The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031263803 Comall University Library arV19274 New elementary algebra 1924 031 263 803 NEW ELEMENTARY ALGEBRA: IN WHICH THE FIEST PEINCIPIES OP ANALYSIS ABE PKOGRESSIVELY DEVELOPED AND SIMPLIFIEDl COMMON SCHOOLS AND ACADEMIES. bt benjamin greenleaf, a. m., AUTHOR OF A MATHEMATICAL SEBIES. ELBOTKOTTPE EDlTIOlf. BOSTON: PUBLISHED BY ROBERT S. DAVIS & CO. NEW YORK: OAKLET S MASON, AND A. B. BABHE8 t CO. PHILADELPHIA: J. A. BANCROFT » COMPANY. BT. LOmS: KEITH AND WOOD8. CHICAGO: B. C. OBIGGS LEs 293 304 311 818 ELEMENTARY ALGEBRA. DEFINITIONS AND NOTATION. 1. Quantity is anything that can he measured; as dis- tance, time, weight, and number. 2i The Unit of quantity is one of the same kind as the quantity, taken as the standard, or unit of measure. 3i Mathematics is the science of quantities and their relations. Letters and other characters used in mathematics, to indicate quantities, and signs used to denote their relations, or operations to be performed, are called symbols. 4< AxGEBBA is that branch of mathematics in which quanti- ties considered are represented by letters or other symbols ; their relations %re imvestigated, and the reasoning is abridged, by means of signs. 5, Addition is indicated by an erect cross, -|-> called plus. ThuSj 10 -f- 4, read ten plus four, signifies that 10 and 4 are to be added. 6, Subtraction is indicated by a short horizontal line, — , called ., mmws. Thus, 10 — 4, read ten minus four, fiignifiep that 4 is to be subtracted from 10. Define Quantity. Unity of Quantity. Mathematics. Algebra. How ia Addition indicated'! Subtraction 1 8 ELEMENTARY ALGEBRA. 7i Multiplication is indicated by an inclined cross, X- Thus, 8X2 signifies that 8 and 2 are to be multiplied together. In Algebra, the inclined cross is usually omitted, ex- cept between two arithmetical figures, separated by no other sign, and the absence of any sign indicates multi- plication. Thus, a b signifies that a and b are to be mul- tiplied together. Note. Sometimes u period is used in place of the inclined cross ; but this should never be done when there is danger of mistaking it for the decimal point. Thus, a . b signifies that a and b are to be mul- tiplied together, and 2.3.5.7 indicates that 2, 3, 5, and 7 are to be mnl- tiplied together ; but 2 . 3 would bo read two and three tenths, unless the connection made it obvious that multiplication was intended. 8. Division is indicated by a horizontal line, with one dot above and another below, -4-. Thus, 8 -;- 2 signifies that 8 is to be divided by 2. Division is otherwise often indicated by writing the dividend above and the divisor below a horizontal line, in the form of a fraction. Thus, f signifies the same as 8 -f- 2. Note. In expressing the ratio of two quantities in a proportion, the line of the sign -r- is omitted, and the two dots (:) are used to imply a division of one quantity by another. 9t Equality is indicated by two short horizontal lines, =, Thus, 10 -f- 6 = 16 signifies that the sum of 10 and 6 is equal to 16. Note. In writing a" proportion, the equality of ratios is usually indi- cated by four dots (:;). lOt Inequality is indicated by the angle >, or <, the opening being towards the larger quantity. Thus, 12 -f- 5 > 14 signifies that the sum of 12 and 5 is greater How is Multiplication indicated? Division? Equality? Inequality? DEFINITIONS AND NOTATION. 9 than 14; and 14 < 12 + 5 signifies that 14 is less than the sum of 12 and 5. 11. A Parenthksis, ( ), or a Vinculum, , is used to include quantities which are to be considered together, or subjected to the same operation. Thus, (4-}-2-f-6) X 3 signifies tha t the s um of 4, 2, and 6 is to be multiplied by 3 ; and 9 — 5 -r- 2 signifies that the difference of 9 and 5 is to be divided by 2. ' Examples. 1. 9 -|- 16 -|- 11 indicates how many? Ans. 36. 2. 17 — 6 -f- 3 indicates how many ? 3. 25x3 + 6 — 2 = how many? Ans. 19. 4. 56-5-7-1-2 indicates how many ? 81 -I— 9 5. -^ [- 20 = how many? Ans. 29. 10 ~25 6. — -J 1 = how many? Ans. 1. ^ 16 Xll 32-1-13 . ,. , , 5, 7. — 7 ^ — mdicates how many? 8. Find the value of (17 — 5) X 8. Ans. 96. 9. Find the value of 108 -|- 12 -=- (16 — 4.) 10. Add ^^ and "^ + ^^ -f 41. Ans. 50. 11 25 11. Subtract ^^^ ~^q ^ ^" from (81 — 63) X 6. 12. Show that ^J -^ 14 > 3 X 4. 13. Show that (1137 — 869) -=- 67 < (101 -}- 37) -i- 23. How is a Parenthesis or a Vinculum used^ 10 ELEMENTAEY ALGEBRA. ALGEBEAIC NOTATION. 12. Algebraic Notation is of a mixed character, con- sisting principally of the figures of Arithmetic and of the letters of the alphabet. 13. Figures of Arithmetic are used to represent known quantities and determined values. 14. Letters are used to represent any quantity what- ever, known or unknown. 15. Known Quantities, or those whose values are given, are generally represented by the first letters of the alphabet, as a, b, c. Unknown Quantities, or those whose values are to be determined, are generally represented by the last letters of the alphabet, as x, y, z. 16. Numerical Quantities are those represented by fig- ures. Literal Quantities are those represented by letters. 17. Factors are quantities which are to be multiplied together. 18. A Coefficient of a quantity is a figure or letter prefixed to it, to show how many times the quantity is to be taken. Thus, in 4:a = a-\-a-\-a-\-a, 4 is the coefficient of a, and indicates that a is taken 4 times ; in bx, b is the coefficient of x, and indicates that x is taken b times ; and in 5cy, 5 may be regarded as the coefficient of cy, or 5 c as the coefficient of y. When no coefficient of a quantity is written, 1 is un- derstood to be its coefficient. Thus, a is the same as 1 a, and xy is the same as Ixy. Of what does Algebraic Notation consist ? What are Figures used to rep- resent? Letters? How are Known Quantities represented? Unknown Quantities ? Define Numerical Quantities. Literal Quantities. Factors. A Coefficient. DEFINITIONS AND NOTATION. 11 As ia indicates that four a'a are to be added to- gether, or that a is to be taken four times, it is evident that a and 4 are to be multiplied together. The expres- sion bx also indicates that x is to be multiplied bj"^ b. Hence, when a figure and a letter, or two letters, are separated by no sign, multiplication is understood, and the quantities are to be used as factors. (Art. 7.) KoTE. This method of expressing multiplication cannot be extended to Jigures separated by no sign. Thus, 82 could not be used to denote the product of 8 and 2, because that form is already appropriated in Arithmetic to eighty-two, or the sum of 8 tens and 2 units. In such expressions as 8 (2 -(- 3], multiplication is understood, for the figures 8 and 2 are separated by a sign. 19i An Exponent is a figure or letter written at the right and above a quantity, to indicate the number of times the quantity is taken as a factor. Thus, x y, x y. X, or XXX, may be written x', in which 3 is the exponent of x, and indicates that x is taken 3 times as a factor. If the minus sign is prefixed to an exponent, it indi- cates that the quantity is to be used as a divisor. Thus flS 1 a* b~^ c~' is the same as j—;, and or' is the same as 3. Unless a parenthesis or vinculum is used, an exponent affects only the single letter or figure to which it is aflSxed. Thus, in the expression Sab'', the 2 affects only the b. If it were to extend its power to the whole expression,, it would be written thus, {Sab)". Note. It will be observed that the coefficient and exponent both sig- nify how many times a quantity is taken. The one, however, denotes that it is taken as an additive quantity, the other as a factor. Thus, 5 a denotes 5 a's added together, or a + a + a + a + a, while a* de- notes 5 a's multiplied together, or aXaXaXaX a. What does the absence of any sign between two qHSiDtities indicate ? Define an Exponent. How far does the power of an exponent extend ? 12 ELEMENTAEY ALGEBRA. 20. A JPowER of any quantity is the product obtained by taking that quantity one or more times as a factor, and is expressed by an exponent. Thus, a X a = ffl^ read a square, is the second power of a ; a X a X a=^a?, read a cube, is the third power of a ; aXaXaXa = a^, read a fourth, is the fourth pow- er of a. When a quantity has no exponent written, it is under- stood to be the firal power. Thus, a is the same as a^, or the first power of a. Note. The relation of a power to a product is similar to that of a product to a sum. The addition of equal quantities, or of a quantity to itself, is multiplication; and the multiplication of equal quantities, or of a quantity by itself, is raising to a power. Thus, 3o is either the sum of three a's, or the product of 3 and a ; and o' is either the product of three a's, or the third power of a. 21. A Boot of any quantity is a factor which, taken a certain number of times, will form that quantity. Thus, a is the second or square root of o^, since a X a = a^ ; a is the third or cube root of a', since a X a X a = a' ; a is the fourth root of a*, since aXaXoXa^ffl*. 22. The Eadical Sign, v'. when prefixed to a quantity, indicates that the root is to be taken. Thus, \/ a indicates the second or square root of a ; V^ a indicates the third or cube root of a ; • 4/ a indicates the fourth root of a. The index of the root is the figure or letter written over the radical. Thus, 2 is the index of the square root, 3 of the cube root, and so on. When the radical has no index over it, 2 is under- stood. Thus, \/a is the same as \/a. A fractional exponent is also used to indicate a root. Define a Power. A Root. What does the Radical Sign indicate ? Define an Index of the root. DEFINITIONS AND NOTATION. 13 Thus a* indicates the square root of a, and a* indicates the cube root of a. The numerator of the exponent de- notes the power, and the denominator the root. Thus, 6* indicates the fifth power of the fourth root of 6, or the fourth root of the fifth power of 6. ALGEBRAIC EXPRESSIONS. 23i An Algebraic Expression is a quantity written in algebraic language. Thus, 2 a is the algebraic expression for 2 times the num- ber a. 3 a' is the algebraic expression for 3 times the square of the number a. ba-\-'lW is the algebraic expression for 5 times a, augmented by t times the cube of 6. 24i The. Teems of an algebraic expression are its parts connected by the signs -\- or — . Thus, a and 6 are the terms of the expression a-\-h; 2 a, b^, and — 2 a c, of the expression 2a-\-b^ — 2ac. 25, The Degree of a term is the number of literal fac- tors which it contains. Thus, 3 a is of the Jirst degree, since it contains but one lit- eral factor. a 6 is of the second degree, since it contains but iwo literal factors. 5 a 6^ is of the third degree, since it contains but three literal factors. The degree of any term is determined by adding the exponents of its several letters. Thus, 3ab^(f is of the sixth degree, since l-|-2-(-3 = 6. What does a fractional exponent indicate 1 Define an Algebraic Ex- pression. The Terms of an algebraic expression. Degree of a term. 2 14 ELEMENTARY ALGEBRA. 26. A Monomial is an algebraic expression consisting, of only one term ; as, 6 a, t a h, or 3 6'' c. 27. A Polynomial is an algebraic expression consist- . ing of more than one term ; as, a + 6, or 3a2 + 6 — 56». 28. A Binomial is a polynomial of two terms ; as, a — h, 2 a + 6^ "-or 3 a c'' — 6. 29. A Trinomial is a polynomial of th^ee terms; as, a-\-b-\-c, or afe-j-c" — W. 30. HoMoeENEOus Tekms are those of the same degree. Thus, the terms a^, 3 be, — 4 x^ are homogeneous. A polynomial is homogeneous when all its terms are homogeneous. Thus, the polynomial a^-\-abc — 6' is homogeneous. 31. Positive Tekms are those having the plus sign; as, -\-a, or -|- a 6^. When a term has no sign written, it is understood to be positive. Thus, a is the same as -j- a. 32. Negative Terms are those having the minus sign ; as, — a, or — 2b(^. This sign should never be omitted. 33. Similar or Like Terms are those containing the same letters, affected by the same exponents. Thus, 2xy and — 1 xy are similar terms ; also, 3 a^ ¥ and 9 a? W are similar terms. 34. Dissimilar or TJnlike Terms are those contaimng different letters or exponents. Thus, o b and a d are dissimilar terms ; also, bx'^y and bxy^ are dissimilar terms. 35. The Eecipkocal of a quantity is 1 divided by that quantity. DeBne a Monomial. A PolynomiaL A BinomiaL A Trinomial. Homogeneous Terms. Positive Terms. Negative Terms. Similar Terms. Dissimilar Terms. A Beciprocal of a quantity. DEFINITIONS AND NOTATION. 15 Thus, the reciprocal of a is -, and of x-X-y i& — -. — . ^ a' ' ^ "^-{-y Note. The recipmcal of a fraction is that fraction inverted. Thus, - is the reciprocal of -. The following examples will serve for an exercise on the preceding principles. Exauplb:s. Put in the form of algebraic expressions: — 1. Three times b, added to two times a. Ans. 2 ffl + 3 6. 2. Three times b, subtracted from five times a. Ans. 5 a — 3 6. 3. The sum of a and 6, diminished by c. Ans. a -\- b — c. 4. The sum of x and two times y, diminished by z. 5. a plus the product of 6 and c, minus d. Ans. -\- be — d. 6. The sum of a and 6 multiplied by the difference of c and d. Ans. (a-j-6) (o — d). * 'I. Five times b, divided by four times c. . 56 Ans. -7—. 8. Four times a, divided by three times c. . 9. a diminished by b, divided by a multiplied by b. 10. a plus 6, multiplied by c into d. Ans. (a -1-6) cd. 11. Two times a, plus the quotient of 6 divided by c. 12. Six times a square into 6 cube, plus three times c square into d cube. Ans. Ga^^-^-Sc'd". 13. a fourth power minus 6 fifth, divided by a minus b square. 16 • ELEMENTARY ALGEBRA. 14. Two a square, into a minus b, into c plus d, plus c cube. Ans. 2a^{a — b) (c-j-d)-{-(f. 15. Fifteen a cube plus 6 fifth, divided by a square minus 6 square, plus two c. 16. The reciprocal of c minus d, plus two a square, minus b cube. Ans. ^^^ -)- 2 a'' — 6^ 17. The reciprocal of a into 6 square, minus the recip- rocal of a square plus c square. 18. The square root of a, plus the square root of 6. Ans. a/ a-\-A/b. 19. The cube root of a, minus b. Ans. \/a — b. 20. The square root of a minus b. Ans. \/ a — b. 21. The cube root of x, minus the square root of so. 22. Write a -polynomial of three terms, with its third term negative. 23. Write a homogeneous binomial of the first de- gree ; a homogeneous trinomial of the third degree, with its second term negative. INTERPRETATION OF ALGEBRAIC EXPRESSIONS. 36i The Interpretation of an algebraic expression con- sists in rendering it into arithmetic, by means of the numerical values assigned to its letters. 37. The Numerical Value of an algebraic expression is the result obtained by substituting for its letters their numerical values, and then performing the operations indicated. Thus, the numerical value of 4 a -\-S bc — d, What is the Interpretation of an algebraic expression 1 How is its Numerical Value obtained ? Ans. 17. Ans. 34. Ans. 54. Ans. 166. DEFINITIONS AND NOTATION. 17 when a = 4, 6 = 3, c = 5, and d = 2, is 4X4 + 3X3X5 — 2 = 59, Examples'. Interpret and give the numerical values of the follow- ing expressions, when a =12, 6 = 3, = 2, d = 4, m = 5, n=9. 1. 0-1-6 — c-\-d. 2. ab-\-c~d. 3. 4a— 56 + 4c — Td. 4. (a — 6)(c + d). 5. (6a + 6=)c4- The Processks of Algebra, in general, are only those of Arithmetic extended, or rendered more compre- hensive by the aid of letters taken in combination with figures. (Art. 12.) ^ The processes of Algebra are employed in the demon- stration of theorems and in the solution of problems. 40. A Theorem is the statement of some relation or property, the truth of whicli is required to be demon- strated. 41. A Problem is a question proposed for solution, or something to be done. 42. An Equation is the expression of equality between two quantities. Thus, X = a — b, is an equation, expressing equality between a; and a — b. 43. The First Member of an equation is the quantity on the left of the sign of equality ; and The Second Member is the quantity on the right of that sign. Thus, in the equation, 5x-\-y=b-{-o, 5 a; -|- y is the first member, and b -\- c is the second. 44. The Solution of a problem or question by Algebra, consists of two parts. 1. In STATING THE QUESTION, by expressing its conditions in the form of an equation. 2. In SOLVING THE EQUATION, hj finding the value cf the unknown quantity. What is said of the Processes of Algebra ? Define a Theorem. A Problem. An Equation. The First Member of an egoation. The Second Metfiber. Solution of a problem. 20 ELEMENTAEY ALGEBRA. Hence, the solution of the equation is the solution of the problem,. 45. The Vbkification of the value found for the un- known quantity, is the process of proving that it wiU satisfy the conditions of the question. Thus, in the equation, 2 a; = 9 + 6, if the value of x be found to be 7, it may be verified bjf substituting T for x, and showing that 2X1=9 + 5. 46. To show some of the simpler algebraic forms and processes pertaining to the solution of problems, there are introduced the following Examples. 1. The sum of the ages of two boys is 21 years, and the age of the older is twice that of the younger ; what is the age of each ? SOLUTION, III tl*^ question, if the age ' ,, of the younger were known, Let X = age of the younger ; ^^ ^^^,^_ l^ ^^^^^^^ .^^ 2 a; = age of the older ; ^^^^^^ ^^^^ „f ^,^3 „lder. 3 a; = 21 years. The age of the younger, x^l years, the younger ; tijgn, may be regarded as 2 a; = 14 years, the older. the unknown quantity. VEKIPICATION. 1 + 14 = 21. We therefore represent the age of the younger by X ; then, as the age of the older is twice that of the younger, 2 x will represent the age of the older ; and a; + 2 a;, or 3 x, will rep- resent the sum of their ages, which, by the conditions of the ques- tion, is 21 years. Hence, if 3 a; equals 21 years, x, the age of the younger, must be one third of 21 years, or 7 years; and 2x, the age of the older, must be 2 times 7 years, or 14 years. Define the Verification of the value of an unknown quantity. Explain the operation. DEFINITIONS AND NOTATION. 21 2. John had 45 cents; after spending a part of them, he found he had twice as many left as he had spent; how many cents had he spent? Ans. 15 cents. 3. James and William together have 56 apples, and one has as many as the other; how many has each? 4. A tree 60 feet high was broken at such a point that the part broken off was 3 times the length of the part left standing ; required the length of each part. Ans. Part left standing, 15 ft. ; part broken off, 45 ft. 5. The greater of two numbers is 5 times the less, and their sum is 126 ; required the numbers. Ans. Less number, 21 ; greater number, 105. 6. My horse and chaise together are worth $ 340, and the horse is worth 3 times as much as the chaise ; what is each worth ? Ans. Chaise, $ 85 ; horse, $ 255. 1. A gentleman divided property, amounting to $2500, between his two sons, A and B, and gave B 4 times as much as he ga;ve' A ; how much did he give to each ? Ans. A, $ 500 ; B, $ 2000. 8. The sum of three numbers is 12 ; the second is equal to twice the first, and the third is equal to three times the first; what are the numbers? SOLUTION. We represent the first Let X = first number ; ^ ''^^f ^^ ^' *''«°' .^= *« _ , ' second number is twice the 2x = second number; £ ^ „ -,. , ,, nrst, 2 X will represent the second ; as the third nrnn- 3x = third number. X :=z 72. ber is three times the first, , a; = 12, first number ; 3 x -Hrill represent the third ; 2x = 24, second number; and a; -(- 2 a: -|- 3 a;, or 6 a;, 3x = 36, , third number. will represent the sum of _ ir. 1 nj I o/> h,n the three numbers, which, by VERIFICATION. 12 + 24 + 36 = T2. , ,. . „ ' '. ■' ' the conditions 01 the question, Explain the operation. 22 ELEMENTABY ALGEBRA. is 72. Hence, if 6 r equals, 72, x, the first number, must be one sixth of 72, or 12; 2 a;, the second number, must be, 2 times 12, or 24 ; and 3 x, the third number, must be three times 12, or 36. 9. It is required to divide $ 300 among A, B, and C, so that B and C may each have twice as much as A. How many dollars will each have ? Ans. A's share, $ 60 ; B's share, $ 120 ; C's share, $ 120. 10. Henry bought some apples, pears, and oranges, for 63 cents ; he paid for the pears 2 times as much as for the apples, and for the oranges -4 times as much as for the apples ; what did he pay for each kind of fruit ? 11. The sum of the ages of A, B, and C is 78 years; but B's age is twice that of A, and C's is eq;ual to the sum of A's and B's ; what is the age of each ? Ans. A's, 13 years ; B's, 26 years ; C's, 39 years. 12. A farmer sold a sheep, cow, and horse for % 180 ; the cow brought t times as much as the sheep, and the horse 4 times as much as the cow ; how much did he get for each? Ans. Sheep, $ 5 ; cow, $ 35 ; horse, $ 140. 13. The sum of three numbers is 350 ; the second is four times the first, and the third is one half the sec- ond ; what are the numbers ? Ans. First, 50 ; second, 200 ; third, 100. 14. John traveled 84 miles in 3 days ; he traveled 3 times as far the second day as the first, and half as far the third day as the first two days together; how many miles did he travel each day ? Ans. First, 14 miles ; second, 42 miles ; third, 28 miles. 15. Three men together contrib^ted for the aid of wounded soldiers, $600. A gave a certain sum, B gave 4 times as much, and C gave an amount equal to the difference between what the other two gave. How much did each contribute ? Ans. A , $ t5 ; B, $ 300 ; C, $ 225. ADDITION. 2a ADDITION. 47. Addition, in Algebra, is the process of collecting two or more quantities into one equivalent expression, called the sum. 48. In algebraic addition there are three cases, de- pending upon the similarity and signs of the terms : — I. When the terms are similar, and have the same sign. II. When the terms are similar, and have different signs. III. When the terms are dissimilar, or some similar and others dissimilar. CASE I, 49. When the terms are similar, and have the same sign. 1. John has 4 books, Edward 6 books, and James T books ; how many books have they all ? It is evident, by Arithme- tic, that the sum of 4 books, 6 books, and 7 books is 17 books. Now, instead of writing the word hooks, we may simply use the letter 6; or we may represent one book by the letter h ; then, 4 h will rep- resent 4 books, 6 h will represent 6 books, and 7 5 will represent 7 books; and since 4 .books -|- 6 books -|- 7 books = 17 books, 4 6-[-6'5+ 76==17J. 2. Let it be required to find the sum of — 4i, — 6 J, and — 7 i. OPERATION. 4 books, 6 books, *l books, 17 books, ■ or. 4 b 6 b 1 b n b Define Addition in Algebra. How many Cases in algebraic addi- tion? Name them. Explain the first operation under Cass I. 24 ELEMENTAEY ALGEBRA. OPERATION. Ii ^^ same manner as in the pre- . , ceding operation, ( — 4 6) -|- ( — 6 J) -|- ( — 7 J) = — 17 6; since, what- ever h may represent, it is taken, in ZZ - the first term, — 4 times ; in the sec- — It ond, — 6 times; and in the third, — 7 dmes; or, in all, — 17 times. Hence, when the terms are similar and have the same sign : RULE. Add the coefficients, and to their sum, with the common sign, annex the com/mon letter or letters. KoTB. It must be remembered, that when n, quantity has no coef- ficient written, I is understood (Art. 18), and that when a term has no sign written, -|- is nnderstood (Art. 31). Examples. (3.) (4.) (5.) . (6.) 2a 4aa: 2xy — 3aJc 3a 2ax xy — aho 5a ax xy — baho a 6ax 1 xy — 2abo la hax 2xy — 8a Jc 6a 2ax xy — 4a5c 24: a 20 aa: 14: xy — 23 a J c i\-) (8.) (9.) (10.) — 4 ix 6mn2 2a + J Zc^d — c^c — Ibx bm,T? a + 5 c^d — c^c — 3bx mn^ 4ta-\-b 2c^d — a'G — 2bx 3mn^ ta+6 bc^d — a^o — 5bx 2mn'^ 3a + & c^d~a^o Explain the operation. Bepeat the Bule. The Note. ADDITION. 25 11. What is ttie sum of — 6n, — 4n, — n, — 8», and — 12 n ? Ans. — 31 n. 12. "What is the sum of 5 a;, 2x, x, Sx, Ix, 6 x, x, aid 8 a;? Ans. 30 a;. 13. What is the sum of 2x-\-3y, x-{-8y, 3x-\-y, 6a; + 2j/, a: + 4j/, and 4a: + 3/? Ans. 17 a: + 19 y. 14. What is the sum of T a^ — b, 3a^~3b, 6 a" — 2 b, 2a^ — b, 4a2— 66, anda= — 46? Ans. 23a.^-^l'Jb. CASE n. 50. When the terms are similar, and have differ- ent signs. 