al ‘ ae) ra Ae Se yay Yi) SS 2 Wed, # University of Illinois. § Books are not to be taken from the Library Room. _ . Qa Pasa Fee _ MATHEMATICS LIBRARY. AN ELEMENTARY ALGEBRA FOR Setat and Academie’, BY JOSEPH W. WILSON, AM, PROFESSOR OF GEOMETRY IN THE CENTRAL HIGH SCHOOL OF PHILADELPHIA. PHILADELPHIA: ELDREDGE & BROTHER, No. 17 North Seventh Street. Entered, according to Act of Congress, in the year 1871, by ELDREDGE & BROTHER, 4 in the Office of the Librarian of Congress at Washington. Mie < $$ — Eas J. FAGAN & SON, Ps XR SF ELECTROTYPERS, PHILAD’A. FERGUSON BROS. & CO., PRINTERS, PHILADELPHIA. PREFACE. 2038300 TINHE attention of Teachers is requested to the follow- ing points, in which the Author has endeavored to make this book a useful one: Clearness. The great aim throughout has been to make everything as plain as the nature of the subject would permit. Every principle has been explained, and unne- cessary verbiage avoided. A very Gradual Advance. Complex subjects have been subdivided as far as possible into their elements, so that the learner may have to take but one step at a time. A Practical Character. There is a continual review and repetition of whatever has been learned. Much more attention than usual has been given to Fractions, as a thorough drill on them is believed to be essential to a full comprehension of Algebraic operations, and a ready facility fn performing them. ‘The examples are very numerous, and are all original, so that this book may be used in connection with others, without fear of any other than accidental repetition. An Analytical Method. ‘This is essentially the same 111 iv PREFACE. as that which has been so successful in Mental Arith- metic. Synthetic arrangement has been subordinated to convenience of teaching. Accuracy of Language. Such expressions as Add the equations together, where the meaning is Add their corre- sponding members, have been carefully avoided. Hun- dreds of these errors have crept into some of the best books. The learner thus insensibly acquires a careless and incorrect use of language, which often clouds his apprehension of principles otherwise clear. In conclusion, an effort has been made to specially adapt the work to those teachers who are not satisfied with anything less than an ability on the part of their pupils to fully explain every operation which they per- form. To such the book is commended, in the hope that it will satisfy a need which the Author has himself fre- quently felt. @ S ENF oi SECTION PAGE I. THE FUNDAMENTAL SIGNS . d : ; : BS etd II. [ILLUSTRATIONS OF THE NATURE OF ALGEBRA . , hg AE III. PROBLEMS TO BE SOLVED BY EQUATIONS CONTAINING BUT ONE UNKNOWN QUANTITY . : ; ; rig IV. DEFINITIONS . : ‘ ; ; , ’ 3 f wea Penton? ne Re eI! : Gh Sot eee VI. PROBLEMS TO BE SOLVED BY EQUATIONS CONTAINING BUT ONE UNKNOWN QUANTITY . : : : . 28 VII. SUBTRACTION q , : : : : F - . 30 VIII. TRANSPOSITION . : ‘ : : : ; Pid d. 06 IX. QUESTIONS PRODUCING EQUATIONS THAT REQUIRE TRANSPOSITION. : ; ; : : ‘ - 40 X. MULTIPLICATION. THE SIGNS + AND — : : » 45 XI. MULTIPLICATION OF LITERAL QUANTITIES : pb ~ ae XII. MULTIPLICATION. EXPONENTS : ; : ; . 48 XIII. MULTIPLICATION OF A POLYNOMIAL BY A MONOMIAL . 61 XIV. MULTIPLICATION OF A POLYNOMIAL BY A POLYNOMIAL 653 XV. EQUATIONS REQUIRING MULTIPLICATION . : : . 58 XVI. QUESTIONS PRODUCING EQUATIONS WITH ONE UNKNOWN QUANTITY ‘ : : : ‘ . ; ; emt 9 XVII. Division. THE SIGNS + AND — : 4 : : a Oe XVIII. Division. LETTERS WITHOUTINDICES . ‘ A « 69 XIX. DIVISION. MONOMIALS WITH INDICES . : : Od (4, 1* y vi SECTION XX. XXII. XXII. XXII. XXIV. 2. a XXVI. XXVIT. XXVIII. XXIX. XXX. 2. @.5 XXXII. XXXITI. De phe Now XXXV. XXXVI. XXXVII. XXXVIII. XXXIX. XL. XLI. XLIT. XLII. XLIV. CONTENTS. PaGE DIVISION OF A POLYNOMIAL BY A MONOMIAL . 5 aH REVIEW OF SUBTRACTION . ‘ : : ‘ . 78 DIVISION OF A POLYNOMIAL BY A POLYNOMIAL x 7 EQUATIONS WITH TWO UNKNOWN QUANTITIES. oe QUESTIONS PRODUCING EQUATIONS WITH Two UN- KNOWN QUANTITIES ~. . , . 94 FACTORING. FUNDAMENTAL THEOREMS 96 FACTORING. PRIME AND MONOMIAL FACTORS. . 100 FACTORING. APPLICATIONS OF THE FUNDAMENTAL THEOREMS : : : ; : : . . 103 FACTORING, (CONTINUED.) . ; ; : P . 106 EQUATIONS WITH TWO UNKNOWN QUANTITIES, (CON- TINUED.) . 108 THE LEAST COMMON MULTIPLE - 113 REDUCTION OF FRACTIONS TO THEIR LOWEST TERMS REDUCTION OF WHOLE AND MIXED QUANTITIES TO FRACTIONS : ; ; F ‘ i : * REDUCTION OF FRACTIONS TO WHOLE AND MIXED QUANTITIES . ‘ é : : 2 " - REDUCTION OF FRACTIONS TO OTHERS HAVING THE LEAST COMMON DENOMINATOR . ‘ : ; ADDITION OF FRACTIONS ‘ ; ; , F - SUBTRACTION OF FRACTIONS MULTIPLICATION OF FRACTIONS . DIVISION OF FRACTIONS ‘ ‘ : x P F REDUCTION OF COMPLEX FRACTIONS TO SIMPLE ONES : : a : . : : CLEARING AN EQUATION OF FRACTIONS . . E QUESTIONS PRODUCING FRACTIONAL EQUATIONS REVIEW ELIMINATION * : : : F ; ; QUESTIONS PRODUCING EQUATIONS WITH Two UN- KNOWN QUANTITIES ; 116 120 125 128 131 « 132 . 134 136 . 138 141 . 144 . 151 154 ae SECTION XLV. XLVI. XLVII. XLVIII. CONTENTS. Vil Pace EQUATIONS CONTAINING MORE THAN Two UNKNOWN QUANTITIES é : : ; ‘ 4 ° : QUESTIONS PRODUCING EQUATIONS WITH MORE THAN Two UNKNOWN QUANTITIES . : - ¥ : PECULIAR EQUATIONS WITH MORE THAN Two Un- KNOWN QUANTITIES . PROBLEMS RELATING TO WORK XLIX. RATIO AND PROPORTION , . z : ; 7 : LXI. LXII. LXIII. LXIY. LXV. LXVI. LXVII. LXVIII. LXIX. LXX. LXXI. . ARITHMETICAL PROGRESSION . GEOMETRICAL PROGRESSION . ARITHMETICAL PROGRESSION, (CONTINUED.) . . GEOMETRICAL PROGRESSION, (CONTINUED.) . ; . QUESTIONS PRODUCING SIMPLE EQUATIONS . ‘ ‘ . INVOLUTION F ‘ . THE BINOMIAL THEOREM j S * 3 EVOLUTION. Roots oF MONOMIALS SQUARE ROOTS OF POLYNOMIALS , . CLASSIFICATION OF EQUATIONS ; 5 ‘ d . . PURE QUADRATIC EQUATIONS AND OTHERS SIMILARLY SOLVED EQUATIONS WITH TWO UNKNOWN QUANTITIES ADFECTED QUADRATIC EQUATIONS MISCELLANEOUS EXAMPLES. REVIEW QUESTIONS ° NEGATIVE AND FRACTIONAL EXPONENTS 7 ; : RADICALS . : F . : P r ‘ - : REDUCTION OF RADICALS TO THEIR SIMPLEST FORMS . 164 168 ay bp anne d 183 - 192 ELIMINATION OF RADICALS FROM THE DENOMINATORS — OF FRACTIONS - . - A : 5 ; : REDUCTION OF RADICALS TO A COMMON INDEX . ‘ ADDITION AND SUBTRACTION OF RADICALS MULTIPLICATION AND DIVISION OF RADICALS AN ALGEBRAIC PUZZLE Now Ready. Saw Pete A KEY TO WILSON’S ALGEBRA, for the use of Teachers,» . . .*% =o. 8° R°e2 Prigegiaaa viii ELEMENTARY ALGEBRA. save Eatea— SECTION IL. The Fundamental Signs. THE sign of Additionis +. Itiscalled plus. Thus,4+6 fs read four plus six; and it shows that four and six are to be added. The sign of Subtraction is—. It is called minus. Thus, 7 — 2 is read seven minus two; and it shows that 2 is to be subtracted from 7. The sign of Multiplication is x. Thus, 5 x 4 means 5 multiplied by 4. The sign of Division is--. Thus, 15-3 means 15 divided by 3. A fraction also indicates division. Thus, 1 means 15 divided by 3. The sign of Equality is =. Thus 4+ 7 = 8 + 3 is read four plus seven is equal to eight plus three. The expression that two things are equal is an equation. Thus, 4 + 7 = 8 + 3 isan equation, Sois16—2=—>3+14 10. Other signs will be explained when there is occasion to use them. 9 10 ELEMENTARY ALGEBRA. SECHGNeTI. Illustrations of the Nature of Algebra. Tuer Mathematics are the sciences which treat of number- ing and measuring. For example, Arithmetic explains how to calculate by means of numbers, and Geometry shows how to measure things, at first by placing them side by side, and afterward in many other ways. Algebra is that branch of mathematics in which numbers are represented by letters. The different ways in which numbers are represented by letters, and the use of doing so, cannot be fully explained until a great deal of progress is made in the study of algebra; but some preliminary ideas of the science may be obtained by considering the following questions: ProsiEM 1. I wish to divide 21 marbles between Henry and Charles, so as to give Henry twice as many as Charles. How many must I give to each? Soxtution. I wish to give to Charles a certain number, and to Henry twice that number. Hence, I must give to both ‘of them three times that number. Then, since both together are to have 21, I know that three times that number are 21. Therefore, once that number is 1 of 21, which is 7. Hence the number which Charles is to receive is 7; and twice that number is 14, which Henry is to receive. If I had used the letter n, instead of the words a certain number, the work would have been somewhat shortened. Thus: Srconp Soxution. Let the number that I wish to give to Charles be called n; and to Henry 2n. Then, since both together are to have 21, I know that 3n = 21. Therefore once n = 7, which is Charles’s share; and 2” = 14, which is Henry’s share. It will be observed that at the beginning of the solution ELEMENTARY ALGEBRA. 11 the number of marbles to be given to Charles is not known. In the first solution it was called a certain number; in the second, it was called n. Now for more than two hundred years mathematicians have been in the habit of using the last letters of the alphabet, x, y, z, u, etc., to stand for numbers that are not known. If we adopt their practice, the solution just given may be changed to the following: Turrp Soturion. Let x represent the number of mar- bles for Charles. Then 2 represents the number for Henry. Hence, oe = 21, _ Dividing by 3, “== 7. Whence, Q2¢ = 14, Therefore Charles is to receive 7 marbles, and Henry 14. PrRoBLEM 2. What number added to itself will make 16? SoLution. Let x represent the number. Then, : e+a2= 16. Collecting, 2a = 16: Dividing by 2, % ==, Os Therefore 8 is the number which when added to itself will make 16. ProsiEeM 38. In a flock of 72 sheep, there are seven times as many white ones as black ones. How many are there of each kind? Sotution. Let # stand for the number of black sheep. Then 7 x stands for the number of white ones. Hence, G iT ae -72: Collecting, 8a = 72. Dividing by 8, ve oe Whence, Te = 63. Therefore there are 9 black sheep, and 63 white ones. 12 ELEMENTARY ALGEBRA. -Prosiem 4, William had three times as much money as James, and the difference between William’s money and James’s was 20 cents. How much had each? Sotution. Let 2 stand for James’s money. Then 8 stands for William’s. Hence, 32—2 = 20. Collecting, 2% == ZU; Dividing by 2, z= 10, Whence, 32 = 80. Therefore James had 10 cents, and William 30. PROBLEM 5. It is required to divide $600 among three soldiers, so that one, who is a lieutenant, may have twice as much as another, who is a corporal, and the third, who is a captain, may have as much as the corporal and the lieuten- ant together. SotutTion. Let x represent the corporal’s share. Then 2 x represents the lieutenant’s share. And 32 represents the captain’s share. Hence, a+2a2+ 32 = 600. Collecting, 6 2 =760U, Dividing by 6, xz == 100. Whence, 22 = 200. And, 82 = 800. Therefore the corporal should receive $100; the lieutenant, $200; and the captain, $300. In the preceding solutions, one of the quantities to be found has always been represented by x. The problems are such that when one of the unknown quantities is found, the others can be very easily obtained by multiplication, or some other simple operation. Such problems can always be solved by using a single letter, and the equations obtained ELEMENTARY ALGEBRA. 13 from them are called equations containing but one unknown quantity. There are some questions which cannot be easily solved without using two or more letters. They will be explained in a subsequent part of this book. In the fifth problem, there are three things to be found, and we might have represented any one of them by z. Thus: 4 SECOND SOLUTION OF PROBLEM 9. Let x represent the lieutenant’s share. Then 4 represents the corporal’s share. And 13 represents the captain’s share. Hence, zx+%tx2+ 142 = 600. Collecting, 8a = 600 Dividing by 3, a = 200, Whence, + £2 = 100. And, 1} a2 = 300. Therefore the lieutenant should receive $200; the corpo- ral, $100; and the captain, $300. THIRD SOLUTION OF PROBLEM 5. Let «x represent the captain’s share. Then 42 represents the corporal’s share. And # represents the lieutenant’s share. Hence, xc+é¢a+ ea = 600. Collecting, 2a == GUE Dividing by 2, x = 800. Whence, +x = 100. And, $2 = 200. These solutions are not quite as easy as the first, because they contain fractions. In equations of this kind, it is gen- erally best to represent the smadlest of the unknown quan- tities by x. By doing so, fractions will be avoided. 2 14 ELEMENTARY ALGEBRA. SECTION III. Problems to be solved by Equations containing but One Unknown Quantity. 1. JoHN and George have each the same number of apples, and both together have 36. How many has each? Ans. 18. 2. If 67 = 30, what is x equal to? Ans. # = 5. 3. Ann has a certain number of books, and Jane has four times as many. They together have 25 books. How many has each? Ans. Ann has 5, and Jane has 20. 4. In a brigade of 2233 soldiers there were 10 times as many private soldiers as officers. How many private sol- diers were there? Ans. 2030 private soldiers. 5. A and B together are worth $2432, and A is worth 7 times as much as B. How much is B worth? Ans. $304. 6. What number added to itself amounts to 88? Ans. 19. 7. William and Robert caught 81 fish, and Robert caught 8 times as many as William. How many did Robert catch? Ans. 72 fish. 8. What number added to 5 times itself will amount to 72? Ans. 12. 9. A farmer bought a horse and carriage for $420, and the carriage cost 3 times as much as the horse. What was the cost of each? Ans. The horse cost $105; and the car- riage, $315. 10. At a county election, there were two candidates for sheriff. One received 3 times as many votes as the other, and his majority was 220. How many votes had each, and how many were polled in all? Ans. One received 330 votes; and the other, 110; in all, 440. 11. Ifa+32+ 52 = 63, whatisthe value ofx2? Ans. 7. 12. Thomas travels a certain distance one day, and twice ELEMENTARY ALGEBRA. 15 as far the next; in all 42 miles. How far does he travel each day? Ans. 14 miles the first day, and 28 miles the second. 13. In a store-room containing 45 casks, there are 4 times as many full as empty. How many are full? Ans. 36. 14. The angles of a triangle always amount to 180 de- grees. If one angle of a triangle is twice a second, and the third is 3 times the second, how many degrees are in each? Ans. 60°, 30°, and 90°. Nore. Let x = the number of degrees in the second angle: then 2x = the number of degrees in the first, and 32 in the third. 15. If 2 angles of a triangle are equal, and the remaining angle is twice their sum, how many degrees are in each? Ans. 30°, 30°, and 120°. 16. If one angle of a triangle is 3 times a second, and the third is twice the first, how many degrees are in each? Ans. 54°, 18°, and 108°. 17. The angles of a quadrilateral, or four-sided figure, always amount to 360°. If they are equal, how many de- grees are in each? Ans. 90°. 18. If one angle of a quadrilateral is twice a second, the third twice the first, and the fourth equal to the third and second together, how many degrees are in each? Ans. 60°, 30°, 120°, and 150°. 19. Two men start from the same point and travel in op- posite directions. The first travels 8 times as fast as the second. How far has each travelled when they are 48 miles apart? Ans. The first has travelled 36 miles; the second, 12 miles. 20. Mary is 14 years older than Jane, and her age is 8 times Jane’s. How old is each? Ans. Mary is 16 years old, and Jane is 2 years old. Notre. Let += Jane’s age: then 8x— Mary’s. We have, by the conditions of the question, 8% —z = 14, 16 ELEMENTARY ALGEBRA. SECTION IV. Definitions. A monomial is any algebraic expression which stands by itself, and is not composed of parts united by the sign + or —. Thus, z isa monomial; — 32 isa monomial; 14a is a monomial. A polynomial is any algebraic expression which is com- posed of parts united by the sign + or —. Thus, a + 6 is a polynomial; 2<— 3y is a polynomial; a —3b + 2n —e isa polynomial. The parts of a polynomial are called terms. A monomial consists of a single term. A polynomial of two terms is called a binomial. One of three terms is called a trinomial. Thus, 2a — 361s a bi- nomial; —8a2+a-+ 4n is a trinomial. A positive term is one having the sign + before it. A negative term is one having the sign — before it. In the polynomial x + 2a — 32 —ce-— aa, the first term is posi- tive, because the sign + is understood before it; the second term is also positive; the other three terms are negative. Sometimes a dot is used between letters as a sign of mul.- tiplication. Thus, a.6 is the same asa X 6. Multiplica- tion is also frequently denoted by writing two letters together. Thus, ab is also the same as a X 6. The same is the case with a figure and a letter. Thus, 3x is the same as 3 X &. The figure before the letter is called a coefficient. Thus, in the. term 7 n, the coefficient of n is 7. It must always be remembered that in algebra, letters stand for numbers. The expression ab means that the number which a represents is to be multiplied by the num- ber which } represents. Sometimes a stands for one num- ber, and sometimes for another. When a number is multiplied by itself oue or more times, ELEMENTARY ALGEBRA. 17 the product is called a power. The second power of 5 is 25, because 5 X 5== 25. The third power of 5 is 125, because 5x5xX5—125. The third power of « is xxx. The fourth power of ais aaaa. The second power is often called the square, and the third power the cube. Instead of writing aaaa, we generally write a‘. The number 4 is used to show how many times a is taken as a factor. Such a number, written to the right of the upper part of any letter, is called an index or exponent. The power 6bbb06 is generally written 6°, where 6 is the exponent. We do not often write 1 as an exponent. Thus ' is just the same as #. It is often said that the exponent of a letter written without any, is 1 understood. Thus, the exponent of x, or a, or c, is understood to be 1. A root is the quantity which is multiplied by itself one or more times to forma power. Thus 2 is the root of the powers x’, x*, x‘, etc. It is the second root of x’, the third root of x’, the fourth root of zx‘, etc. The second root is often called the square root, and the third root the cube root. The square root of 64 is 8, because 8 x 8= 64. The cube root of n? is n, because nnn = n’. The radical sign, /, shows that some root of the quantity placed under it is to be taken. To denote the square root, it is used alone. To denote any other root, a small figure is placed over it. Thus, ./9 means the square root of 9, which is 3; and ~/8 means the cube root of 8, which is 2. When several ‘terms are to be considered together, they are placed within a parenthesis, or a vinculum is placed over them. Thus, (5+ 2+ 7) x 9,or5+2+7 x 9, shows that the sum of 5, 2, and 7 is to be multiplied by 9. If the expression were 5+ 2+7 x 9, the 7 alone should be multiplied by 9 before adding it to the sum of 5 and 2. Operations of multiplication and division must be performed before those of addition and subtraction, unless parentheses in- dicate otherwise. ae B ELEMENTARY ALGEBRA, EXAMPLES. Find the value of each of the following expressions: 1. (8+ 3) x 2. Ans. 22. 2. 8 + 3.2, Ans. 14. 3. (8 — 3) X 2. Ans. 10. 4,.8—38 x 2. Ans. 2. 5. (6 + 4) x 3. Ans. 27. 6. (5 + 3) x 9. Ans. 72. 7. (8 +6+3) x 2. Ans. 34. 8. (8+3—5)x 4 Ans. 24. 9. (10 + 8 — 2) x 7. Ans, 112. 10. 10 + 8—2 x 7. Ans. 4. 11. (8 —3 + 4) x 3. Ans. 27. 12.8—3+4 x 3. Ans. 17. 13. (6 —2 + 3) x 4. Ans. 28. 14.6—2+3 x 4. Ans. 16. 15. 8 x (5 — 2). Ans. 24. 16. 8 x 5 — 2, Ans. 38. 17..9 x (7.—*4). Ans. 27. 18.9 x 7 —4. Ans. 59. 19.5 x (8 + 38 — 7). Ans. 20. 20.5x 8+ 3—7. Ans. 36. 21. 7 x (9.+ 2). Ans. 77. 22.7 xX 9+ 2. Ans. 65. 23. 2X (7 —3-++ 8). Ans. 24. 24.2x 7—8-+ 8. _ Ans. 19. 25.9 x 4+ 2 -+ 8. Ans. 41. 26.9 X (4+ 2 +4 8). Ans. 81. 27. (4+ 3) x (8 — 5). Ans. 21. 28. (4+ 3) x 8 — 5. Ans. 51. 29.4+3 x (8 —5). Ans. 138. 30.4+3 x 8—5. Ans, 23. 31. (2+ 3) x 4. Ans. 20. -82. (2 + 3) x (4 + 8). Ans. 35. ELEMENTARY ALGEBRA. 19 33. (6 — 5) x 4. Ans. 4. 34. 6 x (5 — 4). Ans. 6. 30.6 x 4— 9. Ans. 19. 36. (6 + 4) + 5. Ans. 2. 37.6 + 4 + 5. Ans. 64. 38. (2 +3—4+ 5) x 6. Ans. 36. 3 39. (24+ 38 — 4) +54 5. Ans. 5 40. (2+3—4)+(5+4 5). Ans. 3 41.2+3—4—5+4 5. Ans. 9 42.2+3+4— (5+ 4). Ans. 5 43.2+3—4+4+5-x 6. Ans. 44.2+3 x (—4-+ 5) = 5. Ans. 45.5—2+11 x 3—2. Ans. 46.5—2+11 x (8— 2). Ans. 14. 47. (5— 2) x 114+ 3—2. Ans. 34. 48.7+19—3 x 4—8383. Ans. 11. 49.7 + (19—3) x 4—38._ Ans. 68. 50. 7 + (19 — 8) x (4— 3). Ans. 23. C2 bo co ee Se en eee QUESTIONS ON THE FOREGOING DEFINITIONS. Why is —a + 2 —e a polynomial? What kind of a polynomial is it? What kind of a polynomial is a+ 5? Is ab a polynomial or a monomial? Why? Is 127 atr3y? a poly- nomial or a monomial? Why? How many terms are in the polynomial mn? — 7 abe? + x — 32 yz + w8? Which terms are positive, and which are negative? Name all the coefficients, and all the indices. (Some are 1 understood.) Write ce + ¢+e¢+e¢-+cina shorter way. Write ececece in a shorter way. In the power 79, what is the index of ¥? In the monomial 17 acz*, what is the coefficient of acz4? What is the index of a? (Ans. 1 understood.) What is the index of c? of x? Is za power? What is its fourth root? In the monomial z°7?, what is the coefficient of z5y3? What is the exponent of x? of y? What is 4/z? equal to? What is ~/2zt equal to? What is 4/16? What is %/27? What is the fifth power of 2? of 1? 20 ELEMENTARY ALGEBRA. SECTION V. Addition. In algebra, we may add terms together by writing them down one after another with their proper signs. Thus, if we wish to add x to y, we may write the sum either # + y ory + a. It does not matter which is placed first. The sign + is un- derstood before x or y when no sign is expressed. To add x to — y, we write either x — yor —y+ a. Addition in algebra includes more than addition in arithmetic, since we may find the sum of both positive and negative terms. Similar terms are those which do not differ otherwise than in their coefficients. Thus 3a is similar to 5a: the co- efficients, 3 and 5, are different; but the rest of the terms is the same. So, 46, 60, 6,46, and 120, are similar; but 3a and 5 6 are dissimilar, as the letters differ. Again, w’y and 2x?y are similar; but wy and 2.y’ are dissimilar. Since either xy or yx means the product of x and y, they are not different, and are consequently similar. So 3ad is similar to 14 ba. Similar terms may be added into asingle term. Thus the sum of 8aand 4ais 7a. Yoadd similar terms, perform the operations of addition and subtraction on their coefficients. ProsiEm 1. Add 5 apples, 7 apples, and 16 apples. Prosiem 2. Add 5 dozen, 7 dozen, and 16 dozen. ProsieM 3. Add 5a, 7a, and 16a. OPERATIONS. 1. 5 apples, 2. 5 dozen, 3. 5a, 7 apples, 7 dozen, 74, 16 apples, 16 dozen, 16a, 28 apples. 28 dozen. 28 a. ELEMENTARY ALGEBRA. 21 ProsuiEM 4. John has 6 cents in his pocket, 9 cents at home in a drawer, and his brother owes him 7 cents. What is his financial condition ? OPERATION. EXPLANATION. The sign + is understood 6 cents, before every-term in this operation and the 9 cents, three preceding ones. The sign + is often 7 cents, used before terms which express how muck ——- a person has, or how much is due to him. 22 cents. The result shows that John is worth 22 cents. ProsiEM 5. William has no money, and he owes a grocer 8 cents, a pieman 4 cents, his sister 3 cents, and a playmate 10 cents. What is his financial condition? _ OPERATION. — 8 cents, — 4 cents, EXPLANATION. The sign — is often — 3 cents, used before terms which express how — 10 cents, much a person owes. William owes 25 cents, and has no money to pay it with. — 20 cents. PropiEeM 6, A boy has 16 cents, and owes 9 cents. What is his financial condition ? OPERATION. EXPLANATION. If he pay his debt, he 16 cents, will have left the difference between 16 — 9 cents, and 9cents. His true financial condition is the possession of 7 cents, clear. 7 cents. ProsieM 7. A boy has 11 cents, and owes 17 cents. What is his financial condition ? OPERATION. EXPLANATION. He owes more than he 11 cents, possesses. If he pay as much as he can -— Li cents, of his debt, it will be diminished by all he possesses, but he will remain 6 cents — 6 cents. in debt, without anything to pay it with. p pe ELEMENTARY ALGEBRA. Propiem 8. George has 63 cents in his money-box, 10 cents in his pocket, and William owes him 10 cents. He owes his brother John 7 cents, and a confectioner 14 cents. What is his financial condition ? OPERATION. 63 cents, 10 cents, EXPLANATION. The positive terms 10 cents, amount to 83 cents, and the negative — 7 cents, ones to 21 cents. Their difference is 62 — 14 cents, cents, which George is really worth. 62 cents. PRopLEM 9. James has 6 cents in his pocket, 18 cents at home, and another boy owes him 10 cents. He owes 12 cents to a schoolmate, 14 cents to a stationer for a copy- book, and 11 cents’ to an apple-woman. What is his finan- cial condition? OPERATION. 6 cents, EXPLANATION. The positive terms 18 cents, amount to 34 cents, and the negative 10 cents, ones to 87 cents. Their difference is 3 — 12 cents, cents; and as the negative sum is the — 14 cents, greater, James owes 3 cents more than — 11 cents, he has money to pay his debts with. — o cents. PropuiEem 10. Add —a, —7a, —a,—38a, —7a, —8a, and — 2a. ProsiEeM 11. Add 3a, — 7a,4a, 6a, 9a, —2a,—11a, and 4a. -ProspiteM 12. Add 8a, —4a, —14a, 3a, — 5a, 6a, — 4a, and — 6a. ELEMENTARY ALGEBRA. 23 OPERATIONS. 10. — a, LP 3a, 12. 8a, —= 7 a, — 7a, — Aa, — 4a, 4a, — 14a, — 3a, 6 a, 34, — Ta, 9 a, — 5a, — 8a, — 2a, 6 a, — 2a, — lla, — Aa, 4a, — 6a, — 29 a. 6 a. — l6a. Thus we see that in addition it is convenient to write similar terms in a single column, find the sum of the positive terms, then of the negative terms, take the difference of these sums, and prefix the sign of the greater sum. The operation of counting is performed on the coefficients only. The addition of similar terms is often called collecting them. ProspuiEeM 13. Collect 6a —26 + 5a—36+a— DB. Prosiem 14, Collect a’ — a’y + ax? — 38a°x + ax’y — 4ax’* + a'r. OPERATIONS. 13. 6a— 26, 14. axe—axry+ ax’, 5a — 8b, — 8ar 4+ xy — 4a2’, a— Ob, an 124 — 66. — we — 3 ax’, ProsptEM 15. Add 7Ta—3b, b— 2a, —38a-+ 20, 76 —a, and 206. PrositeM 16. Add 38n—m + 2,2 +n, 3n —m— 2, m + 2, and — 4. 24 15. CONDO > ELEMENTARY ALGEBRA. OPERATIONS. ta 356; 16. 