One gets a good idea of algebraic addition by thinking of a business man receiving during a day pieces of paper ADDITION 13 representing money, as checks, bank notes, promissory notes, due bills, and other evidences of credit and debt. These are "put together " on a pile. The sum is found by adding the debts and credits separately, subtracting the less amount from the greater, and marking the result credit or debt according as the credit or debt sum is the greater. To distinguish credits from debts we will mark the credits with a small + sign and the debts with a small " sign. Thus, + 9^ means a credit of 9^, and is read "plus 9 cents " ; similarly $ ~3 is read " minus 3 dollars." This problem evidently calls for the sum of a credit of another credit of 5^, a debt of 12 X, another debt of 15^, and a credit of 29 SOLUTION. +6^ + +5^ + +29^ = +40^ ; -12^ -f -15 / = -27^ ; +40^ + -27 ?=+13f'. Ans. EXPLANATION. In the solution, the addition of all the credits gives a total credit of 40^, and the addition of all the debts gives a total debt of Then the sum of a credit of 40^ and a debt of 27^ is a credit of CHECK. The correctness of the answer just found can be "checked," or tested, by adding the items in the order in which they come. Thus, +6^ + +5/ = +11^ ; +11/' -f -12^ = ~l? ; ~1/ + ~15^ = - 4- +29 / = +13^, or the same answer as before. Solve the following problems in the same way the problem just given was solved, and check the answer in the same way that answer was checked : 2. + 12 + -3 + -T^ + + 15^ + -2=? Ans. 3. 4. 5. 6. 14 THE ALGEBRAIC NOTATION In the following problems set down the several items with -f- signs between them, marking the credits with small + signs and the debts with small " signs, and find the sum : 7. George has 16 / in his money box and 10^ in his pocket. He owes his brother John 7 '$ and a confectioner 14^. William owes him 10^. How much money has he? 8. William, who has 25^ in money in his pocketbook, owes a grocer 8^, a pieman 4^, his sister 3^, and a playmate 15^. His brother James owes him 17^. How much money will he have left if he collects what is due him and pays his debts? Solve as in the preceding. 9. A man who has $300 in the bank and $125 in his pocket owes A $550 and B $1000. He has due him from one party $400 and from another $640; but he owes as surety on a note $190. Find and mark how he stands financially as in the pre- ceding problems. 10. A ship starting from 1 north latitude goes one day 2 north, the next 3 north, a third 4 south, a fourth 3 south. If north latitude and north sailing are marked + and south latitude and south sailing ~, what latitude is the ship now in and how is the latitude marked ? Solve as in the preceding problems. 11. A thermometer, which stood at 4 P.M. at 70, and which by 8 o'clock had fallen 14, and by midnight 8 more, and by 4 A.M. 3 more, rose from 4 o'clock to 8 o'clock 12, and from 8 o'clock to 12 o'clock noon 25. If above zero and upward movement are marked + , and downward movement and below zero ~, what is the position of the mercury at noon of the second day and how is it marked ? Solve as in the preceding problems. 12. A man deposits in a bank $ 600. He then draws checks at different times for $60, $19, $25.50, $10.19. Later he deposits $75 at one time and $32.12 at another. Afterwards he draws checks for $45.14 and $13.72. How does his account then stand on the bank's books ? Mark deposits + and checks ~, and solve as in the preceding. SUBTRACTION 15 7. Subtraction of Algebraic Numbers. Subtraction in algebra is the process of finding a number which added to one of two given numbers (the subtrahend) produces the other (the minuend). As in arithmetic, the result is called the difference. 1. $+12 --$+7 = ? This problem, it is plain, risks for the difference between a credit of $ 12 and a credit of $ 7. SOLUTION. 8 + 12 - $+7 == 8+5. Ans. EXPLANATION. An algebraic number must be found which being added to the subtrahend 8+7 gives the minuend $ + 12. Clearly $+5 added to $+7 equals $+12. Hence $+5 is the difference sought. 4-4- i A it 4- O "" d O K < + ^5 O Q *? . ^40^ _>?' = : 5. $> o o = r Ex. 5 evidently asks for the difference between a credit of 8 <> and a debt of 83. SOLUTION. $ +6 - $-3 = 8 +9. .4ns. EXPLANATION. A number must be found which being added to 8 ~3 gives the minuend 8+6. Now it takes 8 + 9 added to 8 ~3 to make 8+6. Hence the difference sought is 8 +9. Thus, the difference in fortunes of two boys, the first of whom is worth 8 6 and the second of whom is 83 in debt, is 8 9. The first boy is 89 better off than the second boy. Solve the following four problems in the same way as Ex. 5, and explain the results : 6. $+10-$-4 = ? 7. +25? --lot = ? 8. 22t--16t = ? 9. 12- -5==? Thus, a thermometer stood 12 above zero, and the day before 5 below zero. What was the difference in temperature ? a. REMARK. It is to be understood when a problem asks " what is the dif- ference between two given numbers" that the number first named is the minuend and the other the subtrahend. 10. What is the number of years' difference in time between 26 A.D. (which we will call +26) and 10 B.C, ? 16 THE ALGEBRAIC NOTATION 11. The difference between two algebraic numbers is the num- ber of units of distance between them on the algebraic scale, marked + or ~ to show their relative position. ^6~^5~-4 -3 -2 -1 O +1 + 2 +3 +4 +5 +6 We represent the algebraic scale here in both horizon- _ -) O tal and vertical positions. The numbers are said to be above zero and the numbers below zero, as on a thermometer scale. No matter what position the scale has, from towards the + numbers is called the + direction, and from to- wards the - numbers is the ~ direction. If one number lies in the + direction from another, it is said to be [_- 4 higher up in the scale than the other number. Thus, +2 h - is higher in the scale than ~4. A man who has $2 is better off than another who is in debt $4. If now the minuend is higher up in the scale than the subtra- hend (as it always is in arithmetic), the difference is marked + ; if the minuend is lower down in the scale than the subtrahend, the difference is marked". Thus, a difference in algebra tells two things : first, the number tells how far apart the minuend and subtrahend are ; and, second, the sign + or ~ tells whether the minuend is above the sub- trahend or below it. Find the difference in the following problems by using the dia- gram. See that the answer is marked with the proper sign. The student will find it helpful to think of the numbers as denoting degrees on the thermometer. 12. What is the difference between 12 above zero and 4 below zero, that is, between +12 and ~4 ? Ans- +16. Because + 12 and ~4 are 16 units apart and the minuend is above the subtrahend. 13. +16-+10 = ? 14. +2--2 = ? 15. +11- +15 = ? 16. ~8 -~10 = ? 17. -7- +6 = ? 18. ~6- ~4=? 19. Find the difference in temperature between January 5, when it was 6 below zero, and January 4, when it was 13 below zero. SUBTRACTION 17 20. What is the difference in time between 1600 A.D. and 500 B.C., if A.D. is marked + and B.C. is marked ~? 21. Go over the problems from 12 to 20, working them this time by the method of finding a number which added to the subtra- hend will give the minuend, as explained at the beginning of this article. This time the student should think of the numbers as denoting dollars or cents. The answers are always the same, whichever method is used. 22. The method with the scale which was used in Ex. 12-20 is more satisfactory as an explanation of subtraction, but the method introduced first is better for practical use. We must know how to add in any event, and we can always make subtraction depend on addition. Hereafter, therefore, we will use the rule stated in the next paragraph. 23. To subtract, find a number which added algebraically to the subtrahend gives the minuend. Or, conceive the sign of the subtrahend to be changed, and add the result algebraically to the minuend. Subtract in following : (check, thinking of debts and credits) + 12 REASON. Because -5 and +12 added give +7. CHECK. A credit of $ 12 added to a debt of $ 5 gives a credit of 8 7. 25. +7 26. -11 27. -15 28. +2 "22 +7 2^ 29. +25 30. -48 31. +40 32. -49 +31 -39 -51 +4 ^^^^ ^^ 33. -6 34. -27 35. +25 36. +35 -16 -6 ' -17 +46 37. +25-+19. 38. -30 --36. 39. -12- -40. 40. ~6 - +36. 41. -24-0. 42. ^50 --68. 4.3. + 169-~75, 44. -324-+125, 18 THE ALGEBRAIC NOTATION 45. What is the difference between the fortunes of two men, the first of whom owes $3000, and the second owes $3335 ? 46. What is the difference in latitude between Santiago de Cuba 20 north and Buenos Ayres 34| south, calling north latitude + ? 47. What is the difference in longitude between Berlin 13 24 'E. and New York 74 W., if east longitude is taken as + ? 48. What is the difference between the average July tempera- ture + 65 and the average January temperature ~8 at. St. Vincent, Minn. ? In the following problems + denotes A.D. and ~ B.C., as in the Standard Dictionary. 49. The reign of Augustus Caesar over the Roman Empire ended with his death in +14, having begun in ~31. How long was he sole ruler ? 50. The Roman historian Livy died in the year +17, having been born in the year ~59. How long did he live ? 51. The Roman poet Horace died in the year ~8, having been born in the year ~65. What was his age when he died ? 52. Columbus discovered America in +1492. The earliest accounts we have of the sailors of the ancient world go back to about ~1500. How long a period elapsed from the earlier to the later date ? 53. Greece became a Roman colony "146, two hundred fifty- three years after the philosopher Socrates drank the cup of hemlock. In what year did Socrates die ? SUGGESTION. Mark 253 as -'-253 and subtract it from ~146. *54. Pope Gregory XIII reformed the calendar in +1582. The Council of Nice set Easter +325. Find the difference in time. NOTE. Probs. 49-54 are indefinite. In Ex 49 from Jan. 1, -31 to Dec. 31, +14 is +45 yr. But from Jan. 1, -31 to Jan. 1, +14 is 1 yr. less. Find interval from Sept. 10, -20 to June 15, +50. Solution : Sept. 10, -20 is 19 yr. 3 mo. 20 da. before era, and June 15, +50 is 50 yr. 6 mo. 15 da. after it. Ans. 68 yr. 9 mo. 5 da. Answer may vary 1 day either way. ADDITION AND SUBTRACTION 19 55. The Swiss defeated the Austrian Hapsburgs at the battle of Morgarten in +1315. The Greeks defeated the Persians at the battle of Thermopylae "480. What number of years intervened between these battles ? 56. Titus destroyed Jerusalem in the year + 70. Probably the epoch of greatest glory in Jewish history was at the time Solomon dedicated the temple at Jerusalem ~970. How many years elapsed from the earlier to the later date ? 57. The Roman Empire, as commonly stated, fell in the year + 476. This was 622 years after the destruction of Carthage. In what year was Carthage destroyed ? REMARK. Ex. 54, p. 18 (marked with a star), and Ex. 55-57 following, are to be omitted in the Shorter Course. See 1, p. 6. 8. New Marks for the Series. The sign + is always understood when no series sign is written. Thus 4 means + 4. Now + 7 1 = + 7 ^. (Since adding 7 f to gives 7 $. ) 3^ = ~3^. (Because the sum of a debt of 3^ and a credit of 3^ = 0, 7.) Thus, + 7 can replace + 7, and 3 can replace ~3. But can be dropped, as it has no value. Dropping 0, we have, = + 7and -3 = ~3. Thus, the small + and " signs written above the line of writ- ing, and used to mark the series to which numbers belong can be replaced by large + and signs. It is understood when no sign is written that + is understood. When algebraists desire to indicate that a number belongs to the positive (+) or negative ( ) series, they put it with its sign in parenthesis. Thus, 6 + (4) denotes that 4 in the negative series is to be added to 6. 9. Addition and Subtraction of Algebraic Numbers. By means of the series sign just described, an example in subtraction can be changed into one in addition. 20 THE ALGEBRAIC NOTATION Thus, 6 4, that is, 6 (+4), equals 6 +(4), since each equals +2. Algebraists commonly regard such a problem as 6 + 9-3 + 4-12 as one in addition, thinking of it as written SOLUTION. The sum of the + numbers is -f 19 : the sum of the num- bers is - 15 ; + 19-f (- 15) = + 4. CHECK. (See Explan. G.) 6 + 9 = 15; 15 +(-3) = + 12; +12+( + 4)= +16; + 16 + (- 12) = + 4. Add and check in the problems of this article as in the model just solved. 1. 12-8 + 6-7 + 15. 2. 6-7 + 15 + 2 + 5-9. 3. _3 + 4_9_i2 + 15. 4. -15 + 12-10-8+3-12. 5 . 19 _ c + 21 + 65 -71. 6. 78-44 + 39-25 + 17. 7. On a business man's table there lay at the end of a day's business the following pieces of paper: three ten-dollar bills; a note in his favor for $25; one due bill against him of $12, and a second of $ 23 ; two checks in his favor, one of $ 19 and another of $22; and a memorandum of a note given to the bank in its favor for $40. What is the sum of these items ? 8. The price of wheat in January was 75 ^. During February it rose 7j^; then in March it fell 12^; in April it rose 8/, and in May 10^; and in June it fell 2^. What was the price at the end of June ? Mark upward movements + and downward inside parentheses. 9. Three toy balloons are fastened together, the first having an upward pull of 2 oz., the second of 3 oz., and the third of 1 oz. To them are fastened weights of 21 oz. and 1^ oz., respectively. If upward pull is marked + and downward , what is the sum of these forces ? MULTIPLICATION 21 10. If an elevator starts on the ground floor and goes up 30 ft., then down 15 ft., then up 45 ft., then up 18 ft., and last of all down 33 feet., where is it then, counting from the ground floor ? 11. A man rows 5 mi. an hour in still water and the current of the river he rows in is 3 mi. an hour. If motion downstream is marked -f- and upstream -, what is the man's rate of progress when he rows upstream ? What is it when he rows downstream ? 10. Multiplication in algebra is like multiplication in arith- metic, aside from the question of the sign of the product. This matter we will now consider. (In this article for x read " times.") SOLUTION. Once 7 ? = 7 ^ ; twice 7 ^ = 14 ^, etc.; 6 x 7 / = 42^. Or, 6 credits of 7 ^ each = a credit of 42 ^. A positive number times a positive number gives a positive product. 2. 6 x-7^=? SOLUTION. Once 7 $ = 7 ^ ; twice 7 ^ = 14 ^, etc. ; 6 x 7 Or, 6 debts of 7 ^ each equal a debt of 42 p. A positive number times a negative number gives a negative product. 3. -2x7^=? SOLUTION. 2x7^ = 14 ^ _ Notice that the multipliers decrease by 1 ft 7 ; 2 c = 4 ; then 24 -- 32 + 4 = - 4. 4. Calculate 6 2 # 9 y 2 when a = 4, a? = 2, y = 3. Ans. 111. 5. Calculate 3 a 2 4- 2 c# 6 3 when a = 5, b = 7, c = 15, a; = 3. 6. Calculate 9 a~bc + 3 Vc - & 2 V& when a = 2, 6 = 4, c = 25. 7. As stated before, the rules in 1 and 2 do not hold when symbols of aggregation indicate that parts of given quantities are to be considered as one. Thus, in 2 x (6 3), read "2 times the quantity (6 3)," the parenthesis requires that 3 be subtracted from 6 before multiplying by 2. Similarly, 12 + 4 (5 + 4) -s- (16 - 13) = 24. Explain why. Notice that according to rule 1 of this article 3 a 2 means 3 ao, but (3 of means 3 a x 3 a, or 9 a 2 . Thus, 4 x 3 2 equals 36, while (4 x 3) 2 equals 144. 8. Calculate (a -f &)c when a = 2, b = 3, c = 5. Ans. 25. Evaluate the following quantities when a = 4, 6 = 3, c = 2. 9. 5a(6--c). 10. 5(a6 + c). 11. 3(a + 6c)a. 12. 4(a 2 + 6 2 ). 13. 4(ac 2 -a 2 6). 14. 6(a6 2 -c)-2 a 2 6c. 15. (4a6 2 ) 2 c. 16. 3(2ac) 2 6. 17. 28. Exercise in Writing Algebraic Quantities. The student should fix in mind that if a, 6, and c are three numbers, the sum of a, 6, and c = a + 6 + c, while the product of a, &, and c = a&c. Also, that 3 fo'mes a = 3 a, while the third power of a = a 3 . Thus, if a is 5, then a 3 = 125, while 3 a = 15. 30 THE ALGEBRAIC NOTATION Express the following in algebraic symbols : 1. Six times a, plus three times b. Ans. 6 a + 3 b. 2. Seven times a squared, less four times b cubed. Ans. 7a 2 --46 3 . 3. Eight times x cubed, minus eleven times y cubed, plus three times y squared. 4. Twice a cube, diminished by three times the square of a. 5. The sum of a cubed, three times c squared, and the product of , 6, and c. 6. The sum of a, b, c, and d. 7. The sum of the squares of a and b. 8. The double of x. 9. The difference between a and 5. 10. The product of 3, x, y, and z. 11. Four times the fifth power of m. ADDITION AND SUBTRACTION 29. Addition of Literal Quantities. Solve the easier problems mentally. In the review solve a larger number mentally. 1. Add 6 a, 7 a, and 3 a. SOLUTION. 6 a EXPLANATION. Here a represents some number, r.s o does every letter so used in algebra. The sum of 0, 7, and 3 times a number equals 16 times the number. 10 a 2. Add 4 a, 5 a, 9 a, and 6 a. Ans. 24 a. 3. Add - 3 a, 6 a, - - 2 a, - - 4 a, 11 a. Ans. - 26 a. (See .6.) 4. Add 4 m, 6 m, 9 m, 3 m. 5. Add - 5 b, - 6 b, - 7 b, - b. 6. Add 24 a 2 , 15 a 2 , 13 a 2 . 7. Add - 11 a 3 , - 3 a 3 , - 2 a 3 . ADDITION 31 8. Add 5x, Sx, 14 #, -11 a', and loo?. SOLUTION. 5x + Sx + 15 z = 28 x ; ( 14*) + ( llz) = 26x; 28 x + ( 25 a:) = + 3 x. Ans. CHECK. 5x + 8 x = 13 x ; 13 x + ( 14 x) = x ; - x + ( llx) - 12 x ; - 12 x + 15 x = 3z, which is the same answer as before. a. REMARK. To make the problems of this article seem more real, think of the positive terms as denoting credits and the negative terms as denoting debts. Thus, the sum of $5x, $8z, and $15 a; is $28 a- ; the sum of one debt of $ 14 x and another of $ 11 x is a debt of $25x ; then the sum of a credit of $28 x and a debt of .$ 25 x is a credit of $3 x. Should a student be confused at any time in the process of adding alge- braic numbers, the difficulty will in all probability vanish if he will think of his numbers as denoting credits and debts. Add and check in the following problems as in Ex. 8. 9. 7 x, 3 x, 11 x, 2x, - 5 x, - - 3 x. 10. 2y,7y, 5y, - y, 10 y, --6y, Sy. 11.5 x 2 , 12 x 2 , - 10 x 2 , x 2 , - 16 x 2 , 3 x 2 , 4 x 2 . 12. 3 ax, - 2 ax, 5 ax, ax, 4 ax } - 9 ax. 13. 2bc, --7bc, 3 be, 4 be, 5 be, - - 6 be, 5 be. 14. -4 a 2 15. 7 be 16. 11 a' 2 b 2 17. 20 + 5 a 2 -86c -20-^ +15 - 7 a 2 -96c +11 a 2 b 2 + 30 18. 3a 2 6 19. 6a6c 20. 5 ax* 21. -25 a 3 ?/ 3 4a 2 6 -21a6c -11 ax 3 -12arV a 2 & 3 abc ax 3 39 x^y 3 -10 a 2 6 +12rf&c 7 aa 3 - 10 .T 3 ? 8 22. What is the sum of 3 a 2 b and 5 air ? SOLUTION. Since the two given quantities are not similar ( 23), their co- efficients cannot be added for the coefficient of the sum. All that can be done is to indicate the addition. Thus, the sum is 3 a 2 b + 5 ab 2 . b. Indicating addition is called adding in algebra. 23. What is the sum of 3 a, 5 b, and c ? 32 THE ALGEBRAIC NOTATION 24. Simplify 6 SOLUTION. 6a + (- 3 a)+ (- a) = 2 a ; then the answer is 2 a -f 8 6 2 c. CHECK. Let a = 2, b = 3, c = 1. Then, given quantity, 6 a + 5 6 3 a + 3 6 -- 2 c a = 12 + 15 - G + 9 2 -2 =26; also, answer, 2a + 86 2c = 4 + 24--2=26. Since the answer 2a + 8& 2c has the same numerical value ( 24) as the given quantity, the presumption is that there has been no mistake in the solution. Simplify the following polynomials and check the answers to Ex. 25-29 as in the problem just considered (see 2, p. 6) : 25. 2a + 2b + 2c a 3b + c + 3a. 26. a b + 2c 2a + 3b 4 c. 27. 2x + y 3x + z 3y+5x 28. 7x lly + 4 : z 7z + llx 29. 2 ay 3 ac 4 ay + 4 ac 6 ay -f- 5 ac -f- 11 a?/. 30. 4 a? 2 3 a?;?/ + 5 y 2 + 10 xy 17 2/ 2 11 x 2 - 5 a??/. 31. What kind of terms can be combined into a single term ? Ann. Similar terms (23). How are dissimilar terms added? See Ex. 22, p. 31. 32. Make a rule for simplifying a polynomial ( 21) by answer- ing the following questions : (1) What is done with each set of similar terms ? (2) In adding a set of similar terms, how is the coefficient of the sum found ? (See 6.) (3) What is done with the results obtained ? (See Ex. 22, p. 31.) 33. 2 xy 5 y 2 + x 2 y 2 1 xy + 3 y 1 4 o&y*- -\- 5 xy. 34. 5 am 3 a?m 2 + 4 - - 4 am + 2 m 2 2 + T> + a L 'w 2 . 35. Ila 2 &-7afr 2 + 2ac 2 + 10a&-4ac 2 + 5rr'6-4a& 2 . 36. 3 x 3 - 2.2 or 2 + .4 ar 3 - 7.3 x + .5 or* - 1.5 .r- - .6. 37. 240 a - 306 6 + 205 a - 70 a + 79 6 - 25 & + 100 c. ADDITION 33 38. Add the following polynomials : 7 cr + 5 cm ; - 3 cm 4 d ; 5 cm -+- 3 e 4 d 6 a 2 ; and 4 a, 2 - 7 cm d ; and check answer by letting a = 2, c = 3, d = 1, e = 4, m = 2. SOLUTION. Check. 7 a 2 + 6 cm =28 + 30 = 58 -3cw-4d -18-4 =-22 _ 6 a 2 + 5 cm - 4d + 3 e = - 24 + 30 -- 4 + 12 = 14 4 a 2 - 7 cm - _d _ = 16-42-1 =-27 6 a 2 - 9 d + 3 e 23 CHECK. =20 - 9 -f 12 = 23 EXPLANATION. The first polynomial is set down, and then underneath it the other polynomials are written in such manner that similar terms ( 23) fall in the same columns. As fast as terms appear not similar to any above, they should be written in new columns at the right. Each column is then added as in Ex. 14-21, p. 31. To check the result the numerical value ( 24) of each polynomial is found and the results are added. If the sum found is the same as the numerical value of the answer, the answer is said to check. Thus, the sum 23 of the nu- merical values of the given quantities is the same as the numerical value of the answer. c. That a result checks does not prove that it is correct. Mistakes may have been made in both solution and checking. In rare instances results check and yet are wrong. Putting the letters equal to 1 or is less likely to uncover a mistake than giving them other values. Add, in the following problems, and check the answer in the first four, following the preceding solution as a model: 39. 40. 7a-46 + 2c;6a + 35-5c; -12a + 4c. 41. x + y + z-, x~2y + 2>z; -5x y + z + 2u. 42. 3a6 + 5ac; 6a& Sac; Sab ac + abc. 43. 5x 3a + 6 + 7; 4 a 3 a + 26 9 ; x b. 44. 3 a _5_|_2c; 5a + c 25; 36 a--4c. 45. 20p q+r; 2p + 5q~7r; - 46. 34 THE ALGEBRAIC NOTATION 47. 2ab-3ax 2 + 2a 2 x; 12 ab - 6 o?x + 10 ax 2 ; ax*-8ab-5a?x. 48. 3 a? 2 - - xy + 072; - - 3 y 2 + 4 yz z 2 ; Stfxy xz + Syz-, 6a?-6y-6z; 4=yz - 5yz + 3z 2 -, - Ix 2 + y 2 + 3yz + 3z*. 49. Make a rule for the addition of polynomials telling : (1) How the different polynomials are set down for the operation ; also how similar terms are placed, and where dissimilar terms are written as fast as they appear. (2) What is done with each column. How it is done. (3) How the answer is checked for correctness. 50. 3a 2 b s -7 ab 4 +5axy; -7a 2 b 3 -2ab 4 -axy- ab*-7 axy+8a 2 b 3 ', - 10 ab* + a 2 b 3 + 3axy; - 5 a 2 b 3 + 18 51. 2ab + 3ax 2 +2d 2 x 2 : > 22 ab-6a 2 x 2 +Wa 52. 10 a 2 b - 12 a 8 6c - 15 b'c 4 + 10 ; 8 a?bc - 10 b 2 c* - 4 a 2 b - 4 ; 20 6V - 3 a s bc - 3 a~b - 3 ; 2a 2 b + 12 a s bc + 5 6 2 c 4 + 2. 53. Jaj 2 -2a 8 + |a;-3; i^ 3 - f x 2 + J; 2aj-|x 2 . 54. Of two farmers the first had 2 a? 3 ?/ acres, and the second had x--y acres more than the first. How many acres had the second ? How many had both ? 30. Subtraction of Literal Quantities. We saw in 7 that sub- traction is the process of finding a number which added to one of two given numbers, the subtrahend, produces the other, the minu- end. Solve the easier problems mentally. 1. What is the difference between 6 a and 4 a ? Ans. 2 a. 2. What is the difference between 11 a 2 and 3 a 2 ? 3. 7a 3 -2a 3 =? 4. 20 abc - 14 abc = ? 5. 15 or 5 ?/ - 11 y?y = ? 6. What is the difference between 11 m and 4m? SOLUTION. llm EXPLANATION. 15?w + ( 4 m) = 11 m. -4m CHECK. A credit of $15w and a debt of 15 m amount to a credit of $ 11 m. SUBTRACTION 35 7. What is the difference between 12 x and 10 ./: ? (See 7, a.) 8. What is the difference between 4 x and 6 x ? Ans. - - x. 9. Ux-(+16x)=? 10. -llx-(-ox) = ? In the following subtract the lower quantity from the upper. Check the answer by thinking of the quantities as debts and credits : 11. -15 a 2 12. -21 a 3 13. -9afcc 14. 4crb 5 a 2 - 5 a :i 13 afrc 3 a'b 15. llm 2 16. -15 a 2 17. -6 a 3 18. 20 x 2 -12m 2 -25 a 2 +15 a 3 - ( .U 2 19. 12 y? 20. -17?/ 2 21. 22. 18 a 2 -lly* 12 xy -15??m 23. Make a rule for subtracting one monomial quantity fro-.n another similar quantity. (1) In what two ways can the problem be set down ? (See Ex. 9 and 11.) (2) What quantity must the difference be ? ( 7, 23.) (3) How can the answer be checked ? 24. -12abc 25. -6a 2 6 26. 19 ab 2 27. 7 37/2 + 18 abc - 15 a 2 b -7ab 2 - 11 xyz 28. From the polynomial c~x 2 + 3 ex 2 - 5 ex --4 cc 2 take (?3?--2cx -f 3 c 2 ^* 6 c 8 , and check the answer with c = 2, it* = 1. SOLUTION, c-x 2 + 8 ca: 2 - 5 c.r -- 4 z 2 4 c%2 _ -2cx _ + 8 c-x - 6 c 3 - - 36 3 rx -2 _ 3 cx _ 4 x -2 _ 3 c % + 6 c 3 =: + 32. Check. EXPLANATION. Terms of the subtrahend similar to those of the minuend are placed in the same column ; dissimilar terms are put in new columns. In the second, fourth, fifth, and sixth columns zeros are supposed to take the place of the blank spaces. Notice that in each column, the remainder is chosen so that the subtrahend and remainder added algebraically give the minuend. 36 THE ALGEBRAIC NOTATION 29. From 6 a 2b 5c take 2 a 2 b 3 c. Check. 30. From 7 x 2 8 x - - 1 take 5 x 2 6 cc + 3. Check. 31. From 2 x 2 - 3 a 2 ^ 2 + 9 take or + 5 a 2 x 2 - 3. Check. 32. From X s 3 3?y + 3 x*y 2 take - - x 2 y -f 5 a,* 2 ?/ 2 y*. Clieck. 33. (or - 5 xy - y 2 + 3 yz -2 z 2 ) - (x 2 +6 #?/-fll x*z-4 y 2 -2z 2 ). 34. 35. 36. (a 2 b* - 6 a 2 c + 9 ac 2 ) - ( - 2 crc 2 - 6 acx + 3 a 2 ^). 37. 38. Simplify the following expressions (as in 29, Ex. 24-37) and then take the second from the first : 39. 4 #y 5 cz -{- 8 m - - 4 cz 2 m, and cz -f 2 x 2 y* 4 cz. Check. 40. 4 a? 3 3 x 2 -- 2 x Atfx+l, and 7 x 3 x 2 x 2 Sx 3 11 + 3 x. *41. 3 3 -2a 2 -7a 3 -5a-2a 2 , and a 3 -4a 2 -5a-lla 2 -14. 42. 2 m 3 3 m 2 n -f- 6 m 2 n + ?i 3 , and m 3 + 12 mn 2 - 12 m 3 . 43. From a 3 - '6 a 2 c + 9 ac 2 take the sum of - 2 2 c - 4 a 3 -f 2 ac 2 and - 2 a 2 c + 3 a 3 ac 2 . 44. From the sum of 36 + 2a--4c and 9c + 36--5d subtract the sum of 6 d - - 1 a and 8a 7^ + 96 + 5 c. 45. From the sum of 7 X s 4 a? 2 -f 6 # and 3 x 2 - - 10 x 5 take the sum of 5 3 + 4x + 12 and 8 x 3 11 o^ 2. 46. A man who had four sons gave to the youngest $a, to the next older $ b more than to the youngest, to the third son c + 50 dollars more than to the second, and to the oldest as much as he gave to the three younger sons. How much more did the oldest get than the youngest ? 47. One man has 2 a b + c dollars, and his brother has a 2b + d dollars. What is the difference in their fortunes? * See 1, p. 6. SUBTRACTION 37 48. Lines can be denoted by letters by letting each letter rep resent the number of units of length in its line. (1) Add -3 and +7. SOLUTION. We first measure out 3 units to the left because 3 is negative (page 16). From this point -3 0-3 +0-T) = + 4Ans. we measure 7 units to the right, arriving at -f 4. Thus, 3 + 7 = + 4. (2) Add -5 and +12; > 1 also +5 and +8; also -f- 9 and - 6. (3) Show that -1 + lAns. ' See diagram. (4) Make problems similar to that in last diagram and solve them. Check, solving as in 9. 31. Problems. (Continued from 1.) 1. Find what number x is when 3 x -\- 1 = 13. SOLUTION. If 3# increased by 1 equals 13, 3 x must equal 1 less than 13, or 12. If 3 x = 12, then x = | of 12, or 4. Ans. PROOF. 3 x 4 + 1 = 13. (See 27, 2.) Solve and prove the following in the same way : 5. 7o; + 2 = 44. 6. 10x + 5 = 55. 7. 7oj--4 = 3 SOLUTION to Ex. 7. If 7 x diminished by 4 equals 31, Ix must equal 4 more than 31, or 35. If 7 x == 35, then x = \ of 35, or 5. Ans. PROOF. 7x5-4=31. 8. lloj-3 = 30. 9. 10a?-17 = 33. 10. 15a-23 = SUGGESTION to Ex. 13. The difference between 9x and 4x is 15. 14. 12x 72 = Sx. 15. Sx 14 = #. 16. 14# 24 = * 17. 10# = 5o; + 50. .18. x = 6x 2. 19. 3x 12 = a?, 38 THE ALGEBRAIC NOTATION 20. The difference between two numbers is 6 and their sum is 76. What are the numbers ? SUGGESTION. If x = smaller number, what expression with x in it equals the larger ? Am. x +6. Then x + (x + 6) equals what number ? Now solve 2 x + 6 = 76 as in the preceding solutions. 21. The difference between two numbers is 11 and their sum is 97. What are the numbers ? 22. Two men earn $180 a month, but one earns $10 a month more than the other. How much does each earn ? 23. The sum of the ages of three girls is 37 years. The first is 3 years older than the second, and the third two years younger than the second. What is the age of each ? SUGGESTION. Let x = number of years in age of second. Then what is age of oldest ? Of youngest ? Adding these ages, we have (x + 3) + x + (x - 2) = 37 or, 3x + 1=37. 24. The sum of the ages of three men is 157 years. The oldest is 6 years older than the second, and the second 5 years older than the third. What is the age of each ? 25. A farmer has a doubletree 4 ft. 4 in. long which he wishes to divide so that the longer part will be 1 in, longer than twice as long as the shorter part. How long shall he make the shorter part ? Prove answer. 26. Divide $ 500 among A, B, and C so that B shall have $20 more than A, and C $55 more than B. Prove. 27. Three times a given number diminished by 20 is equal to the given number. Find the number. . See Sug. under Ex. 13. 28. At an election two opposing candidates received together 2150 votes and one received 98 votes more than the other. How many did each receive ? Prove. 29. Three men A, B, and C, keep their cows in the same pasture and together pay $ 56 for the use of it. A has one cow, B has 3, and C has as many as both A and B. How many dollars will each have to pay ? PROBLEMS 39 30. A boy bought an equal number of apples, lemons, and oranges, paying 1^ for an apple, 2/ for a lemon, and 3^ for an orange. How many of each did he buy for 72 ^ ? Prove. SUGGESTION. Let x number of each sort. Then how many cents did the apples cost ? The lemons ? The oranges ? 31. A horse, saddle, and bridle together cost $ 144. The saddle cost 11 times as much as the bridle, and the horse 15 times as much as the saddle and bridle together. What was the cost of each ? a. REMARK. It is important that the word number always come after " Let x = ." Beginners often write : Let x = the bridle, or x = the cow, or x = A, or the like. Such statements are not clear, and are pretty sure sooner or later to lead to confusion of ideas. 32. A gate 49 inches high is made of four boards each 8 in. wide. The second space is to be 21 in. wider *^ than the lowest, and the upper 3 in. still wider. Find the width of each space. 33. Frederick has his money in dimes, nickels, and cents, and he has the same number of each kind of coin. The value of all is 80^. How many pieces of each kind has he? 34. The leaning tower of Pisa in Italy is 179 ft. high, which is 15 ft. more than 12 times the distance it leans over. How many feet does it lean over ? 35. In midwinter at St. Petersburg the night is 13 hr. longer than the day. What is the length of each ? 36. A rectangular field is twice as long as it is wide, and the distance around it is 372 yards. What are its dimensions ? SUGGESTION. Draw a figure and write x beside each short side. Then what shall be written beside each long side? Express the whole distance around by adding the lengths of the sides. 37. At an election there were two candidates, and 1280 votes were cast. The successful candidate had a majority of 40. How many votes were cast for each candidate? 40 THE ALGEBRAIC NOTATION 38. If the total wealth of the United States in 1900 was 88 billions of dollars, and the total wealth exclusive of money in cir- culation was 43 times all the money in circulation, what was the number of dollars of money in circulation ? 39. If there were approximately 4 persons not in public school to every enrolled pupil, and 36 pupils to every teacher, how many public school teachers, pupils, and other persons were there in the United States in 1900 when the population was 76,201,000 (using this instead of the census number, 76,303,387) ? 40. The total population of the United States in 1900 was 76,303,387 and there were 1,815,097 more males than females; how many were males and how many females ? 41. The sum of the exports and imports of the United States during the year ending June 30, '07 was $ 3,315,272,503 and the exports exceeded the imports by $ 446,429,653. What were the imports and exports ? 42. The mill consumption of cotton by the world in 1906-1907 was 19,493,441 bales of an average weight of 500 Ib. Of this amount the United States raised 928,913 bales more than twice the product of all other countries combined. How many bales did the United States produce ? SYMBOLS OF AGGREGATION 32. Removal of Symbols of Aggregation. 1. Simplify (3 x + 2 y) + (4 x - 3 y). SOLUTION. The sum is the same whether the quantities are tied together in groups by parentheses, or are left untied. Thus, (3o; + 2y) + (4a;-3y) = 3 x + 2 ?/ + 4 x By =7 x y. Am. CHECK. Let x 2, y = 1. Then, the given quantity, (3 x + 2 ?/) + (4 x 3 y') 8 + 5 = 13 ; also, the answer, 7 x y = 14 1 = 13. SYMBOLS OF AGGREGATION 41 From this solution and the explanation preceding it, we can see the truth of the following general rule: 2. Symbols of aggregation preceded by +-, or by no sign, can be removed without changing the sign of any term inside. a. It must be remembered here that + is understood before the first quantity within a symbol of aggregation, if no sign is written ( 8). 3. Simplify (3 x 4 y) + (2 x + 7 ?/), and check answer. 4. (3x 2 2x) + (2x-- 1). 5. (b 2c)+-(4c 56). 8. m 2 -3m + S5m-2w 2 |. 9. (6 a; + 2 y) + [4 a; - 11 y]. 1 f\ } j'\ -t / r\ r\ J. U t/ "" r t/ O "T~ tl C/ SUGGESTION. After solving by removing the vinculums, rewrite the problem, using parentheses, and then solve. 11. Simplify (3x + 2y) (a-- b), and check. SOLUTION. This means that a b is to be subtracted from Bx-\-2y. Writing the quantities down for subtraction as in 30, Ex. 28, and seeking a quantity which added to the subtrahend equals the minuend, we have 2y a b 3x + 2y - a + b Or, 3x + 2?/ - (+ a- &) = 3x + 2y a -}- b. Ans. CHECK. Let x = 4, y = 4, a = 3, b = 1, Then given quantity = (3 x + 2 y} (a - b} = 20 2 = 18 ; also, answer =3x + 2y a + 6 = 12+8 3 + 1 =18. From this solution we can see the truth of the following general rule : 12. Symbols of aggregation preceded by can be removed by changing the sign of every term inside. b. Notice that the + or sign before a symbol of aggregation drops out when the symbol is removed. It denotes an operation, and this operation is performed when the symbol is removed as explained in 2 and 12 above. 42 THE ALGEBRAIC NOTATION Simplify the following. Check the answers to the first seven and any other when in doubt about the correctness of the result. 13. (3&4-2c)-(& 4c). Ans. 26 + 6c. 14. (4m-2?i)-(2w 5w). 15. (4 a 2 - & 2 )-(2a 2 + 2 fr 2 ). 16 . (x 2 + 2 xy) ~(x 2 2xy). 17. a& - - (m -- 3 aft + 2 ax). 18. 9m ( 4 ??i + Gn n 2 ). 19. -[a 6] [6 ~ c]. 20. c. We bad just such problems as the preceding ones in 30, commencing with Ex. 33. There the process of solution was different. To show that re- moving the parentheses and then simplifying, as we have just been doing, gives the same result as writing the subtrahend under the minuend and subtracting, we will give the two solutions. Notice that the answers are exactly the same, Solution by removing parentheses and simplifying (adding). (3 a; 2 + 2 xy + y*)- (z 2 -2xy + ?/ 2 ) = 3 x 2 + 2 xy + y' 2 x* + 2 xy - y 2 = 2 x 2 + 4 xy. Ans. Solution by the process of subtraction. 2 x 2 + 4 xy. Ans. In this article the problems are to be solved, of course, by the process of removing symbols. 21. (4a + &) (x + 4:a+b 2y x y). 22. 4?/ 2 - 23. 3a 2 - 24. (5a-46 +3c)-(-3a + 26 - c)-(- a - c). 25. Make a rule for removing symbols of aggregation : (1) What, is done when the symbol is preceded by -f- ? (2) When preceded by - - ? What is done after the symbols are removed ? (See 29, 32.) 26. x 3 + x>- (x - 27. a?- SYMBOLS OF AGGREGATION 43 28. a 6 2r. 13. Make a carefully worded rule governing the exponents in multiplying when the same letter appears with exponents in two or more factors. 14. a 7 x a 2 . 15. s? x or 5 X a 2 . 16. a; 9 x a 5 X x. 17. a 2 X a 2 X a 2 . 35. Multiplication of Literal Quantities. Find the product of monomials mentally. 1. What is the product of 4 a?b and 5 abc ? Ans. 20 a s 6 2 c. EXPLANATION. The sign of the product is because unlike signs give ( 10, 5) ; the coefficients 4 and 5 are multiplied together for the coefficient of the product ; a~b x abc = (a 2 x a) x (b x &) x c by the order and group- ing laws in multiplication ( 13, 3, 4) ; then, a 2 x a = a 3 , b x b = 6 2 , as in the preceding article. 2. 3a-x2ab. 3. 4a 3 6x -3a 2 c. 4. 12 x 2 y x 2 ajy. 5. 3 ac x 5 b. 6. 5 a 2 # x - - 7 ax 3 ?/. 7. ,4 aw x 3 am 3 X am. 46 THE ALGEBRAIC NOTATION 8. - a 4 z x - a s x 2 z. 9. 4 ale 3 x -- 3 ax 2 . 10. 5 Irx* x - - 7 11. 2m 3 ce 2 12. -8a/n/ 2 13. f a 6 raa; 4 x - - 3 m# 4 X - - 5 a?/ 3 x --|fom 3 a? 14. Make a rule for the multiplication of literal quantities, answering the following questions : (1) What sign will the product have if there is an odd number of negative factors ? If there is not an odd number of nega- tive factors? See 10, 25. (2) How is the coefficient of the product found ? (3) What is done with the exponents of a quantity that appears in more than one factor to get the exponent of such quan- tity in the product ? What is done with literal quantities that appear in only one factor ? 15. 4 ax 2 x - - 3 aV. 16.2 mn 2 x -- 3 m 2 n X - - 2 m 3 w x - - 4 mn. 17. - 5 x"y x - - 4 xyz. 18. -5x--3ax--26x--cx4a. 19. - 11 arty X 4 arty 2 z. 20. - G m 2 x -- 3 m 2 x -- m 3 x - - 4 m 4 . 21. X G a-'V;. 22. 3 &c x -- 2 a 2 fcc X ~> /*: 2 X 0. a. By " taking " we can understand removing an object from one place to another. Then taking any quantity no times evidently gives nothing; like- wise, taking nothing any number of times gives nothing. 23. 4a 2 or 3 24. xifz 1 25. forty 26. -5an s y 2 X 3 ax 2 X x 3 ?/ 2 x - - 4- xy 3 X - - 4 a 4 hx 27. -3mn 2 28. 1.3a 8 6 29. 1.7 abc 30. -fa s m 2 X - - 4 mn 4 x .5a6 3 - ,2abd X - - n 31. Multiply m by ?,, the letters denoting number of units of length in lines. Suppose m and /i denote the lines underneath them. Then the n number of units of area in the rectangle formed with m line as length and n line as width equals the product of m and n, inn or mn. MULTIPLICATION 47 32. Multiply a + b by c, when each letter is supposed to repre- sent the length of a line. a a b a + b e x c b c ac be ac + be 33. Multiply 2 ax 2 3cz by 5 az and check with a = 2, c = 2, SOLUTION. 2 ax 2 3 cz = 14 x 5 az = x 30 10a 2 z 2 2 -- 15acs 2 = --420 34. Multiply 8 cr 4 a# by 4 am 3 and check. 35. (4 a 2 - 3 6) x 3 ab. 36. (8 a 2 - 9 ab) x 3 a 2 . 37. 2a(a 2 6 2 -2a&). 38. 2 afy(3 a 2 - 4 / + 5 z 2 ) 39. -5oj 2 2/(3^-2a- 2 z/+2/ 3 ). 40. 2 a&ceZ(a + 6 + c + d). 41. 4 a 2 3ac + 2 42. ??i# + ?iy - - 10 z X 5 ax X wi 2 n 43. Make a rule for finding the product of a polynomial and monomial, telling what terms of the polynomial are multiplied by the monomial, and what rule applies when any two terms are mul- tiplied together. 44. In these problems the distributive law applies. What is it ? 45. -9a6(5a-2a& + 10). 46. (-9 a 5 +3 a*b*-W) X -Sab*. 47. Sx 2y --4 + 5 48. f-aV-- f a^--| x Xx x a?x 49. The student must note carefully the difference in the man- ner of forming the product of two monomials and of a polynomial and a monomial. Thus, 3 x abc = 3 6c, and does not equal 3ax3Bx3c = 27 abc. But 3x(a + &+c)=3a+36+3c, as in the problems just solved. 48 THE ALGEBRAIC NOTATION 50. To multiply the product of several factors by a given quantity, multiply one of the factors by this quantity, leaving the other factors unchanged. 51. To multiply the sum of several factors by a given quantity, multiply each term by the given multiplier, and add the results. 52. 6x2abc. 53. 6 X (a 2 + b 2 + c 2 ). 54. 2 x3 X 4(m + ft+p). 55. 2 m x 3pqr X 4n 2 . 56. 6x7x5 xy. 57. 6(7 + 5 xy). 58. Multiply a + b by c + d and check with a = 2, b = 3, c = 1, c? = 5. DIAGRAM ILLUSTRATION SOLUTION. a - c - -b \-d ^bd - 5 a + & x 6 ac be ac - h be + ad- 30 c ad bd ac A - be 4- ad -i \-bd -30 + ^ EXPLANATION. Notice that each term of the multiplicand is multiplied by every term of the multiplier. (See 13, 5.) 59. Multiply m + n by p + q and check with m = 4, n = 2, p = 5, g = 3. 60. (a? + 5) (a; 4- 10). Check with # = 4. SOLUTION, x + 5 =9 a; + 10 = 14 Z 2 + 5x 126 + 10 x + 50 x 2 + 15 x + 50 = 126 61. (aj + 8)(a? + 8). 62. (x- 7)0 - 10). 63. (a; -12)(a5-l). 64. (2x-3)(x + 5). 65. (a? -15) (-a; + 3). 66. (a; +19) (a; -20). 67. (2 a? + 3) (a? -7). 68. (a? + 3 /) (a? + 4 y). 69. (a --9 6) (a + 5 6). 70. (xy ab)(xy + ab). 71. 