1. Let it be required to add- -f- 8 a, — 5 a, + 7 a, and — 3 a. Since the terms to be addM are some positive and others negative, in finding their sum regard must be paid to their signs. Now, the Signs, 4" and — indicate, not only opposite processes, but may be regarded as -\- '1 a. used to denote opposite qualities, eflfects, or conditions of quantities. Thus, if a merchant's gains are indicated by -|-, his losses will be indicated by — ; if distance north be reckoned -|-, distance south will be '■ — , and so on. Hence, two equal quantities, of which one is positive and the othfer negative, will exactly balance, or cancel each other. Now, in the example, -\-8a-\-7a = -\-15a^ and — 5 a — 3 a, or ( — 5 a) -(- ( — 3 a) = — 8 a. But — 8 a cancels -j- 8 a in the quantity -f- 15 a, which leaves -|- 7 a for the sum of the quantities. 2, A merchant having a certain capital, . In the first quarter of the year gained 6 a dollars, and in the second q^uarter gained 5 a dollars, but in the third and fourth OPERATION. + 8a, — 5 a. ■j-la, — 3 a, *E: xplain the first operation. 26 ELEMENTARY ALGEBRA. quarters lost T a and 9 a dollars. What was the result of the business at the end of the year ? We indicate the gains as positive, and their opposite, the losses, as neg- ative. The sum of — 9 a and — 7 a is — 16 a, and the sum of -|- 6 a and -f- 6 a is -\-ll a. But -|^ 11 a can- — 5 a. eels -^11 a in the quantity — 16 q, which leaves — 5 a, or a loss of 5 a dollars. From the preceding operations, it appears that, The Algebraic Sum of a positive and a negative qiumiitif is numerically the Differenck of the two quantities, with the sign of the greater prefixed. Hence, when the terms are similar, and have different OPERATION. + 6a, + 6a. -T«, — 9 a, RULE. Add the coefficients of the positive terms, and also the coefficients of the negative terms, and to the difference of these sums, vnth the sign of the greater, annex the common letter or letters. Examples. (3.) (^•) (5.) (6.) 3a 4aa; 2Sa;+ SSy Sx' — ix'/ 5 a — 2ax 3bx— 2bp X-+ xf — 2a Sax — -5bx -\- 4:by — 4x^—3x^ T« lax ibx — by 2x'-{-2xf — 4a 12 ax , -6bx-^ Iby -2bx-{-Uby — x^— a; / 9a Y. What is the sum of a — h, — 2 a — 1b,1a — 2b. — 3a + 3 J, — 8fl ' + 5, and a 4- t J ? Ans. — ia-\-b. Explain the second operation. What is the Algebraic Sum of a pbsi- tire and a negative quantity i Repeat the Rule. ADDITION. 2T 8. What is the sum of 5 cd" -f fa^b, — 3 crf^ ^ 5 a^h, 9crf=— lOa^J, _4cd2 + a»J? Ans. 7 c d» — f a" J. CASE ni. 51. When the terms are dissimilar, or some similar, and others dissimilar. 1. "What is the sum of 2 a, 5 J, and — a c ? OPERATION, If the given terms were jimflar, 2o-|-5S— -oc. the addition could be perfonn|d by uniting them into one (Art* 50) ; but the terms being dissimilar, we can only add them by writing them one after the other, with their respective- signs; which gives '2 o 4- 56 — ac. ._ 2. What is the sum of Sa + i, —2 a — 2*, and 6 a 4- 3 i— 2 c? OPERATION. We write similar terms in the 3 a _L 5 same column, for convenience in per- __ a n I forming the operation, _ , „ , „ Beginning at the left, we find oa-4-3o — 2c ,„ „i„ „ ,., -J +3o — 2a+6a= 7a, which we 7a-j-2J — 2c write under the column added ; and -|_6_2&-|-36==+ 25, which we write under the column added;, and there being no term similar to : — 2 c, we write it, with its proper sign, after the other terms obtained, and have as the entire sum, 7 a -j- 2 6 — 2.c. , Hence, since this case clearly includes the two pre- ce' + J''). 6. Reduce the -expression — 6 a" — h — (4 a" -j- ^) to its simplest form. Ans. — 10 a" — S — i". 1. What is the value of a» — i» — (3 a" J — 3 a J^) ? Ans. a» — J* — 3a"J4-3ai". 8. Place the last three terms Qt a^ h -\- x f — ae .—.*l ah — 6 ■\- 9 a? y' in a parenthesis, with the negative sigh prefixed, without changing the value expressed. Ans. a^ h -\- X f — a c — (1 a h -^ 6 — 9 s^ f). 9. What is the value of 10 a^ — {—4. a^ -{-¥ — , and as the sum of any number of positive quantities is positive, the product, a b, must be positive, or -\- ab. (Art. 20, Note.) If 6 = 4, the product of a by 6 may be represented thus, a X. ^^ a -\- a -\- a -^ a = i a. 2. Let it be required to find the product of — a by + h. Here we must take — a as many times as there are units in 6, and as the sum of any number of negative quantities is negative, the product must be negative, or — ab. If 6 = 4, the product of — a by 6 may be represented thus, ( — o) X 4= ( — a) -f- ( — a) + ( — a) -|- ( — a)= — a — a — a — a = — 4 a. 3. Let it be required to find the product of -(- a by — b. Befine the Multiplier. The Product. What are called factors ? Does the order in which factors are taken affect the product? What is the product when factors have like signs ? What is the product when fac- tors have different signs 1 4 38 ELEMENTARY ALGEBRA. Tha negative' multiplier, — i, indicates that a is to be ta&en as many times as there are units in b', but it is to be subtracted, rather than added. Hence, as a positive quantity becomes nega- tive by subtraction, the product must be negative, or — ab. It b = 4, the product of a by — b may be represented thus, a X ( — 4) = — a — a — a — a ^ — 4 a. 4. Let it be required to find the product of — a by — b. Here we must take — a as msmy times as there are units in b, and subtract; and as a negative quantity becomes positive by sub- traotion, the product must be positive, or a J. If 5 = 4, the pro- duct of — a by — 6 may be represented thus, ( — a) X ( — ' 4) = — ( — a) — ( — a) — ( — a) — ( — a)^a-{-a-\-a-\-a = 4a. Note. If any difficulty is experienced in conceiving quantities to be independently additive or subtradive, they may be regarded as added to, or subtracted from, 0, the neutral point, or starting-point, of all positive and negative quantities. 60. From the foregoing discussion it will be noticed, in brief, that in multiplication of algebraic quantities, J/ike signs produce -\-, and unlike signs prodiice -^. 61 • In multiplication of algebraic quantities, there will be three cases : — I. When both- factors are monomials. II. When one factor is a polynomial; m. When both factors are polynomials. CASE I. 62. When both factors are monomials. 1. If a man earn Y a dollars in 1 week, how much will he earn in 2 J weeks ? In Multiplication of algebraic quantities, what do like signs prodiice? Unlike signs ? How many cases of algebraic multiplication 1 MULTIPLICATION. 39 cipERATiON. Since the factors in multiplication h may be taken in any order (Art. 58), 7 a X 2 6 is the same as 7 V 2 V a X 6 ; then, 2 times 7 ^ 14 ; 6 times 14 a J a^^ab; and 14 times a & = 14 a &, the required result. 2, Bequired the- product of 2 a* Itj a". OPERATION. Since the exponent of a quantity 3 indicates the number of times the quantity is^ taken as a factor (Art. ^ * 19), 2 a' is 'the same as 2a aa; and 2 a^ a' is the same as a a ; then, a a times 2aaa= 2aaaaa, or 2a'. The ex- ponent, 5j in the result might- have been obtained at once, by' tak- ing the sum of the exponents^ 3 and 2, of the common letter a. Hence, The exponent of a letter in the product is equal to the sum of its exponents in the factors. Prom the preceding examples and illustrations of mul- tiplication of monomials is derived the following RULE. IMtiply the numerical* coejfficients of the two factors together, and annex to the result the. letters of hoth quantities, giving to each letter an exponent equal to the sum of its exponents in the two factors. Make the product positive, when the factors have like signs, and negative when they have different signs. , Examples. (3.) (4.) (5.) (6.) 3a —4a; —6a 12as 26 — 3y ■j-2b —3a 6ab 12xy — 12aS — 36aa; Explain the first operation nnder Case I. The second operation. To what is the exponent of a letter in the product equal? Bepeat the Rule. 40 ELEMENTARY ALGEBRA. (T-) (8.) (9.) ^ (10.) 4a: -\-11a — 3a — lb X — 26 — 46 -{-4.C (11.) (12.) (13.) (14.) — 2jm' ar" + 2 a» , 5a;y — 6m^ Ix' — 10a« 40a;*y» 15. Multiply 8 a 6 by — 3 c. Ans. — 24 a 6 c. 16. Multiply — a^y by 2 a a;. Ans. — lai^iy. IT. Multiply 20»ira^ by 3otw. Ans. 60>re^w^ 18. Multiply *\ol^c by 2 a 6. Ans. 14a^6c. 19. Multiply — 5 a^ x* by — 3 a« ar. Ans. 15 a» x\ 20. Multiply — 36crfby— 6c When one factor is a polynomial. 1. Eequired the product of a -\- b — c multiplied by c. OPERATION. Since the whole expression a -f- 6 I , is to be multiplied by c, it is evident ' that each term is to be taken c times; c times a = ac; c times 6 = 6 c ; c ac -\-bc — c" times — c = — c" ; and these partial products, connected with their proper signs, give ac + 6c — c*, the required product. Hence the EULB. Muhipli/ each term of the multiplicand separately by the mill- tiplier, and connect the partial products by their proper signs. Examples. (2.) (3.) 1 y -\-b . 4:a-{-4:X la — 3» 6a^^a lb 28ay + 4oi —12 aw — \1nx (5.) ?a: + 4y-}-«' 6aJ- x^ — 4aiK' 2lba? — lab. (6.) — c^x-{-x 1. Multiply la'P-j-Sxy — ac by —a\ Ans. — la*b^—Sa^xy-{-<^c. 8. Multiply 4 a^ 2^ — 6 y 4- 5 a^ by 2. Ans. 8a'y—12y-\- lOar". 9. Multiply a" — ax -\- a^ by a J. Alls, e^b — a'bx '^ aba^. Explain the operation under Case II. Bepeat the Bale. 4# 42 ELEMENTARY ALGEBEA. 10. Multiply —m — n — a — c by —m. Ans. ni' -\- m n -{- a m -\- c^m, 11. Find the product ofSa-^U^ + a'x — b by 4a''. Ans. 12 a' + i a'' ¥-{-4. a* X — 4= aH. 12. Find the product oi 2x^y — ixf — if by 6axy. Ans. Idaa^f—lbax'f — bax^. CASE m. 64. When both factors are polynomials. 1. Eequired the product of 3 o + 2 6 by a 4" *• „„^ „,„„ Since the multiplicand must be OPEEATION. ^ taken aa many timea as thete are 3 a -f- 2 J y^jjg j^ o _|_ J (Art. 57), it is evident, a -]-i that 3 o -|- 2 6 must be taken a times 3 a'' + 2 a i plus 6 times ; a times 3 a + 2 J ^ah-\-2¥ ==3a''-|-2a6; & times 3 a -f- 2J „ „ , . r^^TTu = 3 a J 4- 2 6^ and the sum of these ' ' partial products is 3 o" -)- * * * + 2 o > the required product. Hence' the following BULE. < HMtiply each term of the multiplicand hy each term of the multiplier separately, and add the partial prod'Ucts, n.}|«M Examples. (2.) (3.) 4a + 35 5x + 3y 3 a + * X — 2y 12a^+ 9a5 5x'-{- 3xy 4ai-|- 35^ —10xy~6f 12a»+l3aJ + 352 6 a^ — >lxy — 6f Explain the operation under Case III. Repeat the Rule. MULTIPLICATION. 43 (4.) (5.) (6,) 3 a* — 2y a — h hax-\-Zx X -\- y a — b %ax-\-2x fli-j-a'c' + c* a" — a" -f o" a^ — (^ ' a" — a" — a*(^ — c^c* — c* — a™+" 4- «*" — «'■''" flO — e« o*".^ 20""+" + «''+"' -|- a^" — «"+" 9. Multiply 3 a; + 2y by 2 a; — 32?. Ans. 6 cc" — 5a;y — 6^. 10. Multiply 5 a" 4- 3 a; by 5 a* + 3 as. Ans. 25 a* 4- 30 a» a: + 9 a:*. 11. Multiply n -(^ 2 a; by a — 3 a:. Ans. a' — ax — 6 a^. 12. Multiply 3ffl — a;by2a + 4a;. Ans. 6 a* +10 a a; — 4ar'. - '- 13. Multiply x-\-y by arr^-y. Ans. a? -\- 2 x y -^ f. 14. Multiply X — g by a: -j- y. Ans. a? — f. 15. Multiply a^ -\- ah -{- IF hy a — i. Ans. e? — »". 16. Multiply a' — a+lbya+1. Ans. a' +.1. 17. Eeeiuired the product of a;' — a ar* -f- a^ a; — a' . tod a; -)- «. Ans. a:* -^ «*. 18. Required the product of a* — c^y-\- aV — «3'' + y* and a-\-y. Ans. a? + y°. 19. Required the prodiict of «* + y a^^d ^ + y- Ans. x* + 2a^3r + y*. 20. Required the product of 2 a J — 3 6^ and 3 a J + 4 S». Ans. Ga^U'—ab^— 12 b\ 21. Find the product of a? -\- xy — f hy x — y. Ana. a? — 2xf-{-i/'. 44 ELEMENTARY ALGEBRA. 22. Find the product of «" -- 4 a + 16 by « + 5. Ans. a' 4- a" — 4a + 80. 23. Find the product of 1 — a -j- a^ — a' by 1 + a. Ans. 1 — a^. .24. Multiply a?-\-xy-\-y'\)ya? — xy-{-'f. Ans. V + a^y^ -|- /. 25. Multiply a — hx by c — dx. Ans. ac — (hc~\-ad)x-\-hda?. 26. Multiply Za? — 2xy—f\>y 2a; — 4 y. Ans. Qa?—Ux'y^Qxf^lf. 21. Multiply X — y -\- " by x-\-y — z. Ans. a^ — ^-|-2yz — a'. 28. Multiply 2T a:' + 9 ar'y + 3 aij/^ +y by 3 a: — y. Ans. 81a:* — y*. 29. Multiply l+a; + a:* + a;»by 1— ar + arS — rp'. , Ans. 1 — 3?. 30. Show that a" + y multiplied by a" -\-^ is equal to a''" + 2a"y» + 2'^". 65. The multiplication of polynomistls may be indicated by inclosing each in a parenthesis, and writing, them one after the other. When the operation indicated is actually performed, the expression is said to be expanded, or de/od- oped. ; , , ,. i '. 1. Expand (a — h) {a — b). Ans. a' — 2 a J +■ J*. 2. Expand (a + i) (c + £?). Ans. a c -|- J c -|- a — 2bc — e\ 1. Show that (2a: + 3) (2x — 3) (43:^ + 9) = 16 x* — 81. 8. Fiud the value of the expression (a" -}- 5») (a — 5). Ans. a'+^-f-aJ" — a" b — 6"+\ 9. Find the value of the expression (4 a" -]- 6 5") (a" — ¥) . Ans. 4 a^"" -|- 6 o" 5" — 4 a"" 6= — 6 J"+=. DIVISION. 66i DivisioN, in Algebra, is the process of finding how many times one quantity is contained in another ; Or, it is the process of finding one of two factors, when their product and the other factor are given. The Dividend is the quantity to be divided. The Divisor is the quantity by which we divide. The Quotient is the result of the division. Division is the Converse of multiplication, the dividend corresponding to the product, and the divisor and quo- tient to the two factors. 67i When dividend and divisor have likb siffns, the quotient is PosmvE ; and when dividend and divisor have different signs, the quotient is negative. For the quotient multiplied by the divisor must produce the dividend. Thus, (+ «*) -^ (+S) = + a, for (+ a) X (+S) =+ a6; (+ »b) -i- (- J) a, for (- o) X (- J) = + P6; (_aJ) _^ (+ 6) a, for (—a) X (+ h) = — ab; (—«»)-=-(— ») = + «, for (+a)X(— 6) = — a6. Hence, in division, as in multiplication, Define Division. Dividend. Divisor. Quotient. When is the qnotient positive ? When negative J 46 ELEMENTARY ALGEBRA. lAhe signs produce ~\-, and unlike figns produce — . 68. In division of algebraic quantities, there will be three cases : — \. When both divisor and dividend are monomials. II. When the divisor is a monomial and the dividend a polynomial. Ill, Whea both diyisp? and dividend are polynomials. ■ CASE I. 69. When both divisor and dividend are mono- mials. 1. Let it be required to divide 14 a J by 7 a. OPERATION. Now, the quotient must be a quan- tity which, multiplied by 7o, the di- S. ^26 Tisor, will produce 14 a h, the dividend. 7 a Such a quantity fe 2 6 ; which is ob- tained by rejecting -from the dividend a factor equal to the divisor ; or by dividing 14^ the coefficient of the dividend, by 7, the coefficient of the div^r, and. rejecting from the dividend the factor a, common to both. 2. Let it be required to divide a^ by e^. OPERATION. Since a^ =a a a a a, and (f^aaa, J. it is evident that the quotient, or the — = (1? quantity which, multiplied by the di- «' visor c?, will equal the dividend o', must be a a,, or a'. The exponent 2, in the quotient, which is the result of rejecting from the dividend a factor equal to the divisor j might have "been obtained at once, by taking the difference of the exponents, 5 and a^ Hencp, THe exponent of a letter in the quotient is equal to its exponent in the dividend, diminished ly its exponent in the divisor-. In division what do like signs produce ? Unlilse signs 1 How many cases in division of algebraic quantities? Explain the first operation un- der Case I. The second. To what is the exponent of a letter in the quotient equal? DIVISION. 47 70. When the exponents of the same letter in the div idend and diyisor are equal, the letter may be introduced into the quotient, without affecting the value of the ex- pression, by writing the letter with the exponent 0. For, let o" be any power of the quantity a, then, dividing a" by a", -we have a« a" 5T = «"-" = <">; but— = 1. Therefore (Ax. 7), a" c= 1 ; and as a may have any value what- ever, Anff quantity whose exponent is is equal to 1. Hence, by this notation the trace of a letter which has disappeared in an operation may be preserved, since the introduction of any factor whose value is unity will not affect the value of an expression. Thus, a° b has the same value as b alone. 71. When the exponent of any letter in the divisor is greater than it is in the dividend, the exponent of that letter in the quotient will be negative. For, let it be required to divide a' by cf, and we have (Art. 69), a' 1 1 . Also, "5 "^ ~5 ; consequently, a~* = —,; that is, Any quantity with a neyadve exponent is equal to the reciprocal of that quantity with an equal positive exponent. So, also, ^ — ^a_6 — ^_s. «' 1 But, ^==0*; consequently, —5 = a*; hence. What is the Value of any quantity whose exponent is 0? When may a letter be introduced into the quotient without affecting the value expressed I What will be the character of the exponent in the quo- (jent, when that of the divisor is greater than that of tie dividend'! to what a. a tity with a negative exponent equal % 48 ELEMENTABY ALGEBRA. Any factor may he transferred from the divisor to the dividend, or the reverse, hy changing the sign of its eooponent. Note. As the signs + and — indicate opposite processes, qnalities, or conditions (Art. 50), we sbonld infer, from tlie relations of tiic signs themselves, that, if a positiiie exponent indicates the number of times a quantity is taken as a, faclor, a negative exponent must show the number of times it is used as a divisor. As the negative coefiBcient indicates sub- traction, whether numerically possible or not (Art. 54, Exam. 2), so the negative exponent indicates division. (Art. 19, Note.) If the expressioD in which a^ negative exponent stands is already a divisor, then the quan- tity which it affects is a divisor of a divisor, and may be regarded as a factor of the dividend. The relation of positive and negative exponents to each other, and to the exponent 0, is readily illustrated by such a series as the following, in which the exponents decrease regularly by one, to indicate a division by 3. 3', 3=, 3>, 3", 3-', 3-*, 3-' 27, 9, 3, I, J, i, ^ 72. From the preceding examples and illustrations we have, for dividing one monomial by another, the follow- ing RULE. Divide the numerical coefficient of the dividend by that of the divisor, and to the result annex the literal factors of the dividend which are not found in the divisor. Make the quotient positive, when the dividend and divisor have like signs, and negative when they have different signs. Note. It is evident from the rule that one monomial cannot be ex- actly divided by another: — ' 1st. When the coefficient of the divisor is not exactly contained in that of the dividend. 2d. When the same letter has a greater exponent in the divisor than in the dividend. 3d. When the divisor contains one or more letters not found in the dividend. How may any factor be transferred from the divisor to the dividend! Repeat the Rule. When is exact division of monomials impossible? DIVISION. 49 In each of these cases, the division is to' be indicated by writing the divisor under the dividend, in the form of a fraction, or by the use of negative exponents. Examples. (1) (2.) ah , 12 a* ¥ . (3,) (4,) Sxyz. — tn'n (5.) (6.) = — 3ar „ ,, = 2 a ■' o — 7adx — 8o'i 1. Divide 16 a^ by 8 x. Ans. 2 x. 8. Divide *l mxy hj xy. 9. Divide — xy by xy. Ans. — x"/ or — 1. 10. Divide 15 a^ 5* by 5 a J. Ans. ^aV. 11. Divide — lb a? x^ by baa?. 12. Divide lOanccyby 2ay. Ans. — bnx. 13. Divide Sa^/ by —2x^y. Ans. —4,xf. 14. Divide — a^ by «*. Ans. — a. 15. Divide — 16 0:^2/^2^ by — 4a;«. Ans. 4a:/«. " 16. Divide «""+" by «". ' ' ' Ans. a". IT-. Divide a^-" by a;"- . Ans. af—^". 18. Divide m a^ h^ c" by 9cfWc\ Aria, da^fi^c-". 19. Divide (« + 6)« by (a + 6)^ Ans. (« +-J)'. 20. Divide 4 (a — J)^ by 2 (a — J). Ans. 2 (a — J.) 21. Divide 27 ai^ (a: -j-.y)= by 3J'C^+-y)«. -. Ans. 9aJ-i(a; + y)-^ Note. In the last Jihree examples, the expression in the parenthesis ts to be considered as oife quantity. 5 50 ELEMENTARY ALGEBEA. CASE II. 73, When the divisor is a monomial and the divi- dend a polynomial. 1. Divide 12 a» 6 -f- 24 a' c — 36 a J by 12 a. OPERATION. S''''^^ *^ ^°1^ .^';'d«"d must contain the divisor as 12« )12a''& + 24a°c — 36aft ^^^^ 4;^^^ ^ ^^^ l^tj^, ;, a'h -\- 1c? c — 3& contained in the terms of ; the former, we divide 12o'6, -|-24a'c, and — 36 a 6, respectively, by 12 a, and, connecting the partial quotients by their proper signs, have as the entire quo- tient, 0*6 4- 2a'c — 3 6. HULE. Divide each term of the dividend separately, and connect the resvUs hy their proper signs. Examples. (2.) (3.) (4.) ^a)Qax -\-12ay ah)c?h-\- al? — ah 2xy) 2xy — %xf 2x -\- ^y a-\-h—l ' 1 — 3y (5.) 4 a* e) 12 a* & c + 20 a ^ bc — 8a^c' (6.) bx') 5bx' -{- lOba? — UFa^ 1. Divide 9 a^¥ —Ua^cP hj 3 a. Ans. 3a5^ — 4aV, 8. Divide 12 a^/ — 16 aV by — 4 a*y>. Ans. — 3/ + 4a2^. Explain the operation. Repeat the Hule.' DIVISIOS. 61 9. Divide 25 J c" + 15 x^ _ 5^ by — 5. 4-US. — bh etc. CASE I. 88, To resolve monomials into their prime fac- tors. 1. Find the prime factors of 12 a^b, OPERATION. Since the composite factor 12 is the ■i2__ov'Oy' 3 product of the prime factors 2, 2, and 2 ^, 3, the composite factor a', of a and a, , ,, and is prune, the pnme factors of ^ 12(r'6 are 2, 2, 3, n, a, and 6, or 12aH = 2X 2 X Saab 2X2X3aaJ. Hence the follow- ing By what is the difference of two even powers of the same degree divis- ible 1 By what the sum of two odd powers of the same degree ? Ex- plain the operation under Case I. FACTORING. 61 RULE. To the prime factors of the numerieal coefficient annex each letter written as many times as there are units in its exp&nent. Examples. 2. Find the prime ■ factors oi %aV. Ans. 2 X 2 X 2a$65, 3. Eeeolve 1\rt^n^x into its prime factors. 4. Factor 49 a^Sa^y". Ans. 7 X 1 aahxxyyy. 5. Find the prime factors of b^€?¥»'' -|- «). 91 • To resolve a binomial iato two binomial fac- tors. Any binomial can be resolved into two binomial factors when its terms represent the difference between two squares. 1. Find the binomial factors of a" -^ V. Repeat the Sulc. Qi ELEMENTARY ALGEBRA. Since a' is the square of a, OPERATION. and V is the square of b, we a^ = the square of a, have, by Theorem III, Art. V^ = the square of J. ?8, for the required factors, , I ,s / ,\ the sum and difference of a ,_5^ = (« + J)(«-^)- ,,d6,or(a + 6)(a-J). Hence the a KULE. Take for one of the factors the sum, and for the other the difference, of the square roots of the given terms. 2. Factor as" — c". Ans. (a -{-c)(a — c), 3. Find the factors of ar" — y^. 4. Factor 4 ai^ r- f. Ans. i2x-\-y)(2x— y). 5. Factor Q.a^ — 4 V. Ans. (3 a + 2 6) (3 a — 2 S). 6. Find the binomial factors of 64 a^ 6^ — 16 c'' d^. 1. Factor 1 — 81 a;^. Ans. (1 + 9 x) (1 — 9 a;). 8. Factor c" — a*2^^ Ans. (c''+ «'2/) (<^ — «'y)- ■9. Factor «* — &*. Note, o* — 6* = (o'' + ^) (a^ — 6^) = {«" + 6^) (" + 6> (» - M- 10. Factor 1_— c*. Ans. (1 + c^) (1 + c) (1 — c). 11. Factor Le^/*— 1. Ans. (4 / + 1) (2 ^^ + 1) (2 f- 1). 12. Factor a^ — c'. ■Ans. (a* + c*) (a'' + 0'') (a -}- c) (« — c). 13. Factor a^ — j^. Ans. (x^-^-y^) (x^ -\- y) (x'' —y). 02. Any binomial which consists of the difference of any two equal" powers, or the sum of any two equal odd powers, may be factored by aid of Articles 85, 86, and ' Explain the operation. Repeat the Rule. When may a binomial be factored upon the principles contained in Articles 85, 86, and 87? FACTORING. 65 87; for the quotient and divisor are factors of the div- idend. 1. Factor a» — 1^. OPERATION. (a» — J«) -f- (a — 6) = «!> 4- a J -f ja Since, by Art. 85, a — 6 is a factor of a' — 6', we divide the latter by the foririer, and obtain as another factor, c? -\- al -\- l? ; and thus have a' — 6» = (a — 6) (a= + a 6 + 6"). 2. Factor a» + 5^. Ans. (a + 5) (a^ — a J -f S^). 3. Factor m* — n*. ■ Ans.' (m — n) {m? -\- m^ ri -\- m n^ -\- n^), or (m -\- n) {nfi — m^n-\- mn^ — «'). 4. Factor 1 — a^. Ans. (1— a;) {I -\- x -{- a? -\- a?) ,. ' or (1 -\-x) (1 _ a; -I- ar* — a:'). Note. The second factor, in either answer of the last two examples, may be again resolved into factors, so that either set of answers will re- duce to the same form as those of Examples 9 and 10 in the last Article. 5. Factor 8 a^ — f. Ans. {2x — y) (4.0? -\-2xy -\-y% 6. Factor 8 a^^ + 1. Ans. (2a;4-l) (4a;2 — 2a;+ 1)., T. Factor aP -{-¥. Ans. (a + 5) (a* — a' b •+ aH^ — a J' ,+ h") . 8. Eesolve into factors a' — :¥. A.ns. (a' + JS) (a' — ¥) = (a» + P) (a — b) (a^ + a 5 + b^ = (a -\- b) {a^ — ab -\- ¥) {a — b) {a^ -{■ a b \- 5=) = (a2 — V") (a* -\-a^ ¥ + i*). Note. The factor a^ — 6^ is found by taking tbe product of the fac tors a -\-h and a — 6 ; and either using it as a divisor, or multiplying tlie. remaining factors together, gives the factor a* + a^ 6^ + 6*. By mailing the factors a + 6 and a — 6 divisors, other factors can be obtained. Explain the operation. » . 6* 66 ELEMENTAEY ALGEBRA. 93. In factoring many polynomials, much must depei* upon the skill of the learner, since specific directions cannot well be given to meet every case. Sometimes a portion only of a polynomial can be fac- tored, as when the terms do not all have a common factor. 