3n—m-4 2, —2a+ 8, n + 2, — 3a+ 28, 3n—m— 2, = ted 0; m + 2, vi oe 4, a+90b. Tn — mm. EXAMPLES. . Collect a, a, a, a, a, a, a, and a. Ans. 8a, . Collect — x, — x, — x, — x7, — x, — x, and — a. Avia a . Collect a (a + y), 2a(a@+ y), and 8a (a+ y). Ans. 6a(x% + y). . Collectt5a + 32+22-+ Tx + a. Ans. 182. . Collect —382—42—92x—38a4. #£xAns. —192. . Collect Ty — 2y—5y + 4y. Ans. 4y. . Collect Vy —4y + By + 15y—13y. Ans. 15y. . Collect 15z — 12z + 182 — 52z — 82 — 122. Ans. — 42, . Collect 9a —za2 + 10x — 82+ 38x—Aza. Ans. 9 x. . Collect —x + 2a. Ans. X. . Collect « — 2 2. Ans. — xX. . Collect 82 — lla 4+ 18a—152+4 10a. Ans. 52, . Collect 5a—Ta+ 8a—9a—12a+4a. Ans. — 19a. . Collecta—a+ 2a—2a+ 9a — l6a. Ans. — Ta. . Collect 36 —5b + 76—105—64 38. Ans. — 3b. . Collect — b —2b— 306. Ans. — 6). . Collect — b—26+4 36. Ans. 0. . Collect ax + Yax — Tax + 38axr — ax + Baz. Ans. 10 az. ELEMENTARY ALGEBRA. 25 19. Collect xy —- 8 ay + Tay. Ans. 0. 20. Collect (2 + a) + 19 (# + a) — 138 (% 4 a). Ans. 7 (« + a). 21. Collect x —184a+ 19a—182+4+19%+4 a. Ans. Tx + Ta. Poecolec 2 —106 + Ta — ba — 26+ 5 — 8a. Angi:r— 10-0: 23. Collect y + 11z— 13 y + 21z—15z + 4z—D5y. Ans. —17Ty + 212. 24. Collect 5ba — 2 b’x + 3 b’x — 2 bz. Ans. 3ba + Bx. 25. Collect 18 ay — 2arz 4+ 1342 — 5ay — 11 zz. Ans. 13 ay. 26. Collect 18a? + 9a — 14a — 8a? + 6a + a. Ans. 6@ + a. 27. What is the sum of 6 ab, 6 ab, — 6 ab’, and 5a’b?? Ans. 6a7b + 6 ab — 6 ab? + 5 a’b?. 28. What is the sum of 7 a’2’y, — 8a’xy + 18 a’2*y, and — 8 ax’y? Ans. 17 a’2*?y — 8 ary. 29. Add x —yand x + y. Ans.. 2x. 30. Add y and — y. Ans. 0. 31. Add, as, a’x, ax’, a?x’, and az. Ans. 2ax + aa + az? + a?x’. 32. Add 2a to a’. Ans. 2a + a’. 33. Add ab and ab. Ans. 2 ab. 34. Add abe, — 3 abc, 4 abe, and — 3 aed. Ans. 2abe — 8 aed. 35. Add a — 3y, 6a — Dy, and Ta — 3y. Ans. 14a —11y. 386. Add 10” +4 38y,5a”—2y, and 84 —4y. | Ans. 18a — 38 y. 37. Add 9 be — 7x + z, 8be — 3 a, and 5 be. Ans. 22 be —10 x + z. 38. Add 10 ay + 2 — abr, — 8 ay + 32, Tay + 5 abx, — 8ay —22,and 622—4abzr. Ans. Bay+ 82. 39. Add 7x and 7 y. Ans. Tx + Ty. 38 26 ELEMENTARY ALGEBRA. 40. Add 52+ a,—3a + 6, and—2a—a,. Ans. 6. 41, Add 13 by + 6ac, —11 by + 5ac, and — 2 by — 11 ae. | Ans. 0. 42. Add 21 ax + 18 abe — abe + 2ay, 18 abe 4+ 14 ax — ax — day, and 3 ay — 27 abe + Sax. Ans. 87ax — 2 abe. 43. Add 3ed — 2de + ed — ed’? + ed’, 5ed — e’'d + 3 ed? — ed’, 7 cd — Tde + 8 cd’, 5e’'d — 2de + 3e’d’, and 21ed + 8de,—5ed + 2cd’ — 11 ed’. Ans. 36 ed — &8de + 7 cd? — 8 ed’. 44, Add 13 ab + 2ed and 25a’b + 8ed. Ans. 13ab + 25 ab + 5 cd. 45. Add aba, bex, and bea. Ans. aba + 2 bex. 46. Collect into one sum 8 bex, — 2 bex, 3bex, and — 5 bex. Ans. — bex. 47. Find the value of 3ryz — 2ayz + Sayz — 4axyz + LYZ — yz. Ans. 2 xyz. 48. Add 18atryz + 21a — 38y, —3av + 2y + 19 atxyz, — 3a’ + 2y — 2atxyz, and datryz — 3a — Qy + 12atryz. Ans. 52 atxyz + 12a + 21y. 49. Find the value of 21 a’z*y, 3 atay? — 13 a®aty + 2a°x°y, 13 atxcy*? — 11 aa’y + 8 a®a®y, 12 atay? — 10 a®ay, and 5 a?x°y. Ans. a’a'y + 28 atxy? — 4 a’ay. 50. Add 9 a’b3ct + 8 a3b’ct — 7 atb?c*, 21 a®b’ct — 5a’bict, — ll a’b'ct + 2 a3b’ct, 21 a*b’c®, and 5 a®b’ct — 2 a®b’ct. Ans. — 7 a’b'ct + 34 a8b’ct + 14 a407e*. 51. Add 9(a+ y), — 14 (a+ 2), and — 9 (a+/y). Ans. — 14 (a+z). 52. Add ae and ca. Ans. 2 ae. 53. What is the sum of 3a) — 5 ba + 12 ab — ba. Ans. 9 ab. 54. Add abs, 2 aab, and — bxa. Ans. 2 abz. 55. Add cay, —9 cyx, and —4ayc, Ans. —12 cay. Norse. In the last four examples, the terms are similar in each example. The letters are the same, only differently arranged. ELEMENTARY ALGEBRA. 27 FURTHER ILLUSTRATIONS OF ADDITION. ProsLEeM 1. A mass of iron and wood is sunk in water. The iron weighs 16 Ibs. under water; and the wood would buoy up 13 lbs. What does the whole mass weigh while under water ? OPERATION. 16 lbs. EXPLANATION. Consider weights posi- — 13 lbs. tive quantities. Then what buoys up or diminishes a weight must be a negative 3 lbs. quantity. ProsieM 2. If the iron in the above problem had weighed 24 Ibs. under water, and the wood had been able to buoy up 30 lbs., what would the whole mass have weighed under water ? OPERATION. 24 Ibs. EXPLANATION. The result here shows — 30 lbs. that the mass would not have weighed any- ——— thing under water, but would have been — 6 lbs. able to buoy up 6 lbs. ProsLEM 3. A ship started at the equator and sailed 3 degrees north, then 2 south, then 5 north, and then 9 south. In what latitude was it then? OPERATION. 3 degrees, — 2 degrees, EXPLANATION. Consider north lati- 5 degrees, tude positive. Then south latitude must — 9 degrees, be negative. The above course ends 3 7 degrees south of the equator. — 3 degrees. Prosiem 4, If an ant running up a tree ascends 4 feet, then falls back 2 feet, then ascends 9 feet, and then falls back 5 feet, what progress has it made? Ans. 6 feet. Let the pupil explain it. 28 ELEMENTARY ALGEBRA, SECTION VI. Problems to be solved by Equations containing but One Unknown Quantity. 1. A FARMER had 5 times as many cows as horses, and 3_ times as many sheep as cows. The number of them all was 105. How many of each kind had he? Ans. 5 horses, 25 cows, and 75 sheep. 2. From 64 times a certain number, if 36 times the num- ber be subtracted, 7 times the number added to the remain- der, and 12 times the number subtracted from the last result, the remainder is 46. What is the number? Ans. 2. 3. Four boys have in all 84 cents. The second has three times as much as the first; the third has as many as the difference between what the first has and what the second has; and the fourth has as many as the difference between twice what the first has and what the second has. How much has each? Ans. The first has 12 cents; the second, 36 cents; the third, 24 cents; and the fourth, 12 cents. 4, Two men are 175 miles apart, and travel toward each other. One travels 15 miles a day, and the other 20. In how many days will they meet? Ans. 5 days. Notre. Let x stand for the number of days required. Then the first will go x times 15 miles, or 152, and the second will go z times 20 miles. 5. If 1362 —121a” = 150, what is the yvaluevor a Ye ee 6. In a certain house there are twice as many doors as there are rooms, and twice as many windows as there are doors. There are 21 more windows than rooms. How many doors are in the house? Ans. 14 doors. 7. There are 8 numbers, the second of which is four times the first, and the third is twice as much as the first and ELEMENTARY ALGEBRA, - 29 second together. The difference between the second and third is 86. What are the numbers? Ans. 6, 24, and 60. 8. Two men are 42 miles apart, and travel toward each other. One goes 3 miles an hour, and the other 4. In how many hours will they meet? Ans. 6 hours. 9. Given, « + 16x — 38az = 42, to find the value of z. Ans ane +3, 10. A bankrupt owed A 4 times as much as he owed B. He owed C 3 times as much as he owed A. He owed D the difference between his debts to C and B. The amount of his debts to all four was $28000. How much did he owe D? Ans. $11000. 11. A man bought 3 horses and 4 cows for $600. Each horse cost twice as much as each cow. What did he give for each? Ans. He gave $120 for each horse, and $60 for each cow. Notre. Let z represent the price of a cow; 22, of a horse; 42, of all the cows; and 62, of all the horses. 12. James bought 16 apples at one time and 5 at another, all at the same rate. The first time he paid 33 cents more than the second. What was the price per apple? Ans. 3 cents. 13. Seven men and three boys were hired for a week, each man receiving 3 times as much as each boy. Altogether their wages amounted to $72. What did each receive? Ans. Each boy received $3, and each man received $9. 14. A farmer sold some corn, rye, and wheat; receiving 3 times as much a bushel for wheat as for corn, and twice as much for rye as for corn. There were 20 bushels of wheat, 16 bushels of rye, and 40 bushels of corn. He re- ceived in all $66. What was the price per bushel of each kind of grain? Ans. He received $1.50 a bushel for the wheat, $1.00 a bushel for the rye, and 50 cents a bushel for the corn. . 8 % 30 ELEMENTARY ALGEBRA. Find the value of 2 in the following equations. 15. 852 + 17x — 822 4 384—16x2 = 16 —7 4 19. Ans. & = 4, 16.2+ 22%—8274+42+4+ 54—62+ 72x = 130. Ans. x = 18. WW.24+2r4+ 38a —4xe24+ 5¢ —6e2=74+84 12 — 17. Ans. x = 10. 18.¢@—Ta#+ 8e+4+1l2=7 + 35 + 62 — 24. Ans. x = 10. 19. 2642 —14274+ 32 —1llaxz = 36 + 42 + 12 — 30 + 8. Ans. © = 17. 20. 88a —62 + 262 = 37 — 244+ 103. Ans. x = 2, SECTION VIL. Subtraction. PROBLEM 1. Subtract 42 from 72. OPERATION. (har EXPLANATION. It is evident that the dif- 4g. ference is 3x. This is the same as if we had —— changed the sign of the quantity to be subtracted, 3m. making it — 4x, and then added it to 7x. PROBLEM 2. Subtract 52 from 22. OPERATION. EXPLANATION. It is evident that after es taking from 2 7 as much as we can of the 52, 5 2, 3% will remain to be subtracted. That is what is meant by —3wa. This is the same as if we — 32. had changed the sign of the quantity to be sub- tracted, making it — 5a, and then added it to 22x. ELEMENTARY ALGEBRA. ol PROBLEM 3. Subtract — 2x from 642. OPERATION. EXPLANATION. It is evident that 62 is 6 x, the same as 8x—2z2. If we subtract or take — 22, away the —2., the 8x remains. This is the same as if we had changed the sign of the quan- 8 x. tity to be subtracted, making it + 2, and then added it to 6x. ProstEem 4. Subtract — 3-2 from — 8x. OPERATION. EXPLANATION. It is evident that —8-2x — 8a, is the same as —5x2—8zwx. If we subtract id a, or take away the — 32, the — 5a remains. This is the same as if we had changed the sign — 5x. of the quantity to be subtracted, making it + 32, and then added it to — 8a. PROBLEM 5. Subtract — 7x from — 2a. OPERATION. EXPLANATION. It is evident that — 2 z is — 2x, the same as 5&2 —7ax. If we subtract or take orl a away the — 72z,the 5x remains. This is the same as if we had changed the sign of the sub- 5 x. trahend, making it + 7 x, and then added it to the minuend, — 2 x. PrRoBuLeEM 6. Subtract } from ec. OPERATION. c EXPLANATION. It is evident that the re- b, mainder is c— 06. This is the same as if we had changed the sign of the subtrahend, making e— b. it — b, and then added it to the minuend, ¢. Adding anything with a + sign, and the same thing with a — sign, does not alter the value of a quantity. Thus: a is the same asa + x—wZ. 5c is the same as 5c + 36— 8380. oe ELEMENTARY ALGEBRA. PROBLEM 7. Subtract — 6 from ec. OPERATION. EXPLANATION. It is evident that ¢ is the c same asc + 6 —b, since + b and —b amount — b, to nothing. Subtracting or taking away the —b, the remainder is ec + 6. This is the e+ b. same as if we had changed the sign of the sub- trahend, making it + 6, and then added it to the minuend, c. To subtract anything is to take it away. It cannot be taken away from anything unless it is there. But it can always be got there by adding it to the other with both a + sign and a — sign; for the two balance each other and amount to nothing, as + 6 — bin the example above. If then we take it away, the same thing with the opposite sign will be left added to the minuend. Thus we see that the operation of subtracting is just the same as changing the sign of the quantity to be subtracted and adding it to the minuend. Similar terms should be set down in the same column. It is best not to actwally change the sign, but only to imagine it changed, and then set down the result of adding it to the minuend. ProsueM 8. Take 3a + 5c — 2d — 2y from 7a — y + 2¢— 7d. ProspuEM 9. Take 5ax — 2y + c from 38ax — 2y. PropiEM 10. Take 4cx + ca? from dex + c?x? + 3 cx’. a 11. Take 8ab + 7a) + x from 3a’) — =z. PRoBLEM 12. Subtract 9 ab — 3 ab? from 5 ab? — 6 a’b’, PRoBLEM 13. Subtract 21 y — 4 from 3 y* — 2. aOaNIaark WN ht pe = Oo © a a ee ee Oo © onal & Oo kP Cb bo hd ELEMENTARY ALGEBRA. oa OPERATIONS. 8. Ta— yt 2c— 7d, 9. 8au—2y 3a—2y + dc — 2d, Sax —2y+¢, 4a+t+ y—3e— dd. — 2ax —¢. 10. 5er + cx? + 3 cz2’, 11. 3a’7b — x, 4cx + 2°’ 8ab+7vb+4+ a, Cx + 3cx’, — 8ab — 4a°b — 2x. 12, 5ab?—6a° 13, 8yt—2 —3ab? + 9a’), —4+ 21y, an 00) — 9 ab: 38y4+2—21y. EXAMPLES. . Subtract 8 x from 92. Ans. x. . Subtract 167 + 3y from19xz+ 5y. Ans. 8x 4 2Qy. . Subtract 18 z — 10 from 14z— 12. Ans. —4z— 2. . Subtract 16 y — 15 from 17 y + 38. Ans. y + 18. . Subtract a from a + zx. Ans. x. . Subtract 2 — 10 from z. Ans. 10. . Subtract x — 15 from 15. Ans. — x + 380. . Subtract a + 6 + ¢ from a — 6 —e. Ans. —26—2e. . Subtract a —b—cfroma+6+e Ans. 26 + 2e. . Subtract — 105 from 21. Ans. 126. . Subtract 35 from — 4. Ans. — 39. . Subtract a from — 6. Ans. — a — b. . Subtract 6 from — 6. Ange 2h: . Subtract — b from b. Ans. 2. . Subtract 7 # from — 22. Ans. — 9x. . Subtract — 5a from 122. Ans. 17 x. . Subtract 10 from 5. Ans. '— 5; . Subtract 17 from 6. Ans; = 11; . Subtract 29a from 14a. Ans. — 15a. . Subtract 87 x from 27 x. Ans. — 102 . Subtract — 15 from 5. Ans. 20. 34 22. 23. 24. 20. 26. 27. 28. 29. 30. ol. 32. 9 3. 34. 30. 36. 37. 38. 39. 40. 41. 42. 43, 44, ELEMENTARY ALGEBRA, Subtract — 15 from — 5. Ans. 10. Subtract 15 from 5. Ans. —.10. Subtract 15 from — 5. Ans. — 20. Subtract x + y from x7 + y. Ans. 0. Subtract 63 from — 63. Ans. — 126. Subtract — 6 from 6. Ans. 12. Subtract 23 a’a’y from 5 a’a’y. Ans. 5 a’a*y — 23 a7aty, Subtract 16 amn from 5 amn. Ans. —11amn. Subtract 144 abex from 25 abs. ' Ans. 25 ab2 — 144 abex. Subtract x from y. Ans. y —&. Subtract 12 2 — 3 from 138 y — 4. Ans. 134 —12¢% —1, Subtract 6z + 43 from 122 — 21. Ans. 6z — 64. Subtract 11 a —2¢ from 9a —5e Ans. —2a—3e. Subtract 15 2 —13 y — 2z from 17x —13y 4 z. Ans. 2% + 32. Subtract 13a —26 + & from 23a + 116 —A. Ans. 10a + 136—A—kE. Subtract —15a’a + Tay —15 from 21 a’a% — 18 a’y + 2. Ans. 36 Wa —20a?y + 17. Subtract 192z + 38yz — 14x22’ from 138 #z —12yz + 15. Ans. —6az—1dyz + 1422? + 15. Subtract 2m’n — 5 hn + 6 from 25 m’n —14 mn? + ha — 3. Ans. 23 m’n —14 mn? + 6hn — 9. Subtract 13¢@+ 36 + 5c from 18a—306 + 5e—1. Ans. —6b —1. Subtract 129 a? + 12 a@’x —ay + ax from 189 a? — 36 a'r + ay—3. Ans. 60a? — 48 a’x + 2 ay — ax —3. Subtract 13 azy — 2b —19cx —5 from 17axy —136 1c — 2. Ans. 4axy —11b+ 8lez 4+ 3. Subtract c + « + y —7b from 23mn —7b + e—a —y. Ans. 23mn —22”—2y. Subtract 5 a? —9 pq + 3px —z from 13 pq — 2px + a? —z, Ans. 22 pq — 5d px —4 a’. 45. 46, 47. 48. 49, 50. 51. 52, 53. 54. 5D. 56. 57. 58, 59, 60. ELEMENTARY ALGEBRA. 35 Subtract az’ + 10 ax — 20a from 20 az? — 10 ax + a. Ans. 19 ax? —20 ax + 21a. Subtract 217 amv —18w—3u —5D from 193 amv — 18w + 5u —6. Ans. — 24amv + 8u—1. Subtract 22? —16q — 28m + «'*y’ from 15 a*y? — 29 m + q. Ans. 14 ay’ —m + 17q —22’, Subtract 6 a? —5a + 21 from 2 a?. Ans. —4a? + 5a —21. Subtract —5hk + 3k? — 36 from 3hk — 21. Ans. 8hk —3k? + 15. Subtract 9 a? from — 13 a@—3a + 6. Aus. — 220 —s3a + 6; Subtract 172 2° + 102°y —3ay’ + y* from 250 «® — 120 xy + dry’ — y’. Ans. 78 x —1380a’%y + 82y’? —2 y’. Subtract 1385 c’d + 6my —a’x from 170¢’d —a’x + 5 my. Ans. 35¢’d — my. Subtract 7 aw —10 by from 13 ax + 5 by. Ans. 6ax + 15 by. Subtract —5 ab + a? —0’ from 26 ab. Ans. 31 ab —a? + 0’. Subtract 7 ab? —5a’y + 3a’b from 2a°b — 5a’y + 3 ab’. Ans. —a’b —4 ab Subtract « + y and —a + y from x —y. Ans. « —3y. Subtract 9a —b, —Ta + b, and 6a + 36 from 5a —3b.. é Ans. —3a —6b. Subtract « + y +z and —xz —y +2 from x —y —z. Ans. x —y —32z. Subtract Tay —3ax + ay from 9x —3ay + 2x —5bday. Ans. 144” —16ay. Subtract 8 aty? —6a'y + 3 a‘y’ from 13 ay —9 aty’ — bay. Ans. 14 aby — 20 aty’. Norr. Examples in Subtraction may be proved by adding the subtrahend to the remainder. The result should be the same as the minuend, 36 ELEMENTARY ALGEBRA. SECTION VIII. Transposition. In the equations previously given, all the unknown quan- tities were on one side of the sign =, and all the known quantities werc on the other. For example, we had the equation, 30a” - 17x—382a"+ 34—162 = 16—7 419. Here, in the first member of the equation, every term con- tains x; and in the second member, every term is known. That being the case, we may perform on them the operations which their signs indicate, and obtain as a result, (ae eos. Then, dividing both members by 7, co 4) It is sometimes the case that equations do not contain the unknown quantities thus separated from the known ones by the sign of equality. Then we must separate them, or the equations cannot be solved. For example, suppose 5a2—4= 6. Now, if by taking 4 away from 5 we get 6, the 5x must be 4 more than 6. That is, d”a2=6-+ 4. The equation can now be solved by collecting the terms and dividing : ee ANEE eae ek It will be seen that whereas — 4 was in the left member of the equation at first, + 4 is in the right member of the new equation. This operation is called transposition; and it will be shown that any quantity may be removed from one mem: ber of an equation to the other, if its sign be also changed. ELEMENTARY ALGEBRA. 2 yi PROBLEM 1. What is the value of x in the equation 7 x —9 = 26? SoLution. Here we wish to transpose —9. We can do it by adding + 9 to both members of the equation. (zt —9 = 26 +—+- 9== + 9, 7x = 26 + 9. Thus we have obtained a new equation. The — 9 disap- pears, and + 9 appears on the other side of the equation. Solving now this new equation, 7x2 = 26 + 9. Tx = 35, x= 5, ProBLEM 2. Given 8x2 = 24+ 52, to find the value of x. Sotution. Here we wish to transpose + 5a”. We can do it by adding — 5 to both members of the equation. 8x = 24+ 52. —b)2x= — 5a, 82—bH x2 = 24. In the new equation, the + 52 disappears, and — dz appears on the other side of the equation. Solving now this new equation, ra $. It will be observed, that to transpose a quantity we add the same quantity with a contrary sign. PROBLEM 3. Solve the equation 7x —14 + 3x45 aol. SoLuTion. Here we wish to transpose the — 14 and the + 5, so as to leave the terms containing x by themselves. 4 38 ELEMENTARY ALGEBRA. We can do it by adding these quantities with their signs changed to both members of the equation. Thus, Tae—14+32+4+5= 31. (fy +32 =314+14—5. | Collecting, 102 = ae Whence, c= 4, Now it is plain that adding quantities with their signs changed to that side of an equation in which they are, will always remove them from that side, since the same thing with both a + sign and a — sign amounts to nothing. It is also plain that doing the same thing to the other side of the equation will always introduce those quantities with their signs changed to that other side. And adding the same thing to the equal members must give equal results, thus giving us a new equation. PrRoBLEM 4. Given 7x = 63 —2z, to find the value of x. Sotution. Add 2.2 to both members, and we have (2x = 635 —22. +2z2=> + 2a Tx + 2x = 63. Collecting, 9a = 63. Dividing by 9, co =e ProsiEM 5. Given 132 —2 = 4a -+ 48, to find the value of x. SoLution. Add + 2 and — 42 to both members. 13 2—2 = Ax 43. +2—4%4=—42 + 2. Collecting, Ga = 46. Dividing by 9, Ferma ELEMENTARY ALGEBRA. 39 It is not customary to set down all this operation of ad- dition; but, since we know what the result must be, to set . that down at once, as will be done in the following examples. We must be careful not to leave out any term. It is best first to set down on each side of the sign of equality the terms which are to remain on that side, then those which are transposed to it from the other side. PrositEM 6. Given 1227 —14+4+ 8x2 = 72 4+ 52 +4 4, to find the value of z. SOLUTION. Transposing, 127 + 382—5H%=72+4+4+4 14. Collecting, 102 = 0. Dividing by 10, Cn ProsieM 7. Given —2x% + 41+7=>—2+8-—2 + 14, to find the value of x. SOLUTION. Transposing —2r+er=>—24 84 14—A41 —7. Collecting, — x2 = — 28. And if that is the case,z = 28. EXAMPLES. Find the value of « in each of the following equations: foe tO — 2.0 = 6. Ans. « = 1. 2.52 +11 —2a2 = 20. Ans. x = 38. 3. 4x —6 + 5 = 33 —14. Ans. 2 = 9. 4.12+ 5x2 = 108 —5d52 —6. Ans. x = 9. 5. 14= 60+ 4x2 —8a2 + 2. Ans. x = 12. 6. ¢= 56 —8 — 3m. Ans, 0 212, 72xn+7=24108. Ans. « = 8. 8. 26 —4"% =2x+5—3 —2e. Ans. x = 8. 9.144 32 = 78 —z. : Ans. x = 16. 10. 32 —30 = 41 —2 + 2-. Ans. x = 69. 11. 88 —12 = 14 + 9x2 — 5d. Ans..%@ =+13 40 12. 13. 14. 15. 16. ive 18. 19. 20. ELEMENTARY ALGEBRA. 3x2 —26 = 39 —8. Ans. « = 19, 39 + 2a = 4x —A?7. Ans. x = 48. 27 = 50 —724 61. Ans. x = 12. 8x = 14x” 4+ 14 —55 — 8b. Ans. « = 21. Find the value of y or z in the following equations: 22 —6y + 18 = 100 —14y 4 86. Ans, Y= ie sy—6=y+ 16 —6. Anas, Yi=ae 6+ 2z2=> 5z— 36 + 21. Ans. z = 7. 41 —6z2—17 = 120 —14z. Ans. z = 12. 82 —13z = 92 —15z. Ans. z= 86. Notr. Some equations may be somewhat simplified by merely omitting something which occurs in both members. Thus, in the 17th of the preceding examples, we may subtract the —6 from both members; that is, we may take it away or omit it, leaving the equation By =y+ 16. Then the transposition and the rest of the solution is somewhat simpler. SECTION IX. Questions producing Equations that require Transposition. PRoBLEM 1. What number is that which with 7 added to it amounts to 23? SoLution. Let x represent the number. Then we have the equation, e+ 7 = 23. Transposing, x = 23 —T. Collecting, Das be PropiEeM 2. Three partners, A, B, and C, have a capital of $56000. B’s is $5000 more than A’s, and C’s is $6000 iess than A’s. What is the capital of each? ELEMENTARY ALGEBRA. 41 Sotution. Let x represent A’s capital. Then x + 5000 represents B’s. And «x — 6000 represents C’s. We have the equation x + x + 5000 + « — 6000 = 56000. Transposing, «+ « + « = 56000 + 6000 — 5000. Collecting, 32 = 57000. Dividing by 3, zie 19000. Whence, xz + 5000 = 24000. And x — 6000 = 13000. Therefore A’s capital is $19000; B’s, $24000; and C’s, $13000. ProBLEM 3. B bought two horses and a $50 harness. The first horse was worth the second horse plus the harness. Twice the value of the second was equal to the value of the first plus the harness. What was the value of each horse? Sotution. Let x stand for the value of the second horse. Then x + 50 stands for the value of the first horse. We have the equation Qx2—=>a2+ 50+ 50. Transposing, ° 22 —x = 50 + 50. Collecting, 2 = 100. Whence, zx + 50 = 150. Therefore the first horse was worth $150; and the second, $100. EXAMPLES. 1. From three times a certain number 5 is subtracted, and the remainder is 13. Whatisthe number? Ans. 6. 2. A pole 23 feet long is stuck up in the middle of a pond, going into the mud at the bottom 2 feet. Twice the length in the water multiplied by 3 is the length in the air. How much is in the air? Ans. 18 feet. 3. The difference between two numbers is 1, and their sum is 13. What are the numbers? Ans. 7 and 6. 4% 42 ELEMENTARY ALGEBRA. 4, A man worked three days for $14. The second day he earned $1 more than. the first; and the third day, as much as on both the first and the second. How much did he earn each day? Ans. He earned $3 the first day, $4 the second day, and $7 the third day. 5. I met a man driving a flock of geese, and said, ‘‘Good morning, master, with your hundred geese.” He replied, “T have not a hundred; but if I had twice as many as I now have and 4 geese more, I would have a hundred.” If he spoke the truth, how many had he? Ans. 48 geese. 6. George is 5 times as old as John. The difference of their ages plus 24 years is twice the sum of their ages. What is the age of each? Ans. George’s age is 15 years, and John’s is 3 years. 7. James has 5 times as much money as George, and William has 38 times as much as George. James has 22 cents more than William. How much money has each? Ans. James has 55 cents; William; 33; and George, 11. 8. A and B entered into partnership, A contributing 3 ~ times as much to the stock as B. The sum of their stocks is worth the difference of their stocks plus $8000. What did each contribute? Ans. A contributed $12000; and B, $4000. 9. The head of a fish is twice as heavy as the tail; the body weighs 2 pounds more than twice the weight of the head and tail together ; and the whole fish weighs 20 pounds. What is the weight of the body? Ans. 14 pounds. 10. The head of the same fish is 6 inches long ; its tail is half as long as its body; and its tail and body together are 3 inches more than 4 times the length of the head. What is the length of its body? Ans. 18 inches. 11. John bought a certain number of tops, and 15 times as many marbles. After losing 14 of his marbles, and giv- ing away 30, he had only 16 marbles left. How many tops did he buy? Ans. 4 tops. ; ELEMENTARY ALGEBRA. 43 12. A and B can together earn $40 a week, of which B can earn $4 more than A. What can each earn? Ans. A can earn $18 a week; and B, $22. 13. There are 38 more counties in Connecticut than in Rhode Island, and 6 more in Massachusetts than in Connec- ticut. In the three States there are 2 more than 5 times as many as in Rhode Island. How many are there in each? Ans. In Rhode Island, 5; in Connecticut, 8; and in Mas- sachusetts, 14. 14. Divide the number 43 into two parts, one of which shall be 15 more than the other. Ans. The parts are 29 and 14. 15. Divide the number 43 into two such parts that 3 times the less part shall be 5 more than the greater. Ans. The parts are 31 and 12. 16. Divide the number 48 into two such parts that twice the less part shall be 4 less than the greater. Ans. The parts are 13 and 30. 17. In my largest bookcase are 3 times as many books as in my smallest and 60 books besides. In another, twice as many as in the smallest and 20 books besides. In both, 6 times as many as in the smallest. How many are in each? Ans. In the largest bookcase there are 300 books; in the smallest, 80; and in the other, 180. 18. Three pieces of marble weigh in all 15 times as much as the smallest piece and 19 pounds besides. The second weighs 3 pounds more than 5 times the first, and the third twice as much as the second. What does each weigh? Ans. The first piece weighs 10 pounds; the second, 55 pounds; and the third, 106 pounds. 19. What number is that which will be doubled by add- ing 5 to it and subtracting the sum from 41? Ans. 12. 20. Divide $47 among George, John, and Charles, so that Charles may have $1 more than George, and John $3 more 44 ELEMENTARY ALGEBRA. than Charles. Ans. George’s share is $14; John’s, $18; and Charles’s, $15. 21. A person spends $24 for clothing, and $10 for books, and then has remaining one-third of what he had at first. What has he remaining? Ans. $17. 