2-3r2 + 3r. 72. (- MULTIPLICATION 49 73. Multiply 3 m~ + mn 2 n 2 by m 2 - 6 n 2 , and check with = 3, n = 2. SOLUTION. 3 m 2 -f mn 2 w 2 = 25 m 2 - 6 n z = x -- 15 3 w 4 + m*n 2 m' 2 w 2 - 375 - 18 i2 w -2 - 6 mn* + 12 n* 3 w 4 + w 3 - 20 m 2 n 2 - 6 mn* + 12 n 4 = 375 74. (a 2 - 2 (wj + 4 a? 2 )(a 2 + 2 ax). Check with a = 3,x = 2. 75. (4a 2 + 6.ivy-h92/ 2 )(2a--32/). Check with x = 3, y = 1. 76. (x> - xy + f)(x + y + 1). (See 13, 5.) 77. (a 2 - 78. (_a 5 79. (x 2 3 xy y 2 )( xr -\-xy 80. (2^-3ar + 2x)(2^ + 3 81. Multiply m + ?i +p by r + s + t and explain the result with a graphical illustration as in Ex. 58. 82. Make a rule for multiplying one polynomial by another, telling (1) How to set the problem down. (2) What monomial multiplications are performed. (3) What operation is performed last. (4) How the correctness of the result can be checked. 83. (l + 2a;-3a,' 2 + aj 4 )(^-2a;-2). Check with x = 2. 84. (ct 3 -3a 2 64-3a6 2 -6 3 )(a 2 -2a6 + 6 2 ). Put a = 3, 6 = 1. 85. 86. 87. 88. (aj-2)(a?-3)(aj-4)(- 89. (a 2 - b 2 ) (a 2 + ab + b 2 ) (a 2 - a& + b 2 ). 90. (a 3 + 2 a 2 b + 2 a& 2 -f & 3 )(a 3 - 2 a 2 6 -f 2 a& 2 - 6 3 ), 50 THE ALGEBRAIC NOTATION 36. Powers and Roots of Monomials. (See definitions 17 and 18.) 1. Square 3 a 2 . SOLUTION. 3 a 2 x 3 a 2 = 9 a 4 . Ans. 2. (5a 3 ) 2 =? Ans. 25 a 6 . 3. (mV) 2 . 4. (-4a 3 6) 2 . 5. (-o?bcf. 6. (-2 a 4 ) 2 . 7. (-i<> 2 . 8. (20 a 2 ) 2 . 9. (-^m 3 ) 2 . 10. (^6 6 ) 2 . 11. Make a rule for squaring a monomial. (1) What is done with the given coefficient to get the coefficient of the answer ? (2) By what is each exponent multiplied to get the exponent of its quantity in the power ? (3) What sign does the answer always have ? 12. Cube -4a 2 5. SOLUTION. -4 a 2 b x -- 4 a 2 & x 4 a^b = - 64 6 & 3 . 13. (or 5 ) 3 . 14. (-z 2 ?/ 3 ) 3 . 15. (-2a 2 c 4 ) 3 . 16. (-6?/) 3 . 17. (z 4 /) 3 . 18. (Jm 2 w) 8 - 19. (-5m 4 ) 3 . 20. (-3 a; 4 /) 3 . 21. (-a 10 ) 3 . 22. (a-?/z 2 ) 3 . 23. (-6mV) 3 . 24. (4a^z 2 ) 3 . 25. Make a rule for cubing a monomial. (1) What is done with the given coefficient to get the coefficient of the power ? (2) By what is each exponent multiplied to get the exponent of its quantity in the power ? (3) If the given quantity is positive, what sign will the power have ? If negative, what sign will the power have ? 26. Make corresponding rules for higher powers. 27. (-3 3 ) 3 . 28. (2 a 2 ) 4 . 29. (-3 a 3 ) 5 . 30. (-5aW) 3 . 31. (-3 a; 4 ) 6 - 32. (-15 z 4 ) 2 . 33. (-6 a 4 ) 3 . 34. (-abc) 7 . 35. What is the sign of an odd power of a negative quantity ? Of an even power ? What is the sign of all powers of positive quantities ? 36. V36a 2 = ? Ans. 6 a. a. The sign is used to denote either -f or . Thus, (-f 6 a) 2 = 36 a 2 ; and, also, (-6 a) 2 = 36 a 2 . POWERS AND ROOTS 51 37. VU5 iii 2 . 38. Vl6m 4 . 39. V8a*. Ans. + 2 a. 40. 41. V64 a. 42. V<49a 8 . 43. ^G4~c?'. 44. 225 m?ij -r- - 25 my. 30 ajy -Slabdif m*pW 42 10. ' 1 J. . -- J./a. - * J.O. . . . . - 5 xy 2 3 My mpsr 7 XZ A 14. Make a rule for division of monomials, telling (1) How the sign of the quotient is determined. (2) How the coefficient of the quotient is found. (3) How the exponents of the several literal factors are found. (4) How the solution is proved correct. 15. - 16 x*y 2 + -xy\ 16. - 3 a 5 c 9 -h - 17. 16b 2 x 2 y + -2xy. 18. 63 aW -- 9 aW. 19. (a + b) 5 H- (a + b) 2 . Ans. (a + b)\ 20. 4(m-|-7i) 4 -5-2(w + w) 8 . 21. 8 (a - &)V-=-2(a 22. -50ary-*--5aty. 23. - 35 a 11 -- - 7 a 7 . 24. - 1 x 4 ?/ 7 ^- 1 .T?/ 5 . 25. | m 8 n 7 -. f 26. Divide 9 x 3 - - 12 x 2 y + 3 a by 3 x. SOLUTION. 3 x)0 r 3 - - 12 .r' : ji/ + 8 3 x' 2 4 xy -f DIVISION 53 PROOF. 3 x z 4 xy + z - 12 x' 2 y + 3xz a. Notice that when a polynomial is multiplied by a monomial every term of the multiplicand is multiplied by the multiplier. Hence, when a polynomial is divided by a monomial every term of the polynomial must be divided by the monomial. 27. Divide 15 x 5 25 x* by -5 a 3 . Prove. 28. (10 x 7 - - 8 a; 6 -f 3 x 4 ) -f- or 3 . Prove by multiplication. 29. (a 2 ab ac) -. a. Prove by multiplication. 27 X G - 3G x 5 cf + ab 30. - - ol. 9x a 10 a 2 z -15 z 2 - 25s 8 O/i. - * - OO. xy 3 ab + 12 abx - 9 a 2 b a 8 -V6 34. - ^- 35. Sab a 2 36. ( 3 a 2 + f ab 6 ac) -r- -fa. Prove. 37. (4 #y 8 ary + 6 ct'i/ 3 ) -i- - 2 a? t y. Prove. 38. Make a rule for dividing a polynomial by a monomial. (1) How many terms of the dividend must be divided by the divisor ? (2) What rule have we for dividing one term of the dividend by the divisor ? ( 38, 14.) (3) How can the answer be proved correct ? 39. (5 x?y s 40 a?a?y* + 25 a*xy) -. 5 xy. Prove. 40. (4a6c-24o 2 & 2 -32aM+12a6)-j--4a6. Prove. 41. ( S4aj 8 2/ 2 ~51a;y + 68iBy IT^)-?- ITsB 2 ?/. Prove. 42. [3 a(x + ?/) + c\x + ?/) 2 - 5 b(x + ?/) 3 ] -=- (x -f y). See Ex. 19. A comment like that made in 35, 49, can be made here. 54 THE ALGEBRAIC NOTATION 43. To divide the product of several factors by a given quantity, divide one of the factors by the given divisor, leaving the other factors unchanged. Thus, (30 a 2 x 12 abc) + 6 a=5 a x 12 abc, or 30 a 2 x 2 be, = 60 a 2 bc. 44. To divide the sum of several terms by a given quantity, divide each term by the given divisor, and add the quotients. Thus, (30 a 2 + 12 abc) -i-6a = 5a-f25c. Check the answers found in the four following problems by letting x = 3, y = 2 : 45. (16 a,- 2 x- 24 ay 2 ) -5- 8 a;. 46. (16 a? - 24 a;/) -5- 8 a?. 47. (3x 2 x$xy x6xif) + 3x. 48. (3 ar 5 + 9 it' y + 6 xf) -=- 3 a;. 49. Divide a,* 2 -!- 11 a; + 30 by x + 6, comparing the method of solution with the arithmetical division of 675 by 25. SOLUTION, z 2 -f- 1 1 x + 30 x 2 + Qx x + 6 divisor , 25)675(27 x + 5 quotient 50 5z + 30 175 5 x + 30 175 EXPLANATION. Let the student note how 675 is divided by 25 at the right. He will see that the first part of the dividend is divided at the start by the divisor 25. Then the divisor is multiplied by the quotient figure 2, and the product, 50, is subtracted from the dividend. Then, after another figure of the dividend is brought down, the remainder is divided by the divisor. In algebra, for convenience in multiplying, the divisor is commonly written at the right of the dividend and the quotient underneath it. The first term of the dividend, x 2 , divided by the first term of the divisor, x, gives the quo- tient, x. Hence, x is written as the first term of the quotient. Then the whole divisor is multiplied by the first term of the quotient, the product being placed under the dividend. Next x 2 + 6 x is subtracted from the dividend. The first term of the remainder is then divided by the first term of the divisor and the quotient 5 is written as the second term of the quotient. Last of all, the divisor is multiplied by the second term of the quotient and the product is subtracted from the remainder. PROOF. If the student will compare this multiplication care- x+6 fully with the process of division, he will see that each x + 5 line in the multiplication operation is likewise a line in x- -f 6x the division operation. Evidently the process of division + 5x4- 30 is the reverse of multiplication. x' z + 11 x -f- 30 DIVISION 55 Perform the division required in the following and prove by multiplication : 51. or>+ 15 x + 50. (.r- 52. (r-llm + 30)-s-(m-5). 53. (3a? 2 + 10a;+ 3)-i-(a; + 3). - A / ' f^ ^ I ^^ I f~* *^\ / A.* I ^) n t\ f f C /\ / / /^t I _ / /^ II I * , ^ /** l * ( O* I " 1 1 \ ^\ T 56. (4x 2 + 23:c + 15)H-(4a; + 3). 57. (6x- 7 x 3) + (2x 58. 59. 60. (x* y*}-r-(x x-y 61. (ac ad + Z>c 6d) -3- (a + 6). x-y x-y xy* xif-if ac ad -} - bc- bd a + b ac -4 - be c d a + 4^ + 2^-12 by 7x-2a? + x i -j- 5, and prove by multiplication. SOLUTION. In this problem the dividend and divisor should first be "ar- ranged." See 37. 4 x 5 - 5 x 4 + 20 x 3 -f 50 x 2 + 2 x - 12 4 x 5 - 8 x 4 + 28 x 3 + 20 x 2 x 3 -2x 2 + 7x + 5 4 x 2 + 3 x - 2 2x 15 x 9x 2 --13x--12 4x 2 - Use --10 5 aj2 + x - 2 To prove the work correct, we multiply the divisor by the quotient and add the remainder to the product. If no mistake has been made in either the division or proof, the sum will be the same as the dividend. PROOF, x 3 - 2 x 2 + 7 x +5 - 8 x 4 -f 28 x 3 + 20 x 2 - 2 x 3 + 4 x 2 - 14 x - 10 4x 5 - 5 x 4 + 20 x 3 + 45 x' 2 + x -- 10 5 x 2 + x - 2 4x 5 -5x* + 20x 3 + 50x 2 + 2 x - 12 (Check.) Divide and prove : 69. (2 a 4 + 3 a 2 - 5 a + 2 - 6 a 3 ) -r- (a 2 - 3 a + 1) . 70. (6 s + 4 aW - 13 a 2 6 + 2 fr 3 ) -*- (2 a - 3 6). 71. (2 m 4 + 6 m 2 - 4 ra - 5 m 3 -f 12) H- (m 2 - m + 1). 72. 73. 74. (6 aV - 4 a 3 3 - 4 ax 3 + a 4 -f a 4 ) ~ (cr + x*-2 ax). 75. x 6 - MISCELLANEOUS EXERCISE 57 *76. (ax 2 - ab 2 + b~x - or) -=-(> -f b)(a - a;). 77. (X G - a 6 ) -=- (ar* + 2 ax- 2 + 2 a 2 + a 3 ). 78. (A^-faty + ttay + i^-Kfa + t!* 79. Qr -hpg + 2|>r 2 g 2 + 7 gr ~ 3 r 2 ) -f- (p ~ g + 3 r). 80. 81. (a 4 - 16 & 4 ) -s- (a -2 6). 82. 83. A 8 -A; 8 -r-/r + A- 2 . 84. a^ 1 - 85. Divide 1 + 2 a; by 1 - - 3 a?, getting 5 terms in the quotient. Ans. 1 + 5 a; + 15 or + 45 a 3 + 135 a? 4 + etc. 86. Divide 1 + 3 x 2 by 1 - - 4 #, getting 5 terms in the quotient. 39. Miscellaneous Exercise in Addition, Subtraction, Multiplica- tion, and Division. 1. From or 5 + 3 x 2 + 3 x + 1 take or* -- 3 a 2 + 3 a; - - 1. Check with x = 2. 2. Add3 ajy - 10 y\ - x 2 / + 5 y\ 8 afy 8 - 6 /, 3 a? 4 + 4 afy 2 + 2 ?/ 4 . 3. Divide 8 ar 3 + / by 2 a? -f 2/- Prove. 4. (a 2 4- aa? + a; 2 ) (a 2 - ax + or) = ? 5. (a^ + 6c)a 3 6 = ? 6. 4 a 3 6a? x 7 ftV. 7. 8. (a 4 6 s a; - 5 6 fl ) X i oV. 9. (a; 2 + 2 a 10. 12^ 5 -9^ 3 + 6.T 2 --3x 2 . 11. 12. 13. (ab + cd + ac + bd)(ab 14. From -I a 2 - 4 a 1 take a 2 + a . Check with a = 4. O ^5 O * 15. What expression must be added to 5 x 2 - 7 x + 2 to produce 58 THE ALGEBRAIC NOTATION 16. Subtract 3 3? - 5 x + 2 from 1 and add 5 x 2 6 xto the result. 17. Multiply |- #?/ by - 3 # 2 -f |- or?/. Check with sp = 6, y = 1. 18. (8 x 8 x 2 4- 5 x 3 -f- 7) -r- (5 a? 3). Prove work correct. 19. (30x + 9-71x* + 2Sx 4 -3ox 2 ) + (x*-13x + 6). Prove. * 20. Divide (3 x 4 - I + 3 x + 6 x 2 + 7 or 3 ) (1 + a; 2 - a) by a; + 1 + or. 21. Divide 8 on/ 2 + 3 # 4 5 a 1 ?/ 3 7 # 3 ?/ + ?/ 4 by x 2 -j- 2/ 2 #?/, arranging first according to descending powers of x\ then divide, arranging according to the descending powers of y, and compare the quotients. How are the quotients alike? How different? 22. 5 (a - &) + G(a - b) - 4(a - 6) = ? 23. 3(3* - /) + 7(0* - r) - 2(ar - /) = ? 24. Simplify 2a-2(6 + 3a) -3{6 + 2(a- &)}. In working this and the following problems the student must remember that the absence of sign between quantities denotes multiplication and that the order of performing operations, as stated in 27, is powers and roots first, multipli- cation and division next, and addition and subtraction last of all. However, quantities inclosed in parentheses must be treated as single quantities. SOLUTION. 2 a - (2 b + 6 a) - 3 (6 + (2 a - 2 5)}. (Multiplication performed first.) 2 a 2 b 6 a 3 {2 a 6}. (Parentheses removed and terms collected.) 2 a - 2 b - 6 a - (6 a - 3 6}. (Multiplication first. ) 2a 26 G a 6a + %b = b 10 a. ^.?is. (Simplification.) Simplify the following: 25. a [56--{a--3(c 6)4-2c (a26 c)J]. '-4ns. a. First perform the multiplication inside the brace ; then remove the brace ; next remove the bracket ; lastly simplify. 26. (x + y)(y + z) (z + u)(u + x) ({B + z)(y u). Ans. y z --u 2 . 27. \x(x-\-a) a(x-}-a)l {x(x a) a(a x)\. Ans. a 4 - 2 a 2 .T 2 + a 4 . 28. (a? + 6) (a? + c) - (a + b + c)(x + 6) + a 2 + aft + b 2 + 3 2 a# + a 2 . 29. a + & 2 -a + &a-Z>-2&-2-& 2 -2a. -4wa. 36 2 . MISCELLANEOUS EXERCISE 59 40. Miscellaneous Exercises in Fundamental Operations to be solved by inspection, that is, mentally. 1. From 16 bx 2 take - - 3 bx 2 . 2. Add 6 m z p and 5 m 2 p. 3. From 3x 2y + 5z take 5 y z 8 x. 4 Q r 3 fi - 8 ri*h fi fi r/ 5 r 6 ?/ 9 7 Tt. t/ t// t/ C/ vt \J O. \j \JL *v u I . Subtract: 8. 4 a; 9. -6 a 2 10. 6c 11. 1 y?y 11 x 4 a 2 56c -12ar.y 12. -48 -T- -4. 13. 21 a 10 -f- 3 a 7 . 14. -70a 7 6 8 --14 Add : 15. 5- 76 16. - 6^--16?/ 2 17. -17c/w 2 + 7 4 a + 12 6 9 ar 5 + 3 y -2 a- 66 18. If a (b c) = ab ac, what does 34 (50 - - 1) = ? What is a good way to multiply 34 by 49 ? Multiply in same way 26 by 39 j 45 by 78 j 120 by 151. 19. (5z + 12)7z. 20. 3m(10m 2 -4?0- 21. 2 a (6 a 2 -3a-f-4). 22. (25 a 8 - 15 a 6 ) --5 a 2 . 23. (54 4 & S -GO 7 //'+6a 4 6 4 )--6a 4 6 4 . 24. (5a;-7)(3a?H-2). 25. (2rt-3)(6a-7). 26. 9 w +(- - 4 m + 6 ?i) - (3 m - - n). SUGGESTION. Add m's first ; then w's. 27. 2a?-3- 28. 29 . (5 m + 9 n + 4 a;) + ( - 3 x 7 // - 6 n ) - (10 y - - 8 a? - 2 m) . 30. Subtract 12 # - - 7 ?i 6 // from 11 n + 3 m 8 a?. 31. 5 x 9 / x 9 ?/V x ass 7 . 60 THE ALGEBRAIC NOTATION * 32. - 3 a% x - 5 b s c x - 2 c 3 X - aW. 33. (6 a-V - 9 ab 2 x 2 - 15 a W) -f- - 3 ax*. 34. m# 2 - - [8 y -\- (6 a mx) 2 a]. 35. 5f a 2 m 2 -(-2i 36. Sma-S^ + S 37. 3 a 2 - 6 ab - 8 6 2 -f- 7 a 2 - 3 a 2 + 2 aft - 14 W - 6 06. 38. 4 a 2 - 6 a + 4 - 3 a 2 + a + 1.5 a 2 - 2 - 3.4 a 2 - 3.75. 39. Prom #*,.+ 1 take 1 2 x + # 4 + 3 x 2 4 or 5 . 40. Add 6 n, In, 3 n } 18 w, and 11 n. 41. 4a-2&- 41. Review of Definitions. The best definitions for the student are those which he makes from his own knowledge of words. Let the student write out in his own language from the knowl- edge he has gained of the thing defined, and not by looking up the definitions given in the book, the definitions of the following, and bring them to class for discussion : (a) Addition in algebra. (6) Subtraction in algebra. (c) Multiplication. ( d) Division. (e) Exponent of a quantity. (/) Coefficient, (g) Literal part. (7i) Term, (i) Similar terms, (f) Quantity. (k) Polynomial. (I) Power of a quantity, (m) Eoot of a quantity, (n) Positive quantity, (o) Leading letter, (p) Distributive law. ((/) Symbols of aggregation. (?) Equation (see equations, p. 37), etc. 61b. 6-2 Ib. 6-2 Ib. An equation ( 31, 42) resembles weights balanced on scales, In the figure we see equal weights from equals leave equals. INTEGRAL EQUATIONS Gl INTEGRAL EQUATIONS 42. Solution of Integral Equations. An equation is a statement that two quantities have the same numerical value. Each problem in 1 and 31 led to an equation. In the solu- tion of the equations of those articles, the answers were reasoned out. We will now proceed differently. From the example below we can see that any quantity may be "transposed" (carried across) from one side of an equality sign to the other by changing its sign. Thus, 6 x + 44 = 4 x + 56 (By subtracting 44 + 4 x from each becomes 6x 4 cc = 56 44 side of the first equation. Equals or, 2x = 12 subtracted from equals leave equals.) whence, x = 6 (If 2 x = 12, x = \ of 12.) PROOF When x = 6, 6a; + 44 = 4x + 56 becomes 6x6 + 44-4x6 + 56 or, 80 = 80 To solve an equation, "transpose" so that all terms containing $ are on the left side of the equality sign, and all terms which do not contain x are on the right side. TJien simplify each side. Last of all, find x by dividing the right " member " of the equation by the coefficient of x. (If coefficient is negative, say 4, divide by 4.) Prove the answer right by substituting it in the given equation. 1. 11 + 10 = 8 +-19. 2. 7x + l = 4 a; + 13. 3. 16a?-ll = 7a? + 70. 4. 24^-49 = 19 x- 14. 5. 23 x- 50 = 18 x -15. 6. 3 a + 24 = 79 -2x. 7. 16a; = 64-(12-3aj). 8. 8x-7 -3z = 2-4 x. 9. 5 (x 7) + 63 = 9 x. SUGGESTION. First multiply x 1 by 5. 10. 16aj = 38-3(4-o;). 11. 9(7- a) =6- 3(5 -a). 12. 4(aj-2) = 2a;-7(aj-4). 13. 6o?H-4#- 16 -2a; = 0. 14. Find a number which added to 22 less than itself = 216. 62 THE ALGEBRAIC NOTATION 15. A man who had paid out $700, $400, $200 respectively towards the education of his three sons, directed in his will that this difference should be equalized in the settlement. If his property was worth $ 9200, how much did each son get ? SUGGESTION. Let x dollars, (x + 300) dollars, (x + 500) dollars equal sums the sous got. Why ? 16. A stock fence has nine 4 + 5 a? wires, the lowest three spaces being 3, 3, 4 in. respectively, and the others growing wider upward by the same amount. Find the dis- tances apart to put the wires if the fence is to be 52^ in. in height. 4 + 305 17. How many cents a pound is paid for butter when added to the cost of 2 Ib. gives the same sum as 2/ added to the cost of 4 Ib ? 18. A cashier paid out $26 in $1 and $2 bank notes, and the number of notes of the latter kind was less than the number of the former kind by 5. How many notes of each kind did he use ? 19. A boy has 155 hills of potatoes to dig in 5 hours. He pro- poses to lay out his work in such a way that during each hour he may have 6 hills less to dig than during the hour preceding. To do this how many hills must he dig each hour ? 3t x = number dug first hour. Then how many were dug The third ? etc. 20. Using round numbers, there were in four of the largest libraries in the world, in 1909, 8| millions of volumes (including pamphlets). The Koyal Library in Berlin had ^ million less than the Library of Congress at Washington and 1 million less than the British Museum in London, and the Bibliotheque Rationale in Paris had as many as both the Berlin Library and the Museum. How many volumes had each ? 21. In 1900 had the white population of continental United States been reduced by 4,621,238, there would have been 7 white SUGGESTION. the second hour ? INTEGRAL EQUATIONS 63 persons to every one of negro blood. The population, exclusive of Chinese, Japanese, and Indians, was 75,693,190. What was the white population, and what was the negro population ? 22. In the 573 colleges and universities of the United States in 1906-1907 there were 161,760 students. Of this number the male students were 2,685 less than 2 times the number of female students. How many were there of each? 23. In the colleges and universities of the United States in 1907-1908 there were 24,489 professors and instructors. Of these there were 1634 more than 4 times as many male teachers as female teachers. How many were there of each? 24. According to the census of 1900 the total population of Ohio, Illinois, and Pennsylvania was 15,281,210. The population of Illinois was 664,005 greater than that of Ohio, and the popula- tion of Pennsylvania was 1,480,565 greater than that of Illinois. What was the population of each ? 25. In 1906-1907 there were 367,036 more than 21 times as many pupils in the public elementary schools of the United States as there were in the public high schools. In both there were 16,890,818. How many were there in the secondary schools, and how many in the elementary schools ? 26. The sum of the three dates (all A.D.) on which Jerusalem was destroyed by Titus, the Roman Emperor Constantine recog- nized Christianity, and Constantinople was taken by the Turks, is 1836. From the first date to the second is 243 years, and from the second date to the third, 1140 years. What is each date? Prove answers. 27. The heights of the Eiffel iron tower in Paris, the Washing- ton monument in Washington, D.C. (each the highest in its class), and the twin spires of the Cologne cathedral, added together, give a total of 2051 ft. The Washington monument is 43 ft. higher than the cathedral, and the Eiffel tower is 40 ft. less than twice the height of the cathedral. What is the height of each ? 64 THE ALGEBRAIC NOTATION 28. New York City has the highest office buildings in the world. The total height of three of them, the Flatiron, World, and Times buildings, is 1081 ft. The World building is 89 ft. taller than the Flatiron, and the Times 45 ft. taller than the World building. What is the height of each ? 29. The sum of the heights of the Metropolitan Life and Singer buildings is 1312 ft. and their difference is 88 ft. What is the height of each ? 30. The wheat crop of the world in 1906 was approximately 3423 millions of bushels. The number of bushels produced in other countries was 483 millions of bushels more than 3 times that produced in the United States. How many bushels were produced in the United States? 31. In 1908, among our 464 colleges, universities, and technical schools for men and for both sexes, the N. Atlantic states had twice as many as the Western, 55 less than the Southern, and 101 less than N. Central states. How many had each region ? 32. Of the 464 institutions of Ex. 31, Ohio had the largest num- ber, having 7 more than Pennsylvania and 10 more than New York. Four times the number in these three states and 9 more equals the number in all the other states. What was Ohio's number ? 43. Historical Note. Algebra is not so old a science as either arithmetic or geometry, though it goes back farther than the Arabic notation of arithmetic. The earliest recognized writer on algebra was Diophantus of Alexandria, Egypt, who died about 330 A.D. It should be said, however, that Annies, an Egyptian writer, who lived about 1700 B.C., wrote a book that contained the solution of some simple equations. His work, found on a papyrus now in the British Museum, was translated by Eisenlohr in 1877. Diophantus wrote his work, called Arithmetica, in the Greek language, and introduced the idea of an algebraic equation expressed in symbols. Apparently he did not clearly understand the nature of negative numbers, which he avoided. HISTORICAL NOTE 65 After Diophantus there came the following eminent writers on the subject : Aryabhatta (about 500 A.D.), Brahmagupta (about 630 A.D.), and Bhaskara (born 1114 A.D.), all Hindu writers living in India. Following Aryabhatta and Brahmagupta came Arab authors, prominent among whom was Mohammed ibn Musa, gen- erally called Al-Chwarizmi. Europe got its knowledge of algebra from Arabia, and Europe communicated the knowledge to the rest of the world. More will be said later of the persons just named. It is interesting to find that there have been three stages in the development of algebra: (1) the rhetorical stage, in which all solutions were written out in full, much as in arithmetical analysis ; (2) the syncopated stage, in which abbreviations for important words were introduced; and (3) the symbolic stage, in which symbols are used to denote numbers, operations, and relations. Ahmes and the earliest Italian and East Indian writers used the rhetorical form. Thus Metrodorus, a Greek writer of about 310 A.D., Aryabhatta, and Brahmagupta used the rhetorical form. Diophantus, Al-Chwarizmi, and Bhaskara, on the other hand, used the syncopated form. This syncopated algebra was used up to the sixteenth century. From then on algebra be- came symbolic in form. During the sixteenth and early part of the seventeenth centuries writers were introducing one symbol after another. Thus Widmann in Germany introduced + an( l - ; Robert Recorde in England introduced =, and Oughtred, X ; and so on. The present symbolic algebra in which symbols are almost exclusively employed is considered to have been introduced and used first by Vieta (1540-1603) in France. Since Vieta's day, algebra has been developed by a long line of students and in- vestigators, prominent among whom may be named Sir Isaac Newton in England, Descartes in France, and Euler in Germany. CHAPTER II AB THEOREMS IN MULTIPLICATION AND DIVISION. FACTOR- ING. LOWEST COMMON MULTIPLE 44. A theorem in algebra is a rule or law expressed in words. A formula is a rule or law expressed in symbols, 45. Theorems in Multiplication. 1. Let A and B represent any two algebraic terms, or two lines. Then A +B A B A +B A A* AB A 2 + AB + AB + B 2 A 2 + 2 AB + B 2 From. the multiplication above and also from the diagram, we see that : (A + B)' 2 = A 2 + 2 AB + B 2 . (A + By 2 means " the square of the sum of any two quantities, A and B, of which A is the first and B the second. 1 ' A 2 + 2 AB + B 2 is " the square of the first quantity, A, + twice the product of the two quantities, A and B, + the square of the second quantity, B" Or, we may write (1st number + 2d number) 2 = (1st number) 2 + 2(lst number) x (2d number) + (2d number) 2 . Changing the formula (A + B) 2 = A 2 + 2 AB + B 2 into a theo- rem, we have THEOREM I. The square of the sum of two quantities is equal to the square of the first, plus twice the product of the first quantity mul- tiplied by the second, plus the square of the second. THEOREMS IN MULTIPLICATION 67 a. REMARK. Notice how much shorter the formula is than the theorem. In the formula there are only 16 marks, counting each letter and sign as a mark. In the theorem, counting each letter and comma as a mark, there are 144 marks, or nine times as many as in the formula. In this example we see how wonderfully compact the algebraic notation is. 2. Square the sum of 7 and 4 by the theorem and check result, - 49 + 5G + 16 = 121 or, II 2 =121. Check. 3. (3a 2 + 2m) 2 = ? SOLUTION. (3 a 2 + 2 wi) 2 = (3 2 ) 2 + 2(3 a 2 x 2 m) + (2 n) = 9 a 4 + 12 a' 2 m + 4 m 2 . Ans. PROOF. 3 a 2 + 2 m 3 a 2 + 2 m 9 a 4 + ti a-??* + 6 g?m + 4 m 2 9 a 4 + 12 a-w + 4 m 2 Write the results directly by use of the theorem. Check by actual multiplication until the use of the theorem is well un- derstood. 4. (4cr-f9w) 2 . 5. (2 a 2 + 56)". 6. (5w+3n)(5m+3w). 7. (a + 5)*. 8. (2a + 4) 2 . 9. (a + 7 6)(a -f-7 6). 10. (7z + 3c) 2 . 11. (a 8 + 66)*. 12. (3 a 2 + 5 6 s ) 2 . 13. (l+2a s 6) a . 14. (JoJ + ^y) 2 . 15. (6 + 10 a- 2 ) 2 . 16. 21 2 = (20 -f I) 2 . 17. 72 2 = (70 + 2) 2 . 18. 102 2 = (100 + 2) 2 . 19. 5i 2 = 5- 2 . 20. 12> 2 =12+i 2 . 21. ^4 7? 22. Let ^1 and 5 again represent any two terms. Then from the multiplication in the - margin, we have _ AB 4- 68 THEOREMS IN MULTIPLICATION AND DIVISION Changing this formula into a theorem, we get THEOREM II. The square of the difference of two quantities is equal to the square of the first, minus twice the product of the first quantity multiplied by the second, rjlus the square of the second quantity. 23. (^ SOLUTION, (a - 3 &) 2 = a 1 - 2 (a x 3 ?>) 4- (3 6) 2 = a 2 - 6 ab + 9 b*. Ans. PROOF, a 3 b a-3b a? 3 ab - 3 ab + 9 62 a 2 - a& + & 2 Using Theorem II, write directly the answers to the following : check by actual multiplication until the application of the theo- rem is well understood. 24. (.-4?/) 2 . 25. (2x-3y) 2 . 26. (x-2m)(x-2m). 27. (p*-3q)' 2 . 28. (5z 2 -4) 2 . 29. (4 a - 3) (4 a - 3). 30. (r-f) 2 . 31. (2.r 2 -5?/ 4 ) 2 . 32. (3 3 -6)(3 a 3 - 33. (l-8afy) 2 . 34. (2-i&) 2 . 35. (Ja - J&)(a 36. 99 2 =(100-1) 2 . 37. 147 2 = (150-3) 2 . 38. 995- = (1000 -o) 2 . 39. Let A and J5 represent any two terms. A + B Then from the multiplication in the margin, we A B- see that A 2 + AB Changing this formula into a theorem, we get THEOREM III. The product of the sum and difference of two quantities is equal to the difference of their squares. REMARK. Notice that (A + B} (A-- B} is the product of the sum and difference of two quantities of which A is the first and B is the second. No sign between the parentheses indicates a product. THEOREMS IN MULTIPLICATION 69 40. SOLUTION. (3 a + 5 6) (3 a - o 6) = (3 a)- - (5 b)- = 9 a 2 - 25 6-. Ans. PROOF. Write the answers to the following di- 3 + 5 6 rrctly by use of Theorem III. Check by 3a - actual multiplication until the use of the ; ^ ^ _ 2 - fca theorem is well understood. 9 a 2 _ 25 b- 41. (5a + 26)(5a-2&). 42. (3 a + 11 &)(3a- 11 43. (3 m 2 - 4 p)(3 m 2 + 4. 44. (11 m 2 + 6 n 2 )(11 ??i 2 - 6 n 2 ). 45. (4 3 - 3 6 3 )(4 a 3 + 3 6 3 ). 46. (3 .r/ - 7 .^/)(3 x/ + 7 47. (im 8 + l)am s -l). 48. 49. (a 5 -l)(a 5 + l). 50. 51. (a& + cd)(a6-cc?). 52. 53. (5a 8 6 2 + 2aj)(5a 8 6 2 -2a;). 54. 55. 18x22=(20-2)(20 + 2). 56. 49 x 41 = (45 + 4)(45- 4). 57. 103x97 = (100 + 3)(100-3). 58. 1025x975. 59*. (24 + 4)(24-4) = 24 2 -4 2 = 28x 20=560; then 24 2 = 560 + 16 = 576. 60*. Square in the same way 33 ; 72 ; 124 ; 117, efc. Check by actual multiplication. x 4- a 61. Let .r, a, and 5 represent any three x + b terms. From the multiplication in the &~ + # margin, we see that + bx -\-ab _ (x + a)(x + 6) = or 2 + (a + 6)^ + a6. x> + (a + b)x + ab Changing this formula into a theorem, we have THEOKEM IV. The product of two binomials having a common term equals the square of the common term, and the algebraic sum of the other terms times the common term, and the algebraic product of the other terms. * See 1, p. 6. 70 THEOPtEMS IN MULTIPLICATION AND DIVISION 62. (a? + 6) (a? - 9) = ? SOLUTION, (x + 6)(x + [ - 9]) =x 2 H- (6 -f [ 9])x + 6 x - 9 = x 2 3 x 54 PROOF. x -4- 6 SUGGESTION. Whenever a second term is negative, write it in a bracket preceded by +, as in the problem ff2 I (j y- just solved, until the process is well understood; afterward this will not be necessary. = ~ y? 3 x 54 63. (a? + 2)(a;-6). 64. (a; + 5)(a;-ll). 65. (a; + 6) (a; + 9). 66. (a; + 7) (a; + 8). 67. (m-2)(m + 10). 68. (a; -8)(a?- 12). 69. (z 4 -3)(V-9). 70. ( + i)(-i). 71. (a 2 -5)(a 2 -7). 72. (a 3 + 5)(a 3 - 7). 73. (a 2 + 4~)(V + 5). 74. (aj 2 4)(a? 2 -5). 75. (/-4)(2/ 2 + 5). 76. - 77. (ojy-4)(ajy + 16). 78. 79. (5o-36)(5a-76). 80. (3o-6)(3a 81. (3-aj)(4-a?). - 82. (5 + 6 2 )(7 + 83. (2z + lT)(2z-13). 84. (2m 3 -ll)(2m 3 -12). 85. (3 x 2 - 7)(3 ^ 2 + 16). 86. (4m 4 - 15) (4m 4 + 5). 87. Let A and B represent any two algebraic terms. Then, from the mul- A ~D tiplication in the margin, we have A* AB 2 B\ Put into a theorem, this formula be- A 2 - 2 AB + B 2 comes A - B THEOREM V. The cube of the differ- A * - 2 A'B + AB 2 ence of two quantities equals the cube of A B + -, AB- - B the first, minus three times the square of A* 3 A-B + 3 AB 2 B 3 THEOREMS IN MULTIPLICATION 71 the first times the second, plus three times the first times the square of the second, minus the cube of the second. a. Evidently in the cube of the sum of two quantities, all the signs on the right side of the formula are positive. 88 . ( a s _2 by = ? Expand " and check with a = 2, 6 = 1. SOLUTION, (a* - 2 &)' = (a 2 ) 3 - 3(a 2 ) 2 (2 6) + 3(a 2 )(2 ft) 2 - (2 6)3 = a e ._ (3 4 & + 12 a -b* - 8 6^ ( 27) (4 _ 2)3 = 2* r= 64 - 96 + 48 -- 8 = 8. Check. 89. (a 2 -36) 3 . 90. (?ft-4w) 8 91. (a?- 92. (a 2 + 2/) 3 . 93. (a 2 + 66 3 ) 3 . 94. (2a s -6) s . 46. Miscellaneous Exercise in Applying the Theorems of Multi- plication. As a valuable exercise let the student point to the figures, let- ters, and signs in the formulas, while he says the corresponding words of the several Theorems I-V. 1. (ra + ?i) 2 . 2. (o + 7)(a-9). 3. ( + ra)(a--w). 4. (2a--6) 2 . 5. (2a-36)(2a + 36). 6. (c-7 y)(c+3y). 1. (a 2 -ft 2 ) 2 . 8. (ax + by)(ax-by). 9. (3 xy - 8)(3 xy - 7). 10. (16-aj) 2 . 11. (m 3 -3n 2 ) 2 . 12. (2 a*x + 3 % 3 ) 2 . 13. (c-/0*. 14- 15. (4a + 7x 2 ) 2 . 16. 17. (6jj -5 ^(6^ + 5g). 18. (*.B- 19. (4m 2 -3?i 8 ) 2 . 20. (a 2 -7) (a 2 + 12). 21. (a 8 -7c)(a s + llc). 22. (3 a 2 -I) 2 . 23. (3^ + 2/ 2 )(-^/ + r)- 24. (,.y-a)(^r + 25. (^r 2 -is 3 ) 2 . 26. (2 a 4 - 11) (2 4 + 15). 27. 3a^-5 2 3x 2 -f 7?/ 2 . 28. 72 THEOREMS IN MULTIPLICATION AND DIVISION 29. Find the square of 7J by Theorem I, writing (7) 2 =(7+i) 2 . Check by squaring - 2 ^. 30. (5l) 2 ; (41) 2 ; (201) 2 ; 41 2 = (40 + I) 2 ; 52 2 ; 79 2 =(80-1) 2 ; 99 2 . Cheek the answers by ordinary multiplication. 31. Find the product of 42 and 38, that is of 40 + 2 and 40-2. SOLUTION. (40 + 2) (40 - 2) = 1600 -- 4 = 1596. Check by actual mul- tiplication. 32. 48 x 52. 33. 81 x 79. 34. 93 x 87. 35. 95 x 105. 36. 88JX91J. 37. 257x263. 38. 546x554. 39. 125x135. 40. (ar 5 -l)(ar 5 +l). 41. (a*-7b) 2 . 42. (a-b) 3 . 43. (5|-) 3 . 44. 999 2 . 45. (6.5) 2 . 46. (101) 2 . 47. 242x238. 47. Theorems of Division. 1. Since (A + B)(A-B) = A 2 -B* by 45, Th. Ill, it fol- lows that and Stated as a theorem, these formulas give THEOREM VI. The difference of the squares of two quantities is exactly divisible by either their sum or their difference. Or, The difference of the squares of two quantities equals the product of the sum and difference of the quantities. 2. (A 3 -B S ) + A-B)=? 3. SOLUTION SOLUTION A* - A 9 - A-B A* + AB + B* A* + A-B A + B A* -AB + B 2 A 2 B - A-B A 2 B - AB 2 - A*B - AB 2 AB 2 -- B* + AB- + AB 2 - B s + AB 2 + Thus, (^l 3 -&)-*- (A - B)=A 2 -\-AB+B 2 ; THEOREMS IN DIVISION 73 Stated as theorems, these formulas give THEOREM VII. The difference of the cubes of two quantities di- vided by the difference of the quantities equals the square of the first, pi as the product of the first by the second, plus the square of the second. Or, The difference of two cubes equals the difference of the quan- tities times the sum of the square of the first, the product of the first and second, and the square of the second. THEOREM VIII. The sum of the cubes of two quantities divi-y\ 10. a 3 + (2c) 3 . 11. x^-(3y)\ 12. (2m) 2 - (on) 2 . 13. 8 or 5 - 27 f. SUGGESTION. 8 x 3 - (2 x) 3 , and 27 y* = (3 ?/) 3 . Hence 8 r, 3 27 y* is divis- ible by 2 x - 3 y. The quotient is (2 x) 2 + (2 x) (3 y) + (3 y) 2 , or 4 x 2 + 6 xy + 9 y 2 . Prove by multiplication. 74 FACTORING 14. 8^-125. 15. 8 3 + 125c 3 . 16. 17. a 3 + 125. 18. -216 -if. 19. 9x 2 y 2 -25z 2 . 20. 64 + /. 21. 64 a 3 -343. 22. 64 or 3 -!. 23. 64 a 6 + 125 ?/ 9 . SUGGESTION. The divisor is 4 2 +5 y s. 24. z 3 _64?/ 6 . 25. x + 27y 3 . 26. 27m 9 -64^. 27. 216j9 12 -9 6 . 28. 8z 12 + l. 29. 343 z 3 - 1000 y> 30. A 2 + B 2 + A + B. 31. Show by long division that divisors in Ex. 30, 31 are not exactly contained in the dividends. Make theorems therefrom. 32. (A 3 -B*) + (A + B). 33. (^l 3 + B s ) -=- (A - B). Divide and make theorems as in the preceding exercises. FACTORING 48. Factoring in algebra is separating a quantity into others which multiplied together will produce the given quantity. The same definitions are used in algebra as were used in arithmetic. Thus , SL prime quantity is exactly divisible only by itself and one. We will take up in the following articles only a part of the various cases in factoring; others will be given later on. 49. Factoring of Monomials. A monomial, such as 5 3 6 2 c, is re- garded as being already factored. A monomial, such as 72 xy 1 , is factored when its coefficient 72 is factored. Thus 72 xy' 2 =2 3 3 2 xy 2 . 50. Monomial Factors in Polynomials. A monomial is a factor of a polynomial when it is contained in every term of the polynomial. Type forms : ax + bx == (a + b)x ; 3 mx + 3my 3mz = 3 m(x +y z). MONOMIAL FACTORS 75 1. Factor 8 a 3 2 ax and prove answer by multiplication. DIVISION SOLUTION. 2 g)8 a 9 2 ax PROOF. 4 - - x 4 a 2 x 2 a DIRECT SOLUTION. 8 a 3 2 ax = 2 a (4 a 2 #) 8 a ! - 2 aaj Factor the following and prove the answers correct by multi- plying the factors together: 2. 4 a 2 2 a. 3. 25 m 2 w - - 5 w. 4. or 3 x 2 y. 5. a--atf. 6. 3 or + or 5 . 7. 15a 2 -75ee 4 6 2 . 8. 18 -81 a. 9. 3 a 3 -3 a. 10. 9x-y 2 -9xy 2 . 11. 3 a 4 - 3 a 3 6 + 6 a 2 b 2 . DIVISION SOLUTION. 3 a 2 )3q-3q 8 6 + a-6 2 PROOF, a 2 - ab + 2 6 2 a 2 - a& + 2 Z> 2 3 a 2 3 a 4 - 3 a 3 6 + 6 a 2 6 2 DIRECT SOLUTION. 3 a* - 3 a*b + 6 a 2 6 2 = 3 a' 2 ( 2 - a^> + 2 & 2 ) . 12. 35 a - 21 a 3 + 14 a 4 . 13. 3 m 5 - 12 m s n 2 + 6 mn 4 . 14. 2ar 3 + 6ari/ + 2a 2 . 15. 4 ./ 3 - 4 ary 2 + x 3 ?/. 16. 8z 4 -4x 2 -4. 17. a 3 - 6 orb 4- 10 aft 2 . 18. Make a rule for factoring a polynomial that has a mono- mial factor in it. How is the monomial factor found ? Should it contain every factor which is contained in each of the terms of the given polynomial ? How is the polynomial factor found ? How is the answer proved correct ? 19. 63Z 3 54#. 20. wV + w 3 w 4 --w?i 6 . 21. 12 ax 2 - 3 bx~ + x 2 . 22. tfy* - x*y* + x*y\ 23. 14 be 4 - 21 b 2 c s + 7 b 3 c 2 x. 24. a 6 - 5 a 5 - 2 4 + 3a d . 25. 3a 3 -9a 2 6 + 9a& 2 -36 3 . 26. 24 Z> 4 - 36 6 3 + 185 27. x 12 + x 11 x 10 x 9 . 28. ac be cy ~ abc. 29. SOLUTION. 6 _+ c| q(6 + c)- 2 m(5 + c) a 2w 76 FACTORING 30. 5a(2x-y) + 3b(2x-y). 31. 3 x*(x - 1) - 5(x - 1). a. Throughout factoring the first step to be taken is to remove monomial factors from given expressions if any are present. If the student does not see how to begin the solution of an exercise, the reason will very often be that the problem is disguised by the presence of a monomial factor. BINOMIALS 51. The Factoring of Binomials. Four different cases or classes of problems are taken up in this chapter. I. Binomials that are the Difference of Two Squares. Type form: a 2 -b 2 = (a + b)(a -b). By 47, Th. VI, the difference of the squares of two quantities equals the sum of the quantities multiplied by the difference of the quantities. PROOF 1. Factor 4 a 2 b 2 and prove by actual multi- 2 a + b plication. 2a b 4 a 2 + 2 ab SOLUTION. 4 a 2 - 6 2 = (2 a) 2 - 6 2 = (2 a + 6) (2 a - 6) . - 2 ab - b 2 4 a' 2 ~^b* Factor and prove the following : 2. 9a 2 -6 2 . 3. or 9 -4. 4. x 2 -16y 2 . 5. 25 -c 2 . 6. a 2 x 2 y 2 . 7. m 2 n 2 . 8. a 2 9. 9. m 2 1. 10. 9a 4 -166 2 . Ans. (3a 2 + 4ft)(3a 2 -4 6). Factor the following, proving the first three or four (and any others the student does not feel sure are correct), by actual multiplication : 11. 4m 2 -25n 2 . 12. 16m 2 -49. 13. 9# 4 -49ij/ 4 . 14. a 2 x~-35y\ 15. l-49x 6 . 16. 36 a 4 -!. 17. 36 a 4 -25. 18. a 4 -49?/ 4 , 19. 121 a 2 - 36 b\ FACTORING OF BINOMIALS 77 20. a 2 6 2 -4c 6 . 21. 49- 100 if. 22. 64a 4 6 4 -c 4 . 23. a; 4 -144. 24. 9a?-{ a*. 25. 81 a 6 - ^ ?/ 4 . 26. 5 a b y 5 a?/ 3 . SOLUTION. 5 a 5 y 5 ay 3 = 5 a?/(a* ?/ 2 ) = 5 ay( 2 -I- y)(a 2 - |/). ^Iws. ( 50, a.) 27. 3 a 5 -3 ay 2 . 28. 4a 2 c-9c 5 . 29. 20a 3 6 3 -5a&. 30. 5x 6 5. 31. Sar 5 Sa^ 2 . 32. 6m 4 ?/ -6m 2 ?/. 33. 8z 10 -8/. 34. 15a 5 -15a& 4 . 35. 32 ax 4 - 32 a?/ 10 . 36. Make a rule for factoring a binomial that is the difference of two squares. (1) If a monomial factor appears, what is done? (2) The difference of the squares of two quantities equals ... 37. a 2 m 2 -m 2 n*. 38. aW-a&c. 39. 2mV-8mV. 40. 3a 5 -12a 3 c 2 . 41. 4a 2 c-9c 3 . 42. 16 a 17 -9 II. Binomials that are the Difference of Two Cubes. Type form : a 3 - 6 3 = (a - b)(a 2 + ab + 6 2 ). By 47, Th. VII, the difference of the cubes of two quantities equals the difference of the quantities times the sum of the square of the first, the product of the first and second, and the square of the second. 43. w s -n s . 44. p s -q*. 45. (2a) 3 -6 3 . 46. 8 a 3 -27 ft 3 . SOLUTION. 8 a 3 - 27 6 3 = (2 a) 3 - (3 6) 3 = (2 a - 3 6) [(2 a) 2 + (2 a) (3 5) + (3 6) 2 ]. Thus, 8 a 3 - 27 6 3 = (2 a - 3 6) (4 a 2 + 6 a& + 9 6 2 ). PROOF. 4 a 2 + 6 a& + 9 6 2 2a -36 18 -12a 2 6 -- 18a6 2 -276 3 -276 3 78 FACTORING Factor the following and prove by actual multiplication : 47. 27m 3 -64?i 3 . 48. 125 a 3 -Si/ 3 . 49. 27crV-8. 50. 2/ 6 -64. 51. 8/-2V 52. y 9 -8. 53. 216a-6 3 . 54. 343 - or*. 55. xh/ 3 -512z 3 . 56. a 5 -27 a 2 . (50, a.) 57. 250-2a 3 . 58. 4 a 4 -4 a. 59. Make a rule for factoring the difference of two cubes. How is the first factor found ? How the second ? How is the work proved correct ? 60. m 6 125. 61. m*n mn*. 62. x l5 --6x Q . 63. 64/-729z 3 . 64. aW-729. 65. 512w w -l. III. Binomials that are the Sum of Two Cubes. Type form : a 3 + A 3 = ( a + A) (a 2 -- a6 + A 2 ). By 47, Th. VIII, the sum of the cubes of two quantities equals the sum of the quantities times the square of the first, less the product of the two, plus the square of the second. Factor the following and prove by actual multiplication : 66. x 3 + /. 67. Stf + f. 68. m 3 + 27 n*. 69. 125 + a 3 . 70. aV + 1. 71. 128 + 2 /. 72. 54 a 4 + 250 a. 73. 3m 5 + 375m 2 . 74. 125 x 6 + 27 y 15 . 75. Make a rule for factoring the sum of two cubes. What is the first factor ? How is the second found ? How is the work proved correct ? 76. 64 a 9 + 125 6 6 . 77. 27m 3 + 512w 3 . 78. 729 x 12 + 64 ?/ 15 . 79. 8 x 18 + 27 y 12 . 80. 8 3 ?/ 6 s 9 + l. 81. IV. Binomials One or More of Whose Factors can be Factored. 82. Separate a 6 b 6 into its prime factors. SOLUTION, a 6 -- 6 6 =(a* + & 3 )( 3 -- ft 3 ). (By I p. 76.) = (a + 6) (a 2 - ab + 6 2 )(a - 6) (a 2 + a& 4- & 2 ). (By II and III.) FACTORING OF TRINOMIALS 79 83. m* n*. 84. x G --l. 85. 64 a 6 -/. 86. a? 4 -81. 87. 16 a 4 -!. 88. x-y*. 89. 16 a 4 -- 81 y\ 90. a?y 625. 91. a 6 ?/ G -z 6 . TRINOMIALS 52. The Factoring of Trinomials. L Trinomials that are the Product of Two Binomials having a Common Term. Type form: jr+(a + 6) JT + ab = (x+a} (x + 6). PROOF 1. Factor ic 2 + 12 # + 32 and prove the answer. x + 4 x + 8 SOLUTION. Two numbers must be found whose sum is 12 ^ _^ 4a . and product 32. (See 45, Th, IV.) These numbers are + 8x + 32 4 and 8. Thus x 2 + 12 x + 32 = (x + 4) (x + 8) . ^Iws. x - 2 + 12 x + 32 Factor the following and prove by multiplication : 2. a^+6 +8. 3. ^ + 80; + 15. 1. 2 +16a+48. 5. 6. ^ 2 + 8ic + 16. 7. 8. m 2 + 10 mn + 25 % 2 . 9. or?/ 2 + 22 a^ + 121 z 2 . 10. 0^ + 26^ + 1690;. (50, a.) 11. a^ + a + J. -^2 x 2 10 a/* + 21 SUGGESTION. Sum = 10 ; product + 21. Ans. (x 3)(x 7). Prove by multiplication 13. a 2 -11 a; + 18. 14. aV - 21 a.r; + 80. 15. or 9 -25^ + 1502/ 2 . - 16. m 2 - 28m + 196. 17. aj" - 13 ay + 12 /. 