1. Factor a^ c + 2 a i c 4- i^ c. Ana. (a -\- b) (a -\- b). Note, a^ c + 2 abc + b^c = c {a^ + 2ab + V^), and a^ + 2ab + ^ = (a + 6 (a 4- fr). 2. Factor ab-\-ad-\-cx-\-cy. Ans. a(b-\-d)-\-c(x-\- y). 3. Factor 2 / + 3 a^^ — 9 a^. Ans. 2 y + 3 a^ (3^ — 3). L Factor Qa? -{-llx'y -\- %xf. Ans. %x {x -\-y) (x -\- y). 6. Factor ah -\- ay -\- hx ■\- xy. Ans. (a -\-x) (i -f- y). Note, ab + ay -^-bx-^-xy = a(b+y)-^x(b + y) = (a + x)(b + y). 6. Factor ac — hd -\- be -r- ad, Ans. (a -|- 5) (c — d). T. Factor 6 ax — ■2hy-\- Zhx — iay. Ans. (2 a -f- J) (3 a; — 2y). ' Note. The product of two binomial factors reduces to a trinomial whenever two of the partial products are similar. (Art. 90, Note.) GREATEST COMMON DIVISOR. 94t A Divisor or Measuee of a quantity ia any quanti- ty that will divide it without a remainder. What is said of factoring polynomials ? When can a polynomial be only partially factored ? Define a Divisor or Measure. GREATEST COMMON DIVISOR. 67 95. A CoifMON Divisor or Mkasuks of two or more quantities is a quantity that will divide each of them without a remainder. Hence, any factor common to two or more quantities is a common divisor of those quantities. 96. The Greatest Common Divisor of two or more quantities is the greatest quantity that will divide each of them without a remainder. Hence,^ the greatest common divisor of two or more quantities is composed of all the factors common to those quantities. When quantities are prime to each other (Art, 83), they have no common measure greater than unity. 97.* The greatest common divisor of two quantities is also the greatest common divisor of the least quan- tity and their remainder after division. For, let a and 6 be two quantities, of wMch b is the least. Suppose, now, that b is not contained in a an exact number of times, but m times, vitk a remainder, r. Then, since the ^ideud is equal to the product of the divisor by the quotient, plus the remainder, we have a = mb-\-r. Also, since the remainder is equal to the dividend minus the product of the divisor by the quotient, r = a — mb. Now, any quantity that will exactly divide b will exactly divide m times 6, or mb; and any quantity that will exactly divide 6 and r will exactly divide mb and r, and consequently will exactly divide Define Common Divisor. Greatest Common Divisor. Prove that the greatest common divisor of two quantities is the same as the greatest common divisor of the least, and their remainder after division. • Beginners, at the option of the teaeher, may omit this Article. 68 ELEMENTAKY ALGEBRA. their sum, m 6 + r, or its equal, a. Hence, any quantity that is a com. mon divisor of 6 and r is also a common divisor of a and 6. Again, any quantity that will exactly divide a and 6 will exactly divide a and mJ, and consequently will exactly divide their differ- ence, a — m 6, or its equal, r. Therefore, any common divisor of a and 6 must also be a common divisor of 6 and r. But the converse of this has already been proved ; consequently, the common divisors of a and 6, and of 6 and r, must be identical, and the greatest common divisor of o and 6 must be also the greatest com- mon divisor of 6 and r; which was to be proved. Note. It will be seen that the greatest common divisor of a and h is common to the four quantities a, b, m b, and r, that is, to the dividend, di- visor, product of the divisor by the quotient, and remainder ; but it is not necessarily found in the quotient, m. The divisor, 6, and remainder, r, most nearly approach the common divisor, as they are smaller than either of the others which contain it, or they contain a less number of other factors. Moreover, the greatest common divisor of a and 6 is not, necessa- rily, the greatest common divisor of any other two of the four quantities involved, when taken by themselves. 24 and 9 are convenient numbers to be used for a and b in illustrating these principles! CASE I. 98i To find the greatest common divisor of mono- mials, 1. Find the greatest common divisor of Aa'¥c and OPERATION. 6(^bc^d=3 X2 X a' X bX c^Xd 2a^bc =2 Xa^X* X e Resolving the quantities into factors, we find that 2, a", 5, and c are the only common factors ; and since the product of these, or 2a^bc, is composed of all the factors common to the quantities (Art. 96), it is their greatest common divisor. Hence the Explain the operation. GREATEST COMMON DIVISOR. 09 KULE. Resolve the quantities into their prime factors, and the pro- duct of cM the factors common to the several quantities will he the greatest common divisor. Note. Any letter forming a part of the common divisor will talie tlie lowest exponent with which it occurs in either of the original quantities. Examples. 2. Find the greatest common divisor of Ibc^W The Least Common Multiple of two or more quan- tities is the least quantity that can be divided by each of them without a remainder. Define a Multiple. Define a Common Multiple of two or more quan- tities. Least Common Multiple. 7 T4 ELEMENTARY ALGEBRA. Hence, the least common multiple of two or more quantities must be the product of all their different prims factors, each taken only the greatest number of times it is found in any one of those quantities. 104i If the produict of two quantities be divided by their greatest common divisor, the quotient will be their least com- mon multiple. For, since the greatest common divisor of two quantities is com- posed of all tjie factors common to those quantities (Art. 96), these factors will enter twice into the product of the quantities. Hence, if the product be divided by the greatest common divisor, the quotient will contain only the factors common to the quantities, and those pecu- liar to each of them. Now these are the factors of the least common multiple. (Art. 103.) 105. To find the least common multiple of quantities. 1. Find the least common multiple of 6 a^ 6c and ^aWd. OPEKATION. 6a^Jc =3X2 Xa'^XJ Xc 4ay and the real sign — . 120i If any one of the signs prefixed to the numerator, de- nominator, and dividing line of a fraction be changed, the value of the fraction _ will be changed accordingly. _,, -4-a6 — ah ah ah Thus, ^ = «; -j-=-a; —^ = -a; __=_«. Also, _^ = _a; ___ =_[_„; __ = _|_a; — ah , 121 1 Any two of the signs prefixed to the numerator, de- nominator, and dividing line of a fraction may be changed, withoiU affecting- the value of the fraction. ,T,, ah ah -^-ah — ah , Thus, - = __ = __^- = r-= + «- .. ah — ah __ ah — ah 122i If all the signs prefixed to the terms and the dividing line of a fraction be changed, the value of the fraction wiU be changed accordingly. What does the Sign of a fraction show ? What is the apparent sign of a fraction ? The real sign ? What is the effect of changing one of the signs prefixed to the fraction and its terms ? Of changing two of the signs ? Of changing all the signs ? 80 ELEMENTARY ALGBBEA. Thus, ^== + «, but— -^;^ = — a; ^-= — a, but — = — a, but — '-,- = -\-a. h — b ' REDUCTION. 123. Reduction op Feactions is the psocess of. changing their forms without altering their values. CASE I. 124i To reduce a fraction to its lowest terms. A fraction is in its lowest terms, when its terms are prime to each other. 1. Reduce r-7— to its lowest terms. 9 c We factor both terms. OPERATION. Then, since dividing both 6 ah 3JX2'* 2a numerator and denominator The 8 6 X 3 c ifc by the same quantity does not affect the value of the fraction (Art. 116), we strike from each the common factors 3 and h, or 3 6. But 3 6 is the greatest common divisor of the terms of the fraction, consequently, 2 a and 3 c are prime to each other (Art. 83), and — is the answer required. EULE. Resolve hoth terms of the fraction into their prime fa/don, and cancel all that are common to hoth. Or, Divide hoth terms hy their greatest common divisor. Define Reduction of Fractions. Explain the operation. Repeat the Rule. fractions. §1 Examples. 2. Reduce ^ . to its lowest terms. cfx — a? x(a-\-x) (a — x) x a* — I* (a + a:) (a — x) (u^-{-a?) ~ o'-j-ai" 3. Reduce ■ " " ^ to its lowest terms. 6 m-n^ ar 4. Reduce -rrr — to its lowest terms. 10 o a; 3 a^ u* 5. Reduce ~ — ^ to its lowest terms. »xf 1 Til 71 3? 6. Reduce S7"s - to its lowest terms. 1. Reduce ° , . to its lowest terms. 15 a 6 to" 8. Reduce . ° Ttj^w *» i*^ lowest terms. . a' — ah ~J 1 , X — 1 9. Reduce v.„„._lo,. *° ^*^ lowest terms, Ans. -^-r"- 10. Reduce ^fj^'^^ to its lowest terms. Ans. ^. 11. Reduce ^l2ax+ — . and - to equivalent fractions xy y X ^ having the least common denominator. OPERATION. Since, as has been shown, a common multiple of the {xy -i- xy) X <* = a ; ^ "~ T^ denominators of the given fractions will be a common (xv-i-ii) y. ax =z aa? • "'— ■=:^—- denominator of the required y ^y fractions, the least common Q Q„ multiple of the denominators ixy-i-x) X a =ay\ ^ = ^ will be the least common de- nominator. The least com- mon multiple of the denominators we find to het xy\ it is, conse- quently, the least common denominator of the required fractions. We next divide the least common denominator by each of the given denominators, and ascertain that the multipliers required to change each to the least common denominator are 1, x, and y. As the denominators are to be multiplied by these quantities, respec- tively, the numerators must be multiplied by the same, that the value of the fractions may not be changed (Art. 116), and we thus obtain the new numerators, a, an?, and ay. These, written over the , , . .a ax^ , ay , . least common denommator, xy, give — , — and -^-, the fractions re- xy xy xy quired. EULB. Multiply each numerator hy aU the denominatorB exeq>t its own, for new numerators, and aU the denominators together for a COMMON denominator. Or, Mnd the least commxm multiple of all the denominators for the LEAST COMMON denominator. Divide this midliple hy each denominator, separately, and mniMply the quotients hy the cor- responding numerators for new numerators. Explain the second operation. Repeat the Kale. 88 ELEMENTARY ALGEBRA. Note 1. Entire quantities should be reduced to a fractional form by writing 1 for a denominator to eacli, when required to be reduced to a common denominator with fractions. Note 2. All the denominators, if necessary, should be made positive (Art. 121) before finding a common denominator, and the fractions should be reduced to their lowest terms before finding the least common de- nominator. Examples. 3. Eeduce — and - to equivalent fractions having a common denominator. . 2nx ac Ans. > — • an an 4 3/ 7 7fl M 4. Reduce ^— , -r^r-^, and — , to equivalent fractions having the least common denominator. , i3?y 7mx 5n^ ^®" Wxf' wVy" Wxf' O M- 2 31 771 5. Reduce — , ^r-i, and - to equivalent fractions 2a' 5 a" n ^ having the least common denominator. 6. Reduce 7.-5-, and a + -^ to equivdent fractions having the least common denominator. 45 40a 60 a + 48a; ^^- 60' "eo"' 60 • 2ar •X' 3 5 a: -I- 1 7. Reduce — ^^-- and ^^—- to equivalent fractions having a common denominator. 6a;4-9 5x^-\-x ft «g , 2 , 8. Eeduce a, r, and to equivalent fractions having a common denominator. , ahc -\- ahd ac-^ad hx — 2ft hc-\.bd ' hc\-hd' Jc + ftd' What is Note 1 1 Note 2 ? FRACTIONS. 89 129i Fractions may often be very readily reduced to equivalent ones, having a common denominator, by mul- tiplying both numerator and denominator of one or more of them, so as to make the denominator the same for each, the multiplier being determined by inspection. 1. Eeduce -^ --, and — | — to equivalent fractions having the least common denominator. 2a; 3 _ 2x Z (x — a) i? — a" x + a ~ 3S — a^' s? ~ a" Since we know that (x -\- a) (x — a) = ar" — a', we convert the second fraction into one with a denominator the same as the first, by multiplying both terms of the second by a; — a.. 2. Change ^^^, ^-^, and ^_^^ into equivalent fractions having the least common denominator. A_„ a(^ — y) & (a + y) oni, a-\-x c , be 3. Change ^--^, ^— ^, and ^,_,_^^^^ into equiv- alent fractions having the least common denominator. a-\- X c (a^ -{- a X -\- 1?) he (a — x) a' — 3?' a' — a? ADDITION. 130i Addition op Fractions is the process of collecting two or more fractional quantities into one equivalent ex- pression, called the sum. Since only quantities of the same unit value can be united in one sum, fractions to be added must express fractional units of the same kind, or have a common denominator. How may fractions often be reduced to those having a common de- nominator? Define Addition of Fractions. Why must fractions to bo added hare a common denominator 1 8* 90 ELEMENTARY ALGEBRA. 131. To add fractions. 1. What is the sum of r-, ^ , and ^? OPERATION. Since the fractions, by hav- a . c , d _a + c + d ™g^ ^'"™"°° denominator, r ~r r 'T~ r — % express fractional units of the same kind, we add them by writing the sum of their numerators, a -\- c -\- d, over the common denominator, 6, and obtain a-\-c-\-d ' 6 2. What is the sum of r and -j ? a OPERATION. We reduce the given frao- a ,c _ad ,bc _ad + bc *'°"^ *° equivalent ones hav- h>"d~bd~^hd ~ bd '°g ^ common denominator, and then find the sum as in the preceding example. RULE. Reduce the 'fractions, if necessary, to equivalent ones having a common denominator. Add the numerators, and write the sum over the common denominator. Note. The final result should be reduced to its lowest terms, when- ever any reductions are possible. Examples. 3. What IS the sum of -p; -r-; and -r Ans. -rr-- 4 D o 1^ 4. What is the sum of -. -, and - ? Ans. -rr-- 6. What is the sum of -— and - ? 2a 5 . lbx-\-2ax Ans. -7; 10 a Explain the first operation. The second. Repeat the Rule. FEACTIOIiS. 91 0. -ioa „ 1 .„ ' p. . > and „, . . . 4adm -{- 2icm-f- 3i' mon denominator, 5, and obtain — =; — . 2. Subtract f—, — from 6 -1" c & — c" operation. ax ax ,ahx -\- acx abx — aex r=^ ~~ r+c ~ b' — c' tf — c' ahx -\- OCX — (abx — a ex) 2acz — b^ — c' '~¥ — c' We reduce the fractions to equivalent ones having a common denominator; then, subtracting the numerator of the subtrahend from that of the minuend, writing the difference over the common denominator, and reducing, we obtain ^ 5. Hence the Define Subtraction of Fractions. When can one fraction be subtracted from another? Explain the first operation. The second. FRACTIONS. 93 EULE. Meduce the fractions, if necessary, to equivalent ones having a common denominator. Subtract the numerator of the subtrahend from that of the minuend, and write the difference over the com- mon denominator. Note. When there are «ntire dr mixed quantities in connection with fractions, the former may either be reduced to fractional forms, or sub- tracted lieparatelj. Examples. 3. From '^^ take ^^. 1 , 39 a; Ans. 35 . 4. From - take t- 6 4 Ans. ^. 5. From -j— take — — . . 5ah Ans. — --r-. 4 6. From ^±1 take ^=^. Ans. y. 1. Subtract — r-: from -. X-\- 1 X — 1 8. Subtract from —5 . Ans. s . n — 1 w-^n n' — n 9. From 3x — -^ take x . 2o c . 4hcx4-2hx — ex — 2a5 -^nS. ■ — r ^. 26c 10. From , ' . take -,. Ans. , , . . 1 — ar 1 — ar 1 -[-ar* li. From 4a;+- take 3x — -;. Ans. x-\ \'"' . 'a a 'ad 12. From Ib — ^^^ take 5 + ?. Ans. eb - ''" + '' . Repeat the Eule. The Note. 94 ELEMENTARY ALGEBEA. 13. From i^+^ take 1^1:^. Ans. '-^^. y f y 14. What is the value of 1— il=-?? Ans. -4^. x-\-a x-\-a 15. What is the value of2a: + — — «— — ~ *-" ? . , 9 ex — 2ax-\-2af Ans. X A ^'— — , ' Sac 16. From -?1- subtract -^ a;— 3/ x + 3 Ans. MULTIPLICATION, I34< Multiplication multiplying when one tions. OF Fractions is the process of or both of the factors are frac-- CASE I. 135. To multiply a fraction by an entire quantity. • a 1. Multiply T by c. OPERATION. a , ac Since a fraction is multiplied hy mul- tiplying its numerator (Art. 114), we multiply the nimierator, a, by c, and obtain -i-. 2. Multiply ^ by b. OPERATION. -X J = - Since a fraction is multiplied by dividing its denominator (Art. 114), we divide the denommator, J", by i, and obtain i-. Define Multiplication of Fractions. Explain the first operation. Xb> second. FRACTIONS. 95 KULE. MvMpJy the numerator of the fraction hy the mtire quantity.. Or, Divide Uie denominator of the fraetim hy the entire quan- tity. Note. The second method is prefemMe when the entire quantity will divide the denominator without a r&malnder. Examples. 3. Multiply ^ by a. Ans. ^ .^ y 4. Multiply ~hy ah. Ans '"'* cd 5. Multiply ^ by 3 a:. Ans. ^ 6, Multiply ^i* by ad. Ans. ^l±tlM 1. Multiply f±| by « - c. Ans. "^ '■ ^^*^Ply al> + a\+\c + c' ^y « + - A i — e Ans. j~ — . -\- c '■ ^"'*'P'^ - H^'^-y't+y) '^ ' (" + ^)- Ans "_+Hl<' 136i It is evident, from the second rule of the pre- ceding article, that multiplying a fraction by a quan- tity equal to its denominator cancels the denominator, and gives the numerator for the product. Hence, -5^ a fraction he multiplied hy any muMpk of its denom- inator, the product will he an entire quantity. Bepeat the Rule. The Note. What is the effect of multiplying a fraction by a, quantity equal to its denominator, or a multiple of it ? 96 ELEMENTARY ALGEBKA. 1. Multiply r by 5. Ans. a. 2. Multiply ^^ by 8 by. Ans. 4 a a:. 3. Multiply ^^bya» — J». Ans. iBy= (a + i). CASE II. 137. To multiply an entire quantity, or a frac- tion, by a fraction. 1. Multiply c by ^ . FIRST OPEKATION. 7- X c is equal to -r- • Since the product of two quanti- ties is the same, whichever be taken for the multiplier (Art. 58), c X - is the same as - X c ; and by Case L SECOND OPERATION. c X T- == c X « = ac b~'- = -r- Since r is equal to a 6~' (Art. 126), c X r is equal to c X oS~'i or o c 6-', which, by transferring the factor 6-' to the denomi- nator (Art. 126), gives, as before, -r-. 2. Multiply ^ by |- FIRST OPERATION. X ^ J,,1 bd We first multiply r by-c, and ob- tain -r- i but this result is too great, since the proposed multiplier was not ExpUiiil the operations of the first Example. The second Example. FKACTIONS. 97 c, but c divided by d, or -. ; consequently ^ must be divided by d, a which we do by multiplying its denominator (Art. 115), and ob- . . ac tarn-. SECOND OPERATION. Since I is equal to a 6-' and | to cd-^ (Art. 126), |- X j is equal to o 6-' X c <^""S or a c 6-' rf-' ; which, by transferring the factors 6~' and rf-' to the denominator (Art. 126), gives, as be- « ac fore, -. In the first example, it is evident, since e is the same as — , that the operation might have been performed in the same manner as in the second example. Hence the EXILE. Multiply the numerators together for a new numerator, and the denominators for d new denominator. , Note 1. When either of the factors is an entire or mixed qnt^ntity, it may be best to reduce it to an equivalent fractional form. Note 2. When there are common factors in the numerators and de- nominators, they may be canceled before performing the multiplication, as the result should always be expressed in its lowest terms. EXAUFLES. 3. Multiply '- by ^. Ans. ^. 4. Multiply -g- by -■ Ans. -^^■ 5. Multiply ^ by ^- Ans. ^^ Repeat the Eule. What is Note 1 ? Note 2 ? 9 98 ELEMENTARY AtJGEBBA. 6. Kequired the product of —^ by j' El 7. Multiply 4a;y by Ans. bah. 8. Multiply ?^ by ^. Ans. 1. 9. Multiply IZJ by ,^j. Ans. ^. 10. Multiply ^^ by -^;. Ans. ^^^^--~p. 11. Multiply a + - by ^. Ans. ^^^^. 12. :Kuttiply ^J^ by i^^. Ans. ^-^1+3. ax 13. Multiply -^I— by ^— r-r- Ans. 6 bx ^ 2a;+l Vx a' — V' 3a« a — 6 . 3 14. Multiply "^ by ^rzT^' Ans. ~^- 15. Multiply ^^±^' by -^. Ans. a (^ + y)^ 16. Multiply ^^ by ^^j- Ans. ^.^aafe + y ; l?. teequired the continued product of > ^F-) 18. What is the value of (a — ^ +|) ? , a* — h* Ans. — nr' a' ft 19. What is the value of (^-^) (^^) (^? Ans. 1. 20. Find the product of a -1- - by a; -| . anxy -\- a my -\-bnx-{- btn ny FEACTIONS. 99 DIVISION. 188. DIVISION OF Fractions is the process -of dividing when the divisor or dividend, or both, are fractions. CASE I. il39. To divide a fraction by an entire quantity. 1. Divide -r- by c. OPERATION. ac a 2. Divide - by a. Since a fraction is divided by di- viding its numerator (Art. 115), we divide the numerator, ac, by c, and . . « obtain r ' OPERATION. Since a fraction is divided by mid- a; _ X tiplying its denominator (Art. 115), f ' ay we multiply the denominator, y, by a, and obtaui — . a y BULE. Divide the numereUor 'of the fradion hy the entire qucmti^f. Or, Midtiply the denominator of the fraction by the entire quantity. Note. The first method is preferable when the entire quantity will divide the numerator without « remainder. Examples. 3. Divide — by x. Ans. — . Define Division of Fractions. Explain the first operation. The sec- ond. Repeat the Bole. What is the Kote'? 100 ELEMENTAKY ALGEBRA. 4. Divide — by 5 a;. Ans. — . 5. Divide i|^ by 6 r". Ans. |. 6. Divide ^±^ by x. Ans. 1. t. Divide i ' — by a + S. Ans. — . a; ' a; 8. Divide ■ by a a. 9. Divide 1^^ by a + c. Ans. ^jq^^jf^^^ip- CASE n. 140. To divide an entire quantity, or a fraction, by a fraction. 1. Divide a by -r. FIRST OPERATION. c aX& «& "^6 c c We first divide o by c, and obtain - ; but this result is too small, since the pro- posed divisor is not c, but c divided by 6, or j- > consequently - should be taken 6 times, which we do by multiplying its numerator by 6 (Art. 114), and ob- tain — . c SECOND OPERATION. c , , a ah a-i- ^■= a -i- cir'- ■=: -^^-^ = — co~^ e Since j is equal to c 5"^ (Art. 126), a -j- ^ is equal to a -^ c 5"^, or .j^ , which, by transferring the factor 6~' to the numerator (Art. 126), gives, as before, — . Expliun the dperations of the first Example. FRACTIONS. 101 2. Divide r by ^. FIRST OPKRATION. h ' d ■ cd~^ 6c Since -^ is equal to ah-^ and ^ to cd-^ (Art. 126), ■?■ -^ 4 is equal to a J-* -i- c d-\ or -^ ; which, by transferring factors (Art. 126), ^ves ~. SECOND OPERATION. ' Reducing the fractions. to equiva- a c a d ad lent ones haying a common denomi- b ' d b c he nator, we have r-j to be divided by he i. T. . ^_ e o.d T.T ad . a ^. d a , T-. , which gives, as before, r— . Now, -. — ^ — V — , or -;- is bd a ' 'be • b c 6^c' 6 multiplied by the divisor inverted. Hence the RULE. Mvert the terms of the divisor, and proceed as in multiplica- tion. Note I. When either of the quantities is entire or mixed, it should be reduced to a fractional form before applying the rule. Note 2. After the operation is indicated, the work should be abridged, as far as possible, by canceling factors common to the numerators and denominators, so as to express the result in its lowest terms. Examples. 3. Divide -~ by — . Ans. -j^. 4. Divide I by |f . Ans. f^. Explain the operations of the second Example. Repeat the Rule. What is Note 11 Note 2? 9* %02 ELEMENTARY ALGEBRA. 6. Eequired the qutrtieat of - — divided by -. t. Find the quotient of a 6 divided by - — r. Ans. sd • 8. Divide aw by . Ans. "^^ 9, Divide , by -. Ans 1 — a "4 I— a 10. Divide ^ by i=-*. An,. ^. 11. What is the quotient of a;-|"- by -? Ans. ^^^. 12. What is the quotient of '. J*" by — ^^^- ? Ans !i^±l**. 13. Divide — — ^^ s by a-U-s. Ans. -. an "" ' a « 14. Divide ' by ^. Ans. r. a •* a « — 1 15. Divide -^-^^ by ^^^. Ans. 2^^^- 16. Divide -f-pp by -j-^. Ans. a» + J» •^ a + S" '^"''- a» — a6 + 6'' 11. What is the value of '^"^ -=- ^~"^ ? a;4-2y 3a; + 6y Ans. 3 {x-\-y). 18. What is the value of (1 + a:) -i- - (1 -|- a;) ? Ans. X. lEACTIO.NS. 103 19. What is the value of 13 divided by ^^L+^ _ a 1 Ans. 12a; (^ -\- a X -^ 3?' CASE m. 141. To reduce complex fractions to simple ones. A Complex Fraction is one having a fraction in its, numerator, or denominator, or in both. 142. A con^plex fraction may be regarded as a case in division of fractions. 1. Eeduce -j- to a simple fraction. h FIRST OPERATION. a c a I d h a b e d a c , ad Since -j- is the »ame as ~ "i" r» we regard it as a case in division ; and, reducing the expression hy the rule in Case II., obtain the si^aple fraction —j . ca SECOND OPERATION. a Since multiplying a fraction by any multiple of its denominator cancels that denominator (Art. 136), we multi- ply both terms of the complex fraction by the least common multiple of their denominators, and obtain the simple fraetiqti ^-, ■ ca Hence, to reduce a complex fraction to a simple one, or to simplify it. Xhc a h cd Define a Complex ftactlop. Explain the first operatiain. The second- 104 ELEMENTARY ALGEBKA. RULE. Consider the denominator as a divisor, and the numerator as a dividend, and proceed as in Case 11/ Or, Multiply both terms of the complex fraction hy the least common multiple of their denominators. Note. When the terms of a fraction contain negative exponents, the fraction may be regarded as a complex one. If the letter or quantity which bears the negative exponent is a /actor of either numerator or de- nominator, as a whole, the negative exponent may be removed as in Art. 126; otherwise, both numerator and denominator must bo multiplied by that letter or quantity with an equal positive exponent, in accordance with the above rule. 2. Keduce _IL to a simple fraction. Abs. m n a + & 3. Eeduce X y—a my to a simple fraction. ah4-V Ans. ■ xy — ax 4. Eeduce 5 — c 5.. Eeduce — — ^ — 7 — y a R to a simple fraction. Ans. hx l>y.