22. There are four numbers whose sum is 18 more than twice the third. The second is 2 more than twice the first; the third is 3 times the first; and the fourth is 1 more than the third. What are the numbers? Ans. 5, 12,15, and 16. 23. Three boys spend at a candy-shop 36 cents. The second spends 1 cent more than twice as much as the first ; and the third spends 4 times as much as the first. What does each spend? Ans. The first spends 5 cents ; the second, 11 cents; and the third, 20 cents. 24. A mother’s age added to her daughter’s gives 40 years, and the mother’s is 4 years more than 3 times the daughter’s. What is the age of each? Ans. The mother’s age is 31 years, and the daughter’s is 9 years. 25. At a certain election 2874 persons voted, and the suc- cessful candidate had a majority of 376. How many per- sons voted for each? Ans. 1625 persons voted for the suc- cessful candidate, and 1249 for the other. 26. Divide the number 58 into three such parts that the first may exceed the second by 9 and may exceed the third by 17. Ans. The parts are 28, 19, and 11. 27. Of $120, I spent a part. Three times the remainder are $4 more than I spent. How much did I spend? Ans. $89. 28. A started to overtake B, who was 18 miles ahead of him. Both were on horseback, and A rode 10 miles an hour while B rode but 8. How long was A in overtaking B? Ans. 9 hours. 29. How can an estate of $12000 be divided among a widow, her son, and her daughter, so that the widow shall ELEMENTARY ALGEBRA. 45 have $2000 more than her daughter, and $2000 less than her son? Ans. The widow’s share is $4000, the son’s is $6000, and the daughter’s is $2000. 30. If the difference between 4 y + 20 and 2y —16 is equal to the difference between 84 —3y and 12 — y, what is the value of y? Ans. y = SECTION. X. Multiplication. The Signs + and —. Mourtiety + 3 by + 2. This means that + 3 is to be multiplied by 2, and the result added. Hence, the product is + 6 added, which is + 6. Multiply —3 by + 2. This means that — 3 is to be multiplied by 2, and the result added. Hence, the product is — 6 added, which is — 6. Multiply + 3 by —2. This means that + 3 is to be multiplied by 2, and the result subtracted. Hence, the pro- duct is + 6 subtracted, which is — 6. Multiply —3 by —2. This means that —3 is to be multiplied by 2, and the result subtracted. Hence, the pro- duct is — 6 subtracted, which is + 6. Thus we have had: + 38 multiplied by + 2 = + 6, — 3 multiplied by + 2 = — 6, + 3 multiplied by — 2 = — 6, and — 3 multiplied by — 2 = + 6. In these cases, when the signs of the two factors are the same, the sign of the product is +; and when the signs of the two factors are different, the sign of the product is —. Now this must always be so. To multiply any + quan 46 ELEMENTARY ALGEBRA, tity by a + quantity is simply to take it so many times and add the result; and to multiply any — quantity by a — quantity is to take it so many times and subtract the result. In both cases the product is +. Also, to multiply any — quantity by a -++ quantity is to take it so many times and add the result; and to multiply any + quantity by a — quantity is to take it so many times and subtract the result. In both cases the product is —. Hence, whenever we multiply two quantities together, like signs give +, and unlike signs give —. EXAMPLES. 1. Multiply + 8 by — 2. Ans. — 16. 2. Multiply — 8 by + 2. Ans. — 16, 3. Multiply — 8 by — 2. Ans. 16. 4. Multiply + 8 by + 2. Ans. 16. d. Multiply — 13 by + 21. Ans. — 273. 6. Multiply — 18 by — 5. Ans. 90. 7. Multiply 6 by 7. Ans. 42. 8. Multiply — 6 by 7. Ans. — 472, 9. Multiply 6 by — 7. Ans. — 42. 10. Multiply — 5 by — 18. Ans. 65. 11. Multiply 6 by a. Ans. 6 x. 12. Multiply 6 by — a. Ans. — 62. 13. Multiply — 6 by a. Ans. — 62. 14, Multiply — 6 by — 2a. Ans. 6 x. 15. Multiply —11 by 22. Ans. — 22 x. 16. Multiply 4 by —7a. Ans. — 28%. 17. Multiply — 5 by — 3, Ans. 15 x. 18. Multiply 4 by —a. Ans. — 4a. 19. Multiply —2 by —e. Ans. 2c. 20. Multiply — 5 by 3. Ans. —15¢, ELEMENTARY ALGEBRA. 47 SECTION XI. Multiplication of Literal Quantities. WHEN we write two letters together without any sign between them, we mean that one is to be multiplied by the other. Thus, ad means the product of a and b. So does ba. In the same way, xyz means the product of «, y, and z. This is not the case with figures. Thus, 53 does not mean 5 times 3, but 5 tens added to 3. A figure before a letter, however, indicates multiplication: 4n means 4 times n. EXAMPLES, 1. Multiply 9 ad by 6c. Ans. 54 abe. 2. Multiply 5ay by —n. Ans. — dnxy. 3. Multiply — a’ by — 8’. Ans. a’b’. 4. Multiply — 8a’x by 8 ny. Ans. — 24 a’xny. 5. Multiply — 15 yz by —3 a”, Ans. 45 yzx’. 6. Multiply 6 a? by — 82 ct. Ans. — 192 a’x*. 7. Multiply — 42 abcdef by gha. Ans. — 42 abcdefghi. 8. Multiply — 2 ca*y® by — 5 n’a. Ans. 10 cay*n?a. 9. Multiply 22 mn by 3 p’g. Ans. 66 mnp’q. 10. Multiply —4mn by — 2 az. Ans. 8 mnax. 11. Multiply —9 a by 8y. Ans. —72 xy. 12. Multiply —8z by —9y. Ans. 72 xy. 13. Multiply 4 cn by 4 a. Ans. 16 ena. 14. Multiply 3 2’ by — y’. Ans. —3 x*y', 15. Multiply —2y by 7 nh. Ans. —14 ynh. 16. Multiply — 11 xz by — 6 ws. Ans. 66 xzus. hoe ee Gq y*, 10. Multiply 9a° by 5 a’. Ans. 45 a’, 11. Multiply 21 ab? by 32 ab’c’, Ans. 672 a5b'c’. 12. Multiply 15 2’y° by 3 xy. Ans. 45 ay. 13. Multiply 18 x*y® by 4 ay4z”, Ans. 52 xy'2’, 14. Multiply 12 a*b’c’ by 4 a’ctrt, Ans. 48 a®b’c'x', 15. Multiply 5 a®x’y? by — 6 aba, Ans. — 30 a®a® yz. 16. Multiply — 6 ay*z* by — 4 23732’, Ans. 24 x®y®z*, 17. Multiply — 5 xty’z by 7 a®y’. Ans. — 35 a'aty*z. 18. Multiply 21 az*y” by 3a’xy’. Ans. 63 ax?y”. 19. Multiply — 14 a®x%2? by 2axyz. Ans. — 28 a®xy*z*. . Multiply —4 2/52" by — 5 ayiz’. Ana 20 sfy 22". . Multiply — 38 atatz’ by 9 a®btx®yz. Ans. — 27 a®btx® yz’, . Multiply —7 axry'z by — 38 ary'z. Ans. 21 ax’y%2?, . Multiply — 4 a*m'nt by —9 a’min®=. Ans. 36 am?n™. . Multiply 12 ap*¢’rs by 3 abpqrs’. Ans. 36 a®p'q’r’s”. . Multiply 16 a’m®x%y by 12 atm’y'. Ans. 192 a®m'x®y’. . Multiply 5 a®m®x*z* by — 3 am’a%z*. Ans. — 15 a®m" 2725. . Multiply —0 @ax’y*z by 6 xty'2’. Ans. 0. . Multiply — 6 aty’z" by — 2 a®atz”, Ans, 12 a®aty®2?. . Multiply together 12 a "2°, — 5 a?z, and 6 a’. Ans. — 360 azz’, . Multiply together 5 m’n7p’q°, —3 m'n’, and — 7 p*q’. Ans. 105 mn*p'q™. . Multiply together — 12 a®b’xty’, —5atb’xty’, and — 2 abr y*, Ans, — 120 a*b'x'ty’. . Multiply together 18 aSx’y, 3 a°z’, and — 2 ata®yz’. Ans. .— 108 a®x"y?z', . Multiply together 12 a®b‘c’d’e> and — 9 a’btce*. Ans. — 108 ab®c'd’e’. . Multiply together 11 a®’x’, — 15 az*y*, and — 14 ax’y*z’. Ans. 2310 a" ax'y*2’, . Multiply together — 3 a‘z*, — 17 az‘, and 5 ata*y. Ans, 255 a®x"y, 5. D 50 36. ELEMENTARY ALGEBRA. Multiply together — 12 wx°z, —4 a®xz’, and — 3 axty®z. Ans. — 144 a¥aty®2". . Multiply together — 4 atz’, 6 ax’y%z’, and 7 2’. ‘Ans. — 168 ax?y*z®, . Multiply together 12 m’n’p’, 4 m'n*p*, and 2 mnp. Ans. 96 m'n'p', . Multiply together a’, 12 a‘, 3 a’, a®, and a”. Ans. 36 a, . Multiply together a*x*, 5 a’x*, 3 aa*, 2 2, and a’. Ans. 30 a5z'8, . Multiply together 5 2? and atz’. Ans. 5 atx’. . Multiply together 0 a'a? and 5 ata’. Ans. 0. . Multiply together 7 a’z* and 13 a®y*. Ans, 91 aay’. . Multiply together 15 a‘v'y’ and 4 a®x®y‘z”. Ans. 60 a®a'®y®2?, . Multiply together — 7 a®aty’? and — 11 a*b'cta®y*2?, Ans. 77 aM beta y$2?, Multiply together — 2 a*e’u'd® and — 5 2°7*d¥a’. Ans. 10 a®e®d®xty?, . Multiply together — 18 a°xb*, — 2 b’ctx’, and — 8 2°b'a?, Ans. — 208 ab'cta™. . Multiply together — 5 atz?m®, 3 m*zx’a*, and — 2 a*a’m’. Ans. 30 a?m'*2?, . Multiply together — m,—m, —m, —m, —m, and —m. Ans. m*, . Multiply together — m, m, — m, m, —m, and m. Ans. —m*, . Multiply together — ax, ax, —ax, ax, —ax, and az, Ans. — a’x’, . Multiply together — ax, — ax, —- ax, — ax, and — aa. Ans. — abx', . Multiply together — az, ax, — ax, ax, and — ax. Ans. — aba’, . Multiply together 6 ab, — 2m’, — 9 ab, and 3 m’. Ans. 324 7b’m*, Multiply together 6 ab, — 2 m?, 9 ab, and 3 m?. Ans. — 324 a?b*m‘, _ ELEMENTARY ALGEBRA. 51 56. Multiply together — 6 ab, —2 m’, — 9 ab, and 3 m’. Ans. — 324 a’b’m‘. 57. Multiply together — 6 ab, — 2m’, —9 ab, and — 3m’. Ans. 324 07b’m‘, 58. Multiply together — 6 ab, — 2m’, and — 9 ab. Ans. — 108 a’b’m’?. 59. Multiply together 9 c’d’e*f*g*h' and — 2 c*h®f?dPe*. Ans. — 18 cdSe'f*g*h*. 60. Multiply together — 2 atzx*c’y’ and — 5 &xty?a’. Ans. 10 a®c®x*y, SECTION XIII. Multiplication of a Polynomial by a Monomial. HitHERTO we have only multiplied together single terms. When we wish to multiply a quantity containing several terms, we must multiply all of its terms. To multiply y—z by z, we must multiply y by x and then —z by z. PrositEM 1. Multiply 7 ay—3a + 2¢ by —4 a’ex’, PrRoBLEM 2. Find the value of the following expression: (—9n? —3 cx* —1la’n) X nz. Nore. The parenthesis is necessary here. If it were omitted, the meaning would be that only —11 a?n was to be multiplied by 5 nz, OPERATIONS, +n 7 ay ee Oe + 2¢, — 4 acz’, — 28 atca’y + 12 a®cx*® — 8 ade?x?, va — 9n? — 8ext —I11a’n, 5 nx, — 45 n'x — 15 ena® — 55 a’n?x. D2 12 18. ELEMENTARY ALGEBRA. EXAMPLES. Find the values of the following expressions: .(5a+ 24%—3y) X a. Ans. 5a? + 2ax —3 ay. . (8ax —3a’ + a’) X a’y. Ans. 8 aa®'y —38 aaty + sty. . (Tay —4a’y’ + Say’) x 8ay. Ans. 21 aty? —12 a®y® + 15 a’y*. . 18 aaty — 5 ax’y’? + Tay?) X —8 axy. Ans. — 54 a'aty’ + 15 a’aty’ OOF ax’y*. . (—9ata’y + da’ay’ —2y°) x —42°y*. Ans. 36 ataty® —12 aayt + 8 a’y'. . (5 ab — 207d? + 7 ab® —2b*) x —ab. Ans. —5 ath? + 2 a*°b? —7 a’b*t + 2ab. . (Gat — 5 ab? + 40°b? —7 abt) xX —8ab. Ans. —18a'b + 15 ath? —12 a°bt + 21 ad. . (—21 ming —14b2 + 162?) x —8 amie. Ans. 63 amma? + 42 a&bmtx? — 48 a®mtz3. . (12 aa — day’ — 3a’) x 8abaty. Ans. 36 a’a®'y — 15 a¥aty’ — 9 aay. . (10 a®xiz — Sarat’? + 7 aa®z? — 2a*‘xz) x 0 azz. Ans. 0. . (18 mn? — 11 ab + 5 ty’) xX Oatmiy. Ans. 0. . (abe — 8 ab + 6a —7) X —5Bae. Ans. —45a’be? + 40 a7?be — 30 a7e + 35 ae. . (18¢ —19 a + 16 2’y’) x 2a’. Ans. 36 a’ct — 38 ate? + 82 aeha’y’. . (16 ab —T7 bax + 9 a? — 8 ex) X Sean. Ans. 80 a°be?'x — 35 wbe'x? + 45 a&x — 40 ac®x’. . (12a —5a’b + 6a? —Qay) x —T aye. Ans. — 84atry + 35 a°®bxy —42a°xy + 14a’7ry’. . (19 ay? —13 min'a + 18y) x 5 xm. Ans. 95a&imz*y? — 65 &intntat + 90 may, (10 a3b’e —9 abe? + 38a°b'c? + 2a°bc®) x 8 abe. Ans. 80 aSdtc? — 72 atb’ + 24 abd’ + 16 abd8ct. (Saxy? —Tay + 6aa*y* —2ax) X —2axy. Ans. —10a@’a’*y’ + l4a’xy? —12 aaty? + 4a’a*y 19. 20. yA 22. 23. 24, 25. 26. 27. 28. 29. 30. ELEMENTARY ALGEBRA. 53 (3 e'd?a — 4 d’e’x? + 5 e’f?a*) x 6 efx’. Ans. 18 e’d’efx® — 24 d’e®fa* + 30 e*f*a’. (— 12 a®yz — 5 a?2’*x + 3 a'yz —Qayx) X — 3 ax*. Ans. 36 abx*yz + 15 atatz? — 9 aasyz + 6 abrrty, (— Dd axy’ + 2 a yx” —5 y’za — a’) X A aye. Ans. — 20 a’x’y®? + 8 ata’y? — 20 way?z — 4atvy. (—2aty® + 3a%m™) x 6 2%y%m’, Ans. —12 a m’ay® + 18 a®m7™ x97”, (5 A®k —60°k + 7 bh) x — 8 Dhk. 7 Ans. — 40 bA°k? + 48 b&hk? — 56 b°A7k. (atpq — 7 pqx’ + 5 pq’x — x*p*q?) X 12 axqp*. Ans. 12 p'q?a' — 84 p'q’a> + 60 p'q?c*t — 12 p* qa. (14 b’e? — 8 a®b’ctard'e") x 4 ab cry. Ans. 56 a®b¥ce’xy — 32 a®b™ cde" a*y. (12 —5ctat + 2bex) x — abex. Ans. —12abex + 5 abbe’x — 2 ab’c*x”. (abe — be’a + bea? — abe?) X — abe’. Ans. — ab’ + a?’ — ath’? + atb’ct. (Sak —4ah + 3 —2k’) x 6ahk’. Ans. 30 ahk® — 24 ah?k® + 18 ahk® — 12 ahh’. (12 aca? —5ax’e’b + 4 x yb?) x Tatb’a. Ans. 84 a°b’cx’ — 35 abb’c’x® + 28 atbihx®y’. (5 atb* — 7 b*y* — 6 aty*) x —10 a°d%y’. Ans. —50 ab’y® + 70.a8b°y” + 60 ab y” SECTION XIV. Multiplication of a Polynomial by a Polynomial. To multiply anything by 3 + 5, we must not only take it 3 times, but also 5 times. So, to multiply by x + y, we must first multiply by x, then by y, and add the products. ProstemM 1. Multiplya+odby ez + y. &* 54 ELEMENTARY ALGEBRA. OPERATION. a + 8, co Y; ax + bx == the product of a + band z. + ay + by, = the product of a + 5 and y. ax + bx + ay + by, = the product ofa+bandzty,. PRoBLEM 2. Multiply 30% + 4ca’y —3a°y' by 2c'm —8cx’y —6 xy’. OPERATION. 8cea + 4eay— 3 2°y? 20a — 8cecaxy— 6257 6a’ + 8eay— 6 eaty’ — 24 &a®y — 32 eaxty’? + 24 cay? — 18 Caty’ — 24 ca®y® + 18 x8y4 6 cx? — 16 ea*y — 56 erty’ + 18 xy‘. ProsiEM 3. Multiply 2® —2° + at —a2’+ 2?—2+41 by 2 —a +1. OPERATION. ®o@— O+ wt — P+ wo? — a2 +1 xv—ae+l rmP— 2+ w— 2+ zt— vt 2 Ee at ge — xs xt — ae + a? — xr a— gat gt— Gt 2?— atl ve —220' + 8a —38 24+ 8x —32' + 8a2’—224+1. In the last example, the multiplicand is said to be ar- ranged according to the powers of x, beginning with the highest. So is the multiplier. Hence the operation has a kind of regularity about it, and the product is also arranged according to the powers of z. In the example before it, the factors are arranged according to the powers of-e, beginning ELEMENTARY ALGEBRA. 55 with the highest; and it so happens that they are also ar- ranged according to the powers of x and of y, beginning with the lowest. Such an arrangement is not always possi- ble, but should be made whenever it is possible. EXAMPLES. . Multiply « + y by x + y. Ans. x? + Q2ay + y’. . Multiply e+ d by e+ d. Ans. ’& + 2ed + d’. . Multiply 6+ «by b+ + pq + pd + &. 29. Multiply p®? + q@ by p + q. Ans. pt + p? + pa + o%. 30. Multiply 4+ 5y by 2a + 3. Ans. 8a+ 10ay + 124+ 15y. 31. Multiply 32? + 2y? by 227+ 3y’. Ans. 6 xt + 18 2°y’ + 6 y*. 32. Multiply 2? + 3 a’y + 3827’ + y* by 2 —3 a’y + 3 2y’ —y’. Ans. x® —3 arty? + 8ax7y* — y’. 33. Multiply 3 at —12 ae + 2az* —52* by Ta —2z. Ans. 2145 —90 ata + 24082? + 14 a07x> — 39 ax* + 102°. 34. Multiply 3 a* + d5a’y + y’ by 3a* —5a’y + 7’. 9a? —19 aty? + y*. 35. Multiply 7 aa —2b’y + zx by 5a’ + 3 d* — y’. Ans. 35a'x —10a7b’y + 5a’xz + 21 a®bta —6 by + 3 btxz —7T aay’ + 2 by? —ay’z. 36. Multiply 16 atm? —da’e + 2? by a—z+ 327, Ans. 16 a®’m? —5a®x + az? —16a'm’z + 5arz —2 + 48 atm?a? —15 ax? + 8 x27. 37. Multiply 11 6’e — 5 a’y? + 22 by 14a3m + yz. Ans. 154 0°b’em —T0 amy’? + 28 a'mz + 11 Beyz — 5 ay*z + Qy2*. 38. Multiply 13 7’s —5 rs? + 37s by 67's + 2 rs’ —rs. Ans. 78 rs? —47°s* + 5 73s? —10r’st -L 11 7’s* — 3 75%, 39. Multiply 15 e’'d* — def + 2ed by 16 ef + 4ey —z. Ans. 240 c’dPef — 80 e? f? + 32 edef + 60 ctd®’y — 20c’efy + 8e&dy —15 ed*z + 5 efz —2 edz. 40. Multiply 16 atm + 2 am? —4 mi by 6 a’m — 83 am’, Ans. 96 a9m? + 12 a&m* — 24 a’m® — 48 aim? — 6 a?mi + 12 am’. 41, ELEMENTARY ALGEBRA. 57 Multiply 25 a° + 6a'b + 7 a5b? by 2a —30. Ans. 50 at + 12a5b + 14 a%? —75 a*b —- 18 ath? — 21 ad’. 42. 43. 44, 45. 46. Multiply 1 —3e@ + 32?—a2' by 1 —22 +4 a’. Ans. 1 —5x + 1027 —102° + 5a2t — 2’. Multiply 7a —3b by —5a + 26. Ans. —35 a’ + 29 ab — 60’. Multiply 5a? + 2ab —30’? by —5a-+ 68. Ans. — 25a’ + 2007b + 27 ab? —18 6%. Multiply 3 a? —4 ax + 52? by Ta? —2ax —38 a’. Ans. 21 at — 34 ax + 34072’? 4+ 2ax* — 152%. Multiply 5 a? —4ay + Ty’ by 2a? —3 ay —3y’. Ans. 10 at — 23 ay + 11la’y? —9 ay’ — 21 ¥'. . Multiply m* — n‘* by m* + nt. Ans. m® — n°. . Multiply m* —n‘* by mt —n*. Ans. m®’ —2 mnt + n°. . Multiply 82° —2.x2 + 5 by 4a*—a# + 6. Ans. 1227 —3a° + 182° —823 + 2227 —17x + 30. . Multiply 2 —ay + y —yz—az 4+ ? bye +y+2z. Ans. 22 + y§ + 22 —8 xyz. . Multiply 81 c+ 27¢y + 9c’? + 38cy> + y' by 8¢ —y. Ans. 248 —y’. . Multiply 32 a + 16a‘x + 8a'x’? + 407° + 2 axr* + x by 2a— x. Ans. 64a§ — 2°, . Multiply 1024 b> — 256 be + 64 dc? — 16d’ + 4 bet —éby4b+ 6. Ans. 4096 b§ — &. . Multiply 81c¢ + 54a + 36 ca? + 24 ca + 16 x* by 3¢e—22. Ans. 243 & — 32 2°. . Multiply 75s° + rst + rs + 1 by —7°s* + 1’s’?’ — rs + 1. Ans. — 7°88 + 1. . Multiply m3 + 2 m’n + 2mn? + v3 by m? —2 mn + 2 mn? — n3. Ans. m& — n§, . Multiply a7 — 2 a’y + 3a®y? —4aty® + 5a®yt —6a’y* + Tay’ —8y' by & + 2ay 4+ y’. Ans. a —9ay® — 8 7’. . What is the value of (a? — ax + x”) (a? + ax -+ x”)? Ans. at + a?x? + xt, . What is the value of (« + 1) (1—2) (a + 8) (a —4)? Ans. st —2a° —13a° + 14x + 24. 58 ELEMENTARY ALGEBRA. 60. What is the value of (2 — 5) («a —6) (cw —7) (a —8)? Ans. x* —26 2° + 2512? — 1066 x + 1680. 61. Expand (a + a) («7 + 5b) (x@ + ¢). Ans. 2° + ax’? + ba? + cx’? + abc + acx + dex + abe. 62. Expand (x —a) (4 —n) (« —c). Ans. x° — ax* — na? —ca? + anx + acx + enxz —aen. 63. Expand (a° —2*) (a + 2) (at —a’e + aa*? —az? + 2*). Ans. 0" — a 64. Expand (a + 2°) (a —«x) (at + ade + a2? + aa? + a*), Ans. a” -—2 65. Expand (a? — 2’) (a? + ax + x”) (a —ax + 2’). | ARS te 66. Expand ab (a? + b?) (a+ 6) (a —b). Ans. a’b —ab’. 67. Expand (a + b) (a —b) (a + 0’). Ans. 0 =e 68. Multiply a —b + ¢ by a—b —e. Ans. 2 —2ab + B —e’. 69. Multiply together x + y, y + z, and z+ w. Ans. vyz+ yz + 024 y2 + ayw + yw + xzw + yz. 70. Multiply together 2 + y+ wandy + z+ w. Ans. cy + y? + Qyw + az + yz + zw + ow + w’ SECTION XV. Equations requiring Multiplication. ProsiEM 1. Given (x — 3) xX 10 = 40, to find the value of x. SoLuTIon. (x —3) x 10 = 40. Multiplying, 102 — 30 = 40. Transposing, 102 = 40 + 30. 1032570: x= 7. ELEMENTARY ALGEBRA. 59 ProsLEM 2. Given 5 + x x 11 + 14 = 91, to find the value of x. SOLUTION. 5+a¢x 11+ 14 = 91. Multiplying, 55+ lle+14= 91. Transposing, lla = 91 — 55 — 14. 1M ra ps, eka ProBiEM 3. Find the value of x, when (6—z2z) (13 —2x)—9 =a—8 X 2x4 35. SOLUTION. (5 —ax) (18 —22) —9 = « —8 X 2a + 385. Multiplying, 65 — 23x" 4+ 227 —9 = 22’ —16a@ + 35. Here the 2x? may be cancelled in both members of the equation ; that is, subtracted from each of the equals. We then have 65 —232—9 = —162 + 35. Transposing, —23a2+ 162 = 35 —65 + 9. | —Tx# = — 21. If —7x = —21,then + 7x = + 21. This change may be made by multiplying the equals by —1. Hence, Tx = 21. rk Pes Norse. In problem 2d, 5 + zis multiplied by 11 only. If it were to be multiplied by 11 + 14, there would have been a parenthesis enclosing 11-+ 14. So, in problem 3d, z —8 is multiplied by 22 only. EXAMPLES. Find the value of x in the following equations: 1. (@ +6) x 9—2 = (@# —1) xX 40—1. Ans. ce = 3. 2. (6 — ax) (6 —x) = (7 — x) (8 —x) — 10. | Ans. x = 4, 3. (6 + x) (8 —7) = —9-a + 15. UY Rep ia ELEMENTARY ALGEBRA. . (@ —T) (w@ —8) = (@ — 2) (w@—9) + 1. Ans. = —2. («+ 72+3=> (e—3)r¢—37. Ans. c= —4, .(@—4432)(5—2)=1l27—1. by — 3 m‘nx®. Divide —7 a®b'c’d'e? by 35 ab¥ct, Divide — 93 abate’d by — 36 xd. Ans. oie Ans. — = ; Ans. — ix Ans. = ia se i 3mx Ans. — Ss Ase 31 abc? 47, 48. 49, 50. 51. 52. 08. ‘4, 55. 56. 57. 58. 59. ELEMENTARY ALGEBRA. Divide 21 m‘n*q’a* by 35 ma’. Divide 33 a*y’m?at by — 638 am*a?. Divide — 27 x°m’y* by 36 m’yz*. Divide 13 abcde by — 25 eax. Divide 14 atsx*z?m? by — 35 a®z’q. Divide — 15 f%9°q"* by —5 fig’. Divide — 21 aty’z'w by 65 z°yw. Divide 19 w’z*v by 23 ubw’. Divide 12 x°c°b?y* by a 34 c°b*. Divide 24 abte? by 13 d. Divide —18 m’n’x‘p by — 21 m‘z%p. Divide — 19 y’x by —19 y’px. Divide —15 ab by 50 6’. 7 Ans. — Ans. on 11 an*y" OT eae xy® 4 ~ Ans. — 24 a®bte? 8. i3d . 6 nba Tm ° Ans. A p Ans. oo Ans. — 0s 74 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. ELEMENTARY ALGEBRA. Divide 18 abed by — 21 ed’. Divide — 36 7’s*tt by — 45 rés®x. Divide 34 p’qrs’? by 51 p’q’tz’*. Divide — 48 a*y’z* by 135 m®a’y*q. Divide —12 v’wutx by — 15 wa. Divide 49 a®y?m'n* by — 7 x?min’. Divide — 60 m“n*p” by 125 mp”. Divide 25 gh” by 35 hk, Divide — 13 w’y*z by — 8 a’x’y'z. Divide 15 c*d’e*f by 25 g®e?fe*. Divide — 33 x’y%z*m by — 33 a’y*z*m. 6 ab AN eae ELEMENTARY ALGEBRA, 75 SECTION XX. Division of a Polynomial by a Monomial. WHEN the dividend contains several terms and the di- visor only one, each term of the dividend must be divided. Prosiem. Divide 34a*zx'y —12 axy + 3ay' by 2 a’xy. OPERATION. 2a’xy ) 34 aa'y —12a@7xy + 32y'* : 3y° 17 ax* — 6 + , Pa iy EXAMPLES. . Divide ax + bx + ex by x. Anjwatb+e. 2. Divide —9 aty + 6 a'y —3a’y by —3 xy. 10. Ans. 32° —22? + x. . Divide 20 ac? — 45 aca + 10 a'xy by 5 a’. Ans. 4c? —9cex + Qaaxy. . Divide 15 aby —9 acy’ + 12 axy® by 8 a’y. Ans. 5 ab —8 cy? + 4a*y'. . Divide 36 ab’x*y — 16 2°y? — 20 x*y* by — 427y. Ans. —9a’b? + 4a'y + 5 xy’. . Divide 4 a’z* + 7 art — 9 atxd by — az’, Ans. — 4axn —7 a®a? + 9 a323. . Divide 6 atz*m? —8a*bx?m? + 10 aSerm? —4 abdx?m? by 2 a®a5m?. Ans. 8a —4) -- 6c —2d. . Divide — 15 ata*yt — 9 ata*y’ + 6 a®aty? by — 3 ataty’. Ans. 5y? + 32° —2 at, . Divide 75 ac'x? + 25 a2ce®'a —15 ac’a*yz by 5 eax. Ans. 15 acx + 5 ac —8 xyz, Divide 6 atem — 9 atem? + 12 atem® by — 8 a’, Ans. —2am + 3 am? — 4 ami. 76 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. ELEMENTARY ALGEBRA, Divide 7 ackx + 7 ac’y —7 aéz —T aed by 7 ac’. Ans. x + y—z —d. Divide 45 am‘n? — 27 am’n* + 63 m'n'y by 9 min’. Ans. 5am —8an + 7 n’y. Divide 36 p’q’rs’ — 72 p’q’rt? + 84 pq’ra® by — 12 pq’. Ans. —3 prs? + 6 pr? —T ra’. Divide 42 am?r*st? —63 be’r*st? + 105 7°s*?x by —21 7st”. Ans. — 2am + 3b — 58a. Divide 126 atx’y — 30 a®b’x’y + 18 ax*y’ by 6 xy. Ans. 21 at —5 ab? + 3 ary. Divide 49 vw’st — 35 vw®.— 2 2’yw* by 7 va’. Su Quty as 7 vx Divide 63 e?f*y + 27 etf*a'z — 7 éfax* by 9 efx’. Ans. 7 wx — Tee Ans. es J 48 ef xez — Divide 32 pq*rs — 32 cp’r — 48 p*q’r by — 16 pq’r. Ans. —2q's + e + 3p’ Divide —18 a®y + 16 y® + 27 a®y°z by —3 ay. Ans. 6a — eke — 9 a’y*z. Divide — abex + abe’x? —be by — abex. Ans. 1 —cx + = Divide 5 a’c'x? —7 aa*y + 3 2’cd by a’c’x*, Ans. 5¢ — ee) 2 ac Divide — 21 foh + Tf?g —5 fg’ by —38fogh. Ans. 7 — 23. 24, 20. 26. 27. 28. 29. 30. ELEMENTARY ALGEBRA. 77 Divide 4 y’cx —2 pat —y’ by 2 pe. 9 2 Mngt a po — woe ; Hs 2 px Divide 20 a®atz + 16 a’x’z? — 3 mnb by 5a*x*z. 16 z 3 bmn 5 2 rae Se ko Ans. 4a°a? + 5a? 5 gtxte’ Divide 36 fa’e —5r’s’x + 63 fex*c? by 21 f%x’e. 12 ores: og Ans. 7fi— Weft + 3 f*x’e. Divide 36 am’z* — 54 2’a + 34atm by —18 zam. 3 17 a* Ans. — 2 mz ote ae =. 9 az Divide 16 cfx* + 20 cfa®y —3 fretrtz by 4 efx’. Ans. 4c” + 5exy — pa Divide 18 amz*> + 9m — 15 a’mix’y by 5 a’mtx. 1 Ea yfirs h , Ans. Bat + —— —s8amey. 5a’mx Divide 32 abede —5 axed + 3e by —2 abcde. 5 ax 3 8 Ans. —16 A 5 ta a ae a 2 be 2 abed Divide 25 a®d’e’x + 21 ated — 30 atdex® by 5 abdex. Ans. 5a’de + ae — 6 ax’, 7% 78 ELEMENTARY ALGEBRA, SECTION XXI. Review of Subtraction. BEFORE proceeding any further in division, the student should make sure that he has fully mastered subtraction. The following examples are intended to afford exercise in such difficulties of subtraction as are likely to be met with in division. “Im or > . From a + 2azx take a? + ax. Ans. ax. . From @ + 2ax + a’ takea? + az. Ans. ax + 2%, . From w@ —2 ax + 2’ take a + ax. Ans. —3ax + a”. . From a? —2axz + x’ take a? —ax. Ans. —ax 4+ a”. . From ax + a’ take ax 4+ 2’, Ans. 0. . From —az + 2’ take —ax + 22. Ans. 0. . From 6 a7a? + 4 ac° + at take 6 a2? + 3 az’. Ans. ax + at, . From — 8 aa —5a’*x? + 38a2a° —2* take —8 a’a + 38 a7x2 —38ax°+ 5at, Ans. —8a’7x + 6ax® —6 2+. . From —6a’°y — 3ay —2y*> take —6x’*y — 227’ —4y’. Ans. 8 ee . From —5 arty + 7° take —5 aty + 10 ay’ Ans. — 10 xy? x cot : teh —4Aaa + 6a*x? —4a2z° + a* take at — 2 a®x + a’x”, Ans. —2a?x + 5 a®x? —4aa° + ot, . From —2 ax + 5 ax? —4a2° take —2a*x + 4072’ — 2 ax’. Ans. a?z? — 2 az. . From a’x? —2ax ake a?x? — : F "7 —2Qax*> + a take ax? —2ax* + x Ans. 0. . From —2’y? + y’ take —a’yi—ay*. Ans. zy* + y’. 15. . From —4az + 62’ take —4 ax —9 2” + y. From ryt + y° take ryt + y’. Ans. 0. Ans. 15 x? —y. ELEMENTARY ALGEBRA, 79 17. From — 40 a°b + 57 «2b? + 4ab' —15 b¢ take — 40 a%b + 320°)? + 24ab% Ans. 25 ab? —20.ab® —15 b+ 18. From —2a°b —3a’b? —6ab®? + 40 take —2a°d —-4q'>? —. 8 ab’. Ans. a7b? + 2ab*? + 454, 19. From —5a@’8? + 16 ab? —4 0‘ take — 5 a’?h? —4 0+. Ans. 16 ab’. 20. From 6 ab? — 5 abt + 4ab' take 6 ab? — 4 ab®, Ans. — 5abt + 8ab>. ——.059300—— SECTION XXII. Division of a Polynomial by a Polynomial. WHEN both dividend and divisor have several terms, the process is much like long division in Arithmetic. ProBLEM 1. Divide cx + bx by c + Bb. OPERATION. divisor dividend quotient e+tb)cx+ bx(z cx + bx EXPLANATION. The first term, c, of the divisor, is con- tained in the first term, cx, of the dividend, x times. Mul- tiplying the whole divisor by 2, the product is found to be ex + bx, which we subtract from the dividend, and find no remainder. ProsuLeM 2. Divide mz + my —cx —cy by x + y. OPERATION. x+y)mxe + my —cxr —cy (m—e mx + my — cx —cy — cx —cy EXPLANATION. Here x goes m times into mx; hence we 80 ELEMENTARY ALGEBRA. set down m in the quotient. We then multiply the whole divisor by m, and subtract the product from the dividend. The remainder is —ca —cy. This also must be divided by x+y. Wesee that x goes —c times into —cx; hence we set —c also in the quotient. Proceeding as before, we find no remainder. ProBLEM 3. Divide —3a@’x —a* + 3a2? + a by a + x2? —2ax. OPERATION. a —2Jax + x’) a —3ax + 3ax? —2° (a—ex a —Qax + ax? —avx + Qax? — 2? —ax + 2ax? — x’ EXPLANATION. In this problem, the only letters are a and x raised to various powers. One of these letters should be selected, and both dividend and divisor arranged so that the highest power of that letter comes first, then the next in order, and soon. This is for convenience. In the forego- ing operation, the terms are arranged according to the powers of a. In the divisor, we begin with the second power of a, next comes the first power, and finally a term without a. In the dividend, we have the third power, the second power, and the first power of a, and also a term without a. The process of dividing is the same as before. Let us now perform the same example, arranging the terms according to the powers of x. OPERATION. ow’ —2Qarn +a )—2+ 38ax*°—8ar+a(—x+a Lidar ae ax? —Qa’x + a’ ax’? —2ax + a Nore. It will be observed that the quotients, a —z and —z + a, are precisely the same. ELEMENTARY ALGEBRA, 81 ProsuiEM 4. Divide «* —w* by « —w. OPERATION. x—w) x —w? (at + a?w + aw? + aw? + wt ao’ — aw xtw —w xstw —x*w? rw — up ew? — ow? x’w° —w? x’w®? — rw xcw*t —w awt —w? EXPLANATION. In the dividend, the fourth, third, sec- ond, and first powers of « are wanting. Space has been left for them in the operation. ProsueM 5. Divide x*? + ybyx+y. OPERATION. a a +y(e+eryty x* — xy wy + y xy axe xy’ sy + y° cy Ye 2 y* remainder. EXPLANATION. The quotient may be completed by a fraction whose numerator is the remainder and whose de- nominator is the divisor. The full quotient is x* + ay 2y’ eed aie 4 82 ELEMENTARY ALGEBRA. ProsLEM 6. Divide 1 by 1 + 2. OPERATION. eee yt (1—2 + ata? + at, 1+2 —wzx —r—xz a? ot + x en SP — x — xt at at + a — 7 EXPLANATION. This operation may be carried as far as we please in the same way. When we have carried it as far as desirable, the quotient may be completed as in Prob- lem 5th. Observe that its sign is — in the above operation. 5 ; : A F as The full quotient, then, is 1 —a# + 2? —a* + a* — eras EXAMPLES. 1. Divide @ + 2ax+ a2’? byat oa. Ans. a + 2. 