18. aV - 21 aa? + 90. 19. a 3 - 6 a 2 6 + 9 a6 2 . 20. 5a 4 -10a 2 6 + 56 2 21. a^> - 12 xW + 27 .^6 3 . 22. z 4 - 26 x 2 y + 48 f, 23. o^ + iC 30. SUGGESTION. 6 +( 6)= 1 ; 6 x - 5 = - 30. 80 FACTORING 24. x* + x 6. 25. x 2 z-6. 26. a 2 + 3z-28. 27. o^-Sa^ 28. x 2 + 7x-44. 29. m 2 + 6m--16. 30. z 2 + 5z?/-36?/ 2 . 31. a 2 + 13 a -48. 32. 2a,- 3 -22x 2 -120a. 33. 3 x 8 - 51 aty + 48 34. 3x 6 39z 3 + 66. 35. 5 ar> + 30 a 2 - 35 36. Make a rule for factoring trinomials into binomials having a common term. (1) What should one look for first ? (See 50, a.) (2) How is the common term found ? Ans. By extracting the square root of a term in the given trinomial. (3) What two conditions must the other terms fulfill ? (4) How is the answer proved correct ? 37. x* - 11 mx + 30m 2 . 38. x*z* + 12 xz 13. 39. 2z 8 -4o: 4 -48. 40. 35 + 2^-a; 2 . 41. z 8 - 16 a 4 + 55. 42. mV - 20 wra + 99. 43. 2xy 6 -10tfy s -28o?. 44. a?x* -7 45. a 2 -ay-210y 2 . 46. a 2 2 - 38 aa + 361. 47. m 2 + |m--J. 48. 49. c 2 d 2 -|cde-- 3 /e 2 . 50. or - 23 xy + 132 /. II. Trinomial Squares. Type forms: 51. SOLUTION. Vx^ = x ; V<54 = 8. Ans. (x + 8) 2 . CHECK, (x + 8) 2 = x 2 + 16 x + 64 by Th. I, 45. FACTORING OF TRINOMIALS 81 52. x* + Wxy + 25y*. 53. 9 z 4 + 24 z?y 2 + 16 y 4 . 54. 36 or 2 + 60^ + 25. 55. 64 x 2 - 48 xy + 9 y 2 . 56. 0^ + 40? + 4. 57. (rlr + 2 a&cd + c 2 d 2 . 58. 9/-30/ + 25. 59. 36 a 4 - 84 ofy + 49 /. 60. x 2 -x + i. 61. A/-- 1 ^ 2/4-25. 62. 5a 4 -10a 2 & + 5& 2 . 63. 24 to 3 - 72 bW + 54 & 3 a. 64. Make a rule for factoring trinomial squares. (1) What is to be looked for first? (See 50, a.) (2) How is the first part of the answer found ? The last part? (3) How is the result checked ? (See 45, Ths. I and II.) 65. 12 x 2 --36 a? + 27. 66. a&c 10aW+25aW. 67. 144ar + 600a^ + 625?/ 2 . 68. 49 a 2 + 126 vy 2 + 81 if. 69. 49 n 16 + 84 m s n 8 + 36 m 8 . 70. 25 a w & 10 - 20 a 5 6VcZ 2 + 4 c*d*. 71. 18 a 4 + 60 o?b + 50 a 2 6 2 . 72. 121 a 2 6 2 + 176a6 3 + 64 6 4 . III. Trinomials that are the Product of Any Two Binomials. Type form : ax 2 -f 6jr/ + c/ 2 = (mjr + ny)(px -f y/) = mpx 1 4- (my + /?/?)jr/ + fl?/ 9 . 73. Factor 6 x* - 25 .T?/ + 4 ?/ 2 . SOLUTION. Such problems are usually solved by trying sets of factors until a pair is found which multiplied produce the given trinomial. We will call this the "trial method." It is easy to select two binomial factors for trial multiplication which will give the first and last terms of the given trinomial. Thus, 2x y 2x 4y Qx y 3 x 4y Bx y x -4y 2 - Bxy - 8xy + 4y 2 - 2xy + 4y 2 - 24 xy + 4 6 x 2 - 1 1 xy + 4 */2 6 x 2 - 14 xy + 4 y* 6 x 2 - 25 zy + 4 f 82 FACTORING Each set gives 6 x 2 and 4 ?/ 2 , but only the last pair of factors gives the re- quired middle term, 25 xy. Hence, 6 x 2 - 25 xy + 4 y 2 - (6 x y) (x - 4 y). To factor by this method try different pairs of factors until a combination of signs and coefficients is found which produces the given quantity. The middle terms are called " cross products " ; the others, end products, 74. 9ar + 9z?/ + 22/. 75. 2x? 76. 2x- + 5xy + 2y 2 . 77. 3 x 2 + 13 xy + 12 78. 3 or- 13 x + 14. 79. 6 a? -31 a; + 35. 80. 8x?-3Sxy + 35y 2 . 81. 4m 82. 3x- + 7x-6. 83. 3 m 2 - 19 mn - 14 w 2 . a. If a given trinomial does not contain a monomial factor, evidently wo factor of it can contain that quantity, and there is no use trying factors which contain a monomial divisor. Thus, in solving Ex. 73, it was a waste of time to try the second set of factors, since the first of these, 2 x 4 y, contained the factor 2, while the given quantity did not. 84. 6z 19x + 15. 85. 86. 12x 2 + 7x-l2. 87. *6. In the type form mp 2 + (mq + np)xy + ftQ'?/ 2 given above, observe that mq + np is the swm of two quantities whose product is mp x wg. We look first for two numbers whose sum is the given middle coefficient and whose product is the product of the given first and last coefficients. Thus, in factoring 4 x 2 -- 73 x + 18, (- 72) + (- 1) = - 73, and -- 72 x - 1 = 4 x 18. We choose our coefficients now so that one cross product will be 72, and the other 1. Thus, 4x - 1 * Seel, p. 6, x -18 4z 2 - x -72 3: + 18 4x 2 -73x+ 18 FACTORING OF QUADRINOMIALS 83 88. 3x?-5x~2. 89. 4 ar 8 - o;y 3 # 2 . 90. 4ar J + 16a? + 15. 91. 24a 2 - 29 a*/- 92. 8 a? 4 --19 or 8 15. 93. 40 + 2x-2ar } . 94. 36 it- 4 - 18 x 2 - 10. 95. -10a 4 + 7 96. 40 + 6 a - 27 a?. 97. 16 a 5 + 4 ofy 2 - 30 98. Make a rule for factoring trinomials that are the product of any two binomials. (1) How are the coefficients of the first terms of each binomial factor chosen ? (See Ex. 73, p. 81.) Those of the last terms ? (2) When a set of coefficients and signs are chosen, what is done ? How can selection of coefficients be guided ? (See a, 6, p. 82.) 99. 12 x 2 - 31 x -15. 100. 24 + 37 a; -72 a 2 . 101. 20-9c-20c 2 . 102. 24x* 35xy + 4:y*. 103. 8 m 4 + 38 mV -f 35 n*. 104. 132a 2 + a-l. 105. 12^ + 50^-50. 106. 3 a 6 + 41 a 3 + 26. 107. 15 a? 8 + 224 a; -15. 108. loa 8 - 77 a 4 6+106*. QUADRINOMIALS 53. The Factoring of Quadrinomials. (Polynomials of four terms.) I. Quadrinomials that are the Cube of Binomials. Type form : a 3 - 3 a 2 6 + 3 ab 2 - 6 3 = (a - 6) 3 . 1 . 8 a 8 - 60 afy + 150 xy 2 - 125 y\ SOLUTION. We extract the cAtbe root of the first and last terms, getting 2 x and 5 y. Then (2 x 5 y) 3 is the answer. PROOF. (2x - 5 y)3 = (2 x) 3 - 3(2 z)2(5 y)+ 3(2 x)(5 y)2 -(5 y) 3 . ( 45, Th. V.) = 8 x 3 - 60 x 2 y + 150 z?/ 2 - 125 y s . ( 27.) 2. 646 3 + 486 2 + 126 + l. 3. 216 x 3 - 108 x*y + 18 z/ - 4. 8 a 3 - 36 a*& + 54 a& 2 - 27 6 3 . 5. 27 m 9 + 108 mV + 144 m 8 w 4 + 64 n*. 84 FACTORING II. Quadrinomials that are the Product of Two Binomials. Type form : ac -f- ad + be + bd = (a + b) (c + z 2 -12z 2 . 29. FACTORING OF QUADRINOMIALS 85 III. Qiiadriuoniials, the Difference of Two Squares. Type forms : (1) x- + 2 xy +/ - z 2 = (x +/) 2 - z 2 = [(*+/) + <][(*+/)-*] (% 51, I.) = [a+(6-c)][fl-(6-c)]. (By 51, I.) 30. Factor # 2 6 ax 16 b 2 + 9 a 2 and prove result correct. 1st step (x 2 6 ax + 9 a 2 ) 16 6' 3 . (On putting three terms which consti- tute a square in parenthesis. ) 2d step (x 3 a) 2 (4 6) 2 . (On changing to form of squares, 52, II.) 3d step [ (x 3 a) + 4 6] [ (x 3 a) - 4 6] . (On factoring, by 51 , I. ) 4th step [x 3 a + 4 6] [x 3 a 4 6]. (On removing parentheses, 32.) 5th step. PROOF, x 3 a -f 4 & a- 3a -46 # 2 3 ax + 4 &x - 3 ax + 9 a 2 - 12 a& a 2 - 6 ax + 9 a 2 - 16 Solve the following as the model and prove : 31. a a -4a6 + 46 2 -9c 2 . 32. ar- 33. a 2 + ?/ 2 + 2a,'?/-4ary 5 . 34. a 2 35. 9 2 -a 2 -6a +1. 36. 4ar J 37. a 2 - 36 6V 9 + 126a-l. SUGGESTION Finish solution as in Ex. 30. 38. a 2 + 12 ex 9 y? 4 c 2 . 39. 4c 2 a 4 9& 2 6a 2 &. 40. 9a 2 -a 2 -4a&-4& 2 . 41. 36 & 2 -4 + 20 a -25 a 2 . 42. Make a rule for factoring quadrinomials that are the differ- ence of two squares. Describe five steps. 86 FACTORING 43. 25# 2 -4a 2 + 162/ 2 +40a2/. 44. 49# 2 + 4?/ 2 - 25z 4 -28z?/. 45. (x 5 ?/) 2 4 6 2 . SUGGESTION. Only three steps remain here. 46. (3o;-22/) 2 -36m 4 . 47. 9 x 2 - (4 y - 1 z) 2 . OUTLINE FOR FACTORING 54. General Methods in Factoring. Leaving that case in factor- ing in which, monomial factors are removed from polynomials ( 50) out of consideration, we can say that all the problems of this chapter can be factored in one or other of seven ways : 1st. If they are binomials, by one or other of the three Theo- rems III, VII, VIII ( 45, 47). 2d. If they are. trinomials, by the trial method. 3d. If they are quadrinomials, by Theorem V, or by arranging their terms as follows : either (a) By putting two terms in one parenthesis and the remaining two terms in another parenthesis, and then removing monomials ; or, (5) By putting three terms in a parenthesis to form the square of a binomial. a. More difficult cases in factoring are given in a later chapter, and tri- nomials will be found there which cannot be factored by the " trial " method, as also binomials, that cannot be solved directly by the theorems mentioned above. It should be noted here that while such a trinomial as 9 a 2 + 30 ab + 25 b 2 can be solved by the trial method, it can be solved more easily by observ- ing that it is the square of a binomial. 55. Miscellaneous Exercise in Factoring. The student should carefully memorize the methods of solution for the different kinds of problems, especially the seven cases of the preceding article. In case of trouble, note whether the given quantity is a binomial, a trinomial, or a quadrinomial. Then study to find which case, in the article in which it falls, it comes under. Always look first for a monomial factor, and if any is found put it before or after a parenthesis containing the other factor. MISCELLANEOUS EXERCISE IN FACTORING 87 Separate the following quantities into their prime factors, and by multiplying the factors together, unless the factors are actually multiplied together in the solution itself : 1. x- + llxy + 30y' 2 . 2. a 2 + 16 a; + 63. 3. a 2 -25. 4. 36^-60^ + 25. 5. x 7 + x s y + x*. 6. 4 x 2 + 4 xy + y\ 7. 36 ^-9/. 8. x 2 22 a; + 121. 9. a 4 --ra 4 . 10. 11. crb 4 a 4 b 2 . 12. 13. a 3 + 27. 14. 15. 16 -a 4 . 16. 17. 125 a 3 + 27. 18. 19. x^ Sy 3 . 20. 21. a; 2 -a -2. 22. 23. 125 /- 27 a*/ 3 . 24. 6 + 15 a 2 -19 a. 25. ar 3 -4^-45aj. 26. 169 a 2 + 169 b~ - 338 ab. 27. x 4 - 16 or 2 + 55. 28. 29. mV + 1. ( 47, VIII.) 30. 31. a 6 6 6 . 32. 33. 2m 2 /i--4mw + 2n. 34. 35. x w - 26 a 5 + 168. 36. 37. 27 a 3 + 64 6 3 . 38. aV 5 a 3 # - - 14. 39. 1 x lG . 40. 41. ar 5 x. 42. 43. 0^-49 x + 48. 44. 5 x> + 30 ofy 2 - 35 45. # 2 ax + <./ ac. 46. 47. 250 a;- 2 x 7 . 48. 88 FACTORING 49. 40a 2 +61a6-846 2 . 50. 56 aa?y + 96 axy + 51. a 8 -1692/ 10 . 52. 3 a 8 -51 a 4 + 48. 53. 5 ax 9 5 a. 54. 2 x 4 2 x 3 2 a-x 2 + 55. 110-x-x 2 . 56. 24^ + 5 xy-36y 2 . 57. a + 343 a 4 . 58. x 5 x 4 -- 16 x + 16. 59. 21 a 2 17 a 30. 60. ax 2 - - ex + ax c. 63. 4a 2 + 42/(j/ 2x). 64. 64 x 6 729 ?/ 6 . 65. 8^ 3 +27P 3 . 66. a 3 --a* + GM 2 - 67. c 2 -2c-399. 68. 243 6 5 c 6 - 75 6 7 . 69. x l5 y }5 . 70. 71. M 6 -N 6 . *72. 22 a 2 - 110 f - i ^J C( " I "/j~ \AJ ~^p* I * ^v ^ C/ ^^^ tJ vv C/ I Cv ^"^^ O O * 75. a 9 6 9 . 76. acx 2 + adx -{- box + 77. P 2 -P-72. 78. 49s 4 + 70s + 25 79. a 8 -^-ft 3 . 80. 81. 121 ^ 12 - 324 7i 18 . 82. a; 2 - a + y2 - a;^. 83. a; 6 + 64 & 6 . 84. 1 a 2 + 2 a& b 2 . OO W W />" "" *v ~" J- OO (l/ \jJU ""l " tl' CvA/ " ~" L/C'*// " ~ C/C*e 87. x 12 + i/ 12 . 88. a 2 - QO r'"ir^ /v^ojiO Or^ '>*-' Ot7. J- *~~ ^tJ\J *Hs U * *J\J* *O Q1 ^/'T /Tf"'V* I ft nw" OO r /v?*6 *-14a;-8 = 0. 14. IS a- 2 - 3 a = 36. 15. 49^-35^=66. 16. 40-17z-5r J = 0. 17. Make a rule for solving quadratic equations like the pre- ceding by factoring. (1) What must the second member be ? (2) What is done with the left member ? (3) What is the next step? (4) How many answers are there to prove ? 18. 3x* + 5x = 2. 19. 20. 30* + 10 a; = 57. 21. 22. 4ar = 3z + 27. 23. 24. 6 a; 8 -96 = 0. 25. 3x' 2 -12 = B 2 26. 5a 2 + 19 = 12a 2 -9. 27. 7x 2 - 13 = 2o; 2 -8. 28. Find a number such that its square is equal to 72 more than the number itself. 29. Find a number such that if 17 times the number is di- minished by its square, the remainder will be 70. 30. Find a number such that if you subtract it from 10 and multiply the remainder by the number itself, the product will be 21. 31. What two numbers are those whose difference is 2 and the sum of whose squares is 130 ? 32. Find three consecutive numbers such that the sum of their squares equals 110. SUGGESTION. Let x = first no. ; x + 1 = second ; x + 2 = third. 33. The average height of men who weigh 120 Ib. is a certain number of feet. Three times the square of this number increased by 9 times this number equals 120. What is the average height of 120 Ib. men ? b. In such problems as this, the negative answer has no meaning and is disregarded. 92 LOWEST COMMON MULTIPLE 33 23X 34. The perimeter of a room, that is, the distance around it, is 30 ft., and its area is 54 sq. ft. Find its width. SUGGESTION. Let x = the number of feet in the width. Then 15 x = the X length, and (15 x)x = 54. Why ? 35. A field is 35 rods long and 23 rods wide. It is desired to border it all around with a strip of the same width to contain 11 acres. How many rods wide shall this strip be ? 36. A farmer cuts grain in a field 100 rods bv 80 rods. How tt many rods wide must the strip around the field inside the fence be to contain 4.4 acres ? 37. The number of different calls possible in a small telephone exchange having x subscribers is x(x 1). If the number of calls is 110, how many subscribers are there ? 38. What is the radius r of the base of a cone whose slant height is 11 ft. and total area 660 sq. ft. if it can be found by the formula ?- 2 + 11 r = ^ X 660 ? LOWEST COMMON MULTIPLE 58. Lowest Common Multiple of Quantities. 1. A multiple of a quantity is the quantity multiplied by some integral quantity. Hence a multiple of a quantity is divisible by it. Thus, 6 a 3 is a multiple of 2 a. 2. A common multiple of two or more integral quantities is an integral quantity that is exactly divisible by each of them. The word " common 5: signifies "belonging to each or many." A common multiple of two or more quantities belongs to each in the sense that it contains each quantity. Thus, 36 4 & 3 c 2 is a common multiple of 3 a 2 b, 9 abc, and 2 abr, LOWEST COMMON MULTIPLE 93 3. The lowest common multiple of two or more integral quanti- ties is that quantity of lowest degree which is exactly divisible by each of the given quantities ; its coefficient (Arabic notation) is the least number that will contain all the coefficients of the different quantities. Thus, IScrfrc 2 is lowest common multiple of Sorb, 9 abc, 2 abc 2 . 4. Find the lowest common multiple of a 2 b, 3am, 9m 2 . Ans. 9 arbm 2 , PROOF. 9 a 2 &m 2 -s- a-b = 9 ra 2 ; 9 2 6m 2 -4- 3 am = 3 abm ; 9 2 &m 2 -=- 9 m 2 = a?b. Moreover, if any quantity with a smaller coefficient or with lower ex- ponents of the letters had been taken it would not have contained all of the given quantities. Find the lowest common multiple in the following exercises and prove the answers correct : 5. 3 a-b, 2 ab 2 , 4 a 3 6c. Ans. 12 aVc. 6. 4a, 2a 2 b 2 , 12 a 3 . 7. 6x>y,Sy*z. 8. 12 a 2 x 2 , 9 a 2 /, 4. 9. 2 a, 3 b, c. 10. 8 a V, 30 aV, 4 aV. 11. 3a?b,2ab*. 12. 14 a 3 , 21 a 2 , 5 6, 7 a. 13. 7 a s m 2 , 21 wV. 14. 42 a 3 &* and 60 c + a& 2 , arb-ab*, 3a 2 -36 2 33. ax%T- ?/) 2 , bxy(x 2 - y*\ 34. a; 2 3 a; -54, or-lOa + 9. 35. o? y 2 ,a? 8xy + 7y 2 . 36. a 2 - 6 2 , a 3 - & 3 , a 3 + 6 s . 37. a 2 - f, (x - ?/) 2 , a 3 - /. 38. a? 1, or 2 -!, a? 2, 2 -4. 39. a; 2 ?/ 2 - 4, a 2 - 8 xy + 12. 40. 2^ + ^-1,4^-1,2^4-3^ + 1. 41. a; a, a 2 a? 2 , ic 4 a 4 . Ans. x* a 4 . a. Since the lowest common multiple deals solely with the question )1 divisibility, in getting the 1. c. m. any given quantity can have its sign changed by multiplying it by 1. In the example just given, a 2 oc 2 can be changed to x 2 a 2 . The 1. c. m. x 4 a 4 contains a 2 x 2 without remainder 42. a 3 x s , a 2 2 , x a. 43. 1-205,4^-1,4^ + 1. 44. a? + x 12, -36 + 13 07-a^, a 2 -16. 45. 6 2 -a 2 , 46 + 4a, a 3 - 3 a 2 6 + 3a6 2 - 6 s . 46. 1 - - x, x 2 1, a 2, 4 or*. *47. 4(1 - a) 2 , 8(1 - a), 8(1 + ), 4(0* - 1). LOWEST COMMON MULTIPLE 95 48. 3ar'-5z + 2, 4or J -4ar J -a;+l. (53,11.) 49. ar 1 - ar - 9 x + 9, or + 2 x - 3. 50. a,- 2 - 4 a 2 , ar 5 + 2 ax 2 + 4 a 2 x + 8 a 3 . 51. x 4 - - y 2 , or 3 + x s y -f- xy + ?/ 2 . 52. 2x*-xy-15y*, tf-Gxy + Qy*. 53. or 5 + /, a; + y, xy s? - - if. 54. (x - l)(a? + 3) 2 , (a; + l) 2 (ar - 3), or - 1. 55. (a? - 1) 3 , 7 aj/Caj 2 - I) 2 , 14 a*y(x + I) 3 . Solve the following problems mentally, giving the answer in the factored form : 56. 3 a 2 6, 4 ac 2 ,. 6 & 2 c. 57. 7 x 2 , 2 aj 2 - 6 x. 58. + &, a- + 2 ab -f 6 2 . 59. 6 a6 2 cd 5 , 9 a W, 15 a6 2 c 3 . 60. a 2 + a6, 6 - 6 2 , a 2 -Z> 2 . 61. a; 2 -4a; + 3, a,- 2 -5 a; + 6. 62. x* 4- 8 a; + 15, x 2 + 4 x 5. 63. a? 5 x 84, a; 2 7 a; 60. 64. 6 db(a + b) 2 , 4 cr(a 2 - 6 2 ). 65. a 2 - 9 a? -10, a; 2 - 7 a^-30. CHAPTER III FRACTIONS. HIGHEST COMMON FACTOR 59. A fraction in algebra is an expressed division. The divi- dend, or numerator, is written above a line and the divisor, or denominator, below it. The numerator and denominator, as in arithmetic, are called the terms of the fraction. The operations with fractions are much like those in arithmetic. 60. Reduction of Fractions to their Lowest Terms. This opera- tion depends on the well-known principle explained in arith- metic, that both terms of a fraction may be multiplied or divided by the same number without altering its value. To reduce a fraction to its lowest terms, factor both numerator and denominator into their prime factors and cancel all factors common to both. Evidently canceling common factors is equiva- lent to dividing both numerator and denominator by the factors canceled. Type form : 1. be b 15aW t 25 ab 4 c ' / 25 SOLUTION. ZaWc Y- -^ = ^^-. Ans. 25 a6 4 c 5 6 CHECK. Let a = 2, b = 2, e = 1. Then given fraction equals f , and answer equals f . . Checking here does not show that an answer is in its lowest terms. It merely shows that the answer has the same numerical value as the given quantity 96 REDUCTION OF FRACTIONS TO THEIR LOWEST TERMS 97 Reduce the following to their lowest terms and check several problems : 108 a s b 2 30 oV SaV 110 mx 2 2< 111 m' 3 ' o,. J ' 4 ' "TTJ' 5 ' K~n "2* 144 acr ^ 06 an/ a c or ooO ??I-OT 45 a W 17a 7 6 5 36 a; 2 y 4 V" 4 * 5 ' 195 aV' ' 19a r 6 6 ' ' 60 x*y 2 ' ' arx 7 - A y? 7 icy + 10 y 2 (x^2rjfi (x - 5 ) x 5 10. -/ SOLUTION. -^-^ = - ar 8ic?/ + 12?/ : ' (%^y)(x> ti y) x 6 CHECK. Let x = 3, y =2. Then. a; 2 - 7 yy + 10 y 2 _ 7 a i so x - 5 y - ^L _ 7. x a _ s a:y + 12 y 2 9 x - y 9 9 - 4 a 2 - - 48 aaj + 128 a 63(a 2 - I*) x 2 + 2 fly + ?/ 2 . 16 5 * 17 -. is } -y' 5 )(x + 2/) _ 3 + 3 _ x~y x 6(n 2 -2?i - 22 ' ' ' 23. Make a rule for reducing fractions to their lowest terms. 45 aV - 105 5 ^ 33 ab s x - ' 2 98 FRACTIONS *26. - ' ^ ' SUGGESTION. Square, and add terms in a " the numerator first. One cannot cancel as the problem stands, since neither a + b nor a b is a factor of the numerator. a. The student will avoid errors if he will simplify numerator and denomi- nator (if necessary), then factor, then cancel common factors. 24^-22^ + 5 * ' m 2 +7m + 16^-15 ' 9a 2 -9x'-28 a 2- ' 61. Highest Common Factor. 1. A common factor of two or more integral quantities is a quantity that will divide each of them. In the preceding article the factors canceled were common factors of both numerator and denominator. i 2. The highest common factor of two or more given integral quantities is that integral quantity of highest degree and largest numerical coefficient which will divide each of them exactly. It is the product of all the prime factors that are common. 3. A rule that can be used in reducing a fraction to its lowest terms is : Divide both numerator and denominator by their highest common factor. 4. Find the highest common factor of the numerator and de- nominator in each of the exercises of the preceding article. 5. Find the highest common factor of x* a 4 , a^+a 3 , and x 2 a 2 . SOLUTION, x 4 - a 4 =(x 2 + a 2 ) (a; -f a}(x a) x s + 3 = (x + ) (X 2 ax + a 2 ) H. c. f. = x -f a (By definition in 2 above.) SIGNS IN FRACTIONS 99 6. 7. l-lla + 18a 2 , 8a 3 -l, 18o-'-oa-2. (See 58, 41.) 8. 10. a 2 -3a--18, 2a 2 - a - 21, 3a 2 + 4a- 15. 11. 2a 2 + 17a-t-36, 4a 2 -4a-99, Ga 2 + 25a-9. For additional exercises use the quantities of 58. NOTE. Highest common factor and lowest common multiple are com- monly put in adjoining chapters. As the processes of finding them are somewhat alike, students may confuse the two, particularly if they fail to grasp what the terms mean. For this reason highest common factor is here treated in connection with reduction of fractions to their lowest terms - the only place where this subject finds any application in elementary algebra. Indeed, so far as elementary algebra is concerned the student can get along very well without h. c. f . by factoring. 62. Allowable Changes in the Signs of 'the Numerator and Denomi- nator of a Fraction. There are three signs connected with every fraction : that of the numerator, that of the denominator, and that of the quotient which stands before the fraction and is called the sign of the fraction. Thus ^ ^ has + (understood) for the sign of its nu- - (c - d) merator, for the sign of its denominator, and for the sign of the fraction. 1. The signs of all the terms of both numerator' and denominator of a fraction may be changed, because this amounts to multiplying both numerator and denominator by 1, which does not change its value. ( 60.) Th a + ba b. 2a 3b-)-c2a+3b c c d c + d m + r a a--m r ( 60, 13, 1.) 100 FRACTIONS 2. The sign before a fraction and the signs of all the terms of its numerator can be changed without altering its value, for this amounts to multiplying it by 1 twice, and 1 X 1 = + 1. Thus, m 4- n m -f- n 3. The sign before a fraction and the signs of all the terms of its denominator may be changed without altering its value, for this amounts to multiplying it by -- , or 1, as ( - j = (!)=!. ~~ Thus, c b b c a. If one or all three of the signs of a fraction are changed, its sign is changed ; if any two are changed, its value is not altered. Factors and terms of the numerator and denominator should not be confounded. (See 4 and 5 below.) 4. The signs of an even number of factors of the numerator, or of the denominator, or of both, can be changed without changing the value of the fraction, since an even number of Vs multiplied or divided give +1. Thus (b a)(d~c) = (a - fr)(c - cf) m)(q--p) (m n}(p 5. TJie signs of an odd number of factors of the numerator, or denominator, can be changed if the sign before the fraction also is changed. (See 2 and 3 above.) [ c 2 c 2 ' (a? - y) (z - y) (x-y)(y-z) b. It is important that the sign of a fraction should always be written on the line. Beginners are frequently careless about this, and are led into con- fusion as a result. Thus, a + - is bad form. It should be written a + - c ADDITION AND SUBTRACTION OF f FACTIONS U'l* 63. A mixed quantity is composed of an integral quail tity. and a fraction. Thus, a and ^ + r are mixed quantities. c p + q a. It is desirable that the student should note here a difference between the Arabic and the algebraic notations. In algebra, when no sign is written between quantities, multiplication is understood ; in arithmetic, additicn is understood. Thus, a - must mean in algebra a times - ; in arithmetic, on the other c c hand, 3| means the sum of 3 and |- I-* If one desires to denote the sum of a and _ , he must write + between ft c them ; thus, a H c 64. Addition and Subtraction of Fractions. Type form: = AjB. _ a . b A a-\-b 3. - + - Ans. 2 c c c CHECK. Let a = 2, 6 = 3, c = 6. Then 2 + $ = M = 6. also c c 66 6 c 6 6 a b c K a b c d 4. -H 1 5. m m m x x x x 2x 1 1 + a: . x 2 ru^z, 6. Ans. - Check. 1 071 OS 1 q-f-6 rr / \ ' p^ 102 FRACTIONS 453 4 J b 8 4 3 3 6 5 25 1 a; x 4 37 43 6 SOLUTION. Fractions in algebra are added by the same process as in arithmetic. We first find the lowest common multiple of the denominators, calling it the lowest common denominator (I.e. d.). Then the 1. c. d. is divided by the denominator of the first fraction and both terms of this fraction are multiplied by the quotient. The same is done with the other fractions, thus yielding equivalent fractions, since multiplying both terms of a fraction by the same quantity changes its form but does not alter its value ( 60). Last of all, the numerators are added and the sum placed over the common denominator for the answer, which is to be reduced to its lowest terms if not already so. The 1. c. d. in this problem is 12. Thus, 5_ 43 6 becomes * - *~ -. * , (60) 12 12 12 8s-(4s-16)-.C2s+10) t (As in Exs. 1-8, p. 101.) \a 3% 4x 4- 16 2 .x 10 _ 6 3 x (On removing parentheses QjV .j-- __ - - * 12 12 and adding.) But, = n * x > = ?^=-Z. (On reducing to lowest terms.) 1. *j r~r^ * 4 / CHECK. Let re = 4. Tben --^-=^-^-^ = 1 ---? = -!; 43 6 322 2 -=-^ = = =--. (62,2.) A A A O ^ ** / ADDITION AND SUBTRACTION OF FRACTIONS 103 a. The student will do well, in solving the problems of this article, to fol- lc/w the model pretty closely, beginning solutions by writing down the 1. c. d. Later he can shorten the process somewhat by omitting one or more of the steps. He should make use freely of checking in fractions, as this is likely to be his only means of knowing whether the answer is correct. He should be careful to see that his answers are in their lowest terms, as the checking does not test for this. If at any time a letter is given such a value that it makes any denominator zero, such value of the letter must be rejected and another must be selected in its place. 13. is x-3 3x + 2 23.T-9 2-x 3: V 1 iC 1 2 3 7 v ff f < & JC, JC 30 17. 19. 5 ' 5a-76 2 10 3 a 5 6 8 12 4 4 6 2a 4a Ga b + d 20 . + _. xy xz yz 01 be ,ac ab & \. ~j~ ' ^j " a b c ett . a b c a.b c **& : 1 h ab ac bo 23. 4a 3a-7b 5 be 3 jc 7 ab \ *. 3 a 46 2c 25 . 6 a a 26. m n m n m + n 27. x-- 1 2 + ^ X 2 -l 60. [ 3 a 2 ax a x a -f x or x 2 29. a a b a- b 2 a -{-b 30. 32. a+6 a 6 x2 xl 81. 4 x 5 a a-- 3x(a-b) -i L. c 1 n to x 6 x-\-b x 2 b 104 FRACTIONS + 2) 2 I I SUGGESTION. L. c. d. is ac(a +c). 36. H - To divide 1. c. d. by denoins., see 1-vO I O Cv "t " Ct/O -f~l -i / /"vr* Ex. 16, p. 93. 37- ~^~~ ~~r H : r^ g TV 38. - -| - 39. Write a rule for the addition (including subtraction) of frac- tions. (Apply 60 to each fraction before beginning.) (1) What is found first ? (2) What change is made in the several fractions ? How ? (3) What is done with the resulting fractions ? (See Exs. 1-8.) If precedes a fraction, what is done with its numerator (in parenthesis if it consists of more than one term)? If 4- precedes a fraction, what sign precedes its numerator ? (4) What is the next operation ? (5) What reduction is very often made in the answer ? (6) How is the answer checked for accuracy ? 11 21 x+6 40 . 41 3x-2y 5x-2y x-2 x + 2 or' + 4 40 ^ I 43 _ ^-t. [^ ^-*J x i. x x + a 3 7 4 20 x SUGGESTION. Multiply both terms A^ . __ - . l 2x l + 2x 4 x 2 1 of the last fraction by - 1. ( 62, 1.) 45. zz a_a-_-. > _ __. ( 62 a_ 6 ^6 ft 2 -* 2 3_ a a 2 -9 2 x 61 (i 9 47 i . 48. h C 3 a; 4- 6 x 2 4-2x 3 a a 2 3 a ADDITION AND SUBTRACTION OF FRACTIONS 105 49. - 4 ~ + p -^77-+^ * , 50. 2-20 SUGGESTION. As in arithmetic, in- i [ 9 ^2 51. 3 # tegral quantities are given the denomi- x nator 1. SOLUTIO,. XX X X CHECK. Let x = 5. Then 3 x - 1 + 2a;2 = 44 j also = 44. x x 52 . 3a! _ 2a , 7 , - 54. a + o 55. la a + b 5 - 6 a 56. L-JL+a 1. 57. . 2a a 1 58. X 2_ a . + 1 -- 1 . 59 ^_ a _ a? + 1 a; + a 60 . 4^-2-i^. 61. x*-3x- 3x ( 3 - 2x+l x2 62. a 2 -2aa;--^--f4a5 2 . 63. l 64. l-[a-^_). 65. Moj,a;.2 66. 5 # 1 \-x 2 . 67. H- 2 ^v ** ^ 7 9 x-1 x+1 68 . _ 3 + 69> _ 2> oj-f- 1 a? 8 - -1 (a + b) 2 a + b 70. _ - _ H-- _ - 71. _ 4-- 2 9 106 FRACTIONS * 72 . ___ 73 m , y . 1+m a; 2 - -2 a? 8 a^ + 5aj-4-6 y(xy) m(yx) my 74. + +_ __ 75. b 2 ax ab a 1 a 3 76. __ + =i_^- 77. 3a 1-a 3 xy 4 a- - 9 b 2 9bd-6ad 78. _ . - /^^A 79 ' ^3 -- i --- ^3 -- i -- ( oO. or a 6 , or + a 6 80. 81. 2 ^ +^-i 82 83. 84. 85. , '-- + 4 a; +2 1- 4-7a-2a 2 4-5a-6a 2 5a 2 a + 3 & 4a 9 a 2 - 25 6 2 6 86. ^- ^ SUGGESTION. Add 1st and 5th terms, and 2d and 4th terms, first. REDUCTION OF IMPROPER FRACTIONS 107 88. ^ '- 1 (a b~)(x a) (b a)(xb) b 4- c SUGGESTION. Change 2d fraction to See 62, 5, 3. (a - b)(x-b) 89. , 2a +y .. + + 6 +v +*- - (x a) (a b) (x b) (b a) (x a) (x b) 90. H \. Decomposition of a Fraction. It is sometimes desirable to reverse the operation of adding fractions. adn 4- ben bdm i. bdn SOLUTION " ~~ 4. bdn bdn bdn bdn = 1 + 3--' Ans ' ( 60 -) bdn 6 a 2 3 b 2 + 10 c 2 a5c -f 6c^ + adc + abd z. - o. - - 30 abc abed 66. Reduction of Improper Fractions to Integral or Mixed Quan- tities. An improper fraction is one that can be reduced to an in- tegral or a mixed quantity ( 63). A proper fraction is one that cannot be so reduced. In exercises 51-69 of 64 we reduced mixed quantities to im- proper fractions by writing 1 for the denominator of the integral quantities and adding. In this article we reverse the operation of adding. Since a fraction is defined ( 59) as an expressed division, an im- proper fraction can be reduced to an integral or a mixed quantity by simply dividing its numerator by its denominator. 108 FRACTIONS 1. a 2 +26 2 a b 2. SOLUTION. 3 ^ _ 7 3. _ SOLUTION. a 2 - 6 a 5 2 a + 64- 3b< ab + 2 & 2 a& 6 2 c* ( c/ (- a 6 3a 3a 2 -7z- 10 2 O v t t/ tJU x-3 4 3r | Q X -f- ^ -f 2a -10 2x - 6 x- 3 362 _ 4 The remainder is written over the divisor and the resulting fraction added to the quotient. (See 63, .) In the second example, after 4 is written over the denominator, the sign of the numerator is changed from to + and the sign before the fraction from -f to . ( 62, 2.) Algebraists regularly do this. In reducing fractions to mixed quantities, the rule is to continue the divi- sion as long as the divisor is contained an integral number of times in the remainder. The answers can be proved correct by addition. 3. 5. 7. 9. 11. 3 a; 2 + 2 x + 5 x-3 2x 2 -7x-l x-3 a 4 -16 a? + 2 a 2 ax 3 x 2 '' . m 4 a-j- x 1 15. 17. 4. 6. 8. 10. 2a-3b 2x-3 5 a; + bx /v.2 a a 14. 16. 18. 10 a 2 -17 ax + 10 a; 2 5 a x 2x-3 MULTIPLICATION AND DIVISION OF FRACTIONS 109 67. Multiplication and Division of Fractions. , A AC M P MQ MQ Typeforms: X- = 35 _ 2 v 7 25 ' 4 X f 2 '5 X 8 4 S x - SOLDTION - v x= Ans - a CHECK. Let a = 4, 6 = 3, c = 5. 2 a 6 6c 8 90 , 4 c n a cx 2 3 qfi 3ac 3a 2 5 14c-q * - 7 ^ -f a: 2 70 9 7 14 7 CHECK. Let = 8. Then, x = - - ; also = - 8 80 32 64 32 a: 2 - 25 a 2 5 a + 4 a 2 xy ab-2b 2 3 -^ ^* a 3 110 FRACTIONS 22. Make a rule for multiplication of fractions. (1) What operation is performed first, if necessary, with thu terms of each fraction ? (2) What is done with factors common to any numerator and any denominator ? (3) After canceling what is done ? (4) What is the last step if there is any uncertainty as to the correctness of the answer obtained ? 23. On what principle does canceling factors common to any numerator and any denominator depend ? Notice that the prin- ciple applies only after the numerators are multiplied together for a new numerator and the denominators for a new denomi- nator. But it is more convenient to cancel before multiplying. 25 * ex (x -y)(x 26 x - 27. xx a 6 & 6 a-* -- a 2 + 4 x 2' x3 X-- 29 x ^ + 3a^ . a? + 3 a + 1 . a'-l. - - -- : -- - - - - * OX. -r- a; + 4. a^ + 4o; a a 32. ?!nA 2 -5-(a-6). 33. ^ -*- (a + c). a; + y b + c 34 8 a 3 . 4a 2 a 2 -121 a ^ a 8 --4 q? ' 8a; + 5 a 8 -! -7 x 2a;-l m 2 ?i + 2 m?i 2 + n 3 m + n 12 aj 3 -+ 24 x 2 x? + 2x a 4- X 4 t g* x + flJ rt g + 2o-15 . a 2 + 9a + 20 a^"" a 8 -a? 8 ' cr + 8a-33 ""a 2 + 7 a--44' POWERS OF FRACTIONS 111 42. Make a rule for dividing fractions. j3 + y/V/ . p 2 r.s s 2 r -- s 4 ar 9 4 ar 1 a-s _ 6 or 9 + 36 a? a; 4 + 21G a; -- 45. : x 4 -17or 9 +16 a*~3x-4: ^z D 9a 4 -34 2 + 3.2 _ 3. _ 12 x 2 2x 3 x 2 -\-x SUGGESTION. Invert 47. X : -f- - divisor and multiply rpZ 11 nt- / /v> V y I '/ three factors together. a 2 ax a + x a 3 8 8 a 3 48. - x - r-- 2 3 a + a 2 a?- -a? a 2 + a# + ar 49 -6 3 3 2ab-2U 2 a? 6xy + 9y 2 . (x* 9y 2 . a; 2 4- %y6y' y' 2 \x 2 + y~ x* - .'// - - 6 ?/ 2 51 -. x x ' ' / -xy + y x* + xy + y' 2 x y 8*> HO -i / 9 / h 1 ' H/>9 x- 2x 1 ox 1 6 ar/ ox L bar 2 m 2 - -m 6 v 2 ??i 2 - m 3 2 m 2 - m -- 3 5o. - X m 2 - 5 m + 6 6 m 2 - 11 w + 3 3 m~ - 10 m + 3 2a 2 -5a-12 . 2a 2 -7a-4 X ' * 55. ^_ v^ w_^ 68. Powers of Fractions. Type form : ( I = /T>* V bj 6** / +j OLO C \ 4 4 Ct '.' _ / * wu \ rt 1. I 1 Ans. r 2. f -3- - 3. ' '' / 112 FRACTIONS / 2a*x\* (3 wm) 4 2 by 4. I -- I 5. I -- I 6. * ^ 7. - V 7b 3 c 2 J \ 5byJ 7 bo (4c 3 ) 3 (x 2 I) 3 SOLUTION. Notice that if (6) = a*b*, then 8 ' (x-iy' (* 2 -l)3=(z+l)3(* Thus, Og^ = (*+l) 3 (*-l) 3 = (*+l)*. ^ . #12. O 4 ~ .V 4 ) 8 15 x 16 ' 69. Exercises involving Additions and Subtractions First, and Mul- tiplications and Divisions Afterwards. As a rule, before beginning the operation of multiplication or division of fractions, each factor should be reduced to a simple fraction, that is, to a fraction with integral quantities for its numerator and denominator. SOLUTION, c 2 + ^ - 2 c^ c-^. c 2 + d 2 c c d c tf-cd CHECK. Letc = 4, d = 3. Then j 2 a? \ . /., d\ 1.1 4 c 2 + 1? 2 / \ c/ 25 4 25 c 2 + (Z- 25 x \y~ y x X \ X\ f X 1 * ^ : -- r 1 + # a; y V 1 + x x ADDITIONS AND MULTIPLICATIONS 113 5. 6 14 25 x - 29 2-ajJ (5-4a?)(2-a;) 6- --8 a/ 4?^ 5 a 2 6 b 9 y| 5. 26 . o ora 3 x 2 y . c 7 cm/ 7n 2 4a% + & ^- ? / 2 a 3 - 6 3 i+i a b x a + & a; 6. *t o , * a-- b X + ?/ a 2 6 2 1 1+ _ a + o y a 2/ SOLUTION to Ex. 9. *I - ^+_1 = I x __!_ = ^_+_y . ^ ng> x y a: y + 1 xy + x a- ^ 1 + S m X 10. C 11. 6 n y a + - 1 +^ - + m c a ?i y a a + 6 i -3y a; - 2 x + 3 13. 6 a -- b i + ?/ X 1 ^ 3-> JU + 4 b a -- b 1 -I- "* a a + 6 ;--2/ a;+La; COMMON ERRORS IN FRACTIONS 115 6 a-\-b b a; -r 1 a?-- 1 /y _ 1 I _ J ^1^ __ L 1 x 6 b a + b , as - - 1 a? -f 1 *lo. ! 3 S-2 + ^7 +, x o a o 1 2 m 3 -f wi 971 X -^ Cl 1 REMARK. Such a fraction as this is often called a 22 _ _ _ . i continued fraction. To simplify it we commence at ~T the bottom and replace each complex fraction by its - equivalent simple fraction. x SOLUTION. - - - = - - - = - ^ - = x + l. Ans. x + l-x x a. Common Errors in Fractions : Certain errors are made so often and by so many pupils that it is desirable to describe them, and caution against them. Checking serves to find such errors. 1. Never cancel single terms in the numerator and the denominator of a fraction when either is a polynomial. Only factors of the whole numerator and the whole denominator can be canceled ( 00). 2. In subtraction of fractions, and later in equations, when a fraction is preceded by , students must change the signs of all the terms of the nu- merator in simplifying. Thus, q_ zL d = a-c + d b b b ?>. To change the signs of both numerator and denominator, the signs of all the terms of each must be changed. Thu 3& - a _a 36 (This amounts to multiplying both terms '2d^~ /^'/> /i <>9 o O7ir ji 6a + 6 6-6a 3a 2 -3 26+2 46-4 6-66 3 22. x- + /--... 23. + ?/ a 4 if 1 \ / r 24. (a 4 - :: ]-i-[0-V 25. nA I \ nt Jj \ A/, RATIO AND PROPORTION 117 72. Ratio and Proportion. 1. The ratio of two numbers is the quotient of the first number divided by the second. The ratio of 3 to 8 is f . We see from this that a ratio is & fraction. 2. What is the ratio of 3 to 4? of 6 to 8? of $7 to $14? of 20 to 10 ? of 9 men to 3 men ? of 20 Ib. to 5 Ib. ? of 12 ft. to 1 ft. ? of i pt. to 1 qt. ? of I to | ? of a dollars to b dollars ? of c to w + n? of cr-6 2 to a-b? of m 2 -9 to m 2 + 6 m + 9? 3. The first number in a ratio is called the antecedent, and the second the consequent; both are called the terms. A ratio is an abstract number, since it is always the quotient of one number divided by another of the same kind. A colon is often used to denote a ratio. Thus, 4 : 7 signifies the ratio of 4 to 7. 4. A proportion is an equation ( 42) the members of which are ratios. Thus, f = f j or, 3 : 4 = 6 : 8 is a proportion. 5. Fundamental Tlieorem. In a proportion the product of the "extremes" or first and last terms, equals the product of the " means." PROOF : If - = - is any proportion, fiw a ^ HA c M (Because, equals multiplied by bd give men, x />a x oyc, p ft equals.) (a and d being the extremes, and b and c or, ad = be the means or middle terms.) 6. Solve for x in = , or 16 : x = 12: 27, using theorem. x 27 Q i P 9 6 45 x x $12 7. Solve for x :-= ; = --: = ; x 24' 60 24' 21 $36' 12 Ib. : 20 Ib. = 15 cents : x cents ; $ 6 : $ 15 = 75 yd. : x yd. (a) As ratios are abstract, drop $, ^, etc. , before solving. 8. If 12 T. of hay cost $ 132, how much will 16 T. cost ? SOLUTION : x = f x $ 132 = $ 176. (By reducing 16 : 12 to 4 : 3.) Notice that the answer x will always equal the number correspond- ing to the answer multiplied by the proper ratio of the other two. 118 FRACTIONS 9. If 24 bbl. of flour are eaten by a garrison in 9 weeks, how many barrels will be eaten in 4 weeks ? w iHiswBwBi ' p 10. A lever is a bar having two ^ forces, the power P and weight W, ap- plied to it, which counterbalance each other. The point of support, F, is called the fulcrum. " 11. It is always true of a lever that the ratio of the power to the weight equals the ratio of the weight's distance from the fulcrum to power's distance from the fulcrum, or P : W= b : a. 12. Find P in a crowbar when W= 90 lb., a = 4 ft., b = 1 ft. 13. Find Win an oar when P = 25 lb., a = 10 ft., 6 = 6 ft. Fulcrum is where blade is in water, and TF"is at rowlock. 14. Two horses pull on a lever 44 inches long with forces equal to 300 lb. and 350 lb. Where shall fulcrum pin be put so that the forces balance ? Let x and 44 x equal lever arms. 15. Two weights of 12 lb. and 8 lb. respectively at the ends of a bar 14 ft. long (supposed without weight) balance each other. How far is the fulcrum from the 8 lb. weight ? Let x = number. 16. One quantity is said to vary as another when as one changes, the other changes in the same ratio. For example, the total pay a man gets varies as the number of days he works. If he works twice as long one time as another, he gets twice as much money. If two quantities vary so that the ratios of old to new values are the same, they are said to be in proportion. Thus, if 5 bu. of wheat are worth $5.75, 8 bu. are worth $9.20, since 5 bu. : 8 bu. = $ 5.75 : $9.20. The word " vary " is also often used in a loose way to denote that as one quantity changes another quantity dependent on it also changes, though not in the same proportion. Thus the prices of articles vary with the seasons, etc. GRAPHICAL RELATIONS 119 * 73. Graphical Relations. If two quantities are related, the change in one as the other changes can be pictured to the eye by means of " squared paper." Squared paper is paper accurately divided into little squares. (See page 172.) 1. To show graphically the price of butter throughout a year. SOLUTION. "We take a piece of squared paper and write Jan. 1 at its lower left-hand corner (see figure below) ; Mar. 1 at the right of Jan. 1 and below the second heavy vertical line ; May 1 below the third heavy vertical line ; and so on to the following Jan. 1. (Two-month intervals are taken instead of one-month intervals to save space.) We now count upward from the lower left-hand corner 83 spaces (for 33^, the price of butter on Jan. 1 see table of prices given at the left of the diagram below), and put a point at a. On the second heavy vertical line we count upward 34 spaces (for 34 ^, the price on Mar. 1), and put a point at b; and so on for each month. Through these points we now draw a running line, abcdefg. This line gives us very quickly an idea of the changes in the price through the year. If we suppose the changes in the price are made gradually, we can find the prices at any intermediate dates. Thus, on June 1 the price was 2G/-. Wholesale Price of Butter In New York City, 1907-1908 b -^- - \ ( \ \ 3( (' \ \ - n. 1 Mar.l May 1 Julyl Sept.l Kov.l Jan. 1. 120 GRAPHICAL RELATIONS and then construct from the table a diagram, using graph paper: Jan. 1, 36^; Mar. 1, 23^; May 1, 21^; July 1, 21^; Sept. 1, 30 ^ 5 Nov. 1, 45 ^ ; Jan 1, 34 3. Make a table and the corresponding diagram from the fol- lowing prices of wheat in 1907. Jan. 1, 73 / ; Feb. 1, 79 ^ ; Mar. 1, 79^; April 1, 81^; May 1, 96^; June 1, 97^; July 1, 94^; Aug. 1,93^5 Sept. 1, 100^; Oct. 1, 104^; Nov. 1, 93^; Dec. 1,96^. 4. Make a table and diagram from the following Chicago prices of oats (1907-1908): Jan. 1, 34^; Feb. 1, 37^; Mar. 1, 40^; April 1, 42/; May 1, 45^; June 1, 49^; July 1, 42^; Aug. 1, Sept. 1, 54^; Oct. 1, 52^; Nov. 1, 49^; Dec, 1, Jan. 1, / / 1 X. T it- -. n N. / \\ V 1. / / / ^ X 1 S. K n T 111 IS CHAPTER IV SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 74. An equation in algebra is a statement that two quanti ties have the same numerical value ( 24). 75. Equations are of two general kinds, identical and conditional. 