-\-a' to a simple fraction. Ans. 5 a — ac 7 X — xy 6. Eeduce ^ , ^ ^ to a simple fraction. Ans. dny — dm dnx -\- an How may we reduce a complex fraction to a simple one, or simplify it? What is said of fractions containing negative exponents ? SIMPLE EQUATIONS. 105 *!. Beduce ^ , ^ to a simple fraction. ' .ah Ans. b — x b + x Ans. ^ 1 — ar" ix' — b 5 • a—b i A 8^- -4J -56 a^ ' X Ans. a? ■ ' X .1 aa; -(- 1 I 8. Simplify the expression 9. Simplify the expression 10. Simplify the expression Note. The last example furnishes a good opportanity for the nse of neg- ative exponents. Dividing ar* — x-^ by a: + a:-i gives ifi — x-\-x-^ — x-^ as a quotient. The answer given above may also be obtained by divid- ing a:* — — j by a; H , or by simplifying the fraction according to the role, and then reducing the fraction to a mixed number. SIMPLE EQUATIONS. 143i An Equation is an expression of equality between two quantities. Thus, a; + 4 = 16 is an equation, expressing the equality of the quantities a; -}- 4 and 16. 144i The quantity on the left of the sign of equality is called the Jlrst member, or side, and that on the right, the second member, or side, of the equation. Define an Equation. Members or sides of an equation. 106 ELEMENTARY ALGEBRA. 145. The Degree of an equation containing but one unknown quantity is denoted by the exponent of the highest power of that unknown quantity to be found in the equation. (Art. 15.) Thus, An equation of the first degree is one that contains no higher power of the unknown quantity than its first pow- er ; as, a; -f 14 = 28 — 4, or ex=.a^-\-hd. An equation of the second degree is one in which the highest power of the unknown quantity is the second pow- er, or square ; as, 3 aji* — 2 a; = 65. In like manner, we have equations of the third degree, fourth degree, and so on. NoTK. The degree of an equation must be distinguished from the de- gree of its terms (Art. 25). The former depends altogether upon the un- known quantity, without any reference to the latter. Thus, the equation ex =^ 0? -Y hd \s of the fir A degree, while each of its terms is of the second. 146. A Simple Equation is an equation of the first degree. ^ 147. A Numerical Equation is one in which all the known quantities are expressed by figures ; as, 2 a; — a; = It — 5. Note. The degree of a numerical equation corresponds with the high- est degree of any of its terms. 148. A Literal Equation is one in which some or all the known quantities are expressed by letters; as, 2 a; + a = a;'' — 10. 149. An Identical Equation is one in which the two Define the Degree of an equation. Equation of the first degree. Sceond degree. A Simple Equation. A Numerical Equation. A Literal Equar tion. An Identical Equation. SIMPLE EQUATIONS. 107 membera are %hB same, or become tbe same on performing the operations indicated ; as, x — y = x — 2f, or 2a-f26c = 2(a4-5c). TRANSFORMATION OF EQUATIONS. 150i The Transformation of an equation is the process of changing its form without destroying the equality. 151. The transformation of an equation depends upon the axioms (Art. 38), and we may, without destroying the equality, — 1. Add equal quantities to both members (Ax. 1). 2. Subtract equal quantities from both members (Ax. 2). 3. Multiply both members by the same quantity (Ax. 3). 4. Divide both members by the same quantity (Ax. 4). 5. Raise both members to the same power (Ax. 8). 6. Take the same root of both members (Ax. 8). In the transformation of simple equations, there are iwo principal cases : — I. Transposition of terms. II. Clearing of fractions. CASE L 152. To transpose terms of an equation. Transposition is the process of changing terms from one member of an equation to the other, without destroying the equality.- Define the Transformation of an equation. Upon what does the transform mation depend ? What are the two principal cases ? Define Transposition. 108 ELEMENTAEY ALGEBRA. 1. Let it be required, in a; — a = J, to transpose —a to the second member. Since we may add an equal quan- OPERATION. tity to both members- of an equation, X — a = without destroying the equality (Art. a =^ O' 151), we add a to each member, and a; = 6 -|- a obtain x = h -\- a. 2. Let it be required, in x -\- a =^ b, to transpose a to the second member. OPEKATION. Since we may subtract an equal I , quantity from both members of an ~'~ equation, without destroying the equal- ity (Art. 151), we subtract a from : a X = b -^ a gach member, and obtain a; = 6 — o. Now, the result is the same, in each of the above oper- ations, as if we had transferred a from the first to the second member, and changed its sign. Hence the RULE. Any term may be transposed from one member of an equation to the other, provided its sign be changed. Note. It also follows, that the signs of all the terms of an equation may be changed, loithout destroying the equality. Transpose the unknown terms to the first member, and the known terms to the second, in the following Examples. 3. 2x — a = b. Ans. 2x^a-\-b. 4. lla;4-9 = 6a; + 34. Ans. 11 a; — 6 3; = 34 — 9. 5. 5a;-l-3 = 2a; + 24. Explain the first operation. The second. Bepeat the Rule. The Note. SIMPLE EQUATIONS. 109 6. 3h-\-2x — 25 =ax. •Ans.2x — ax=z25 — 3l. 1. Sac — cd-\-xy =: Gad — Tat. Ans. 1 x-\-x^ = Gad — 3ac-\-cd. CASE II. 153i To clear an equation of fractions. I. Clear the equation -^ 26 := \-2 of fractions. OPEKATION. ^'""^ multiplying a frac- P^ tion by any multiple of its — ^- 26 = 1- 2 denominator will give for the pro,duct an entire quantity 2 a + 12 — 104 = 5 X 4- 8 , ■ (Art. 136), we multiply each term of the equation by the least common multiple of the denominators, or 4 (Art. 151), and, canceling each denominator, obtain 2a; -(-12 — 104 = 5a; -|- 8. Hence the RULE. Multiply each term of the equation hy the least common multi- ple of ike denominators, and reduce fractional to entire terms. Note 1. Also, an equation may be cleared of fractions by multiply- ing each njimerator by all the denominators except its own. Note 2. It must be observed, that when u fraction is preceded by — , the sign requires the value of tlio fraction to be subtracted, so that, on removing its denominator, all the signs of its numerator must be changed. (Art. 55.) Note 3. A negative exponent, is used to indicate that the single letter or quantity to which it is attached is a divisor, or denominator (Art. 71) ; hence, negative exponents are removed from an equation in the same way as fractions. To clear an equation of a negative exponent, we should then multiply each of its terms by the letter or quantity which bears that negative exponent, the multiplier taking an equal positive exponent. If there are two or more letters which have negative exponents, we should multiply by their least common multiple, with the signs of the exponents changed. Explain the operation. Repeat the Rule. What is Note 1 'i Note 2 1 Note 3? 10 110 i:rj::MENTARY algeeba. Clear the equations of their fractions in the following Examples. 2. - = b -\- c, Ans. 'X = ab-\-ac. a ' 3. a: + f + | = T- Ans. 4a; + 2a; + a:=28. 4. a: — f — 13=^. Ans. 6 a; — 2 a;— T8 = 3a:. 5. a; + f — 40=-^. Ans. 10 a: + 5 a; — 400 =vT a;. Ans. ax-\-l = b X -\- e X -\- d. "T. m.-\-ocr^ = n — px~^. Ans. m a; -j- 1 = ra a; — p. 9. a; — i^i? = 8. Ans. 6a; — 4a: — 8 = 48. b 22 J* -4- 4.9 10. -v-^-i- = T. Ans. 22 a; + 42=21 « 4- 49. Sa;-|- 7 ' ' 11. a + a; = — L_^ ' a-\-x Ans. a2 + 2aa; + a:2 = a:2-|-2aJ. 43 — Ty 69 — 9y 5 "~ 11 Ans. 473 — 17 y = 345 — 45 y. 12. Ans. 6a:+9a;— 15 = 'J2 — 4a: + 8, ■.A a — x ia — x 14. — J = a — J-^. 9 c Ans. ac — ex — i,ab -\- bx =z abc — c. 15. a: — 12 = I (44 — a: + 12). Ans. 8 X — 96 = 132 — 3 a; + 36. SIMPLE EQJJATIONS. Ill 16/^-6§=i(30-x). Ans. 6 a: — 60 = 120 — 4a!. Ans. ar* — 4a; + 4 + a:« + 4a: + 4= 14= (ar" — 4). SOLUTION OF SIMPLE EQUATIONS CONTAINING ONE UNKNOWN QUANTITY. 154i The Solution of an Equation is the process of finding the value of the imknown quantity in the equation. 135. The Root of an equation is the value of its un- known quantity. 156. The root of an equation is found by bringing all the terms containing the unknown (Juantity into one mem- ber, and freeing it from all connection with known quan- tities. 157. The root of an equation is verified, or the equa- tion sedisfied, when, the root being substituted for its sym- bol in the equation, the members are found to be equal, and the equation is thus reduced to an identical one. 158. The unknown and the known terms of an equa- tion may be combined in various ways': — 1. By addition ; as, a; -)- 6 = 18, or a; -j- a = J. 2. By subtraction ; as, x — 3 = 10, or a; — a = d. 3. By multiplication ; as, 3 a; =? 24, or ax:^ c. 4. By division ; as, -r ^:; 3, or - = a. 4 C 5. By a combination of two or more of these ; as, -— -|-10:=2a; — 5, or -j^-j-c = ca; — d. What is meant by the Solution of an equation 1 Boot of an equation f How is it found? When verified? How may the unknown and the known terms of an equation be combined ? 112 ELEMENTAJIY ALGEBKA. 159. To solve simple equations of one unknown quan- tity. 1. In the equation a; -[- 9 = 20, find the value of x. OPERATION. a; -f 9 = 20 a; = 20 — 9 a: =11 VERIFICATION. 11 + 9 = 20 20 = 20 Transposing the known quantity in the first member to the second (Art. 152), we have a; = 20 — 9 ; and unit- ing the terms of the second member of this equation, by performing the subtraction indicated, we obtain 11 as the value of x. This value of x we verify, and find it satisfies the equa- tion (Art. 157). 2. In the equation x — 7 = 11, find the value of x. X OPERATION. Y = ll ar=ll-fT X ■■ 18 VERIFICATION. 18 — T = ll 11 = 11 Transposing the known quantity in the first member to the second (Art. 152), and, in the equation obtained, performing the addition indicated, we have 18 as the value of a;. This val- ue of X we verify, and find it to satisfy the equation (Art. 157.) 3. In the equation 6 a; -|- 24 = '72, find the value of x. •*• Transposing 24, and unitmg the known terms by subtraction, we have (2) ; and dividing both members of (2) by 6, the coeflScient of x, we ob- tain 8 as the value of x. This value of x being substituted in the original equation, and the terms of the first member united by addi- tion, we have (2), an identical equar tion; therefore, the value is verified and the equation satisfied. OPERATION. 6 ir + 24 = Y2 6 a; = 48 a;= 8 VERIFICATION. 48 -f 24 = 12 12 = t2 (1) (2) (3) (1) (2) Explain the first operation. The second. The third. SIMPLE EQUATIONS. 113 3a; OPERATION. X — X . 3x " 2 "T T ~ :20 + 5 (1) 4 X — 2a:-f 3a; = ; 80 + 20 (2) 5x = :100 (3) x = :20 W VERIFICATION. 20 20 . 60 2 + T = 20 + S ) 20 — 10 + 15 25 = 20 + S = 25 . \ 4. In the equation x — |-l_-£ = 20-|-5, find the value of a;. Clearing the equation of fractions (Art. 153), we obtain (2) ; uniting similar terms, we have (3); and di- viding both members of (3) by 5, the coefficient of x, we obtain for the value of x, 20. This value, by verification, we find satisfies the given equation. From the preceding operations, it vsdll be noticed that, when the unknown quantity is combined with known quantities by addition or subtraction, it may be separated by transposition; when combined by multiplication, it may be separated by division ; and when combined by division, it may be separated by nvukiplication. It will be observed that the first and last of these cases have been fully treated under the heads of transposition (Art. 152), and clearing effractions (Art. 153). The second case most frequently oc- curs as the last step in the solution of an equation, when the coef- ficient of the unknown quantity is to be removed by division. It is usually, therefore, very simple, and has not been treated under a separate head. If at any time, however, all the terms of an equation can be exactly divided by any quantity, the equation may be thus simplified. Combining the principles illustrated by, the foregoing examples, we have, for the solution of simple equations containing only one unknown quantity, the following general Explain the the fourth operation. How is the imknown quantity sep- arated from known quantities ? 10* 114 ELEMENTARY ALGEBRA. RUIiB. Clear the equation of fractions, if it has any. Transpose the unknown terms to the fir^ member, and the hmwn terms to the second member, and reduce each member to its sim^st form. Divide both members by the coefficient of the unknown quan- tity, and the second member of the resultinff equation will be the value of the unknown quantity. Note. If the coefficient of the unknown quantity is negative, in di- Tidiiig ijt must be rememhered that like signs produce -|~ And unlike eigns produce — (Art. 67). Thus, — Zx — — 24, divided by — 3, the coef- ficient of X, becomes x = 8. The negative sign may also be removed by changing the signs of all the terms of the equation (Art. 152, Note). The positive coefScient would then be used as a divisor. Examples. 5. Given 5a!4-43 — 5 = 100 — 27, to find x. Ans. a; = t. 6. Given t a; + T -|- 1 = 96 — 11, to find x. Ans. X = 11. t. Given l5a;-|-8 — 9 = 212 + 87, to find a;. Ans. ap =i= 20. 8. Given 9 a; + 9 = a; — 71, to find x. Ans. x = — 10. 9. Given 4 a; — 15 = 2 a; -f 13, to find x. Ans. X = 14. 10. Given 4 (a; — 12) — 2 (12— a;), to find x. Ans. x = 12. NoTB. Performing the multiplication Indicated, the given equation bfr comes ix — 48 = 24 — 2x. Repeat the Rule. The Note. SJMl'LE EQUATIONS. 115 11. Given 9 (a; + l) = 12 (x — 2), to find x. Ans. a;= ll" 12. Given 3 (a; -^3) +2 a: = 3 (40— a:— 19), to find a;. Ans. a: = 9. 13. Given S (2 a;+ 3 a:) — 15 = 72 — 4 (ar — 2), to find x. Ans. a; = 5. 14. Given 2 (a; — 6) + 3 (2a;4-5) = 3 (3a;.^2) — 1, to find X. Ans. x = 10. 15. Given x — - — | = 30, to find x. Ans. x = 90. 16. Given l~8x-^~^-\-8 sr', to find x. Ans. x = 20. Note. We may either free the equation of ita oegative ezpanen^. or find the value of x~^ and take its ireciprocal (Art. 7)). IT. Given ^ a: + 12 = - :« + 6, to find re. Ans. X ^ 45. 18. Given | -j- if -f- i? = 158, to find x. Ans. X = 105. 19. Given | -|- 1 -f 5 = 28, to find x.. Ans. « = 32. 20. Given ax-\-b=^cx-{-d, to find x. OPERATION. Transposing, we obtain «» + a = ca: -f- rf (1) (2> '< factoring 1^ 6xst mem- ax—ex=zd—b (2) ^'^ °^ ^^>' '"' ''^''^ (*> ' ""^ , , , , )„. dividing by a — c, the coef- ^ ' ^ ' ficient of X, we obtain for the «f — 6 /j.\ , „ d — h X = (41 valae of x, . a — c ^ ^ 'a — c 21. Given — == 2 6 48. Given ax ~ ^r±^ — a¥ = bx -\- ^^^-^<^ - *-l±l^, to find X. Ans. :. = l^li^) * 4a — 3J ■ SIMPLE EQUATIONS. 119 PROBLEMS LEADING TO SIMPLE EQUATIONS CONTAINING ONE UNKNOWN QUANTITY. 160t The Solution of a Prpblem by Algebra, as has been already shown (Art. 44), consists of two distinct parts : — 1. The Statement of the problem in algebraic language. 2. The Sehuion, which determines th« values -of the un- known quantities. The Statement is Usually in the form of an equation, and the Solution is, then, that of the equation. 161 • Problems often include in their solution the con- sideration of ratio and proportion, especially in expressing relations of algebraic quantities. 162t Ratio is the relation, in respect to magnitude, which one quantity bears to another of the same kind ; or the quotient arising from the division of one quantity by another. Thus, r is the ratio of a to h. Ratio may be written in the form of a fraction, as t, or with two dots ( : ) be- tween the two terms, as a : 5, to be read a is to b. 163< Proportion is an equality of ratios. Thus, 4:2 = 6:3, or a : h = c : d, is a proportion. It may be written either with the sigh of equality ( = ), or with four dots ( : : ), between the ratios ; as 4 : 2 : : 6 : 3, to be read 4 is to 2 as 6 is to 3. 164. Any four quantities, then, are said to be propor- tional to each other, when the first contains the second as many times as the third contains the fourth. Of what parts does the Solution of a Problem consist t Define Ratio. Ptoportiott. When are any four quantities said to be proportional to each other! 120 ' ELEMENTAEY ALGEBRA. Thus, 9, 3, 12, and 4 are proportional, since 9 contains 3 as many times as 12 contains 4. 165. The first and last terms of a proportion are called EXTREMES, and the middle terms means. Thus, in a : b : : c : d, a and d are the extremes, and b and c the means. 166. In any proportion, the product of the extremes is equal to the product of the means. Let a : b : : c : d; then a X d=b X ". For, since the quantities are in proportion, a c _ b~d'' and clearing of fractions, ad = hc. But ad ]s the product of the extremes, and he the product of the means. Hence, To convert a proportion into an equation, place the product of the extremes equal to the product of the means. Thus, a; : 16 : : 20 : 4 may be converted into the equar tion 4 a: = 320. 167. It is impossible to give any general or precise rule for stating or solving every problem ; yet the follow^ ing directions may furnish some aid. 1. Denote the unknown quantity or quantities by some of the final letters of the alphabet. 2. Form an equation, by indicating the operations required to verify the answer, were it already obtained. 3. Determine the value of the unknown quantity in the equa- tion thus formed. What terms are called the extremes ? What the means ? Show that the product of the extremes is equal to the product of the means. How is a. proportion converted into an equation 1 What directions are given for solving problems? SIMPLE EQUATIONS. 121 PROBLEMS. ' 1. There are two numbers, whose difference is 9, and whose sum is 43 ; what are the numbers ? SOLUTION. Let X = the smaller number, and aj -[- 9 = the larger number. Their sum, x -{- x -{- 9 = 4:3 Transposing and uniting, 2 a; =^ 34 Dividing by 2, a; = If , the smaller number. Then, a; + 9 = 26, the larger number. TEEIFICATION'. 26 — IT = 9, and 26 + IT = 43. 2. It is required to find two numbers whose sum shall be 40 and their difference 16. Ans. 12 and 28. 3. At a pertain election, 1296 persons voted, for two cau- didates, and the successful candidate had a majority of 120 ; how many voted for each ? Ans. 688 and T08. 4. Find two numbers whose difference is 13, and which are such that if IT be added to their sum, the whole will • amount to 62. Ans.. 16 and 29. 5. A bankrupt owes B twice as much as he owes A, and C as much as he owes A and B together ; out of $ 3000, which is to be divided among them, what should each receive ? Ans. A, $ SOO ; B, $ 1000 ; and C, $ 1500. 6. A company of 266 persons consists of men, women, and children ; there- are four times as many men as chil- dren, and twice as many women as children. How many are there of each ? Ans. 38 children, T6 women, 152 men. Explain the solution of Problenl I. 11 122 ELEMENTAEY ALGEBKA. Y. Two trains of cars start at the same time towards each other, the one from Albany, running 26 miles per hour, and the other from Boston, 24 miles per hour ; in what time will they meet, the distance by railroad being supposed to be 200 miles? SOLUTION. Let X = number of hours required. Then 26 a; = distance run by one, and 24 a: = distance run by the other. Their sum, 26 a; -|- 24 a: = 200 Or, 50a; = 200 Whence, a; = 4, number of hours required. VERIFICATION. 26 X 4 + 24 X 4= = 200. 8. If two persons start at the same time from places 396 miles apart, and travel towards each other, the one at the rate of 36 miles per day, and the other 30 miles per day, in how many days will they meet,^and how far wDl each have traveled ? Ans. In 6 days ; the one will have traveled 216 miles, the other 180 miles. 9. A person starts from a certain place, and travels at the rate of 4 miles per hour ; after he has been traveling 10 hours, a horseman, riding 9 miles per hour, is de- spatched after him ; how many hours must the horseman ride to overtake him ? Ans. 8 hours. 10. A house and garden cost $ 850, and five times the price of the house was equal to twelve times the price of the garden ; find the price of each. Ans. House, | 600 ; garden, $ 250. 11. Two shepherds owning a flock of sheep agree to Explain the solution of Problem 7. SIMPLE EQUATIONS. X28 divide its value equally ; A takes 12 sheep, and B takes 92 sheep and pays A $35. Eequired the value of a ^'^^^P- Ans. $3.50. 12. Divide a line 21 inches long into two parts, such that one may be three fourths of the other. SOLUTION. Let X = length of one part, and — = length of the other part. Then, :c -f if == 21 Clearing of fractions, 4 a: -f- 3 a; = 84 * Or, Y a; = 84 "^Jience, a; = 12, length of one part, "■^^en, _f = 9^ length of the other part. 13. John's age is once and three fifths the age of James, and the sum of their ages is 39 years ; required the age of each. Ans. John's, 24 years ; James's, 15 years. .14. A, B, and have altogether $ 145 ; A's share is two thirds, and B's three fourths, as great as C's ; what is the share of each ? Ans. A's, $40; B's, $45; C's, $60. 15. A man being asked his age, replied that, if it were increased by a half and a third of itself, it would be 44 years ; what was his age ? Ans. 24 years. 16. A person spends one fourth of his yearly income in board, and one seventh in other expenses, and saves $ 85 ; what is his income ? Explain the solntion of Problem 12. 124 ' ELEMENTAKy ALGEBRA. SOLUTION. Let 28 a; = the number of dollars of incomfl. Then :|- of 28 a: = t a; = what he spends in board, and I of 28 a: := 4 a; :^ what he spends in other expenses. Then, ta;4-4a;+85 = 28a: Or, —17 a; = — 85 Whence, a; = 5 Then, 28 a: = 140, number of dollars of income. To avoid fractions, we resort to the artifice of supposing 28 x to be tbie nttmber of dollars of his yearly income, 28 being chosen because it is divisible by both 4 and 7, the denominators of the given fractions; then, by the question, he spends in board tx dol- lars, and in other expenses 4 x dollars, and 7 a; -f- 4 a; -f- 85 equal 28 X, or the yearly income. Thus, when fractions are foreseen to enter an equation, it will often be better to use, instead of x, such a multiple of a; as will preclude their entrance. 11. There is a pole standing one half and one. third of its length under water, and 4 feet above ; required the length of the pole. . Ans. 24 feet. 18. A man having completed two fifths of a journey, finds that, after traveling 30 miles farther, only three sevenths of the journey remain ; required the length of the journey. Ans. It5 miles. 19. From a cask one third full of oil, there leaked out 21 gallons, when there was found to be just half the oil left ; required the capacity of the cask. Ans. 126 gallons. 20. There are three brothers whose ages together amount to 24 years, and their birthdays are two years apart. What is the age of each ? Ans. Youngest, 6 years ; next, 8 years ; oldest, 10 years. 21. A and B have together a dollars, but B's share is n times as great as A's ; what is each one's share ? Explain the solution of Problem 16. SIMPLE EQUATIONS. 125 SOLUTION. Let a; = A's share, and nxt='B's share. Their sum, x-\-nx =^a Or, (1 + ra) a: = a Whence, x = — ?— , A's share. 1 -|- ra Then, nx=z r^~, B's share. 22. A man bought the same number of pounds each of coffee at a cents, tea at b cents, and sugar at c cents ' per pound, and the whole amounted to d cents ; required the number of pounds of each. . d a-\- b -j- c 23. Twice my age, increased by b, is equal to a ; what is my age ? i a — b Ans. — - — years. 24. At a certain election, a persons voted, and the suc- cessful candidate had a majority of b ; how many votes- did he receive ? , a + S Ans. — —-• 2 25. My carriage is worth 1^ times as much as my horse, and both together are worth c dollars ; what is the value of each ? a -cr 2 c . 3 c Ans. Horse, — - ; carriage, —-• 26. A courier left this place n days ago, and goes a miles each day. He i® pursued by another who goes b miles daily. In how many days wiU the second, start- ing to-day, overtake the first? ^ na ' b — a ^ ' 27. I have a certain number in my mind. I multiply it. by 1, add 3 to the product, and divide the sum by 2 ; I then find that if I subtract 4 from the quotient, I get 15 ; what number am I thinking of ? Ans. 5. Explain the solution of Problem 21, 11* 1^6 ELEMENTARY ALGEBKA. 