2. Divide a? — 2 ax + 2? by a — «. Ans. @ —®&. 3. Divide a —2’ by a + x. Ans. & —%. 4. Divide m? — n? by m —n. Ans. m+ n. 5. Divide ae — be — ad + bd by a — 6. Ans. ce —d. 6. Divide —7r’? + s’ by —r+ s. Ans. r+ s. 7. Divide at + 2a’x? + ax by a 4+ a”. Ans. a? + 2. 8. Divide a* —y* by x —y. Ans. x + xy + y’. 9. Divide zt —y* by x —y. Ans. 2° + a’y 4+ ay? + y’. 10. Divide a’ — 0° by a — 2b. Ans. a + ab + a®d? + a5? + abt + BE 11. Divide a’ — 3 a‘b? + 8 abt — B by a& — 30°) + 3ad’ — 6, Ans. @& +307b + 38ab? + BF. 12. 13. 14, 15. 16. AVE 18. 19. 20. 21. 22. ELEMENTARY ALGEBRA. 83 Divide m’ + m’n + mn? + n3 by m + nr. Ans. m’ ok rn’, Divide m* + m’n + mn® + nt by m + n. Ans. m® + ni’. Divide a + 1 bya +1. Ans. @ —a-+1. Divide y® — y by y’ — y+ y® — y. Ans. y* + 1. Divide 6° —1 by 6? —1. Ans. 6§ + 66+ 6? + 1. Divide 36 a’ — 36 B' by 9a — 908. Ans. 4a® + 40°) + 4 ath? + 405? + 40°b* + 48° + 4 b% Divide 2* — 256 by 2? —4. Ans. #6 + 4a* + 162’ + 64, Divide 16 «* — 81 y* by 24 —3 y. Ans. 8a + 12 a’y + 18 ay? + 27’. Divide 16a*+ 8a-+ 38 by 4a? —4a-+ 3. Ans. 47 +4a+1. Divide 81 2* — 81 x2? — 36 ry —4y’ by 92?+ 9u + 2 y. Ans. 92° —9x —2y. Divide 64 c&& — 128 ed*® + 64 d* by 2¢ —2d. Ans, 82+ 32ctd + 32 ed? —32 cd’ — 32 edt — 32 d°. 23. 24. 25. 26. 27. 28. 29. Divide 18 2’y? —15 xy + 2 by 6 xy —1. Ans. 8 ae — 2. y? Divide 2? 2b A — . + y yret+y Ns. X dare Divide 2° + y°> by x + y. Ans, 2 — Ty -- ¥ Divide «* + y* by x + y. 2 v4 Ans, 2 —2*y + ay?—y + Divide x® + a by x + a. Ans. at — ax? a5 ax? — va + at Divide a® + b& by a + 0. Ans. a —a‘b + a®b? —a’b® + abt —b® + —_. =; b Divide c’ + d’ by ¢ + d. Ans. & —ed + cfd? — ed + ed‘ —ed® + d’ o+ty 84 30. 31. 32. 33. 34. 30. 36. 37. 41. 42. ELEMENTARY ALGEBRA, Divide m$ + n° by m + n. Ans. m' —m'n + mn} — mn? + mnt — mn? + mné 2 n* m + n Divide 96 xy? + 12.a°y* — 24 aby’ — 48 ay’ —6 ays + 12 ay" by 6 ay —3 zy’. -Ans. 16 aty + 2ry? —4 yy. Divide 32 — 80x + 8027 — 40 a#* + 102+ —=25 by 4 —4a+ x’, Ans. 8 —12a + 62? — a". Divide — a®b® + 1 by —a*b* + ab? —ab + 1. Ans. a®b® + abt + ab + 1. Divide 312 xy? — 32 zy’ + 20 wy? — 160 2’y* + 88 xy’ — 12 0°y? by 12 a°y + 8 ry’ —2 ay. Ans. she — 20 xi" a 6 xy. Divide x? —9 ay® — 8 y? by ay? + 2ary’? + y’. Ans. x'y’ —2 ay’ + 3 aby’ —4 vty’ + 5 ay’ —6 xy" + Tay’ —8y'. Divide 243 m'n§ — mn” by 81 mint + 27 min? + 9 mon® + 8mm" + mén’, Ans. 3mn— mn’. Divide a? —b*® by a + 0. an anaes 2 SS ee Ans. a —ab + b ai ; | ig ae Pale ee ae . Divide 1 —22’> by 1 +2. Ans. 1 —x Tae y? 2 ‘ 2 . Divide x? by x + y. Ans. x vt ae . Divide 1 + 4a by 1— 6-2. Ans. 1+ l0e+ > 60 x nb —62 Divide x* + 32° by 2? —2a 4 1. 9a” —5 Ans. x + 5 ae 3D eee Divide 182 — 44a? — 24a + 3a° + B by 6— 2a’. E Ans. 22—4a + : oe 6—2a? 43. 44. 45, 46. 47. 48. 49. 50. 51. 52. 53. 54. 5d. 56. o7. ELEMENTARY ALGEBRA, 85 Divide 56 y? — 4 by — Bb’ by Ty — b. 4 by —B’ Ans. 8y + a ear Divide 162° — 962’? + 40% —240+4+ 2° + y' by 22 oa 19, Ans. 80° + 20 + et Divide #* + y°+ zbyx+y. 2 y'* tie Ans, or zy”? — n y + ry’? —y + aT; Divide 20 a7b — 25 a —18b* + 27ab? by —5a + 66. Ans. 507 + 2ab —3 & Divide 382’y? + 48 cid’x’? 4- 5eay + cy? + 16 ed? —dceur —16 cd’y —y’ — 15 ea’ by c—y4+ 82. Ans. 16 cd? — 5e’x + y*. Divide — 9 a’y? + 25 a’y’? by 5a®y + 38 aty. Ans. 5 ay —3 aty. Divide —3 ed® + 4ed —9c’d? + 6ct+ 2d‘ by — ed + 3¢ —2 ad’. Ans. 2c? + 2cd — d?. Divide 12 x’2? + 5y4 —32 2°’ + 48 ct —3 y?2? by 42° —y’. en 122°— by 4+ 327, Dividea + b by a+ 8. Ans. 1 Divide ay + by + ax + ba bya+6. Ans. yt oa. Divide 6 cz? —16 ca’'y — 56 c’aty’? + 18 «y* by 2 cx — 8 cu’y — 6 xy’. Ans. 80’ + 4 cx’y — 3 xy’, Divide 28 —2a7 + 32° —3a° + 32* —32° + 32? —2¢+1 by #’—2+4+1. Ans. c& —a> + ot —a? + a’? —o + 1. Divide 1 by 1 —a + 2’. . Ans. 1 + «@ —2? — ot 4+ of + a’ — Ke. Divide 1 + 6z by 1—2z. Ans. 1 + 8x + 162? + 322° 4+ 642+ + 12825 + &e. Divide 1 by 1 —4z. Ans. 14+4a24+162?4+ 642° + 256 a + 1024284 &e. 58. Divide 1 + 7x by 1+ 6z. Ans. 1 + «1 —6 2? + 36 2° — 216 xt + 1296 a — &e. 8 86 59. 60. 61. 62. 63. 64. 60. 66. 67. 68. 69. 70. 12 72. 73. 74. 70. ELEMENTARY ALGEBRA. Divide 82* —8a’y? + 38272 + byt —38y'2 by 2 —y’. Ans. 32? —5y’ + 32’, Divide 4 «* — 25 ay* + 20 ry’ —4 y® by 2 2° + 5 xy’ — 27’. — Ans. 243 —d ay? + 2y’. Divide 6 x#’y — 12 ryz + 6y2 by 2xy —2 yz. Ans. 8% —3z. Divide m? — mx — x —1 by m —a —1. Ans. m + 1. Divide 4p* — 25 p’¢* + 20 pq’ —4¢q° by 2p? — 5 pq’ + 2 q’. Ans. 2p? + dpq’ —2 4. Divide ax + cx + ay + bx + bz + by +z + cy t+ azbyx+y+z. Ans atb+e. Divide ath? — 2 a®b®? + a’b* by a’*b — ab’. Ans. a’b — ab’. Divide ab? + ab? + a’b + ab’ by ab? + ab. Ans. a + b. Divide — 16 a° + 36 x*y’ by 4a? + 627y. Ans. —4a® + 6 x’y. Divide a® — a’ by a? —1. Ans. a’ + a’, Divide 6x* + 4a°y —9 ax’y’ —3 ry? + 2 y* by 2x? + 2xy —y’. Ans. 3 x —ay —2y’. Divide « + 32a) by x + 2a. Ans. xt —2a°a + 4x70? —8 xa? + 16a. Divide m + n by a? — mn. Ans. Baers a> —mn Divide acy —pey by a —p. Ans, cy. Dividea+ aby a+ a. Ans. 1. 2 Divide n? + x’ by n? — 2. Ans. 1 + pee Divide a® — b* by a” —b. Ans. a® + ab + ab? + 6%. Note. Examples in Division may be proved by multiplying the divisor by the quotient. The result should be the same as the divi- dend, ELEMENTARY ALGEBRA, 87 SECTION XXIII. Equations with Two Unknown Quantities. PROBLEM 1. There are two numbers whose sum is 43 and whose difference is 11. Find them. Sotution. Let x stand for the greater number, and y for the less. Then, according to the conditions of the problem, x+y = 43. And L—y= Ui Adding equals to equals, 2 x = 54, Dividing by 2, x aie Putting 27 instead of x in the first equation, we have 27 + y = 48. Transposing, y = 48 — 27. y = 16. Hence the numbers are 27 and 16. When we have two unknown quantities, we must always have two equations. It will be seen that if we did not know the difference of the two numbers above, but only that x + y = 43, the numbers might be 40 and 8, or 23 and 20, or any others that amounted to 43. When we also know that x —y = 11, we can find the numbers to be 27 and 16, as above. The two equations must also be independent. Thus, if they were « + y = 43 and 2x + 2y = 86, we would not know what the numbers were. They might be any two numbers that amounted to 43; for twice that amount would be 86. The second equation would not be independent of the first; for it could be produced by multiplying both members of the first by 2. When we added the equations « + y = 48 ae x —y = 11, we produced the new equation 2x = 54, which con- 88 ELEMENTARY ALGEBRA. tains but one unknown quantity, z. Thus we eliminated the other unknown quantity, y. By subtracting, we might have eliminated x. Thus. x+y = 43 e—y=tili Subtracting equals from equals, 2y == 32, f= 16; The result is the same as before. The method of elimina- tion by addition or subtraction will.now be further explained. PropiEmM 2. Given the equations 1a ay 3 s 6 i to find the numbers which & and y stand for. Sotution. In this case, adding or subtracting does not eliminate either x or y, because the coefficients are different. Adding, we would have 8a + y = 44; subtracting, we would have —2% + 7y = 18. Neither x nor y would be eliminated. When this is the case, the equations must be prepared for elimination. Multiply both members of the first equation by 5, and both members of the second equa- tion by 3. We shall then have, Multiplying by 5, 15a + 20y = 155. Multiplying by 8, 1lba— 9y= 389. Now the coefficients of x are the same, and by subtract- ing, we may eliminate x. It is easily seen that we must subtract. By adding, we would get 30” + Illy = 194, so that there would be no elimination. Performing this oper- ation of subtraction, 15a” + 20y = 155. 15a— 9y= 39, Subtracting equals from equals, 29 y = 116. y= 4. ELEMENTARY ALGEBRA. 89 Putting 4 times 4 instead of 4 in the first of the original equations, we have 32+ 16 = 31. Transposing, 32 = 31 — 16. 5 ra a Eth e Hence, x = 5 and y = 4. In the preceding solution, we multiplied the members of the first equation by 5, which was the coefficient of # in the other equation. We multiplied the members of the second ‘equation by 3, which was the coefficient of x in the other equation. We can always make the coefficients of a letter the same in this way. If the coefficients have a common factor, it is best not to take the trouble of multiplying by that, but by the other factors only. : 32 +2z= 16 PrRoBLEM 3. Giyen ee egrets 30 f to find the values of x and z. SoLuTion. It is easily seen that if we multiply the mem- bers of the first equation by 3, the coefficients of x will be the same. Multiplying by 3, 9x + 6z = 48. Leaving the second equation un- changed, 9x2 — 8z = 20. Subtracting equals from equals, 142 == 28, Zu 2, Putting 2 times 2 instead of 2 z in the first equation, we have 382+4=—> 16. Transposing, 3x2 = 16 —4, 50 == 12, cea 4, Hence x = 4 and z = 2. 8 * 90 ELEMENTARY ALGEBRA. PROBLEM 4, Given | Oar 4 ax o ; to find « and y. Sotution. Putting 4 times 4 instead of 4y in the first equation, we have 82+ 16 = 25. 3x = 25 — 16. S@= YQ, c= 3, Hence x = 3 and y = 4. PRoBLEM 5. Given hes ely area ; to find x and y. Cine s ated EXPLANATION. Here, if we multiply both members of the second equation by 4, it becomes the same as the first. Hence by adding we would get 82 + 8y = 56, and by subtracting we get 0 = 0. There is no elimination. This is because the equations are not independent of each other. It is just the same as if we had but the one equation x + y = 7; for then we could multiply both members of it by 4 and get the other. Hence any numbers which amount to 7 would do for the values of x and y. The problem is inde- terminate. EXAMPLES. Bt to find x and y. Ans, 2 = 639 =a i} to find x and y. Ans. x = 8; y = 4. a to find # and y. Ans. & = Ot eae ie to find x and y. +4 bo <&S I Ans, = 33 =o ELEMENTARY ALGEBRA. 91 le ff to find x and z. ADgi dT ea ei an Mota g { to find a and 2. Ans. «© = 13; 2 = 4, y—2x2=1 Ye = 5 to find « and y. Anse Go == ly yi=s5. . 2y—xz#=7 8. Given ey, ; to find x and y. Ans. £ == 3; y= 8: 17y —15x = 49 9. Given Hoe 37 | | to find 2 and y. Anat se — Ls yaa 10. ee fee, & 7 tt find x and y. Ang 2 = 2-47 = — EE Nore. The members of the first equation can be divided by 11 and the equation reduced before eliminating. 17y + 5x2 = 101 11. Given | Goh maae e to find x and y. Ats. 0 == 1h). ia ig en BY . Given } 16% 4 26y — 162 to find z and y. Ans. © = 2; y= 5. 3} 13. Given | a = Bea \to find y and z. ANS -Y == 10 (2 =a (4. Given 17 Fiat ie he iy to find @ and y, ANG. Disa) ant i5. Given cite 23 tare 3 | to find x and y. ANS, es 2 epee 92 . 16. Ve 18. 19. 20. en 22. 23. 24. 25. 26. 27. ELEMENTARY ALGEBRA. 1342+ 2y= 180 doe ED 192 ¢ # find x and y. Ans. « = 12; y =z Given Agee ane ar 335 f 1 find x and y. Ans. © = 2; y= 9. | to find x and y. Ans. x =1; y = 60, 89 291 ae bo © 8 ll il aa + Ans, «= 1934) eee Given ay nf iy mii 36 ¢ find x and y. Ans. © = "15 Yee a ai a Ans. © == 2s yee ri a Given | ,* "Y= fo { to find « and y. Ans. w= 7; y=. Given 37 3, = at to find @ and y. Ans. 2 = 5; y = 2. Given O71 gy = — 19 f tofind « and y. Ans. © = 38; y= 58. Given } 457 t 99% Bi eerie find x and y. Ans. «x = 10; y= 9. Given ce a 5 to find x and y. Ans. 2 = 174 28. 29. 30. dl. 32, 33. 34, 35. 36. 37, 38. 39. ELEMENTARY ALGEBRA. 93 . 1091 « — 576y = — 969 Given J _ 10912 + 596y = 889 to find x and y. Ans. «= —38; y = —4, ° G—y = § Given 12 9 = 6 | to find a and y, Ans. m= 4; y= —4, ; 2 3y = 29 Given J rt 4y = 93} to find «and y, Ans. = 4; y= 7. Given oF TY = 28) to find andy Ans. c= 9; y= 9, . 2” —32=—15 Given | Re 57 to find x and z. Ans. «= 3;2= 7. Given Dnt 5s = 31 to find @ and z Ans. x = 2; z= 5, Given } eels “40 ft find x and w. Ans. © = 20; w= 1. Given lates Pe to find x and w. Ans. & = 11; w= 22. Given fut y= to find w and y. Ans. w= 2; y = —2, 33 to find p and q. Ans. p= 4; q =4, ee Ans. § = —4; t= 4, Given ee hee hs to find s and ¢. Ans. s = 100; ¢=1. 94 ELEMENTARY ALGEBRA, SECTION XXIV. Questions producing Equations with Two Unknown Quantities. ProspLEM 1. One man buys 5 collars and 7 handker- chiefs for $3.80; another buys at the same rate 12 collars and 2 handkerchiefs for $3.20. How much are the collars and handkerchiefs apiece ? Soutution. Let x represent the price of a collar, And y the price of a handkerchief. Then we have, by the statement of the question, 5a + Ty = 380. 122+ 2y = 320. By solving these equations, it will be found that the col- lars were 20 cents each, and the handkerchiefs 40. PROBLEM 2. A person has ten-cent notes and twenty- five-cent notes, 100 in all, whose value is $13. How many has he of each? SotutTion. Let x = the number of 25-cent notes he has, And y = the number of 10-cent notes he has. Then, by the sas of the question, ee: 25 2 v 10y = 13800. By solving these, it will be found that he had 20 twenty- five-cent notes and 80 ten-cent notes. EXAMPLES. 1. A gentleman bought an encyclopedia in 24 volumes, and a history of England in 6 volumes, for $132. The en- cyclopedia cost $3 more per volume than the history. What was the cost of each per volume? Ans. The encyclopedia cost $5 per volume, and the history $2 per volume. ELEMENTARY ALGEBRA, 95 2. A merchant has some wine worth $5 a gallon, and some worth $10 a gallon. He wishes to make a mixture of 20 gallons worth $7 a gallon. How much of each must he take? Ans. 12 gallons at $5, and 8 gallons at $10. Notr. The whole 20 gallons will be worth $140. 3. How many gallons of wine at $1.25 and at $1.00 must be mixed to fill a cask of 63 gallons at $1.10? Ans. 251 gallons at $1.25, and 374 gallons at $1.00. 4, Find two numbers whose sum is 27, and their difference 3. Ans. 15 and 12. 5. A hatter sells to one person 7 hats and 11 caps for $39; and to another, at the same prices, 5 hats and 1 cap for $21. What was the price of each? Ans. $4 fora hat, and $1 for a cap. 6. At what price are horses and oxen sold, when 4 horses and 5 oxen cost $300; and 9 horses and 10 oxen cost $650? Ans. $50 for a horse, and $20 for an ox. 7. Eight Spanish doubloons and ten pistoles are worth $163.12. Two doubloons and five pistoles are worth $50.49. What is the value of each? Ans. A doubloon is worth $15.535, and a pistole, $3.884. 8. Find two numbers such that their sum is 34, and five times the greater minus three times the less is 58. Ans. 20 and 14, 9. Find two numbers such that double the greater minus the less is 22, and double the less minus the greater is 1. Ans. 15 and 8. 10. Divide 38 into two such parts that one may exceed the other by 6. Ans. 16 and 22. 11. A and B together own property amounting to $120,000, and seven times*A’s property is equal to five times B’s. How much has each? Ans. A has $50,000, and B has $70,000. 96 ELEMENTARY ALGEBRA. SECTION XXV. Factoring. Fundamental Theorems. 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 and second, plus the square of the second. a+ 06 Proor. Let a and 6b standforany a + 6 two quantities. Then, multiplying a a 4 The + b by itself, we have what the theorem ab 4. # states. ————___—____—_— v+ 2ab+ B Theorem II. The square of the difference of two quantities as equal to the square of the first, minus twice the product of the first and second, plus the square of the second. a— 6b Proor. Let a and 6 stand for any a — 6b two quantities. Then, multiplying a (2 op — 6 by itself, we have what the theorem __. ae states. _t a? —2ab + 0 Theorem III. Zhe sum of two quantities multiplied by their difference is equal to the difference of their squares. a+ 56 Proor. Let a and 6 stand for any a — 6b two quantities. Then, multiplying a at sabe + b by a —), we have what the theo- _2 oh aoe rem states. 2 A Theorem IV. The difference of the same powers of two quantities is always evenly divisible by the difference of the quantities. ELEMENTARY ALGEBRA. 97 Thus, we find by performing the operation of division, (a? —b*) ~ (a—b) = a+ Bb. (a? — 6°) + (a —b) = a’ + ad + OB. (at — bt) + (a —b) = a® + ad + ab? + OD. (a —b’) + (a —b) = a + a + ad’ + ad? + OF. (a —b°) + (a—b) = @ + ad + ab’ + a’? + alt Any other expressions might be used instead of a and 6. Thus, we may find in the same way, (mn? — c?) + (mn —c) = mn + ¢. (m'n3 — c®) + (mn —e) = m’n’? + mne + ©. (mint — ct) > (mn —c) = min? + m’n’e + mne’ + &. &e., &e. Also, ( 9 —4) + (8 —2) = 34 2. (27 —8) + (8 —2) =9+644, &e., &e. The complete proof of the theorem is too difficult for insertion here. As far as the student needs to use it, he can prove it by per- forming the operation. EXAMPLES. Write down, where you can, the values of the following expressions, without ‘actually performing the work. State what theorem you use in obtaining your result. Thus: What is (x + y) (« —y) equal to? Ans. To 2 —y’, by Theorem III. The sum of two quantities multiplied by their difference is equal to the difference of their squares. Note. Let the student tell why parentheses are needed wherever they are used in this section. 1. (e + dy’. Ans. 2 + 2ed +. a’ 2. (e —d)’. Ans. ¢ —2ed + d’. 3. (a + 1)”. Ans. @ + 2a4+ 1. 4. (ec — x)’. Ans. ce? —2cx + x’. 5. (1 +. y)*. Ans. 1+ 2y+ y’. 9 G ELEMENTARY ALGEBRA. 6. (ab + c)?. Ans. ab? + 2abe + @’. 7. (8@¢@ + a)’. Ans. 25 at + 10 a*x + a. 8. (8¢+ 2y) Ans. 9 + 12 cy + 4y’. 9. (ed + y’)’. Ans. ced? + 2cdy’? + y'. 10. (ex — 3 y’)’, Ans. a? —6 ery? + 97. 11. (Sax + 37°)’. Ans. 25 atx’? + 30 a@zry> + 97. 12. (6cedy + 208’). Ans. 36 e'd*y’ + 24 ab’cedy + 4 a’b*. 13. (2 + 3) Ans. 4+ 12+ 9 = 25. 14. (6 —5)’. Ans. 36 —60 + 25 = 1. 15. (6¢ — 0)’. Ans. 36c? —0 + 0 = 362. 16. (3 ab + ed)’. Ans. 9 ab? + 6abed + ed’. 17. (4can + 3m’ar’y’)’. Ans. 16 c?n?x? + 24 em’nx*y? + 9 mitxty*. 18. (ab — ed)’. Ans. ath? —2 abe'd + cd’. 19. (52+ — 6 2°), Ans. 25 2§ — 602" + 362% 20. (ce? + 2ax). Ans. ct + 4ac’a + 47x”, 21. (a + ©) (a—*2). Ans. a? — x’, 22. (ec —y) (c+ y). Ans. ? —y’. 23. (2a + 36) (2a —3D). Ans. 4a? —9 6. 24. (5m.+ x?) (6m —2?). Ans. 25m? — x*. 25. (8a + c) (8a—ce). Ans. 64a? —c’. 26. (3 abet — 5 mn®) (8 abc! + 5 mn’). Ans. 9 a*‘b’c® — 25 m?n™. 27. (Qaminx + y) (Ya'mnx —y). Ans. 81 atm'n’?x? — y’. 28. (5 —3) (5 + 3). Ans, 25 —9 = 16. 29. (6 — 2) (6 + 2). Ans. 36 —4 = 32. 30. (9¢ + 1) (Qe —1). Ans. 81 c? —i. 31. (8ab + 0) (8ab —0). Ans. 64 a’b? —0 = 64a7b”. 32. (6y + 2am) (5y —2am). Ans. 25 y? — 4a?m’, 33. (5 + 4) (5 —4). Ans. 25 —16 = 9. 34. (6 + 4)”. Ans. 36 + 48 + 16 = 100. 35. (6 — 4)’. Ans. 36 —48 + 16 = 4. 36. (m® —n*®) + (m —n). Ans. m' + mn + min? + min? + mint + mn? + mn® + n. ELEMENTARY: ALGEBRA. 99 37. (c& — b*) + (ec — bd). Ans. & + eb + cb? + cb? + cbt + 0B. 38. (7° —s*) + (r —s). Ans. rv? + rs + 8”. 39. (mi —r") + (m —7r). Ans. m® + mr + mtr? + mr3 + mrt + me 4+ 40. (a7b? — x’) + (ab — 2). Ans. ab + x. 41. (ea? — y®) + (cx — y). Ans. cat + ea®y + ea*y? + cxy® + y'. 42. (a*b* —c*d*) + (ab — cd). Ans. ab? + ab’ced + abe’d? + ed’. 43. (a? —1) + (a—1). Ans. a + 1. 44, (at —1) + (x —1). Ans. vo + 2? +2 +4 1. 45. (1 — ec) + (1 — 0). An. 1l+e+e4+e. 46. (25 —9) + (5 —8). Ans. 5+ 3= 8. 47. (125 — 27) + (5 —83). Ans. 25+5x3+9 = 49. 48. (a® — B®) + (a —D). Ans. a® + ab + a®d? + add? + att + a®d® + a’! + ab’ + BD. 49, (m® — x*) + (m — 2). Ans. m® + mtx + mia? + m2? + mat + 2. 50. (g* — cht) + (g —ch). Ans. g& + gch + gch? + &h’. QUESTIONS ON THE FOREGOING SECTIONS. What is the difference between (a + xz) (a —zx) anda -+ za —z, oratzxXa—zx? Is a + xa monomial, a binomial, or a trinomial? Why? Is a binomial a polynomial? Write three monomials, each containing four letters and a number. Write three binomials, three trinomials, three polynomials of seven terms each. Name all the coefficients you have just written. What is the difference between addition in arithmetic and addition in algebra ? How is any term subtracted from another ? 100 ELEMENTARY ALGEBRA. What is the rule for the signs + and — in multiplication and di- vision ? How do you multiply together powers of the same letter ? How do you divide one power by another power of the same letter? SECTION XXVI. Factoring. Prime and Monomial Factors. A measure or factor of a quantity is anything which di- vides it without a remainder. Thus, 5 isa measure or factor of 15; ais a measure or factor of a’ or of ax; ¢d’e is a fac- tor of cd’ex. A prime quantity is one which has no factor but itself and unity. Thus, a, 2,7, and a + & are prime; but az is not, as it has a and « for factors. . The prime numbers between 1 and 50 are 2, 3, 5, 7, 11, 1O2 14,19, 20;-20; 01, Olo oleae ys The prime factors of a are a and a. The prime factors of 24a’c’d are 2, 2, 2, 3, a, a, ¢, ©, ©, and d,. The prime factors of 7‘ are y, y, y, and y. Polynomials, it must be remembered, are expressions composed of two or more terms, connected by + or —. They often have a monomial factor which may be separated from the rest. Thus, in the polynomial ax + a’, both terms may be di- vided by the monomial « Performing the division, we get a+ <2. Hencethe factors of aw + x? arexanda+z2x. In other words, av + 2? = 4 (a+ @). Again, in the polynomial 24 a’cx — 36 abe’x’ 4. 8 a’ex%, each term may be divided by 4, by a’, by ¢, and by «. That is, the monomial 4 a’cx is a factor of the polynomial. Per- forming the division, we get 6ct—Qac’x + 22°. Hence ELEMENTARY ALGEBRA, 101 the polynomial is the same as 4 a’ex (6 ct —9ac’a 4+ 22”), The parenthesis is used to show that each term within it is multiplied by 4 a’ex. In the same way we may find that the factors of 35 a‘x —2ax+ x” are « and 38a*—2a+ 2. In other words, 38a%—2JZaxr+2?=—-2 (8Bat—2a+ 2). EXAMPLES. 1. What are the prime factors of a°? Ans. 4, @, a, a, a. 2. What are the prime factors of a*a*y? Ans. 4, Q, A, £0, 2, Yy. 3. What are the prime factors of 5 a*x*y? Ans, .5,&, 0; 0, X,Y. 4. What are the prime factors of 24 a’z’? Anse 352522, a, a, @, @, G;.0;.0;,%,2. 5. What are the prime factors of 18 m‘n? Ans. 3, 8,2,m,m,m, m,n. 6. What are the prime factors of 21 p*q’x? Ans. 3,7, ),D; D5, J, 2: 7. What are the prime factors of 135 abbx’y ? Ans. 5, 3,3, 3, a, a, a, a, a, 6, X, X, ¥. 8. What are the prime factors of 235%’? Ans. 5, 47, k, k. 9. What are the prime factors of 37 ? Ans. 37 is itself prime. 10. What are the prime factors of n? Ans. n is itself prime. Separate each of the following polynomials into two fac- tors, one of which is the greatest monomial factor in the ex- pression. 11. 2a+ 2-2. Ans. 2 (a + x). 12. 13a—138. | Ans. 13 (a —6). 13. 24a? + 24. Ans. 24 (a? + 1). 14. 32 a® + 48 a®b — 24 ab’. Ans. 8a’ (4a? + 6 ab — 30’). 15. 5a + 10y. Ans. 5 (a + 2y). Q* 102 ELEMENTARY ALGEBRA. 16. 14ac —7d. Ans. 7 (2 ac — d). 17. ab — ax, Ans. a (6b —2). 18. ax + Qay. Ans. a(x + 2y). 19. 10 cy — 10 xz. Ans. 10” (y —2). 20. 10 xy + 15 xz. Ans. 5x (2y + 32). 21.) ab — 20 ar, Ans. 5a (b —5 2), 22. ab + a. Ans. a (6 + 1). 23. ab + 0. Ans. b (a + 1). 24, 30 ab — 54 b. Ans. 6b (5a — 9). 25. 25 a’y — 25 az. Ans. 25 @ (y —2). 26. 3a + 3. Ans. 8 (a + 1). 27. 840° — 34 a5. Ans. 34a° (a —1). 98. 27 m? + 9 mn. Ans. 9m? (3 + n). 29. 477 —16 7°s + 2475s? 4+ 12 7°s°, Ans. 47° (1 —48 + 67°s’ + 31°s*). 30. 8aty + 8a’y — 18 ay’ + 33 aay. Ans. ay (8a + 3—18y + 38 2a’). 31. 36 b'z — 21 ba — 9 bY 2% y’. Ans. 3 6° (12 b’2 —7 « —8 b’a*y’). 82. 15 am?yz? — 12 b'm'n?y? — 9 a*b?min’. Ans. 8 m? (5 abyz’ — 4 b?mn y? — 3 a7b’m'n’). 33. 216 a®btetdéa®y — 96 aed®x’. Ans. 24 a’ctd’x*® (9 abty — 4c). 34. 56 ad?’mniatz + 224 bd'mxz. Ans. 56 d?maz (a’e?m®n'a? + 4 bd*2*). 39. 216 m’n®p’q’s + 36 m?n'p’*qs’. Ans. 86 m’n®p’qs (6 mq’ + 8°). 36. 185 a?y?z? — 87 a®bed?y’z. Ans. 87 a’y’z (5z —abed’). 37. 25 ate?x*yt2® — 25 abe’aty'z5. Ans. 25 a®c’x*y*2’ (a — 2). 38. 12 a®e’atyz® — 5 ale’atyz'. Ans. aeatyz (12 a — 9). 09. 8 ab’e’+ 5 ac’d —5ad’e + 8 ae’*f. Ans. a (8b'e + 5d —5d’e + 8 ef). 40. 2ay + 2y? + 2yz. Ans. 2y (a@ + y+ 2). Nore. The foregoing examples may be proved by reversing the operation, that is, by multiplying the factors together. ELEMENTARY ALGEBRA. 103 SECTION XXVIII. Factoring. Applications of the Fundamental Theorems. PrRoBLeM. Factor r? + 27s + 8’. Soxtution. The first term is the square of 1, the last term is the square of s, and the middle term is plus twice the product of rand s. Hence, by Theorem L., it is the square ofr+s. Thus, 7? + 2rs+ 3? =(r+s)% PrRospLEM. Factor 4n? —12n + 9. Sotution. The first term is the square of 27, the last term is the square of 3, and the middle term is minus twice the product of 2m and 8. Hence, by Theorem IL., it is the square of 2n —3. Thus, 4n? —12n + 9 = (2n — 3). ProspuEM. Factor 25 x*/* —4n’. Soturion. The first term is the square of 5 x’y’, and the second term is minus the square of 2. Hence, by Theorem IIL., it is equal to the sum of 5 x’y? and 2 multiplied by their difference. Thus, 25 aty8 —4n? = (5 2’y® + 2m) (5 a?y® —2n). It is an excellent exercise to tell of a trinomial whether it is a square or not, by observing whether or not Theorem I. or Theorem II. applies to it. Thus, 25 a? + 40 az + 16 2’ is the square of 5a + 42, by Theorem I.; and 25 a* — 40 az + 16 2’ is the square of 5a —4z, by Theorem II.; but 25 a? — 20 az + 162’ is not a square at all, as the middle term is not twice the product of 5a and 42. In the following examples, when the trinomial is not a square, tell why it is not. 104 ELEMENTARY ALGEBRA. EXAMPLES. Tell which of the following trinomials are squares, and factor those which are. 92? + 24 az + 16 27, N25 we 16 x2 + 9 23, . 8627? + 12 427-4 2. . 062? + 12424 42% 9y + 80yz 4+ 22”. v+b6ab+ 90%. . 25m? + 20mn + 4n2, . 81 p? — 86 pq + 4¢q’. 64 ¢? — 12 cr + 9 x’. . 640 —48 cx + 9 2? 164 .cC'—o2e0 4+ 425 . 36 & — 54ed + 49d’. . 067? — 84 rs + 49 8”. . 9a7bt + 21 a?b’exr + 49 a’e*z’. . 9a7bt —42 c7b’er + 49 ae?x”. . 144 mn? — 72 m?nx® + 9 mtx. . 121 p’s* + 32 r*ps* — 25 7’. . 81 mn + 33 m’?n? + 25 min, . ary? + 48 2°y'z + 64 a? yt”, . 9 a®y® + 82 a y'z + 36 x*y'2”, . 9 aby? + 86 a y®z + 86 a’y'2”. . 9 ay + 36 a y’z + 16 x’y'2”. . 9 ar8/h + 24 ay®z + 16 x’y*2”. . 9 xy? + 18 xyz + 9 xy*2?. . 9 xy? — 12 aryfz + 4 ax yt2?, . 64073 — 48 aca + 9 x”. . 64 act — 48 ac’a — 9 2”. . 64 ee + 48 ac’x — 9 x. . 64 act — 48 wey’? + 9x. . 64 ac —48 aca + 9 x”. ~ Imint + 12 mint + 4 mint. Ans. (3x + 42). Ans. Not a square. Ans. (6x + z)’* Ans. Not a square. Ans. Not a square. Ans. (a + 36)’. Ans. (5m + 2n)?. Ans. (9p —2q)’. Ans. Not a square. Ans. (8e —382)*. Ans. (8¢ —2«)’. Ans. Not a square. Ans. (6r — 7s)’, Ans. Not a square. Ans. (3 ab? —7 acx)’, Ans. (12 mn — 3 m’2°)?, Ans. Not a square. Ans. Not a square. Ans. (3 xtyt + 8 xyz)”, Ans. Not a square. Ans. (8aty* + 6 ay’z)’. Ans. Not a square. Ans. (3 xty* + 4 xyz). Ans. (3 aty* + 3 xy’z).. Ans. (3 xty* — 2 xy’z)’. Ans. Not a square. Ans. Not a square. Ans. Not a square. Ans. Not a square. Ans. (8 ac? —32x)?. Ans. (3 mn? +- 2 m?n?)