76. An identical equation either contains no letters, or its let ters may have any values. The sign for an identical equation is =. Thus, in (x + 2) (a? + 3) = x 2 + 5 x + 6, show that the equation holds true for x = l, x = 2, x = 5, x = 0. etc. All given quantities in what is called literal arithmetic set- equal to their answers make identical equations. By literal aritli metic is meant addition, subtraction, multiplication, division,, powers, roots, factoring, and fractions in which letters as well an figures are used to represent numbers. The student will recall that we have been checking our answers in literal arithmetic by assigning any values to the letters. Algebra is often defined as the science of the equation. By this definition literal arithmetic belongs rather to arithmetic. REMARK. The sign = is commonly used for identical equations as well as for conditional equations ( 77). Thus, so far, we have always used =.. Whenever it is desired to draw special attention to the fact that an equation is of the identical kind, or that one letter is to be used to take the place, o^i some combination of other letters to save space, the sign = will be used. 77. A conditional equation is one that is true only for cer* tain values of the letters involved. Conditional equations appeav in the solution of problems. 122 DEFINITIONS 123 The equations obtained from the problems of 1, 31, and 42 were conditional equations. Ordinarily when we speak of equa- tions we mean conditional equations. 78. A simple equation is one which, when reduced to its simplest form, contains only the first power of the unknown quantity or quantities ( 57). 79. Known and Unknown Quantities. The custom in algebra is to represent unknowns, as a rule, by the last letters of the alpha- bet, and known quantities by the others. In physics and other sciences the custom is to denote quantities known and unknown by initial letters, as v for velocity, p for pressure, t for time, etc. 80. The root of an equation is the value of the unknown quan- tity, that is, the answer. This use of the word should be dis- tinguished from root of a quantity ( 18). The student may try to explain why the answer is called the root of the equation. 81. To solve an equation is to find its root. The word " solve ' : means literally to "loosen' or "untie." When one solves an equation, he loosens or unties the unknown quantity x from the others, and gets it by itself on the left side, or " member," of the equation. 82. The verification of an equation is accomplished by substi- tuting the root found in the original given equation and getting an identity. If a value of the unknown quantity, upon being sub- stituted in the given equation, makes the two sides reduce to the same number, after simplying, this value is said to satisfy the equation. 83. An axiom is a truth assumed to hold in a study or in- vestigation. The axioms have been called self-evident truths because they are so simple that the mind accepts them as true directly. The use of axioms changes the study of equations into a science. 124 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY AXIOMS 1. ADDITION AXIOM : The same quantity, or equal quantities, added to equal quantities, gives equal quantities. (1) 4 + 8 = 12 (2) 5 + 11 = 10 + 6 2 = 2 3+ 5= 7 + 1 4 + 8 + 2 = 12 + 2 5+11 + 3 + 5 = 10 + 6+7 + 1 2. SUBTRACTION AXIOM : The same quantity, or equals, sub- tracted from equals, leaves equals. Thus, (1) 3 + 9=5+7 (2) 6 + 11 = 2 + 15 6=6 2+ 3 = 4 + 1 3 + 9_6 = 5+7-6 6 + 11-2-3=2 + 15-4-1 3. MULTIPLICATION AXIOM: Equals multiplied by the same quantity, or equals, give equals. 15 + 10 = 30 - 5 x2= x2 30 + 20 = 60 - 10 4. DIVISION AXIOM : Equals divided by the same quantity, or equals (not zero), give equals. 5. POWER AXIOM: Equals raised to the same power give , equals. 6. ROOT AXIOM : Corresponding roots of equals are equal. 7. Quantities equal to the same quantity are equal to each other. 8. GENERAL AXIOM : If the same operation is performed on two equal quantities, the results are equal. The last axiom holds true in general only for arithmetical numbers. The axioms will be referred to as Add. Ax., Sub. Ax., Mult. Ax., etc. SOLUTION BY MEANS OF AXIOMS 125 84. Signs for Words. As solving equations is reasoning, we shall need signs denoting cause and conclusion. Mathematicians use .'. for "therefore" or "hence," and v for "since" or "because." 85. Solution of Equations by Means of Axioms. Very simple equations can be solved by ordinary reasoning without using the axioms; but when the equations become at all complicated, it is much easier to make use of the axioms. 1. Solve - + ^ ? = - 3 and verify. 326 SOLUTION. 1. c. d. = 6. .-. 2 x + 3 x - 6 = 7 x - 18. (Mult. Ax. The equals, f + 2Ln2 and 7 x 32 3, multiplied by the same quantity, 6 6 (the 1. c. d.), give the equals 2 x -f 3 x 6 and 7 x 18). .*. 2x + 3x 1x = 18 + 6. (Sub. Ax. (see p. 61). --The same quan- tity, 7 x 6, subtracted from the equals, 2 x + 3 x 6 and 7 x 18, gives the equals, 2x + 3xlx and 18 + 6). .-. _ 2 x = - 12. (By addition, 29, Ex. 24-32.) .*. x = 6. (Div. Ax. The equals, 2 x and 12, divided by the same quantity, 2, give the equals, x and 6.) VERIFICATION. + _ 3; or, 2 + 2 = 7-3. Q A O REMARKS, (a) The first operation, in which the equation is multiplied through by the I.e. d., is called clearing of fractions. Notice that in the new equation the denominators have disappeared. (&) The second operation, in which all the unknowns are placed on the left side of the equation and the known numbers on the right side, is called transposition. (See page 61.) To transpose any term, that term with its sign prefixed is subtracted from both members of the equation. Any term can also be transposed by adding the term with its sign changed to both members. (c) Simplifying members by adding is often called '"collecting.''' 1 (d) The last operation is called dividing through by the coefficient of x. 126 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY Because it is a little difficult to describe the use of the axioms on the preceding page, we will explain this matter more in detail. It is very important that the student should understand this sub- ject, as it is fundamental in algebra, and his ability to solve equations depends in no small degree on his knowledge of the axioms, and of how they are applied. (1) Explanation of Clearing of Fractions. - + ^^ = I* - 3 (equals) 326 multiplied by x 6 = x 6 (the same quantity) give 2x + 3x 6 = 1 x-- 18 (equals). (Multiplication Ax.) (2) Explanation of Transposition. 2x + 3x -6 =7318 (equals) diminished by _ 7 x 6 =7 x 6 (equals) give 2x + Sx-~7x -18+0 (equals). (Subtraction Ax.) (3) Explanation of Dividing through by Coefficient of x. - 2 x = 12 (equals) divided by -+--- 2 = -4--- 2 (the same quantity) give x = 6 (equals). (Division Ax.) Solve and verify, and explain the operations, especially those based on the axioms in the following. Write all the reasons at the right margin, as is done in the model. In oral explanations the student should point to the appropriate members of the differ- ent equations as he repeats the axioms. x 2 x -f- 3 x 2 x + 7 x 7 = - H -- o. -- -- = 2343 341 4-l 3_ 3 + 5 4x 5 3 8xQ ~ ~~ ~' 5> ~~ = ~~' 6 . <_: = 10 + =- 7. . AXIOMATIC CHANGES 127 8. _ A 10. x + 2 x 3_ 3 4 .2-*- 1 2 9. a; = 2 a; 11. 65 3 a; + 2 -13. 86. Explanation of the Axiomatic Changes made in Equations by Means of the Equation Balance. It is very desirable that the stu- dent should get the idea that an equation is very different from a quantity, though made up of two quantities. An equation is very much like two weights balanced on scales. In the one case the numbers on one side added give the same sum as the numbers added on the other side ; in the other case the weights on one side added give the same total weight as the weights added to- gether on the other side. Axiomatic changes in an equation can be easily and simply illustrated by means of a special kind of balance, called an equa- tion balance, one form of which is shown herewith. F is the base, a thick board ; ON is a vertical post ; PQ is a cross arm fixed rigidly at right angles to ON. BC is the scale beam moving about pivot at A. At P and Q are frictionless pulleys fastened at equal distances from N. The letters h, m, n, k, mark scale pans. For weights whole pieces of crayon can be used. Let us call the scale pan m on the 1 jft side positive ; then h is negative, since it pulls in the opposite direction. On the right side n is positive, and k negative. 128 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 1. To explain transposition. Place weights on the pans so that the apparatus balances, as 4 on h, 8 on m, 6 on n, 2 on k. Thus, If now we remove 2 on & from the right side to the left side, we must place it on m if we wish the apparatus to continue balanced. Thus, 2 on the right side must become + 2 on the left side. In other words, when we transpose we must change the sign. 2. To explain clearing of fractions. Place weights on the scale pans so that the apparatus balances, as 3 on h, 5 on m, 4 on n, and 2 on k. Now double the number on each scale pan. Thus, -3+5=4-2 becomes 6 -f 10 = 8 - - 4. The student should learn from this that in clearing of fractions he must be careful to multiply each term of the given equation by the lowest common denominator. The other axioms can be explained in a similar manner. Interest will be added if this apparatus is actually made and used. It can be employed to explain - times - gives + by de- scribing putting on weights as positive, and taking off as negative. Thus, if 4 crayons are on the h pan, and two are taken off twice, to balance the scales 4 must be put on the n pan on the other side. Hence, -2x-2 = + 4. 87. Exercise in Solving Equations containing One Unknown Quan- tity. Solve the following, writing the reasons at the right margin as is done in the model, 85. Be sure also to verify the answer in the given equation. 1. 9 x - 2 = 7 x + 16. SUGGESTION. As this equation does not contain fractions, the first step in the solution is to transpose (see 86, b). 2. 6x-25=7 -2x. 3. 8a-7 + 3a SOLVING EQUATIONS CONTAINING ONE UNKNOWN 129 4. 6a; + 4a;-13-2a;-3 = 0. 5. 42 -21 a; + 15 = 57 + a; +22. 6. SUGGESTION. Perform the multiplications, and then transpose the un- known terms to the left side, and the known terms to the right side. The x* terms cancel. The equation is now like those we have solved. 7. 3(x - I) 2 - 3(x 2 - 1) = x - 15. 8. 9. 2(a; + 2)(a;--4)=a;(2aj 10. 9(13 - a?) -13 a? = 5(21- 2 a?). - ^ Ju i JU Ju **j ~\~ 10 ~~ -^ ~~~~ *-^ "~"~ ^ U ' 2 + 3~4 = ~2~' 12 ' "2 4~ = ^' ( 64,39, (3): 85,1.) a 2 a; 3 a^-^7 x_ x 2_ x x 9 ~5~ 4 " 10 " 6 5 "5 4 3r4- 1 *> 7- r 1 1711 15. + f^ = 10 + - i. 16. ; + i=- .w O O O *v O *v 17. ^j^ = 3 . 18. z^ + 24 = I 1 + 1 o 3y 180-5y_ 29 2Q 60 -a; 3.T-5_3a; 19 ' T + 6~~~ y * ^4" 7 " 4 21. = a; ~ 5 f ,- 22. E 1 + 1 = 1. 2a;~5 2^-2 ar + 1 x v q/v, ^4.i'^--Sr1--iB 23. = ^L_2. 24. 2- P = ^ it' + l a; + 2 6 12 1 + a; ^ / _l_ Q 9/y ^ 3ft 4^ 25. ^i2 + ^Zl3_ 6 . 2 6. -^ -= - O^ -^ 2 O* - 7/ ^ ?/ 130 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 27. Make a rule for solving simple equations in one unknown quantity answering the following questions : (1) If the given problem calls for multiplications (excepting those in denominators), what is done first ? (2) If the equation contains fractions, what operation is per- formed ? Upon what axiom does this change in the equation depend? (3) After removing parentheses, in case any remain, what changes are made in the terms of the equation ? Upon what axiom does this change in the equation depend ? (4) After transposing, what usually has to be done with the terms ? Does this depend on an axiom ? (5) What is done last of all to complete the solution ? Upon what axiom does this step depend ? (6) How is the answer verified ? 28. *1_* = 3 = . 29. x i x + 3 x a? + 1 x 1 3x-l 9x-31 321 oO. - --- - - - . ol. 874 z + 1 x+2 x+3 32. - + = -. 33. 6o;+4 2x+3 2x--l 3x- 6 34. 3x+5 6 2x+3 & \ 9 36. _l_ = 7a , 37. - 9 3 6|-3z S(x-2) 38. _ = _ 2 _ 3 w 2 40. 3.2^-^^^55=8.9. 41. . .. . .5 .5cc .4 2 a; .1 42. S-a-- x) 2 43. SPECIAL PROCESSES OF SOLUTION 131 8 2,-e-l 2x-l 4ar -1 1+2* 44. = 2 v j J_ X J_ X -L 45. ~~ fl? + rrL: a; 46. 3 + = 4 2 3/6 3 47. t_l_*_4 = t- *48 - I ^ = ^ I j r-v 1 *~* /^ \ I 49. 50. 4-2o; 8(1-*) 2-# 2-2o? 6 :(a?4 Y" 2 x 2 1+3 a? 2 x 2- a - 15 14 (#-1) 21 6 105 51. (a? 5)(jB-2)-(o? 6)(2 6) + (a? + 7)(a? 2) = 0. Special Processes of Solution. The regular method of solu- tion is : (1) clear of fractions ; (2) transpose ; (3) collect ; (4) divide through by the coefficient of x. But any method is per- missible which solves, or loosens, x from the other quantities, provided the several steps can be justified. 89 Problems and Exercises. By a problem in algebra is com- monly meant a question that involves first the forming of an equa- tion and afterwards its solution. See 1, 31, and 42. Other questions are usually called exercises, though any question calling for a solution or simplification can be called either a problem or an exercise. The word " exercise J; is also used as a collective noun, to denote a series of questions. 132 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 90. Steps in the Solution of Problems. It is important that the student recognize clearly the several steps in the solution of prob- lems like those in 1 and 31. These steps are : 1. Looking for the unknown number and letting x equal it. 2. Constructing the functions of x, that is, the quantities con- taining x described in the problem. 3. Turning a sentence in the given problem into an equation. 4. Solving the equation by the method of 87. In the following articles training is given in constructing func- tions and in changing sentences into algebraic language. 91. Exercise in constructing Functions. To test answers, sub- stitute for x any value, and work as an arithmetical problem. See whether the answer agrees with the result obtained by substi- tuting the same value of x in the algebraic answer obtained. Express the answers to the following in algebraic language : 1. What is the value in cents of x dollars ? Ans. 100 it' ^. CHECK. Let x = 5. Then the question reads what is the value in cents of five dollars ? Ans. 500 ^. Also putting x = 5 in the answer above, we get 100 x 5 = 500, or 500 f . 2. What is the value in cents of x two-dollar bills? Of x quarters ? Of a? dimes ? Of x nickels ? 3. If a man sold a horse for 3 a; -f 6 dollars and lost x 1 dol- lars, how much did the horse cost ? 4. If a house is worth 5 x + 10 dollars and the lot it stands on 3 x 1 dollars less, how much are both worth ? 5. What is the distance around a rectangular field whose width is x rods and length x + 10 rods ? 6. What part of 9 is 5 ? What part of 9 is x ? 2 is what part of x ? 7. A room is x ft. long and x 5 ft. wide. How many square yards are there in its area ? 8. By what number must 8 be multiplied to produce Sx 5 ? FUNCTIONS 133 9. The smallest of four consecutive numbers, that is numbers that follow one after the other, as 7, 8, 9, 10, is x. What are the others ? 10. Write down five consecutive numbers of which the middle one is x. 11. The sum of two numbers is 15 and one is x 1. What is the other ? 12. What is 15 per cent of x + 5 ? 25% of 3 x 7? 13. The number x + 5 is 20 % of what number ? 14. Express in cents x dollars and x -f- 5 cents. 15. A boy had x dollars and spent x + 1 dimes and x -f 2 cents. How many cents had he left? 16. If the length of the day is x + 4 hr., what is the length of the night ? 17. A boy is #-f-4 yr. old to-day. How old was he 6 yr. ago ? 18. A man is 40 yr. of age. How old will he be in x + 2 yr. ? 19. The divisor is 4, the quotient is # + 1, and the remainder x: what is the dividend ? 20. "Rate, time, and distance " problems occur very frequently in algebra, and the student ought to be skillful in constructing the functions needed in solving them. By rate is meant the number of miles per hour, or meters per second, or the like ; by time is meant the number of hours or minutes or seconds ; by distance, the number of miles or meters, etc., in the whole distance. It is possible to generalize from a single problem. If a man rides for 5 hr. in a carriage that moves at the rate of 6 mi. per hour, how far does he go ? Keeping this problem in mind, answer the following questions : (1) When time and rate are given, how is the distance always found ? Ans. Multiply the time and rate together to get the distance. 134 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY (2) When the distance and time are given, how is the rate found ? (3) When the distance and rate are given, how is the time found ? 21. If a man travels x mi. per hour, how far does he go in 10 hr. ? 22. If a carriage travels x mi. going 4 mi. an hour, how long is it on the road ? 23. If a train travels 180 mi. in x 4 hr., how many miles per hour does it go ? 24. If a horse ran 2ic+3 minutes at the rate of 40 rods per minute, how many miles did it go ? 25. If 60 3x is the distance and 2xl is the time, what is the rate ? 26. If J x + 18 mi. is the distance one train goes and 33 mi. is its rate, and ^ x 18 mi. is the distance and 21 mi. is the rate of another train, what is the number of hours each is on the road ? 27. If A travels at the rate of 7 mi. in 5 hr. and goes x mi., how long does it take him ? 28. If x is the distance to the top of a mountain, and a person walks up at the rate of 1^ mi. per hour, and down at the rate of 4i mi. per hour, how long is he away from home ? 29. A man travels x mi. an hour for a hr., and then 2x 3 mi. an hour for b hr. How far does he travel in all ? 30. What is the velocity in feet, per second, of a train that travels 90 mi. in x hr. ? Of a train that travels x mi. in 90 min. ? 31. How many miles can a man walk in 30 min. if he walks 1 mi. in x min. ? How many feet can he walk in a minute if he walks x mi. in 1 hr. ? FUNCTIONS 135 32. " Number of articles, price, and cost " problems are often given. By price is meant the cost of a single article ; by cost the value of all. Thus the price of sugar is 6 ^ a pound, while the cost of 11 Ib. is 66^. (1) When the number of articles and price are given, how is the cost found ? (2) When the cost and number of articles are given, how is the price found ? (3) When the cost and price are given, how is the number of articles found ? 33. A man bought x tons of hay at $12 a ton. How much did he pay ? 34. Fifty yards of cloth cost x dollars. What is the price per yard ? 35. Each of x 2 persons paid $4 for a trip. How much did they all pay ? 36. A man paid $2 a day for board for 26 x days. How much did his board cost him ? 37. If 2 x 5 Ib. of tea cost $9, what is the price per pound ? 38. What is the total cost of x dozen oranges at 40 ^ a dozen, and 25 x dozen at 35 a dozen ? 39. If S.E was the cost of a camping trip, how much more would each man have to pay if there were only 8 men instead of 10? 40. If x eggs cost $ 1.05, how many cents will 7 cost ? How much more will 9 cost than 7 ? 41. If maple sirup sells at $1.40 a gallon one month and $ 1.60 another month, how many more gallons could you get for $# the first month than the last month? 42. Length, breadth, and area problems occur frequently. (1) How is the area of a rectangle found when its length and breadth are given ? 136 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY (2) How is one side found when the area and the other side are given ? 43. What is the area of a field x -+- 1 rods long and x 3 rods wide? 44. What is the area of a field whose length is x ft. and width | x ft. ? Of another field whose length is x + 10 ft. and width | x + 5 f t. ? 45. What is the width of a field containing x acres whose length is 100 rods ? Whose length is a rods ? 46. What is the width of a garden whose area is 165 sq. ft. and length 2 x - 9 ft. ? 47. " Work and cistern J:> problems are solved by using recipro- cals. If A can do a piece of work in 7 days, what part of it can he do in 1 day ? In x days ? If a cistern can be filled by a pipe in x minutes, what part of it can be filled in 1 minute ? In 5 minutes ? 48. If a job of work can be done in x -f 1 da., what part of it can be done in 3 da. ? 49. If a piece of work can be done by 1 man alone in 15 days and by another man alone in 20 days, how much can be done by both men in x days ? In x + 2 days ? 50. If a cistern can be filled by one pipe in x minutes and emptied by another in x -f- 5 minutes, what part of the cistemf ul runs in in one minute if both pipes are open ? 51. If a man can do a piece of work in 20 days, and a boy can do the same piece of work in x days, what part of the whole can both do in 5 days? 52. If A can do a job in x days, B the same in | x days, and C in Jo; days, what part of the whole work can all working together do in one day ? 53. Write down a number which when divided by x gives a quotient b and a remainder c. EXPRESSING SENTENCES AS EQUATIONS 137 54. If a bill is shared equally among x laborers and each pays , how many dollars does the bill amount to? 55. A man makes a journey of c miles. He travels x mi. an hour and a hr. by boat, x + 10 mi. an hour and b hr. by train, and the remainder on foot. How far does he travel on foot ? 56. A floor is a ft. long and x + 2 ft. wide. How many square yards are there in it ? 57. A horse eats x bu. and a cow b bu. of corn a week. How much will be left out of c bu. if they feed n weeks ? 58. A boy is x yr. old and 5 yr. from now he will be half as old as his father. How old is his father now ? 59. What is the interest on $1000 in b years at c% ? 60. How much does a man save in a year if his salary is $# a month and he spends $b each month for 6 months, and $c a month for the remaining 6 months ? 61. How long will one man require to mow a acres if x men mow & acres in a day ? 62. How long will it take a person to walk b mi. if he walks 20 mi. in x hr.? 63. A man drives m hr. at the rate of x mi. an hour. How long will it take him to walk back at the rate of b mi. an hour ? 64. A man bought b yd. of cloth at $c a yard and gained $#. What per cent did he gain ? 92. Exercise in expressing Sentences as Equations. Two kinds of sentences may be distinguished : (1) those in which the equality of two described quantities is expressly stated; (2) those in which an equality is implied. 1. The double of x is 24. Ans. 2^ = 24. 2. Three times the sum of x and 4 exceeds b by 12. Ans. 3( Sx + 2x + 600 = 12z+12. (Mult. Ax.) _ 1x = _ 588. (Sub. Ax. and Collecting.) x = 84. (Div. Ax.) VERIFICATION. Half of 84 is 42 ; one third of 84 is 28 ; 42 + 28 + 100 = 170 ; also twice 84 and 2 more is 170. a. Do not verify problems by substituting the answer in the equation as was the rule in 87, since a mistake may have been made in getting the equa- tion. Instead, test the answer by seeing whether it satisfies the reading of the problem. The student should make it his regular practice to test all answers in this way. 2. What number is that whose half, third, and fourth parts together equal 65 ? 3. What number is that which being increased by its six sevenths and the sum diminished by 20 equals the result 45 ? 4. What number is that to which if 1H be added, 4 times the sum will be equal to 10 times the sum of the number and one ? PROBLEMS 143 5. A certain number multiplied by 5, 24 subtracted from the product, the remainder divided by 6, and this quotient increased by 13 results in the number itself. What is the number ? 6. Demochares has lived a fourth of his life as a boy; a fifth as a youth ; a third as a man ; and has spent 13 yr. in his do- tage : how old is he ? (From a collection of questions by Metro- dorus, 310 A.D.) 7. Divide the number 181 into two such parts that 4 times the greater exceeds 5 times the less by 49. SUGGESTION. Here there are two unknown numbers. Let x = the greater number. What then is the lesser number, if their sum is 181 ? 8. "Heap, its seventh, its whole, it makes 19," that is, find a number such that the sum of the number and one seventh of the number equals 19. (From Ahmes Collection of Problems, made in Egypt 1700 (?) B.C.) 9. What part of the day has disappeared if the time left is twice two thirds of the time passed away ? (Palatine Anthology, 300 A.B.) SUGGESTION. Let x = number of hours of the 24 that have passed away. Then how many remain ? 10. Diophantus of Alexandria, the reputed first writer on algebra, had the following engraved on his tombstone : Diophantus passed i of his time in childhood, yL in youth, and 1 as a bachelor ; 5 years after his marriage was born to him a son who died 4 years before his father at half his father's age when he died. How long did Diophantus live ? SUGGESTION. Let x = number of years D. lived. Then the sum of all the periods named (including the age of his son) =? 11. The British House of Commons in 1908 had 670 members. Ireland had 5 less than three halves of Scotland's number and A 144 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY England had 20 less than 5 times Ireland's number. How many members came from each country ? 12. A person at the time of his marriage was three times as old as his wife, but 20 years later he was only twice as old. What were their ages on their wedding day ? See 73, Ex. 22. 13. A father aged 54 years has a son aged 9 years; in how many years will the age of the father be just 4 times that of the son? 14. A father is now 40 years of age and his daughter 13. How many years ago was the father's age 10 times that of the daugh- ter? 15. The Pennsylvania Railroad from New York to Chicago is 912 miles long. From Philadelphia to Pittsburg is 6 miles less than 4 times the distance from New York to Philadelphia, and from Pittsburg to Chicago is 24 miles more than from New York to Pittsburg. What are the distances from New York to Phila- delphia and from New York to Pittsburg; also the distance from Pittsburg to Chicago ? 16. The difference between two numbers is 12 and the greater is to the less as 11 : 5. What are the numbers ? SUGGESTION. Let llcc=the greater number and 5x the less number. Follow a like course in solving Ex. 17-19. See 94, b. 17. In Georgia in 1900 there were not far from 8 white persons to 7 negroes. The population of the state was very nearly 2,216,100. How many were whites and how many negroes ? 18. In Tennessee in 1900 there were not far from 16 whites to 5 negroes and the population of the state was very nearly 2,020,410. How many were whites and how many negroes ? 19. In 1900 in the United States for every 104 persons engaged in agriculture there were very nearly 71 engaged in manufactures, 57 in domestic service, 48 in trade and transportation, and 13 in professional pursuits. If the same ratio holds and the total popu- lation is taken, as estimated by governors of the states Jan. 1, PROBLEMS 145 1908, to be 88,912,022, how many persons were engaged in each form of industry ? 20. J. Nilsson made a skating record February, 1895. In Feb- ruary, 1887, on a straightaway course with the wind at his back, Tim Donoghue made the mile in 23.4 seconds less time than Nilsson. Nilsson's time was to Donoghue's as 20 is to 17. What were the two times ? 21. The needle of a talking machine follows a spiral (see figure) ; suppose we think of the spiral as five circumferences whose sum is 75.57 in., the smallest equal- ing 15 in. What is the common distance between them if it is the common difference between two ad- jacent circumferences, x, divided by 6f(27r)? 22. A vote was taken in a debat- ing society arid the motion was car- ried 5 to 3 ; on reconsideration 110 affirmative votes deserted to the negative and the motion was lost 3 to 4. What was the number of affirmative and the number of negative votes on first ballot? 23= The number of representatives in Congress by the appor- tionment of 1900 was 386. Of this number the Southern had 5 less than 6 times as many as the Western states ; the North Central had 6 less than Western and Southern together, and the Eastern 28 less than the North Central. How many had each region? 24. The area of New York state is 8164 sq. mi. more than that of Ohio, and Texas has 4836 sq. mi. more than 3 times New York's and Ohio's areas together. The area of all three is 356,140 sq. mi. What is the area of each state ? 25. Mt. Shasta is 203 ft. higher than Pike's Peak, and Mt. McKinley is 6114 ft. higher than Mt. Shasta. The sum of the three heights is 48,961 ft. What is the height of each? 146 SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY 26. In the National League in 1905 Pittsburg won 9 games less than New York, Chicago 4 less than Pittsburg, Philadelphia 9 less than Chicago, Cincinnati 4 less than Philadelphia, St. Louis 21 less than Cincinnati, Boston 7 less than St. Louis, and Brooklyn 3 less than Boston. In all 612 games were played. How many games did each team, win ? 27. A rectangular field is 6 rods longer than it is wide ; and if the length and breadth were each 4 rods more, the area would be 120 square rods more than it is. Find the length and breadth of the field. 28. The length of a rectangular floor exceeds the width by 6 ft. If the width is increased by 3 ft. and the length by 2 ft., the area is increased by 134 sq. ft. Find the length, breadth, and area of the floor. 29. The length of a room exceeds its breadth by 8 ft. If each had been increased by 2 ft., the area would be increased by 60 sq. ft. Find the original dimensions of the room. 30. The total number of persons speaking English, German, and French in 1801 has been estimated at 82 millions. Of this number the German language was spoken by 9 millions more than the English, and the French by 1 million more than the German. How many millions spoke each language ? 31. The total number of persons speaking English, German, and French in 1901 was 266 millions. Of this number the English language was spoken by 46 millions more than the German, and the German by 32 millions more than the French. How many spoke each language at this time? 32. A man invested $1600, a part at 6 % and the rest at 5 %. If his total annual income was $ 90, how much did he invest at each rate ? IMPORTANT NOTE. Heretofore the unknown number has usually been denoted by x. The student should now accustom himself to using other letters instead, preferably initials. Thus, use n (number) in Ex. 32-35, 38 ; r in 36 j v (velocity) in 37 ; / (franc) in 39 ; and so on. PROBLEMS 147 33. A man has two thirds of his property invested at 4 \ to find values 4 of x and y that will satisfy both equations ( 82). a. Observe the equations are numbered (1) and (2). We need to mark the equations in some such way so that we can refer to them readily. Whenever, in what follows, either of the given equations has its form changed, it will no longer be marked (1) or (2), but these figures will have what are called subscripts, that is, small figures written under and after them. Thus. (li) means eq. (1) has been changed in form; (1 2 ) means (li) has been changed ; and so on. SOLUTION. (1) 2 x + 3 y = 13. (Sub. Ax.) (Div. Ax.) JU * If so desired , Chapter VI can be taken up before Chapter V. 154 ELIMINATION BY SUBSTITUTION 155 We now substitute this value of z, , in place of x in the second a equation. (See definition of simultaneous equations, p. 158.) Thus, (2) 3 x + 4 y = 18 becomes 3 / 13 ~ 3 ?/\ + 4 y = 18. \ 2 / ,-. SJL^^J _|_ 4 y = 18. (Multiplication of fractions.) tt 39-9?/ + 8?/ = 36. (Ax.?) -y=-S. (Ax. ?) y = 3. Ans. (Ax.?) To find the value of x, we substitute the value of ?/ just found in (12), the most convenient to use of the equations which contain both x and y. Thus, putting 3 for y, we have, 13 ~ 3 x 3 x = = 2. VERIFICATION. (1)2x2 + 3x3 = 13; (2)3x2 + 4x3 = 18. (2) < ' l2a; f 12 5 a? - 3 (a? - y) = 1 3. 12 a? - - 1 7 = 4. (2) 156 SIMULTANEOUS EQUATIONS SUGGESTION. First, we clear each equation of fractions. Next, taking the second equation, we transpose so that the y terms will be on the left side and all the other terms on the right side, and then the value of y is easily found in terms of x, that is, the value of y contains x. We now substitute this value of y in the first equation, getting an equation containing only one unknown, like those in 87. Solving this equation, we substitute the value of x found in one of the given equations to get the value of y. Ans. x = |, y = . 15. 2' 16. 13 7 S+21.1T. b. The student has probably observed that we can solve either equation for either unknown and substitute the value found in the other equation. As a rule we take the simpler equation and solve it for the unknown with the smaller coefficient. It is usually best to have the other equation in its simplest form before substituting. 17. 19. 'x -\- x T x + y x y 18. x + y =1-* 2 = 8, 20. i 4 3ft 2 __3j/ + 7 x 21. I 2 a? + 5 y = 35. 1 5 ft - 33 y = 0. 23. Make a rule for solving simultaneous .equations containing two unknowns, using elimination by substitution. : 24. 3ff+4?/ + 3 2ft + 7-?/__ 5 ?/--8 10 15 5 ' 9 7/4. 5 ft --8 ft + ?/ _ 7 ft + 6 12 4 11 "' ELIMINATION BY ADDITION AND SUBTRACTION 157 25. 26. 7 + x 2x-y = _ 3 - 5?/- ^ 4a;-3 J8 1.5 4.2 -.7. .5z-^2 41 "5 1.6 27. 6# + 9 3 a; + 5 y = 4 4 a; 6 16 + 4 10 7 6a?-3y " - 4) -4 = 100. Elimination by Addition and Subtraction. 1. Given (1) 2x - 9 y = 11 ; (2) 3 a; - 12 y = 15. SOL. (2i) 6 x - 24 y = 30 (Mult. Ax.) (li) 6 x - 27 y = 33 (Mult. Ax.) 3 y = - 3 (Sub. Ax. Explain.) 2/ = 1, vlns. (Div. Ax.) (1) 2 x - 9(- 1) = 11 ; or 2 x + 9 = 11 ; whence x = 1. .4n*. VERIFICATION : (1)2x1- 9(-l) =11, or 2+ 9 = 11; (2) 3 x 1 - 12(- 1) = 15, or 3 + 12 = 15. EXPLANATION. Observe that equations (1) and (2) were multiplied through by such numbers as made the coefficients of x the same in the two equations. If these coefficients (as and 6 in (2i) and (li) above) have like signs, one equation is subtracted from the other to eliminate the unknown. If the two arithmetically equal coefficients have unlike signs, one equation is added to the other. Strictly speaking, it is not the equations that are added, or subtracted, but only their corresponding members, though this language is often used. Evidently if it is more convenient to do so, instead of making the coeffi- cients of x the same, one can make the coefficients of y the same. 2. f4+9y=61, > I Qx y = l. 6. 4. 7. 158 SIMULTANEOUS EQUATIONS g flloj+6y=64, 9 | 7 a?- 9^ = 7, 1Q f 1.5a?-3.7y=-56, il3a;+102/=96. ' la?-2y = 3. u . (i) +2y + SOLUTION 12- (11) 9a;+122/-6a: + 4y = 252 (2i) 45- 20^=000+3 a; (Mult. Ax.) (1 2 ) ax + 16y=262 (2 2 ) 42 a;- 20?/ = 600 (Ax.?) (1 3 ) 42 a; + 224 y= 3528 (Ax. ?) 244^=2928 (Sub. Ax.) ?/ = 12 (Ax. ?) (1 2 ) 3 x + 192 = 252, whence x = 20. VERIFICATION. (1) %- + 24 - 20 + - 2 /-=42, (2) GO - -^ = 40 + ^>-. 3a?-5y__2a; + 4 , - 17. Make a rule for eliminating by addition and subtraction by answering the following questions : (1) If the given equations contain fractions, what is done ? (2) When is transposing necessary and how is it done ? (3) How is the unknown to be eliminated selected ? (4) What is the coefficient of the unknown to be eliminated made in each of the next pair of equations ? Ans. The least common multiple of the two coefficients of this unknown in the two equations just found. ELIMINATION BY ADDITION AND SUBTRACTION 159 (5) How is the number by which to multiply the first equation found ? The number by which to multiply second equation ? (6) After the coefficients of either x or y are made the same in the two equations, what is done if these coefficients have the same sign ? If they have opposite signs ? (7) After the value of one unknown is found, how is the other found ? (8) In what equations is the verification performed ? 18. -13 = 19. = _; 9a; _ = x-y 8 7 20. = . ? 14 2 i. ii O ! 7 O / 11 -# , 4 a; -f 8?/ 2 22 -- = . --_. 5 6 3 "G 3 6 23. ^-20- = - ' '23 -x 2 a? -18 3 x+5 6x-2. 2y-3 , . . " -~~' ~~ + = ~~' ( 25. (1) - = 2; (2) + = . ; x y ' ^ xy 3 In the solution which follows we treat - and - ('called the reciprocals of x x y and / as the unknowns. 160 SIMULTANEOUS EQUATIONS (li) -- =6 (Mult. Ax.) x y (2i) + = (Mult. Ax. ) x y 3 29 58 , c , . , - = - (Sub. Ax.) y 3 87 = 58 y. (Mult. Ax.) VERIFICATION. 58 2 (1) - = 2, whence x = 1. x 4 0) ?= 26. 28. 30. 1 + 1 - 1 x + ^"2' .05 14 0? L = l /~6' 25 = 7 49 _50__ o . x y a; 9 y 22 8 = 15* _ 3 Z 25 ?/ 27. 9 5 y 10 -- = 1. *29. [21.1 , 12 1 -p -iw = 5 X y 1 i i "#~~42* 31. 2 ?/ x = 4 #?/. 101. Elimination by Comparison is performed by finding the value of the same unknown from each of the two given equations and setting these values equal to each other. (See Ax. 7.) This method is rarely better than one or other of the two methods just learned, and exercises on it are not given. 102. Miscellaneous Exercise in Elimination. The choice of the method in particular problems is left to the student, but he should not confine himself to one. EXERCISE IN ELIMINATION 161 1. 3. 5. / J fl o; 1 Jl Y 3 + 5 ' 4 ' (> 2* 2. / O* ?/ ~ i~\ Ox' y 1 I 3 5~ ' [3 8 2 x 4 y 4- 4 r u + 3 5 a,. _3 y + 7' V + 3 ~6' 4. . l; + 5_ 2 /-l u -3 2 U + 2 y-2 .V -3 3 (2x 3y 00 r 5 7 3 4 X 4-2 y 2 x + y 6. y _|_ 1 00 7 5 ^/ 1 J-V/V7 .3x-2 Gy-7 7. 4 ^ + 5 ?/ 25 = 3, 8 x + ?/ 4- 6 -L^- - = o. 8. 23 -a? 4- - a; -18 = 73 -3 ( 88.) 9. 2y 5-z^41 2y l 3 2 : 12" 4 *10. 11. ^ 2a; + 8 10 15 5rc--15 a; y + 1 _ 7 a; 4- 12 12 3 11 12, 18 x y 4 r=l( ;+: : J+iV ( 100 ' 25 > a? ?/ oVa; ?/y 12 162 SIMULTANEOUS EQUATIONS 103. Special Methods of solving Simultaneous Equations. Systems of Equations. Instead of eliminating immediately, it is often better to derive one or more new equations from the given ones, and let such equation or equations take the place of one or more of the given ones. 1. (1) 9 a + 130 = 184, (2) 13^ + 190 = 268. SOLUTION. (h) 27x + 39y = 552 (Ax. ?) (2i) 26 x + 38 y = 536 (Ax. ?) (3) x+ y= 1(J (Ax. ?) (3i) 9 x + 9 y == 144 (Ax. ?) (1) 9a+13y== 184 4y- 40 (Ax. ?) y= 10 (Ax. ?) (3) +10= 16;.-. X = 6. VERIFICATION. 9 x 6 + 13 x 10 = 184; 13 x 6 + 19 x 10 = 268. 2. (1) 140-170 = 159; (2) 29 x ~ 37 y = 324. SUGGESTION. Multiply (1) by 2 and subtract from (2). *3. (1) 1390+152^=1377; (2) 35 x + 37 y = 348. 4. (1) 755 x - 564 # = 2074 ; (2) 1133 x- 847 y = 3113. 5. (1) x+ 500 = 557; (2) 500 + = 361. SUGGESTION. First add the two equations and divide by 51 ; then subtract (2) from (1) and divide by 49. a. In each of the preceding problems (1) and (2) constituted the given "system of equations." In exercises 1, 2, 4, the given system is replaced by the system composed of equations (1) and (3). In exercise 3, it is better to replace the given system by (2) and (3). In exercise 6, the given system is replaced by (3) and (4). The new systems in each case have the same values of x and y because they were obtained by reversible axiomatic processes. These new systems are formed because in such problems as the preceding, they greatly shorten the labor of calculation. Thus, by actual count, to solve exercise 2 by regular process would require 99 figures ; the above method calls for only 46 ; besides, the labor is easier to perform. 6. (1) 590 + 730 = 390$; (2) 73 x + 50 // = 379J. 7. (1) 23 x - 41 y = - 2y, ; (2) 39 - 2,1 y = 6|. 8. (1) 2S0-350 = r>G; (2) 29 -x- 1, //= 151. PROBLEMS --TWO UNKNOWN NUMBERS 163 104. Problems involving Two Unknown Numbers. The sugges- tions given in 94 apply equally here. 1. A grocer has two kinds of sugar, one worth 5 4 and the other 6^ a pound. How many pounds of each sort must be taken to make a mixture of 100 Ib. worth 5J^ a pound? SOLUTION. Let x no. of Ib. of 5 ^ sugar, and y = no. of Ib. of 6 ^ sugar ; then 5 x no. of cents in cost of 5 ^ sugar, and 6 y no. of cents in cost of 6 / sugar. Now by the conditions of the problem we have two equations : (l)x+y=lQQ. (2) 5x + 6ij = 575 (li) 5 x + 5 y = 500 (Ax.?) y= 75 (Ax.?) (1) .-.x= 25 VERIFICATION. 5 x 25 + 6 x 75 = 100 x 5f ; also 25 + 75 = 100. 2. Find two numbers whose sum is 196 and whose difference is 8. 3. In a meeting of 324 persons a motion was carried by a majority of 34, all voting. How many voted aye, and how many no? 4. A father said to his son, " After 3 years I shall be 3 times as old as you will be, and 7 years ago, I was 7 times as old as you then were." What were the ages of father and son ? 5. A farmer paid 4 men and 6 boys $10.80 for laboring one day, and afterwards he paid 3 men and 9 boys $12.15 for one day. What were the wages of a man and what the wages of a boy? 6. If 1 is added to the numerator of a fraction, the resulting fraction will be equal to one fourth ; but if 1 is added to the denominator, the resulting fraction will be equal to one fifth. What is the numerator, and what is the denominator, of the fraction ? 164 SIMULTANEOUS EQUATIONS 7. What fraction is that whose numerator being doubled and denominator increased by 7, the value becomes f; but the de- nominator being doubled and the numerator increased by 2, the value becomes -| ? O 8. Thirty feet more than ^ of the present height of the highest of the Egyptian pyramids is equal to 1 of its original height; 300 ft. less than 6 times its present height is equal to 5 times its original height. Find the original and present height. 