28 From one end of a rod is cut away a fifth part of it, and from the other end 3 inches more than a sixth part' and there remains 16 inches ; required the length of the rod. ^°s. 30 inches. 29. A and B had equal sums of money ; A lost $ 50 more than a quarter of his, and B gained as much as A lost ; then B had twice as much as A ; what sum had each at first ? SOLUTION. Let X = what each had at first. Then x — 7 — 50 = what A had after losing, 4 and a; 4- - + 50 = what B had after gaining. ' 4 ' Then, a;-l-|+50 = 2(z-|-50) Or, x + j+60 = 2a:— | — 100 Or, _a: + | + |=-150 Whence, —a: =—600 Or a; = 600, what each had at first. 30. I buy four houses ; for the second I give half as much again as for the first, for the third half as much again as for the second, and for the fourth as much as for the first and third together; I pay for the whole $8000. What is the cost of each ? Ans. First, $ 1000 ; second, 1 1500 ; third, $ 2250 ; and fourth, $3250. 31. A father is three times as old as his son, but five years ago he was four times as old ; what are their ages now ? Ans. Son's age, 15 years ; father's, 45 years. 32. A vessel holding 120 gallons is partly filled by a Explain the solution of Problem 29. SIMPLE EQUATIONS. 127 spout which delivers 14 gallons in a minute ; this is then turned off', and a second spout, delivering 9 gallons in a minute, completes the filling of the vessel. How long did each spout run, the time occupied by both being 10 minutes ? Ans. The first, 6 minutes; the second, 4 minutes. 33. A can do a piece of work in a ■ days, which it requires h days for B to perform ; in how many days can . it be done if A and B work together ? SOLUTION. Let X = the number of days required. Then — = what A can do in one day, and — = what A can do in x days. Also J = what B can do in one day, and -r = what B can do in x days. Then, f 4- -5 = 1 Clearing of fractions, ax -\-hx-=ah Or, (a-\-h)x=.ah Whence, x = number of days required. Let X be the number of days, and 1 the entire work ; then, in 1 day A can do - of the work, and B -j- , therefore, in z days, they can do - and r of the work. Hence, by the conditions of the question, ^ -| — =1. 34. A can mow a field in 3 days, which it takes B Explain the solution of Problem 33. 128 ELEMENTAEY ALGEBEA. 1 days to mow ; in how many days can it be mown by A and B working together ? Ans. 2iV days. As a and 6 may have any value whatever, and retain their iden- tity in the final result, the solution of Problem 33 furnishes a for- mula which can be used for the solution of any similar problem. Thus, to obtain the required result in Problem 34, we have only to substitute 3 for a and 7 for 6, which gives ab 21 „ , ^ — Sq^— 10— -^Tir- A problem is said to be generalized when letters are, in this manner, used to represent its known quantities. The above formula may be expressed as an arithmetical rule, thus : When the times are known in which two agencies, act- ing separately, can accomplish a certain result, the time required for them conjointly to accomplish the same result may be found by dividing the product of the given limes by their sum. The principle demonstrated by any other general problem may be drawn from the formula in a similar manner. ' 35. A can perform a piece of work in a days, B in S days, and C in c days ; in how many days will they ac- complish the ■^ork, if they all work together ? A ab c . Ans. -^j—. .— £— days. an -\- ac -\- be •' It will be seen that, when three agencies are employed, the re- quired time is the product of the given times, divided by the sum of their products, taken two and two. 36. A cistern can be filled by tliree pipes ; by the first in 2 hours, by the second in 3, and by the third in 4 ; in what time can it be filled by all the pipes running together? Ans. 56 min. 23 ^V sec. SI. How many pounds of sugar at 9 cents a pound must be mixed with 20 pounds at 13 cents, in order that the mixture may be worth 10 cents a pound ? When is a Prohlem said to be generalized? SIMPLE EQUATIONS. 129 SOLtTTION. Let x = number of pounds at 9 cents, and a; -|- 20 = number of pounds in the mixture. Then, 9x = value of a: pounds at 9 cents, and 10 a; + 200 =: value of a; -f- 20 pounds at 10 cents. Also 260 = value of 20 pounds at 13 cents. Then, . 9 a: + 260= 10x4-200 Whence, — a; = — 60 Or, X = 60, number of pounds at 9 cents. 38. How much rye at four shillings and sixpence a bushel must be mixed with fifty bushels of wheat at six shillings a bushel, that the mixture may be worth five shillings a bushel? Ans. 100 bushels. 39. A liquor agent has 40 gallons of superior wine, worth $ 7 a gallon ; he wishes, however, so to reduce its quality, by the addition of water, that he may sell it at $4.50 a gallon; how much. water must he add? Ans. 22| gallons. 40. A banker lets three fifths of his money at 5 per cent, and the remainder at 6 per cent, and at' the end of the year receives $ 1080 interest. What is the amount I6t? SOLUTION. Let 5 a; = amount let. Then 3 a; = amount at 5 per cent, and 2 a; = amount at 6 per cent. Then, 3a.Xy^o+2-Xiro = l««« Or lif _|_ 11? = 1080 ^^' 100 ' 100 Clearing of fractions, 15 a; + 12 z = 108000 Or, 21xz= 108000 Whence, x = 4000 and 5 a; = 20000, amount let. Explain the solution of Problem 37. Problem 40. 130 ~ FXEMENTARY ALGEBRA. 41. A capitalist has two thirds of his money in United States 6 per cent stocks, and the balance in 8 per cent railroad bonds ; his yearly income from both is 1 1200 ; required the amount in each investment. Ans. In United States stocks $ 12000, and in railroad bonds $6000. 42. The rent of an estate this year is $ 1890, which is 8 per cent greater than it was last year ; what was it last year? Ans. $1750. 43. A merchant adds yearly to his capital 40 per cent of it, but takes from it, at the end of each year, $ 3000 for expenses. After deducting the last $ 3000, at the end of the second year, he finds his original capital has been increased 60 per cent. What was that capital ? Ans. $20000. 44. Of my income, ^ is derived from bank-stock, -J from a farm, J from a -factory, and the aggregate from these sources is $3800. Required my entire income. SOLUTION. Let X = the entire income, and a = ^00. > Then. | + | + f = « Clearing of fractions, 4a;-|-5a:-|-10a:=20a Or, 19 a; =20 a Whence, x = lj\ a Or, X = 4000, the entire in- come. We here represent the numeral 3800 by a letter, and in the result restore its value. An artifice of this kind may often be advantageously used, in order to avoid the use of large numbers. Explain the solution of Problem 44. SIMPLE EQUATIONS. 131 45. A young man, by putting three sevenths of his earn- ings in the savings' bank and one eighth into government stocks, found at the end of the year that he had thus laid by $ 930. Required the amount of his yearly earnings. Ans. $ 1680. 46. Divide the number a into two parts that shall have to each other the ratio oi m ton. FIRST SOLUTION. Let X = one part, and a — a: = the oth er part. Then, x : a — x-:=m : n Whence, nxz=ma — mx Or, mx-^nx = ma And , {m -\- n) X =: m a Whence, x = — ■. — , one part. and m-\-n j^» the other part. SECOND SOLUTION. Let mx = one part, and n a; = the other part. Then, mx -{- nx = a Or, (m -\- n)x = a Whence, x ^ — ■. — m-\-n ma mx = — i — ', one part, m-\- n ^ and nx = — ;— , the other part. m-|- n ^ 47. Divide 34 into two such parts, that the difference between the greater and 18 shall be to the difference between 18 and the less in the proportion of 2 to 3. Ans. 22 and 12. Explain the solution of Problem 46. 132 ELEMENTARY ALGEBRA. 48. A person has 264 coins, dollars and eagles ; the number of dollar pieces is to the number of eagles in the ratio of 9 to 2 ; how many of each coin has he ? Ans. Dollar pieces, 216 ; eagles, 48. 49. The ages of two persons are in the ratio of 3 to 4, but 5 years ago the ratio of their ages was that of 2 to 3 ; what are their ages ? Ans. 15 and 20. , 50. Two pieces of cloth were purchased at the same price per yard, but as they were of different lengths, the one cost 1 5, and the other $ 6.50. If each had been 10 yai-ds longer, their lengths would have been as 5 to 6. Eequired the length of each piece. Ans. 20 and 26. 51. A market woman bought sotoe eggs at 2 for a cent, . and as many more at 3 for a cent ;' she sold them all at the rate of 5 for 2 cents, and found she had lost 4 cents, How many did she buy of each sort ? Let X = the numb&r of each sort. Then - = the cost of the first sort, and — := the cost of the second sort. But ^ 2 a; = the entire number. and 2 i x' 2 a; X 5 = ^ — amount received for whole. Then, 2^3 5 ~ Clearing Whence, effractions, 15 a; -|- 10 a; — 24 a; = 120 X = 120, number of each sort. 52. Two merchants, A and B, traded in company, with a joint stock of $6300. A's money was employed 12 Explain the solution of Problem 51. SIMPLE EQUATIONS. 133 montks, and B's 8 months; and, on dividing profits, each had gained exactly the same sum. How much capital did each furnish? Ans. A, $2520; B, $ 3180. 53. A workman was employed for 60 hours, on condi- tion that for every hour he worked he should receive 15 cents, and for every hour he was idle he should forfeit 5 cents ; at the end of the time he received $ 2.40. Ee- quired the number of hours he worked, and the num- ber he was idle. SOLUTION. Let X = number of hours he worked, and . 60 — a: = number of hours he was idle. Then 15 a; = his pay for working, and 5 (60 — x) = his forfeiture for being idle. Then, 15 a; — 5 (60 — a:) =!= 240 Or, 15a; — 300 + 5a; = 240 Whence, 20 a; = 540 And X = 2*1, number of hours he worked. Then, 60 — a; = 33, number of hours he was idle. .54. A workman engaged for 48 days at the rate of $ 2 per day and his board, which is estimated at $ 1 per day. At the end of the time he receives $42 only, his employer having deducted the cost of his board for every day he was idle. How many days did he work? Ans. 30 days. 55. Two casks contain equal quantities of vinegar; from the first 34 quarts are drawn, and from the sec- ond 80 ; the quantity remaining in one vessel is now twice that in the other. How much did each cask originally contain ? Ans. 126 quarts. Explain the solution, of Problem 53. 12 134 ELEMENTARY ALGEBRA. 56. Two thirds of a certain number of persons re- ceived 18 cents each, and one third received 30 cents each. The whole sum received was $6.60. How many persons were there ? Ans. 30. b1. There is a fish whose head weighs 12 pounds, his tail weighs as much as his head and half the weight of his body, and his body weighs 26 pounds more than his head and tail both. Eequired the weight of the fish. Ans. 1*74 pounds. 58. A boatman who can row at the rate of 9 miles an hour, finds that it takes twice as long to row his boat up river a certain distance, as to row it down river the same distance ; at what rate does the river flow ? 4.ns. 3 miles per hour. 59. The paving of a square court with stone, at 40 cents a yard, will cost as much as the enclosing it with an iron fence, at $ 1 a yard ; what is the length of the side of the square in yards ? Let X = length of the side in yards. Then ix = number of yards. of fence, and aP = number of yards of pavement. Hence 4 a: X 100 = 400 = cost of fence, and a:^ X 40 = 40 a:^ = cost of paving. Then, 40 a;^ =£ 400 x Dividing by x, 40 a; = 400 Whence, x = 10, length of the side in yards. 60. A farmer has hogs worth $ 12.50, and pigs worth $ 2.50, each ; the number of hogs and pigs being 35, and their value $197.50. Eequired the number he has of each. Ans. Hogs, 11 ; pigs, 24. Explain the solution of Problem 59. SIMPLE KQUATIONS. 135 61. A lady being asked her age, replied, that if a half of her age were taken from it, and also a half of that remainder were taken away, she should be 19. Eequired her age. Ans. 76 years. 62. A gentleman hired a servant for 12 months, at the wages of $90 and a suit of clothes. At the end of 1 months, the man quits his service and receives $33.15 and the suit of clothes. At what price were the clothes estimated? Ans. $45. 63. A gentleman having $ 12000, employs a portion of the money in building a house. One third of the money that remains he invests at 4 per cent, and the other two thirds at 5 per cent ; fi-om these two investments he ob- tains an income of 1 392. What was the cost of the house? Ans. $3600. 64. A person desirous of giving some children 3 cents apiece, found he had not money enough in his pocket by 8 cents ; he therefore gave them each 2 cents, and had then 3 cents remaining ; required the number of-children. Ans. 11. 65. Three towns. A, B, and C, raise a sum of $11800; for every $ 20 which A contributes, B contributes $ 12, and C $ 18. What does each contribute ? Ans. A, $4720; B, $2832; C, $4248. 66. A newsboy gains during one day as much money as he had in the morning, but spends 16 cents at night ; the next day he gains as much as he had that morning, and spends 16 cents at night ; and so on, each day doubling his money, but spending 16 cents at night. At the close of the fourth evening, he finds that he has nothing left ; how much money had he at first ? Ans. 15 cents. 67. What number is that to which, if we add its fourth, fifth, and eightU, the sum will be 126 ? Ans. 80. 136 ELEMENTARY ALGEBRA. 68. A and B find a puise containing gold dollars. A takes out two dollars and one sixth of what remains, then B takes out three dollars and one sixth of what remains, and they find that they have taken out equal shares ; how many dollars were in the purse, and how much did each take out ? Aus. 20 dollars in the purse ; 5 dollars taken out by each. 69. If from three times a certain number we subtract 8, half the remainder will be equal to the number itself diminished by 2 ; required the number. ' YO. A man and his wife usually consumed a bag of flour in 12 days ; but when the man was from home, it lasted his wife 30 days ; how many days would it last the man alone ? Ans. 20 days. 71. At 12 o'clock both hands of a clock ai*e together; when will they next be together ? Ans. At 5^V minutes past 1 o'clock. 72. I have 90 sheep. If I would divide them into flocks, such that, if the number in the first be increased by 2, the number in the second diminished by 2, the number in the third multiplied by 2, and the number in the fourth divided by 2, the results will all be equal ; how many must I put in each flock ? Ans. In the first 18, second 22, third 10, and fourth 40. 73. A certain article of consumption was subject to a duty of 6 cents a pound ; in consequence of a reduction in the duty, the consumption increased one half, but the revenue fell one third ; what was the duty on a pound after the reduction ? Ans. 2§ cents. 74. A hare is 50 of her own leaps before a greyhound, and takes 4 leaps to the greyhound's 3, but 2 of the gi-eyhound's leaps are equal to 3 of the hare's ; how many leaps must the greyhound take to catch the hare ? Ans. 300. SIMPLE EQUATIONS. 137 15. A general arranging his men in the form of a solid square, finds he has 21 men over, but attempting to add one man to each side of the square, finds he wants 200 men to fill up the square ; required the number of men. Ans. 12121. Let X = the number of men on a side at first, then a;* -|- 21 = the whole number of men. SIMPLE EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES. 168i Independent Equations are such as cannot be made to assume* the same form. If they relate to the same problem, they must, there- fore, express essentially difierent conditions of that prob- lem. Thus, 4a;-|-10y='r2 + 4y and &x-\-9y= 108— a; are not independent equations, because each reduces to the form of 2 a; -{-Sy = 36. 169i When a problem requires two or more unknown quantities to be determined, it is necessary that there should be as many independent equations as there are unknown quantities. For, if ye have an equation containing two unknown quantities, X and y, as x — y = l, transposing y, we have x = l+y. (1) But the value of y is not known ; consequently, from this equation alone, the value of x cannot be determined. If, however, we have a second equation, as a; + ^ = 7, or, x=7—y, (2) in which the value of x and y are the same as in the first, the sec- A/VTiat is an Independent Equation ? Show that there should be as many independent equations as there are unknown quantities. 12* 138 ELlCMli-NTAKY ALGKCllA. ond members of (1) and (2) being equal to the same quantity, x, and conskiuently equal to each other (Art. 38, Ax. 7), give or, 22, = 6. Whence, y ^ 3- Substituting 3, the value of y, for y in either equation (1) or equation (2), we obtain 4 as the value of x ; and the values ob- tained for the two unknown quantities satisfy the two equations. 170. Simultaneous Equations are those in which the unknown quantities are satisfied by the same values. Two unknown quantities require for their determina- tion, as shown in the preceding Article, at least two inde- pendent, simultaneous equations. When by means of these we cause one of the unknown quantities to disappear, we are said to eliminate it. ELIMINATION. 171. Elimination is the process of deducing, from two or more simultaneous equations having two or more un- known quantities, a single equation having only one un- known quantity. There are three methods of elimination, and conse- quently as many cases : — - I. By comparison. II. By substitution. III. By addition or subtraction. CASE I. 172. To eliminate by comparison. 1. Given 3a; + 4y = 20, and 4a; — 2y=12, to find the values of x and y. Define Simultaneous Equations. How many simultaneous equations are required to determine two unknown quantities 1 Define Elimination. SIMPLE EQUATIONS. 139 OPERATION. 3 X + 4y = 20 (1) By transposing 4^ in equa- tion (1) and dividing by 3, also transposing 2y in equa^ 4a: — 2y = 12 (2) tjo^ (2) and dividing by 4, 20 — iy ,„, we have (S) and (4), equa- 3 tions in which the value of 12-)-2y a; is expressed in terms of 4 ^ ' y. Then, since each of the 12 + 2y _ 20 — 4y , two quantities ^A+Jl and ' r — 3 ' ^ J * 36 + 6y = 80 — 16y (6) ^^^^ '" ®1"*^ *° ''' *^>' 22 « = 44 /'J') are equal to each other (Art, y= 2 (8) "" ' "^ ' 20 — 8 3 a; = 4 (10) 38, Ax. 7) ; and placing them equal the one to the X = ^^^^-5 — - (9) other, we obtain (5), an equa- tion with only one unknown quantity, y. Reducing, we have (8), or y ^ 2. Substituting, now, 2 for ^ in equation (3), and reducing, we have (10), or a; = 4. Hence the EULB. Find an expression for the vcdue of the same unhnoum quantity in each of the equations, and form a new equation, by placing these values equal to each other. Note 1. The equation thus formed is solved as we would solve any equation containing one unknown quantity. Note 2. The value of the remaining unknown quantity may be de- termined in the same way as that of the first, thus making two in- dependent solutions, one for each unknown quantity. When, however, the value already determined is a simple number, it is best to sub- stitute that value for its symbol in some one of the equations, and thus obtain the value of the remaining unknown quantity. Note 3. It is usually most convenient to reduce each equation to its simplest form, by clearing of fractions, transposing and uniting, &c., ' before attempting to eliminate, by either method. Explain the operation. Repeat the Rule. Note 1. Note 2. Note 3. 140 ELEMENTARY ALGEBRA. Examples. 2. Given 2a;+3y = 23, and 5a; — 2y=10, to find the values of x and y. Ane. a; = 4 ; y=^&. 3. Given 4.x,-\-y = M, and x-\-4cy^lQ, to find the values of x and y. Ans. a; =: 8 ; y = 2. 4. Given bx — 3 y = 9, and 2 a; + 5 y ^ 16, to find the values of x and y. - Ans. a; = 3 ; y = 2. 5. Given Y a/ + 3 y = 13, and 5 a; + 2 y = 9, to find the values of x and y. 6 Given %x—*ly=. — 15, and 3y — 6a; = — 9, to find the values of x and y. Ans. a: = 6 ; y ^ 9. t. Given 14a;-)-6y=0, and 6a; — 46=4y, to find the values of x and y. Ans. a; = 3 ; y = — '7. 8. Given ^- -f | = Y, and ^ + | = 8, to find the CO U.J values of x and 2/. Ans. a; = 6 ; y = 12. 9. Given a; -1-2^=17, and 3 a; — y=2, to find the values of x and y. Ans. a; = 3 ; y = 1. 10. Given - — y ^ 1, and x — | = 8, to find the values of x and y. Ans. sc = 10 ; y = 4> 11. Given^ + ^=0„and^+^^ = l, to find the values of x and y. Ans. a; = 4 ; y = 6. . 12. Given ^^^ + 6y = 21, and ?^ = 23 — 5 a:, to find the' values of x and y. Ans. a; = 4 ; y = 3. CASE IL 173. To eliminate by substitution. 1. Given a; + 2^ = 17, and 3 a; — y = 2, to find the values of x and y. SIMPLE EQUATIONS. 141 OPERATION. By transposing- 2y in equa- a; + 2y= It (I) *'°° ^^^' ^® ^^^® equation - ^ ■' (3), which gives the value "^ ~" w of a; expressed in terms ofy. a; = lY — 2y (3) Substituting this value of x, 3(11 — 2y) — y= 2 (4) or 17 — 2 2^, for a; in equa- 61 — 1r/= 2 (5) tio° (2), we obtain (4), an Tm= 49 Cgl equation with only one uii- pt /hv known quantity, y. Reduu- ^= It- 14 (8) ^g.-ehave(7),or,= 7. '' ' substituting, now, 7 for y in ^ C^J equation (3), and reducing, we have (9), or a; = 3. 2. Given 2 a; + 5^^ = 23, ^and 3 a: — 2y = 6, to find the values of x and y. OPERATION. , By transposing 2 y in equa- 2a: + 5y = 23 (1) *'<>" (2) and dividing by 3, we 3a:-2y= 6 (2) •'^^e (3)> ''tich gives for the — ; value of a;, „ " Substi- 6 4-2y ,„, '3 x = g— (3) tuting this value of a: in eqna- 2(6+2y) , g _oq (AS ^'""^ 9^' "^^ ''^''^ ^*^' ^" 3 -J- o 2r — ^o (4) equation with only one un- 12 -j- 4y -(- 15y = 69 (5) known quantity, y; and re- 19y = 51 (6) ducing, we have (7), or «;— -3 Ci) S = ^- Substituting 3 for y 6 -j- 6 . . in equation (3), and reduc- 3 *• ■' ing, we have (9), or x ==4. a; = 4 (9) Hence the RULE. I^nd an expression for the valoe of one of the unknown quantities, in either equation, and substitute this value in the place of the same unknoym quantity in the other equation. Note. This method may be advantageously used when either of the unknown quantities has 1 for a coefficient. Explain the operation of Example 1. Of ExamplS 2. What is the Rule? The Note? 142 ELEMENTARY ALGEBRA. Examples. 3. Given a;-|-4y=16, and 4a;-|-y = 34, to find the values of x and y. Ans. x = 8; y = 2. 4. Given x-|-2y=18, and 2x — y = 1, to find the values of x and y. Ans. a; = 4 ; y = 1. 5. Given a;-(-y=13, and a; — y = 3, to find the val- ues of X and y. Ans. a; = 8 ; y = 5. 6. Given - — ^ = 1, and a; — ^ = 8, to find the val- nes of X and y. Ans. a; = 10 ; y = 4. 7. Given 3 a; -|- 5 y = 40, and a; + 2 y = 14, to find the values of x and y. 8. Given 5 a; -j- 3 y = 0, and x — y = 8, to find the val- ues of X and y. Ans. a; = 3 ; , y ^ — 5. 9. Given 6 a; -j- 5y = Tt, and 4a; — 3y=.Y, to find the values of x and y. Ans. a; = 7 ; y = 7. 10. Given | + ?! = 6, and -/ + f = 6, to find the values of x and y. Ans. a; = 6 ; y = 10. 11. Given ^i^-f 8y=31, and?^ + 10ar = 1^2, to find the values of x and y. Ans. a; = 19 ; y =; 3. 12. Given | -f | — 5 = 0, and 2 a; + | — 17 = 0, to find the values of x and y. Ans. a; == 6 ; y = 15. 13. Given ^ _ ^ = 0, and ^:i? + "-^ = 1|, to find the values of x and y. Ans. x = 5 ; y = 3. CASE ra. 174. To eliminate by addition and subtraction. 1. Given 6 a; -f- 4y = 66, and 4 a: — 3 y = 9, to find the values of x and y. SIMPLE EQUATIONS. 143 OPERATION. 6 X -\- 4,y = 56 4a: — 3y= 9 12a:+ 8y= 112 12 a:— 9yi= 27 Multiplying both members of equation (1) by 2, and of equation (2) by 3, we obtain equations (3) and (4), in which the coefficients of x are the same. Now, since the coefficients of x in these equations have like signs, we cancel the terms containing X, by subtracting (4) from (3), member from member (Art. 151), and obtain (5), an equation with only one unknown quantity, y. Reducing, we have (6), or y = 5. Substitut- ing 5 for y in equation (2), and reducing, we have (9), or a; = 6. 2. Given 6a: + 4y = 32, and 4a: — 2y=12, to find X and y. lYy = y = 4 a;— 15 = 4a; = X = 85 5 9 24 6 (1) (2) (3) (*) (5) (6) 0) (8) (9) OPERATION. 