*. 32. 30. 34. 39. 36. ov. 38. og. 40. ELEMENTARY ALGEBRA. 8a? + 16ab + 402 mn? + 2mn + 1. mn? + 2mn + 4. w+ 2Qae+ x. a’ —2Qax + x”. 4a’? + 4ab + 0% 4a’? —4ab-+ b* 927+ 24x + 16. 64a? — 80a -+ 25. Factor the following: 41. 42. 43. 44, 45. 46. 47. 48. . 9at — 25 Dt. % a} — gh, : a? _ q?®, . wa? — §”, ciao. 20 0”, - 42° —162t . . 16 a*b’c*d? — 9. . 81 — 86. . 64— a’. . 2? — 49. . 9 —4. . at — cde’, Note. a —2Z’, A9 a? — 64 y’. 4y’ —9 2", 16 2’ — y%. 25 x — 6", 64 a& — 9 yx. 9 ee ues 4 Ce ah 25 a’® — 49 b8y¥, 105 Ans. Not a square. Ans. (mn + 1)?. Ans. Not a square. Ans. (a + 2), Ans. (a —x)’. Ans. (2a + 6)*. Ans. (2a —b)’. Ans. (3x + 4)’. Ans. (8 a—5)*. Ans. (a + x) (a — 2). Ans. (Tx + 8y) (7x —8y). Ans. (2y + 32z) (2y —32z). Ans. (42 + y’) (4z—y’). Ans. (5 2*® + 68) (528 — 6°), Ans. (8a + 3y’) (8a —3y’). Ans. (3 y® + 2 2q’) (3 y® —2 zq’). Ans. (5 a® + 7 b’y") (5 a —7 By’). Ans. (3a + 56?) (8 a? — 5D’). Ans. (a5b® + a®b*) (a®b> — a®d®). Ans. (a® + a’) (a® —a’). Ans. (a® + b*) (a® — B°). Ans. (5 a® + 56°) (5a® — 50°). Ans. (2x* + 427) (2a*—42’). Ans. (4abed + 3) (4 abed — 3). Ans. (9 + 6) (9 —6). Ans. (8 + a) (8 —a). Ans. (x + 7) (x1 —T). Ans. (3 + 2) (38 — 2). Ans. (a? + e’d'et) (a? — c'd'et). The foregoing examples may be proved by reversing the operation, that is, by multiplying the factors together. 106 ELEMENTARY ALGEBRA. SECTION XXVIII. Factoring. (Continued.) WHEN it is desired to separate any algebraic expression into factors, it should be first divided by any monomial which will divide each of its terms; then by any other ex- pression which by the preceding four theorems, or by ob- servation, may be seen to be contained in it. Thus, taking the expression 28 bex’ — 84 bexy + 63 y’be, it may easily be seen that every term is divisible by 7, every term contains b, and every term contains c. Hence we may divide by 7 bc, decomposing the expression into Tbe (4a? —12 ay + 9y’). The last of these factors is seen by Theorem II. to be the square of 2a —3y. Hence, finally, the original expression is equal to 7 be (22 —38 y) (2x2 —3y). Again, taking the expression, 17 a’ —17 a’y’, we may divide each term by 17 a’, decomposing the expression into 17 x? (a® —y’). By Theorem IV., the last of these factors is divisible by « —y. Performing the division, we have, finally, 17 a? (wv — y) (at 4 ay + ay? + ay? + y'). EXAMPLES. Separate each of the following expressions into factors, one monomial, and the polynomial ones factored as far as possible. 1. 9a —92’. Ans. 9 (a—x) (a+ &). 2. 20 y? —45 2. Ans. 5(2y —32) (Qy+ 32). 3. 1762 —11 y’. Ans. 11 (427 — y) (42? + y). 4, 52 mt — 1300 nt Ans. 52 (m? —5 n’) (m? + 5n’). 5. 822 —8 x’. Ans. 8x (2—2x) (2+ z). 6. 19 a°b — 19 ab’. Ans. 19 ab (a? + 6°) (a + b) (a —B). 7. 25 ab — 25 bt. Ans. 25 b (a —b) (a + ab + 0°). 15. ELEMENTARY ALGEBRA. 107 . sey —3 xy’. Ans. 3ay (x&—y) («© + y). . ar*® —2 ar’z + arz’. Ans. ar (r —2z) (7 —2). . dbx? + 10 Be’x + 5 bc’. Ans. 5 be’ (x + b) (a + b). Pigaea — 38 a'b’s* + 19 a7b.°. Ans. 19 a*b'x* (a —2x) (a — 2x). . dl aa —62a@bx + 31 ab ?x. Ans. 31 ax (a —b) (a —D). . das? + 10 atzt + 5 aPr', Ans. Sax (a + x) (a+ @). . d8ax’? —18 ax + 27 a. Ans. 3a (x —3) (a — 8). 45 ay — 380 ay? + 5ay®. Ans. Say (8 —y) (8 — y). .387b4+ 18ye+ 9yxz Ans. 8y’? (b+ 6¢e4 822), . 8m’nx —18 mnx’ + 2mn'>y. Ans. 2mn (4mx —9 x? + nty). . 15 ay — 25 b’y + 1dc’y. Ans. 5y (80 —50B? + 3). . Bax® + 6an’y® + 42 ax’. Ans. 6 ax? (at + 4° + 7a). 26 ax’ — 26 ay’. Ans. 26 a (« —y) (2° + ay + aty’? + a5y? + a’yt eye eh . 6a2t — 96. Ans. 6 (x? + 4) (a + 2) (x — 2). . 64 atd® — 25 a?b®. Ans. ab (8a —5b) (8a + 55). . 2477s —127°s*t —67rs. Ans. 6rs (4r —27’st —1). Soa —— Oy Ans. 5 (a? + y*) (fe + y) (« — y). . aent —atert. Ans. ate? (nr? + 1) (n + 1) (n —7). . ax? —a. Ans. a (« + 1) (« —1). at a2". Ans. a? (x + 1) (x —1). . a —a, Ans. a (a? + 1) (a + 1) (a—1). ~la’y—22aey+ 11. Ans. 11 (zy —1) (zy —1). . 04am’? —- 17 atm + 51 a’mt. Ans. 17 a’m (2 atm — a? + 3m’). anche 108 ELEMENTARY ALGEBRA. SECTION XXIX. Equations with Two Unknown Quantities. (Continued.) : 9x—2=10y41 PROBLEM 1. Given [Fy 4 Say ode find the values of # and y. Sotution. In this case, transposition is necessary to bring the equations into the proper forms for eliminating — one of the unknown quantities. Transposing, we have 9a — 10y 1 + 2. bye 3a—2xe4—y ob Ett | 92—10y= 3. } x+ 4y= 81. The easiest method of elimination here is to multiply the members of the second equation by 9, and subtract those of the first. Thus, Collecting, we have 92+ 36y = 279. The first equation is, 9x—10y= 4. Subtracting, 46 y = 276. y= 6. Instead of 4y, put 24, in the second equation. Then we have, x+ 24 = 31. Transposing, x = 31 — 24. Hence x = 7 and y = 6. PROBLEM 2. Given [er uD + Dy oe 22 +1 ey ee to find the values of x ae y. Sotution. In the first equation, performing the oper- ations of multiplication, we have xy + dy—Tx—385 = ay +y—I9Iaex—9+4+ 18. ELEMENTARY ALGEBRA. 109 Here we cancel xy, by subtracting it from each member. Then the equations are nites 10 =y—9x—9+ a 2¢+10=y-+4. Transposing, lama Sa 9x=— 9+ mt 242 —y= 4—10. Collecting, and arranging, Sane ah ea as i Eliminating, and continuing the solution, it will be found tastes =), and y = 1(). : EXAMPLES. 1. Given § +182 —ay = (14—~2) (11+ y)+4+ 138 12x+9y= 82+ 1l4y4+3 to find the values of and y. Ans. x = 12; y= 9. 2. Given ‘i 78 nw a aaah (y¥ +1) +5) to s find the values of and y. Ans. x= 4; y= 1. 3. If $500 be taken from A’s property, and $1500 added to B’s, they will both have the same sum. If A’s property were five times as great as it is, and B’s six times as great as it is, they would together have $120,000. How much has each? Ans. A has $12,000, and B has $10,000. 4, A number consisting of two digits is four times their sum. The number consisting of the same digits in an in- verse order is 18 more than the number itself. Required the first number. Ans. 24. Notre. Any number of two digits is 10 times the first, plus once the second. Thus 23 is 10 times 2, plus 3; inverting, it becomes 382, which is 10 times 3, plus 2. If z and y stand for the digits, 10% + y will be the number, and 10 y + z will be the number formed by in- verting the digits. 5. If 21 years be added to the sum of A’s age and B’s, the result will be double A’s age. If 9 years be added to 10 110 ELEMENTARY ALGEBRA, their difference, the result will be double B’s age. Required their ages. Ans. A’s age is 36 years; B’s, 15 years. 6. Find two numbers such that the greater minus 2 is double the less, and the less plus 22 is double the greater. Ans. 14 and 6. 7. If A give B $100, B will have twice as muchas A. If B give A $200, A will have twice as much as B. How much has each? Ans. A has $400, and B has $500. 8. If the greater of two numbers be multiplied by 5 and the less by 10, the sum of the products is 140. If the less be multiplied by 5 and the greater by 10, the sum of the products is 160. What are the numbers? Ans. 12 and 8. 9. Ten years ago, A’s age was 10 times that of B. Six years hence it will be double that of B. Required the age of each. Ans. A’s age is 30 years; B’s, 12 years. 10. Find a number consisting of two digits which is equal to 8 times the sum of the digits, the number consisting of the same digits in an inverse order being 18 more than their sum, Ans. 72. 11. Find a number consisting of two digits which is 7 times the sum of its digits, and is 27 greater than the num- ber consisting of the same digits in an inverse order. Ans, 63. 12. A grocer has two kinds of coffee. One pound of the first kind and one pound of the second are together worth 48 cents. Four pounds of the first plus nine pounds of the second are worth $2.92. What is the price of each per pound? Ans. The first kind is 28 cents per pound, and the second 20 cents. 13. A gentleman has two horses and two saddles. One saddle is worth $40; the other, $20. The best horse with the best saddle is worth $80 more than the worst horse with the worst saddle. The best horse with the worst saddle is worth $10 less than twice the value of the worst horse. ELEMENTARY ALGEBRA. 111 What is the value of each horse? Ans. The best horse is worth $150; and the other, $90. 14, A farmer, having 9 bushels of wheat at $2 a bushel, would mix with it rye at $1.40 a bushel and barley at $1 a bushel, so as to make a mixture of 50 bushels at $1.38 a bushel. How many bushels of rye and barley must he take? Ans. 25 bushels of rye, and 16 bushels of barley. 15. Find a number consisting of two digits which is equal to 7 times the sum of its digits, and is equal to 9 more than 12 times the difference of its digits, the unit digit being the less. Ans. 21. 16. Find a number of two digits, the first being the greater, such that the number itself is 3 more than 7 times the sum of its digits, and 7 more than 16 times the difference of its digits. Ans. 52. 17. Divide the number 149 into two such parts that the greater diminished by 5 may be double the less. Ans. 101 and 48. 18. A miller mixes wheat costing $1.25 a bushel with rye costing 75 cents a bushel. It amounts to 68 bushels, cost- ing in all $79. How many bushels of each kind does he take? Ans. 56 bushels of wheat, and 12 bushels of rye. 19. Divide $3000 between A and B, so that A shall re- ceive a half-eagle as often as B does a dollar. Ans. A’s share is $2500; B’s, $500. 20. Divide $5000 between A and B, so that 5 times A’s share shall be 4 times B’s. Ans. A’s share is $22222; B’s, $2777 4. 21. A vintner wishes to fill a puncheon of 84 gallons with wine, so that it may have cost him $100.80 in all. He draws his wine from two casks which cost $1.00 and $1.42 a gallon respectively. How much of each kind must he take? Ans. 44 gallons at $1, and 40 gallons at $1.42. 22. I purchased 20 pounds of sugar and 25 pounds of 112 ELEMENTARY ALGEBRA, coffee for $10; but the price of each having fallen 2 cents a pound, I purchased 29 pounds of sugar and 25 of coffee for $10.72. What was the first price of each? Ans. For sugar, 20 cents per pound; for coffee, 24 cents. 23. I bought 100 yards of linen and 150 yards of muslin for $82.50. I afterward bought 50 yards of linen and 200 yards of muslin for $55, the linen costing 10 cents a yard more, and the muslin 5 cents a yard less, than at first. What were the first prices, and what the second? Ans. At first the linen was 60 cents, and the muslin 15 cents; after- ward the linen was 70 cents, and the muslin 10 cents. 24. Find a number equal to 5 times the sum of its digits, such that if 9 be added to it, the digits will be inverted. Ans. 45. 25. Find a number equal to 8 times the sum of its digits, such that if 45 be subtracted from it, the digits will be in- verted. Ans. 72. 26. Divide the number 111 into two such parts that the difference between the greater and 120 shall be 9 more than 50 minus the less. Ans. 86 and 25. 27. A laborer engaged for 100 days, on condition that for every day he worked he was to have $1.25, and for every day he was idle he was to forfeit 50 cents for board. At the end of the time he was entitled to $55. How many days did he work, and how many was he idle? Ans. He worked 60 days, and was idle 40 days. 28. Divide 15 into two such parts that if the less be mul- tiplied by 5 and the greater by 7, the sum of the products shall be 93. Ans. 6 and 9. 29. A gentleman owns two houses. If he should spend $900 in improving the better house, and $1000 in improving the other, the two would be equal in value. But if he should spend $2000 i in improving the better house, and $500 in im- proving the other, the first would be worth twice as much ELEMENTARY ALGEBRA. 113 as the second. What is the value of each? Ans. One is worth $2000, and the other $1500. 30. If a merchant mixes brandy and Port wine, putting in twice as much of the former as of the latter, the mixture is worth $22 a dozen. But if he puts in twice as much of the latter as of the former, the mixture is worth $24 a dozen. What is the price per dozen of the brandy and of the Port wine? Ans. The brandy, $20 per dozen; the wine, $26 per dozen. Nore. Three dozen of the first mixture are worth $66; of the second, $72. In other words, 2 dozen of the brandy plus one dozen of the wine are worth $66; and one dozen of the brandy plus two dozen of the wine are worth $72. SECTION XXX. The Least Common Multiple. A multiple of a quantity is that which contains it exactly. Thus 63 is a multiple of 7 and of 9. Also a is a multiple of a; xy is a multiple of x and of y; a? — 0? isa multiple of a —b and ofa + b. A multiple of two or more quantities is said to be a com- mon multiple of them. Thus az is a common multiple of a and 2; so are a’x, ax’, a’2*, a’x*, etc., common multiples of aand #; and az is the least of these. The least common multiple of several quantities is found in precisely the same way in Algebra as in Arithmetic: one or two examples will serve to show the process sufficiently. ProsieM 1. Find the least common multiple of 9 a’z, 4 ax*, 12 ax’, and 6 a?z*, 10* H 114 ELEMENTARY ALGEBRA. OPERATION. 3 ) 9 ax 4 ax? 12 a®a? 6 a723 2 ) 3a°x 4 ax? 4a°x” 2 a? 2 ) Bae 2 ax? 2 ax? ax a ) I ee ax? ‘ih: a*a* a ) Sax oc or ax 7 ) aa x a x? x x ) 3 ie ax Me 3 1 a L Continually divide the quantities by any prime factor of two or more of them, setting down the quotients and un- divided numbers in a line below. When no two of them have a common factor, multiply together the divisors and final quotients. The product is the least common multiple. The product of 3, 2, 2, a, a, x, x, 3, a, and & is 36 a‘z*. Hence the least common multiple of 9 a’#, 4a2?, 12 a*a’, and 6 a’z? is 36 a’x’*. ProstEM 2. Find the least common multiple of (a+ x)’, a —2Qax + x’, a® —a’ax, and a? + a’x. OPERATION. a) (ata)? w@—2ar4+2? &b—aae ata a)(a+a) w@—2ar+2* a'—ar a+ae a+ax)(a+x)' a —2axr+a* a—s@ a+ a a— x ) atx @—2ar+2”? a—z 1 a+ 2x a—x 1 1 The product of a, a,a + %,@—%,a + #, and a —Z is a’ —2 atx? + a’a*. Hence the least common multiple of (a+), a —2 ax + 2’, a —a’a, and a+ a’e is a’ — 2 ata? + axt, ELEMENTARY ALGEBRA. 115 EXAMPLES. . Find the least common multiple of 9 2’, 63 a’x, 21 a'x%, 14 ax’, 18 az, and 6 a’. Ans. 126 a’z*. . Find the least common multiple of aand b. Ans. ab. . Find the least common multiple of ad, be, cd, ac, ad, and bd. Ans. abed. . Find the least common multiple of a’, 6’c, ed, a’e, a’d, b’d, ab’, bc”, ed’, ac?, and ad’. Ans. a’b’c'd’. . Find the least common multiple of 3 and xy. Ans. 3 xy. . Find the least common multiple of a and a + 8. Ans. aw + ab. . Find the least common multiple of a’z, ax’, a? + a, and a+ x’. Ans. aba? + atat + abx® + a5, . Find the least common multiple of az + 2’, a + 2, @& — 2x’, and ax — x’, Ans. va —x', . Find the least common multiple of « — 3, 7 ay, « —4, and 7 y. Ans. T x®'y —49 a’y + 84 xy. . Find the least common multiple of 2? + y’, « —y, x + y, and 2? —y’. Ans. ‘x — yf. . Find the least common multiple of r? — s’, (r —s)’, r —s,andr-+ s. Ans. 7? —7’s — rs’ + 8°. . Find the least common multiple of (a — x)*, a? —2 az +27,¢0—2’7,a+ 27, and a—z2. Ans. at —2a®x + 2 az’ — x, . Find the eet common mu iple of a? + Zax + 2’, (a+tax,anda+a. Ans. a + 8a’x + 8a2? + 23, . Find the least common multiple of 3 (a —b)’,4a —4, and 6 a —6b. Ans. 12a? — 24 ab + 12 6’. . Find the least common multiple of 24 (a —b), 12 ab’, © and 8ab. - Ans. 24 a’b? —24 a7b*. . Find the least common multiple of 9 a? —9 b?,4 (a+ b), AG + b)’,and4a—4ob. Ans, 360° + 36 a’b — 36 ab’ — 36 B*, 116 ELEMENTARY ALGEBRA, 17. Find the least common multiple of 15axz — 5 az’, (3 — ax)’, 15a —5aa, and 15 « — 52’. Ans. 45ax — 380 a2? + 5 az. 18. Find the least common multiple of a? — x’, 5a? + 10 az + 52’, v« —a’x’, and 5a’x', Ans. 5a’x*? + 5atxt — 5 a®x’® — 5 a? x*, 19. Find the least common multiple of ab’ + abe, ab? — abc, and ab® — abe’. Ans. ab’ — abe’. 20. Find the least common multiple of x —3, «7 —62 + 9,3b4—9b6,and3b. Ans. 3ba? —18 bx + 275. SECTION XXXI, Reduction of Fractions to their Lowest Terms. THE rules for operating on fractions should be proved in Arithmetic before the study of Algebra is commenced. Hence we shall generally only give examples of their appli- cation. | 215 st wx 265 stwaty? SoLutTion. W2 find by observation that 5, s‘, w®, and x will measure both numerator and denominator. Dividing them both by 5 s‘w*®x, we have pesca RIN which is the fraction 53 s°a8y? PROBLEM 1. Reduce to its lowest terms. in its lowest terms, as nothing else will measure both terms. 5 abc’ — 15 a7c® +- 20 a®ax* 10 a’2° + 30 ax? — 20 ata? ProspueM 2. Reduce to its lowest terms. SoLution. We find by observation that 5, a’, and 2+ will ELEMENTARY ALGEBRA, 117 measure both numerator and denominator. Dividing them gan0 6 $ both by 5a’x*, we have id ee Sa raledlelha Be nothing 22° + 6ax — 4 a’x* else will measure both terms, this result is the fraction in its lowest terms. a + aa Peoriem 5. Reduce ———"~ aé + atx? + a’xt to its lowest terms. SotuTion. We find by observation that a? will meas- ure both terms. Dividing them both by a’, we have w+ 2 at + aa? + xt asa—2,anda-+ x. It will be found that a —~2 will not measure either term, and a + a will measure the numerator only, giving as a quotient a?—ax-+ 2. Try that as a divisor of the denominator. It will be found to measure it. Then a? —ax + 2’ is a divisor of both terms. Dividing them both by it, we have __“ + __ which is the ori- at ax t+ x ginal fraction in its lowest terms. For further divisors, try such quantities Norg. If divisors do not readily occur to the pupil, let him try to factor both numerator and denominator. Then common factors may show themselves. Thus, ay mri when factored, becomes ax? — ay 24/2 aig zy (4 —¥")> The common factor may be cancelled, giving a a (x? — y?) a EXAMPLES. Reduce the following fractions to their lowest terms. 5 aX yz A 3 5a Chiara sige 25 ata®y?2 Say eS atay P = mee 12 min'p qe Ans 3 n'p’g 28 m'naxy® ay 118 ELEMENTARY 16 a’bictd*e® 24 aSbctd’e? Dery z 49a oytz 106 a'mn?x* 24 ma‘cx’yd 18 p'a*baye 121 abcde 22 fghij 35 x°y2°ep* 45 may var yZp 27 c’yzma 36 2 yze'm Tac? + 21 atet — 5 abe® ae + dare? +12 act 121 ad’ex + 55 ate? — 33 a? C 11 ae 34 Me — 12 ieee 121 aed — 5atex? 3aee + Qare’x —5atexr 14axy —1l4axy + axy 14axry —l4azxry + any x —y v—2xey+ y’ A oa ree a? —y? 18,00 3 + y® xy? —y! 19. ALGEBRA. 7 2 ax? " 3 minty 4a’dma Ans. 308 pas 11 abede Qfohi Ans. 7 ep 2 9 mx Ans. 3 4 a 7 + 21 ave? — 5 ac ao + Dae ll@« + 5dac—3 : =o Te — 6a? ' 136+ 140% 121 ed —5 ax? a e Ans. Ans Ans Ans Ans. 1 Ans. =. ns. i a: z—yY a+ wy ty xe+ey Ans. x — aly + + xy’ ayy — ea ty 2 Ans. ¥.. x Ans. Ans 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. dl. 32. 33. ELEMENTARY ALGEBRA. 119 _. Jj Ans a baat a —xy + zy’? —y Se 10a + 200% + 100° Ne 15a* + 30 a5d + 15a°d® eo a’ — x" Ans Li at§ — a oe + 2x 3 are i yee las ga aut Hic xv —y a fe ARO Ree eat Sg i Seti iia Ana olkeoers iy a — aly + xy? —y Dp eg. Ans eee foo Bs rs —rs* r-+s oo 2 Sc EES Ans. cont eS rs — 3r°3? + 37r’s® — rst fas r —2rs + 3? xy — xy? Aneto —$———$ ———_—___.___.., ns. ° ay + 2aty* + xy? wate an” 7h ary Ane 1 ax? — ary™ ie ee Ang een as—y’ aty $. peat al ;- Ans. The fraction 7s in its lowest a—2ax+2 terms 12 m‘n* —12 min? + 24 mn’? yer 2mn “18 mn? —18 mn? + 36 mn- bk ax + ay Anan Zax + ay 2x + y at y a Ans. - tips ae ASR ONS ab — x® | SY a ea 120 ELEMENTARY ALGEBRA. SECTION XXXII. Reduction of Whole and Mixed Quantities to Fractions. PRoBLEM 1. Reduce ac + bx to the form of a fraction. Soutution. This can be done by merely placing 1 under ac + bx iia it as a denominator; for ac + bx is the same as ProsLEM 2. Reduce ¢ to a fraction having 6 as a de- nominator. SotuTion. A unit = - Then ¢ units are equal to ¢ b times ;-, or 3 Hence c = ae bi Sib b PRoBLEM 3. Reduce x + y to a fraction with the de nominator « — y. Sotution. A unit = ——# . Then x + y units = “—y _ ay eae. (a + y). Hence & + y = - EF, We see by the above reasoning that any quantity may be reduced to a fractional form by multiplying it by the pro- posed denominator, and placing the result over that de- nominator. PROBLEM 4. Reduce 54 to the form of a fraction. Sotution. Multiply 5 by 7, producing 35. Hence the 5 = 3. Add 4 to %, and we have 39. Hence 54 = 32, PROBLEM 5. Reduce a + ~ to the form of a fraction. ELEMENTARY ALGEBRA. 121 Sotution. Multiply a by n, producing an. Hence a = an ca, an an + ca =, Add = to Fi, and we have ——— n } PROBLEM 6. Reduce x —y — to the form of a fraction. s .Soutution. Multiply « —y by x + y, producing 2’ 2 —y’. Hence x —y= -—-—. As the fraction is nega- 2 tive, subtract ae from ¥. The operation of sub- dr Y y traction is this: “t—y sake}, ioe 2 a5 os which is the same as —~ ee. : the minus sign before the whole fraction means fae each term of the numerator is to be subtracted. The result is , as — by Prope 7. Reduce — sgh) aes + y to the form of a fraction. Sotution. Multiply — ax + y by a — 6b, producing —ax + ay + bx — by. — ax + ay + bx — by aa a—b ; As the fraction is positive, add ant to this. Add the ba —2 numerators. The result i is — oo ai ca ete Dey Hence — x + y = EXAMPLES. 1. Reduce ac to a fraction whose denominator shall be cz”, 3 2 ackx Ans. “ Cc ex? 11 122 10. dt, 12. ELEMENTARY ALGEBRA. Reduce » + a to a fraction having the denominator n+ a. Ans n+a Reduce a? — 0’ to a fraction having the denominator a 0, Ans. Reduce x? + y’ to a fraction having the denominator : x ees Ans, C1 See Reduce n*x?y* to a fraction having the denominator an®a’y*t — en3x? a—c. Ans. UE panic ih! y* a—e. Reduce n’y’ —a’b? to a fraction having the denomina- 2474, ,2 4},2,,2 a’nty’ — atb’n tor n’a’. Ans. aga ; a Reduce n’y —a to a fraction having the denominator ney? pala 8 eS n'y + a. Ans. ea Reduce x? + 1 to a fraction having the denominator at —] vee AL Ans. ee |: Reduce n? —a to a fraction having the denominator a’n? —a — n? + an a —n a? — n. Ans. Reduce 2° —y + ¢ toa fraction having the denomina- ax® —ax*y + ame tor az’. Ans. az* Reduce a@ + a to the form of a fraction. ab + ax Ans. ae. Reduce a + : to the form of a fraction. Ans 90a ' c 13. 14. 15. - 16. bi: 18. 19. 20. 21. ELEMENTARY ALGEBRA. 123 Reduce ax + < to the form of a fraction. 2,3 Reduce ea? — — to the form of a fraction. $8. as Ans. as sabes x Reduce m?n? — eae Y to the form of a fraction. Ans. — Be x Reduce ab — aes — to the form of a fraction. 23.2 Ans. Cs ed 2 2 Reduce a — x — in sol to the form of a fraction. a+ or 22 Ans. auita. Reduce a + «& + aed to the form of a fraction. 204+ 2ar+ 2x Ans. ___ a+2zx 2 2 Reduce a + Rauk ree to the form of a fraction. a+b 2 ab Ans. —— mau 3 2 Reduce ec + d — - ale : to the form of a fraction. vA he Ans. pres Eee Reduce 32? —2a + eet to the form of a fraction. 2 Ans. (hie, 2 124 ELEMENTARY ALGEBRA. 22. Reduce 3a — os to the form of a fraction. Ane ee a 23. Reduce 5ab — i oe g to the form of a fraction. 15 ab? —Ta —2 Ans, — 3h —. - —8y7’ 24. Reduce 82 —2y — eae Sy to the form of a frac- : 62°+ 47’ tion. Ans Sane j : 25. Reduce 1 te i oa oe to the form of a fraction. 28 Ans. ao+e 26. Reduce 1 — Z ee ze ile the form of a fraction. ; 2 Ans. a + BF 27. Reduce ; + 1 to the form of a fraction. Ans aie Y, y 28. Reduce * — 1 to the form of a fraction. Ans. See 29. Reduce 1 — 2 to the form of a fraction. Ans. yee 2 » 2 y 30. Reduce — ee oe aa ed + 1 to the form of a fraction. a + a’ 2a + Qax + 2at Ans. i. fae ELEMENTARY ALGEBRA. 125 SECTION XXXIII. Reduction of Fractions to Whole and Mixed Quantities. 2 2 PrRoBLEM. Reduce a Boe Gg to a whole or mixed quantity. SotuTion. Divide the numerator by the denominator as far as possible. e—d)?’+ed+d@(e+2d ce —cd- Qed -+ d? 2cd — 2d? 3a Set down the quotient as a whole quantity, and if there is a remainder, set it down over the denominator. In this i he e—d problem the answer is ¢e + 2d + EXAMPLES. Reduce the following fractions to whole or mixed quan- tities : (Soon iaege Ans. weep ee a a 24 2. a a ; - Ans. n—m + f. 2 3. aad Ans. 6? + - 2 ee eee rt i ee 2 ltd a) 5. a. Ans. 1. 11* 20. 21. ELEMENTARY ALGEBRA, SEG mo Ans. on ee 3m 3 25 mp — 9 mp R Ane Ae mp y Aare Fire 18 aba —12 ax -5a Ane 8 bo ee 6 ax 6x 20 abe — 3 ae — 2e 38ac + 2e Ser a oe ee a} Ans. 4e¢ ——-: eos ih ° 5 5 7 +. Ans. x* —a*y + a’y’? —ay® + y'. 3 3 i Ans. x? —xy + y’. Bae. 8 2 2 5 ae Ty Ans. a® —a’b + ab? — B® + pe a’ + B i 2 Aereny Ans. a Ot ae 5 5 aa Ans. xt + xy + xy? + xy’ + viet a dd — ed? Ans. ed — ed?. e+d wan Ange + 1. y y Somkd Ans. 1 — 4 c c ao eRe Ans, 46120 + 2—© 2ab + 26" Ans, 2b. a+ 6b a” + ax + a a eee Ans. a + ae ax + by + 2bx by + bx bisists ace Dia Ans Ce 22. 23. ELEMENTARY ALGEBRA, 137 2 2 ; oft 2ed + dl ees ht Some e—d Praha 2° +ed+a@ -p ae Seige appa Ans. ¢ + aa Te a + 30% + 3800°+ Fh +e+4+d | Sg aera oe 2 : e+d Ans. a? + 2ab + 6 ea m:* aT 2 mn +n? —m—n i ee ee Ans. m+n —1. ary + axy’ + axy* | axy . Ans. 1 a y + ee a+t+ab+e : merED br Ans. a + wea a —ab —e ; Se Ry Ans. a — ee ety te + dry ty’ Bei. ane ie Anz. 1+u+y. ee 2 : : Es Ane oe ee x—y Ria br oe 5 aires a Ans. «* + ax + —[——<. Wie oy a is + ve Ans. ax. an 4 ctx —5 pele 2 cx —2 Ans. 2¢? + ea: 3 nht = es we Anahi ee ab a 2 —— . ae Ans. axy + ay. 128 ELEMENTARY ALGEBRA. SECTION XXXIV. Reduction of Fractions to others having the Least Common Denominator. a ac c Se2 ph ne ee ee and ite i ees to a*—c ac+e a” —ac equivalent fractions having the least common denominator. PROBLEM. Reduce Soxtutron. The fractions must first be in their lowest terms. The second is not so, both terms being divisible by c. Performing this, division, the fractions become a a q Seeley) ee . av—ce ate a” —ac We find the least common multiple of the denominators. a+c)ai—e, a+, a? — ae. GreC:) a—e, 1, a? — ae. 1k ie a. Multiplying together a + c, a —c, and a, we find the least common multiple of the denominators to be a* — ae’. This is the least common denominator. The first denominator is contained in it a times; hence both terms of the first fraction must be multiplied by a. 2 Performing the operation, we obtain = The second denominator is contained in the common de- nominator a? —ac times. Multiplying both terms by this, en the second fraction becomes aoe a’ — ac The third denominator is contained in the common de- nominator a + ¢ times. Multiplying both terms by this, e+e ; a the third fraction becomes = Rane Hence the fractions, reduced to others having the least common denominator, are, ELEMENTARY ALGEBRA, 129 a? a’ — are ac + ¢ ——_~ ~——— and ———.. a—ac a—ae’ a — ac’ EXAMPLES. Reduce the following fractions to others having the same values, and having the least possible common denominators. a b ac b? 1. b and ~ Ans. 7 and Bey Joys and g Atevene= ee ae bat) oid z LYZ UYzZ xyz 4 3. ee = and as Ans. tides ie and Le ax cx ac ace ace acx ax dx xy vx dx xy” A, Her ay’ and a Ans. ady’ ady’ an ady a+b a—b 2 a+2 ps oo Gira —— be and — meee f Oe ale ag ee a+b a—b eee ss . Ang, CTL +B ae — Bab +B ns. Pane ae ey Te : ee a a + 53 a’? — P? a® +- 5° it e+ and a + b* Ans. ao +O and Pah BF 8 we 2 5 ene ee vx ay —aby d a’x + abx a—b” a—bh’ Fed 9. s and “4 Ans. a ae an oe. 10. 