9. Five dry quarts and 10 liquid quarts equal 15 liters, and 7 dry quarts and 4 liquid quarts equal 11^ liters. How many liters are there in a dry quart and how many in a liquid quart ? 10. One metric ton and 400 kilos equal 3080 lb., and 4 metric tons and 100 kilos equal 9020 lb. How many pounds in a kilo ? 11. Fifty laborers were engaged to remove an obstruction on a railroad ; some of them by agreement were to receive $ 1.80 per day and others $3. There was paid them just $ 96. No memo- randum having been made, it is required to find how many worked at each rate. 12. A merchant received from one customer $26 for 10 yd. of silk and 4 yd. of cloth ; and from another customer $ 23 for 7 yd. of silk and 6 yd. of cloth at the same prices. What was the price of the silk and of the cloth ? 13. A mechanic and an apprentice together receive $40. The mechanic works 7 days and the apprentice 12 days; and the mechanic earns in 3 days $7 more than the apprentice earns in 5 days. What wages per day does each receive ? 14. I have 7 silver balls equal in weight and 12 gold balls equal in weight. If I place 3 silver balls in one pan of a balance and 5 gold balls in the other, I must add to the gold balls 7 ounces to maintain equilibrium. If I place in the pan 4 silver balls and in the other 7 gold balls, the balance is in equilibrium. What is the weight of each gold and of each silver ball ? 15. When the greater of two numbers is divided by the less, the quotient is 4 and the remainder 3 ; and when the sum of the PROBLEMS TWO UNKNOWN NUMBERS 165 two numbers is increased by 38 and the result divided by the greater of the two numbers, the quotient is 2 and the remainder 2. SUGGESTION. To find the quotient, subtract the remainder from the divi- dend and then divide by the divisor. 16. Henry expended 95 ^ for apples and oranges, paying 5 f for each orange and 4 ^ for each apple. If he had 22 of both, how many of each did he buy ? 17. The admission to an entertainment was 50 f for adults and 35 ^ for children. If the proceeds from 100 tickets amounted to $ 39.50, how many tickets of each kind were sold ? 18. An apple woman bought a lot of apples at 1 f each and a lot of pears at 2 ^ each, paying $ 1.70 for the whole : 11 of the apples and 7 of the pears were bad, but she sold the good apples at 2^ each and the good pears at 3 $ each, receiving $2.60. How many of each fruit did she buy ? 19. A boy bought 570 oranges, some at 16 for 25^ and the remainder at 18 for 25 p. He sold them all at the rate of 15 for 25 ^ and made a profit of 75 ^. How many oranges at each price did he buy ? 20. Up to 1908 the highest speed for a short distance on a railroad was made March, 1901, by a train on the Plant System between Fleming and Jacksonville, Fla. The next highest was made July, 1904, on the Philadelphia and Reading, between Egg Harbor and Brigantine. The time being taken the same for both, the distance run by the first train was to the distance run by the second train as 25 is to 24, and the difference in rates in miles per hour was 4.8. What was the rate of each train in miles per hour? SUGGESTION. The time being the same, distances run are proportional to rates. 21. A gasoline launch made a trip up a river and back in 4 hr. Had it made a trip 1 mi. more in total length on a lake, it would have been gone only 3| hr. The rate of the launch in still water was 3 times as great as the rate of the river current. What was the round trip distance, and what the rate of the launch ? 166 PROBLEMS 22. The sum of the two digits of a number is 8, and if 36 is added to the number, the order of the digits is reversed. Find the number. SUGGESTION. In order to represent numbers in the Arabic scale by letters, we must proceed as follows : Let t = the number represented by the figure in tens' place, and u -. - the number represented by the figure in units' place. Then 10 + u = the number, and 10 u + t the number when the figures are reversed. 23. Find a number which is greater by 2 than 5 times the sum of its digits, and if 9 is added to it, the digits will be reversed. 24. If a certain number is divided by the sum of its two digits, the quotient is 6 and the remainder 3; if the digits are interchanged and the resulting number is divided by the sum of the digits, the quotient is 4 and the remainder 9. What is the number ? 25. Find that number of two digits to which if the number found by changing its digits is added, the sum is 121 ; and if one of the two is taken from the other, the remainder is 9. 26. Two persons 27 miles apart, setting out at the same time, are together in 9 hr. if they walk in the same direction ; but if they walk in opposite directions towards each other they meet .in 3 hr. Find their rates. 27. Two trains set out at the same moment, the one to go from Boston to Springfield, the other from Springfield to Boston. The distance between the two cities is 98 miles. They meet each other at the end of 1 hr. and 24 min., and the train from Boston travels as far in 4 hr. as the other does in 3 hr. What is the speed of each train ? 28. A pound of tea and 10 Ib. of sugar cost $1. But if tea were to rise in value 10% and sugar 20%, they would cost $1.15. Find the price per pound of each. 29. A boat's crew can row 15 mi. an hour downstream. The crew can row a certain distance in still water in 15 min. and require 20 min. to row the same distance up the stream. Find the rate of rowing in still water and the rate of the stream. PROBLEMS --TWO UNKNOWN NUMBERS 167 * 30. If the sides of a rectangular field were each increased by 2 yd., the area would be increased by 220 sq. yd. If the length were increased and the breadth diminished each by 5 yd., the area would be diminished by 185 sq. yd. What is the area? 31. A rectangular field has the same area as another which is 6 rods longer and 2 rods narrower, and also the same area as a third which is 3 rods shorter and 2 rods wider. Find its dimensions. 32. The circumference of the large wheel of a carriage is 55 in. more than that of the small wheel. The former makes as many revolutions in going 250 ft. as the latter makes in going 140 ft. Find the number of inches in the circumference of each wheel. 33. A man invests $10,000, part at 41% and the rest at 31%. He finds that 6 yr. interest on the first investment exceeds 5 yr. interest on the second by $1098. How much does he invest at each rate? 34. The report of an explo- o- = vekc. of .sound sion traveled 5560 ft. in 5 sec- - onds with the wind, and 3204 ft. y= Vl ' loc - of Avin<1 in 3 seconds against the wind. ./ + //= vi-i...-. with win.i. Find velocity of sound in still > air and velocity of wind. *~ y= veloc ' affalnst wiml 35. Two men can do a piece of work in 30 hr. ; they can also do it if the first man works 251 hr. and the second 324 hr. In o O how many hours can each alone do the work ? 36. Portland cement, when pure, contains 65 % lime and 35 r / silica and aluminium, that is to say, 10 cu. ft. of it would contain 6.5 cu. ft. lime and 3.5 cu. ft. silica and aluminium. If made from limestone rock containing 60% lime and 24% silica and aluminium and clay containing 2% lime and 95% silica and alu- minium, how many cubic feet of limestone and how many cubic feet of clay should be mixed and burnt together to make 10 cu. ft. of cement? Ans. 10.8 cu. ft. and .95 cu. ft. 168 SIMULTANEOUS EQUATIONS SIMULTANEOUS EQUATIONS WITH THREE UNKNOWNS 105. Simultaneous Equations containing Three or More Unknown Numbers. Such equations are solved by repeating the process of elimination. 1. (1) x-2y + 3z = 6, (2) 2.4-3?/-4z (3) 3x-2y + 5z On inspection we perceive that of the three unknown numbers x can most readily be eliminated. (2) 2x+3y- 4z 2 .c - 4 ?/ + 6 z = 12 (Mult. Ax.) (4) 7y-10z= 8 (Sub. Ax.) (3) 3x-2y+ 50 = 26 (1 2 ) 3 x - 6 y + 9 z = 18 (Mult. Ax.) (5) 4?/- 4z = 8 (Sub. Ax.) y - z= 2 (Div. Ax.) Taking now equations (4) and (5j), we proceed as in solving simultaneous equations containing two unknowns. ( 100.) (5 2 ) 7 y - 7 z = 14 (Mult. Ax.) (4) 7y-10z= 8 3 z = 6 (Sub. Ax.) z 2 Ans. (Div. Ax.) (5^ y 2 = 2 (Substituting its value, 2 ; for 2.) ?/ = 4 ^4??s. (Sub. Ax.) (1) x - 2 X 4 + 3 x 2 == 6 (Substituting their values for a 8 + 6 = 6 y and z) x = S Ans. THREE UNKNOWN M .MI5EKS 169 VERIFICATION. Since the value of x was just found from (1), it is scarcely necessary to verify in (1). (2) 2xS+3x4-4x 2= 20, or 16 + 12 - 8 = 20, (3) 3 x a- 2 x 4 + 5 x 2 = 26, or 24 - 8 + 10 = 26. Observe tha.t we (1) Begin by choosing the unknown that can be most easily eliminated. (2) Eliminate this unknown between two of the equations, getting equation (4), and then eliminate the same unknown between another pair of the given equations, getting equation (5). (3) Take equations (4) and (5) and solve as in 100, getting the values of both unknowns. (4) /Substitute the two answers found in one of the given equations, getting the value of the unknown first eliminated. (5) Verify in the other two equations. 2. 4. 4z=29, 3. x + 3y + 2z = 25. x + y + z = 6, 3x-2y + z = 2, 5. x 3y + 3z = 25. 2y z-- 6, 6. -42 = 98, 9 x 15 y + 6 z = - - 6, 80? 3 9=12. 7. 5 x - - 11 ?/ z = 7, 8. 9. -- = -20, -8 =-27. 106. Special Processes of Elimination. Yl) 3 x + 4 11 = 25 SI-GGESTION. Combine (1) and (2), 1 (2} 5 ?/ -I- f>"z = 50 eliminating y, and getting equation (4). Then taking (-4) and (3) together, solve J 107. Review. 1. Sim. Eq's.; 2. Elimination; 3. Methods of Eliin. 170 PROBLEMS IN THREE UNKNOWNS 108. Problems involving Three Unknowns. 1. Determine three numbers such that their sum is 9 ; the sum of the first, twice the second, and three times the third, 22; and the sum of the first, four times the second, and nine times the third, 58. 2. A and B together possess only -| as much money as C ; B and C together have 6 times as much as A ; and B has $ 680 less than A and C together. How much has each ? 3. Three men bought grain at the same prices. A paid $5.70 for 2 bu. of rye, 3 bu. of wheat, and 4 bu. of oats ; B paid $ 7.40 for 3 bu. of rye, 5 bu. of wheat, and 2 bu. of oats ; and C paid $ 6.20 for 2 bu. of rye, 4 bu. of wheat, and 3 bu. of oats. What was the price of each ? 4. Divide 800 into three parts such that the sum of the first, of the second, and -| of the third shall be 400 ; and the sum of the second, f of the first, and ^ of the third shall be 400. 5. A quantity of water sufficient to fill three jars of different sizes, will fill the smallest jar four times ; the largest jar twice with 4 gal. to spare; or the second jar three times with 8 gal. to spare. What is the capacity of each jar in gallons ? 6. The average age of three persons is 41 yr. The average age of the first and second is 37 yr., and of the second and third is 29 yr. Find their ages. (See 153.) 7. A number is expressed by three figures whose sum is 11. The figure in the place of units is double that in hundreds' place, and when 297 is added to this number the sum obtained is ex- pressed by the figures of the number reversed. What is the number ? SUGGESTION. Let x = figure in hundreds' place, y = that in tens' place, z = that in units' place. Then the number is 100 x + 10 y + z. (See 104, 22.) 8. The middle digit of a number of three figures is one half the sum of the other two digits. If the number is divided by the sum of its digits, the quotient is 20 and the remainder 9 ; PROBLEMS 171 and if 594 be added to the number, the digits will be reversed. Find the number. 9. In round numbers in 1904 New York City, Boston, and Chicago paid out to the public together 20.4 millions of dollars merely as interest on the indebtedness of these cities. Three times Boston's amount increased by twice Chicago's equals 2 millions more than New York's ; and 3 times Boston's and Chicago's together is 4.4 millions more than New York's. What was the interest paid by each in round numbers ? 10. For running expenses in 1904 the cities of Washington, D.C., Minneapolis, Minn., and Indianapolis, Ind., paid together in round numbers 12 million dollars. Twice Indianapolis's amount increased by Minneapolis's was $800,000 more than Washing- ton's, and the Washington and Minneapolis sums together were $ 1,200,000 less than 5 times Indianapolis's. What was the running expense of each? 11. If a circle is inscribed in a triangle as shown in the figure, it is proved in geometry that /( \B AF = AD, DB = BE, CE = CF. Now if AB = 12 in., EC = 8 in., and AC 16 in., find the lengths of AD, BE, and CF. 12. A circle is inscribed in a triangle ABC, as in the figure, whose perimeter (distance around it, or AB -f- BC -f- CA) equals 42 ft. Now AF exceeds DB by 6 ft., and CF exceeds DB by 3 ft. Find the sides of the triangle. 13. The total value of farm property of continental United States according to the census of 1900 was in round numbers 20.5 billions of dollars. The land exceeded double the value of the buildings added to the value of the live stock and implements by 2 billions ; and the land and buildings together exceeded 4 times the live stock and implements by ^ billion. What was the value of the land, buildings, and live stock and implements respectively ? CHAPTER VI uRAPHS AND GRAPHICAL SOLUTION OF SIMPLE EQUATIONS 109. Graph Paper is paper ruled very accurately into small squares or parallelograms. (See graph paper ruling in the dia- gram of the next article.) Formerly such paper was very ex- pensive, but of late years it has been manufactured in increasingly large quantities, and is now sold very much cheaper. The paper is commonly ruled in units of the metric system, the centimeter being the standard unit. A centimeter is one hun- dredth of a meter, just as a cent is one hundredth of a dollar. A millimeter is one tenth of a centimeter or one thousandth of a meter, just as a mill is one thousandth of a dollar. Now a meter is somewhat longer than a yard, since it equals nearly 39.37 inches. Hence a centimeter is .3937 inch, or very close to A inch. A milli- meter is one tenth of A inch, or .04 inch, or -^ of an inch. The larger squares on the graph paper are square centimeters, and the little squares are therefore 2-millimeter squares. 110. Location of Places by Reference to Two Standard Lines. As an example of such location of places take the case of townships, or towns, as they are often called. In the figure, p. 173, let the centimeter squares denote town- ships, XX 1 being a "base line," and TY' a " principal meridian." Then any township, as the one marked with the letter A, is described as Town 3 North, Range 2 East; the township marked B, as Town 4 North, Range 3 West ; the township marked (7, as Town 2 South, Range 1 West; etc. Townships are divided, of course, into 36 sq. mi., and not into 25 squares, as in figure. 172 AXES 173 o A Y- We desire here, however, to locate points, and not squares. In the West, roads usually run on section lines. If a village is at a cross-roads, any other cross-roads can be located by saying it is so many miles east or west, and so many miles north or south of the village. Similarly in a city in which the streets run east and west, and north and south, a street corner can be located in the same way by refer- ence to some square. As another exam- ple, the location of places on the earth's surface can be taken. Every such point can be located by saying it is such and such a number of degrees north or south of the equator, and such and such a number of degrees east or west of the standard meridian. 111. Axes. Two lines are taken at right angles as lines of ref- erence, and are called axes. The horizontal one, XX' (see figure), is called the X axis, and the vertical one, FF', the F axis. Dis- tances measured to the right of FF' are positive ; those to the left of FF' are negative. Distances above XX' are positive; those below XX ' are negative. The point where the axes cross is called the origin. The axes divide the diagram into four parts called quadrants. A is in 1st quadrant, B in the 2d, C in 3d, and F in 4th. 174 GRAPHS 112. Location of Points. Any point in the plane of XOY, as P 1 (see figure below), is located by measuring the perpendiculars, or shortest distances, from this point to the reference axes. Dis- tances measured parallel to the X axis, as 2\B, are denoted by x, -X Unit -IV*- Y- Pr fl -A- p; and are called abscissas; and distances parallel to the Faxis, as PiAj are denoted by y, and are called ordinates. The abscissa and ordinate of any point are called the coordinates of the point. Centimeter being unit of measure, the coordinates of PI are 2, 1 (x = 2, y = 1) ; that is PB = *2, PA = 1 ; of P 2 are - 2, .4 (x - - 2, y = .4) ; that is PoO = - 2, P 2 D = A ; of P 3 are - - 1.6, --\A (x= -- 1.6, y = - 1.4) ; for signs see 111 ; of P 4 are 3, -- 2.8 (x = 3, y = - 2.8) ; for signs see 111 ; of P 5 are 0, - 2 (x = 0, y = 2) j for sign see 111. LOCATION OF POINTS 175 1. Write the coordinates of the points located by the letters a, b, c } d, e, /, g, h on the diagram. 2. Take a piece of squared paper and draw two heavy lines (on two centimeter division lines) to represent the two axes, just as was done in the two diagrams already given. Now locate the following points by a dot on the diagram, putting a small letter beside the point located, writing a beside the first point, b beside the second, and so on. a. (x=2,y=4:). * b. (# = 4, ?/ = 4). c. (x=-3, 2/=3.4). d. 0=7, y= --1). e. (x = 3, y = 4). /. (x = 2.2, y = 0). g. (#= 1.8, #= .6). h. (x = Q, 2/ = 2.5). (a; = 2, y = 0). 3. Make rule for locating points, their coordinates being given. (1) Where does one start from to count out the units in x? Ans. From the origin at 0. (2) If x is given positive, in which direction is it measured ? If x is negative, in which direction is it measured ? (3) Having arrived at the extremity of x, from there in which direction is y measured if it is positive ? If negative ? (4) Where is the point sought now found ? 4. Make another diagram by drawing axes at another place on the sheet of squared paper, and locate the following points, understanding that the first number inside any parenthesis gives the value of the abscissa, or x, and the second number the value of the ordinate, or y. a. (1, 1). 6. (2, 4). c. (- 2, 1). d. ( - 3, - 3). e. (-3, 6). /. (- 2, - 4). g. (- 5, 0). h. (7, - 2). i. (0,0). j, (0, -3). k. (-2, -8). 1. (10,11). 113. Graphs. If a series of points is " plotted " (that is, located and marked on a diagram) representing values of a quan- tity that changes, and these points are joined by a running line, this line is called the graph of the law or data that determined the points. (See 73. It can replace 113, if teacher so desires.) 176 GRAPHS 1. Graphs can be used to show to the eye changes in tempera- ture. The following graph is obtained from the data contained in the table at its right. The hours counting from 12 noon are abscissas, and the number of degrees at these hours are the ordi- nates of the points marked on the figure. in- -12J LE..M, r > I. \ \ M, 12 M. 15 3 P.M. 17 6 P.M. 14 9 P.M. 11 12 A.M. 9 3 A.M. 6 6 A.M. 4 9 A.M. 9 12 M. 11 2. Construct a similar graph being given the following data : 12 M., 28; 3 P.M., 20; 6 P.M., 14; 9 P.M., 9; 12 P.M., 7;3 A.M., 5; 6 A.M., 3; 9 A.M., 11; 12 M., 18. 3. Construct the graph for the following readings actually taken at a certain place in the month of December : 12 M., 28 ; 3 P.M., 10; 6 P.M., 4; 9 P.M., -1; 12 P.M., -5; 3 A.M., -6; 6 A.M., -7; 9 A.M., -3; 12 M., 5. 4. The graph, p. 185, made from the data below, shows changes in Public Debt of the United States from 1830 to 1907. 1830, $49 millions. 1864, 1816 millions. 1894, 1632 millions. 1840, 4 millions. I860, 2773 millions. 1900, 2136 millions. 1850, 63 millions. 1870, 2481 millions. 1907, 2492 millions. 1860, 65 millions. 1880, 2120 millions. 1862, 524 millions. 1890, 1552 millions. GRAPHS OF STATISTICS 177 In this problem it is convenient to let a centimeter along the X axis represent 10 years, and a centimeter along the Y axis repre- sent 500 million dollars. Mill ons \ 200(1 \ 1500 1000 X- Y' 1830 J840 1850 1860 1870 1880 1890 1900 This graph tells the story of the Civil War in a most vivid way. 5. Construct the graph corresponding to the following table of figures, representing expenditures in round numbers of the United States government for every fifth year, 1870-1905. 1870, $294,000,000. 1885, .$260,000,000. 1900, $488,000,000. 1875, 275,000,000. 1890, 298,000,000. 1905, 567,000,000. 1880, 265,000,000. . 1895, 356,000,000. 6. Construct graph for Post-office Expenditures, 1865-1905. 1865, $15,000,000. 1880, $37,000,000. 1895, $ 90,000,000. 1870, 23,000,000. 1885, 49,000,000. 1900, 110,000,000. 1875, 36,000,000. 1890. 09,000,000. 1905, 169,000,000. SUGGESTION, Put one T centimeter = $20,000,000. 178 GRAPHS 7. Compare by the graphical method the growth of population of the United States and the United Kingdom (England, Ireland, and Scotland), 1800-1900. STATISTICS OF POPULATION OF UNITED STATES IN ROUND NUMBERS 1800, 5,300,000 1840, 17,100,000 1880, 50,200,000 1810, 7,200,000 1850, 23,200,000 1890, 62,600,000 1820, 9,600,000 1860, 31,400,000 1900, 76,300,000 1830, 12,900,000 1870, 38,600,000 STATISTICS OF POPULATION OF UNITED KINGDOM IN ROUND NUMBERS 1800, 15,700,000 1841, 27,000,000 1881, 35,000,000 1810, 17,600,000 1851, 27,000,000 1891, 38,000,000 1820, 21,000,000 1861, 29,000,000 1901, 42,000,000 1831, 24,000,000 1871, 31,000,000 Put both graphs on the same figure, using the same axes for both. 8. Construct a graph for the movement of train No. 25 on the Boston and Maine which ran on. the time card annexed. hr hi, 2 Miles 10 20 30 40 50 MILES STATIONS Boston 4 10 20 Cambridge Waltham Concord 36 Ayer 50 Fitchburg A.M. 4.40 4.49 5.00 5.17 5.58 6.37 Where the graph consumes time faster than distance, the train is going slower ; where the reverse is true, it is going faster. 9. A train on the Pennsylvania Railroad ran on the following schedule : Pittsburg, 8 A.M. ; Irwin, 21 mi., 8.43; Latrobe, 40 mi., RENE DESCARTES (1596-1650) GRAPHS OF SIMPLE EQUATIONS 179 9.18 ; Johnstown, 76 mi., 10.17 ; Cresson, 99 mi., 11.01 ; Altoona, 114 ini., 11.36; Huntingdon, 148 mi., 12.34 P.M.; Newport, 218 mi., 2.16 ; Harrisburg, 245 mi., 3.00. Construct a graph for its movement, letting 1 cm. = 25 ini. and 1 cm. = 1 hr. How do you account for its irregularities ? 10. Scientists construct all sorts of apparatus to register graphs automatically. A good illustration is the thermograph (see drawings in Century and Standard Dictionaries). A cylinder covered with paper is revolved by clockwork, and a pen rules a line on it as it revolves. The position of the pen is controlled by the temperature of the air. As a piece of metal to which the pen is attached expands or contracts, it moves the pen up or down. The graph is a continuous wavy line, the high parts of which show a higher temperature and the low parts a lower temperature. The hours of the day are printed on the paper. Other graph apparatus that may be mentioned are the baro- graph, the seismograph, the anemograph, etc. 11. Graphs are used by writers on history and economics to show changes in population, expenditures, production, etc. ; by physicists, scientists generally, engineers, etc., to express laws of nature, the working of machinery, etc. ; by business houses to show changes in prices, sales, cost of production, etc. In short, graphs have a wide range of uses, and the student should learn well how to construct and read them. 114. Graphs of Simple Equations containing Two Unknowns. An equation containing two unknowns, as 3 x -+ 4 y 9, can have a graph, because for every value of x the equation gives a cor- responding value of ?/, and for every such pair of values we get a point located ; joining several such points, a portion of a graph results, called the graph of the equation. The graph of an equa- tion has been called its geometric picture. The idea of constructing graphs of equations was conceived by the French philosopher Descartes and published to the world in the year 1637 A.D. The full development of this idea is called 180 GRAPHS analytic geometry, which is a college study. In analytic geome- try a graph is called the locus of a point. 1. Construct the graph corresponding to the equation 2 x + 3 y = 6. SOLUTION. We first solve the equation for y in terms of x : y = 6 ~ 2x (Axioms ?) We next find values of y which correspond to assigned values of x in this way: Let x 4 ; then y=- - .67 (to nearest hundredth). o o (\ 9 v 3 x = 3 ; then y = - = 0. o In the same way we get the value of y corresponding to other assigned values of x. The results of all these calculations are put in the following table, and the points located by the sets of coordinates in the table are marked on the graph diagram, and a line is drawn through them. Each decimal value is taken to the nearest hundredth. TABLE s Y // "S x X ^ x y x s s X I ^ ~^ S II s *N \ (' \ \ s \ X s \ ? *s \ / X \ V \ \ s \ (' s x \ X /> \ V, \ K / I \ o ', X \ s s \ X \ 5 \ 7 k, A N \ \ i \ ( *, y) PT. ( 4, -.07) a ( 3, ) b ( 2, .67) c ( 1, 1.33) d ( > 2 ) e (-1, 2.67) f (-2, 3.33) 9 (-3, 4 ) h GRAPHS OF SIMPLE EQUATIONS 181 a. Notice that all the points lie in a straight lino, or would do so if the values of their coordinates were found to more decimal places. It w r ill be found that whenever the given equation is of the first degree (see 57), all the points lie in a straight line. The student should test this statement until he is satisfied of its truth. For this reason equations of the first degree are often called linear equations. Since two points determine a straight line, it will be necessary to find the coordinates of only two points, and not of more, as was done in Ex. 1, p. 180. As a rule the two points where the line crosses the axes (e and 6 in the pre- ceding figure) are found. This is done by letting x= and finding the cor- responding value of y, and then letting y = and finding the corresponding value of x. To check the result, locate a third point at the extreme edge of the diagram, and see whether the line drawn through the first two points goes through it. The student should each time make a table, writing its equation above it. 2. Construct the graph of the equation x + y = 1. y =-x- 1 (, >/) PT. (0, - 1) k (-1, 0) m (-3, 2) n SOLUTION. The values are assigned as suggested in paragraph just given. The graph is constructed on the diagram to Ex. 1 of this article. Notice that the three points k, m, n lie in a straight line. Construct the graphs of the following equations, following Ex. 2 as a model. 3. 7. = 4. 4. 8. 5. x 9. = = 2. 6. 2x 3y = 6. . 10. 4 a;- 11. Make a rule for constructing the graph of 'a simple equation. (1) What value is assigned to x to get corresponding value of y? (2) What value is next assigned to y to get corresponding value of ? (3) What is now done with the pairs of values found ? (4) After the two points are located, what line is drawn ? (5) How is the work checked ? 182 GRAPHS b. When the two points located by putting x and y in turn equal to zero are close together, they do not give a good determination of the graph. In this case get the check point first and draw the graph through it and the more remote of the other two, using the intermediate point to check the result. 12. 15. = 2x 9. 13. 2 a; -3 # = 16. 14. 4 a; + 5 #=24. 17. a; + # = 20. 12a?-15# = = 2 x. SUGGESTION. Here two other points besides (0, 0) are needed. 18. # = --4 --7 a. 19. 05 = 5 #+3. 21. # = 4. SUGGESTION. Locate any two points whose ordinate is 4. The graph is a line parallel to axis of X. -3. 23. x = y. 24. 2a?4-#=20. 25. lla?4-13#= 6. 115. Solution of Simultaneous Equations containing Two Un- knowns by Means of Graphs. This is accomplished by construct- ing the graph of each of the two given equations on the same dia- gram. The coordinates of the point of intersection of the two graphs are the answers sought, that is, are the values of x and # which satisfy both equations ( 98). 22. 1. Solve (1) 3 x -f 4 it = 8 ^ ' (A) 4:X u # = 5 , by the graphic method. (1) 3aj + -x'- ;? X N -x (0,2) a (2.7, 0) 6 (4, - 1) c (2) 4aj~6# = 5 (0, -8) d (1.2, 0) e (4, 1.8) f SIMULTANEOUS EQUATIONS SOLVED BY GUAI'IIS 183 The coordinates of the point of intersection, P, of the two graphs are x = 2, y = |, which are the values of x and y sought. To check the result, we solve the two given equations in the usual way. (See 100.) (li) 12 x + 16 y = 32 (Ax.?) (l)3x + 4x| = 8. (2i) 12 a- 18 y == 15 (Ax.?) .'. x = 2. (Check.) 34 y=-. 17 (Ax. ?) y = \. (Ax. ?) Solve the following by the graphical method and check by one of the algebraic methods ( 99, 100). (If this chapter is studied before Chapter V, this way of checking, of course, cannot be used.) 2. { 2 *+32/=12, 3 (2x-y=3. 5. 8. \ llx-2y=2l, 20+4 y=-18. 1 3 a; 5 #=19. n 6. 9. 12. (3x-2y=S. 7. j X+ ' (x v 3. f4aj-5y=10, l2 I /i /Y* /~i /^ 1 /^ 7^ o* I O /v o* n ^ /v '> ^ / 11 tt*/ n C/// CiO J. V/ C/4*/ "t* . tt 1 *// tl/ O tc/U/ O A X SOLUTION OF LITERAL EQUATIONS 189 12. a -l=' J -9. 13. !+* = . 14. * + * = <;. xx 1 -- x b a b a b a-\-b 3 ac bc 15. - = - Ans. x = - x c x + 2c 26 VERIFICATION. a ~ h = a + b (Substituting its value 3 ac - be Sac -be for x m lven et l ua - 2b 2b tion -) a - b a -f b 3 ac be 2 be Sac be + 4 be . ) (Adding in the denomi- 26 2b 2 b(a -- 6) _ 2 b(a + 6) (Reducing the complex frac- 3 c ( -- 6) 3 c (a + 6) tions to simple ones.) 6bc=6bc. (00, and Mult. Ax.) 1 4- 2 a a; + a 1 x -f- a ft # 5 a 16. - -- = - 17. = 18. - ~+T* oxo xom b a a b J.J7 & + ~ ' L. REMARK. It is often convenient, instead of using new letters, to use the same letters as before, but to mark them in some way. In this problem, instead of using three new letters, d, e, /, we have written in their place a', 6', c', read ' a prime,' ' b prime,' ' c prime.' The a' means a number entirely different from a. The marks a", a'", etc., are also used, read, ' a second,' ' a third," 1 etc. Also ai, ao, s are used in the same way and read, * a snb ewe,' 'a snb two," 1 etc., or merely ' a one,' l a two," 1 etc. SOLUTION. The same method of procedure is followed here as in 100, 1. (20 aa'x + ab'y = ac f ( (2) x a, Mult. Ax.) (li) aa'x + a 1 by = a f c ( (1) x ', Mult. Ax.) ab'y a' by = ac' a'c (Sub. Ax.) _ gc' a'c (Factoring left member, and Div ~ ab'-a'b' Ax.) SIMULTANEOUS LITERAL EQUATIONS 191 Instead of substituting the value of y just found in one of the preceding equations to find the value of oj, it is just as easy to start again with the given equations and this time eliminate y. (li) ab'x + bb'y = b'c ( (1) x &', Mult. Ax.) (2i) a'bx + bb'y = be' ( (2) x 6, Mult. Ax.) (ab 1 - a'b}x --= b'c - be'. (Ax. ?) _ 6/c _ &c' 'ab'-a'b Verification in (1) a . & ' c ~ &c ' + & . qc ' ~ q ' c = c, ab'-a'b ab'-a'b or, a&'c &c' + a&c' a'bc=ab'c a'bc. {mx + ny = p. faa? + &y = l. faic + &v = 2a&. 9. J 10. J 11. c + c??/ = 1 . ( ox-\-ay = a~ o # + a?/ = 6. f ax by = 0. 14. \ ( mx nil = 0. 12. \ 13. ax by = c. ( ex dy = 1. 15. ! 16. a? a +l_a+l 2/+1" 17. a, y 18. a; y SUGGESTION. Solve without clearing of fractions. 19. 22. c ^ see J lu, 25. L 1 ~~~ l# 2/ f- + - = c, a 1 T, - + - = 6, ( a? v 20. \ 21. :/-j /y 1 7^**?/ ^1 J m n 5 = a. T 11 L jc y .a 2/ 'x a a b flO 9 ' a? -(-_?/ a ~i ~~~ ^) y a a-\-b 23. U V 24. xy o c 37 O - . /I* 5 4 +l. TRINOMIALS 128. Factoring of Trinomials Continued. See 52. I. Trinomials that are the Product of Two Trinomials. Type form : mW paW + /? 2 6 4 . TRINOMIALS CONTINUED 195 (1. (9 a 4 + 12 dW + 46*)- 16 a 2 & 2 . (Twice the product of the square roots of 9 a* and 4 6 4 is obtained and placed between these terms (see 116) ; then the last term is chosen so that the given trinomial is unchanged. Thus, 12 a 2 b 2 16 o 2 6 2 = 4 a 2 6 2 .) (2. (3 a 2 + 2 6 2 ) 2 - (4 a&) 2 . ( 52, II.) (3. [(3 a 2 + 2 & 2 ) -f 2 aft] [(3 a 2 + 2 6 2 ) - 2 a&]. (On factoring, 53, III.) (4. [3 a 2 + 2 ab + 2 6 2 ][3 a 2 - 2 a& + 2 6 2 ]. ( 32, 13, l.) (5. PROOF. By actual multiplication. 2. 9 x 4 + 2 x 2 y 2 + y*> 3. z 4. 25 a 4 - 36 aty 8 + 4 ?/ 4 . 5. 9 a 4 - 34 a 2 6 2 + 25 & 4 . 6. Make a rule for factoring trinomials that are the product of two trinomials, giving steps. 7. 9 a 4 + 26 cr& 2 + 25 b\ 8. 4a 4 -13a 2 9. 16 x 4 - 17 x*f + 2/ 4 - SUGGESTION. This problem can be solved in two different ways : (1) by making the middle term of the trinomial square positive ; (2) by making it negative. The reason such a quantity can be solved in two ways can be illustrated in this way : 60 = 2 x 2 x 3 x 5 = 4 x 15, or 6 x 10. 10. 25a 4 -41zy + l62/ 4 . 11. 4a 4 -29a 2 a,' 2 -f 25 12. 4a 4 -5a 2 l. 13. 25 a 4 + 71cr II. Trinomials, the Product of Binomials and Trinomials. 14. 4 z 3 - 43 a- 21. SOLUTION. The trial method is used. 2 x 2 + 7 x + 3 EXPLANATION. As in 52, III, the first coeffi- 2 x 7 _ cients of the factors must give the first coefficient 4 x 3 + 14 x 2 + 6 x of the given quantity, and the last coefficients of 14 x 2 -- 49 x 21 the factors, the last coefficient of the given quan- 4x 3 43 x 21 tity. The middle coefficient of the first factor must be so chosen that the x 2 term disappears in the product, and the middle 196 SUPPLEMENTARY FACTORING term of the given trinomial is obtained. It may cause the student some trouble to find the desired coefficients ; but if he persists he will get a satis- factory combination, provided, of course, the given quantity can be factored. 15. 9z + 5z + . 16. 17. 8ar 5 -24z 2 + 25. 18. 21 x 3 + 26 or + 25. QUADRINOMIALS 129. Factoring of Quadrinomials Continued. See 53. Quadrinomials the Product of a Binomial and Trinomial. m *~** ~ " ~ mxz Type forms : ( ax 3 + 6jr 2 + ex + d. The first type is factored by putting two terras in one paren- thesis and the other two terms in another, and proceeding very much as in 53, II. The second type is factored by the trial method. (1. (4x 2 -25*/ 2 )-(4z2- 10?/2). (See 33.) (2. (4x 2 -25?/ 2 )-22(2z~5?/). (50.) (3. 2x-5?/|(4x 2 -25?/ 2 )-2 g (2x-6y) f c 50 29 ^) 2x + 5y-2z (4. (2 x + 5 y - 2 2) (2 x - 5 y) = 4 cc 2 - 25 ?/ 2 - 4 xz + 10 ?/2. 2 . m 2 -9rt 2 + 21ra?i + 63?? 2 . 3. 4 a 2 6 2 - 169 c 2 + 6 6d + 39 cd. 4. a? 3 6 X 2 -j- 6 a? 1. SUGGESTION. Write as (x 3 -- 1) (6x 2 6 x). 5. m 3 + 5m 2 + 5m + l. 6. 8 a 3 + 6 a 2 ?/ 2 -9 ay*-'27y & . 7. SOLUTION. 2 x 2 + 3 x + 4 (See suggestions under 52, 73.) 5x - 7 - 14 x 2 21y-28 10 r s + x 2 - x - 28 POLYNOMIALS 197 8. 4z 3 -21ar + 44z-30. 9. 6 or 3 -3 a 2 -33 a? -6. a. Quantities like the preceding and those at the end of 128 can often be factored by the method explained in 190, 70. POLYNOMIALS 130. The Factoring of Polynomials of Five or More Terms. I. Powers of a Binomial. II. Square of a Polynomial. These quantities can be readily factored by the student when he becomes familiar with the forms these powers take. This subject of the powers of quantities will be studied in the chapter on Involution. III. Polynomials that are the Difference of Two Squares. (1. ( 2 + 2a&+ &2)_(4 C a_. i2cd+9eP). ( 52, II, 33.) (2. (a + &) 2 - (2 c - 3 dy 2 . (3. [(+ &) + (2c-3d)][(a + 6)--(2c- (4. [a + 6 + 2 c - 3 d] [a + 6 -- 2 c + 3 d]. (5. Proof by actual multiplication. 2. aj 2 --2 xy + y 2 - - m 2 + 2 mn--n 2 . 3. x~-2 4. 4 m 2 - - 12 mn + 9 n 2 -x? - kpq - - 4 q 2 . 5. (a + b + c) 2 rf 2 . SUGGESTION. Factor as J. 2 B 2 . 6. 7. (2 a& + 2 erf) 2 - (a 2 + 6 2 - c 2 - d 2 ) 2 . SUGGESTION. After factoring, each factor can be factored. 8. 4a; 198 SUPPLEMENTARY FACTORING IV. Polynomials Separable into Two Trinomials or a Trinomial and Binomial. 9. 6x 2 llxy + 3y 2 xz + 5yz-- 2 z 2 . SOLUTION. We will use the trial method. 3x y 2z 2x By + g 6 x 2 - 2 xy 4 xz - 9xy + 3xz 6 x 2 11 xy - xz + 3 y 2 + 5 ye 2 z 2 EXPLANATION. The coefficients of x in the factors are chosen to give 6 ; the coefficients of y to give 3 ; the coefficients of z to give 2. If the factors selected do not give the intermediate terms of the given quantity, another combination must be chosen. 10. 2 11. 6a 12. a* 13. 2a 14. 2a 15. x 2 SOLUTION. (1. (x 2 2 xy + ?/ 2 ) + (5 x 5 /) (2. (x-y) 2 + 5(05-0). (3. (a-lOLXs-lO+6]. 16. # 2 + xy - - 12 y 2 + 2 o;z + 8 ?/z. 17. 3ic 2 + 2a;?/--4/ 2 9a; 2 131. Miscellaneous Exercise in Fractions involving the use of the preceding cases in factoring and 60. Simplify the quanti- ties, and check, if testing is desired, with figures. m 4 - ' 6a; 9 oj 8 oj 6 4. ? MISCELLANEOUS EXERCISE IN FRACTIONS 199 7. 10. 12 4 x 4 - -4 a- 2 // 2 + 9?/ 4 11. 13 6-fc-a ' 14. 15. 16. 17. 18. O O f~\ x 2 -- xy xz a? 1- x (a? - ? /) 2 - (5 - a) 2 . (a. + a)* -(y + b) 2 (x - a) 2 - (6 - y)* _(a - b) c x c frfZ I ,,,2 ^,2\2 I *C' ' ~ */ ~ ^ ) 4 ?/ 2 19. (12 OJ 2 + 5a?/ - 2 2 xz 2 a; 4 5 2 132. Historical Notes. Algebra got its name in Arabia. The full title of the first Arabic work on algebra was " Al-gebr we'l mukabala." It means that the same quantity may be added to or subtracted from both sides of an equation. (AL means the, as in other Arabic words, such as almanac, alchemy.) Thus, the thought in the minds of the first algebraists was that algebra is the axiomatic science. 200 BIOGRAPHICAL NOTES The equations of 124 are often called Diophantine equations, after Diophantus, the early writer on algebra ( 43). It is a curious fact that Diophantus spent more time in solving these indeterminate equations than in solving the determinate equa- tions in which other writers have been most interested. When Descartes invented the method of constructing graphs from equations (see 114), he is said to have applied algebra to geometry. Thereafter these two subjects, which had before been separated, became, so to speak, married. The analytical geometry of Descartes, an extension of the study of graphs, is a powerful instrument for the study of geometry, just as algebra enables us to solve more difficult problems than those of arithmetic. 133. Biographical Notes. It may be worth while for the student to interest himself a little in the mathematical studies of some of the world's eminent men. Short biographies usually pass over with a few words the education of the person described, but the larger ones often contain interesting details. Washington had no training in languages, but considerable in mathematics. In his earlier years he had a little schooling in reading, writing, and arithmetic. When fourteen years of age he wanted to go to sea, but his mother at the last moment failed to give her consent. He then attended a school kept by Mr. Williams at Bridges Creek, Virginia, where he continued for two years, paying special attention to mathematics, as he intended to fit himself for a surveyor. There are still extant some frag- ments of his school exercises which show that his work was neat and accurate, and that he tried to master his subjects. After he left school, and when at his brother Lawrence's and at Lord Fairfax's, he kept on with his mathematical studies. The alge- bras in current use in England at the time, one or more of which he probably studied, were Harriot's, Oughtred's, Wallis's, and Sir Isaac Newton's (Universal Arithmetic). Jefferson also was strong in mathematics. His biographer says : " Jefferson was at his best in mathematics, and he could read off BIOGRAPHICAL NOTES 201 the most abstruse processes with the facility of common dis- course. This study he kept up as long as he lived ; and he delighted in applying its principles to anything and everything." How Lincoln came to study mathematics beyond the arithme- tic he learned in early life is an interesting story. In Nicolay and Hay's life of Lincoln we find this passage: "It was at this time that he gave a notable proof of his unusual powers of men- tal discipline. His wider knowledge of men and things, acquired by contact with the great world, had shown him a certain lack in himself of the power of close and sustained reasoning. To remedy this defect, he applied himself, after his return from Congress, to such works upon logic and mathematics as he fan- cied would be serviceable. Devoting himself with dogged energy to the task in hand, he soon learned by heart six books of the propositions of Euclid, and he retained through life an intimate knowledge of the principles they contain." General Grant was good in mathematics and at one time hoped to get a detail as assistant in mathematics at West Point. " Grant," says a schoolmate of his, " was strong in mathematics. I studied algebra with him, and I remember that he would never let Carr White or me show him the way to do problems, but always wanted to work them out himself. He had a way of solving problems out of rule by the application of good hard sense." In one of his letters home from West Point he speaks of having long and hard algebra lessons to get. Napoleon Bonaparte was distinguished for his application to the study of mathematics and took prizes in it. Peel and Gladstone, two of England's greatest prime ministers in the nineteenth century, took double firsts in languages and mathematics when they finished at Oxford University. The seven characters just named stand among the greatest of all the men of modern times, and we have seen each of them interested in mathematics and some of them masters of the sub- ject. If the lives of lesser yet successful men are read, some- thing like the same facts will be found true of many of them. CHAPTER VIII INVOLUTION AND EVOLUTION I. INVOLUTION 134. Involution, as a term in algebra, refers to raising quantities to powers. An integral power of a quantity may be denned as a continued product arising from multiplying unity by the given quantity as many times as the exponent of the power indicates. Thus, a 3 = lxaxaxa. 