6a: + 4y=32 (1) 4a: — 2y=12 (2) 3a: + 2y = 16 (3) 7 a: = 28 (4) a;= 4 (5) 12 + 2y=16 (6) 2y= 4 (7) .y= 2 (8) Dividing equation (1) by 2, we obtain (3), in which the coefficient of y has been made the same as in (2). Sirfce the coefficients of y in these equations have diflferent signs, we can cancel the terms con- taining y, by adding (2) and (3) together, member to member (Art. 151), and thus obtain (4), an equation Seducing, we have (5), or with only one unknown quantity, a;. ^, „ a: = 4. Substituting 4 for x in equation (3), and reducing, we have (8), or y = 2. Hence the RULE. Multiply or divide one or both of the equations, if necessary, by such fz number or quantity that one of the unknown quanti- Explain the operation of Example 1. Of Example 2. What is the Rule 1 144 ELEMJiNTARY ALGEBRA. ties shall have the same coefficient in. both. Then, if the signs of the terms havitig the same coefficients are alike, subtract one equation from the other; or, if unlike, add the two equations together. Note. If the coefficients of the quantity to be eliminated are prime to each other, each equatioa must be multiplied by the coefficient found in the other equation. In general, such a multiplier may be used for each as will produce the least common multiple of the coefficients. If we wish to avoid fractions, it is convenient to divide only when one of the equations is not reduced to its simplest form, that is, when all its terms are exactly divisible by some quantity, as in Examples 3 and 4, below. Examples. 3. Given 4a;4-3y = 25, and 12a; — 6y = 30, to find the values of x and y. Ans. a; = 4 ; y = 3. 4. Given Zx — y = 22, and 2 a; -|- 4 ^z = 24, to find the values of x and y. . Ans. a; ^ 8 ; y=^1. 5. Given a; + 8 y ;= 44, and 6 a; + j/ === 29, to find the values of x and y. Ans. a; = 4 ; y = 5. 6. Given 23a;— 8^/= TO, and 8a; — 2y = 40, to find the values of x and y. Ans. a; = 10 ; y = 20. *l. Given 4a; — 6y=0, and x — y=l, to sblve the equations. 3 a; 9 w 8. Given a;-|-y = 35, and — + -j = 18, to solve the equations. Ans. a; = 21 ; j/ = 14. 9. Given ^ + ^ = 29, and ^_^=T, to find x and y. Ans. a; =; 24 ; y = 18. 175. Find the values of the unknown quantities in each of the following equations, by any of the methods of elim- ination. What is the Note ! SIMPLE EQUATIONS. 145 1. Given j^+/ = 'n. Ane. 1^ = 12- „ ^. (5y — 5a;:=15) (x= 1. t4y-3x+l = 0r ^^"-lyz^i. + 3. Given 10. X = 12. 16. 5. Given 2y-f6a;= 4 6. Given 1. Given flOa;= 24-22^1 I 42^ = 20 — 4a;j 8. Given 9. Given fto: — 3y = 26| l2a; — 2y = 6fr 10. Given ^-^ = 9.). (2a; — 2y=16) 11. Given' 4 ^5 (¥+.= 18)' 1^ = Ans. r ^^; Ans. \ a; = 40. ^ = 15. Ans. -j . ly =4, a;= 1. r =4. a + 6 2 Ans. -^ _ , J a 2 a; = 4. u = Ans. , a; = 20. ^°^- ■! 12. ,x=12. Ans. ■{ jo_ (2v+19z=5x ) , (x= 51. 12- (^^---^ I3.I T=4 + a;+J- ^^" 1^ = 103.- 13 146 ELEMENTARY ALGEBRA. 1,2 11 j^ + P H. Ans. j* = ^- is , 4_ 9j (2^ = 5. 13. Given Note. Eliminate before clearing of fractions. ' X y 5 14. Given ], _iT„*_, ,„\ ■ Ans. \ „ .. _. (l + l = ") , fz=:18a-24J. 15. Given ■{ ^ • Ans. \ oc i o^ 1^4-1— h\ (y=366 — 24a. 2a; . 32? 9 16. Given J.. , „, ^ • Ans. J , ^ A I K 1 OA ' 120 ,4a; + 11. Given PEOBLEMS LEADING TO SIMPLE EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES. 176i In a problem expressing two conditions, two re- quired quantities may be so related, that, one of them being found, the other may be readily derived from it; in which case the solution can be effected by means of a single letter. There are, however, certain problems whose solution requires each of the ULiknown quantities to be represented by its own proper symbol, and the formation of as many independent equations as there are unknown quantities. SIMPLE EQUATIONS. 147 1. A merchant sold at one time 3 hats and 4 caps for $ 23, and at another time 2 hats and 1 caps for $ 24 ; what was the price of each ? SOLUTION. ^®* « = the price of a hat, *°*^ y = th e price of a cap. Then, 3a; + 4y = 23 (1) and 2 a: -j- T y = 24 (2) Transposing and dividing (1), x= ^^—■^V /gx Transposing and dividing (2), x= ^ — ''V a\ Equating, 23-4,^24-^ ^5j Clearing of fractions, 46 — 8 y = '72 — 21 y (6) Eeducing, 13y=26 (Y) Whence, y = 2 (8) Substituting 2 for y in (4),^ x = ^^ ~ ''^ (9) Whence, a; = 5 (10) 2. The sum of two numbers is 133, and their difference is 4T ; required the numbers. Ans. 90 and 43. 3. A farmer pajd 4 men and 6 boys '?2 shillings for laboring one day, and afterwards, at the same rate, he paid 3 men and 9 boys 81 shillings for one day; what were the wages of each ? Ans. Men's wages, 9 shillings ; boys', 6 shillings. 4. The value of my two horses is such that, if the value of the first be added to four times the value of the second, the sum is % 580 ; and if the value of the second be added to four times that of the first, the sum is % 620 ; required the value of each. Ans. The first, % 100 ; the second, $ 120. Explain the Goludon of Problem 1. 148 ELEMENTARY ALGEBRA. 5. Find that number, consisting of two figures, to which, if the number formed by changing the place of the figures be added, the sum is 121 ; and if it is subtracted, the remainder is 9. Let and Then and Therefore and Hence and Then, and Dividing (1), Dividing (2), Adding (3) and (4), Whence, Subtracting (4) from (3), Whence, Therefore, SOLUTION. X = the first figure, y = the second figure. 10 a; = the first in tens' place, 10 y ^ the second in tens' place. 10 a; -j- y == the number required, 10 y -f- a; = the number formed. 11 a; -|- 11 y = the sum of the numbers, 9x — 9 y = the difierence of the numbers. (1) lla;+lly: 9x — 9y. 121 9 X 11 1 12 6 10 5 65 (2) (3) (4) (5) (6) 0) (8) (9) 2x = X = 2y = 10a;-f-y = 6. There is a number consisting of two figures, which is equal to four times the sum of those figures ; and if 9 be subtracted from twice the number, the places of the figures will be reversed ; what is the number ? Ans. 36. T. A gentleman asked a lady her age ; she replied : " 1 years ago I was three times as old as you, but if we live Y years longer, my age will be twice as great as yours " ; what were their ages ? Ans. Lady's age, 49 years ; gentleman's, 21 years. Explain the solution of Problem 5. SIMPLE EQUATIONS. 149 8. A said to B, "If ^ of your money were added to ^ of mine, the sum would be $6." B replied, "If J of yours were added to ^ of mine, the sum would be | 5f ." What sum had each ? Ans. A, $ 12 ; B, $ 16. 9. I have in two purses $ 84 ; and if the sum in the purse containing the most be divided by the sum in the other, the quotient will be 13. Eequired the sum in each purse ? Ans. In one, $ T8 ; in the other, $ 6. 10. The ages of a father and his son added together equal 140 years ; and the age of the father is to that of the son as 3 to 2. Let and Then, and Or, SOLUTIOlf. a; = the age of the father, y= the age of the son. x + y = 140 (1) x:y = 3:2 (2) 3y = ■.2x (3) y = 2x ' 3 (*) ). . 2x ^ + ^ = :140 (5) bx = :420 (6) X = : 84 0) y = : 56 (8) Dividing (3), 2 X Substituting -r- forj^i ■Reducing (5), Whence, From (4), 11. The age of James is to that of John as 3 to 4 ; but 6 years hence their ages will be in the ratio of 5 to 6. What are their ages ? Ans. James's age, 9 years ; John's 12 years. 12. Find two numbers, the greater of which shall be to 24 as their sum to 42, and the difference of which shall be to 6 as 4 to 3. Ans. 32 and 24. Explaiu the solution of Problem 10. 13* 150 ELEMENTARY ALGEBRA. 13. If 3 be added to the numerator of a certain frac- tion, its value will be -J ; and if 1 be subtracted from the denominator, its value will be I- What is the fraction? SOLUTION. Let X = the numerator, and y = the denominator. Therefore - = the fraction. ^ Then, ^ = i (1) and z^.=\ (2) y 3 X 1 r^T 5 Clearing (1) of fractions, 3 a; -f- 9 = y (3) Clearing (2) of fractions, 5 a; = y — 1 (4) Subtracting (3) from (4), 2 a; — 9 = — 1 (5) Or, 2 a: = 8 (6) Whence, a; = 4 (1) From (3), y = 21 (8) Hence, ^ = 21 ^^ 14. Divide 12 into two such parts that 3 times the greater shall exceed twice the less by 121. Aus. 53 and 19. 15. Fifty laborers were engaged to remove an obstruc- tion on a railroad ; some of them by agreement were to receive $ 0.90, and others, $ 1.50. There was paid them just $48, but no memorandum having been made, it is required to find how many worked at each rate. Ans. For f 0.90, 45 ; for % 1.50, 5. 16. The wages of 5 men and t women amount to $16.40, and *l men receive more than 6 women by $4. What does each receive ? Ans. Men, $1.60; women, $1.20. Explain the Eolution of Problem 13. SIMPLE EQUATIONS. 151 1*1. If 4 be added to the numerator of a certain frac- tion, its value will be J- ; and if t be added to its denom- inator, its value will be \. What is that fraction ? Ans. ^. 18. A sum of money was divided equally among a cer- tain number of persons ; had there been three more, each would have received 1 1 less, and had there been two fewer, each' would have received $ 1 more than he did ; required the number of persons, and what each received. SOLUTION. Let X = number of persons. and y = no. dollars each received; also, a;y = sum divided. Then, (x-\-3) (y — 1) = xy (1) and (x—2)(y-\-l)=xy (2) Prom (1), xy-\-3y — x — 3 = a;y (3) From (2), xy — 2f/-\-x — 2 = xi/ (4) Transposing in (3), 3 y — a: = 3 (5) Transposing in (4), x — 2y = 2 (6) Adding (6) and (6), y = 5 a) Prom (6), a; = 12 (8) 19. My income tax and assessed tax together amount to $ 30 ; but. if the income tax were increased 20 per cent, and the assessed tax were decreased 25 per cent, the two together would amount to $ 32^ ; required the amount of each tax. Ans. Income tax, $ 21jV i assessed tax, $ 8|^. 20. Eequired two quantities such that, if the first be increased by a, it will become m times the second ; and if the second be increased by b, it will become n times the first. ^^g_ a+i"^ ^„^ h + an mn — 1 mn — 1 Explain the Solution of Problem 18. 152 ELEMENTAEY ALGEBEA. 21. A has I as much money as B ; but if A should gain $ 10 and B lose the same sum, they will have equal amounts. How much has each? Ans. A, $16; B, |36. 22. A man and his wife can consume certain provisions in 15 days ; but after partaking of them for 6 days, the woman consumed the remainder in 30 days. In what time could either consume the whole ? Ans. The man, in 21f days ; his yfite, in 50 days. Eliminate before clearing of fractions, if the unknown quantities appear as denominators in each equation. Or, use negative expo- nents. (See Ex. 13, 14, Art. 175). 23. A merchant has sugar at a cents a pound and at b cents a pound ; how much of each must he take to make a mixture of d pounds, worth c cents a pound ? Ans. At a cents, — ^^- — ~ ; at 5 cents, — ^^ r^. 24. A composition of copper and tin, containing 100 cubic inches, weighed 505 ounces ; how many ounces of each metal did it contain, supposing a cubic inch of cop- per to weigh 5|- oz., and a cubic inch of tin to weigh 4^ oz. ? Ans. Copper, 420 oz. ; tin, 85 oz. 25. There is a rectangular garden of a certain size ; if it were 5 feet broader and 4 feet longer, it would contain 116 square feet more ; and if it were 4 feet broader and 6 feet longer, it would contain 113 square feet more. Eequired its dimensions. Ans.- Length, 12 feet ; breadth, 9 feet. 26. A person possesses certain capital which is invested at a certain rate per, cent. A second person has $ 1000 more capital than the first person, and invests it at one per cent more; thus his income exceeds that of the first person by $ 80. A third person has $ 1500 more capital than the first, and invests it at two per cent more ; thus SIMPLE EQUATION 168 his income exceeds that of the first person by $ 150. Re- quired the sum of each person, and the rate at which it is invested. ("Sums, $3000, $4000, and $4500. (, Rates, 4, 5, and 6 per cent. SIMPLE EQUATIONS CONTAINING THREE OR MORE UNKNOWN QUANTITIES. 177. Any of the methods which have been given for the solution of simple equations containing two unknown quantities may be extended to those containing three or more unknown quantities. ( 3^+ y+ «= ^) 1. Given < x -\- 2 y -\- S z = li \ to find x, y, and z. (Sa;— y-\-4.z—U) By multiplying equation (1) by 3, we obtain equa- tion (4), and by subtracting (1) from (2), and (4) from (3), we have (5) and (6), equations containing only two unknown quantities. Multiplying (e) by 2, and subtracting the product (7) from (5), give_ (8), an equa- tion containing only one un- known quantity, y. Dividing (8) by 9, we have (9), or y =2. Substituting 2 for y in (5) gives (10), and re- ducing, we have (12), or z i= 3. Substituting 2 for y, and 3 for z in (1), and re- ducing, we have (14), or a; = 1. Explain the operation, OPERATION. «+ y-\- "■= 6 (1) a; + 2y-f-32: = 14 (2) 3a;— y-\-A.z=. 13 (3) 3a;-f-3y-f-3z = 18 W y + 2z = 8 (5) — 4y+ z = - - 5 (6) — 8y + 2« = - -10 0) 9y = 18 (8) y = 2 (9) 24-2» = 8 (10) 20 = 6 (11) z = 3 (12) a; + 2 + 3-= 6 (13) X = 1 (14) X 53 y — ■ z X = lot 2.V- Zz X = IST 3y- ,4z 154 ELEMPNTAEY ALGEBRA. 2. Given U + 2y + 3z = lOt j ^^ ^^^ ^_ OPERATION. a;+ y+ «= 53 (1) a; 4-23^ + 32= 101 (2) a;-f-3y-i-^^=13'7 (3) (^) (5) (6) 53— y— z=101 — 2y—3z (T) lOY— 2y — 32 = 13t — 3^^ — 4» (8) 2^= 54 — 2 » (9) y = 30 — « (10) 30— z= b4t — 2z (11) «= 24 (12) y= 6 (13) a;= 23 (14) By transposing terms in (1), (2), and (3), we obtain equations (4), (5), and (6). Equating the second members of (4) and (5), and those of (5) and (6), we have (7) and (8), equations containing only two unknown quantities. Transposing terms in (7) and (8), we have (9) and (10). Equating the second members of (9) and (10) gives (11), an equation with only one unknown quantity, which reduces to (12), or s ^ 24. Substituting 24 for z in (10), and reducing, we have (13), or y = 6 ; and substituting 24 for z, and 6 for y, in (4), and reducing, we have (14), or a; = 23. From the preceding examples and illustrations, we deduce the following RULE. Deduce from the given equations, ly elimination, a new set of equations containing one less unknown quantity, and condnite Explain the operation.. Repeat the Rule. SIMPLE EQUATIONS. 155 the pmeess until an equation is oUained containing hut me un- known quantity. Find the value of the unknown quaraity in this equation. By substituting this value in either one of the set of two equa- tions containing, two unknown quantities, find the value of a second unknown quantity. Then, by substituting these values in either of the equations which contain three unknown quan- tities, find the value of a third; and so on, till the values of aM are found. Note. Upon the good judgment and discrimination of the learner in selecting the quantity to be first eliminated, and the method of elim- ination suited to the particular case, will depend the simplicity and ele- gance of the solution. Examples. Find the values of the unknown quantities in the fol- lowing equations. f x-\-2y-\- « = 24) 3. Given J2a;+ y-j-3«=38[. Ans. (3a:+3y-}-2« = 46) !4:X-\-2y— e=z26\ 5x-{--2y — 3z=1GV. Ans. 2x— y-{-2z = 23) 5. Given 6. Given x-\-y-{-z=33 y — x-\-z=i23 z — X — y =■ 1 M -)- a; -|- y = 6 " u -\- X -\- z = Q «* + 3'-i-« = 8 x-\-y + z = *l J Ans. Ans. (X = 4. p = 6. Kz = 8. (X = 6. y = 5. (« = 8. (X = 5. .y = 11. 1. = 17. u = 3. X := 2. y = 1 .z znz 4 The solution may here be abridged, by the artifice of assuming the sum of the four unknown quantities to equal s. What is the Note ? Explain the operation of Example 6. 156 KLEMENTAKY ALGEBRA. Thus, U-\-X-^1/-\-Z= s Then the first equation is s — z= 6 (1) The second is s — y= 9 (2) The third is s — x= 8 (3) The fourth is S — M = 7 (4) By addition, 4s — s = 30 (5) Whence, s= 10 («) Substituting the value of s in (4), (3), (2), and (1), and reduo- ing, we have u = 3, x = 2, y = 1, and 3 = 4. Ans. ■x-\-y-{-z = 131 T. Given ■ u-\-x-\-i/=11 u-\-x-{-z=l8 .M + y+2 = 2lJ 'tx — 3y— 2=12 8. Given a; + 2y + 3z=lt '4a;— y + 2z=13 r a: 4-^ — = \ 9. Given [y-j-z — X = i) a ' h 10. Given i ^ +i = l a ' c -• 1 + - = 1 1.6 ' c J Ans. Ans. Ans. Note. Eliminate before clearing of fractions. ^ X -^ y=a) (x = ^(a-\-h — c). 11. Given X -\- z ^= b \y -\-z = e Ans. \ y ^ 5- (a + c — J), (z =^(5 +c— a). 4 12. Given { -— + 3 T^ 2 ~ " ^ + - = 11 2^4 — -^J^ 3x L 4 "T" 3 2 ~ Ans. SIMPLE EQUATIONS. 157 PROBLEMS LEADING TO SIMPLE EQUATIONS CONTAINING THKEE OR MORE UNKNOWN QUANTITIES. 178. Problems leadings to simple equations contaimng three or more unknown quantities require precisely analo- gous processes in their solution to those required by prob- lems leading to simple equations containing two unknown quantities. 1. Three boys, James, Henry, and Arthur, bought fruit at the same prices. James paid for 3 oranges, 1 apple, and 2 pears, 14 cents ; Henry paid for 4 oranges, 3 apples, and 1 pear, 11 cents ; and Arthur paid for 1 orange, 4 apples, and 3 pears, 13 cents. What was the price of each ? Ans. Oranges, 3 cents ; apples, 1 cent ; pears, 2 cents. 2. A gentleman divided $ 100 among his four daugh- ters, Mary, Isabel, Jane, and Ellen, in such a manner, that twice Isabel's part added to three times Ellen's part was $ 160 ; three times Mary's part added to twice Jane's part was $ 90 ; twice Mary's part added to Ellen's part was $ 60. What sum did each receive ? Ans. Mary, $10; Isabel, $20; Jane, $30; Ellen, $40. 3. I have three ingots, composed of different metals. A pound of the first contains Y ounces of silver, 3 ounces of copper, and 6 ounces of tin ; a pound of the second contains 12 ounces of silver, 3 ounces of copper, and 1 ounce of tin ; and a pound of the third contains 4 ounces of silver, .'T ounces of copper, and 5 ounces of tin. How much of each of these three ingots must be taken in order to form a fourth, each pound of which shall contain 8 ounces of silver, 3|- ounces of copper, and 4j- ounces of tin? Ans. Of the first, 8 ounces ; of the second, 5 ounces ; and .of the third, 3 ounces. 14 158 ELEMENTARY ALGEBRA. Let X, y, and 2 denote the number of ounces that must be taken of each of the three ingots, respectively. Then, since there are 7 ounces of silver in a pound, or 16 ounces, of the first ingot, in 1 ounce of it there are =-j of an ounce of silver, and, consequently, in 16 X ounces there are 7^ of an ounce of silver. In like manner, we 16 may find that — ^ and -^ denote the number of ounces of silver •' 16 16 to be taken of the second and third ; but, by the problem, one pound of the fourth ingot is to contain 8 ounces of silver ; hence we have for the first equation, tf.4. ]lK^ — = 8 16 "T" 16 "T" 16 Proceeding in like manner with respect to the copper and tin, we have for the other equations, 3_£ . 3^ . 7j 15 16 "I" 16 "1" 16 ~-4 ' 6a; 16"' 16 ' 16 1 IB 1" IB 4. ■ From these equations, the results given above are readily ob- tained. 4. A gentleman purchased a chaise, horse, and harness for $400. He paid four times as much for the chaise as for the harness, and one third as much for the harness as for the horse. How much did he pay for each? Ans. Chaise, $ 200 ; horse, $ 150 ; harness, $ 50. 5.- There are three numbers whose sum is 324 ; the second exceeds the first as much as the third exceeds the second ; and the first is to the third as 5 to t. What are the numbers ? Ans. 90, 108, and 126. 6. A man speaking with his wife and son respecting their ages, said that his age added to that of his son was 12 years more than that of his wife ; the wife said that her age added to that of her son was 8 years more than that of her husband, and that their ages together amount- ed to 92 years. Kequired the age of each. Ans. Husband, 42 years ; wife, 40 years ; son, 10 years. SIMPLE EQUATIONS. 159 7. A bin holding 146 bushels is filled with a mixture of wheat, barley, and oats. The barley exceeds the wheat by 15 bushels, and there are as many bushels of oats as of both wheat and barley. What is the quantity of each ? Ans. Wheat, 29 bushels ; barley, 44 bushels ; and oats, 13 bushels. 8. A and B can perform a piece of work in 8 days, A and C in 9 days, and B and in 10 days ; in how many days can each alone perform it ? Ans. A, in 14ff days ; B, in llf J days ; and 0, in 9. A certain number consists of three digits, whose sum is 9. If 198 be subtracted from the number, the remainder will consist of the. same digits in a reverse order; and if the number be divided by the digit at the left, the quotient is 108. What is the number? Ans. 432. Let a;, y, and z denote the digits, respectively, beginning^ at the left ; then, \Wx -\- 10y-|-z = the number. 10. I have three horses, and a carriage, which of itself is worth $ 220. If I put the carriage with the first horse, it will make the value. equal to that of the second an.d third ; but if I put it with the second horse, it will make the value double that of the first and third ; and if I put it with the third norse, it will make the value triple that of. the first and second. What is the value of each horse ? Ans. First, $ 20 ; second, $ 100 ; third, $ 140. 11. A and B can reap a certain field- in a days, A and C in 5 days, and B and C in c days ; in what time can each alone. reap it? Ans. A, m p-j 5 days ; B, m , , , ^ — — - days; ' ac-{-bc — ab "' ' ao-]-bc — ac _ . 2abc , V, m — J— i j— days. ' ab -j- ac — be " 160 ELEMENTARY ALGEBRA. 12. Find three numbers such that J of the first, ^.of the second, and i of the third shall be equal to 62 ; i of the first, i of the second, and ^ of the third shall be equal to 4T ; and i of the first, -^ of the second, and J of the third shall be equal to 38. Ans. 24, 60, and 120. 13. Three boys, A, B, and C, owe, together, $2.19, and no one of them has so much money. But by uniting, it is found that it can be paid in several ways ; first, by ^ of B's money and all of A's ; secondly, by |- of C's money and all of B's ; or, thirdly, by f of A's money and all of C's. How much money has each ? Ans. A, $1.53; B, $1.54; and C, $1.1T. 14. Find four numbers, such that the first, together with half the second, may be equal to 35t ; the second, with -J of the third, equal to 476 ; the third, with ^ of the fourth, equal to 595 ; and the fourth, with | of the first, equal to '114. Ans. First number, 190 ; second, 334 ; third, 426 ; fourth, 61Q. 15. A merchant has three kinds of sugar. He can sell 3 lbs. of the first quality, 4 lbs. of the second quality, and 2 lbs. of the third quality, for 60 cents ; or, he can sell 4 lbs. of the first quality, 1 lb. of the second quality, and 6 lbs. of the third quality, for 59 cents ; or, he can sell 1 lb. of the first quality, 10 lbs. of the second quality, and 3 lbs. of the third quality, for 90 cents. Eequired the price of each quality. Ans. First quality, 8 cts. per lb. ; second, 1 cts. ; third, 4 cts. 16. A, B, and C engaged in a squirrel hunt, and killed 96 squirrels, which they wish to share equally. In order to do this. A, who has most, gives to B and C as many as they each already had; then, B gives to A and C as many as they each had after the first division ; and, lastly, C gives to A and B as many as each had after the second division, when it was found that each had the same num- ber. How many had each ? Ans. A, 52 ; B, 28 ; and 0, 16. DISCUSSION OF PROBLEMS. 161 DISCUSSION OF SOME PROBLEMS LEADING- TO SIMPLE EQUATIONS. 179i The Discussion of a problem consists in attribut- ing various values and relations to the known quantities entering into the general equation, and in interpreting - - the results. INTERPRETATION OF NEGATIVE RESULTS. 180. The interpretation of negative results obtained by- means of simple equations is illustrated in the problems which follow. 1. Let it be required to find what number must be added to the number a, that the sum may be h. Let X = the required number. Then, a-\- x = b, whence, x = b — a. Here, the value of x corresponds to any assigned values of a and b. Thus, for example, • Let a = 12 and 5 = 25. Then, a;= 25 — 12 = 13, which satisfies the conditions of the problem, for if 13- be added to 12, or a, the sum wiU be 25, or 6. But suppose a = 30 and b == 24. Then a; = 24 — 30 = — 6, which indicates that, under the latter hypothesis, the problem is impossible in an arithmetical sense, though it is possible in the alffe- , ^braic sense of the words "number,'' "added," and " sum." In what does the Discussion of a problem consist? Give- the discus- sion of the first problem, stating what the negative result points out, and correcting the enunciation. 14* 162 ELEMENTAEY ALGEBRA. The negative result, — 6, points out, therefore, either an error or an impossibility. But, taking the "value of x with a contrary sign, -we see that it will satisfy the enunciation of, the problem, in an arithmetical sense, when modified so as to read : What number must be taken from 30, that the difference may be 24? 