2 S and = ye =e oe an ae ee car! aan sae ™ ate I 130 12. 14, —— 15. 16. 18. ELEMENTARY ALGEBRA, db ad Cx en ee “Or 9 2 an eo parts acx’y cba’y” axsy’ Pedy od ee wvbexy® a?bera*y” abexy® 13 am em da ‘ a'masy” acy” azxtm acm’ am*x*y edy® Ans. comizty® aemiaty? *"* emiaye Ta 9a 6a l4ax 9a 8 x? 9 ’ Ao” and oye Ans. Awe Ag” and ae at+a Ax d 9a a—xata’ oe at i 2a7x? + 4ax’® + 2 xt 8 ax’ —8 a hee 2 a’? — 2 a 9 ax? —Qa% d 9a — 9 ax’ a Fiat — 2 aot ha C bdh adk bck iy B and i Ans. bak’ bdk’ and bdk eA Se Mn x EO ys RO erat Ane, 2%, 08 Bedm 5 4 aba ' Pbed wbed wbed * wbed 2 2 4 2 2 “| ees has and AS, Ang. oe hs and —. nn n n nn n n ax+b a b ME Pir at ones and oa a ax + b ac + ad be — bd me of —d? of —d?’ et e —d? s 2 2 ae ab qv Ang owe b a 20. BB wo ODS oe ab ab? ape 2nd oe ELEMENTARY ALGEBRA. ales f SECTION XXXV. Addition of Fractions. Frrsr reduce the fractions to their lowest terms, then to the least common denominator; add the numerators, and place the result over the common denominator. EXAMPLES. Find the values of the following expressions : 2 2 zs - ~ Ans. 2 a 2 2 4 wyz2 ab’y d’xyz xyz. 8x 9a 6a 2427+ 9a _ 9a 3. Or ae we Ans. Sey ee etry 6 + 4y° = ei ae Ans. 1 ete 2 P -F-g a+b. a—b 2a? + 206? 2 eg os An a? —}? a—sx@ a+z 2a 6. = Pe + eae Ans Aas | 1 ee 1 2x Fe a + ey Ans. ey ) a aa ap es me Ans. 0 10 9a", 2abee | Tax “ Tabe 5a'b’xe 8ab'a ) | vee 300 ba? + 112 be? + 245 a’e 2 Es edt + amx + ee a Eee Ans. 4ama. 20. ELEMENTARY ALGEBRA. 1 1 il 2a+1 oe Rae a eT Ant ay aad thes ay® + y'2* + xia “2? se x? ap y” Ans. ea) fe "s iis a Ans. Ye eshte x y Zz LYz a+ b 4a b—a 4 a? AE Oo Ge Bl es Ans. Cu 7 8 3 Tf+8g+3e —+—+ >. Ans. ~ v_>? og of | ig fy pole Di asy + bse + bry oF ak ta) iia Ans. bsy . a—b ¢#+0 a + b 38a + 3B Bers ah aie phan An. Cte a2 cat Zan ee ae ‘ata + 3 oe Aye eO+ar+art+est+2ad+ 22x 8 rn EE Aiea) Ans. x. SECTION XXXVI. Subtraction of Fractions. PREPARE the fractions as in addition. Subtract the nu- merator of the subtrahend from the numerator of the minu- end. Set down the result over the common denominator. EXAMPLES. 1. Subtract — _ from si , Ans. & 2. Subtrace 18 ¢ from = a Ans 20 &. 14 By “i ADT 16. Li ELEMENTARY ALGEBRA. 133 rR a —b a+b 4ab_ . Subtract - + ; from aeocaegs Ans. my . Subtract u from : Ans. wT . Subtract Pe be from ; Ans. = 2y : ary ef x —y . Subtract — ab from — od Ans. 0. ab cd 2 abex" pate a’bex 7 abema —4 abcx* 9 mn 18 mv’ a 18 mn Subtract — ” fr rom a Ans wim age . Subtract > a Saat Mestre . Subtract SSE from ba 2 Ans. bce a oe? 2a a hee ene 2 . Subtract Carte from eats: Ans. Was 4 poe a—Z ata a? — zx —b a+b ars from Mic Dab ake Ans. 0, 1 1 a . Subtract ae from Ee Ans. — ce 1 yi 1 . Subtract a 6 from as Ans. pam Fea i af ec—b . Subtract “a from eer Ans. ico ghee 2 1l+a 1 . Subtract — TC. - from Eee, Ans. sparmyticsr pees 2 — Subtract ~ from ea Ans. z rt! : z z Zz Subtract =e from laid Perms Yat ee ee 12 134 ELEMENTARY ALGEBRA. 2 18. Subtract sh sa from ——— Rating Chutes | Ans. 0. a—b a 4-6 | 19. Subtract ” from ~. Ans. De ees ; J at xy a a a 20. Subtract ore j from = Ans. Poe —0892.00——. SECTION XXXVII. Multiplication of Fractions. THE process is the same as in arithmetic. Multiply the numerators together for a new numerator, and the denomi- nators together for a new denominator, cancelling common factors of any numerator and any denominator. By such cancellation, the result will be in its lowest terms. EXAMPLES. Find the values of the following pee” 1 18 a*bex cach . 25 aberx ve Be * 35 mn 27 ake’x 30 a®ca?" " 8a'nx 12a°* 18 aby’ ab*c’y 2 Tipe ae pee = mg 2 Bae tae Ans. = 5 nae Y The Pie ac 4, — b x A Ans ate es 24c’m’n 121 ama 5 ya? veux : 132 aba’y ©" BS m 5 * {2 mn’ ae 6b- 18 6. —, of 5 atu. Ans. 90 a’ex. a’e ELEMENTARY ALGEBRA. 135 3,2 7. 11 amin?x x ASE mica eee ; y’ mica “* 144 ate’x’ pee V1 ay? 4) 51 ay! ene of — cx. 31 ary 5b 1 10. 21abe x 7 Be of 93ay x iF 1 11. —abe x ay xX any X 7. 1g, Bam’ IL my’ 2 * 15 ¢’my n> 302 na" 13 Zor ye llutzy® | 24 cux Rie” 492° * "132 mée’q i 50 a*y > se a’ : ab 28, Coe of OT op OO’ Ans. 2. a9; SE + Bw Sat oe en 30. — = ze aes snes ee _ SES ae SECTION XXXVIII. Division of Fractions. THE process is the same as in arithmetic. Invert the divisor and proceed as in multiplication. If the divisor is a compound fraction, invert every component fraction of it. EXAMPLES. Find the values of the following expressions: Say — 15 aPecy’ a4 i Q2abea ~ Lata?’ AM, 6 a®y 13 a’bed’ . oe 13 bd? 2. OT aides. + 121a exy ° Ans. 2541 a abeny® Li. 18. Ley 32 abex®y © ELEMENTARY ALGEBRA, 25 a*bex® edz __ 33 minp”? 125 aFyw +5 —a--. a 21 e’'dax® LL ated? | 13 mny ° 225 r'sdx . 51 s*pq __ 31 mnop* ‘5 abp’c 18 ghk* 5 g’e x — y? 32+ 3y ° m? — n? Sat ee? + ed | of 18 = "ee — ed hay at bh am’r —amy . a+b a—b Oat 12* cco 3 Tabey © 117 mcd 99 na’ax ade, 150%8 | Silas Oe as Zab e _m—n “mtn abex. -- 121 mnpg. ees = hom ae 2 ae'dy * 3 ac’dxz? 16 by’ — -- 3a'cd'. 22 3 cx” 3 acmnop 25 xty 162. ity 9 e+d e—1 ex? — cy? " vera + vey Ans. 137 Ans. 5a 3 q° 1375 xyu 6 aetdx?y? . ne aes Ans. — Ans Ans. 1. cae " 3a'n?y eee " 34 pqrs® __ 31 boty 25 acp 2 hk? Beg® Ans Ans a’ + 2ab + B? a — 2 aberan 2 Ans am on 138 ELEMENTARY ALGEBRA. Ed aeons est COW 19. Pe ” of eB See Eas SNP Ans. 1. ae +182+4+9 | Fre | | 20. >a ea | Se 9a? —1824+9 Ans. 81x oh 81. SECTION XXXIX. Reduction of Complex Fractions to Simple Ones. THE process is the same as in arithmetic. First reduce mixed quantities to improper fractions, compound fractions to simple ones, etc., so as to have the numerator of the com- pound fraction nothing but one simple fraction, and the de- nominator nothing but one simple fraction. Then multiply the upper numerator by the lower denominator for a nu- merator; and multiply the upper denominator by the lower numerator for a denominator. The resulting fraction should be reduced to its lowest terms. a ne £4 PROBLEM. Reduce to a simple fraction. ab a ne ne SoLurion. The numerator, — of ——, reduces to ——- C atx* aba* 2 The denominator, a + cass reduces to — ot a, The com- ab ab ne eee hott ie plex fraction then becomes —— Baha tetig ab We now multiply ne by ab for a numerator, producing abne. Then multiply a’b + 4c by act for a denominator, producing a®ba* + 4 acct. abne a’bxt + 4a3er* ELEMENTARY ALGEBRA. bne it to atba*t + 4 a’%ex*’ EXAMPLES. 139 Hence the resulting fraction is Both terms can be divided by a, reducing Reduce the following complex fractions to simple ones. Lg nS Sia} ore Slaloe x Ans. =: b 140 10. WE 15. 16. 3 Pay Ans. ke a 4a 3 1 Hed 3 Ans. = a amc Tm? ac'dx? Ans EE 53 66 c'dx 144 PoE Ans iam — 1 a 2 a 2 Ans, 000% eA Hie, 9987 2mnr 3 mv 4 mn Ans, 1 6 m’n 1 a Ans. = oe Fy Be oy a+ 2x a? — Ghee - 5 Ans. - i a+cx a + Ey a’ — o° 1 a GE ey +a Ans. naan m—n of oe Ans. ds ELEMENTARY ALGEBRA. 7. dors atn a mn ELEMENTARY ALGEBRA. 141 a 1 c 18 @ aveta 18. ‘ Ans. -—— a c = OL = Cc a Lae : ac + 19. ss Ans Porat a+ - ‘ Cc 1 a a Ans ie 7 i alah Nigk a —079400——. SECTION XL. Clearing an Equation of Fractions. If equals be multiplied by the same number, the results will be equal. Take 2 = 2. Multiply both of the equals by 3. Then we have 6 = 6. Take 5 = 3+ 2. Multiply both of the equals by 6. Then we have 30 = 18 + 12. Take} = 4+4. Multiply both of the equals by 4. Then we have 2 = 1+ 1. The equation has been cleared of fractions. — In order to clear an equation of fractions, multiply every term by the least common multiple of the denominators. : SP ice AE Sete ; PROBLEM 1. Given 7 dame tek Diet, + 42, to find the value of x. Sotution. The mixed number musi first be reduced to an improper fraction. Then we have - + emai: oth 142 ELEMENTARY ALGEBRA, a The least common multiple of the denominators is 84. Multiply by 84. As it is a multiple of the denominators, every denominator may be cancelled in the multiplication, giving 21x + 28x —Tx = 12% + 360. Transposing, 2lx + 28x2—T7a# —122 = 3860. 30x = 3860. oii t AE : 1 Ses Os cael a PROBLEM 2. Given 9 ao ra +e rere: to find the value of x. SotutTion. Here, the least common multiple of the de- nominator is 36x. Multiplying every term by 36 x, we have 4x+6x2-+ 9x = 86. 19. v= 36: x= 11%. EXAMPLES. Find the value of x in the following equations: 52 32 1 OO Loyal a> thig) da agen Ans. © = 24. ye 2. 29% —28ax + Rar 8 —4}. Ans; a 3: we 4440 = 88 —4e. Ans. « = 11. 4, Sf 10 == —2 42 —32. Ans. « = 5. 16 x 23 10x ; ie Dee rih chude Or ee Ans. x = 9. 6. stl +2e=—5. Ans. « = —2J, 1. E+ e=4a— >" + 80. Ans. 2 =e Nl 8 ELEMENTARY ALGEBRA. x x 32 ete A & g = 1 + 5. +25 = —15 Tx a, “ : mee 8 Oe + 10g + 76}. Seerent. 993 ee 5 ot zx+6 x—6 2 5 ao 4 ao oO = ae ae ae PL ie Va re ae te. We let 5 55 2a. 50 75 125 __ a, Bae Sx 8e 27 ‘i Se a+ 5 le 5 25 2x2—T 38x2—2 Beat a = x7—7 Tx 5x (ms et 30 Ee a ve a 162 —68 47 143 ANS Meee: Ang oi 01, Ans. x = 40. Ans. x = — 80. ANS. el Ans, I= 24 Ans. x = 86. Ans. & = 6, Ans. % cay). Ans, & = 2b, Ans. x = 61, Ans. x = — 358. Ang: Ciel in Ans. % = 93, Ans. x = — 16. Ans. x = 54. 144 - ELEMENTARY ALGEBRA. 24. 30” = 293. Ans. x = 138. 25. een = wali saat Ans. 2 = 1 2 3 eo Deh alent ae, 26. gy Aor aS ees od. Ans. e ==: 94, 27. : 27 «2 + = = — 24,1). Ans. x = 33 5x2 10x—21 mer 28. EL ae — 82, Ans. x =, 82. 29. 5a — 86 = « —171. Ans. & = 17%. x—8 Tax—A4 30. ab er prt rel Ser et a ee Ans. x = 12. —059400——_ SECTION XLI. Questions producing Fractional Equations. PrRoBLEM 1. John has z as many marbles as George, when George gives John 36 marbles. Then 7, of John’s, plus z of George’s, amount to 50. How many had each at first ? | SoLution. Let 2 stand for the number of George’s mar- bles. Then : of x, or ake is the number of John’s. 6 When George gave John 36, George had left x — 36. John then had oe + 36. ELEMENTARY ALGEBRA. 145 Now, we must find ‘ of John’s and . of George’s. 7 5x : 3852 , 63 ig f (| + 36) is aur 2 A 2 x g of (x —36) is See rat Then we have the equation, Bie eed SS Ae Fe a a Clearing of fractions, 105 x + 4536 + 64 « — 2304 = 14400. Transposing, 105 + 64a = 14400 — 4536 + 2304. 169 « = 12168. x= 72. Then ee = 60. Hence John had 60, and George had 72. 1 h3 ae ProspitEM 2. At what o’clock does 9 of F of 7 of the time from noon equal the time to midnight? Sotution. Let x equal the time since noon, (in hours.) Then : of : of - of x is the time to midnight. te oe LD x Now, 5 of 5 of of x = 7 Then, since the whole time from noon to midnight is 12 hours, we have the equation, e+ 7 = 12. Clearing of fractions, Tx+ax= 84. 8x2 = 84. x = 103. Hence it is half-past ten in the evening. 18 K 146 ELEMENTARY ALGEBRA. EXAMPLES. J, Find a number to which if its third part be added the sum will be 48. Ans. 36. 2. Find a number such that if it is increased by 3 of it- self, 4 of the sum will be 22 less than the sum. Ans. 22. 3. Find a number which being added to its fourth part, the sum will equal half the number, added to 15. Ans. 20. 4, One-fifth of a cistern was filled with water. By add- ing 5 barrelfuls 1 of the cistern was filled. How many bar- relfuls of water would it contain in all? Ans. 100 barrel- fuls. 5. A spent } of his money, and then received $1. He then spent 3 of what he had, and $8 remained. What had he at first? Ans. $20. 6. Find a number whose half, third, fourth, and fifth parts added, are equal to 154. Ans. 120. 7. Find a number whose sixth part exceeds its ninth part by 5. Ans. 90. 8. B’s age is 211 times A’s. The sum of their ages is 47 years. What is the age of each? Ans. A is 12 years old, and B 35. 9. Divide $60 among A, B, and C, giving B 3 as much as C,and A 2.as much as C. Ans. A must receive $12; B, $18; and C, $30. 10. Divide $60 among A, B, and C, giving B 3 as much as C,and A 4 as much as B. Ans. A must Ben $10; B, $20; and C, $380. 11. A merchant lost 4 of his capital during the first year. The second year he gained 2 as much as he had left at the end of the first. The third year he gained ;3 of what he had at the close of the second, making his cannes $7000. What was it at first? Ans. $5000. NotE. Let z = his capital at first. Pe lost ~ and had te re- ELEMENTARY ALGEBRA. 147 4 f 4x maining at the end of the first year. Get 3 of ee and add it to —, 5 5 to get what he had at the end of the second year. Get then 5% of that result, and add it to that result to get what he had at the end of the third year. 12. My capital is $2220, being 11 per cent. greater than it was last year. What was it last year? Ans. $2000. Nore. 11 per cent. is +4. 13. Divide 32 into two parts such that 1 of one may equal 4 of the other. Ans. 12 and 20. 14, Find a number whose fourth plus its third less 12, equals its half less 6. Ans. 72. 15. At what hour does } + 1 + 3 + ,5 of the time past noon equal the time to pass before midnight? Ans. 4 P.M. 16. Divide 325 into two parts, one of which is 12 times the other. Ans. 125 and 200. 17. If you multiply a certain number by 3, add 4 to the product, divide the sum by 5, and subtract 7 from the quo- tient, the remainder will be 10. What is the number? Ans. 27. 18. If you add 6 to a certain number, divide the sum by 9, add 16 to the quotient, and multiply the sum by 3, the product will give the number itself. What is it? Ans. 75. 19. A’s capital was at first 3 of B’s. A gained $100 and B lost $100. After this 14 of A’s capital added to the whole of B’s amounted to $6500. What had each at first? Ans. A had $3300, and B $4400. 20. A man spent $14 more than 3 of his money, and had $6 more than $ of it left. What had he at first? Ans. $84, 21. A and B have the same income. A saves 10 per cent. of his. B spends $75 a year more than A, and saves $125 in 5 years. What is the income? Ans. $1000. | 22. A merchant sold his stock, good-will, and fixtures for $50,000, receiving 5 times as much for the good-will as for 148 ELEMENTARY ALGEBRA, the fixtures, and 34 times as much for the stock as for the good-will. What did he receive for each? Ans. $38,000 for the stock, $10,000 for the good-will, and $2000 for the fixtures. 23. From a certain sum I took its fourth part, and then added to the remainder $29. I then took away the tenth part of this, and afterward added to the remainder $25, which made it $70. What was the original sum? Ans. $28. 24. A young man spent } of his yearly income for board and lodging, 4 of the remainder for clothes, and 4 of what was left for amusements. The rest, $400 per annum, he saved. What was hisincome? Ans. $1600. 25. A laborer was engaged for a year at $440 and a suit of clothes. He left at the end of 5 months, receiving $160 and the suit of clothes. What was the value of the suit of clothes? Ans. $40. Nore. What he received was ;; of a year’s wages. 26. A bought stocks amounting to $3000. B bought stocks amounting to $2000. A sold a certain amount of them, and B sold half as much as A. A then had 1} times as much as B. How much did each sell? Ans. A sold $1500 worth, and B, $750 worth. 27. A quantity of pure water being decomposed, 3 of it by weight plus 125 ounces was oxygen, and ,'; of it by weight less 15 ounces was hydrogen. What was the quan- tity of water? Ans. 90 ounces. Note. If 2 represent the quantity of water, the two parts of it added together equal z. 28. A father divided a certain sum of money among his four children at a fair. To the oldest he gave 65 cents less than } of it; to the second, 25 cents less than 4 of it; to the*third, 9 cents more than | of it; and to the fourth, the remainder, which was 70 cents. What was the whole amount? Ans. $3.30. ELEMENTARY ALGEBRA, 149 29. Divide the number 22 into four parts such that if the first be increased by 4, the second multiplied by 4, the third diminished by 2, and the fourth increased by 2, the results shall be equal. Ans. 4, 2, 10, and 6. Notre. Let x stand for any one of the equal results. Then x —4, x + 2, and x —2 are the parts. 30. In a certain quantity of Chinese gunpowder, the nitre was 19 lbs. more than } of the whole, the charcoal was 163 lbs. more than 3 of the whole, and the sulphur was 29 lbs. less than 1 of the whole. What was the amount of gun- powder? Ans. 162 lbs. 31. Divide $870 between two men, so that 34 of what the first receives will be equal to = of what the second receives. Ans. The first receives $370; the second, $500. 32, A merchant buying goods spent at one store } of his money and $300 more, when he had $3000 left. How much had he at first? Ans. $4400. 33. Divide 63 into two such parts that if the less be di- vided by 5 and the greater by 4, the sum of the quotients will be 15. Ans. 15 and 48. 34, An estate is to be divided among three children so that the first shall have $1400 less than half its value; the second, $1300 more than } of its value; and the third, $840 more than } of its value. What isits value? Ans. $14,800. 35. A man has a lease for 39 years, and 2 of the part of it which has expired is equal to .5, of the part yet to run. How long before it expires? Ans. 24 years. 36. What number is that whose } added to its 1 makes 382? Ans. 60. 37. A, B, and C gave a certain sum of money to the poor. A gave i of it; B,4; and C, $50. What wasgthe sum? Ans. $120. 38. A lived in his house 24 years, during which time an 13 * 150 ELEMENTARY ALGEBRA. addition was made to it. One-third of the time he lived in it before the addition was made, is equal to 4 of the time he lived in it afterward. How long is the latter time? Ans. 15 years. 39. After paying away 4 of my money and § of what re- mained, I had still $75. How much had I at first? Ans. $120. 40. What number is that from which if 21 be subtracted 3 of the remainder will be 60? Ans, 121. 41. After paying away } and 3 of my money, I had $84 left. How much had [ at first? Ans. $144. 42. A gamester staked 3 of his money, which he lost, but afterward won $3, when he had $42. How much had he at first? Ans. $52. 43. A gentleman spent in one year 2 of his income for the support of himself and his family, and 2 of the remain- der for a piano. ‘There was $200 left. What was his in- come? Ans. $1500. 44. A horse and a chaise are together worth $800, and the chaise is worth 13 as much as the horse. What is the value of each? Ans. The chaise is worth $509, and the horse, $300. 45. A house rents for $924, which is 10 per cent. more than last year. What was it last year? Ans. $840. 46. A and B have the same income. A contracts an annual debt amounting to 3 of it. B lives on 3 of it. At the end of 3 years B lends A enough to pay his debts con- tracted in that time, and has $135 of his savings left. What is the income of each? Ans. $540. Nore. Interest is not to be counted either on debt or savings. 47. What number is it whose 1, 3, and 2 are together equal to 92? Ans. 105. 48. A person, after spending $300 more than 4 of his in- ELEMENTARY ALGEBRA. 151 come, had remaining $200 more than $ of it. What was his income? Ans. $900. 49, Find a number such that if it be increased by 9, the sum shall equal ? of the number plus 13. Ans. 16. 50. From a purse containing a certain sum there was taken $10 more than its half, and from the remainder, $3 less than its fifth part, when $55 were left. How much did it contain at first? Ans. $150. SECTION XLII. Review. EXAMPLES. 1. Divide 21 ay by 35 a*y'n. Ate oe 2. Add together abe, acb, and cba. Ans. 3 abe. 3. What are the factors of a? — tee 6. Ans, © =e What is the value of the expression 3 + 6 x 9 —5? Ans. 52. . What is the value of the expression (81 — 4) + (4 + 3 xX 9 —A4)? Ans. 1. . Multiply 6 —1 by 6 — 2, and subtract 5 from the pro- duct. Ans. b?’ —4b + 2. | What is the value of © — 72 4 % _ — ee ax cr 6, am 3T 4 Ans. 4 7b 9,2 8 aw? . Reduce oe 4 ats to its lowest terms. Ans. aa . Divide a® + aa + ax” + x¥ by a? + 2”. Ans. a + a. a+b @ eed a” — ? a? = ae | . Multiply together — i pt AL Gab ee and ee 2 _ a 2ab + 6 Ase a —b Be ka pias 4 ELEMENTARY ALGEBRA, 153 22. Find the value of x when x aioe = (0, Ans. x = 5. 23. Find the value of (9 abe —16y’ + 3rs) —(12abe + 5y? —rs + abc). Ans. —4abe —21y’? + 4rs. .., a —F a—bd a+ b 24, Divide TTS by ciiaiie Ans. aM: 29. 30. . What is the number whose fifth part divided by 3 is equal to 7? Ans. 105. . Divide a’ + 2acx + ex’ by a + ex. Ans. a + cx. . If 40a” is equal to 16 x (2x + 1), what is the value of x? Ans. x = 2, . Reduce x — y — — to the form of a fraction. Py sma Y yeas til yp ities kad Aik Brac a -2 Y x? 2 z 3 F Reduce Se to its lowest terms. 1 Ans. : a—y Add 5 m'n'p + 3 m’n'p? — 11 m‘ndp, 3 m'n'p + min'p — dm’n*p’, 3 min*p + 16 m?n3p? — 12 m‘nip, and 3 mn'p*? + 21 min'p — 5 min'p. Ans. 32 m’n'p + 17 m’n*p? — 27 m‘n'p. REVIEW QUESTIONS. What is an exponent ? a coefficient? Name the exponents in the expression 24 a8i?74. What do they show? What is a monomial? a binomial? a trinomial? a polynomial? Is a binomial a polynomial ? Write two monomials, two binomials. Isa+<«capolynomial? Is 36 az? a polynomial? In what ways is multiplication expressed ? In what ways is division expressed ? Explain the use of a parenthesis or of a vinculum. What is 4/25? /xty?? W238? 154 ELEMENTARY ALGEBRA. What are similar terms? How are they added? Explain the process of subtracting — a from e¢. What is the reason that a quantity which is transposed from one side of an equation to another, must have its sign changed ? Explain the multiplication of — 4 by 6. What do we mean by multiplying by a negative quantity ? What is the rule for signs in multiplication and division? What is the rule for indices in multiplication? in division ? Explain the division of —z by —y. What is meant by elimination? When is it necessary ? What is the square of the sum of two quantities equal to? the square of the difference? the product of the sum and the differ- ence? What is meant by factoring? What is a measure of a quantity ? What is a prime quantity ? What is a multiple of a quantity ? SHOLION elas: Elimination. Ir has been already explained that when we have two equations, containing two unknown quantities, one of them must be eliminated before the equations can be solved. Three methods of doing this are in common use. One of them, elimination by addition or subtraction, has been already ex- plained. They will, however, all be brought together pee for the sake of a comprehensive view. EXAMPLE. Given ie ay eteg 1 5x S.A eel (x —10) (y —6) = 5 + (& —5) (y —8) to find the values of 7 and y. The equations must first be brought into the form proper for elimination. ELEMENTARY ALGEBRA. 155 Clearing the first equation of fractions, we have, 48a —45 = 8y —72 + 452. Transposing, 482 —45x —8y = —72 + 46. 38a —8y = —27. Performing the operations of multiplication indicated in the second equation, zy —6x —10y + 60 = 5+ czy —8x2—5y + 40. Cancelling xy, and transposing, —6x2—10y 4-8x+ 5y = 5 + 40 — 60. 2% —d5y = —15. The equations are now reduced to forms suitable for elimination ; all the unknown quantities are on the left of the sign =, and they are collected into as few terms as possible. This must be done before any method of elimination is used. We will now take these equations and show how the values of x and y may be found in three different ways. 1, Elimination by Addition or Subtraction. 382—8y = — 27 2x—5y = — 15 Multiplying the mem- bers of the first by 2, 62—16y = — 04. Multiplying the mem- bers of the second by 3, 62—1dy = — 45. Subtracting, —y = — 9. Or, Gist o, Substituting —5 times 9 for —5¥ in the second equation, 2x2 —45 = — 15. Transposing, 2x2 = —15 + 45. ESN U3 Hence x = 15and y = 9. 156 ELEMENTARY ALGEBRA. 2. Elimination by Substitution. 8a—8y = —27. 2x2—dy = —16. In this method, find the value of one of the unknown quantities in one equation, just as if the other quantity were known. Thus in the above equations, 2 2 being the simplest term, find the value of x in the equation that contains 2 x. Thus, 2x —dy = —10. Transposing, 22 = Sy — 165. teas 5y —15 Dividing by 2, 2 ea Now the other equation contains 3x. Multiplying the value of x just found by 3, we have of = em Substituting this instead of 32 in the first equation, we have irae —8y = — 27. This equation contains but one unknown quantity, and can be easily solved. Clearing of fractions, 15 y —45 —16y = — 54. Transposing, 15y —16y = — 54 + 46. Collecting, —you PD, Whence, ys: We found before that C= oY oe : Putting 45 for 5 y, we have 45—15 30 Hence x = 15 and y = 9, as before. This method is generally longer than the first, though when the coefficient of x is unity, and in a few other cases, it may be the shortest. ELEMENTARY ALGEBRA. 157 3. Elimination by Comparison. BS nee, Diese Bafa 15; In this method find the value of the same unknown quan- tity in both equations, just as if the other quantity were known. Thus, 32 —8y = — 27. 2x2 —b5y = —15. 3a = 8y —27. 2x = 5y —15. eS ee 0 ure ty anaes aes sss TS A Now, we have here two values of x; and as they are values of the same thing they must be equal. Hence we — 5y —15 have a new equation, SB Beg CLIN Ire 2s 3 2 Clearing of fractions, 16 y —54 = 15y — 45. Transposing, 16 y —1dy = 54 — 45. Yy = Now, as before, 7 = ee Putting 45 for 5 y, we have 45—15 30 Se a 15. Hence x = 15 and y = 9, as before. Rule for Elimination by Addition or Subtraction. Multiply or divide the members of each equation by such numbers as will make one of the unknown quantities have equal coefficients. Then add equals to equals, or subtract equals from equals, so as to cancel or eliminate that unknown quantity. 14 158 ELEMENTARY ALGEBRA. Rule for Elimination by Substitution. Find the value of one of the unknown quantities in one of the equations. Then substitute it for the unknown quantity in the other. fule for Elimination by Comparison. Find the value of the same unknown quantity in both equations. Then place these two values equal to each other. EXAMPLES. Solve the first five examples by each of the three methods. 1. Find the values of and y, when y+ wv = 6 and y—x = 4, Ans, 2 =e 9. Find the values of x and y, when 4a + 2y = 20 and 3”2—dy = —l11. Ans. 2 == 5 5 nee ae 3. Given 139 ENP z a at to find the values of y and z. Ans. y = 13 fee 4, Find the values of x and y in the following equations: eee eee Ans. © = 8; y= 2. 