135. Raising Monomials to Powers. Type form : (xy b ) n = jf a y bn . EXAMPLES. (3 m*n 8 ) 2 = 1x3 m 2 w 8 x 3 m 2 n 3 = 9 m*n & . (-2a 3 ) 3 = lx -2a 3 x -2a 3 x -2a 3 =-8a 9 . 1. Square 4 a 2 , 3 a 3 , - 2 m\ 5 x 2 y\ - 11 b s c 2 , - 4 xy\ -Sy 5 . Check results in the preceding and in the following problems whenever there is any doubt of their correctness. Thus, to test (4 a 2 ) 2 = 16 a 4 , put a = 2. Then, 4 a 2 = 16, and 16 2 = 256 ; also the answer, 16 a 4 = 16 x 16 = 256. 2. Cube2aa;; 3 a 2 ; -6 3 d; c 4 ; 5cZ 5 ; -36 2 c 3 ; -7 xy z ; -4 arty 6 . 3. Raise to 4th power a 2 ; 3m 2 ?i; 2a 4 6 3 ; -c 5 d 7 ; -2 a 12 . 4. Raise to 5th power - 2 a; - 3 6 3 ; - 2 c 2 d 4 ; - c 7 c? n ; - m 6 . 5. O 3 ) 5 ; 0V) 6 ; (-3 a;?/ 2 ) 4 ; (-2mV) 3 ; (-6 2 2/V) 2 ; (-abc) 5 o\ 3 / 2\ 5 202 5 RAISING MONOMIALS TO POWERS 203 7. Make a rule for raising monomials to powers. (1) What sign does the power always have if the given quan- tity is positive ? \Vhat sign has the power if a negative quantity is raised to an even power ? To an odd power ? (2) How is the coefficient of the power found ? (3) How is the exponent of a quantity in the power found from the exponents it has in the given problem ? Ans. Multiply the exponent of the quantity by the index of the power for the exponent of the quantity in the answer. (4) State how the result can be checked for accuracy. 8(9, r/ 2 r a ?/ 4 V ( 2 T- 2 */ 3 ?^ 3 ( X nh 2 r 3 \ s ( - 10 o^ 3 > ( 1 "\99 . I *J (Aj JU U J . I ~J JU fj 6 I j I tJttC/U) j I -LV*t/) . I J_ ) . 9\ 9 f rt \ 1 /r\ 4\ Q / Q \ c ' / n"vn \ JU \ I fj \Ai \ I U III U \ I Ui I Ik \ o / ' I ^2,/'\ o s J'lova J' o y v or?/y \ 9. ( a# m ) n = ? Show that the solution of this problem gives a general formula or rule for solving any problem. Regard a as the coefficient of x m . If n is odd and the given sign is , what sign has the power ? Otherwise, what sign has it ? 10. Cube 3 a 4 ; - - b 4 y 2 ; a 13 x a 10 ; 2 a n ; a y ; raise Wy to the fourth power ; c 3 x to the fifth power ; 2 m s n 2 to the sixth power. 11. . /2a&cV. 12. The student should now turn back to 27 and study that article anew. In 2 a 2 + 5 (3 a 2 6 2 ) 3 -*- 4 a 4 6 6 , which operation must be performed first ? Which comes next ? Which next ? Which last ? 13. Simplify 3 <26 2 m 3 ) 3 ; 2a(4a 2 ) 2 ; 4(-4opV') 2 ; (_3n) 4 x4(-2cm 4 ) 3 ; a 14. Simplify 5(3 a) 4 (2 6) 2 ; 10 (f a) 3 (f) 2 ; 9(3 a 2 ) 4 ( - 2 b 2 ) 3 ; 12(2 a:) 3 (- y) 3 ; T ?,(4 m 3 ^.) 4 (3 n) 3 ; 10(2 z) 4 (3 y) ; Kl 3 204 INVOLUTION AND EVOLUTION 136. Raising Binomials to Powers. Newton's Theorem. 1. By actual multiplication show that (A - ) 2 = A 2 - 2AB + B 2 . (A - BY = A* - 3 A 2 B + 3 AB 2 - B 3 . (A - B)* = A* - 4 A*B + 6 A*B* - 4 AB* + B*. (A - ) 5 = A* - 5 A*B + 10 A*B 2 - 10 A*B* + 5 AB* - B 5 . - 20 A 3 B* + 1 5 A*B* - 6 .4 5 + B*. Examining eacft of these products we find that (1) If A B is raised to any power, the signs of the terms are alternately positive and negative. Evidently if A -f- B is raised to any power, all the signs are positive. (2) The exponent of A in the first term of the right side is always the same as the exponent of the binomial on the left side. How are the exponents of A in the remaining terms always found ? Which is the first term containing B on the right side in each case ? How are the remaining exponents of B found ? (3) What is the coefficient of the second term on the right side each time as compared with the exponent of the binomial on the left side ? (4) The student will not readily see how the remaining co- efficients may be found. They can always be obtained in this way : Each coefficient can be found from the preceding term by multiplying its coefficient by the exponent of its leading letter and dividing the product by one more than the exponent of its other letter. Thus, 10 in 10 A S B 2 in the value of (A - ) 5 comes from tMis : : - = 10, where 2 is one more than the exponent of B understood. 12 2. Expand by Newton's theorem (a b) 4 . SOLUTION, (a - 6)* = a* -- 4 a 3 & + a% 2 - - a& 3 + = a 4 - 4 a 3 6 + 6 a-b 2 - 4 ab* + & 4 . CHECK. Let a - 3, 6 == 1 . Then, (3 -- 1) 4 = 2 4 == 1C ; also - 4 aW + 6 4 = 81 - 108 + 54 - 12 + 1 = 16. RAISING BINOMIALS TO POWERS 205 3. Expand (a b) 5 by Newton's binomial theorem and compare the result with that given on p. 204 in Ex. 1. Similarly expand (x y) 6 and compare the coefficients and exponents with those given on p. 204. Expand in the following problems and check as in Ex. 2 until ';he process of checking is well understood. 4. (a + 6) 4 ; (a-6) 7 ; (x + yf; (w+w) 9 ; (p-q) w . 5. Develop (3 m 2 2 n) 4 by Xewton's binomial theorem. SOLUTION. Notice that 3 m 2 takes the place of A and 2 n that of B. To preserve the exponents of A and B for use in getting coefficients it is neces- sary to write 3 m 2 and 2 n in parentheses. (3 w 2 - 2 7i)4 = (3 ?n 2 ) 4 - 4 (3 m 2 ) 3 (2 n) + ^-^ (3 m 2 ) 2 (2 w) 2 ^^ (3 m) (2 ) + (2 n)* ^ o = 81 w8 216 m*n + 216 m*n* 96 m%8+16 n 4 . ( 27) To check the result let m = 2,n = l. Then, (3 m 2 - 2 )* = 10,000; also 81 m 8 - 216 w% + 216 w 4 2 - 96ra 2 n 3 + 16 w* = 20736 -- 13824 +3456 - 384 + 16 = 10000. a. In the preceding solution we think of (2 m 2 3 n) as written [2m 2 (+ 3n)]*. If we wrote it [2w 2 + ( 3 )] all the signs between terms would be +, and 3n instead of + 3 n would appear in parenthesis each time. The results would be the same. 6. (2a + 56) 4 . 7. (2.T + 3?/) 6 . 8. 9. (4a 2 c 2 +5c 3 ) 3 . 10. (3a 2 -56 2 ) 4 . 11. (a 2 - 12. (2 a 2 b - 7 c 3 ) 4 . 13. (9 a 7 - 4 /) 3 . 14. b. Let the student write clown the values of (a + &) 2 , (a + ft) 3 , (a -f 6) 4 , (a + &) 5 , (a + &) 6 , and memorize the binomial coefficients, so that he will not have to calculate them each time. Thus, the binomial coefficients for the fourth power are 1, 4, 6, 4, 1. 15. (5 -4 a?) 4 . 16. (Go 3 -/) 4 . 17. (2 a 2 -I) 8 . NOTE. When either term is 1, as in Ex. 17, retain the 1 in parentheses with its different exponents, using these to get coefficients by Ex. 1, (4), p. 204 206 INVOLUTION AND EVOLUTION 3 2x\ 6 18. 9 > SUGGESTION. - - -6- - + 16 - -etc. 3\4/O r \2 ; -) (^ ) - 2a; / V 3 / - etc. 729_243 + 135_ etc< 64 x 6 8 x 4 4 x 2 19. x-. 20. z-S. 21. * *- ar 2/ 22. (Qp-tf 1 )*. 23. (l + a n ) 5 . 24. 25. Find first four terms of (a + b) 15 ; of (x y) 20 . 26. Calculate 99 3 by writing it (100 I) 3 . Check by actual multiplication. SOLUTION. (100 - I) 3 = 100 3 - 3 (100) 2 + 3 (100) - 1 = 1000000 - 30000 + 300 - 1 = 970299. 27. Calculate in the same way 998 3 =(1000-2) 3 ; 197 4 ; (12i) 4 = (12 + i) 4 ; (29i) 2 ; 389 4 ; 205 4 ; 1010 3 . 137. Raising Polynomials to Powers. This is done by changing the polynomials to binomials by inclosing terms in parentheses. 1. (x 2 -x+2) 3 . SOLUTION. Change given trinomial to binomial. [(a;2 _ X ) + 2] = (x 2 - x) 3 + 3 (x 2 - x) 2 (2) + 3 (x 2 - x)(2) 2 + & = ( X 2)3 _ 3 ( X 2)2(a;) + 3 (x 2 ) (x) 2 - x 3 + 3[x 4 - 2 x 3 +z' 2 ] x 2+3[4 x 2 -4 x] +8 = x 6 - 3 x 5 + 3 x 4 - x 3 + 6 x 4 -- 12 x 3 + 6 x 2 + 12 x 2 - 12 x + 8 = x 6 - 3x 6 4- 9 x 4 - 13 x 3 + 18 x 2 - 12 x + 8. Check with x = 1. EXTRACTING ROOTS OF MONOMIALS 207 2 . (l-a + or 9 ) 3 . 3. (m 4 -n 2 -3) 3 . 4. (a + 6 + c) 4 . 5. (2a 2 -3a-4) 3 . 6. (a -1- b + c + d)\ SUGGESTION. Write thus [(a -f 6) + (c + d)] 4 . 7. (a-6 2 + 2c-3cf) 3 . 8. (i-f 2n- 9. (1 + a + ary. 10. (2 cT - b n 11. 1799 3 = (1000 + 800 I) 3 . Verify answer by multiplication. 12. (a + 5 + c) 2 . 13. (a + & + c + d) 2 . 14. By generalizing from the results of Ex. 12 and 13 we get this theorem: The square of the sum of any number of quantities equals the sum of their squares plus the sum of twice the product of each quantity by each quantity that follows it. Write directly the answer in the following by using the theo- rem just enunciated. Check Ex. 15, 16. 15. (x 2 -3a + 2) 2 . 16. (xy + yz+zx) 2 . 17. (a 3 -- a 2 -- a -- 3) 2 . 18. (2 a 8 - -4 x 2 - -3 19. (15 a - 9 y - 12 z) a . 20. (3m + n II. EVOLUTION 138. Evolution in Algebra means the extraction of roots of quantities. A root of a quantity is one of its equal factors ; or it is a divisor which being divided into the given quantity and then into the quotient, and so on, gives the quotient unity. 139. Extracting Roots of Monomials. Type form : ^/a n b n = ab. 1. What sign should be given to the square root of a number, as 4 or 9 ? Prove 2 2 = 4 ; also that (-- 2) 2 = 4. What sign, then, should be placed in front of the root? (See 36, a.) What sign should the cube root of 8 or 27 have ? What sign should the cube root of 8 or 27 have ? What sign should the fifth root 208 INVOLUTION AND EVOLUTION of 32 have ? The fourth root of 16 ? Ans. to last question . Prove the answers correct in each case. What sign should be given to an odd root of a positive quan- tity ? To an odd root of a negative quantity ? To an even root of a positive quantity ? 2. Show that V- -4 is neither + 2 nor 2. The square root of a minus quantity is commonly called imaginary. 3. Write the square root of 36a 2 6 2 ; of 9aW; of 16 4. -v/ 27aV. 5. A/ -8 x 6 u 12 . 6. -J/16 a 4 6 8 c 7. A/ -64 a 12 . 8. -x/lGa 5 . 9. ^/1728 a 12 ?/ 3 - 10. V-16a 4 . 11. V2W. 12. A/-243a u V 5 . 13. 125a 21 . 14. tf. 15. - / 27 a^v'" ' \- -64^' 25 c 4 V'a (i! 6 14 \ 64 19. How is the coefficient of the root found from the given coefficient ? How is the exponent of a quantity in the root found from the given exponent of the quantity and the index of the root? 20. J/^ri*. 21. VlSV 1 . 22. -v / -2 10 a 5 6 10 c 20 . 23. V81x 64x121. 24. V(64a 8 ) 2 . 25. 481 a? 128 4181 12 " 1 26 ' Simplify in following problems, taking only + value of even roots : 29. V25 a 2 6 4 c 2 + A/' - 8 aW - -\/81 a W - >/ - 32 a 30. V - 27 ar 3 / x V 243 ?/V X Vl6 SQUARE ROOT OF POLYNOMIALS 209 140. Square Root of Polynomials. 1. The method of extracting such roots is obtained from a study of the familiar formula/ It is required to find a process to obtain A + JB from A 2 + 2 AB -f B' 2 . A + B A 2 + 2 AB -f B 2 A 2 2A+ B 2 AB + B 2 2 ^45 -} B 2 derneath the line obtained ? SOLUTION. How is A in the root found from A 2 + 2 AB + B 2 ? From which term is it obtained and how? ( 116.) How is the A 2 under A 2 obtained ? Ans. By squar- ing A in the root. How is 2 AB -f B 2 un- How is 2 A (called the trial divisor) , in the divisor at the left, obtained from A in the root? Ans. By doubling A ( 116). With the divisor 2^1 and the dividend 2 AB + B 2 , how is Bin the quotient found ? Where else is the quotient of 2 AB -^-2A written be- sides in the root ? Ans. At the right of the divisor 2 A . 2 A + .B is called the complete divisor. How is the operation completed ? 2. Extract the square root of SOLUTION. Arranging this ( 37) with reference to the powers of x (and y and 2), we have - 4 4- 12 xz* - 6 ~ ay+ 3z* 4 x ay 4r+ fc v * 4 /7^"?/ 1 1 9 O" ?*2 1 /y2?/2 ^1 LtiX^y ^^ A *il/v \ \Jv ti 4 r?/ + -y 2 ?-4 -' ^ 5 , + 3, 2 12 xz 2 - 6 a^ 2 4- 12^ 2 -6^ 2 + How is 2 # in the root found ? How is 4 x in the first divisor found ? How is ay in the root found ? In what two places is ay written when found ? What are the next operations ? How is 4 x 2 ay found ? By dividing what by what is 3 z* found ? Where is. 3 z' 2 written besides in the root ? What is last operation ? How can result be proved correct ? Ans. By squaring the result. (See 137, 14. ) The answer to this problem is (2 x - ay + 3 z' 2 ) since squaring either 2 x ay + 3 2 2 or ay 2 x -- 3 2 2 produces given poly- nomial. Is it important to arrange the terms in a given problem and also the remainders as they are obtained? ( 37, 38.) 210 INVOLUTION AND EVOLUTION 3. a 4 + 6or ) -h9 2 . 4. 49 a 4 42 a?b + 9 fc 2 . 5. a! 4 - 6. 4ce 8 7. 25 a 4 - 30 oaj 8 + 49 ah? - 24 a B x + 16 a 4 . 8. l_ 9. a;- 10. a 4 a 2 6 3 6 6 a 2 c 4 c 8 12. 13. !- 14. a? 6 + 4 ax 5 10 a V + 4 a 5 a? + a 6 . In the two following problems extract the 4th root by extract ing square root twice : 15. ^(16 x 4 - 96 tfy + 216 x 2 y 2 - 216 xy 5 + 81 ?/ 4 ). 16. ^/(a 4 + 8 o 8 6 + 24 a 2 6 2 + 32 a6 8 + 16 5 4 ). 17. 6 ab 2 c - 4 a 2 bc + a 2 & 2 + 4 a 2 c 2 + 9 6 2 c 2 - 12 18. 5 19. Write a rule for extracting the square root of polynomials. 20. l- JL.L + - ^.i^L 8 . i^- + ' 25 25 36 25 5 49 35 15 4 g 4 ?i 4 2 m 3 ? ~35~ ~2T SQUARE ROOT OF ARITHMETICAL NUMBERS 211 22. Obtain three terms of the square root of /y.2 /v4 SOLUTION. a* + x 2 2a + - 2a 2a + ^- r 2 _L a 8 a* X 4 X 4 _ _^_ +_ n a 2 8 a 4 64 a* PROOF. 8 a* 64 s x3. (137,14.) 2a 8aV 8 23. Obtain three terms of the square root of a?b 2 c 2 . Prove. 24. Obtain three terms of the square root of 9 a 4 4 & 4 . Prove. In the following get the square root three times in succession : 25. -fy(x 8 + 8x 7 + 2Sx & + 56x : + 70x* + 56x 3 + 2Stf + 8x + l). 26. ^(266 a 8 + 3072 aj r + 16128 a 6 + 48384 x 5 + 90720 a 4 + 108864 a 3 -f 81648 aj 2 +34992 a;+6561). 141. Extraction of the Square Root of Arithmetical Numbers. 1. Extract the square root of 625. Since 625 lies between 400 and 900, its square root lies between 20 and 30, that is, between 2 tens and 3 tens. For guidance in the solution we use the formula ^ + u y = t z + 2tu+ U 2 = p + (2 t + ?Q u in which t stands for number in tens' order and u for number in units' order. The process at the right below is just like that at the left, which was explained in the last article. t + u t 2 + 2tu + u 2 \t + u 6'25 120 t 2 P = 4 00 u Trial divisor = 2 t = 40 u = 5 2 tu + w 2 Complete divisor = 2 t + u = 45 225 225 212 INVOLUTION AND EVOLUTION If a number has an even number of integral orders, or none, each pair of its figures gives one figure in the square root : if it has an odd number of such orders, its first figure gives the first figure in the root; and after that each pair of figures in the number gives one figure in the root. For this reason, in extracting the square root, numbers are separated by marks into " periods " of two figures each commencing at the decimal point. In the following solutions ciphers not actually needed are not written. Instead of writing 40 + 5, as on p. 211, it is enough to write 4 only and then annex 5 to it when 5 is found. 2. 3445.69. SOLUTION. 3. .0003627. 34'45'.69[58.7 25 SOLUTION. '.OQ'03'62'70[. 01904+ .0001 1167 29 38( 262 261 )4 1 70'00 15216 1784 Ex. 2. Into periods of how many figures is the number pointed off ? At what point does one start in pointing off ? How is 5, the first figure of the root, found ? Why is it not 6 or 4 ? How is 25 found ? 10 in the divisor at the left ? 8 in the root ? Where is 8 placed besides in the root ? Ans. At the right of 10. How is 864 found ? 116 ? 7? 8169 ? Ex. 3. What is the first thing done? Where does one start to point off'? Why is the cipher annexed to 7 ? What does the first period of two ciphers give in the root ? When the cipher after 9 is placed in root, where else is it placed at the same time ? Remember that the trial divisor in square root always has understood. PROOF. By actual multiplication (58. 7) 2 = 3445.69 ; also (.01904) 2 + .0000001784 = .00036270. 4. 484. 7. 4624. 10. 494209. 13. 74.1321. 5. 1024. 8. 18769. 11. 328329. 14. 60.3729. 6. 7396. 9. 106929. 12. .165649. 15. 4076.8225. 16. .004761, 17. 150.0625. 18. .00459684. R S' CUBE ROOT OF ALGEBRAICAL QUANTITIES 213 Find answer in Ex. 19-26 to three decimal places, Ex. 27-31 to four, and Ex. 32, 33 to five places. If the fourth figure in Ex. 19-26 is one of the figures 5-9, then in giving the answer the third figure should be made one greater. In proving the problem keep last figure unchanged, square, and add remainder. 19. 3. 20. 5. 21. 7 T 4 T . 22. 2. 23. 10. 24. 8J. 25. 0.3. 26. 0.5. 27. .0021. 28. .723. 29. .062. 30. .000067. 31. f. 32. f. 33. -fa. 34. To explain square root by means of a geometrical figure, let us take 729 and extract its square root. _ ? Suppose there are 729 little square units in a given square. Find the number of linear units on one side of the square. How is 20, one side of, the square S in the dia- gram, found ? Why is one side of it not 30 ? When S is found, what is the length of rectangle R and of R ? How many square units are left out of the 729 to be put around 8 into R, R', and S'? Ans. 329. Neglecting S' as small compared with R + R' + S', how can the width of R + R> be found from their total length, 2 x 20, and their area 329- ? A ns. By dividing 329- by 40. (329- means some- thing less than 329.) What is 2 x 20 called in the process of extracting the square root? Ans. Trial divisor. If the length of #' (found by dividing 329- by 40) is added to R + R' , what is the total length of R + R + S'? What is this 47 called ? Ans. Complete divisor. 35. Explain geometrically the extraction of the square root of 576; of 3364; of 6068.41. *142. Extraction of the Cube Root of Algebraical Quantities. 1. By Newton's theorem (see also 45, V), 20 R' 20 * See statement about starred articles on p. 6. 214 INVOLUTION AND EVOLUTION It is required to frame a process to obtain A + B from A* + 3 A 2 B + 3 AB* A* 3 A 2 B + 3 AB 2 + How is A the first term of the root found ? How is 3 A 2 B + 3 AB 2 + B" on the third line found? How is 3 A 2 (called trial divisor) found from the first term of the root ? Ans. By squaring the first term and multiply- ing the result by 3. How is B, the second term of the root, obtained from 3 A 2 and the first term of the remainder, 3A 2 B? Tell how 3AB + B 2 (what is added to 3 A 2 for the complete divisor) is constructed from A and B, the first and second terms of the root. Ans. To the trial divisor .are added three times the product of the first term of the root by the secondhand the square of the second. What operation is performed last ? 2. -3Qx 5 + 54**- 27 +25 60z 4 -180z 3 +285;c 2 -225+125 60 a 4 - 180 x 3 + 285 y? - 225 x + 125 EXPLANATION. The first 12 se* = 3(2 z 2 ) 2 ; the - 18 x 3 = 3 x 2 yt x - 3 a; ; x 2 = (-3^) 2 ; underneath, the 12 x* 3Gx 3 + 27ic 2 = 3(2x 2 -3x) 2 j etc. Apply the questions under 1 above, to this problem. 3. 8w 3 -60m 2 >i-r-150m^ 2 -125w 3 . 4. 27 a 8 ^ - 108 a 2 6 2 c + 144 abc* - 64 c 3 . 5. a 6 - 6. a 7. 6 + 12 a^ + 63 a 4 + 184 a^ + 315 a 2 + 300 a; + 125. 8. 125 x 9 - 300 x? + 465 a; 7 - 424 a 6 + 279 x 5 - 108 x* + 27 a 3 . 9. a^ + 6aS* + 15a? + 20 + i5 + ^+i. or or ic 6 1 0. 60 cV + 48 cz 5 - 27 b 6 + 108 + 6 2 is symmetrical, because if a and 6 change places, the quantity has same value as before. Simi- larly (a + 5) 3 , (a 4- 6) 4 , (a-ffr-f-c) 2 are all symmetrical. Notice a 5 + 5 a*b 4- 10 a s b 2 + 10 a 2 6 3 + 5 atf + b 5 is symmetrical in the sense kh&k first part is like last part. 145. Miscellaneous Exercise in Involution and Evolution. 1. 2a(-35 2 ?i 3 ) 3 ; a(5a) 3 ; 5(m' 1 )" ; 7(7 p 3. One angle of a triangle is m times another, and the third angle is 4 times the latter. How many degrees in each angle are there ? 4. THEOREM. In any right-angled triangle, the square of the number of units of length in the side opposite the right angle (called the hypotenuse} is equal to the sum of the squares of the number of units of length in the other ttco sides. Stated algebraically by denoting the lengths of the sides by letters as in the margin, the theorem becomes c 2 = a 2 + b 2 . a To test the truth of this theorem, measure 3 inches along one side of the sheet of paper and 4 inches along the adjacent side from the right angle. Now measure the distance apart of the extremities of the two distances measured. It will be found to be 5 inches because 3 2 -f4 2 = 5 2 . Again, measure 2 inches along one side and 5 inches along the adjacent side and see if the distance is V2- + 5 2 = of + . Meas- ure still other distances, making the same test. 5. Solve c 2 = a 2 + b 2 (1) for c j (2) for a ; (3) for b. 232 APPLICATIONS OF ALGEBRA Solve by substituting in the formula just found the following problems dealing with right-angled triangles. 6. If two sides of a right-angled triangle are 6 and 8 inches respectively, what is the length of the longest side ? 7. Given a = 7, c = 12, to find &; a = 16, 6=22, to find c; b = 12, c = 40, to find a. 8. A ladder leaning against a vertical wall stands on a hori- zontal pavement 10 ft. from the wall. If the ladder is 22 ft. in length, how high up does it reach on the vertical wall ? 9. A carpenter makes mortises for a brace 41 ft. out from the right angle along a beam and the same distance along a post, meeting the beam at right angles. What length must the brace have ? 10. How far is it from one corner of a farm 80 rd. by 120 rd. to the opposite corner ? 11. If a, b, and c are the adjacent edges of a rectangular box and d the diagonal from one lower corner to the most distant upper corner, show that d 2 = a? -f- b 2 -f- c 2 - SUGGESTION. Denote by m the diagonal of the lower face or base of the box. Then what does m 2 equal ? Consider now the right triangle whose base is m and perpendicular one of the vertical edges of the box. 12. Solve d 2 = a 2 + b 2 + c 2 for (1) d ; (2) a ; (3) 6 ; (4) c. (See H7, 1.) 13. What is the length of the longest straight rod that can be placed inside a box 1.4 m. by 1.2 m. by 9 m. ? 14. A tree standing vertically on level ground is 60 ft. high. Upon being broken over in a storm, the upper part reached from the top of the trunk to the ground just 30 ft. from the foot of the trunk. What was the length of the part broken off ? (See 97.) 15. A rectangular room is 9 ft. high and is 2 ft. longer than wide. If it were 1 ft. longer and 2 ft. wider, it would contain 270 cu. ft. more. What are the dimensions of the room ? SOME THEOREMS OF GEOMETRY 233 16. Given the base b and the difference d, between the hypote- nuse and perpendicular of a right-angled triangle, to find the perpendicular. 17. If the base of a triangle is 6 and the difference between the hypotenuse and perpendicular is 2, what is the perpendicular ? 18. THEOREM. A line drairn parallel to one side of a triangle divides the other two sides into parts which are in proportion; also, the corre- sponding sides of the two triangles formed are proportional. (See 72.) Thus, if the line n is parallel to the side m of the triangle in the margin, a + b __ c + d a a then, = ; also b d' and c a m n 19. THEOREM. If the parts of two sides of a triangle are in pro- portion, the line dividing these sides is parallel to the third side. The truth of these theorems can be tested by drawing figures and making measurements. 20. Solve - - for a: for b: for c; for d. b d 21. If a = 2 inches, c = 1.75 inches, rZ = 4.5 inches, find b by substituting in the formula. 22. To find the distance across a stream by theorem in Ex. 18 above. To find the distance AB across a stream, measure a, 6, and c as repre- sented in the figure, b being parallel to x. Then x is found from the pro- a f* portion - = - by solving for x. u X 23. If = 40 ft., ^ = c = 540 ft., what is x ? ft., and 234 APPLICATIONS OF ALGEBRA 24. To find the height of a tower by measuring its shadow and that of a pole at the same time of day. SUGGESTION. Let AB or x in the figure to Ex. 22 denote the height of the tower, 6 that of the pole, a the shadow of the pole and c that of the tower. Find x. Make measurements of the height of different objects in this way. 25. What is the height of the Washington Monument at Wash- ington, D.C., if the monument casts a shadow 90 ft. long at the time a flagpole 37 ft. long casts a shadow 6 ft. long ? 26. THEOREM. TJie circumference of a circle equals TT times its diameter, or 2?r times its radius (TT = 3.1416, nearly). Divide the circumference by the diameter in various sized circles and compare the results. 27. THEOREM. The area of a circle is equal to one half of the circumference times one half of the diameter. To show this cut a circle into ten equal pieces as a pie is cut. These pieces can be arranged as in the figure below, mak- ing what resembles a parallelogram whose base is half the circum- ference and altitude half the diameter. If the circle had been cut into 1000 pieces instead of ten, one could not distin- guish by the eye the lower figure from a rectangle. SOME THEOREMS OF GEOMETRY 235 28. If r=the radius of a circle and a its area, show by eliminat- ing c, from a = rx Jc (Ex. 27) and c=2vr (Ex. 26), that a=^. 29. If d = the diameter of a circle and a its area, show from Ex. 28 by eliminating r that a = -J Trd 2 , or a = .7854 d 2 . 30. Calculate the area of a circle whose radius is 14 inches, (1) by using the theorems of Ex. 27 and 26; (2) by using the formula of Ex. 28 ; (3) by the formula of Ex. 29. 31. Calculate in the shortest way the areas of circles whose radii are 5, 4, 6, 24; whose diameters are 4, 11, 35. 32. If r is the radius of a circle, d its diameter, c its circum- ference, and a its area, make a formula to find a in terms of c, that is, get a value of a which has no letters except c and TT in it. 33. Find a from the formula just obtained, when c = 7j 9; 12. 34. Solve the formula of Ex. 32 for c.. 35. Find c when a = 25, using the formula just obtained. 36. Show that the area of a circular ring, whose outer radius h R and inner radius is r, is w(/2 2 --r 2 ). 37. Calculate area when R = 20, r = 10 ; also when R 100, r = 25. 38. The point C is the middle of the straight line ACB(=4r). On AC, CB, and AB semicircles are con- structed with these lines as diameters, all on the same side of AB. If x is the radius of a circle which touches the small circles externally and the large circle internally, show that x = f r. SUGGESTION. Express FC, EF, and CE, using x and r. What kind of figure is EOF? REMARK. Such a construction might appear in a large window, or in laying out a park, or the like. F 236 APPLICATIONS OF ALGEBRA III. APPLICATIONS WITH SQUARED PAPER 155. Solution of Simple Practical Problems by Squared Paper. The solution depends on 154, 18. 1. Given the price of an article, to construct a graph from which may be obtained the cost corresponding to any given num- ber of units (or articles), or the number of units corresponding to any given cost. SOLUTION. Thus, suppose 1 Ib. of tea costs 75^. To construct a graph with these conditions, take sixteen 2-millimeter units on the Y-axis above the origin to denote 16 oz. or 1 Ib., and fifteen 2 -millimeter units along the JT-axis to the right to represent 75^, thus making one 2-milliineter length Y Oz. 30 25 20 15 10 2 P 23 50 75 1.00 1.25 1.50 X represent 5/'. Join the point P located by these coordinates with the ori- gin 0. Then the cost of any number of ounces of tea at this price can be read off the graph by starting at the given point on the Y-axis, running across on a horizontal line to the graph line, and then down on a vertical line to the X-axis, remembering that each 2-millimeter length ou the JT-axis is 6^. SOLUTION OF PROBLEMS BY SQUARED PAPER 287 Thus, to find the cost of 23 oz. , start at 23 on the F-axis, pass along the horizontal line 23 to the graph, then down vertically to $ 1.08. Again, to find the number of ounces corresponding to any given cost, start with cost, go up vertically to the graph, and then along a horizontal line to the required number of ounces. Thus, find how many ounces can be bought for $1.25, and check answer by an arithmetical solution. 2. Construct a similar graph for the cost of butter at 35^ a pound. Then find by the graph the cost of 21 oz. ; of 27 oz. ; of 11 oz. How many ounces can be bought for 50^? for 30^? Check in each case by an arithmetical solution. a. It was stated at the beginning of this article that these graph solutions depend on 154, is, which says a line drawn parallel to one side of a tri- angle makes the corresponding sides of the given triangle and the new tri- angle formed proportional. Now, if we take the X-axis, the graph line, and the 1.50 vertical line in the figure in this article, we have a triangle. The $1.50 line is thought of as one side and we know the other vertical lines, as the $1.00 vertical line, are parallel to it. Thus, we see that any no. of oz. : any other no. of oz. = cost of 1st no. : cost of 2d no. Of course, ail such problems can be solved by proportion in arithmetic. 3. Construct a graph for reading off interest, taking 6 % as the rate. SUGGESTION. Locate a point, using 60 2-millimeter units for the abscissa (representing 360 days), and 6 centimeters representing .$6 interest on $100, as the ordinate, and join it with the origin. Then the interest on $100 for any number of days can be read off by starting at the given point on the X-axis, going vertically to the graph, and then horizontally to the F-axis, where the answer is read off. Evidently a 2-millimeter unit on the X-axis means 6 days, and a 2-millimeter unit on the F-axis 2 4. Find the interest by the graph just constructed for 96 days at 6 % on S 100. For 36 days. For 50 days. How many days' interest is $ 4.60 ? $ 7.20 ? $ 2.80 ? 5. Construct graphs for other per cents and use them in the same way. 6. Suppose an automobile travels 15 mi. an hour. Construct the graph for this motion by using 15 2-inillimeter units as 238 APPLICATIONS OF ALGEBRA abscissa and 6 centimeters as ordinate of the point determining the graph. Find from the graph corresponding intermediate values of time and distance. Thus, how long does it take the automobile to go 11 mi. ? 19 mi. ? How many miles has it gone in 38 inin. ? In 73 min. ? Etc. 7. Construct a graph for reducing centimeters to inches. SUGGESTION. Join origin to point (100, 39.37). 8. Construct a graph for reading off circumferences from radii, or radii from circumferences. SUGGESTION. Join the origin to (5, 31.416). 9. Construct graphs for other similar reductions as they may suggest themselves. *156. Solution of Problems by Squared Paper. 1. Two men set out from two towns A and B 30 mi. apart at the same time, the one from A walking toward B at the rate of 4 mi. an hour, and the one from B riding toward A at the rate of 13 mi. an hour. How far from A will they meet, and in how many hours will they meet ? SOLUTION. Take a piece of squared paper and select two points A and B on the same horizontal line and 30 units apart. Each horizontal unit dis- tance denotes 1 mi. In this problem we will take 5 vertical units to denote 1 hour. The diagonal lines drawn on the figure are graphs. The graph AK M 2 Hours IHour A 7- / / / / -i / V < / ^* > ^* ". -> * T r / ^. 1) / \ >. N, / ^ "-> / "^> > / ^ / ^. -> H Miles 10 15 20 25 30 SOLUTION OF PROBLEMS BY SQUARED PAPER 239 for the man walking is constructed in this way : We count 4 units to the right from A for the number of miles the man walks in an hour, and then go up 5 units for 1 hr., thus getting the point K. Through A and K the line AK is drawn. In the same way we count 13 units to the left from B for the number of miles the man from B rides in an hour, and then go up 5 units for 1 hr., thus getting point L. Through points B and L the line BL is drawn. Where AK and BL meet is the point C. This point <7, so to speak, is the graphical answer to the problem. It is 7 units to the right of A, that is to say, the men met 7 mi. from A, and it took them If hours to meet. The equation for graph AC is y = , and for BL is y = - VERIFICATION BY AN EQUATION SOLUTION (see 94). Let x = number of units from A to where they meet. Then - = = , whence x 7^ 7 mi. This shows that the squared paper solution is not perfectly accurate, but gives a very good approximation to the answer. 2. Two travelers start from two places 60 mi. apart, going toward each other, the one by automobile going 18 mi. an hour and the other by train going 40 mi. an hour. How long after they start will they meet, and how far will their meeting place be from the first man's starting point? Verify by algebra. 3. A man made a trip to the country on his bicycle ; but it broke down and he had to walk back. He rode 10 mi. an hour and walked 4 mi. an hour, and was gone 3 hr. How many miles out did he get ? Solve with graphs and check answer by an algebraic solution. SUGGESTION. Draw one graph from upwards to the right, using 10 miles as the rate, and the other downwards and to the right from the 3-hour point above as origin, using 4 miles as the rate. Where these graphs meet is the answer. 4. One man walking 3 mi. an hour starts to go from Wash- ington to Baltimore. A second walking 5 mi. an hour starts 4 hr. after the first to make the same trip. How far from Washington will the second man have to go to catch the first man ? Check with an algebraic solution. SUGGESTION. Draw the second graph upward and to the right from the 4-hour point above as origin, using the 5-hour rate. 240 APPLICATIONS OF ALGEBRA 5. Suppose, using the data of 113, 9, that a train starts from Pittsburg at 9 A.M., traveling 30 mi. an hour, and another leaves Harrisburg at 10 A.M., traveling 35 mi. an hour. Where will they meet ? SUGGESTION. Put the stations at the proper distances on the X-axis. The figure will be like that in Ex. 1, p. 238, except that the second graph will be drawn from 1 hr. above B instead of from B, In a general way this is the train dispatcher's problem. When a new train is put on or when the schedule is changed, he finds where the trains meet by the graphical method. 6. Using the data of 113, 8, suppose a train leaves Boston at 4 P.M. going 40 mi. an hour, and another leaves Fitchburg at 4.30 P.M. for Boston going 30 mi. an hour. Where will they meet ? 7. A train leaves Harrisburg, traveling 45 mi. an hour, two hours and three quarters after a freight which runs 20 mi. an hour. Near what station will the passenger catch the freight ? 8. Two pipes fill a cistern, the first in 3 hr., the second in 4 hr. How long will it take both to fill it ? Solve by graph method and prove by algebraic solution. SUGGESTION. Let any horizontal distance AB (say 10 cm.) represent 1, or a cistern full. Then draw graph from A to 3 cm. above B, and one from B to 4 cm. above A. Where they cross gives time for both. 9. It takes A 8 da. to do a job, and B 11 da. How long will it take both? Check answer by algebraic solution. 10. It takes A 6 da. to reap a field, B 8 da., and C 13 da. How long will it take all working together ? Check answer. SUGGESTION. Get time for A and B as in preceding problem. Then take the result for A and B together along with C as in the preceding. IV. APPLICATIONS IN PHYSICS 157. Exercise in solving Some Formulas of Physics. It is ex- pected that the teacher will use no time in explaining the mean- ing of these formulas. The names which some of the letters represent are given to show the student that the formulas have APPLICATIONS IN PHYSICS 241 a meaning, so that he may not be repelled by their unfamiliarity. The formulas are given also partly to show that algebra has value in other branches. Begin by reviewing 118. p 1. Given (7 = , to solve first for R, and then for E. R REMARK. To solve for any letter means taking that letter to represent the unknown number, or the , in the solution. The formula in Ex. 1 ex- presses Ohm's law in electricity. SOLUTION. CR - E (Mult. Ax.) ; R = , Ans. (Div. Ax.) ; E = CR, Ans. O Tfi 2. Solve C = - for E. R, and r in turn. R + r SOLUTION. Ans. CR + Cr = E (Mult. Ax.) ; CR - E- Cr (Sub. Ax.) ; - Cr . _ E- R = C 3. Solve - - = - + - f or R, r, and r' in turn. R r r' SOLUTION TO IST. rr = Rr' + Rr (Mult. Ax.) ; Rr' + Rr = R(r' + r) = rr' ; R = -^ (Div. Ax.). r' + r 4. Solve s - 7 (formula for specific gravity) for w and I. i on V X P ll r, 5. Solve - = - for x. V - x + v P SUGGESTION. Clear, transpose unknown numbers to left, known numbers to right. Factor and divide by coefficient of x. 6. Solve E- = - (a formula in electrical work) for h. t PV P'V 7. Solve : - = - for each letter in turn (Combination of Boyle's and Charles's law). -pi 8. Solve J - (a formula for one electrical cell) for R. B + g + R 9. Solve = - (a formula for lenses) for /. 242 APPLICATIONS OF ALGEBRA 10. Solve a = for v and t respectively. 11 Solve C= -- for E, R, r, and N (Ohm's law in series). J\i j~ _i.V / E 12. Solve C ' = - for E, R, r, or N (Ohm's law in parallel). 13. Solve = for R, h. and r in turn. R 2 7TT 14. Solve Pd = Rs + i for each letter in it. This formula ex- presses a law for all machines. 15. Solve Rs = i mv 2 for R, s } and m in turn. 16. Solve g : G : : - - : - for g. What is the value of A~ (R -f- g when h 2 (but not Ji) is put equal to zero ? ( 60.) 4. 17. Solve e = - for P. Also for 6, and for m. bh s m 18. Solve g=2x + ?^-^xf or #. s V. FORMULAS 158. General Exercise in framing and using Formulas. We have seen derived by induction the formulas for percentage, interest, cost of merchandise, etc. In this exercise we will study formulas in still other subjects. Consult the dictionary freely for the meaning of technical words used. 1. It is taught in arithmetic that the area of a rectangle is equal to the product of its base by its altitude. If b is the number of units in the length, h the number in the altitude, and a the number of square units in the area, make a formula for the area. Ans. a = bh. FORMULAS 243 2. What is the area of a rectangle whose base is 20 ft. and altitude is 12 ft. ? Solve by substituting in the formula. 3. Make a formula for the area of a triangle if its area is equal to half the product of its base and altitude. 4. What: is the area of a triangle if b = 16, and h . = 10 ? 5. Make a formula for the volume of a box (or right paral- lelepiped) if I is its length, b is its breadth, h is its height, and v is the number of units of volume in it. 6. What is the volume of a box if 1 = 12 inches, 6 = 8 inches, h = 6 inches ? 7. Show if s = the total surface of a box that s = 2 (Ib -\-lh-\-bh). 8. What is the total surface of the box described in Ex. 6 ? 9. Make a formula for the area of a trapezoid if its area, a, equals half the sum of its parallel sides, b and b' } times its altitude h. 10. What is the area of a trapezoid whose & = 15, &'=12, h = 7? Draw a figure and mark the letters and dimensions on it. 11. Make a formula for the sum of the interior angles A, B, C, etc., of a convex polygon. SOLUTION. Join the vertices A, B, (7, etc., to any point O within the polygon. Then each of the triangles formed, AOB, BOC, COD, etc., contains 2 rt. angles (by 154, i). Then, if the polygon has n sides, all the angles within all the triangles will amount to TTI T^ 2 n right angles. Now the angles at which are evidently equal to 4 right angles are not to be included in this sum, since all we want is sum of ^4, B, (7, etc. Hence, if s = number of right angles in the sum of the interior angles of any polygon, s = 2 n 4. 244 APPLICATIONS OF ALGEBRA 12. Find by substituting in the formulas just obtained the sum of the interior angles of (1) a quadrilateral or 4-sided figure (ti = 4) ; (2) of a hexagon or 6-sided polygon ; (3) of a decagon or 10-sided figure ; (4) of a 13-sided figure. 13. How many degrees are there in one angle of a pentagon, or 5-sided figure, all of whose angles are equal ? In one angle of a heptagon, or 7-sided figure, all of whose angles are equal ? 14. The volume of a pyramid equals ^ the product of the area of its base and its altitude. Make a formula for the volume when base is a rectangle, letting v = volume, I = length of base, w = width of base, and h = altitude or height. 15. Make a formula for the distance s a body travels, if v is its velocity, or rate, and t is its time. 16. Make a formula for the weight w of a body if the weight w is equal to the product of its volume v and density d. (Density = weight of a unit of volume.) 