2. Let it be required to find the epoch at which A's age is twice as great as B's, A's age at present being 35 years, and B's 20 years. Let us suppose the required epoch to be after the present date. Then X = the number of years after the present date, and S5-\-x= 2 (20 + k) ; whence, x = — 5, a negative result. On recurring to the problem, we find it is so worded as to admit also of the supposition that the epoch is before the present date, and taking the value of x obtained, with the contrary sign, we find it will satisfy that enunciation. Hence, a negative result here indicates that a wrong choice was made of two possible suppositions which the problem allowed. From the foregoing examples and illustrations we may infer : — 1. TTiat negative results indicate either an erroneous enuncia- tion of a problem, or a wrong supposition respecting the quality of some quantity belonging to it. 2. That we may form a possible problem, analogous to that which involved the impossibility, or correct the wrong supposition, by attributing to the unknown quantity in the equation a quality directly opposite to that which had been attributed to it. 3. Hiat the true answer of the corrected problem wiU be fouiii by simply changing the sign of the negative result obtained. Give the discussion of the second problem, and show what the neg- ative result indicates. . What may be inferred from the examples and illustrations t DISCUSSION OF PROBLEMS. 163 Interpret the negative answers obtained, and modify the enunciation so as to give positive results, for the fol- lowing PROBLEMS. 3. What number must be taken from 10, that the re- mainder shall be 15 ? Ans. — 5. 4. What number is that whose fifth part exceeds its fourth part by 4 ? Ans. — 80. 5. A man at the time of his marriage was 40 years old, and his wife 36 years ; how many years must elapse before his age will be to hers as 6 to 5 ? Ans. — 16 years. 6. The length of a certain field is 8 rods, and its breadth 5 rods ; how much must be added to its length that its contents may be 30 square rods ? Ans. — 2 rods. 1, What number is that, the sum of the third and fifth parts of which, diminished by T, is equal to the original number? Ans. -^15. 8. If 2 be added to the numerator of a certain frac- tion, its value is |- ; but if 2 be added to its denomina- tor, its value is j. What is the fraction? . — 5 •— 12 9. A father has lived 45 years, and his son 15 years. Find in how many years the age of the son will be one fourth of the age of the father. Ans. — 5. 10. A man worked 12 days, his son being with him 8 days, and received $ 22, besides the subsistence of himself and son while at work. At another time he worked 10 days, and had his son with him 4 days, and received $ 19. What were the daily wages of each ? Ans. The father's wages, $ 2 ; the son's, — 25 cts. That is, the father earned $2 a day, and was at the expense of $ 0.25 a day for his son's subsistence. 164 ELEMENTAKY ALGEBRA. ZEKO AND INFINITY. 181. Zero, which is represented by the symbol 0, not only denotes absence of value, or nothing, but may, in Algebra, stand for a quantity le,S3 than any assignable value. 182. Infinity, which is represented by the symbol co, denotes a quantity greater than any assignable value. In comparison with infinities, finite values may be con- sidered as all equal to one another. INTEEPEETATION OP ^, -, -^, AND ^. to A 183. In order to explain the meaning of these symbols, let us take the fraction r- 1. SupptJse the numerator, a, to remain constant, while the de- nominator, 6, continually decreases. Then, since the value of a frac- tion depends upon the relative value of its terms (Art 113), the fraction must increase as the denominator decreases ; consequently, when 6 decreases below any determinate limits, the value of the fraction must exceed any determinate or assignable quantity. Hence, representing any finite quantity by A, we have f = »- That is, If a finite quantity is divided hy zero, the quotient is infinity. 2. Suppose the numerator, a, to remain constant, while the de- nominator, 6, constantly increases. Then, the value of the fraction must decrease as the denominator increases ; consequently, when S increases beyond any determinate limits, the value of the fraction must be less than any determinate or assignable quantity. Hence we have ^ = 0. 00 A A Define Zero. Infinity. Interpret the symbol -^. The symbol — . DISCUSSION OF PROBLEMS. 165 That is, Jf a finite quantity is divided hy infinity, the quotient is zero. 3. Suppose, now, the denominator, 6, to remain constant, while the numerator, a, constantly decreases. Then the value of the frac- tion must decrease as the numerator decreases; consequently, when a decreases below any determinate limits, the value of the fraction must be less than any determinate or assignable quantity. Hence we have ! = «■ That is, If zero is divided hy a finite quantity, the quotient is zero. 4. Suppose, next, a and h both to decrease, at the same time and in the same ratio. Then, the value of the fraction will not be changed ; but when a and 6 decrease below any determinate limits, the terms of the fractions each become zero, and the fraction itself becomes -. As - may have any value, - will represent any finite quan- tity. Hence, If zero is divided hy zero, the quotient may he any finite quantity. • • Note. If « is the result of an expression whose numerator contained more zero factors than its denominator, its value is ; and if its de- nominator contained more zero factors than its numerator, its value is' co. Sometimes ^r- , by canceling a common factor in the terms of the frac- tion from which it originates, is fonnd to have a definite, finite value. 184. From the foregoing discussion we draw the fol- lowing inferences : — 1. That a prohlem whose result appears under the form of r- is impossihle, or carmot he satisfied hy finite ■ quantifies. 2. That a problem whose result appears under the form of - is generally indeterminate, or can be satisfied by any finite quantities whatever. Interpret the symbol j. The symbol ~ . What two inferences are drawn? 166 ELEMENTARY ALGEBRA. Interpret the results which may be obtained in the following PROBLEMS. 1. Three railway companies issue a, h, and c shares, respectively, and the price paid. on each is the same sum per share. But the second and third afterwards call for p dollars and q dollars per share, respectively, in addi- tion to that originally paid, whereby the total capital paid up on the first and second together becomes double that paid up on the third. Required the price per share originally paid up. Let X := the price required. Then, ax -\- h (x -\- p) ^ 2 c (x -{- q) \ whence, x = — —-j — &- • If a = 9000, 5 = 11000, c = 10000,;? = 19, and 5= 11, 11000 ■?rhich shows that the problem, according to the conditions, is impos- sible. Again, if a = 9000, 5 = 11000, c = 10000,^ = 10, and q = 5\, ^ = 0' which shows that the conditions are satisfied without reference tp the sum originally paid up ; and that the unknown quantity may have any finite value whatever. 2. A is 60 years old, and B 40 years ; when will they both be of the same age ? Ans. 00. 3. A person buys 400 sheep in two flocks ; for the first he pays $1.50 per head, and for the second $2. Of the first he loses 30, and of the second 56. He then sells the remainder of the first flock at $ 2 per head, and of the second at $ 2.50 per head, and finds he has lost nothing. Required the number in each flock. Ans. 2. INVOLUTION. 157 INVOLUTION. 185. A Power of any quantity is the product obtained by taking that quantity one or more times as a factor. Thus, a = c^ is the first power of a, aa = a^ " second power, or square or a, aaa=.a^ " third power, or cube of a, aaaa = a* " fourth power of a ; and so on, the exponent (Art. 19) of the power denoting the number of times the quantity a is taken as a factor. If the exponent is n, the power is the product of n factors, when n is any entire quantity whatever. 186i Intgltjtion is the process of raising a given quan- tity to any required power. This may be effected, as is evident from the definition of a power, by taking the given quantity as a factor as many times as there are units in the exponent of the required power. 187. When the quarvtity to he involved is positive, aU'the powers wiU he positive. For, any positive factor taken any number of times must always give a positive result (Art. 59). Thus, (+ a) X (+ a) = + a» (+ «) X (+ a) X (+ a) = + a*, and so on. 188. When the quantity to he involved is negative, aU the even powers will he positive, and all the odd powers negative. For, a negative multiplier causes the sign of the product to be the opposite of that of the multiplicand (Art. 59), and therefore eaxih Define a, Power. What does the exponent of the power denote 1 If the exponent is n, what is the power? Define Involution. "When the qnantity involved is positive, what sign do the powers take ? When the quantity is negative? 168 ELKMENTARY ALGEBRA. additional negative factor changes the sign of the result. As the first power is negative, each odd power will be negative, and the even ones positive. Thus, — a (—a) X (— a) = + a» (—a) X (-«) X (-«) = (+ cf) X (— a) = — a' (—a) X (—a) X (—a) X (—a) = (- a') X (— «) = + a* and so on. 189. The involution of a polynomial, or of a monomial composed of several factors, is indicated by inclosing the quantity in a parenthesis, and writing the exponent at the right and a little above the expression. Thus, (a -f- by indicates the second power of a-\-b; (2 ah cY indicates the fifth power of 2 a 5 c. POWEES OF MONOMIALS, 190. It is evident that the rules for involution musli be based upon those for multiplication. 1. Let it be required to raise 3 a" 6 to the third power. OPERATION. (3 a^ 5)8 = 3 a2 J X 3 a^ S X 3 a^ J = 3X3X ia^a^anhh = 27 a» W. Since the required power is equivalent to the given quantity- taken three times as a factor (Art. 185), we proceed, by the rule for multiplication (Art. 62), to find the product of 3 a" 6 X 3 a* J X 3a*6, or 27a°6'; from which it appears, — 1. That the coefficient 3 has been raised to the third power. 2. That the exponent of each letter has been multiplied by 3, the exponent of the power. How is the involution of a polynomial, or of a monomial composed of several factors, indicated ? Explain the operation. INVOLUTION. 169 Hence, for raising a monomial to any power, we have the following RULE. Eaise the numerical coefficient to the required power, and midtiply the exponent of each letter hy the exponent of the re- quired power. Note. If the quantity involved is positive, all its powers will be pos- itive {Art. 187) ; but if it is negative, all the ecen powers will be pos- itive, and all the odd powers will be negative (Art. 188). Examples. 2. Find the cube of a h. Ans. c? W. 3. Find the square of a a?. Ans. a^ x*. 4. Find the fourth power of a? y. Ans. a? y*. 5. Find the third power of a 6 a;". Ans. c?Wa?^. 6. Find the with power oi c^'f. Ans. c'^ a?^ f^. 7. Find the fifth power of ia^a?. 8. Eaise — 3 a; to the third power. Ans. — 2t a?. 9. Raise — 4 a;^ to the' second power. Ans. 16 a;*, 10. Raise a^Vcd^ to the fourth power. Ans. a^e^c*d\ 11. Required the cube of — 4:a^¥x*. Ans. —64:aH^o^. ' 12. Required the fourth power of 5 a^ J' c*. Ans. Q2&c^m^\ 13. Required the square of — 3 a J^ a;. Ans. ^a^¥x^. 14. Raise lalx? to the sixth power. •Ans. 64aH»<^. Repeat the Rule. The Note. 15 170 ELEMENTARY ALGEBRA. 15. Raise —2 as? to the fourth power. 16 Eequired the fifth power of 4aa^y. Ans. 1024o*a^V- It. Eequired the nth power of — Go? 5*. Ans. ± 6" a** 5*"- NoTB. Since n may be any number whatever (Art. 185), the nth power of the given negative quantity may be either, even or odd, and therefore either positive or negative, as is indicated by the sign ±. POWERS OF FRACTIONS. 191. Fractions, like entire quantities, are involved by multiplication. 2 a x' 1. Let it be required to find the third power of r-r — . OPERATION. •2aa^ ,2aa;'..2aa». X a !„ X \ybc) ~ 3T7 ^ 3 67 ^ 3 6c 2oar'X2aa;''X 2 a a;* 8 a" x? 3 6c X 3 5 c xTsTc WW^' Since the required power requires the given quantity to be taken three times aa a factor (Art. 185), we find it by multiplying, as in multiplication effractions (Art. 137). Hence the following RULE. Raise both the numerator and the denominator to the re- quired power. Examples. n n- J ii. „ 3 0= 6 . 9 a* 6' 2. Find the square of --=-. Ans. lirJl, ' a 49 d' J ^^^^^^ ^ Why does the answer to Example 17 have the sign ±? Explain the operation. Bepeat the Rule. INVOLUTION. 171 3. Find the cube of ® *' c A ^ 27 6" L Find the square of ^"f. 6 , 4a«ai* A°«- 25 6- 5. Required the fifth power of ^. A «'«"' 6. llequired the fourth power of -j^ • Ans. j/. 1. Required the third power of ^^. Al^^- 8 3^' 8. Required the fifth power of -^-vj- • 3 9. Required the fourth power of - a' e^. 10. Required the fiecond power of — - — —- • 81 We''""'' ^°«- 121^- 192t llhe rules already given hold true when any of the exponents are negative. For and (a»)-" = ^ = ± = cT^. (Art. Tl.) 1. Required the third power of 5 ar^ b~\ Ans. 125 cr*b-». 2. Required the fourth pbWei* of -- 2 c" eT^ Ans. 16c"d-'. 3. Required the wth power of — 6 a ar^)/^. Ans. ± 6»fls"ar-'^y»". 4. Develop the expression ( — ar^^zy. Ans. — a^-»*'^^'"^^ 5. Develop the expression (^x* if~^ sr^)~^- Ans. x-*y'i'. 172 ELEMENTARY ALGEBRA. 6. Develop the expression (—mr^rr^yK Ans. — m' »'. 1. Develop the expression (— 2 a"' 6-^)-*. Ans. ^sa"5«. a"" c" 8. Eequired the sixth power of ^ni • Ans, -jis-- 9. Raise — _a» .^» to the third power. . . 125o-°c'd' 10. What is the mth power of ^^^^ ^ n-a ? POWERS OF BINOMIALS. 193. Binomials, like monomials, may be raised to any power by the process of successive multiplications. Thus, a-{-b raised to the second power is (ffl + J) (a + S) = a^ + 2 a J + J^. And a — b raised to the third power is {a — b) (a — h) (a — b)=a' — 3a'b-\-3a¥ — W. But this process of involving binomials by actual mul- tiplication must be very tedious, when high powers are required. There is, however, a much abridged process, discovered by Sir Isaac Newton, called THE BINOMIAL THEOREM. 194. The Binomial Theorem expresses a general method of developing any power of a binomial. How may binomials be raised to any power 1 What does the Bi- nomial Theorem express? INVOLUTION. 173 195i With a view of elucidating the principles govern- ing the development of Newton's theorem, we shall, by actual multiplication, find a few of the powers of a bino- mial, when both terms are positive ; and also when one term is positive and the other negative. 1. Let a -|- 6 be raised to the fifth power. a -\-b 1st power. a +b o^-f ab + ab +i a^ + 2 a 6 + i a +6 a»-|-2a^J + a 6^ 2aJ» 4-i^ a^^3aH + « 4-* saV" -\-y . . . . + «'* + 3 aH^ + a¥ 3aH^-\- 3al? -\-b* a 4-6 6aH^-{- i.al? -fJ* . + «*6 + Q^ . . . . . • a — h a^ — 1an-\- a J'" — a" 5 + 2a J'' —J^ a» — 3a2J + ^aV" —V . . a —h a* _ 3 aH + 3 aH^ — o J» — d'l-\- 3a2J3_ SaJ' +S* a* — 4a'J + 6a2j2_ 4a J? ^ J4 . . a —h aB ^ 4 ai J _|_ 6a»62_ 4a''J'+ «6* — an-\- 4aH2— 6a=J' + 4ffli*- -5" 3d power. 4th power. a6 _ 5 a* 6 + 10 a» 6^ — 10 aH» + 5 « 6^ — 5' 5th powey, In like manner, the higher powers may be developed. It will 1)6 seen that the number of multiplications is uni- formly less by one than the number of units in the expo- nent of the power. It will also be seen on examination, that certain invariable laws hold with regard to five other things : — 1. The number of terms. 2. The signs of the terms. 3. The letters in the terms. 4. The exponents of the letters. 5j The coefficients of the terms. What is seen with regard to the number of maltiplicationg ? What five other things follow invariable laws % INVOLUTION. 175 NUMBER OF THE TEKMS. 196i By examining either of the examples, we observe that the first power has two terms ; the second power, three terms ; the third power, four terms ; the fourth power, five terms ; and the fifth power, six terms. Hence, The number of terms is always one more than the exponent of the power. SIGNS OF THE TEEMS. 197t By examining the two examples, we observe that all the terms of the powers oi a -\-h are positive ; and of those of a — h, all the odd terms, reckoning from the left, are positive, and all the even terms are negative. Hence, When both terms of the binomial are positive, all the term^ of the power are positive. When the second term of the binomial is negative, cdl the odd terms, reckoning from the left, are positive, and all the even terms negative. LETTERS IN THE TEEMS. 198. From the examination of the several powers, it is evident that The leading letter or quantity enters all the terms of the power except the last; the following letter or quantity enters all the terms except the first ; and the product of some powers of both letters compose all the intermediate terms. EXPONENTS OF THE LETTERS. 199. By observing the diiferent powers of a -{- b, and of a — b, we shall find that the exponents of the letters What is the number of terms in any power of a binomial? What are the signs of the terms 1 In what manner do the letters enter into the terms ? What is the law governing the exponents of the letters ? 176 ELEMENTARY ALGEBRA. of the several terms follow an invariable order. Thus, in the fifth power of each of the binomials, the exponents are, Of a, 5, 4, 3, 2, 1, ; Of 6, 0, 1, ?, 3, 4, 5 ; whose sum in each term is 5, or the same as the expo- nent of the power. Hence, The exponent of the hading letter in the first term is the same as the exponent of the power, and decreases ly one in each successive term to the right. The exponent of the following letter in the second term is one, and increases hy one in each successive term to the right, until the last, where the exponent is the same as that of the power. The sum of the exponents in any term is the same, as the eooponent of the power. COEFFICIENTS OP THE TERMS. 200i It will be observed that the coefiScients of any poTyer in the examples, as the fifth power, are, Of the first term, of, 1 ; Of the second term, 5 a* b, the same as the exponent of the power, or 5 ; Of the third term, 10 a' ¥, the product of the coeffi- cient of the preceding term by the exponent of the lead- ing letter in that term, divided by 2, the number which marks the place of the term, or C' = 10 ; and, in like manner, the coefiicient of any term. Hence, The coefficient of the first term is one; that of the second term is the same as the eacponent of the power; and, in gen- eral, the coefficient of any term is found hy multiplying the What is the law governing the coefficients of the terms ? INVOLUTION. 177 coefficient of the preceding term h/ the exponent of the leading ktter of the same term, and dividing the product h/ the num- her which marks its place. Note 1. When the number of terms is even, there will be two terms in the middle, having the same coefficient; and since the same coefficients are repeated in an inverse order after passing the middle term or terms, most of the coefficients may be obtained without actual calculation. Note 2. It will be seen that Theorems I. and II., Arts. 76 and 77, are only special cases coming under the Binomial Theorem. EXAUFLES. 1. Eaise x — y to the third power. OPERATION. Coefficients and signs, 1 — 3 -|- 3 — 1 X and its exponents, 3^ x' x y and its exponents, y 1^ ^' Combining, x* — Sa^y + Sa:^^ — / After a little practice, the learner can write out the final form at once. 2. Eaise x-\-y Xo the second power. Ans. a?-\-2xy-\-f. 3. Expand {c — d)*. Ans. c« — 4c»rf+6c2. 13. Find the third power of J a; — § y. Ans. i^ — ^x'y + %xf — ^f. 14. Required the sixth power of a;" — 2 x. Ans. a;>«— 12a;» + 60aJ»— 160a;»4-240a;8 — 192a;»4-64a;^ ' Note. The Binomial Theorem may be applied to the development of the powers of any polynomial whatever. Thus, by changing the form of a + 6 + c to {a -\- b) -\- c, or the form of a + b — c + rfto (a + b) — (c — d), they may be treated as binomials ; and so of any other polynomial. Repeat the Note. 180 ELEJIENTAEY ALGEBRA. EVOLUTION. 202. A EooT of any quantity is a factor taken a cer- tain number of times to form that quantity. Thus, a is the second or square root of a" ; a is the third or cube root of c^. 203. Eoots are indicated either by the radical sign, or by a fractional exponent (Art. 22). Thus, /\/ a, or a*, indicates the second or square root of a ; /^ a, or a^, indicates the third or cube root of a ; ^ a, or a^, indicates the fourth root of a ; 1. i^a, or a", indicates the wth root of a ; 2 A^ c?, or a^, indicates the third root of the second power of a ; and so on, the index of the radical, or the denominator of the fractional exponent, denoting the .de- gree of the root. 204. Evolution is the process of extracting any re- quired root of a given quantity. It is the reverse of involution. 2C5. Any quantity whose root can be extracted is called a perfect power, and any quantity whose root cannot be extracted, an imperfect power, A quantity, however, may be a perfect power of one degree, and not of another. Thus, 8 is a perfect cube, but not a perfect square. Define a Boot of any quantity. How are roots indicated ? How is the degree of a root denoted ? Define Evolution. What is a perfect power ? An imperfect power ? EVOLUTION. 181 206> The odd roots of a positive quantity are positive. For, a positive quantity raised to any power is positive (Art. 187) ; but a negative quantity raised to any odd power is negative (Art. 188). Thus, 4^cF= -\-a, or 4nf= +a. 207. The even roots of a positive quantity are either positive or negative. For, either a positive or a negative quantity raised to an even power is positive (Arts. 187, 188). Thus, x/ a^ == ± a, or 4^~c? = ± a. 208. The odd roots of a negative quantity are negative. For, a negative quantity raised to an odd power is negative (Art 188) ; but a positive quantity raised to any power is positive (Art. 187). Thus, i^~^= — 3, or '^'^^^^ == —a. 209. Eeen roots j>f a negative quantity are not possible. For, no quantity raised to an even power can produce a nega- tive result (Arts. 187, 188). Thus, . V — 4, 4^—64, and \/ —a\ or indicated even roots of negative quantities, are called impossible, or imaginary quantities. SQUAEE ROOT OP NUMBEES. 210. The Square Eoot, or second boot, of a number is a factor which must be taken twice to form that number. Thus, -v/9 = 3, because 3X3 = 9. Why are the odd roots of a positive quantity positive? Why are the even roots either positive or negative ? Why arc the odd roots of a neg- ative quantity negative ? Why are the even roots Impossible 1 What are indicated even roots of negative quantities called? Define the Square Boot, or second root, of a number. 16 182 ELEMENTAEY ALGEBRA. 21 li A Perfect Square is any number or quantity that can be resolved into two equal factors. (Art. 205.) 212. The square of any integral number consists of twice as many places of figures as the number itself, or of one less than twice as mcmy. For, the first tea numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and their squares are 1, 4, 9, 16, 25, 36, 49, 64,' 81, 100; also, the square of 99 is 9801, of 100 is 10000, of 999 is 998001, of 1000 is, 1000000, and sq on. H^nce, 213i If a point be placed over every second figure in any integral number, beginning with the units' place, the number of points will show the number of figures in the square root. 21 4 • The square of any number ^ consisting of more than one place of figures, is equal to the square of the tens, plus twice the product of the tens by the units, plus the square of the units. For, if the tens of a nunaber be denoted by a, Jind the units by 6, the, number ■will be denoted by a-\-b, and its square by (a + by = a'+,2ab + V. Then, by this formula, if a ^ 3 tens, or 30, and 5 = 6, we have 3. tens 4- 6 nnits = 30 + 6 = 36 ; and 36» — (30 -f 6)' = 30' + 2 (30 X 6) 4- 6* = 1296. Again, since every number, consisting of more than one place of figures, may be considered as composed of tens and units, the f(M> mula is general, and applies equally whether the root has two places of figures or more than two places. (Nat. Arith., Art. 524.) Define a Perfect Square. Of how many places of figures does the square of a number consist 1 How may the number of figures in the square root of a number be shown f To what is the square of any number consisting of more than one place equal ? EVOLUTION. 183 CASK I. 215. To extract the square root of entire numbers. 1. Let it be required to find the square root of 4366. OPEKATION. 