3 \ to find what x and y are equal to. Ans. © = 133 ¥ =a: In the following pairs of equations find the values of x and y by whatever method seems most convenient. Ei eaanannoca set Ans. «= 6; y =2. 4 2A == 2. if Eyairsomen ‘ Ans. 2 Sy ae 8. sia aa eal Ans. c= 6; y = —6. 10. 12. 13. 14. 15. 16. ELEMENTARY ALGEBRA. 159 Gay — 244 af . 2 Ceo ; Ans. © = 4; y= 6. by —3x = 18. oui ay 13 y — 80 oe gee ety a ee PRT : Ans. Gi 9: y = 16; Bet 4 Leys esis Sen | Ans. © = —2;y=1 Sa+4 38y—a2 | Ty —62—5 [oars Ce Bes CE oo TY 39 —y—15 Ans 2=7;, y = —3 2xa—3y 8x—D9y es oa Se ae 7 1 me ane — Soegiciaek Shae Ans, x=, 44; y= Gat+4y 2_1 FE Ree fe 175 = Oy + 62. Ans. x = 174; y = —17}. LEU —38y+4e—11. 252+ 2y Pho ae Ans. = 43.4 == 2. llxw+9y 9 99 aie Soe +10y=1lxe + 12. ANS. (0 ==. Ois yee 1 T. 160 ELEMENTARY ALGEBRA. 17. 8a2—8 ob ety = 34. ee eee ee 18. Pred be Ans. 0.%=\21 4) ee te, 19. 20. 22. 23. 24. 25. | Han “i i ies tn >< 2x—dsy= 0.| Ans. © = 8; y = 2. ELEMENTARY ALGEBRA. 161 SECTION XLIV. Questions producing Equations with Two Unknown Quantities. 1. Fryp two numbers such that { of the first plus 3 of the second equals 9, and } of the first plus 1 of the second equals 8. Ans. 20 and 15. 2. Find two numbers such that the greater plus } of the less equals 23, while the less minus | of the greater equals 11. Ans. 20 and 15. 3. Find two numbers such that 5 of the first less 3 of the second equals 8, and 4 of the first less ; of the second equals 48. Ans. 720 and 512. 4, There is a number represented by 2 digits whose sum is 9; and double the number added to 18 gives another number represented by the same digits in an inverted order. What is the first number? Ans. 27. Notr. See the note to Example 4 of Section 29. 5. A man has two horses; also a saddle worth $50. The value of the best horse and the saddle is double the value of the other horse. The value of the latter horse and the saddle is 4 of the value of the first horse. Required the value of each. Ans. The first horse is worth $250; and the other, $150. 6. Two casks contain wine; and 3 of the contents of the first is 20 gallons less than § of the contents of the second, while 4 of the contents of the first is 2 of the contents of the second. How much does each contain? Ans. The first con- tains 84 gallons; and the second, 63 gallons. 7. Find a fraction whose value is 7 if 19 be added to its numerator, and is ;1 if 19 be added to its denominator. Ans. 2. Note. Represent the fraction by 7 14* L 162 ELEMENTARY ALGEBRA. 8. Find a fraction such that 5 being subtracted from both terms, its value will be 4; and 3 being added to both terms, its value will be ;43. Ans. 3}. 9. A and B together have $12,000. If A lose 4 of his capital and B gain } of his, they will have the same. How much has each? Ans. A has $7000, and B has $5000. 10. A boy bought 46 cents’ worth of balls and tops, giv- ing 4 ee for each ball and 2 cents for each top. He then sold =; of the tops and } of the balls for 20 cents, clearing 6 or on them. How many of each did he buy? Ans. He bought 6 balls and 11 tops. Nore. Let x= the number of balls, and y the number of tops 4 2 which he bought. Then4z-+ 2y = 46 cents. Also, > a aie what he gave for those which he sold. But he cleared 6 cents on + 2 these. Hence, ze of =+ + 6 = 20. 11. A grocer bought $1.23 worth of watermelons and cantaloupes, at the rates of 12 cents and 3 cents. He then sold + of the watermelons and 4 of the cantaloupes for 50 cents, clearing 17 cents on them. How many of each did he buy? Ans. He bought 8 watermelons and 9 cantaloupes. 12. A owes $700, and B owes $600. A’s money plus 3 of B’s would pay A’s debt. B’s money plus 7% of A’s would pay B’s debt. How much money has each? Ans. A has $650, and B has $250. 13. A merchant bought two casks of wine, the first at $1.25 and the second at $1.50 a gallon, the whole costing $147. He sold 4 of the first and § of the second for $140, gaining $29 on what he sold. How many gallons were in each cask? Ans. There were 42 gallons in the first, and 63 gallons in the second. 14. Five years ago, A’s age was 21 times B’s. One year hence it will be 14 times B’s. How old is each now? Ans. A is 12 years old; B, 8. “ELEMENTARY ALGEBRA, 163 15. Find a fraction such that if the numerator be dimin- ished by 5 and the denominator by the numerator, its value will be 1; but if the numerator be increased by the denom- inator and the denominator by 33, its value willbe 4. Ans. ote i fg 16. If A should give to B as many books as B has, and then B should give A + as many as A had remaining, each would have 250. How many has each? Ans. A has 350 books, and B. has 150. 17. Divide $100 between A and B, giving B 3 times as much as A. Ans. A must have $25; and B, $75. 18. Find two numbers whose difference is 6; and if 4 of the less be added to 4 of the greater, the sum will be equal to 4 of the greater diminished by 4 of the less. Ans. 18 and 24, 19. A man bought a house and lot for $3500. Two- thirds of the price of the lot was equal to 4 of the price of the house. What was the price of each? Ans. The lot cost $1500; and the house, $2000. 20. Divide 189 into two parts so that 4 of the greater minus 3 of the less shall be equal to ;% of the less minus 2 of the greater. Ans. 84 and 105. 21. A man bought two pianos for $880. If of the price of the first be subtracted from 4 of the price of the second, the remainder will be 2 of the difference of the prices. What was the price of each? Ans. The first was bought for $540; and the second for $340. 22. What fraction will equal 4 if 2 be added to its nu- merator, and 4 if 12 be added to its denominator. Ans. 4§, 23. A certain number consisting of two digits, on being divided by 5 gives a quotient equal to 4 times the unit digit, and a remainder of 3. The same number on being divided by 18 gives a quotient equal to half the digit in the tens’ place, and a remainder of 9. What isthe number? Ans. 63. 164 ELEMENTARY ALGEBRA. 24. If A were to give B $300, B would have 3 times as much as A has left. If B were to give A $300, A would have 3 as much as B has left. How much has each? Ans. A has $700; and B, $900. 25. Two brothers had just sufficient money to purchase a certain house. The elder could purchase it with his own money, 3 of his brother’s, and $1000. The younger could purchase it with his own money, i of his brother’s, and $2000. How much had each? Ans. The elder had $4000; the younger, $3000. SECTION XLV. Equations containing more than Two Unknown Quantities. x+t2Qy+32z = 382 PrRoBLEM. Given {22+ y+4z = 37> to find 3x+5y+2z2 = 49 the values of x, y, and z. SoLutTion. Here there are three unknown quantities. If we eliminate one of them, we will have only two. But when we have two unknown quantities, we must have two different equations. To get these we will have to perform two processes of elimination. Here the coefficients of x are smallest, and such as to make it best to eliminate x. This will be best done, first, from the first and second ete and then from the first and third equations. Multiplying the members of the first equation by 2, 2x+4y+ 6z= 64. The second equation is, 2% + y+ 4z= 87. Subtracting, 3y +2z= 27. ELEMENTARY ALGEBRA. 165 Multiplying the members of the first equation by 3, 32+ 6y+ 9z= 96. The third equation is, 8x2+ 5y+2z = 49. Subtracting, . yt+tz= 47. By these two eliminations of x, we have obtained two equations with two unknown quantities: oy +22 = 27. y tz = 47. These can be solved as has been previously done. Multiplying the members of the last equation by 3, The equation preceding it is 3y +22 = 27. Subtracting, LO 2c: z= 6 Substituting 2 times 6 for 22, we have 3y + 12 = 27. 3y = 27 —12, dy = 15. y = 5. In the first equation, substituting the values of y and z, x + 10+ 18 = 82. x = 32 —10 — 18. © = 4, Therefore « = 4, y = 5, and z = 6. General Rule for Elimination. Eliminate one of the unknown quantities from as many different pairs of the given equations as there are other un- known quantities. Do the same with the new equations obtained, till finally but one unknown quantity will remain in one equation. Find the value of this, and then of the others by substi- tution in previous equations. Nore. There must be as many equations as there are unknown quantities, and each equation must be used in one of the elimina- tions. The equations must be independent. See page 90. 166 ELEMENTARY ALGEBRA. EXAMPLES. Find the values of the unknown quantities in the follow- ing equations. e+ y+2z= 115. 1. xrt2Qy— z= 70. =e, Ans. % = 35; y = aUsieenee Ans. = 20; y= 8; 2=3. Ta—Qy+2z= 23. 4, x— yt+z= 12. Qea+ yt2z2= 29. Ans. 7 = 3; y = 77 = 21. x—9y + 8z= — 6. —— 6. Ans. 2 = 383; y= 7; 2=6. ce hae or ctiaya eens 11, Sat a ee 6. Ania 9. a +f +2 = 183 Ans. & = 12; 4 =e Y : TROD oh 259 y iA TR TF oe 4. L+y tz = 39. Ans. x's 24 34) =O eee ELEMENTARY ALGEBRA. 167 PiU es ihm p Ati | hl Mie ae — tae PE ety + z=2137. 11. 4¢a—ytz= 7. Sty t= 145 Ans. = 73; y= 15; 2 = GH. Norse. In this example, add the members of the second equation to those of the third, and find z; subtract the members of the sec- ond from those of the first, and find y; subtract the members of the third from those of the first, and find z. et yt4z= 12. 2} xt Qy+ = 2e+ 8y4+42= 24. is Cee 4 of ae Aaa Yy Zz 13.4 2+2y+2z=10 gtyti2z= 9 AR ee et) coed a cee, 168 ELEMENTARY ALGEBRA. sy+2z= 28. Ans. 2 = 5; yo le Zee ( elytra alt 15. 4—ax2—ytz=—1. —a+y—z=—3. Ans. 0 == 2: eee z= 4, Notr. This example may be solved in a similar manner to that pointed out for the 11th. SECTION XLVI. Questions producing Equations with more than Two Unknown Quantities. 1. A Boucut three houses. The price of the first plus } the price of the other two was $2200. The price of the second plus 4 of the price of the other two was $2300. The price of the third plus 3 of the price of the other two was $3000. What did he pay for each? Ans. He paid $1100 for the first, $1900 for the second, and $2500 for the third. 2. A and B together have 2 as much money as C. A and C together have 43 times as much as B. A has $6600 less than B and C together. How much has each? Ans. A has $900; B, $1500; and C, $6000. 3. A’s money exceeds ,;% of B’s and C’s by $100. B’s money exceeds 4 of A’s and C’s by $100. C’s money ex- ceeds ;% of A’s and B’s by $100. How much has each? Ans. A has $800; B, $900; and C, $1000. 4, A number expressed by three digits, gives 96 for a quotient when divided by the sum of the digits minus 9. ELEMENTARY ALGEBRA. 169 The middle digit is half the sum of the first and third. If 396 be subtracted from the number, the remainder will consist of the same digits as the first, but in an inverted order. What is the number? Ans. 864. Note. Let the digits, in order, be represented by 2, y, and z. Then 100 z + 10y + z represents the number. 5. A number expressed by three digits is 22 times their sum. If 99 be added to it the result will consist of the same digits as the number, but in an inverted order. The middle digit is equal to the sum of the first and third. What is the number? Ans. 132. 6. Find four numbers such that once the first, } of the second, 3 of the third, and } of the fourth shall be 23; twice the first, once the second, 3 of the third, and 3 of the fourth shall be 41; three times the first, twice the second, once the third, and three times the fourth shall be 111; and the sum _ of the numbers shall be 51. Ans. 8, 12, 15, and 16. 7. Find three numbers such that the first with ;3, of the difference of the other two shall be 42; the second with ¥, of the sum of the other two shall be 120; and the sum of the three shall be 204. The first number is the smallest, and the third is the largest. Ans. 24, 60, and 120. 8. A certain sum of money was divided among A, B, and C, so that A’s share exceeded ;4, of the sum of B’s and C’s shares by $200; B’s share exceeded # of the sum of A’s and C’s shares by $300; and C’s share exceeded ;, of the sum of A’s and B’s shares by $400. What was the share of each? Ans. A’s share was $600; B’s, $500; and C’s, $800. 9. If A’s age be added to B’s, the sum will be C’s age. If A’s be subtracted from B’s, the remainder will be 3; of C’s age. The sum of their ages is 140 years. How old is each? Ans. A’s age is 34 years; B’s, 36 years; and C’s, 70 years. 10. A’s money added to ,*, of the sum of B’s and C’s 15 170 ELEMENTARY ALGEBRA. equals $55,000. B’s added to ;8 of the sum of A’s and C’s equals $80,000. C’s added to ;3; of the sum of A’s and B’s equals $60,000. How much has each? Ans. A has $30,000, B has $40,000, and C has $45,000. 11. A certain sum of money is to be divided among A, B,andC. A is to receive $1000 less than 4 of it; B, $1000 more than } of it; and C, $1000 less than 4 of it. What is the sum, and what is each to receive? Ans. A is to receive $5000 ;- B, $4000 ; and C, $3000: in all, $12,000. 12. A farmer has sheep in three pastures. The number in the first, added to 4 of the number in the other two, makes 46. The number in the second, added to 4 of the number in the other two, makes 40. The number in the third, added to ;4; of the number in the other two, makes 30. How many are in each pasture? Ans. There are 35 sheep in the first, 30 in the second, and 25 in the third. 18. A certain loaf of bread is made of flour, rice, and water, and weighs 11 lbs. The weight of the rice augmented by 1 lb. is + of the weight of the flour and water together, and the weight of the water is 3 of the weight of the flour and rice together. Required the weight of each. Ans. The rice weighs 1 lb.; the flour, 5 lbs.; and the water, 5 lbs. 14, There is a number consisting of three digits, the first of which is } of the sum of the other two. If 99 be added to the number the digits will be inverted. The middle digit is 1 more than the sum of the other two. What is the num- ber? Ans. 142. 15. A merchant sells 2 lbs. of his best tea, 5 lbs. of his best coffee, and 7 lbs. of his best sugar for $7.40. He sells 1 lb. of the tea, 2 lbs. of the coffee, and 5 lbs. of the sugar for $3.80. He sells 3 lbs. of the tea, 2 lbs. of the coffee, and 4 lbs. of the sugar for $7.60. What is the price of each arti- cle? Ans. The tea is $2 per lb.; the coffee, 40 cents; and the sugar, 20 cents. : 16. Divide 50 into three such parts that 3 times the first ELEMENTARY ALGEBRA, bil shall be twice the sum of the second and third; and the second plus 14 shall be equal to the sum of the first and third. Ans. 20, 18, and 12. 17. A’s age is twice the sum of his son’s and his daugh- ter’s. The three ages amount to 60 years; and 3 times the daughter’s plus 4 years is equal to the difference between her father’s and her brother’s age. What is the age of each? Ans. A’s age is 40 years; his son’s, 12 years; and his daughter’s, 8 years. 18. A drover has oxen, sheep, and hogs; in all, 196 ani- mals. Jour times the number of oxen is equal to the num- ber of sheep, plus the number of hogs, minus 16. Four times the number of hogs is equal to the number of sheep, plus the number of oxen, plus 4. How many has he of each kind? Ans. He has 36 oxen, 120 sheep, and 40 hogs. 19. Four boys bought a foot-ball for $2.00. The first paid $1.00 less than the other three together. Two-fifths of what the first and second paid, plus 35 cents, is 3 of what the third and fourth paid. Two-thirds of what the first three paid, minus 65 cents, is 5 of what the fourth paid. What did each pay? Ans. Each paid 50 cents. 20. The population of four towns amounted to 7000. The population of the first and third together was equal to that of the second and fourth together. The population of the first and second together was 1000 less than that of the third and fourth together. The population of the first and fourth together was 2000 more than that of the second and third together. What was the population of each? Ans. The population of the first town was 2000; of the second, 1000; of the third, 1500; and of the fourth, 2500. 172 ELEMENTARY ALGEBRA. SECTION XLVII. Peculiar Equations with more than Two Unknown Quantities. Prosiem 1. Given 4 ~ ty + w= 22 to find the values of x, y, z, and w. Sotution. It may be observed that one letter is wanting in each equation. There is no w in the first equation, no z in the second, no y in the third, and no 2 in the fourth. Adding the corresponding members of all the equations, 8x2+ 38y+ 3824+ 38w = 90. Dividing by 3, “ ety+tz2+uw = 30. Now by subtracting the members of each of the original equations separately from the members of this last, the values of the unknown quantities can be found. Thus: xty+z+w= 30. x+y+z =. 21. 9; cety+tz+w= 30. x+y + w= 22. Zz ra ety+tzt+w= 380. x +z+w = 23. y asi Ji xatytz+w= 30. ytez2t+u= 24. x = 6, Ww ELEMENTARY ALGEBRA. 173 1 1 1 Zhao anes ede ome VA Ud Big goeat tot Ne: ; 3 RAD £9 | PRoBLEM 2. Given < —- + —-+ -= 34 7 to find the Bi tt eee 4° § 2.3 Gi gueae s values of x, y, and z. Soturion. Here it will be seen that clearing of fractions would give quantities containing zz, ry, yz, and xyz, making the solution difficult. The equations can be best solved without clearing of fractions till an equation is found with only one unknown quantity. Multiplying both members of the first equation by 4, we have A ae AS Soy ee ae The second equation is 2 -- -}- semees tt tpi RIERA AS 2 : i gay 5 Subtracting, - ie ; = = Sar. 5G. Oe The third equation is aD f na ae Multiplying both members of 3 4 3 rp 3 13 the first by 3, py cy ah PAs Thriaid 13 Subtracting, S83 15 Now, taking the two equations that contain x and y¥ only, this iis age api $9: i Lae a 5 . fon Subtracting, es Clearing of fractions, a Or, y= 4, 15* 174 ELEMENTARY ALGEBRA. | Substituting 4 for y in the equation : + : = a we have, Lie 5 Pay bs: T ; 1 op 2 ransposing, 5 oR ee ee? Bic Clearing of fractions, a= Mel Or, v= 38: Substituting the values of x and y in the first equation, lL) doa we have Siar + ete) Transposing 1 - : z° : 1295050 1 ae 2 aan Clearing of fractions, Das eee Or, z=s 2, xyz = 40 Prosuiem 38. Given oh a a to find the values of yzw = 8 X,Y, 2, and w. Soxtution. Multiplying the corresponding members of both equations together, we have, «x*y*z’w* = 64000. Extracting the cube root of both members, axyzw = AO. Dividing by the members of the first equation, es Dividing by the members of the second equation, Z == 2 Dividing by the members of the third equation, y = 4, Dividing by the members of the fourth equation, 7 2s ELEMENTARY ALGEBRA. 175 EXAMPLES. Find the values of the unknown quantities in the follow- ing sets of equations. (atytz+itutv+w= 28. etytz+tt+utv—wel4 etyt+z+titu—v= 9. l + 6 a°b? + 4ab* + b¢ = the fourth power. So to raise any polynomial to any power, take that poly- nomial as a factor as many times as the exponent of the power indicates. EXAMPLES. 1. Square 3 a‘e; that is, raise it to the 2d power. Ans. 9 a®c?. 2. Square — 3 ate. | Ans. 9 a®c*, 21. Ans. x? + 2xy +7? = the 2d power. ELEMENTARY ALGEBRA, 211 . Cube 7 a’xyz*; that is, raise it to the 3d power. Ans. 3438 abc? y2"4, . Cube —7 a’ayz’. Ans. — 348 abr? y*z”4, . Raise 8 ac’x*y*z* to the 4th power. Ans. 4096 ate®x?y'62™, . Raise —2acx to the 5th power. Ans. — 32 a’cbx', . Raise 2 acx to the 5th power. Ans. 32 aécia°, . Raise 7 m to the 2d power. ° Ans. 49 m’. . Raise a to the 27th power. Ans. a™. . Raise — a to the 27th power. Ans. —a”™, . Raise — 2 xty*z to the 7th power. Ans. —128 xyz", . Raise — 3 ab*c*u’e to the 6th power. Ans. 729 ab*cld'e°, . Raise ab to the 23d power. Ans. ab”, . Raise — a’b to the 23d power. Ans. — ab”. . Raise mn’c to the 8th power. Ans. m'n'®e°, . Raise — axz"y'™ to the 9th power. Ans, —a®a”y™, . Raise —10 aay" to the 10th power. Ans. 10,000,000,000 a1, . Raise 4 c'n™y? to the 5th power. Ans. 1024 c*n®y”, . Find the 3d power of — 8 cy’. Ans, — 512 cy”. . Find the 2d power of — 8 cy’. Ans. 64 c%y". Raise 2 + y to the 2d, 3d, 4th, 5th, 6th, and 7th powers. eo + 82°y + 8ay’? + y* = the 3d power. att 4a%y + 6a’y’? + 4ry> + y* = the 4th power. a + Saty + 10a*y? + 1l0ey’ + baytt+y = | the 5th power. a + Gaby + 15 aty? + 20 a2°%y? + lda’y* + 62xy° + y® = the 6th power. x’ Taby + Wary’? + 85 aty® + 35 a*y* + 21 xy? + 7 ary® + y' = the 7th power. yyy ELEMENTARY ALGEBRA. 22. Raise « — y to the 2d, 3d, 4th, 5th, 6th, and 7th powers. Ans, 2? —2xy + y’ = the 2d power. x* —3 ay + 38axy’? —y* = the 3d power. at —42°y + 62°y? —4ay’ + y* = the 4th power. aw —b ay + 10 2y? —102’y> + dry* —y = the 5th power. a —6ay + 15 2ty? — 20 ay? + 15 x’*yt —6 27° + y° = the 6th power. x? —Ta’y + 21 ay? — 35 aty® + 35 x8y* — 21 ay? + 7xy° —y' = the 7th power. 23. Raise 2ab + 3d? to the 3d power. Ans. 8 a*°b* + 36 a’b’d’ + 54 abd* + 27 d®. 24. Raise 2ab — 3d? to the 3d power. Ans. 8 a*b? —36 a’b'd? + 54 abd* — 27 d*. 25. Raise 5 ed + n to the 5th power. Ans. 3125 ed’ + 3125 ctd'n + 1250 cd'n? + 250 c’d’n® + 25 edn*t + n’. 26. Raise 5 cd —n to the 5th power. Ans. 3125 ed’ — 3125 ctd'n + 1250 cd*n? — 250 c’d?n* + 25 ednt — n°. 27. Raise x + y —z to the 3d power. Ans. 24+ 382°y —82%4 y4 382ay —38y2z2—24 3x2’ + 3 yz”? —6 xyz. 28. Raise a + 6 —ec + d to the 2d power. Ans. a + 2ab + 0? —2ac + 2ad + &—2be + 2 bd —2ed + ad’. 243 cx’ my” Scones 29. Raise —- to the 5th power. Ans. my 30. Raise —— to the 4th power. 625 at — 2000 akr + 2400 a2x? — 1280 ax® + 256 xt Ans. “5 — Said + Mai’ — Bae? + 16a, ELEMENTARY ALGEBRA. 213 SECTION: LVI. The Binomial Theorem. THE BrnoMIAL THEOREM gives a short way of writing down, without actual multiplication, the powers of any bi- nomial. Let us take + y and x —y as specimen bino- mials, and consider their powers, as found in Examples 21 and 22 of the preceding section. First. As regards the Sieans in these powers. We ob- serve that in Example 21, when the connecting sign of the binomial is +, all the signs of the powers are +. In Ex- ample 22, when the connecting sign of the binomial is —, the even terms (the 2d, 4th, 6th, and so on) are —. Sreconp. As regards the Exponrents. The first term of the binomial, x, has its leading exponent in each power the same as the number of the power. Thus, the 2d power begins with z?; the 3d power, with 2°; the 4th power, with xz*; and so on. The powers of « then regularly decrease. Thus, in the 5th power of the binomial in both examples, we have first 2°, then x‘, then x’, then x”, and then 2, in the suc- cessive terms. The powers of y, on the contrary, increase. They begin in the second term with y, then we have y’, then y*, then y‘, and then y’, ending as high as 2 began. Leay- ing out the coefficients, the terms would stand 75 + xty + - dra + rs -|- xy + a 4 ago Ly +4- xy? ws xy + xy* ss y’. Tuirp. As regards the Corrricients. The first coeffi- cient in each of the powers may be omitted, as it is 1. In the 2d term, the coefficient is the same as the number of the power. In the 2d power it is 2; in the 3d power, 3; in the 4th power, 4; and so on. This is the case both in Example 21 and in Example 22. Now, it will be found that if we take any term, multiply its coefficient by the exponent of its leading letter, and divide by the number of the term, we shall 214 ELEMENTARY ALGEBRA, obtain the coefficient of the next term. Thus, in Example 22, take the 7th power of « —y. Its 2d term is 7 xy, the coefficient being 7. Multiply 7 by 6 (the exponent of the leading letter, x). The product is 42. Divide this by 2, as we are at the 2d term. The quotient, 21, is the next co- efficient. Continuing in the same way, we may find all the succeeding coefficients. PROBLEM 1. Raise a —n to the 6th power. SoLturion. Since the connecting sign is —, the even terms (2d, 4th, &c.) will be —. The first term will be a®, and the powers of a will decrease, thus : a&—- a+ adé— &+ v— a. Inserting now the increasing powers of n, we have: a— an+ aiv— ani+ a’nt— ani+ n'. As to the coefficients, that of the 2d term must be 6, giv- ing thus far: a® — Gain. Multiply this 6 by the exponent, 5, of the leading let- ter, a, and divide by 2, as we are at the 2d term. 6 x 5 = 30. Dividing by 2, we have 15, giving thus far: a’ —6a'n + 15 a‘n?. Multiply this 15 by the exponent, 4, of a, and divide by 3, giving 20. So we have thus far: a’ —6a'n + 15 a‘n? — 20 abn’. Multiply this 20 by 3, and divide by 4, since we are at the 4th term. We get 15. So we have thus far: a’ —6a'n + 15 atn? —20 a'n®? + 15a’nt. Multiply this 15 by the exponent, 2, of a, and divide by 5. We get 6. So we have thus far: a’ —6ain + 15 atn? — 20 a'n? + 15 a’nt —6 ani. Multiply this 6 by the exponent of a (1 understood) and divide by 6, as we are at the 6th term. Weget1. This need not be set down as a coefficient, since it will be under- stood without actually being there. So we have finally: ELEMENTARY ALGEBRA. 215 a’ —6a'n + 15 a‘n? — 20 a'n3 + 15 a’nt —6 an> + n° This is the 6th power of a —n. It may be observed also that the coefficients vary from the end just as from the be- ginning. So that when half of them are known, the others can be written down. This mode of finding the powers of a binomial was discovered by Sir Isaac Newton. He left no proof of its correctness. Whenever it. has been tested by actual multiplication, the results have been found to be correct. It has also been proved that it must be correct in all cases. The proof may be found in most treatises which include the more advanced and difficult parts of Algebra. PROBLEM 2. Raise «* + 2 y° to the 4th power. Souution. In this example, we must perform the same operations on the terms as if they were single letters whose coefficients and exponents were 1 understood. We have explained how to use the Binomial Theorem in raising a binomial with such simple terms to the 4th power. Thus, we may take a + 6, and find its 4th power. It will be found to be at + 40° + 6a’b? + 4ab? + Ot. Now, instead of a put x‘, and instead of 6 put 2’, en- closing them in parentheses, so as to show what operations are to be performed on them. We then have (at) + 4 (at)? (2y*) + 6 (at)? (2y*)? + 4) ZY?) uj act 2 Y)' It is now seen that to obtain the true first term, 2* is to be raised to the 4th power; to obtain the second term, x‘ is to be raised to the 3d power and the result multiplied by 2 y° and by 4; to obtain the third term, 2* is to be raised to the 2d power, so is 2 y*, and the results are to be multi- plied together and by 6. So on with all the terms. Per- forming these operations, the result is : x 4+ Say + 2aey + 3822ty? + 16 y”. Thus, if each term of a binomial is not simply a single 216 ELEMENTARY ALGEBRA. letter, the Binomial Theorem may be applied to it just as if each term were so; but the terms must be enclosed in paren- theses, and the operations which are indicated must be per- formed on them afterward. EXAMPLES. 