17. Fahrenheit-Centigrade Formula. In the Fahrenheit ther- mometer, in common use, the freezing point is 32 above the zero of the scale, and the boiling point is 180 above the freezing point. The Centigrade thermometer, used mostly in scientific laboratories, has freezing for its zero point on the scale and 100 for its boiling point. If /denotes the number of degrees in any reading of the Fahrenheit thermometer, and c the number of the Centigrade at the same time and place, let it be required to ex- press/in terms of c. SOLUTION. 100 Centigrade degrees = 180 Fahrenheit degrees. 1 Centigrade degree = ^- or f Fahrenheit degrees, c Centigrade degrees = c x f Fahrenheit degrees. But as/ is counted from 32 below freezing, /=fc + 82. 18. Solve the equation just found for c. 19. Using the formula, find / when c = 50 ; 4 ; 200 ; - 10. FORMULAS 245 20. Using the formula, find c when/=60; 98; 400; -32; 0. 21. Taking a whole sheet of graph paper, construct the graph of/ = -|c-f 32. Let /denote X distances, and c, Y distances. 22. Use the graph diagram as in 155, 1, to check the results of the calculations in Ex. 19 and 20 above. 23. Use the diagram to reduce from Fahrenheit to Centigrade, being gi ven /= 35; -16; 48; 70; 100; 90. 159. General Exercise in finding Numerical Values from Formulas. 1. The formula for the surface of a sphere is s = 4 irr 2 . Using 3.1416 for TT, find the surface of a sphere whose radius r is 6 inches. 2. The volume of a sphere is | Ti-r 3 . Find the volume of a sphere whose radius r is 12 feet. 3. The volume of a sphere is also 1 ircfi. Find the volume of a sphere whose diameter d is 2 meters. 4. The volume of a pyramid having an altitude a and a rectangle for base whose length is / and width w is Find o the volume of a pyramid in which a = 16 ft., I = 12 ft., and W = 4:.S ft. 5. Using the formula of 158, 17, find c when /= 20; change 25, 38, 95, - 15, 219, - 273 from each scale to the other. 6. A body falling from a state of rest in t seconds moves through a distance s, where s = ^ gt 2 , provided the resistance of the air is supposed to not retard the body. Find the distance an object, as a stone, moves through in 3 seconds; in 5 seconds; in 10 seconds. Use g = 32.16. 7. A stone dropped from the top of a building reaches the ground in 3J seconds. How high is the building ? 8. Find the time in which $324 will draw $64.80 interest at 5% interest, using the formula prt = i. 246 APPLICATIONS OF ALGEBRA 9. A man sold a horse for $161.57, on which he gained 7 What was the cost of the horse ? Use formula a p (1 + r). 10. Find p from the formula a 2 = b 2 + c 2 2 pb when a = 7, 6 = 6, c = 5. 160. Exercise in translating Formulas into Rules. 1. Translate formulas in Ex. 1-9 of preceding article into rules. Solution of 1st. The surface of a sphere is equal to four TT times the square of the radius. 2. Translate the formulas in 154, 4, 11, 28, 29, into words. 3. If V is the volume of a cylinder, V= ?rr% where r is the radius of its circular base and I is its length. Translate this for- mula into a rule. 4. If V is the volume of a cone, V= \ -rr^h, in which r is the radius of its base and h is its height or altitude. Translate this formula into a rule. 5. If Fis the volume of a hollow sphere, V= f 7r(R 3 r 3 ), in which R is the radius of the sphere and r the radius of the hollow sphere. Translate the formula into a rule. *VI. APPLICATIONS IN MANUAL TRAINING AND DOMESTIC SCIENCE 161. Problems in Manual Training, Domestic Economy, etc. 1. A boy in manual training making a bookshelf finds the dis- tance from the top of the lowest shelf to the under side of the top shelf is 4 ft. 6 in. He desires to put between these four other shelves of inch boards in such a way that the book space will di- minish one inch for each shelf from the bottom to the top. What will be the several spaces between the shelves ? SUGGESTION. Let x = no. of inches between two lowest shelves. Then x + 1 = no. of inches to top of second shelf ; x -f 1 + (x 1) + 1 = no. of inches to top of third shelf ; etc. MANUAL TRAINING AND DOMESTIC SCIENCE 247 2. The height of a writing desk from the floor is 29| inches. Four drawers one above the other are to be inserted, the lower three being of the same height and the topmost drawer one inch lower. The drawers are separated by three strips of J-inch stuff, and the distance to the floor from the under side of the lowest drawer is one inch less than the height of the drawer ; also the distance to the top from the upper part of the upper drawer is one inch less than the height of the upper drawer. Required the several heights described. 3. In a bookstand whose shelves are adjustable, the distance from the top of the lowest shelf to the under side of the top shelf is 4 ft. 1^ in. If four shelves are inserted each f in. thick, and if the second and third spaces from the bottom are each 11 in. narrower than the one below it, and the two top spaces are each one inch narrower than the one next below it, how far apart shall the several shelves be put ? 4. A farmer noticed that when he had 100 Ib. of milk and 20 Ib. of cream he got out of both 12 Ib. of butter. At another time when he had 150 Ib. of milk and 25 Ib. of cream he got 16 Ib. of butter. What part of a pound of his milk is butter, and what part of a pound of his cream is butter? 5. How many gallons each of cream containing 33 % fat and milk containing 6 % fat shall be mixed so as to produce 10 gal- lons of cream containing 25% fat? 6. How much 95 % alcohol, 5 % alcohol, and water shall be mixed to make 15 gallons of 25 % alcohol, if the number of gal- lons of water used is one half of the total number of gallons of alcohol used? 7. The following formula has been obtained by experiments. If b = number pounds of butter and /= number pounds of fat in 100 Ib. of milk, b = 1.16 (/- 0.2). Find by formula number of pounds of butter in 70 Ib. of milk testing 4.4 % ; also, find per cent of butter fat when 500 Ib. milk yields 19.7 Ib. butter. 248 APPLICATIONS OF ALGEBRA The statistics in the following table were taken from Bulletin 28 of the United States Department of Agriculture. It should be stated here that the beef per cents given are for porterhouse steak. The cuts from the different parts of the animal contain, of course, varying per cents. FOOD PEK CENT OF PROTEID PER CENT OF FAT PER CENT OF CARBOHYDRATE -Ot-'t/I. * 22 20 Bread ...... 9 1 53 Butter 1 85 Cabbage 2 .3 6 Corn bread 7 5 46 Eggs 15 10 Fish (bass) 19 3 Milk 3 4 5 Oatmeal ...... 1(3 7 68 Peas (green) 7 1 17 Potatoes. 2 .1 18 Rice . ... 8 79 Wheat 14 2 72 8. A normal daily ration as given by one authority consists of 4.5 oz. proteid, 2 oz. fat, and 18 oz. carbohydrate. How many ounces each of bread and butter are needed to contain the right amounts of proteid and fat? SOLUTION. Let x = no. oz. of bread, y = no. oz. of butter. Then, using the per cents given in the table, we have (1) .09 a; + .Oly = 4.5 (2) .Olar+ .85^ = 2 (2i) . 09 a; + 7.65 y ==18 7.64y = 13. y = 1.8 x = 50 (Mult. Ax.) (Sub. Ax.) (approximately) Ans. (approximately) Ans. NOTE. This result shows that bread itself approximates to a normal ration, since only a little butter needs to be added. PROBLEMS IN MANUAL TRAINING 249 9. To contain 4.5 oz. of proteid and 2 oz. of fat, how many ounces each are needed (1) of bread and beef ? (2) of bread and eggs? (3) of bread and milk? 10. To contain 4.5 oz. of proteid and 2 oz. of fat, how many ounces each are needed (1) of beef and eggs? (2) of bread and fish? (3) of oatmeal and milk? a. Notice in these problems that an excessively large number of ounces of any food would be out of the question, and that a negative answer denotes impossibility of fulfilling the requirements. 11. To contain a daily ration of 3.1 oz. of proteid food and 18 oz. of carbohydrates, how many ounces each are needed (1) of rice and milk? (2) of bread and cabbage? 12. Dinner for an English canal laborer consists of bread, beef, and tea. If the food contains 2.48 oz. of proteid matter and 1.68 oz. fat, how many ounces of each are used? 13. If an agricultural laborer in Ireland gets .68 oz. proteid food, and 5 oz. carbohydrate from a supper of potatoes and corn bread, how many ounces of each does he eat? 14. The daily ration in the German army in war contains 4.7 oz. of proteid, 2 oz. fat, and 17.5 oz. carbohydrate. How many ounces each are needed of bread, peas (with bacon added to make fat 5 %), and eggs to make up this ration? 15. If a daily ration for a girl of nine years is desired of cereal (ground wheat), milk, and eggs consisting of 2.44 oz. proteid, 1.62 oz. fat, and 7.21 oz. carbohydrate, how many ounces of each ingredient are needed? 16. If Portland cement contains 25 % of silica, 10 % of alumin- ium, and 65 % of lime, and limestone rock containing 60 % of lime, 24% of silica, and 4% aluminium, chalk containing 70% lime, 9% silica, 2% aluminium, and clay containing 3% lime, 50% silica, and 45% aluminium, are mixed to make the cement, how many cubic feet of each are needed to make 10 cu. ft. of cement ? CHAPTER X FRACTIONAL EXPONENT QUANTITIES AND RADICALS I. FRACTIONAL EXPONENT EXPRESSIONS 162. Formulas giving the Laws of Exponents in the Fundamental Operations. The laws for integral exponents were given in 34, 36, and 38. They can be expressed in a few formulas. 1. Addition and Subtraction Formulas : ax m + bx m = (a + b)x m -, ax m - bx m = (a - b) x m . 2. Multiplication Formula : x m x x n = x m+n . yin 3. Division Formulas : x m -f- x n = x m ~ n j - = x m ~ n . x n 4. Power Formulas : (x m ) n = x mn ; (xyz) n = x n y n z n . m m p 5. Hoot Formulas : -tyx = x n ; \J x m y* x"y\ Evidently m in the fractional exponents of the last line denotes the power to which its quantity is to be raised, and is called the exponent or index of the power; and n denotes the root to be taken, and is called the index of the root. 6. It may be proven that any letter in formulas 1-5 may have any value except when a divisor becomes zero. Thus, any letter in the preceding formulas may have any value, positive, negative, integral, fractional, or irrational, as V3. There are limitations, however, that have to be put on the meaning in cer- tain cases, as will be explained later. 250 FRACTIONAL EXPONENT EXPRESSIONS 251 Simplify the following, noting which formula applies, and how. 7. 3x 7 + 4x 7 =? 8. x*xx 5 =? 9. 2z 4 -- 2 . 10. 9x 3 x4ar>. 11. 12 a 4 -=-3 a. 12. 5m 2 2m 2 . 13. (ra 2 ) 3 . 14. ^a 5 . 15. 7a^-2a 16. 3a 3 x7a 2 x2. 17. a 4 x 3 a,- 3 -=- a 2 . 18. S^ 12 -h (a 2 ) 2 . 19. V^y. 20. (a& 2 ) 3 . 21. 24a 3 6 2 -6a6. 163. Rules giving the Laws of Exponents, whether Integral, Frac- tional, or Negative, in the Fundamental Operations. 1. In addition add the coefficients of similar quantities for the coefficient of the common part in the sum. 2. In multiplication add the exponents of the same quantity in the factors for the exponent of this quantity in the product. 3. In division subtract the exponent of a factor in the divisor (denominator) from the exponent of the same factor in the dividend (numerator) for the exponent of this quantity in the quotient. 4. In raising to powers multiply the exponent of a factor quantity by the index of the power to which it is to be raised for the exponent of this quantity in the answer. 5. In extracting roots divide the exponent of a factor quantity by the index of the root for the exponent of this quantity in the answer. The formulas 162, 1-5, and their translation into words, 163, 1-5, should be thoroughly memorized. 164. Exercise in dealing with Quantities containing Fractional Ex- ponents. 1. We have Vc? = a 2 ; -\/o~ 6 = a 2 ; v / a i2 = a 3 . Then, if we fol- low the same rule in solving vV, we get -\/a 7 = d s . Similarly, Va? = cA Thus, as stated in 162, 5, the numerator of a frac- tional exponent denotes the power to which its quantity is to be raised, and the denominator denotes the root to be extracted. 252 FRACTIONAL EXPONENT QUANTITIES AND RADICALS 2. Calculate 8*. SOLUTION, v^ = 2 ; 2 2 = 4. Thus, 8' = 4. It shortens the labor in such problems to get the root first and the power required afterwards. 3. Calculate in the same way : 4^ ; 16- ; 27^ ; 64^ ; 8* ; 4^ ; 16'; 125*; (i)*5 (^) J (-27)* 5 (-216)*; 1*; 0*. 4. SUGGESTION. What does the exponent \ denote ? ( 162, 5.) 5. (4a 2 -12aZ> + 96 2 ) = ? (a 3 - 3a 2 6 + 6. Express with fractional exponents : Vaj 6 (-4ns. / 7. Express with radical signs m (^4?is. v^i 3 ); c 3 ; r 3 ; ?/ 71 . HI _ 3 s. 5Vm 3 ); 2 a 5 ; 3a^j 5?/ 4 ; aa;*. 8. Simplify 2 a^ x 3 a^. Ans. 6 a 5 . EXPLANATION. The coefficients 2 and 3 are multiplied together, and the exponents \ and | are added for the exponent a in the product, just as in the multiplication of quantities with integral exponents. 9. 2ox5o 10. 3ax4c 11, 3ax-- 12. 2 m* X 5 mi 13. aM x a-fri 14. M x a 15. ^-j-aji 16. -4a^-j--a'. 17. 9 18. m i x ^- 19- (w*). 20. (3 a 1 ) 3 . wi* (^Ins. 27 a*). 21. (5a ! ) 5 . 22. (a*)*. 23. FRACTIONAL EXPONENT EXPRESSIONS 2.38 24. Multiply x + x*+2 by x-\-x*--2 and check product by letting x = 9. The 9 is chosen so that its square root can be extracted. SOLUTION. x + x* + 2 14 - = x 10 140 , -f a; + 2 33. Divide a 3 6 a 3 b 3 16 b 3 by SOLUTION. as 6 as&s - - 16 &5 at + 2 - 8 _ 8 a - 16 Prove by multiplying divisor and quotient together. ,4 _ 2,1 34. x 2 + 2 2 + x - 4 = 140 25. (y* + y* -f 2/*-f-l)Q/^ -1). 26. 27. (a 3 -f a"3& + b 3 )(a 3 - - a 3 b 3 -f 6^. 28. 29. (x^y* + a' 5 ?/ 3 + y 7 ) x 3 y 3 . 30. (.1^ -f- 5) (V s - 4). 31. (a-\-a 2 b^a 3 b T )(a-\-a^b r2 -\-a^b 3 ^). 32. (a 3 -- 4)(a 5 + 3). 35. (8 m + 12 w 3 + 6 mn^ + ?l 2 ) H- (2 m + 36. (a 2 - 3 a + 6 a* - 7 a + 6 a* - 5 a^ + 1) -r- (a - a^ + 1). 37. (x* - xy* + x^y -- y*-) (x + x%y% + ?/). 254 FRACTIONAL EXPONENT QUANTITIES AND RADICALS 165. Quantities involving Zero and Negative Exponents. a 5 1 1. We have - - = by dividing both terms by a 5 ( 60) ; a 7 or i a 5 _ _ 2 (By subtracting the exponent in the divisor from aiso - el . a 7 the exponent of the same letter in the dividend for the exponent of that letter in the quotient.) a~ 2 = ' (Things equal to the same thing are equal to o each other, by Axiom 7.) It is clear that by taking other exponents than 5 and 7 the negative exponent could be made any number. Hence, any quantity ivith a negative exponent is equal to one divided by the quantity with the sign of its exponent changed. ab~ z 1 1 In ;-, if we replace b~ 3 by and c~ 2 by , we get or. (70.) Hence, -=. T~ Wd c~-d b s d - 2 Xd cr Tims, we see that any factor of the numerator or denominator can be transferred from the numerator to the denominator or from the denominator to the numerator of a fraction by changing the sign of its exponent. Integral quantities having negative exponents are supposed to have 1 for denominator. Change the following expressions to equivalent ones with posi- tive exponents, simplifying when possible : . Ans. 3. T^T- 4. a~ 3 . ab 2 a~-b- 6. .. .. 7. 8. o^xcr 4 . 9. a 2 -=-a- 2 . 10. (a 3 )- 4 . 11. (m~ 2 ) ZEKO AND NEGATIVE EXPONENTS 255 3 b 2 c 12. To study zero exponents take the problem 3c Dividing both terms by b' 2 , we get -- ( 60.) Cv But if we follow the rule of division and subtract the exponent of b in the denominator from that in the numerator for the ex- 3 bc ponent of b in the quotient, we get - -- Thus, 6 means that b is \AJ used no times as a factor of the result, that is to say b has dropped out of the result. b 2 b 2 We have - = 1 ; also - - = 6 ; therefore b = 1. Or. tr cr ^4?i?/ quantity whose exponent is zero is equal to unity. (See p. 274.) 13. What is the value of a n ~ 2 when n 2 ? 14. Simplify ; . 15. 1000' = 2 2 r> 12 16. 3 3 =? 3 2 =? 3 J = ? 3 = ? 3-*=? 3" 2 = ? 3~ 3 Express with positive exponents : 17. or 4 /. Ans. %- 18. a;- 1 ?/- 2 . 19. m~%*. a; 4 20. Sm- 1 ^. 21. 4cf*. 22. a-^V 3 . 23. 8 a~ V. Write the following fractions without denominators : 24. -^-- ^%s. 3a-W. 25. -4-- 26. (* ** n -/i- 1 ns* ** y C(/ L/ *v 2 3 1 Q 1 27. -^-- 28. ? - 29. ^ - 30. a* 166. Miscellaneous Exercise in the Use of Fractional and Negative Exponents. It is best to begin by changing any quantities written in the radical sign notation to the fractional exponent notation; also, as a rule, to change negative exponents to positive ones as ex- plained in 165. However, in many exercises it is shorter to 256 FRACTIONAL EXPONENT QUANTITIES AND RADICALS follow the simple rules for exponents, 163. Thus, in such an expression as (2cT*)~ 3 9 it is better to follow the rule of multiply- ing the exponent of each factor by the index 3 of the power. 1. 25*. 2. 4- 2 . 3. 16~i 4. a 3 x a~ 2 . 2. 4- 2 . 3. 16~*. 4. 6O"~T o . 7. (-8)*. 8. 10. (a*) 2 . 11. (a- 4 ) 2 . 12. 14. ( a^) 4 . i <=> /'S"^ 4 J. O. I O 1 16. 18. (x*)^. 1Q -\ / 7^ _s_ -\ //r J.J7. V w ^*^ \ " 20. 22. (8i~M. 9^ ^TM-^'tN-S A .) ' //(/ IV I m 24. 5. a 2 -*- a~ 4 . 6. 8~i 7. ( S)i 8. x* + x 9. x% -f- af i 10. a* 2 . 11. a~ 42 . 12. aM x 13. 17. (A/3) 8 . 18. (a;*)*. 19. #?-*- V. 20. ^2 X \/2~ 3 21. 2-5-V2. Solve the following fourteen problems mentally) quoting in each case the theorem used. See 45- 47. 25. (a* + &*) (a* b^). 26. (a 6)-s-(a* -- 6^). 27. (a^-6*) 2 . 28. (a + 1 - cr 1 ) 2 . (137, 14.) 29. (a;* - 6) (a* + 5). 30. (or" - 3 Jr 2 ") 3 . 31. (^ + 2) 3 . 32. 33. (x + 27)-5-(x3+3). 34. (16 - 4 -24a- 2 35. Factor 9 or 2 . 36. Factor 27 - - b~ 3 . 37. Factor a 2 + 2 + a~ 2 . 38. Factor 12 -- a; l - x~ z . 39. 40. (4 a 2 - 9 + 30 a- 1 - 25 a~ 2 ) -5- (2 + 3 a' 1 - 5 a~ 2 ). 41. (3a- 5 -4a- 4 +5- 3 )(2a- 2 -a- 1 ). 42. y(9a-12rt* + io_4a^ + - 1 )- (140.) 43. J/(x% - 6 oj + 15 ^ - 20 + 15 af ^ - 6 a- 1 + a;-*). 44. (a?* - 2 7/3-)* ( 136) ; (2 ar 1 + 2/~^) 4 ; (3 af * -- 5 ?/- 2 ) 3 . RADICAL EXPRESSIONS 257 II. RADICAL EXPRESSIONS 167. Radical Quantities. An indicated root of a quantity is called a radical quantity, or simply a radical. Thus, V4, A/7, 5-. In 3 V2a 3 , or 3 (2 a 5 )"*, 3 is called the coefficient of the radical, 4 is the index of the root, and 2 a 5 the radicand. Radical quantities may be either rational, as V}, or irrational, as V5. A rational quantity is the ratio of two integral numbers. An irrational (or surd) quantity cannot be the ratio of two inte- gral numbers. To show this by illustration, we note I = -625, [ f = .428571428571428571 ., while V5 = 2.2360679775 .... Notice that values of surds neither end nor repeat figures in sets. 168. Principles used for Reductions in Radicals. We have seen ( 163) that the same rules hold for fractional as for integral exponents, ; If, then, (abc) 2 = aW, (a&c) 8 = aW, etc., it ought to be true that (abc)% = aMc^, (a&c)* = a*6*c*, etc. ' Thus, (4 x 9)* = 4* x 9^ ; (8 x 27)* = 8' x 27* ; We have (4 x 9)* = 6, and 4^ x 9 * = 6. If, now, the sign on one side of (4 x 9)* = $ X 9^ were taken -K and that on the other side , the equation would not be true. To avoid this difficulty, only positive roots are used. C hanging formula, (abc---) 11 = a n b n c u into a ruUj we have 1. Fundamental Principle. A root of a product is equal to the product of the same root of the several factors ; and conversely. 258 FRACTIONAL EXPONENT QUANTITIES AND RADICALS By means of this principle any factor within a sign denoting a root may be removed outside the sign provided the desired root of the factor can be extracted. Thus, 12* = (4 x 3)* = 4* x 3* = 2(3)*, 16* = (8x2)* = 8^x2* = 2(2)* 2. If both terms of a fractional exponent are multiplied by the same number, the value of the quantity is not changed. Here, again, we find it necessary to limit ourselves to arith- 1 2 metical numbers. For, while 4 T = 4, 4* may equal either 4. 3. A quantity with a fractional exponent may have its value cal- culated either by raising to the power first ( 162, 5) and extracting the root afterwards, or vice versa. Thus, ' 8* = (8 2 )* = (8*) 2 = 4. This principle also holds true only for arithmetical numbers. Thus, 4* = 16* = 4, while (4^) 2 = + 4 only. 169. The Double Notation for Radicals. We have already seen ( 164) that fractional exponents as well as radical signs denote roots. The radical sign to denote roots came into use (1551 A.D.) a century before fractional exponents were thought of. Unfor- tunately, the existence of this duplicate notation has increased for students the difficulty of learning the subject of radicals. Change the following quantities from the radical sign to the fractional exponent notation, or vice versa : 1. -\/S. Ans. 8*. 2. V162. 3. 4. "V^f- /T nr Vra' 2 x 32 68 69. 71. w . 74. 276 75. 3(f)i / . 9 " ~ 78. a 262 FRACTIONAL EXPONENT QUANTITIES AND RADICALS 172. Miscellaneous Exercise in Simplifying Radical Expressions. 1. V80. 2. V288. 3. 2(175)1 ' 4. VlOO tfy 6 . 5. v'aW. 6. (36)* 7. A/289. 8. 9. ^64a. 10. -N/r 11. 4V27 a 3 6 5 . 12. ^v /3000 13. \- - 14. -V^ - 15. V8--206 2 . 16. 3(18a?-9)* * a- a V 3 6 2 Ans. to Ex. 15. 2 V2 5 6 2 . How is it obtained ? 17. Describe the three kinds of simplifications we are using. 18. Make rules for solving each kind. 19. 3645. 20. Sm 21. 22. V5 ar - 10 xy + 5 if. 23. V432 a 3 6 5 . 24. V50 ab' 2 c 2 28. f--1 29. -[--] . 30. b Sa* cb 31. . 32. . 33. 34. V(a 2 -6 2 )(a + &). 35. V189. 36. vVfrV 4 ". "a 3 " . 38. (xyy b\d J 40. a- ~ 41. .. , 97\i- 43. ( - ] 44. 3a 2 A / 648a 71 . 45. V196 x 98. or ADDITION AND SUBTRACTION OF RADICALS 263 46. V75 or 5 - 100 x s . 47. V4 a 3 - 24 a?x + 36 I 2 a . \/- \2-- W625 z 48. \/- - 49. a-f-36 50 . . 51. \ 512 a 6 (a 173. Putting Coefficients of Radical Quantities under the Root Sign. It is sometimes desirable to reverse the operation of the last articles, and, instead of removing factors from under the root sign, to put coefficients back into the radicand. 1. 3(5)*. SOLUTION. 3(5)* = 9* x 5* = (45)*. ( 168, 1.) 2. 4(3)1 3. 5VT4. 4. 6a(36)s 5. a 6. 1(4)1 ^n*. (!)*=(*. 7. 4(3)* 8. T * r 9. 3aVH. 10. 2SX5. 11. T 12. aV6. 17. Which is greater, 3 V7 or 4 V4? 3V3or2V6? 3S/2or2\/7? 174. Addition and Subtraction of Radical Quantities. When radical quantities are simplified, as explained in 170, they can be added and subtracted by adding and subtracting their coeffi- cients, provided the quantities are similar, that is, have the same radicand and same root index ( 23). 1. Simplify 3 V45 + V20 7 V5 ; and check answer to two decimal places. SOLUTION CHECKING 3(45)*= 9(5)* ( 170, 2) 3V45= 20.12+ + (20) = + 2 (5) ( 170, 2) V20 = 4.47 -7 (5)* = -7 (6)* _ -7V5 = --15.65 Sum= 45)^ = 8.94+. 8 - 94+ - Check. 264 FRACTIONAL EXPONENT QUANTITIES AND RADICALS REMARK. The student may not have seen the object in simplifying radicals as explained in 170, while studying that article, but he ought to see it now. How much easier it is to calculate 4 V5 than to calculate 3\/45+ V20-7 V5~! Simplify and check to two decimal places the eight following problems. The checking here serves two purposes : it checks the accuracy of the radical reduction, and also of the root extractions. If desired, check those problems which contain letters by assign- ing values to the letters. 2. V27 + V12 + V75. 4. 50* + S* - 98*. ' 6. 600* -32-* + 108*. 3. 8V125 + 2V80-3V45. 5. 44 -2 7. 250* - 16* - 54*. 9- 10. 2V3--1V12 + 4V27. 11. 40* + 25* + 2 (625)* 12. 12 (32)* + 5 (162)* -512*. 13. ^192-7-^9 14. 15. 17. 3/T !b_ ac ab ax- by 2 4 ax? by* 20. T 21. (4 a 22. 3 + 6(75 a y + V3 a (a - 9 &) 2 . - ^/375^ + -v/54^. 23. 18 5 9 ' 24. Make a rule for solving exercises in the addition and subtraction of radicals. How can the answers be checked? 25. 26. MULTIPLICATION AND DIVISION OF RADICALS 265 tr. 28. SUGGESTION. When simplified, are all the radicals similar ? See 29, 22. 29. 5 V363~a 3 F- 3 V242 3 /. 30. V2^-4o^-f-2a - V2 c^ + 4 00; + 2 a. SUGGESTION. Factor quantities under the radical signs. Ans. 2 V2 a. 31. \ / *32. (5 a 5 + 30 a 4 + 45 a 3 ) * - V5 a 5 - 40 a 4 + 80 a 3 . /a -I- a? 34. (a fljj-v^ 11 ^ a(a- 35. ^p^ + -- Va 3 6 - 4 a 2 6 2 + 4 a& 3 . 36. VHJ+3V8- V^O-iVMS-SV 11 ^"- V24f. 37. 38. 2 7 + 4 SUGGESTION. Reduce decimal to common fraction. 175. Multiplication and Division of Radicals having a Common Index. Before working, change problems given in the radical sign notation to the other, and then change the result back. 1. Multiply 2 V 14 by 3 V21. SOLUTION. 2 (14)* x 3 (21)^ = 6 (14 x 21)^ ( 168, 1) = 6 (2 x 7 x 3 x 7)* CHECK. 7.48+ x 13.75 = 102.85. = 42 V6 = 102.88. Steps in the solution : (1. Multiplying coefficients for coefficient of product. (2. Multiplying radicands for radicand of product. Result merely indicated. 266 FRACTIONAL EXPONENT QUANTITIES AND RADICALS (3. Separating each number in resulting radicand into its prime factors. (4. Simplifying the result as in 170. In division of radicals substitute the words " divide ' : for " multiply," and "quotient' 1 for "product" in the foregoing steps. 2. 3(8)^x5(6)1 3. 2Vl5x5V27- 4. V20 x V30. 5. 2(6)* x (18)1 6. V} X Vf 7. 4 Vf X V|. 8. 8V70-iV63. Ans. 16 ()* = - 1 / VlO. 9. 3(2)^-2(3)*. 10. (15)*-i-(f)* n. 2V6-SX|. 12. V21-S-VJ. 13. 9 (7)* -i- 2 (21)* 14. 5Va^2V3o~. 15. (72)* -i- (81)* . 16. (*)*x(|$)*. 17, aV^X&Va. 18. tycMftf + -\/cfl>y. 19. A/n-iA/i. 20. V55 #?/ X V 66 21. a6xcxca. 22. (x~yf X 23. -\/168 x ^147. 24. (48)* -s- (56)* 25. 7V75^5V28. 26. 6 (7)* -s- 126* 27. ^a-j-Ca 28. 29. Make a rule for the multiplication and division of mono- mials having a common root index. 30. SXI6 x 5 v/4 x 3\/|. 31. ^343 x ^- 1568. 32. VjxVJxV}. 33. 34 (1 J y )^ -5- (5f)*. 35. J/3Tc X 36. Va 6x Va 37. (|m 8 )H-(|m) 38. A8a 2 - (2 a MULTIPLICATION AND DIVISION OF RADICALS 267 39. Multiply 2V2 + V3by4V2--3 V3. SOLUTION CHECKING 2(2)*+ 3* 4.56 4 (2)* --3(3)* x .46 8 x 2+4(6)* 2.0076 _ 0(6)* - 3 x 3 16 - 2(6)* - 9 = 7 -- 2 V6 = 2.1. 40. (2V5- V7)(3V54V7). 41. [2(6)* -3(5)*] [3(3)* -f 2(5)*]. 42. (L>V6-7V7)(4V34-2V5). 43. (4V6-3V745V8) x6V2. 44. (7 + 3V7)(2V7-7). 45. (2a43V#)(3a-2 46. (Vl5-V2043V21)x V5. 47. (VI5- V3) -5-V3. *48. (V6-2V?>4 V7)-f-V2. 49. (a-6Vc)(d-ev7). 50. (or - ab V2 4 ^ 2 ) (o? 4 &6 V2 4 & 2 )' 51. (2* 4- 3* -5*) (2* + 3* + 5*). 52. [2(8)*4-3(5)*-7(2)*][72*-5(20)*-2(2)*]. 53. riVi-Wi-f 54. a?a?-- a?y 4- 55. [a? 4 2/^][ Vx 4 . Multiplication and Division of Radical Quantities which do not have a Common Index. Radical quantities that do not have a common index can be reduced so that they will have a common index, and can then be multiplied or divided. The operations depend on the principles of 168. 268 FRACTIONAL EXPONENT QUANTITIES AND RADICALS i, SOLUTION (2 x 3)^ X ( 32 x 2)* = (2 x 3)6" x (3' 2 x 2)* ( 168, 2) = (2=2 x 3*)* x (3 6 x 23)^ ( 168, 3) = (2 2 x 3* x 36 x 2 3 ) ( 168, i) = (3 8 x 2 5 )* = 3(25 x 3*)* =3 #288. ( 170, 2) Steps in the solution : (1) Separate each of the radicands into its prime factors ( 49), and if not already so written, put in the fractional exponent notation. (2) Reduce the fractional exponents to a common denominator. (3) Raise each radicand to the power denoted by the numerator of its fractional exponent. (4) Change the product or quotient of the same roots to the like root of the product or quotient by converse o/ 168, 1. (5) Simplify the result by 170, and, if called for, change, back to the radical sign notation. 2. V2 x \/4. 3. VI2 X ^14 4. ^I^F x V7 x. 5. V6 x -v/9. 6. A 3 /4 X -\/6. 7. A/32 - 8. 6-V2. 9. A/4 x-if -r- ^2^. 10. 11. 10 -s- 6*. 12. 20-5-3(10)*. 13. 14. +f 15. afn-ii. 18. 17. -\/10x4-v / 2. 18. 2^x3^x5^ 19. 6(54) -3(2)i 20. 2 S/:5 - 3 VL2. 21. V^-^-V^. 22. -x/iX^fWf. 23. (5 + ^4 + 2 v / 5)( V 6+ V5). ( 35, 82.) 24. RATIONALIZATION OF DENOMINATORS 269 177. Rationalization of Denominators. (See 170, 4.) To ration- alize the denominator of a fraction is to cause the irrational quantity in its denominator to disappear. This is accomplished by multiplying both terms of the given fraction by some quantity. T 2 2 32 2(3)^ 2V3 1. - SOLUTION. x - - = - V3 i i 3 3 CHECK. A = _L_ = 1.155- ; = - = 1.155-. V3 1-732 3 3 The advantage in this reduction consists in its replacing one root extrac- tion and one long division, by one root extraction and one short division. 3 4 V2 _ V^ A Vaj- -1 2. . 3. . 4. 5. 6. V7 V5 V7 V& 7. - SUGGESTION. Multiply both terms by \/3. 8 Q * in 10 11 12 8. --9. - 10. 11. -tyx 2 3 A/25 5 V3 13. Rationalize the denominator of = - = Vo+ V2 SOLUTION. When the denominator is a binomial, the multiplier is found in a different way from that when the denominator is a monomial. In the case of a binomial denominator both terms of the given fraction are multiplied by the denominator with the sign between its terms changed. This multiplier is called the conjugate surd to the given denominator. 32 5? _ 2^ 15^ - 6^ VT5 - Vd x = 5-2 3 n A ~. - ,. CHECK. -=.47+; = .47+ V5 + V2 REMARK. Here by the rationalization we replace three root extractions and one long division by two root extractions and one short division. 270 FRACTIONAL EXPONENT QUANTITIES AND RADICALS 14 V3 + V2. 15 1 + 2V3. 3(5)1-2(2)* V3-V2 5--V3 7*;-l 2-V3 5-2V5 178. Powers and Roots of Radical Quantities. See 166. III. IMAGINARIES 179. Imaginary Quantities. A pure imaginary is an even root of a negative number, as (- - 1) 2 . The sum of a real number and a pure imaginary, as 3 + 2 V- - 1, is called a complex number. 180. Exercise. 1. (V- - 1) 2 = - 1, by the^definition ( 18). The letter i is often used to denote V--1. Then 2 = 1. Hence, i 3 = i 2 x i = * ; ^ 4 = r X *' 2 = 1 x 1 = -f- 1. 2. Add 3 + 2 V : - 1 and 5 + 7 V : - 1. Ans. 8 + 9 V- 1. 3. 4. V : 25 + V36 = ? ^w*. lit. 170,2. 5. V-4a 2 6. 3V-20a V 80a=? ^4s. 2V- a. 7. (6 + 2V--l)(3-4V- -1)=? 9. (6 + 3) (7 + 4 S). ^3 = V3 i 10. (4-6V6) 2 . 11. 6V3^-2V5 = ? Ana. f V5. IV. EQUATIONS CONTAINING RADICALS 181. Solution of Equations Involving Radicals. After putting one radical by itself on one side, apply power or root axioms. 1. Solve x% = 27 for x. SOLUTION. (x 2 ) 3 = 27. x 1 ~ 3. (On extracting root of both sides denoted .. x = 9. by numerator of fractional exponent.) SOLUTION OF EQUATIONS CONTAINING RADICALS 271 2. 0* = 8. 3. Vz = 4. 4. 4ajf = 5. J 3^ = 25. 6 - \ / ? = 81. 7. oH==5. 8 Vis - 4> 7 Q . ^/ I . i/. 11. 3 Vie- 4 = 11. SOLUTION. 3 Vx = 15. (Transposing 4, Sub. Ax.) 9 x = 225. (Squaring both sides, Power Ax.) x - 25. (Ax. ?) VERIFICATION. 3V25 4 = 11, or 15 - 4 = 11. 12. l+Va = 5. 13. l+Va?--l=3. 14. 8 V3a; 6 = 14. 15. 16. a+3 = 2a;-10. 17. 18. l + 2v / = 7--VaJ. SUGGESTION. Transpose Vx and combine it with 2 Vx. 19. 2-\/x 3& = x--7^/x. SUGGESTION. After squaring, divide through by x. Then x = is a root. 20. Vo + # -f- V# = 5. SOLUTION. V5~+le = 5-Vx. ( One radical quantity on one side. Axiom used ?) 5 + x = 25 10 Vx + x. (On the right side we have the square of the difference of two quantities (5 and Vx) equals the square of the first, minus twice the product of the first by the second, plus the square of the second. What axiom is used ?) 10 Vx = 20. ( Tne ra( ji ca i quantity put by itself on one side. Ax. ?) vx = 2- (Dividing equals by 10. Ax. ?) x - (Power Ax. Quote it and ex- plain how it is used.) VERIFICATION. V5 -f 4 + V4 = 5, or 3 + 2 = 5* 272 FRACTIONAL EXPONENT QUANTITIES AND RADICALS 21. a? 4- 10 - - = 2. 22. 2 23. o?* + 2 = (a; + 3l > )-. 24. (15 + x)? = 3(6)* - a?*. 25. (a; + 25)* = 1 + 3*. 26 - ( + 13)* = 13 - oj*. SUGGESTION. Raise both members to the 3d power . 3 _ 27. v --1 = 2. 28. -v/9 a; - 4 + 6 = 8. 29. 30. V3~^ - A/3~+^~ 2 = 0. 31. 3+V-9a? = a?. 32. -\/x + 6 = a. 33. Va# + 2 a6 a = 6. 34. VaT+ar J --aj---J- = 0. 35. ?i = 9 - - Vn* + 9. 36. a + 2-Vl(> + a; 2 =0. 37. V* - 4 + 3 = V* + 11. 38. V4 a; 2 + 6 a; + 36=2 z+4. 39. V2(l - a;) (3 - 2 a,-) +1=2 40. (4a? l -7aj+l)*=2-lf 41. (3 x- 5)* + (3 x + 7)* = 6. 42. SOLUTION. 4 a + x = 4(6 + a) + 4 V(6 + *) + * (Ax.?) + a; 2 = 4 & + 4 a - 4 a. (Ax.?) x 2 = 6 + x - a. (Ax.?) bx + x 2 = 6 2 + x 2 + a 2 + 2 6x - 2 a& - 2 az. (Ax. ?. 137, 14.) 2 ax -- bx = a 2 -2ab + b' 2 . (Ax. ?) 43. Make a rule for solving radical equations covering the cases where there are (1) One, (2) Two, and (3) Three radicals each containing the unknown. So long as the radical quantities containing x remain in the equation, what plan is followed ? (See suggestions above.) How is the equation solved after it is cleared of radicals ? 44. V# a Vcu; + a 2 = Va, 45. aV# & = c V# d. HISTORICAL NOTES 273 46. In the simple pendulum, (heavy weight hung by string, swung slightly back and forth) the time t in seconds equals very approximately TT times the square root of the length in feet, I, divided by the acceleration of gravity, g, or t = %/-. Solve this equation, first for I, and then for g. 47. What is the length of a pendulum that vibrates in | sec- onds ? Use g = 32.2 and TT = 3.1410, substituting in the formula just found. 48. Solve v = a \/-- for r. 49. Solve v = V- yh for h. \ r _ _ 50. Solve = -y - for R. 51. Solve u = ~Vv- + 2/s for s. 182. Historical Notes. In the very earliest stages of the de- velopment of algebra, as already stated in 43, all demonstrations and solutions were written out in full in words, so that an algebra looked like any other book, as a reader or history. Later, abbre- viations were introduced, but grammatical endings and agreements were still employed, just as students nowadays often write ='s. Later still, symbols were used freely. To show how the notation of algebra has changed we will give Diophantus's syncopated notation. Following the usual Greek notation for numbers, using a for 1, b for 2, c for 3, etc. (see 203), he represented the unknown by a character somewhat like s (plural ss), its square by d y (for dynamis, power), its cube by c", known terms by m (for monad, unit), addition by juxtaposition, subtraction by /p, and equals by i. Coefficients were written after instead of before quantities.* Thus, c u d ss c v dy b m e i s a represents (4 x 3 + 3 x) (2 x- + 5) = x. * Instead of using the Greek letters, which are unfamiliar, the correspond- ing Roman letters are used. 274 FRACTIONAL EXPONENT QUANTITIES AND RADICALS As an example of a notation used by a writer of a time not long before our present symbols came into use we give that of Regio- montanus in a work prepared in 1464 and published in 1533. 16 census et 2000 aequales 680 rebus, that is, 16 y? + 2000 = 680 things. Our present symbols, as already stated, began to be used in the sixteenth century. As showing in a convenient form how they gradually came into use, we give on the opposite page in tabular form a list of these symbols, with information about them. The references in the right column are to Chrystal's " Text-Book of Algebra," Ball's "A Short History of Mathematics," and Cajori's " A History of Elementary Mathematics." This table shows clearly how we came to duplicate symbols for the same operation. Thus, apposition, x , and all came to be used to denote multiplication. Similarly, the fraction sign, -;- , and / all denote division, and the radical sign and fractional ex- ponents both are used to indicate roots. Sir Isaac Newton, in a letter to Oldenburg of 13th June, 1676, writes : " Since algebraists -I o , write a 2 , a 3 , etc., for aa, aaa, etc., so I write a 2 , a 2 for Va, V 3 , etc. : and I write or 1 , or 2 , etc., for -, , etc." The radical sign, a act as the table shows, had been in use for over a hundred years before mathematicians thought of using fractional exponents. The symbol oo denotes an exceedingly large number, greater than any that can be named. It is often associated with 0, the being thought of as the limit of an exceedingly small number that differs from by a number less than any that can be assigned, however small. These two kinds of quantities are called infinites and infinitesimals, and are largely dealt with in a branch of math- ematics called the calculus, which was developed by Newton and Leibnitz about 1670. A quantity that is neither infinite nor in- finitesimal is called & finite quantity. The theorem on page 255, strictly speaking, should read, any finite quantity whose exponent is zero is equal to unity. HISTORICAL NOTES 275 TABLE SHOWING THE ORIGIN OF THE SYMBOLS IN COMMON USE SIGN, OR CONVENTION YEAR INTRO- DUCED GEN- ERAL USE BY WHOM INTRODUCED REMARKS Division. Dividend over Divisor, Very early, Hindus and Arabs, This is the oldest sign. Oblique line also used. See Ball, pp. 213, 148. Chrystal, p. 24. Apposition fur Multiplication, 1200 See Chrystal, p. 24. + and - 1525 1630 Rudolff and Stifel, Diff . of opin. as to orig. See Cajori, p. 141. Ball, p. 185. = 1557 1680 Robert Recorde, Whetstone of "\Vitte. Ball, p. 191. v~ 1551 1633 17th C. Rudolff, Stifel, Girard, V V= = 4thrt., VVV = cube rt., etc. Girard used indices as we do. Letters and Yinculum, Letters, 1591 1637 Vieta, Descartes, Used capitals. Vowels for unknowns. Used small let. Last let. for unknowns. Parentheses, 1629 18th C. Girard, See Chrystal, p. 24. X, : : 1631 1686 Oughtred, See Cajori, p. 234. x, >,< 1631 Harriott, Caj., p. 234, Ball, p. 213. -f- 1630 Pell, Ball, p. 214. Pos. Int. Exp's., 1637 Descartes. In analytic geometry. Dot for Mult., 1(537 18th C. Descartes, Ball, p. 213, C., p. 194. Frac. Exp's., oc , 1658 Wallis and Newton See Chrystal, p. 201. ; 1760 Clairaut, Ball. p. 213. Asso. Com. and Distrib. Laws, 1840 Peacock, Gregory, and De Morgan, Cajori, p. 244. =,>, < Recent Ball, p. 214. CHAPTER XI QUADRATIC EQUATIONS 183. A quadratic equation is one which, when reduced to its simplest form, contains the second, but no higher, power of the unknown quantity ; or, a quadratic is an equation of the second degree (see 57). Quadratic equations containing but one unknown quantity are of two kinds, complete and incomplete. A complete quadratic equation contains both the second and first powers of the un- known, as 3 x 2 -f- 5 x -f- 7 = 0. An incomplete or pure quadratic contains only the second power of the unknown, as 6 x 2 24 = 0. I. INCOMPLETE OR PURE QUADRATICS 184. Solution of Incomplete Quadratics. The type form of such equations is mx 2 = n. To reduce to this type form it may be necessary to clear of fractions, transpose, and combine terms con- taining $. 3 _4s* ~ SOLUTION. 9 z 2 - 27 = 4 x" 2 + 18. (Ax. ?) 9 y? - 4 x 2 = 27 + 18. (Unknowns to left member; ,, 2 45 knowns to right. Ax. ?) z 2 = 9. (Ax. ?) x = 3. (Root Ax.) VERIFICATION. 9 3 = (36 + 18) *- 9, or 6 = 6. a. It is not necessary to write x 3, for x = 3 has the same roots as x = 3. Show that this is true by writing -f x = -f 3, -f- x = 3, - x 4- 3, - x = 3, and multiplying the third and fourth equations through by - 1. 276 APPLICATIONS OF INCOMPLETE QUADRATICS 277 2. 7o 2 -2o=5a- 2 + 73. 3. 9a 2 - 53 = 6^ + 94. 4. (o? + 4; 2 = 8a; + 25. 5. 3ar J + 12 = 7 2 -88. 6. (3-7)(3a + 7)=32. 7. (. + 1) (a? - 1) = T V 8. .c + 2# + 3=2a,- 2 + 5a. 9. a? + 3aj- 3= 66- 2x 2 . - t , or - a c a c 10. aa; 2 b = C.JT + d. ^4/is. \- or - - V ; -> c^Z --be cd. * a # __ a& - 3 a a; -f 5 6 11. -j 1/2. --- - u. x a x x 5 b 3 a + 10 b 13 , _ ., 14 a; 3 __ ' ' x-2 x 2 15. Make a rule for solving incomplete quadratic equations, stating (1) What is done when fractions appear in the equation. (2) What the next step is. (3) How the terms are collected. (4) What the last step is. _ x a 2 ,x + a? ^ x + a x a =1. 19. 0; 2 . a 2 a? a- a; + 6 x b x 2 b 2 20. V25 -- 6 a; + V25 + 6 x = 8. 21. V + a= 22. 205^+3 + 2+2=1. 23. 185. Applications of the Solution of Incomplete Quadratics. 1. The area of a circle is 3.1416 r 2 . What is the radius of a circle (to three decimal places) whose area is 30.1 sq. ft. ? 2. The surface of a sphere is 4 x 3.1416 r 2 . What is the radius of a sphere whose surface is 91 sq. meters ? 3. Solve s = \gt 2 for t, reducing the answer to the form re- quired by 170, 3. 278 QUADRATIC EQUATIONS 4. In the formula s = gt 2 , s is the distance a body falls, t the number of seconds it takes, and g, the acceleration of gravity, is 32.2 ft. Find by substituting in the answer to the preced- ing exercise the number of seconds it takes a body to fall 64.4 ft. 5. The volume of a cylinder is .7854 d 2 x I in which d is the diameter of one end and I is its length. Find to three decimal places the diameter of the end of a cylinder 14 inches long which contains 155 cu. inches. 6. If d and D are the diameters of the pistons of a hydro- static press (see dictionary for description) and p and P are the pressures respectively on the small and large pistons, it is shown in physics that cl L_ P D 2 ' Solve this equation for D, having p, P, and d given. Solve also for d when p, P, and D are given. Find by substitution in the answer just obtained what D is when p = 2 lb., P= 150 lb., and d = 1 in. mv 2 7. Solve /= for v. r 8. Solve /= p for t. Simplify result. ( 170.) t V 2< 3 9. Solve^=- for V and d respectively. Simplify. 8 10. How many rods of fence will inclose a square garden whose area is 2^ acres ? a acres ? 11. The diagonal of a square is 16 ft. Find one side of it. 12. Find the diameter of a down-spout for a roof 40 X 20 ft., using 1 sq. in. of pipe area for every 150 sq. ft. of roof area. 13. Let p and p' be the periods of revolution of two planets r> 2 d 3 about the sun, and d and d' their mean distances. Then E- = . p' 2 d' 3 Solve this equation for p and also p', simplifying as in 170. SOLUTION OF QUADRATICS BY FACTORING 279 14. What is the diameter of a pipe that will carry the same flow as four 1-inch pipes ? 15. A side of an equilateral triangle is 8 ft. Find altitude x. Generalize by putting a for 8, and - for 4. 16. The side of an equilateral triangle is 12 8 ft.. Find its altitude by substituting in for- mula of Ex. 15. Find also its area. Find the side when the altitude is 20 inches. 8 4 17. A regular hexagon is formed out of six > ^ 'I i til J_L VJ v Cl> ^ y equilateral triangles having a common vertex at its center. Find one side x of the hexagon if its area is 15 sq. ft. Draw diagram. 18. One side of a square whose four corners fall on the circum- ference of a circle is 2 ft. Find its radius. Generalize. 19. Radius of circle is 12 rods. Find side of inscribed (ver- tices on circumference) equilateral triangle. Generalize. SUG. Geometry shows distance from center to one side = radius -4- 2. II. COMPLETE QUADRATICS. SOLUTION BY FACTORING 186. Solution of Quadratics by Factoring To Construct a Quad- ratic, its Roots being given. Properties of Quadratic Equations. Whenever the roots of a quadratic are either whole numbers or simple rational fractions, the factoring method ( 57) is usually the easiest and quickest. 1. Solve and verify in x- -f 8 x = 33. SOLUTION, x 2 + 8 x 33 = 0, (Ax. ? Eight member made 0.) or, (x + 11 ) (x 3) = 0. (On factoring.) .-. x + 11 = 0, whence, x = 11, (On setting each factor equal to zero.) and x 3 = 0, whence, x = 3. VERIFICATION. (-- II) 2 + 8 x 11 = 33; 3 2 + 8 x 3 = 33. 2. x 2 --6x = 16. 3. x 2 -f 12 x = 64. 4. x 2 + 1 7 x = 0. 5. 2o?-5x = 3. 6. 6x 2 + l = 5a;. 7. 8a? 280 QUADRATIC EQUATIONS 8. Construct the equation whose roots are 3 and 4. SOLUTION. [* - 3] [x - ( - 4) ] = 0, or, y? + x 12 0. Ans. PROOF. (x 3) (x + 4) = 0. (On factoring.) .. x 3 = 0, whence, x 3, and x + 4 = 0, whence, x - 4. Evidently to form an equation ivhose roots are two given numbers, one subtracts each from x, multiplies the remainders together, and sets the product equal to zero. 9. Form the equations whose roots are : 4 and 6 ; 2 and 5 ; 11. 2 . 1 . _ 3. JL 2", ^, ^, 4 , 5, 3. 71. Q ft . 1 1 ^ 90 ft. '> J- > -*> - ) 1U, 10, ZU, - - D, 10. Form the equation whose roots are a and b. 11. In the equation or 2 -- 7 # + 12 = 0, what are the two roots ? What coefficient of the given equation is their product ? What coefficient with its sign changed is their sum ? 12. Examining 11 and- 10 above we see that if the coefficient of 3? in a quadratic equation is unity and the right member is zero, the coefficient of x with its sign changed equals the sum of the two j and the term which does not contain x equals their product. III. "COMPLETING THE SQUARE" METHOD 187. Solution of Complete Quadratics by Completing the Square. Whenever it is not easy to solve an equation by factoring, it can be solved by a process called " completing the square." (See 117.) To complete the square, we think of the well-known expression a 2 + 2 ab + 6 2 , the square of a + b, and consider that we have the first two terms given to find the third. (See 116.) To find 6 2 from a 2 + 2 ab, divide the second term by twice the square root of the first term and square the resuU. Thus, 2 ab -r- 2 Vtr' = b ; and squaring 6 gives b 2 . COMPLETING THE SQUARE 281 1. Solve and verify in the equation x 2 +- 14 x +- 45 = 0. SOLUTION. z 2 -f 14 x = 45. (Sub. Ax., known quan. to right side.) We have 14 x + 2 Vz 5 = 7 ; 7 2 = 49. Then, a;2 + 14 x _j_ 49 - 49 _ 45 - 4. (Add. Ax., 49 added to each side of the preceding equation.) x + 7 = 2. (Root Ax.) x = 7 + 2, or - 7 2, i.e. 5 or 9. ^4s. (Sub. Ax.) VERIFICATION (-5) 2 + 14 x --6 + 45 =0 25 - 70 + 45 = 0=0 (-9) 2 + 14 x -9 + 45 = 81 -- 126 + 45 = = 2. z 2 -6z-91 = 0. 3. x 2 - 4. aj"-4a;-77 = 0. 5. x- 2 +- 60 a? + 891 = 0. 6. a* + 8 a; = 768. 7. 23 = 120-0* 8. 24 = 11 a; -a 2 . 9. 225 = 30 x -x 2 . 10. oj* _ 341 = 20 a. 11. x = -34. x SOLUTION. 3 x' 2 = 57 - - 10 . (On clearing of fractions, Ax. ?) 3 x 2 + 10 x = 57. (Unknowns to left side, Ax. ?) The coefficient of x 2 must now be made a perfect square like a 2 in the for- mula p. 280. This is accomplished by multiplying the equation by 3. (It might be done by dividing through by 3, but this would introduce fractions into the solution, and it is usually better to avoid fractions.) 9 a 2 + 30 ae= 171. (Ax. ?) We now complete the square. Thus, 30 x -f- 2 V9 x 2 = 5 ; 5- = 25. Then, 9 x 2 + 30 x + 25 =: 196. (Ax. ?) 3x + 5 =: 14. (Root Ax.) Bx= - 5 14- (Sub. Ax.) x = 3, or L 9 . (Div. Ax.) VERIFICATION. 3 = ^ - - -^ ; also - - ^ = - - 3 . 3 282 QUADRATIC EQUATIONS 12. 2or J -7a + 5 = 0. 13. 9or 2 + 4z = 5. 14. 7 g 2 -f 26 = - 19. 15. a 2 -fa; = 32. 16. 2 2 -z = i. 17. igx = $ x*. 18. 8a 2 +-z = 30. 19. 3^ + 35 = 220;. 20. 18 a?- 27 a; == 26. 21. 18 a,- 2 + 27 x = 26. 22. 3x* + 10 = 17x. 23. (2 a? - 3) 2 = 8 x. 24. 2aj-3 = 3aj + 2aj-3. 25. a; + l(2a? + 3= 4 a; 8 -22. ofi 1 _ 27 = 4z+-7 : "oj + 7* 7-5"2aj-13 9. ^-4-.^ 28. aj 2 3 a; + l --! 2 ,+l = 13 " 31 a; a; " 6 aj + 1 ^' + 4 32. Solve the preceding by the factoring method, and compare the length and difficulty of the two solutions. It will be found that some are easier by one method and some by the other. 33. 5x 2 + 7x + l = 0. SOLUTION. 100 x' 2 -f 140 x = 20. (Mult, and Sub. Axs.) 100 x 2 -f 140 x + 49 = 29. (Completing square, Ax. ?) lOz + 7 = V29. (Ax. ?) 10 10 = _. 1615 , or _ L2385+. VEBIPIOATIO*. 5 7 ' + 7 29 - 14 V29 + 49 + 14 \/29 - 98 + 20 = 0. (On canceling and clearing.) a. In the verification only the positive root was used. That of the nega- tive root - x/21) 7 would differ from the other only in certain signs. COMPLETING THE SQUARE 283 Solve and verify in the following as in the model just given : 34. a^ + 3:c+l = 0. 35. 3xr-2x = 2. 36. 2^-7 37. x*-9x+12=0. 38. 7tf + l = 7x. 39. 3ar'-ll 40. 0^ 4^ + 7 = 0. See 179. 41. 2^5 #+4 = 0. 42. Make a rule for the solution of quadratic equations. (1) What is the first step when fractions appear in the given equation ? (2) What terms are transposed and where ? (3) How are the terms collected ? (4) What is the equation multiplied or divided through by (when necessary) ? Ans. By some number which will make the first term a perfect square and positive, and at the same time make the subsequent part of the solution as short as possible. See b, p. 284. (5) How is the square completed ? (6) What is the next step ? (7) The next ? (8) The last ? (Roots containing irrationals should be calculated to three decimal places.) (9) How is the answer verified ? In verification use radical form of the answer, as in Ex. 33. Give the axiom used for each step except the third and (9). What it is best to multiply or divide through by is learned from practice. Solve and verify in the following, using the suggestions : 43. In 7 tf + 21 x - 70= divide through by 7. 44. In 3 x 2 + 2 x - 33 = multiply through by 3. 45. In 2 x 2 17 x + 21 = multiply through by 8. 46. In 15 x 2 + 5 x 70 = divide through by 15. 47. In 18 x 2 - 12 x - 48 = divide through by 2. 284 QUADRATIC EQUATIONS 48. In 18 x 2 - 17 x - 111 = multiply through by 72. b. What is called the Hindu rule for solving quadratics avoids fractions. It is, Multiply the equation through by 4 times the coefficient o/x' 2 . This was done in the third and last of the equations just given. But it is not always necessary to multiply by so large a multiplier to avoid fractions. Thus, frac- tions were avoided in Ex. 44. 14 , + - -, = x-l. 9 2x-3 #+22 4__9a? 6 3 Ix 15 -5#- V2-7?/ = 49. 51. 53. 55. 56. 57. ax 2 + bx + c = 0. SOLUTION. ax 2 + bx 4 a' 2 2 + 4 abx 4 a 2 x 2 + 4 aftx + & 2 2 ax + b ert 50. - 3x 52. 4 o ^ 3 x -2 x + 1 X ~r~ O X O O 9 ^ " c, 4 ac, b' 2 - 4 ac, 2 4 ac, x 6 V6- -- 4c - (Ax. ?) (Hindu rule, Ax. ?) (Comp. Sq., Ax. ?) (Ax.?) (Axioms ?) Tr VERIFICATION. a . , + b A + c = 0. - 4 ac - 2 b Vfr 2 - 4 ac + b 2 4 a' 2 V62._4 ac _fr2 4a C ^ c. Careless pupils sometimes want to put b V- 4 ac, for VJfl - 4 ac, or the like. Now the square of a monomial, as 3 a, is a monomial, and the square of a binomial, as a + 5, is a trinomial. Hence, a binomial, as b' 2 4 ac, cannot be the square of a rational quantity. The square root of such binomials can only be indicated. See 29, 22. COMPLETING THE SQUARE 285 58. mx 2 + nx + p = 0. 60. tf-3bx = 62. x 64. 12x*-cx-2Q I C/ */ Lt'tC' -- \Ai\j \jvU 77. aj2--4aaj--10a; = --40a. 70 a _ ^ I C7 . 76. mx 2 -f- HO; = pic 2 - - g. 78. cic 2 2 cdx = ax 2 cd 2 . 80. a; a 4 lo 81. 83, 85. 82. 2s4- 1 t 84. V = 86. c Vl c 2 = n 286 QUADRATIC EQUATIONS Solve and verify in the following : 87. 3aj 2 + 8 + 3=0. 88. 5y a - 89. 2 a; 2 - 11 a? + 16 = ( 179). 90. 4^-12^ 91. 9aj 2 -6a?-2 = 0. 92. 1 111 93. - = ---- 1 (After simplifying, divide out by a 6.) a b -\-x a b x IV. THE FORMULA METHOD 188. Solution of Quadratic Equations by Means of a Formula We solve the equation ax 1 + bx + c = 0, regarding it as a type of any or every quadratic, getting (see Ex. 57, 187) 2a Now, having a given quadratic to solve, we compare its coeffi- cients with those of ax 2 -f bx + c = 0, thus getting values for a, b, and c. These values for a, b, and c are then substituted in the value of x just given. The result simplified is the answer sought. The student should practice solving the equation ax 2 -t-bx+c=Q until he is sure he can remember the process. One should know how to derive the formula before using it. 1. Solve 2X 2 9 a + 10 = by the formula method. SOLUTION. Writing the given equation underneath the type form, we have ax 2 + bx + c = 0, 2x*-Qx + 10 = 0. We see now that in this particular problem a = 2, b = 9, c = 10. Substituting these values for a, 6, and c in - 6 Vb' 2 4 ac jj 2a _C_9) V(- 9Y- 4 x 2 x 10 we get a = ^ '- ! or x- - -80_01_5 2 4 4 2' VERIFICATION. 2 x 2 2 - 9 x 2 + 10 = ; 2(|) 2 - 9 X f + 10 = 0. SOLUTION BY MEANS OF A FORMULA 287 Solve and verify in the following, using the model above : 12. o2_7a;_|_6 = 0. 3. a? + 8x-2Q = 0. 4. 6jf 2 -3oaj-6=0. 5. 2 a? + 13 a; + 15 =0. 6. 6^ + 37 a? + 56 = 0. 7. 4ar>- 4a?- 3 = 0. 8. + = <. 9. 934 2* + l oj-3 2x-\-l 5 #-8 --- = - 1-2* 7 2 SUGGESTION. Equations 8, 9, 10 must be first cleared of fractions and then the terms transposed so that they have the form ax 2 + bx + c = 0. 11. Make a rule for solving quadratic equations by the formula method. (1) Jn what form is the given equation put if it is not already in that form ? (2) How are the values of a, b, c in the formula obtained ? (3) What is done with these values ? (4) How are the two roots now found ? 12. 5oj 2 -4i=12. 13. 2^ + 7a;-4 = 0. 14. ar + 2_ SUGGESTION. Put in the formula 1 for a, 2p for 6, 3 q for c. 15. 4ar-6f?.r = 4rf 2 . 16. G or + 8 m 2 = 49 ma. - rv 9 I -| -| *? -I O - y\ 99 7OO y!7 9 T 9 17. or + 11 w = 12 ??i#. 1 8. rr.r crar 4 ctoaj = cr 0. SUGGESTION. Put in the formula d~ 6 2 for , --4 ab for 6, and -- a' 2 + 6 2 for c. 19. Solve problems 2-10 by factoring, and also by completing the square, and compare with the formula solutions. 20. Solve maj 2 4-wa?+p = 0. Then, from the value of a? found make a rule for solving an equation of the form mx 2 -\- nx +p = 0. ~ Hence, 288 QUADRATIC EQUATIONS V. PROBLEMS 189. Problems involving the Solution of Quadratic Equations. 1. A man bought a quantity of cloth for $120. If he had bought 6 yd. more for the same sum, the price per yard would have been $ 1 less. What was the price ? SOLUTION. Let x = number of dollars in cost of one yard, Then, - = number of yards bought. Qu x -- 1 = number of dollars in supposed price. = number of yards from supposed price. x 1 120 120 _ 6 x -- 1 x 120 x - 120 x + 120 = 6z 2 - Gx. (Mult. Ax.) 6 x 2 -Gx- 120 = 0. (Sub. and Mult. Axs.) x i - x 20 = 0. (Div. Ax.) (x 5) (x + 4) = 0. (Factoring.) .-. x 5 = 0, or x = 5, Ans. x + 4 = 0, or x = 4. a. The second answer, 4, can have no meaning in this problem, since the nature of the problem is such that negatives are inadmissible. Let us substitute for the given problem another which has the same numbers in it and gives rise to the same equation, but which admits of the entrance of negatives. Thus : The exchange account of a banker amounted in a certain number of days to $120, during which time exchange remained the same for each day. Had the period differed from what it was by 6 days and in his favor (that is 6 more if premiums or 6 less if discounts), it would have made a difference of\per day. What was his daily premium or daily discount ? To this problem 4 is as satisfactory an answer as + 5. The 4 means a daily discount of General Note. The solution of quadratics by factoring is not feasible when the roots are irrational or imaginary. When the roots are rational, as they commonly are in problems and exercises, the factoring method (review for it 52, 57) generally gives the quickest and easiest solution, even when the coefficients are large. When it fails, one of the others must be chosen. The formula method is shorter, but harder to remember than that of 187. PROBLEMS INVOLVING QUADRATIC EQUATIONS 289 2. Find a number such that if you subtract it from 10 and multiply the remainder by the number itself the product is 27. SOLUTION. Let x = number. Then, (10 x)x = 27, or, 2 10 x = - 27. a: 2 10 x + 25 = 2. (On completing the square.) x-5 = V~^2. (Ax, ?) x = 5 V^2. (Ax. ?) VERIFICATION. [10 (5 V^2) ] (5 V^) = 27. &. The imaginary answer 5 V 2 ( 179) shows that there is no real number which being multiplied by the remainder obtained from subtracting it from 10 gives 27. Thus, trying 8, we get 16 for product ; trying 7, we get 21 ; trying 6, we get 24 ; trying 5, we get 25 ; trying 4, we get 24. Neither will any negative number satisfy the conditions. To make this problem "real" it must be stated as follows: What complex number on Argand's diagram is it which multiplied by the difference between 10 and itself gives as product 27 ? c. Algebra is a formal science made to cover all cases, and without any reference to particular problems. In problems which by their nature admit of negative numbers, a negative answer has meaning ; in others negative answers are inadmissible. Moreover imaginary values for the unknown in ordinary problems denote that the conditions are impossible of fulfillment. 3. A railway train traveled 5 mi. an hour slower than usual, and was 1 hr. late in making a run of 280 mi. How many miles per hour did it travel? Solve first, and later generalize; solve and test generalized answer. Let a = number miles per hour slower, b = number hours late, c = number miles in whole dis- tance. See 123. 4. A company of gentlemen engaged a supper and agreed to pay $ 80 for it. Four of the gentlemen failed to attend, and each of the rest paid $1 more than he expected to pay. How many were present at the supper ? Generalize. 5. A man worked a certain number of days for $ 30. If he had received $1 a day less than he did, he would have been obliged to work 5 days longer to earn the same sum. How many days did he work ? Generalize. 290 QUADRATIC EQUATIONS 6. A picture that was 8 in. by 12 in. was placed in a frame of uniform width. If the area of the frame was e^ual to that of the picture, what was the width of the frame ? Draw diagram. Generalize. Check. 7. Two men, A and B, can do a piece of work in 20 da. If B requires 9 da. more than A to do the whole work, how many days would it take each to do the whole work ? Generalize. 8. A body of soldiers can form a hollow square 4 men deep. By re-forming the company with 36 men less on each outer side, the soldiers can form a solid square. Find the number of soldiers on one side of the solid square and the total number of soldiers. 9. The perimeter of a rectangular field is 180 ft. and its area is 1800 sq. ft. Find its dimensions. Generalize. Check. 10. The two digits of a number differ by 1 ; and if the square of the number be added to the square of the number with its digits reversed, the sum is 585. Find the number. 11. Find the price of eggs when 2 less for 30 / raises the price a dozen. Generalize. Check. 12. A crew can row downstream 18 mi. and back again in 7|- hr. Their rate upstream is 1|- mi. an hour less than the rate of the stream. Find the rate of the stream and of the crew in still water. 13. A boatman rowed 8 mi. upstream and back in 3 hr. If the velocity of the current was 2 mi. an hour, what was the rate of rowing in still water ? Generalize. Check. 14. Divide a straight line a inches in length into two parts so that the ratio of the whole line to the larger part may equal the ratio of the larger part to the smaller part. This was called by the Greeks dividing the line in "golden section." 15. The circumference of the fore wheel of a coach is 5 ft. less than that of the hind wheel. If the fore wheel makes 150 more revolutions than the hind wheel in going a mile, what is the cir- cumference of each wheel ? Generalize. PROBLEMS INVOLVING QUADRATIC EQUATIONS 291 16. Two points start out together from the vertex of a right angle along its respective sides, the one moving m ft. per second, the other n ft. per second. How long will they require to be c ft. apart ? 17. The hypotenuse of a certain right-angled triangle is 8 ft. greater than the perpendicular, and 4 ft. greater than the base. What are the lengths of its sides ? Generalize. Check. 18. Find two numbers whose difference is d and product is p. Verify answers. 19. The product of two numbers is p and their quotient is q. What are the numbers ? Verify. 20. It is required to find three numbers such that the product of the first and second equals a, the product of the first and third equals 6, and the sum of the squares of the second and third equals c. Verify answers. 21. A farmer sowed one year a hectoliter of wheat; the next year he sowed what he harvested the first year less b hectoliters, and reaped c fold of what he sowed and d hectoliters besides. Assuming a like fruitfulness both years, what did he reap the first year ? SUGGESTION. Let x = number of hectoliters reaped first year. Then, he reaped second year x times the number of hectoliters he sowed ? Why ? 22. A set out from C towards D and traveled 7 mi. per day. After he had gone 32 mi., B set out from D towards C and went every day -jL of the whole journey, and after he had traveled as many days as he went miles in a day, he met A. Required the distance from C to D. Verify. 23. In a school field meet one runner won over another in the 100-yd. dash by half a second, and his speed was .4 yd. greater each second than his competitor's. What was the time of each ? SOLUTION. Let x = number of seconds the winner took. x 4- .5 = number of seconds the other runner took. Then, -- - = .4, whence a? + .5 x = 125, giving x = 10.933+. x x + .5 VERIFICATION. 100 -- 10.933 = 9.146; 100 -- 11.433 = 8.744; 9.146-8.744 = .4, nearly. 292 QUADRATIC EQUATIONS 24. From New York to Chicago over the New York Central and Lake Shore lines is 960 mi. The 20th Century Limited covers the distance in 5} hr. less time than the Fast Mail Limited going at a speed which is 12.9 mi. an hour faster. What is the rate and time of each train ? SUGGESTION. After clearing and transposing make the coefficient of a; 2 unity by dividing through by its coefficient. Keep three or four decimal places. 25. In the Vanderbilt cup race for automobiles on Long Island, in 1905, Hemery on a French Darracq beat Tracy on a Locomobile by 22 min., the total distance covered being 283 mi. Hemery's rate was 41 mi. an hour faster than Tracy's. What was Hemery's rate ? 26. A man proposes to cut 3-inch squares out of the four corners of a piece of tin 2 inches longer than wide, turning up the sides so as to make a box containing 72 cu. in. How long shall he cut the piece of tin? 27. An ancient problem from India reads as follows : One fourth of a herd of camels was seen in the forest; twice the square root of the number had gone on mountain slopes ; and 15 remained on bank of the river. How many camels were there ? 28. The dimensions of a rectangular garden are 3 a and 4 a. Running along three sides of it, as in the marginal figure, is a walk of the same width throughout, whose area is b. What is the width of the walk ? x 30 4 a 29. Suppose a 12 ft. and b = 448 sq. ft. Find by substituting in the formula just derived the width of the walk. Again, sup- pose that x= .5 rod and b = 2 sq. rods, and find the corresponding value of a. SIMULTANEOUS QUADRATICS 293 30. The length AB of a rectangle, ABCD, exceeds its width AD by 119 ft., and the diagonal, BD joining two opposite vertices, equals 221 ft. Find AD and AB. Draw diagram. 31. A circular basin is surrounded by a path 6 ft. wide, and the area of the path is -J of the area of the basin. Find the radius of the basin. 32. E. A. Waltbour rode a bicycle in a mile race against time in 1904, making a record. Had he ridden the mile in 2.2 seconds less time, he would have covered 2.74 ft. more each second. What was his time to the nearest tenth of a second ? 33. The distance from Chicago to San Francisco over the Northwestern, Union Pacific, and Southern Pacific railroads is 2280 mi. Discarding small fractions, the Overland Limited covers the distance in 15^ hr. less time than the China and Japan Fast Mail traveling 5 miles an hour faster. What is the time of each train ? 34. In the Olympic games at Athens, April, 1906, the Mara- thon race of 42 kilometers (nearly 26 mi.) was won by the Canadian Sherring. Had he covered the distance in 6 min. less time, he would have had to increase his rate .5334 kilometer per hour. What was his time ? VI. SIMULTANEOUS QUADRATICS 190. Simultaneous Quadratics. (See 98.) Degree of an Equa- tion. ( 57.) Simultaneous quadratic equations may be sepa- rated into two classes : those in which there is only one quadratic equation ; and those in which there are two or more quadratics. Tlie degree of an equation is the greatest exponent or sum of exponents of the unknown quantities in any one term. Notice that by this definition xy = is a quadratic equation as well as We may think of the x 2 , xy, and y 2 terms as denoting areas, while x and y terms denote merely lengths. 294 QUADRATIC EQUATIONS 191. Simultaneous Equations one of which is Quadratic can always be solved by the Substitution Method. (See 99.) . (1) (2) 9Q 4. ii SOLUTION. (2i) x = - ^. (Sub. and Div. Axs.) o ^54, (99.) or. 841 - 232 y + 16 ya 87y-12yg ,. 9 3 then, 20 f- - 29 y = 355, (Mult, and Sub. Axs.) whence, solving, y = 5, or |. On substituting these values of y in (2), x = 3, or - 7 3 2 -. VERIFICATION. (1) 9 + 3x3x5=54. (2) 3x3 + 4 x5 = 29. 5184 ~~~ 3x 20 2. 4. 6. 8. 10. f (1) xy x = 0, 1( 1 a? y = 5. 12. xy = 45, a; y = 4. x y 3. 5. 7. 9. 11. f (1) 9 (2) 3x 1 -2y = 9 t 9 -i orv ry- + a- = loO, s - q = 8. 13. ?/ o; QUADRATIC EQUATIONS SOLVED BY SQUARING 295 14. 16. a+^= ~ y - iL. x 2 r , 3s o 3_ 12 r ' 3s -j + = 2. ' o 15. 17. 7 236 a; x _ _ 40 a? y 21 2 = 1. 18. 20. t 29 -t = -2. [_ 4 xii = a 2 b 2 . 19. 32 15 a?-3 = 1. r 22. 23. o? 2 -f- dy 2 = q. 24 26. 1 ; +?/ = 5 25. 27. = . 192. Quadratic Equations Solved by Squaring, Adding or Subtract- ing, and Extracting the Square Root. Problems in which one equa- tion is quadratic and the other simple (as in the preceding article), or in which both are quadratic, can be solved more elegantly by the following method, provided their coefficients are so related that the sum and difference of two quantities can be found. 296 QUADRATIC EQUATIONS i I (1) a 2 + f = 250, 1(2) x-y=4. SOLUTION (2i) x' 2 2ry-\- y' 2 = 16 (Squaring Ax.) (3) 2xy =234 (Sub. Ax.) (1) a; 2 + ?/ = 250 (4) x 2 + 2 xy + y 2 = 484 (Add. Ax.) (4 X ) + ?/ = 22 (Ax.?) 02) x - y = 4 2x =26 or -- 18, (Ax. ?) a; = 13 or - 9. (Ax. ?) 2y = 18 or -- 26, (Ax. ?) y = 9 or -- 13. (Ax. ?) VERIFICATION. 13 2 + 9 2 = 250, 13 -9 = 4. (_. 13)2 = 250, ( _ 9 ) -(-13) = 4. 2. 3. 4. i 5. - = 36. I a? + y = 8. r ^ 2 _i_ ^ ? / _i_ ? /2 _. g^ SUGGESTION. From (1) subtract (2) 8 squared. Then get (3) xy = 18. Add (3) ~y~ to (1), and extract square root. * = 21 9. 10. QUADRATIC EQUATIONS SOLVED BY SQUARING 297 (2) 4 a? + ^ = 25. SUGGESTION. Square (1) and subtract (2). Now to get 2 x y find the middle term of the trinomial square whose end terms are 4x 2 +# 2 (116, 20). Subtract 4 xy = 24 from (2). Knowing the values of 2 x + y and 2 x y it is easy to find x and ?/. 15. 9 2? = 241. 16. 18. (1) a 2 + a# + y 2 = 13, (2) x-y = 2. 20 . { 17 \i x 19 21. . I 4 ? 2 = 21. 22. 24. 26. (9x>- -13ajy + 9/ = 101, { 3 xy = 12. r y? + xy 4. f = a , = 6. 23. 25. 27. or xy -\-y 2 = 19. [ icy = a a; 4- y = a, 4 a? = a 2 6 2 . 28. 1-1 = 12, a? y SUGGESTION. Solve as in Ex. 1 without clearing either equation of fractions. (See 100, 25.) 29. 36 ' 30. - + - = 8, a y i + 1 + 1 , = 49. a?" 0$ ?/" 298 APPLICATIONS OF ALGEBRA IN SCIENCE 31. The course of a yacht 30 mi. in length is in the shape of a right triangle one leg of which is 2 mi. longer than the other. What is distance along each side ? 32. A can fold 3600 bills in 54 min. less time than B, and both can fold 7200 bills in 4 hours. How many can each fold in one hour ? 33. H and K run a race of 1 mi., K winning by 1 min. If H increases his speed 2 mi. an hour, and K lessens his by the same amount, H will win by 1 min. Find former rates of each. NOTE. Certain equations of higher degrees, as x 4 13 x 2 + 36 = 0, or, 2/ 6 + 7 y 3 8 = 0, can be solved as quadratics. ALGEBRA USED IN TRIGONOMETRY 193. Exercise in solving Formulas in Trigonometry. In the fol- lowing s stands for a number called the sine of an angle, c for the cosine, t for the tangent, k for the cotangent, and z for the secant of the angle. o 1. Solve t = - f or s and c. c 2. Solve t = - for k. k 3. Solve s 2 + c 2 = 1 for s and c. 4 Solve 2 + 1 = z 2 for t and z. o 5. Given t - and s 2 -f- c 2 = 1, to find t in terms of s. c 6. Given c = -, and t 2 + 1 = z 2 , to find c in terms of t. z 7. Given s = , to find first a and then h. h 8. Given t = -, to find first a and then b. b 9. Given a 2 = 6 2 + c 2 2 6m, to solve for m. INDEX* (The numbers refer to pages.) ABSCISSA, 174. Addition, of algebraic numbers, 12-19. of equations, 157. of fractional exponent quantities, 251. of fractions, 101. of imaginaries, 270. of literal quantities, 30. Aggregation, symbols of, 28, 40, 44. Algebra, 9, 12, 122. Antecedent, 117. Approximation, 228. Arithmetic, 9, 12, 138, 220. Arithmetical numbers, 12. Arrangement of letters, 51. Associative law, 24, 25. Average, 228. Axes, 173. Axioms, 123, 124. BINOMIAL, any power of, 204. cube of, 70. factoring, 76, 194. square of, 66, 68. Binomial coefficients, 204. Binomial quantity, 27. Biographical notes, 200. Brace, bracket, 28. CHECK, 13, 33. Circle, 234, 235. Clearing of fractions, 125. Coefficient, 27. of radical, 257. Common factor, 98. Common multiple, 92. Commutative law, 24, 25. Comparison, elimination, 160. Complete quadratic, 276, 279, 280, 286. Completing square, 280. Complex fraction, 114. Complex number, 270. Conditional equations, 122. Conjugate surd, 269. Consequent in proportion, 117. Consistent equations, 183. Continuation symbols, 12. Coordinates, 174. Cost problems generalized, 228. Cube, 26. Cube root, 213-218. DECOMPOSITION of a fraction, 107. Degree, of a term, 27. of an equation, 90, 293. Develop, 205. Difference, 15. Distributive law, ?5-, Divisibility, 72, 'li. Division, 23, 52. axiom of, 123. See Multiplication. Domestic science problems, 246-249. ELIMINATION, 154. by addition and subtraction, 157. by comparison, 160. by substitution, 154. in quadratics, 294-297. special methods, 162, 169, three unknowns, 168. Equal, 61, 141. Equation, 60, 61, 90, 122. balance, 127. compatible, 183. conditional, 122. construction of, 280. graphical solutions, 182. identical, 122. incompatible, 184. * This can be used to advantage in reviews by having students ignore all topics the page numbers of which are greater than the last page studied by the class. 299 300 INDEX Equation, in physics, 241. linear, 18]. literal, 188. quadratic, 90, 186, 276-298. radical, 270. simple, 90, 123. simultaneous, 154, 293. systems of, 102, 184. Equivalent equations, 162. fractions, 102. Errors in algebraic operations, 115. of calculation, 228. Evolution, 207. Exercise, 131. Expand, expansion, 204. Exponent, 26. in division, 52, 251. in multiplication, 45, 251. zero and negative, 254. Exponents, 219. Expression, algebraic = quantity. Extremes, 117. FACTORS, 74, 194. common, 98. in multiplication, 21, 45. order of, 51. prime, 74, 89. zero, 46. Figures, 9, 12, 166. Formulas, 66, 72, 204, 242, 246. Fraction, 96, 251. proper, 107. Function, 132. Fundamental Laws, 24. GENERALIZATION, 193. Geometry in algebra, 230. Graphs, 119, 172-184, 236. HIGHEST common factor, 98. Hindu rule, 284. Historical notes, 64, 199, 273. Hypothesis = what is given. IDENTICAL equation, 122. Imaginaries, 270, 289. Improper fraction, 107. Incomplete quadratic, 276. Indeterminate equations, 193. Index, 26, 257. Infinity, infinitesimal, 274. Interest, 222. Involution, 202. Irrationals, 257. LAWS, 24, 28, 250-256. Leading letter, 51. Letters, 51, 125. Like signs, 22. Literal equations, 188-192. Literal quantity, 27, 219. Locus, 180. Lowest common multiple, 92. MANUAL training problems, 240. Mean, Arithmetical, 228. Means, in proportion, 117. Mechanics, formulas of, 118, 273, 278 Member of an equation, 61. Minus, 12-19. Mixed quantity, 101. Monomial, 27. Multiple, 92. Multiplication, axiom, 124. fractions, 109. imaginaries, 270. literal quantities, 45, 250, 251. radicals, 265-268. simple algebra numbers, 21, 25. NAUGHT. See Zero. Negative exponent, 254. Negative quantities, 12, 19. Negative series, 12, 16. Negative solutions, 193, Newton's binomial theorem, 204 Notation, 9-65. radicals, 257. Numbers, algebraic, 12. complex, 270. irrational, 257. natural or arithmetical, 12. rational, 257. real, 289. Numerical value, 28, 245. OMISSION symbols, 12. Ordinate, 174. Origin, 173. INDEX 301 PARENTHESIS, 28, 40, 44. Percentage, 220. Physics, 240, 244, 273, 278. Plotting points, 175. Plus, 12, 19. Polynomial, 27. factoring, 197. square of, 207. Positive, 12, 19. Power, 26, 50, 202. of quantity, 50, 111. of radicals, 270. Precedence of operations, 28. Prime factor, 89. Prime marks, 190. Prime quantities, 74, 89. Principles, 24. Problem, 131. Proofs, 55, 77, 117, 125, 254. Proper fraction, 107. Property, of identical equations, 122. of quadratic equations, 280. Proportion, 117, 141. Pure quadratic, 276. QUADRANT, 173. Quadratic equations, 90, 186, 276. complete, 279. incomplete or pure, 276, 277. simultaneous, 293. Quadrinomial, 83, 19(>. Quantity, 26. known and unknown, 123. RADICALS, 257. Radicand, 257. Ratio, 9, 117. Rational numbers, 257. Rationalization of denominators, 269. Real quantity, 289. Reciprocal, 159. Reduction, of fractions, 96. of radicals, 259, 269. Root, of equation, 123. of quantity, 26, 50, 207. SATISFY an equation, 123. Scale, algebraic, 16. Science, 240, 244, 246, 273, 278. Series, 12. Signs, 13, 19, 21, 23. Similar radicals, 26;!. Similar terms, L'7. Simple equation, 90, 123. Simple term, 27. Simultaneous equations, 154. quadratic, 293. three unknowns, 168. Solution, 123. negative, 193. Square, 26. Square root, 209-213. Squared paper, 119, 172, 236. Substitution elimination, 154, 294. Subtraction, 15, 34. Surd, 257. Symbols, of aggregation, 28. of omission, 12. of operation, 13, 15, 25, 26. of quantity, 9. Symmetry, 218. System of equations, 162, 184. TERMS, 27. of a fraction, 96. of a proportion, 117. similar, 27. Theorems, 66-74, 194, 204, 207. Transposition, 61, 125. Trigonometrical expressions, 298. Trinomial, 27. factoring, 79, 194. square of, 207. UNITY, as coefficient, 27. as denominator of integral quantity, 105. as exponent, 26. Unlike signs, 22, 23. terms, 27. VALUE, numerical, 12, 28. Verification, 123. Vinculum, 28. ZERO, as exponent, 254. as factor, 46. PLANE GEOMETRY DEVELOPED BY THE SYLLABUS METHOD Bv EUGENE RANDOLPH SMITH, A. M. : Head of the Department of Mathematics, Polytechnic Preparatory School, Brooklyn, N. Y. Formerly Head of the De- partment of Mathematics, Montclair High School, Mont- clair, N. J. $0.75 THIS is the only complete and thoroughly worked-out plan of teaching geometry by the Syllabus Method. It makes the study more profitable. It makes the pupil do the thinking, not the teacher or the author. It teaches pupils how to discover the proofs for themselves, instead of memorizing those given in the textbook. It gives the maxi- mum of mental training with the minimum waste of energy. ^j The book contains the definitions, axioms, etc. ; the state- ments of theorems; a discussion of methods; four kinds of exercises; and classified summaries of the theorems. Except for a few specimen proofs and some necessary hints, the de- monstrations are omitted, but before the work is undertaken, propositions having any difficulty are fully discussed in class with the teacher, who guides the investigation by questions which do not imply their own answers. The pupil is made familiar with the laws of logic used in plane geometry, and undesstands the "what'' and "why" of each step. Much aid is given by the oral and review questions. ^[ The work is sufficient to prepare for any college, although the total of propositions to be proved is about fifty less than in the average book. The Syllabus Method is as well adapted to large as to small classes, and takes no more time than any other plan. It produces tremendous class enthusiasm, which acts as .a strong incentive and encourages individuality. It en- ables the teacher to become better acquainted with his pupils and their difficulties, and hence to aid them more intelligently. AMERICAN BOOK COMPANY ROBBINS'S PLANE TRIGONOMETRY By EDWARD R. ROBBINS, Senior Mathematical Mas- ter, William Penn Charter School, Philadelphia, Pa. THIS book is intended for beginners. It aims to give a thorough familiarity with the essential truths, and a satisfactory skill in operating with those processes. It is illustrated in the usual manner, but the diagrams are more than usually clear-cut and elucidating. ^| The work is sound and teachable, and is written in clear and concise language, in a style that makes it easily under- stood. Immediately after each principle has been proved, it is applied first in illustrative examples, and then further im- pressed by numerous exercises. Accuracy and rigor of treat- ment are shown in every detail, and all irrelevant and ex- traneous matter is excluded, thus giving greater prominence to universal rules and formulas. ^| The references to Plane Geometry preceding the first chapter are invaluable. A knowledge of the principles of geometry needed in trigonometry is, as a rule, too freely taken for granted. The author gives at the beginning of the book a statement of the applied principles, with reference to the sections of his Geometry, where such theorems are proved in full. Cross references in the text of the Trigonometry to those theorems make it easy for the pupil to review or to supplement imperfect knowledge. ^| Due emphasis is given to the theoretical as well as to the practical applications of the science. The number of ex- amples, both concrete and abstract, is far in excess of those in other books on the market. This book contains four times as many exercises as most books, and twice as many as that having the next lowest number. AMERICAN BOOK COMPANY (313) THE GATEWAY SERIES of English Texts for Admission to College Henry Van Dyke, General Editor ADDISON'S Sir Roger de Coverley Papers (Winchester) $0.40 BURKE' s Speech on Conciliation (MacDonald) . . .35 BYRON, WORDSWORTH, SHELLEY, KEATS, AND BROWN- ING Selections (Copeland and Rideout) ... .40 CARLYLE'S Essay on Burns (Mims) .35 COLERIDGE'S Rhyme of the Ancient Mariner (Wood- berry) .30 EMERSON'S Essays Selections (Van Dyke) . ... .35 FRANKLIN'S Autobiography (Smyth) .40 GASKELL'S Cranford (Rhodes) .40 GEORGE ELIOT'S Silas Marner (Cross) .40 GOLDSMITH'S Vicar of Wakefield, and the Deserted Village (Tufts) .45 IRVING'S Sketch-Book Selections (Sampson) . . . .45 LAMB'S Essay s of Elia Selections (Genung) .... .40 MACAULAY'S Addison (McClumpha) .35 MACAULAY'S Milton (Gulick) .35 MACAULAY'S Addison and Johnson (McClumpha and Clark) .45 MACAULAY'S Life of Johnson (Clark) .35 MILTON'S Minor Poems (Jordan) .35 SCOTT'S Ivanhoe (Stoddard) .50 SCOTT'S Lady of the Lake (Alden) .40 SHAKESPEARE'S As You Like It (Demmon) .... .35 SHAKESPEARE'S Julius Caesar (Mabie) .35 SHAKESPEARE'S Macbeth (Parrott) .40 SHAKESPEARE'S Merchant of Venice (Schelling) . .35 TENNYSON'S Idylls of the 'King Selections (VanDyke) .35 TENNYSON'S Princess (Bates) .40 Teachers' Outlines for Studies in English, with refer- ences to the Gateway Series (Blakely) .... -50 AMERICAN BOOK COMPANY (99) A HISTORY OF ENGLISH LITERATURE By REUBEN POST H A LL E C K, M. A. (Yale), Louisville Male High School $1.2.5 HALLECK'S HISTORY OF ENGLISH LITER- ATURE traces the development of that literature from the earliest times to the present in a concise, interesting, and stimulating manner. Although the subject is presented so clearly that it can be readily comprehended by high school pupils, the treatment is sufficiently philosophic and suggestive for any student beginning the study. ^j The book is a history of literature, and not a mere col- lection of biographical sketches. Only enough of the facts of an author's life are given to make students interested in him as a personality, and to show how his environment affected his work. Each author's productions, their rela- tion to the age, and the reasons why they hold a position in literature, receive treatment commensurate with their importance. ^[ One of the most striking features of the work consists in the way in which literary movements are clearly outlined at the beginning of each chapter. 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It is just the kind of book that should be given to a beginner, because it will give him a clear idea of what to read, and of the relative importance of the authors he is to read; yet it is much more than merely a book for beginners. Any student of the subject who wishes to do good work hereafter must not only read Mr. Matthews's book, but must largely adopt Mr. Matthews's way of looking at things, for these simply written, unpreten- tious chapters are worth many times as much as the pon- derous tomes which contain what usually passes for criticism; and the principles upon which Mr. Matthews insists with such quiet force and good taste are those which must be adopted, not only by every student of American writings, but by every American writer, if he is going to do what is really worth doing. There is little room for division of opinion as to the excellence of Mr. Matthews's arrangement as a whole, and as to the soundness of his judgments. 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The characteristics of each group are presented by means of a typical example, which is described in unusual detail. Many realistic exercises are in- troduced to direct the student hov/ to study the thing itself, whenever practicable, or some experimental or pictorial repre- sentation of it. These exercises include both field and labora- tory work, and should be made fundamental rather than supplemental. ^| The order of general topics is the Planet Earth, the Land, the Sea, the Atmosphere, and Life, and each topic is treated with such fullness that it enables the teacher who has not had a special course in geography to teach the subject in- telligently. At intervals throughout the book there are in- troduced discussions of the consequences which follow the conditions described, and chapters upon Life, containing a full treatment of the controls exerted by geographical conditions upon plants, animals, and men. ^| The book is eminently readable. The style is less formal and dogmatic than is usual in a scientific text-book, and approaches that which a teacher uses in conversation. The appendix contains directions for laboratory exercises, full in- formation in regard to the best material for the equipment of a geographical laboratory, and a reference list of the available literature upon the subject. The book is profusely illustrated. AMERICAN BOOK COMPANY ESSENTIALS IN HISTORY ESSENTIALS IN ANCIENT HISTORY . . $1.50 From the earliest records to Charlemagne. By ARTHUR MAYER WOLFSON, Ph.D., First Assistant in History, DeWitt Clinton High School, New York. ESSENTIALS IN MEDIAEVAL AND MODERN HISTORY $1.50 From Chanemagne to the present day. By SAMUEL BANNISTER HARDING, Ph.D., Professor of Euro- pean History, Indiana University. ESSENTIALS IN ENGLISH HISTORY