1. Raise x + a to the 5th power by the Binomial Theo- remo: | Ans. x + 52ta + 10 aa? + 10 xa? + 5 rat 4+ a’, 2. Raise a — «x to the 6th power by the Binomial Theo- rem. Ans. a& —6a5x + 15 atx? — 20 a3 + 15 a?2* — 6 aa5 + 7°, 3. Raise c —a to the 9th power by the Binomial Theo- rem. Ans. &€ —9 &a + 36 cla? — 84 8a? + 126 bat — 126 ca + 84 ca’ — 386 ca + 9 ca® — a’. 4. Raise a + n to the 10th power by the Binomial Theo- rem. Ans. a + 10 an + 45 abn? + 120 an’? + 210 a®nt + 252 an? + 210 atn® + 120 a'n™ + 45a7n® + 10 an? + n”™. 5. Raise m —n to the 7th power by the Binomial Theo- rem. Ans. m’ —Tm*n + 21 min? —35 mn? + 35 mint —21 mn + 7 mn? —n’. 6. Raise 8 x —6 y to the 3d power by the Binomial Theo- rem. Ans. 27 28 —1622’y + 324 xy? — 216 y'. 7. Raise 2? —y’ to the 8th power by the Binomial Theo- rem. Ans, x —8 x'ty? + 28 c?yt — 56 xy + 70 zy? — 56 ey + 28 xty? —8 x7y"* + "6, 8. Raise 5 « — y to the 5th power by the Binomial Theo- rem. Ans. 3125 a5 — 3125 «ty + 1250 ay? — 250 a’y? © + 22ry* —y’. 9. 10. Ls. ELEMENTARY ALGEBRA. 217 Raise « + ¢ to the 2d power by the Binomial Theo- rem. Ans. x? + 22e + ¢. Raise a + 6 to the 3d power by the Binomial Theo- rem. Ans. a + 3076 + 3 ab’? + 5B. Raise 52 + 3y to the 6th power by the Binomial Theorem. Ans. 15625 a® + 56250 ay + 84875 xy’? + 67500 zy’ 12. 13. 14. 16. ive 18. + 80375 2’y* + 7290 xy® + 729 y’. Raise 2m’? + 5n‘* to the 4th power by the Binomial Theorem. Ans. 16m” + 160 mnt + 600 mén® + 1000 min?? + 625 n’*. Raise 5 p> —2q*° to the 4th power by the Binomial Theorem. Ans. 625 p® — 1000 p*q* + 600 pq* — 160 p'q° + 16q”. Raise 2 ct —3m’n to the 5th power by the Binomial Theorem. Ans. 82c” — 240 c®m’n + 720 c?m*tn? — 1080 c8m*n? + 810 ctm®n*t — 243 mn'. . Raise 3 m’n —2c* to the 5th power by the Binomial Theorem. Ans. 243 mn — 810 ctm®nt + 1080 c&mén3 — 720 c?’m‘n? + 240 cm’n — 32 &”. Raise 1 —v to the 9th power by the Binomial Theo- rem. Ans. 1 —9v + 86v? — 840° + 126 0 — 1260 + 84 v§ — 36 0’ + 9v* —v°, Raise 2 « —1 to the 6th power by the Binomial Theo- rem. Ans. 64 «®§ —192 2 + 240 xt —160 2* + 60 2? —122 + 1, Raise 6 + 1 to the 3d power by the Binomial Theo- rem. Ans. 216 + 108 + 18 + 1. 19 218 ELEMENTARY ALGEBRA. 19. Raise 3 — 3 to the 5th power by the Binomial Theo- 1 (5 . 10. 10) See rem. Ans. Sag + gy > op 20. Raise 5 = ; to the 4th power by the Binomial Theo- yi x ns ad (epee Z 2xn3 n* rem. rvs. 16 6 The “97, a1 Notre. After performing the 18th example by the Binomial Theo- rem, add all the terms together and see if they are equal to the third power of 7. They should be so, since 6 + 1 is the same as 7. SECTION LVII. Evolution. Roots of Monomials. Evolution is the extraction of roots. Hence its processes are the reverse of the processes of Involution. Roots of Monomials. As the 2d power of a’ is a®, the square root of a® is a’. As the 8d power of a? is a®, the cube root of a is a’, And in all cases, as we multiply the index of a monomial by the number expressing the power to which we wish to raise it, so we must divide by the number expressing the root which we wish to extract. Thus the fourth root of a” is a’. To extract the cube root of a monomial composed of sev- eral factors, such as 64 #®y'®z”, is the same as to extract the roots of its factors separately. The cube root of 64 is 4, the cube root of x* is x’,and so on. The result, the cube root of the whole monomial. is 4 x°y5z’. oon oar WN & — he wo © fm OO 14. 15. 16. 17. 18. 19. 20. ELEMENTARY ALGEBRA. 219 EXAMPLES. . Extract the square root of 4 a’z’ . Extract the cube root of a®x"y’. . Extract the 4th root of 16 many’. . Extract the 5th root of 32 cd®e” . Extract the 3d root of 8 m<%d™. . Extract the 8th root of z*y**z®. . Extract the 10th root of np! . Extract the 3d root of 27 din . Extract the 4th root of 81 «*y"z*, . Extract the 5th root of 248 a7y'5z™. . Find the 3d root of 125 2®y*. . Find the 2d root of ne ral Te . Find the 2d root A ie ae b By 16 Find the 4th root of ns Find the 3d root of Hee Zu; ah Find the 5th root of —— ay ye Find the 10th root of ~ a x : : 1 Find the 6th root of 64 ty” Find the 3d root of 27 i Find the 2d root of = ays Ans. 2 ax. Ans, a’xty. Ans. 2 m'xy’, Ans. 2 cde’. Ans. 2 mic*d***, Ans. x*y*z°, Ans. np. Ans. 3 c’d’m® Ans. 3 xy*z. Ans. 3 27782", Ans. 5 x®y", Ans. 10 x*n”™. 220 ELEMENTARY ALGEBRA. SECTION LVIII. Square Roots of Polynomials. To take the square root of x? + 2ay + y’, we must first find the square root of a part of it, for instance, of «. The square root of x’ is x. This is the first part of the root. We must next consider 2xy + y’, the rest of the quan- tity whose square root we are finding. If we double the first part of the root, obtaining 2, and divide the first term of 2xy + y’ by it, we get y, the next term of the root. This may be proved true by actual multiplication: «x + y multiplied by z + y will give 2 + 2ay + y’. We must observe that this square consists of two parts, x? and 2xy + y’. The latter part may be obtained thus: To the 2 a, the partial divisor by which we got the term y, add that term, making 2x + y. Multiply the divisor thus completed, by its last term, y, and we obtain 2 xy + y’. Now what we have said is appiicable to any quantity which has a square root. The first part of the root may be operated on as x was above, and the next term found as y was. This must be true; for x and y may stand for the two parts of any square root, the square of which must then be the same as x? + 2ay 4+ y’. Before extracting the square root of any quantity, its terms should be arranged as in Division. PROBLEM. Extract the square root of 4a* —12a°5y + 29 aty? — 30 ay? + 25 ay. OPERATION. Root. 4 x6 —12 x5y 4 29 xty2 — 80 x3y3 + 25 atyt ( 223 —3 xy + 5zxy? 4 x6 4733 a) —12 ay + 29 xty? — 30 x3y3 4 25 x2y4 —12ay+ 9xty? 4a —6 ay + d xy?\ 20 xty? — 30 x3y3 4 25 x24 20 xty? — 80 x3y8 4. 25 x2y4 — ELEMENTARY ALGEBRA. 921 EXPLANATION. Here the terms are already arranged according to the powers of x, beginning with the 6th, then the 5th, then the 4th, and so on. The square root of 42° is 2%. We set down that much of the root, and subtract the square 4° from the whole power. We then double 2 2’, giving 4°, and divide that into the first term of the remainder. It gives —38a’y. We set down —3zx’y in the root, and also after the partial divisor 42°, and multiply the divisor thus completed by —32’y. It gives —12a°y + 9 x*y’, which we subtract from the part of the power which we are considering. We then double 2 x? —3 2’y, the part of the root already found, and set down as before the product, 4 2* —6 2x’y, as a partial divisor. Dividing, we get + 52y’, which we set down in the root and after the partial divisor. Multiplying the completed divisor by 5 xy’ and subtracting it from the remainder of the power, we find that nothing remains. EXAMPLES. Find the square roots of the following quantities. 1a2+4ay+4y’. Ans. x + 2y. Daa + Jake -- a%s'. Ans. a? + ax. 3. a + Qaix -- x’. Ans. a® + 2, 4. at —6a*xr + 9 2’. Ans. a — 382. 5. a®a? —2 a's’ + ax’. Ans. atx —ax*. 6. ct —10c’d + 25 d’. Ans. & —5d. 7 1+10c + 25e’. Ans. 1+ 5e. 8. 225 + 90m + 9m’. Ans. 15 + 38m. 9. 36 +1241. Ans. 6 +1 = 7. 10.27 + 2ay+2az+yY+2yz4 2. Ans. «© +y +4 19 * 222 ELEMENTARY ALGEBRA. 11. 4a* + 20 ae —4am + 256 —10em + m. Ans. 2a + 5¢—m 12.14+ 684+ 2:4+9%+68st+8. Ans. 1+3848 13.9+ 6m+ 48n + m’? + 16mn + 64n7. Anz. 3+ m+ 8n. 1441+4+104+ 1249. Ani 14+2+3=6. 15. 4074+ 44+ 4ee¢* +14 22 4+ x. ‘ Ans. 2a +1 + 2%. 16. 4é—4é +46é—34é—2é+ 3¢€—2et lL. Ans. 246—e +e—1. 17. a2 + 2ab + 2ace—2ad + & + 2he —2dbd +e —2ed + d’. Ans. a+6+e¢e—d. 18. 2° + 227’ —2*§ + Bet — 224+ 2, Ans. o*+ FP —#+4+ 2. 19. 252° + 202 + 142+*—6e°—3xr—22r+1. Ans. 52° + 227 + « —1. 20. 49 2*—1l4axr* + 15 es? —2e2 + at. Ans. 7 x? —azx + at SECTION LIX. Classification of Equations. Simple equations are those in which the unknown quanti- ties are all of the first power. Simple equations are also called equations of the first degree. The equations thus far given in this book have been of the first degree. Quadratic equations, or equations of the second degree, are those which contain the second power or the second root of an unknown quantity, or the product of two unknown quan- tities; but do not contain any higher power or root, ora ELEMENTARY ALGEBRA. 223 product of more than two unknown quantities. They may also contain the first power of an unknown quantity. If they do, they are called complete or adfected quadratic equa- tions. If they do not, they are called incomplete or pure quadratic equations. Cubie equations contain the third power or root of an un- known quantity, or the product of three unknown factors; but no higher power or root, or product of more than three unknown quantities. Cubic equations are of the third degree. Biquadratie equations contain the fourth power or root of an unknown quantity, or the product of four unknown fac- tors; but no higher power, root, or product. They are of the fourth degree. ; Equations containing higher powers, roots, or products, are of the fifth, sixth, seventh, &c., degrees. A simple equation, x—6 = Tx —50 —2z. x? = 36 Pure quadratic equations, nn estye xy = 48 x? + xy = 52 ( om Am =i Adfected quadratic equations, | os i fe a a0 Liy + %at = 408. A eubic equation, et 24+ x= 39. A biquadratie equation, z*—5aS+ 3 = 64. An equation of the fifth degree, a es eee An equation of the sixth degree, Fa? —32' + ot—47 —2' + 2—4= 1, 224 ELEMENTARY ALGEBRA. SECTION LX. Pure Quadratic Equations, and others Similarly Solved. PROBLEM 1. Given x’? = 49, to find the values of a. SotutTion. Any square number, such as 49, has two square roots; for, + 7 multiplied by + 7 gives 49, and — 7 multiplied by —7 gives 49. Consequently, the value of x may be either + 7 or —7, not both at once, but either one. The operation is generally set down thus: a? = 49. The double sign should be read plus or minus. ProsLEeM 2. Given (x + 3)? = 6a + 34, to find the values of x. Soxtution. This is not strictly a pure quadratic equa- tion, as it contains the first power of x. But, as will be seen, it can be easily reduced to one. (2 + 8)? = O@ + 34. Squaring thew +3, 27+ 6x2+9= 62 +4 34. Cancelling, x? +9 = 34. Transposing, x? = 84 —9. Chee Taking the square root, est 8. Hence x may either equal + 5 or —5 in the given equa- tion. By substituting either + 5 or — 5 for # in that equation, the results will be found to be equal, thus proving the answer. The true value of an unknown quantity, when substituted in an equation for that quantity, is said to satisfy the equa- tion. That is, the equality is preserved. ELEMENTARY ALGEBRA. 225 : a PROBLEM 3. Given 5 — a = le =; aS : + - to find the values of 2. Sotution. As this stands, it is a simple equation. By clearing of fractions, it becomes a pure quadratic equation. ee a [We ae Saa eee GAS 36 mem Gre OF Clearing of fractions, 4%? — 2? = 2592 — 9x’ + 6x* + 42x’. Transposing, Ag? — x? + 9x? — 6a* — 42? = 2592. 2a* == 2592. we == 1200: Propiem 4. Given /9 4+ 2? = 5, to find the value of x. SoLuTION. J/9 + x2? = 5. Squaring both members, 9 + a2? = 25. a = 25 —9. a ea 16: x= +4, It must be known that to square the square root of a quantity is to obtain the quantity itself. Hence, to square a radical quantity, remove the radical sign, provided it is the square root which is indicated by that sign. /104—a’ —2x PropiEeM 5. Given a 4, to find the values of a. S V¥ 104 — 2’ — 2 OLUTION. *¢ceeier srapmmpr re ra 4, Clearing of fractions, V¥104 — 2’? —ax = 4a. It will not do to square yet, as the radical quantity does not stand by itself. P 226 ELEMENTARY ALGEBRA. Transposing, Vv 104 —a” = 4x + ov, /104 —a? = 5a. Squaring now, 104 —x? = 252’. Transposing, — 25x? — x2? = — 104. Changing all the signs, 25x? + a? = 104. 26.4 =i: Ere Pf EV he Propiem 6. Given Y 4 + 2V¥ 160 — 82 = x — 2, to find the values of x. Sonution. “4+ 2160 — 8a = x —2. Squaring, 4+ «2/160 — 8x" = a? —4r + 4. Cancelling 4, x/160 —8xr = 2 —4z. Dividing by 2, V160 — 8% = « —4, Squaring again, 160 — 8x = x’ —8xr-+ 16. Cancelling — 82, 160 = a? + 16. —x’ = + 16 —160. a? = —16 + 160. x? = 144. xt use EXAMPLES. 1. Given «2? —15 = 49, to find the values of 2. Ans. c= + 8. 2. Given 5x? + 5 = 50, to find the values of =. Ans. & = = 38. 3. Given 8x? — 12 = 8 + 372’, to find the values of 7. Ans. @ == + 2. 4, Given 7x? —90 + 8 = 45 —15, to find the values of be Ans. & = 4s 5. Given 9x? —4 = 0, to find the values of z. Ans. & == re q. ELEMENTARY ALGEBRA. 227 6. Given 3x? —72 = 22? + 72; to find the values of 2. Ans. x = + 12. 7. Given 28x27? —7 = 28x? — 52, to find the values of z. Ans. « = +3. 8. Given sain + islet = 12, to find the values of x 2—x 242 ‘ k Ans. © = £1. 9. Given 3a? —4 = 121 — 2z’, to find the values of za. Ans. c= t5. 10. Given pie -- bode = 8 to find the values of x . 1l+2z2 -l1—c« 38 : Ans. a — i ze 11. Given (x + 4,? = 8x +4 25, to find the values of x. Ans. © = + 8. 2 12. Given e x? — (52? —16) = es x uo to find the values of x. Ans. © = +5. : Bt 8x 13. Given 3z Tea Se to find the values of z. Ans. x = + 9. 14. Given (4c + 1)? = 2x + 1,4, to find the values of 2. Ans. © = t1 4° ; bai + 3° Zz? xe+2 15. Given 9 ae ma = 4a* + 19” to find the values of 2. Ans. © = +3. 627+ 3 Note. First perform the subtraction re 3 pee Bee , and re- duce the resulting fraction to its lowest terms. A 1 3 108 16. Given Dan? + 8 = gi 1 find the values of 2. Ans. Ci oo ys 17. Given eae + surat B20 —, to find the values rej” gt} of 2. Ans. x = + 8. 228 ELEMENTARY ALGEBRA. 18. Given oe = 45, to find the values of z. Ans. © = + 6. 19. Given “32? + 6 = 9, to find the values of z. Ans. % = + 5. 20. Given “a? + 27 = 2,/9, to find the values of a. Ans. «x = + 8. 21. Given x + 7x = 22%, to find the values of 2. Ans. 2 = + 2, 22. Given “2? + 15 —5 = 3, to find the values of 2. Ans. x = + 7, In the following examples, the equations, when reduced, are not quadratic, and x has but one value. 93. Given “3x + 10 = 5, to find the value of z. Ans. x = 5. 24. Given V 21+ x = 7 — // 2, to find the value of z. Ans. % = 4, 25. Given Vx —5 = ./x —1, to find the value of z. Ans. «© = 9. 26. Given Va —11 = /x + 1, to find the value of 2. Ans. x = 36. 27. Given Vx —9 = ./x —1, to find the value of z. ; Ans. x = 25, 28. Given /3x —11 = 3x —1, to find the value of z. Ans. x = 12. 29. Given “4 +2 Va? —80 = « —2, to find the value of x. Ans. x = 12. 30. Given « + VW 16x + a? = 8, to find the value of z. Ans. x = 2. ELEMENTARY ALGEBRA, 229 SECTION LXI. Equations with Two Unknown Quantities. WHEN pure quadratic equations contain two unknown quantities, various artifices are employed for elimination, the most general of which will appear from the following examples : Propiem 1. Given 2” ey 3 i} to find the values of x and y. SoLution. Squaring the members of the first equation. e+ dey+y’ = i. Subtracting 4 times the members of the second, Ary = 56. x* —Qey + y? = 25 Taking the square root, xe—y = +5. Adding the members of the first equation, e+ ys 9. 22 = 14 or4 x eax OF aw Substituting the values of x in the equation, 7 + y = 9, we have y = 20Y /. Wrosec that if a2 = 7, y = 2; but, if 7 = 2, y =. 7. Prosiem 2. Given ‘a BE de ul } to find the values L—y = of x and y. Soxtution. The first equation is, oe + y? = 65. Squaring both members of the second, a? —Qry + y? = 49. Subtracting, Qxey ‘ox 16, Adding the members of this to the members of the first equation, a? + Qey+ yy’? = 8l. Extracting the square root, at-y = +9. 20 230 ELEMENTARY ALGEBRA. With this equation and the second of the original equa- tions, the values of x and y may be found as in the preced- ing problem. It will be found that z may be equal to 8, in which case y = 1, or x may be equal to — 1, in which case y= —8. | EXAMPLES. 1. Given ‘i La =e . to find the values of # and y. Ans. x = 7 or —1; y = 1 or — 7. 2. Given ie ate cs on to find the values of x and y. Ans. «x = 50r3;y = 8 or 5. 3. Given Fay a " rs ibe to find the values of x and y. ge ke Ans. « = 9 or 3; y = 2 or 36. Norse. This is similar to on two preceding it. 4, Given se re ge a to find the values of x and y. Ans. « = 4or —2; y = 8 or —6. Norts. This also is similar to the preceding, only it is not 4zy that is to be added to the middle term after squaring the members of the first equation, but 24zy, as that is double the middle term. Rips, Sao) 5. Given iy «x "+ to find the values of # and y. v+ y= 82. Ans, © = 2243 y = 4, Nore. Clear of fractions, and then find what 2zy equals. 6. Given 13, ihe is 1 to find the values of x and y. Ans. «== 19 See a+ y’ = 10. gah y meee. t to find the values of x and y. 7. Given | Ans. & = 380r1; y=1lor3. ELEMENTARY ALGEBRA. 231 8. Given he ms = 240.) 14 find the values of 2 and y. tr me $25 Ans. © ese 435 y= & 2, Note. First divide the members of the first equation by those of the second. 9. Given 1 =| sid 3. to find the values of x and y. | Ana 2226: Y= 2. : c —y’ gs WA 10. Given | oy 8. to find the values of x and y. Anas 2 22:6) .y =2'7. SECTION LXII. Adfected Quadratie Equations. IF we square a monomial, the result isa monomial. Thus, if we square 7a’x*, we get 49a‘tzx”. If we square a binomial, the result is a trinomial. Thus, if we square x + a, we get az? + 2ax + a’. Reversing these operations, we see that we can take the square root of a monomial or of a trinomial. We cannot, however, take the square root of a binomial. Let us consider the equation w -+- 62 = 16. We cannot take the square root of x’ + 62, since it has but two terms; but we can add something to it which will make it a complete square, and then take the square root. By Theorem I., page 96, the square of the sum of two quantities as equal to the square of the first, plus twice the product of the first and second, plus the square of the second. Now here we may consider x? the square of the first, and 6x twice the product of the first and second. Then 3z will be once the product of the first and second. Dividing by x, which is 932 ELEMENTARY ALGEBRA. the first, we find that 3 must be the second. The square of this is 9, which must be added to x? + 6z, and also to 16, to preserve the equality. 7 a? + 6x = 16. 9= 9. xz’? + 62 + 9 = 25. Taking the square root, £ +o eee c= +5 —8, x = 2 or —8. The operation we have just explained is called completing the square. Prospiem. Find the value of x in the equation a+ x—l1 3 re eel a So.tution. This equation is not quadratic as it stands, but becomes so when cleared of fractions. Clearing of fractions, . 227 + 4a + 2 = 2x? —4a + 2 + 382? — 8. Collecting, — 3x7 + 8x4 = — 3. Changing signs, 382° — 8& = 3. Now, as the first term is not a perfect square, it must be made so. This can always be done by dividing by the coef- ficient of x’. Performing this division, go — ba = I, In order to complete the square, consider that 2 must be the first term of the root, and —8z twice the product of the two terms. Then — 4x must be once that product. Divid- ing by x, which is the first, we find that — #4 must be the second. The square of this is 4°, which must be added to x” — 8x, and also to 1, to preserve the equality. x ee 8a =_ i 6 Taking the square root, ELEMENTARY ALGEBRA. 2aG emts + FZ. xv = 3 or —}#. These operations should be proved by substituting each of the values of x separately in the original. equation. EXAMPLES. Find the values of x in the following equations: 10. th 12. 13. 15. 16. 17. 18. 19. . 2+ 8r = 20. . a —2r = 3. a+ 4a = 45. ee — 4x = 12. x? —10x = 24. mae 2 = 90. a+ a= 10 “+e = 3 .a—x=6 xz’ — 102 = 56 xe? + Tx = 60. x? — bx = 24. xr+l x—d peeeey 3°10 x—3 2x£—24 dy. = 7800 ha? — 3x0 + T = 42? + 5. 3x0? —9x = 12. x? —13x2 = — 40. x4+2 8 iegecinmet 4x? —9a = 385. ers ig, 2 20. 1 2 3 4 5. 6 (| 8 9 20 * Ans. x« = 2 or —10. Ans. x = 3 or —1. Ans. x = 5 or —9. Ans. x = 6 or —2. Ans. « = 12 or —2. Ans. x = 9 or —10. Ans. x = % or —8. Ans. x = 4 or —3. Ans. x« = 3 or —2. Ans. x = 14 or —4. Ans. « = 5 or —12. Ans. x = 8 or —3. Ans. x = 20 or 1. Ans. « = 18 or 2. Ans. x« = 2 orl. Ans. x = 4or—l. Ans. x = 8 of 5. Ans. x = 2 or — 5h. Ans. x = 11 or — ¥. Ans. « = 5or0, 234 ELEMENTARY ALGEBRA. Nore. In eee 20, if the value 0 be substituted for x, we shall have e 0+0 oT =0+6. This reduces to . ~ = 6. This equation cannot be said to be false, as 6 times 0 are equal to 0. It is, however, meaningless, as far as the science of Algebra is explained in this book. In more advanced treatises, a meaning is found for it, and the fraction $ is said to have any value whatever, since any number of times 0 are equal to 0. SECTION LAXIII. Miscellaneous Examples. Review Questions. 1. Drv1pE a line 54 inches long into two parts differing by 14 inches. Ans. The parts are 20 inches and 34 inches. 2. Divide $118 among A, B, and C, so that B shall have $i2 more than A, and C $10 more than B, Ans. A must have $28; B, $40; and C, $50. 3. A laborer was engaged for 25 days. For each day he worked, he was to receive $1.00 and his board. For each day he was idle, 50 cents were to be deducted for his board. At the end of the 25 days, he received $17.50. How many days did he work, and how many days was he idle? Ans. He worked 20 days, and was idle 5. 4, A hare is 25 of her own leaps in advance of a grey- hound, and takes 11 leaps while the greyhound takes 10. Five of the greyhound’s leaps are equal to six of the hare’s. How many leaps must the greyhound make to overtake the hare? Ans. 250. Notre. If 5 of the greyhound’s leaps are equal to 6 of the hare’s, ELEMENTARY ALGEBRA. 235 10 of the greyhound’s must be equal to 12 of the hare’s. Thus the greyhound gains one of the hare’s leaps while taking 10 of its own. 5. Divide 45 into 3 parts, so that the second may be three times, and the third five times the first. Ans. 5,15, and 25. 6. Divide 45 into 3 parts, so that the second may be double the first, and the third three times the second. Ans. 5, 10, and 380. 7. A’s age is half of B’s. B’s is three times C’s. The sum of their ages is 110 years. What is the age of each? Ans. A is 30 years old; B, 60; and C, 20. 8. A, B, C, and D have $100,000. B has twice as much as D. C has as much as A and D together. 9 bined alt 5 Scie 11 . Add 8z7*a—*, -— —~, 9=,-, and a x akx 0 . Add 15a°x-5, oe shee ——, and #—*, a? 2 a Ans. “= 242-, From 2a’n—'— y~*ax + 1 take 1 — 9axy—? — 3n~'a’. Ans. Bann! 4 Bary? = n y From 162—?— 5y~? take 152-7 + > ‘ 2 ol ees Ans. x~?—8y Bea Ta 5a —3 —4 MEADS, Siem ee From az—* + ay~‘ take Sorry 6a , 6a a =8 obese Sly Ans. —6aa—* + 6ay = + 7 From 162~*y—* take i iiss z rag) 3 16 Ans. 16a-*y—* — 5abys = pe Multiply 25 wWe&y’x by — 3 a*zy°ex. Ans. — Th wv exy’. Multiply a—* by a-*. Ans, a~* = as 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. ELEMENTARY ALGEBRA. 243 Multiply 5 a%x—*— 3 a’e—*a? + 15 ¢%a* by 2a "a —*. Ans. 10a71cetx—* — 6 ace—24—* + 304a7—'e%2? _10¢ 6a , 30¢%" att eat us a Divide a~* by a~*. Ans. a. Divide c~*x—*y? by e’a—*y, Ang. 6 Ry == Z, Divide x” by x”, Ans, 2*—" = =e. Divide x” by x”. Ans, 2 =I, Add x?,—17 27, 9/2, 5x?, and 13,/z. Ans. 11 a? ae ae Add 13a*, —53/a,—T72/a, 214/a, and a’, Ans. 8a = 234/a, Add / az and ata. Ans. 2/ ax = Qatx?, Add /az and phe Ans. Jax + on From 29 atxty? take 16 Jaixy. Ans. 13Jaay = 13a'x'y?, Multiply —17 ata? by — Qata8, Ans. 34a°x® = 34 aad Multiply at —atat ao ot by at + aot Ans. oe + ot = Ya + 2%, Divide ax? by —5a%x?, $ | Lr Ans. a a ‘a= — = —; a 5a ° 244 ELEMENTARY ALGEBRA. 29, Divide —17a'x* by —2a%a?. 17 3 ae Ans. aa = i ae 80. Divide at —2* by at —2’. Ans. a+ atat + ata? + ata? at a ee Ans. 9a? — 30a'*x® 4+ Gata + 25 atat —10 ata + at. + 2x. 31. Square 3at —5aé od | —1l1 32. Square ee a*"— 2x i "SO 9a? + at m (a —2 ax? + at) 33. Find the sq. root of a7? + 2a—'b—* —2a7*e-* + b-§§—2b-%e-* + e-8, Ans. a-! + b-§—e-+, 34. Raise — Bat bic? to the 2d, 3d, and 6th powers. Ans. 25ab%c®, —125a2b%, and 15625 a°b°ct 35, Find the 2d and 3d roots of 64a7a—*. t 4 Ans. Bata = oe and bgt = an SECTION LXV. Radicals. A radical expression is an expression containing a radical 2 sign or a fractional index. Thus, ./x and c* are quadratic radicals or radicals of the 2d degree; Wa* and 3a* are cubic ELEMENTARY ALGEBRA. 245 radicals or radicals of the 3d degree; 4/y and an* are bi- quadratic radicals or radicals of the 4th degree. Any algebraic expression may be reduced to a radical form, as the following examples will show. PROBLEM 1. Reduce mn + y to a quadratic radical. Sotution. The square of mn + y is m’n? + 2mny + y. Hence the square root of mn’ + 2mny + y’ is mn + y. That is, mn + y=V mn? + 2mny + y’. PROBLEM 2. Express 3x+/ a—-x entirely under the radi- cal sign. Sotution. The 3d power of 3x2 is 27x%. Hence the given radical is the same as 7/272°x Ya—zx. Multiply- ing together 272* and a—z, we have 27az*— 272+. To take the cube root of this is the same as to take the cube roots of its factors and multiply them together. That is, 3/27 ax */a—= is the same as 3/27 ax® — 27 x. GENERAL PRINCIPLE OF EyouutTion. To take any root of a quantity is the same as to take the same root of each of its factors and multiply the results together. EXAMPLES. 1. Reduce 8 to a quadratic radical. Ans. / 64. 2. Reduce 4 to a cubic radical. Ans. - Ans. 2/15. 3 5 eo afm Ans. qv. vy) J Ans. 2/7. 6 ~. Ans. aie . mn \ =, Ans. m /amn. . Vabe \/ a - Ans. 5 abe, 2 ; f=. Ans. Sn. 5a ss /35 ax ‘ 72 ns. nda 3/1 ; 1 ; P \ 2° Ans. a V4 8 2 1 P 5° Ans. 5 00. ; 2 Ans. ave 8 1325 1, we Ans. 9 v 680. tf 4 28 at ee: Ans. y Set cy cy /*. ule: Mea nee Zs [p—q Ane VOR ODIED Be ae 3 7— V3 PN pita enn 0 5—/5 Pee 38— V9 252 ELEMENTARY ALGEBRA. 24. o4+ 73° Ans. 11 —6 Jd. 5+ /2 | 25. Bey 97 Ans. 19 -b 13 Jas 1+ V3 26. Raa Ans. 2 + Jd. Sih tee TA Ans li a— f/x a—ax i 3— fT er ey Hoi! Pewee 28. Ji-3 ns 5 5+4/2 30 + 28 f/2+15/5+4 12 410 29. fas 75, Ans. ———— a —_—_———__—., ee Ju, 2 m+ /n m—n —0$-¢.0-0-——_——_ SECTION LXVIII. Reduction of Radicals to a Common Index. ProspitEM. Reduce v/a, c’, and 4/5 to a common index. SotutTion. Expressing the radical quantities with frac- 1 tional indices, we have a, ce’, and BF. Reducing the frac- tional indices to others having the least common denomi- nator, we have xu, #3, and 7, instead of 4, 4, and +. The 4A radical quantities thus become a’’, oft, and Bt, which may be expressed ¥/a', ¥/c®, and 4/125. » EXAMPLES. Reduce the following radicals to others having a common index. | Lisd/a, /a. Ans.