120 + 6 4356 60 + 6 3600 r or, ■ 4356 36 156 156 126 156 156 66 By pointing the given number according to Article 213, it ap? pears that the root consists of two places of figures. Let, now, a + 6 denote the root, -wheae a is the value of the figure in the tens' place, and h of that in the units' place. Theq a must be the greatest multiple of ten whose square is less than 4300; this we find to be 60. Subtracting a', that is the square of 60, from the given number, we have the remainder 756, which must contain twice the product of the tens by the units, plus the square of the units, or 2 a 5 -|- 6*. Dividing this remainder by 2 a, that is by 120, gives 6, which is the value of 6. Then (2 a + 6) 6, that is, 126 X 6, or 756, is the quantity to be subtracted; and as there is now no remainder, we conclude that 60 + 6, or 66, is the required square root. In the vovk %3 it stands at the right, the eipheu^ are on^itted. Had the root consisted of three places of figures, we coul4 h^6 let a represent the hundreds, and h the tens ; then, having obtained a and h as before, we might let the hundreds and tens together be Considered^ as a new vs^ue of a, an(} find a new value of b for the units. RULE. Separate the given number into periods, hy pointing every second figure, beginning with the units' place. Find the greatest square in the left-hand period, and plac6 its root on the right ; subtract the square of this root from the first period, and to the remainder bring doinn the next period for a dividend. Explain the operation. Repeat the Rule. 184 ELEMENTARY ALGEBRA. Divide this quantity, omitting the last figure, hy doulk the ■ part of the root already found, and annex ths result to the root, and also to the divisor. Multiply the divisor as it now stands by the part of the root last obtained, and subtract the product from the dividend. If there are more periods to be brought down, continue the operation in the same manner as before. Note 1 . If a root figure is 0, place at the right of the divisor, and bring down the next period to complete the dividend. Note 2. If there be a final remainder, the given number has not an exact root ; but we may continue the operation, by annexing an even number of decimal ciphers, and thus obtain a decimal part to be added to the integral part already found. Note 3. In pointing a number having a decimal part, we begin at the units' place, and point both to the right and left of it ; and, if the deci- mal has no exact root, we may continue to form decimal periods to any desirable extent. Note 4. The root of a decimal without an integral part may be ■ found as though the decimal were an entire number, care being taken to make the number of decimal places even, by annexing a cipher, if necessary. , ( Examples. 2. Eequired the square root of 365, to four decimal places. OPERATION. 36 5.0 0000000 1 29 265 261 38T 38204 382089 400 381 190000 152816 37 18400 3438801 2T9599 19.10 4 9 + What is Note 1 1 Note 2 f Note 3 1 Note 4 ? Explain the operation. EVOLUTION. 185 It will be observed that four periods of decimal ciphers were annexed to the given number, to correspond with the number of decimal figures required in the root. On obtaining the fourth deci- mal figure of the root, there is stiU. a remainder, and to show that the root obtained is an approximate one, we annex the sign -|— 3. Kequired the square root of 611524. Ans. 782. 4. Kequired the square root of 56644. Ans. 238. 5. Kequired the square root of 6561. 6. Extract the square root of 2116. Ans. 46. 7. Extract the square root of 10246401. Ans. 3201. 8. Extract the square root of 16.2409. Ans. 4.03. 9. Extract the square root of .9409. Ans. .97. 10. What is the square root of .0081 ? Ans. .09. 11. What is the square root of .006, to four places of decimals? Ans. .0774+. 12. What is the square root of 12, to six places of decimals? Ans. 3.464101-|-. 13. What is the square root of .0000012821 ? Ans. .00111. CASE II. 216. To extract the square root of fractions. 1. Required the square root of -f^. „„„„ . »„„„ Since to square a iraction we OPEKATION. ^ square both its numerator and de- /_^ __ \/ ^ * nominator separately (Art. 191), we Y 16 y^ 16 4 g^^ ^]^Q square root of the given fraction by taking the square root of its terms for the corresponding terms of the root. Hence, When both terms of a fraction are perfect squares, its square root may he obtained by extracting the square root of both nu- merator and denominator. Explain the operation. How is the square root of a fraction obtained when both its terms are perfect squares? 16* 186 ELEMENTARY ALGEBRA. Note. If the fraction has not both terms perfect squares, and can- not be reduced to an equiralent fraction having such terms, its root cannot be exactly found. It may, however, be reduced to a decimal, and the root can then be found as provided in Art. 215, Notes 3 and 4. 2. What is the' square root of '^^^? Ans. -j^g-. 3. What is the square root of ^|| ? Ans. f f . 4. Extract the square root of 2j^. Ans. 1^^. KoTE. Reduce the mixed number to an equivalent common fraction. 5. Kequired the square root of -f^. Ans. §. KoT£. Neither term of the given fraction is a perfect square ; but on 4 reducing it to its lowest terms, we obtain - , 6. Eequired the square root of tVWf- ^^^- ?• 'I. Kequired the square root of y^g-. Ans. .832-1— 8. Eequired the square root of xw§t- Ans. .1246, nearly. CUBE ROOT OF NUMBERS. 2I7i The Cube Root, or third root, of a number is a factor which must be taken three times to form that number. Thus, /^^ = 3, because 3 X 3 X 3 = 27. 218i A Perfect Cube is any number or quantity that can be resolved into three equal factors. 219. The cube of any integral number consists of three times as many places of figures as the number itself, or of one or two less than three times as many. How is the square root of a fraction obtained when its terms are not perfect squares ? Define Cube Root. A Perfect Cube. Of how many places of figures does the cube of. a number consist ? EVOLUTION. 187 For, the first ten numbers are 1. 2, 8, 4, 5, 6, 7, 8, 9, 10, and their cubes are 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000; also, the cube of 99 is 970299, of 100 is 1000000, of 999 is 997002999, of 1000 is 1000000000, and so on. Hence, 220. If a point he. plaoed over every third figure in any integral number, beginning with the unit^ place, the number of points will show the number of figures in the cube root. 221 • The cube of any number, consisting of more than one place of figures, is equal to the cube of the tens, phis three times the product of the square of the tens by the units, plus three times the product of the tens by the square of the units, plus the cube of the units. For, if the tens of a number be denoted by a, and the units by I, the number will be denoted by a-\^h, and its Qube br (o 4, 6)» == a» + 3 o* 6 4- 3 a 6» + 6'. Then, by this formula, if a = 3 tens, or 30, and & = 6, we have 3 tens -\- 6, unit? = 30 -|- 6 = 36, and 36» = (30 -|^ Gf = 30» -f 3 (30» X 6) + 3 (30 X e^) + 8' = 46656. Again, since every number, consisting of more than one place of %ure9, may be considered as composed of tens and units, the for- mula is general CASE I. 222, To extract the cube root of entire numbers. 1. Let it be required to find the cube root of 405224. How may the number of figures in the cube root of » number be shown ? To what is the cube of any number consisting of more than one figure equal ? 188 ELEIIKNTAKY ALGEBRA. OPEEATION. 405224 343000 14T00 840 16 15566X4 = tO+4 62224 62224 By pointing the given number according to Article 220, it ap- pears that the root consists of two places of figures. Let, now, a-{-b denote the root, where a is the value of the figure in the tens' place, and b of that in the units' place. Then a must be the greatest multiple of ten whose cube is less than 405000 ; this we find to be 70. Subtracting a', that is the cube of 70, or 343000^ from the given number, we have the remainder 62224, which must contain three times the product of the square of the tens by the units, plus three times the product of the tens by the square of the units, plus the cube of the units, or S a'b -\- S ab^ -\- b'. Divid- ing this remainder by 3 a', that is by three times the square of 70, or 14 700, we obtain the value of 6, or 4. Then (3a?-^3ab-\-l')b, that is, (14700 -\- 840 -|- 16) 4, or 15556 X 4 = 62224, is the quan- tity to be subtracted; and as there is now no remainder, we con- clude that 70 -f- 4, or 74, is the required root. For brevity in the operation, instead of writing 70 in the root, we may simply write 7 in the tens' place, and for the cube of 70 write the cube of 7, or 343, without the ciphers, observing to place the figures under the proper period. Had the root consisted of three figures, we could have let a rep- resent the hundreds, and 6 the tens ; then, having obtained a and h as before, we might let the hundreds and tens together be consid- ered as a new value of a, and find a new value of 6 for the units. From the preceding example we deduce the following Explain the operation. EVOLUTION. 189 RULE. Separate the given number into periods, hf pointing every third figure, beginning with the units' place. Find the greatest cube in the left-hand period, and place its root on the right ; subtract the cube of this root from the first period, and to the remainder bring down the next period for a dividend. At the left of the dividend write three times the square of the root already found, for a trial divisor ; divide the dividend, omitting the last two figures, ly it, and write the quotient for the wex! figure of the root. Add together the trial divisor, with two ciphers annexed; three times the product of the last root figure by the rest of the root, with one cipher annexed; and the square of the last root figure; and the sum will be the complete divisor. Multiply the complete divisor by the last root figure, and sub- tract the product from the dividend. If there are more periods to bring down, continue the operor tion in the same manner as before. Note I. The observations made in Notes 1, 2, 3, and 4, under the rule for the extraction of the square root (Art. 215), are equally ap- plicable to the extraction of the cube root, except that two ciphers must be placed at the right of the trial divisor when it is not contained in its corresponding dividend, and in pointing off decimals each period must contain three places of figures. Note 2. As the trial divisor is necessarily an incomplete divisor, it is sometimes found, both in cube and in square root, that after completion it gives a product larger than the dividend. In such a case, the root figure last found is too large, and the one next less must be substituted for it Examples. 2. What is the cube root of 8.144865728 ? Bepeat the Rule. Repeat Note 1. Repeat Note 2. 190 ELEMENTAEY ALG3BEA. OPERATION. 8.1448657 28,2.012 8 120000 600 1 120601 X 1 = 12120300 12060 4 12 1323 64 X 2 = 144865 120601 24264728 2426472 8 It will be observed that, in consequence of the in the root, we annex two additional ciphers to the trial divisor, 1200, and bring down to the corresponding dividend another period. 3. What is the Cube toot of 941192 ? Ans. 98. 4. What is the cube root of 389017 ? AnS. 73. What is the cube root of 37259704 ? Ans. 334. What is the cube root of 251289591 ? Ans. 631. What is the cube root of 46268279 ? Ans. 359. 8. Kequired the cube root of 1481.544. Ans. 11.4. 9. Eequired the cube root of .008649. Ans. .2052+. CASE n. 223. To extract the cube root of fractions. 1. What is the cube root of f|f ? OPERATION. J343 V 729 ^j/343 f343 7 729 "~ ^729 ~ 9 Since to cube a fraction w6 cube both of its terms Separately (Art. 191), we find the cube root of the given fraction by taking the cube root of its terms for corresponding terms of the root. Hence, Explain the operation of Example 2. Of Example 1, Case IL KVOLUTION. 191 WTten both terms of a fraction are perfect cubes, its cube root may he obtained by extracting the cube root of both numer- ator and denominator. Note. If the fraction has not both terms perfect cubes, and cannot he reduced to an equivalent fraction having such terms, it may be re- dnced to a deeiraal| and the root found as provided in Art. 222. 2. Find the cube root of ^ff. Ans. f . 3. Extract the cube root of ffH^. Ans. 1^. 4. Eequired the cube root of f-g. Ans. .4t2 -\-. EOOTS OF MONOMIALS. 224. The rules for evolution must be deduced from those for involution, for the oae is the reverse of the other. (Art. 204.) 1. Let it be required to extract the cube root of 2T a^ V. OPERATION^ A^WcFW = 2'?^ X «* X ** = 3 a^ b. Since to cube a monomial we cube the coefficient and multiply the exponent of each of its letters by 3, the exponent of the re- quired power (Art. 190), to find the cube root of the given mono- mial, we reverse the process, and extract the cube root of its coef- ficient, 27, and divide the exponents of each of its letters, a and &, by 3. The result, 3 a' b, being an odd root, is positive (Art. 206). 2. Let it be required to extract the square root of OPEKATION. . ! — Q2 V «t v/ Alf — ^T¥V = 9^ X a* X 5^= ± 3a«5. Since extracting the square root is the reverse of the foitnation When both terms of a Traction are perfect cubes, how is the cube root obtained? How, when its terras are not perfect cubes? Explain the op- eration of Example 1. Of Example 8. 192 ELEMENTARY ALGEBRA. of the square, we extract the square root of the coefficient 9, and then divide the exponents of the letters a and 6 by 2. The re- sult, being an even root of a positive quantity, may be either posi- tive or negative (Art. 207), and therefore is written with the double sign ±. From these examples is deduced the following EULE. Extract the required root of the numerical coefficient, and divide the exponent of each letter hy the index of the root. Note 1. Prefix to odd roots of positive quantities -|-, to odd roots of negative quantities — , and to even roots of positive quantities ±. Note 2. The root of a monomial fraction may be found by extracting the required root of each of its terms separately. Thus, V*' "^ 7P = ^ * ' ^°' V=^ M = 6^- Examples. 3. Find the square root of 16 x'. Ans. ± 4 a. 4. Find the cube root of 21 a'. Ans. 3 a. 5. Find the fourth root of 16 a* a^. Ans. ±2aK^. Note. The fourth root of a quantity is one of its four equal factors, or it is the square root of its square root, since the fourth power of a quantity may be found by squaring its second power. 6. What is the square root of 144 a*¥c^? 1. What is the cube root of 125 a^a^f Ans. 5 a'' a;. 8. What is the fifth root of — 32 a^" a^ ? Ans. — 2a'x. 9. What is the square root of ^-^? Ans ± ^^. 9 a* If' Sa'y • . Eepeat the Rule. Note 1. How may the root of a monomial frac- tion be found ? What is the fourth root of a quantity ? EVOLUTION. 193 10. Required the cube root of -^^ Ans — 729 f 3 f 11. Required the square root of 25 a 5' c^. Ans. ± 5 a* J^ c. Note. As we cannot extract the square root of either a or 6', we in- dicate the division of the exponents by 2. Hence the propriety of indicat- ing roots by fractional exponents (Art. 203). 12. Required the cube root of — T29 ar^ b-". Ans. — 9er^b-^. 13. Required the value of 4^ 24:3 afi i/. Ans. 3xy^. 14. Required the value of (169 a" J-^ c-^i. n 1 Ans. ± 13 a^ b~'^ c'K 15. Required the seventh root of — ^-4— 128 a;' y' , Ans. ^ . 1 2a:7y 16. Required the value of (a"5"'c2" rf'-^")" m Ans. ab^c' dr^. SQUARE ROOT OF POLYNOMIALS. 225. The' manner of forming the square of a polyno- mial must, by reversing the process, lead to the discov- ery of its root. If we take any binomial, as a-\-b, we have (a-]-5)2 = a24-2aJ + 52; and the last two terms of -this expression factored give (2 a + 5) 6. 1. Let us now reverse the involution, and discover How may we discover the process of finding the square root of a polynomial 1 IT 194 ELEMENTARY ALGEBRA. how the root a-\-i may be derived from the a^ + 2 a J + 5^. square a'' + OPERATION. 2aJ + J^ 2 « + b 206 + 6= 2a6-i-i" + 5 The square root of -the first term, a°, is a, which is the first term of the required root. Subtracting the square of a from the given polyno- mial, we have 2ab -\-V, or (2 a -{- J) 6) for a remainder or dividend. Dividing the first term of the dividend, 2ab, by 2a, which is double the first term of the root, we obtain 6,^ the other term of the root, which, connected to 2 a, completes the divisoi-, 2 a "l" 6. Multiplying this divisor by the last term of the root, 6, and subtracting the product, 2a 6 -1- 6", from the remainder, we have nothing left. By a like process, a root consisting, of more than two terms may be found from its square, since aU such roots can be expressed in a binomial form. Thus, a -\- b -\- c = (a -{- V) -\- c, and its square, a«-(- 2a6 + 6^ + 2ac + 26c + c^ = (a-(-6)'' + 2 (a + 6)c + c", which, factored, gives a^ _|_ (2a 4- J) 6 + (2a + 26 + c) c. 2. Let it next be required to find the square root of a'i^2ab+¥-^2ae-{-2bc-\-c\ a' 2 a 6 + 6» -1- OPERATION. 2ac + 26c + c'' 2a + 6 2 a 6 + 6^ 2 6 4-6" 2 a-|-26 + c 2ac4-26c4-c'' 2ac4-26c4-c" a + 6 -|- c Explain the operation of Example 1. How may the process be ex- tended to square roots consisting of more than two terms 1 Explain the operation of Example 2. EVOLUTION. 195 We find a -)- J of the root as in the preceding exaniple, and have, on Bubtracting and bringing down the remaining terms, iac-\- 2bc-{-c', for a remainder or dividend. Dividing the first term of this quantity, 2 a c, by 2 a, which ia. double the first term of the root, we obtain c, the third term of the root, which, con- nected to 2 a -|- 2 6, or double the part of the root already found, completes the divisor, 2 a -|- 2 b -\- c. Multiplying this di- visor by the last term of the root, c, and subtracting the product, 2ac-\-2bc-\- c', from the dividend, there is nothing left. From these examples and illustrations we derive the RULE. Arrange the terms according to the powers of some letter. Find the square root of the first term, write it as the first term of the root, and subtract its square from the given pohf' nomial, hg bringing down two or more terms for a dividend. Divide the first term of the dividend by double the part of the root already found, ancC annex the result to the root, and also to the divisor. Multiply the divisor as it now stands by the term of the root last obtained, and subtract the product from the dividend. If there are other terms remaining, continue the operation in the same manner as before. f Note 1. Since all possible even roots may be either positive or neg- ative (Art. 207), the square root obtained by the rule will remain a root, when all its signs are changed. Note 2. The fourth root may be. obtained by taking the square root of the sqaare root. Examples. 3. Find the square root of 4a;*— 12 a:'-}- 5 a:" + 6 a; +1. Repeat the Rule. What is Note 1 ? Note 2 1 196 ELEMENTAKY ALGEBRA. 4 a;* OPERATION. Ix^ — Sx — Ux' + bx' — 12 a;* + 9 a:" 4a;« — 6 a:— 1 — 4 jc" + 6 a; + 1 — 4a;»-j-6a:-|-l 2 ar" — 3a: — 1 4. Find the square root of a* -|- 4 a" 6 -|- 4 ff". Ans. a" + 2 5. 5. What is the square root of 9 a;* — 12 a;' + 16 a;' — 8a; + 4? Ans. Sar" — 2a; + 2. 6. What is the square root o{ x^ -\- 4: b x -\- 4: h^ ? 1. What is the square root of a* -\- 4: c^ b -\- 10 a" V -\- 12 a 6' + 9 6* ? Ans. a^ + 2 a J + 3 J^. 8. Eequired the square root of a* — ^ 2 o' -[- 2 a^ — 'f + i- Ans. a' — « -f- i- 9. What is the square root of a;* — 2 a:'' -}- 1 ? Ans. a^ — 1. 10. What is the square root of a'' — 2 -\- er^? Ans. a -^ cT^. 11. What is the square root of ia" — 12 a 6 -\- 4:ax _j_ 9 52_6 Jar + ar*? Ans. 2 a — 3 J + a;. 12.* Eequired the fourth root of «* + 8 a* 6 + 24 a" S" + 82ffl5'+16J*. Ans. a + 2i. CUBE BOOT OF POLYNOMIALS. 226i An investigation of the formation of a polynomial cube, by reversing the process, must lead to the discovery of its root. If we take any binomial, as a -}- *, we have Explain the operation. In what way may we discover the process of finding the cube root of a polynomial t EVOLUTION. 197 (a + S)« = o» + 3 «H + 3 a 5= + W, and the last three terms of this expression factored give (3 a^ 4- 3 a S + 52) I. 1. Let us now, by reversing the involution, discover how the root a-\-l may be derived from the cube d -\- 3 a= i + 3 a 6" -j- J». OPERATION. 3a» + 3aS + J2 « + * 3a2s_|_3„5!!_|„j? The cube root of the first term, a', is a, which is the first term of the required root. Subtracting the cube of a from the given polynomial, we have 3 a« J + 3 a S'' + 6', or (3 a* + 3 ah -\- W) b, for a remainder or dividend. Dividing the first term of the dividend by the trial divisor, 3 a', which is three times the square of the first term of the root, we obtain 6, the other term of the root. Adding, now, to the trial divisor, 3 a', three times the product of the first term of the root by the last, and the square of the last term of the root, we have for the complete divisor, Sa' -\- Sab -\- b'. Mul- tiplying this by b, the last term of the root, and subtracting the product from the dividend, there is no remainder, and the root is obtained. By a like process, a root of more than two terms may be found from its cube, since all such roots can be expressed in a binomial form. Thus, a -\- b -\- c == (a -\- b) -\- c, and its cube, (f-\-Sa'b-\-3aV + h'-\-3a'c-\-Gabc-\-3¥c-\-3ac'-{-3bc^ + <^ = (a + 6)' + 3 (a + S)» c + 3 (a + 6) c« + c' which, factored, gives a'-\-(3a''-\-3ab-\-¥)b+i3a'+6ab+3V'-\-3ac-\-3bc-\-c')c. Explain the operation. How may a cube root consisting of more than two terms be obtained from its power ? 17* 198 • ' ELEMENTARY ALGEBRA. 2. Let it now be required ta derive the cube root a -\-b -\- c from its power. OPERATION. I a + 6 -[" "> root.* a'-\-3a'b-\-3aV+b'+3a'c-[.6abc-\-SlPc-\-3a<^-\-3bc'-\-c' a' 3a« + 3ab-\-l Sa'b-^-SaV + b' 3a?b-{.3ab^ + ¥ 300 + 360 + 0^ 3a»c+6oJc-f-36V+3ac«+36c»+c» 3a?c+6abc-\-3¥c-{-3ac'+3bd'-\-i^ We find a -\-b o{ the root as in the pr.eceding example, and have, on subtracting and bringing down the remaining terms, 3a^c-\-Sabc -\- 3l^c -{■ Sac* -\- Sbc' -{- c?', for a remainder or dividend. Di- viding the first term of the dividend by the first term of the trial divisor, 3 a', we obtain c, the third term of the root. Adding to- gether three times the square of the first two terms of the root, which is the trial divisor, three times the product of the first two terms by the third, and the square of the third, we have for the complete divisor, 3a« + 6a6-f-36"-}-3ac-|-36c-fc«. Multiply- ing this by c, the last term of the root, and subtracting the pro- duct from the dividend, there is no remainder, and the root is ob* taiued. Hence, we deduce the following RULE. Arrange the terms according to the powers of some letter. Find the cube root of the first term, write it as the first term of the root, and subtract its cube from the given polyno- mial, by bringing down three or more terms for a dividend. Take three times the square of the part of the root already Explain the operation. Repeat the Rule. * The root is written, in this case, above the power, and the divisors each on two lines, to economize space. EVOLUTION. 199 found for the tnal divisor, divide the^ first term of the dividend bt, it, and write the quotient for the next term of the root. Add together the tried divisor, three times the product of the first term hy the last, and the square of the last, for a complete divisor. Multiply the complete divisor by the last term of the root, and subtract the product from the dividend. If there are other Urms remaining, form a new dividend, and continue the operation in the same manner as before. Note 1. The terms of each new dividend must be arranged, if neces- sary, according to the powers of the leading letter of the root. NoTB 2. If there are three terms in the root, the first two terms must take the place of the first term in obtaining the third. The trial divisor will strictly contain three terms, but only the first need be used, tiU the divisor is completed. , Examples. 3. What is the cube rootofa;«-f6a;= — 4:0a:'-|-96a; — 64? OPERATION. 4-6a:5 — 40a:'4-96a; — 64 a^-\-2x — i 3a:«-f-6a^-|-4a;2 6 a;' — 40 a^ 6a;5*-f 12 a;* + §0^ 3a:*+12a;' — 24a;+ 16 12 x* — 48 a^ + 96 a; — 64 12 a:* — 48 a? -I- 96 a: — 64 We here bring down only two terms at each time, instead of three, since in the given expression two terms, those containing X* and a?, are wanting. In the last complete divisor, 12ar' and — 12 2" cancel each other. 4. What is the cube root ef a^ -f 3 a;''^ -}- 3 a:/ -f- / ? Ans. X -\- g. • Bepeat Note 1. Note 2. 200 KLEMENTARY ALGEBRA. 5. Find the cube root of / — 3/ + 5/ — 3y — 1. Ans. 'if: — y — 1. 6. Find the cube root oi 21 x" -{-bi.x'y -\- Z^xf-\-iy\ 1. Find the Cube root of »i*-|-6?»^ — 4,Qm^-\-QQm — 64. Ans. ni^ -\-2m — 4. 8. Kequired the cube root of a' -|- 3 a -j- 3 a~^ -j- ar^. Ans. a -\- a~*. EADICALS. 227. A Kadical is a root of a quantity indicated ei- ther fey a radical sign or by a fractional exponent ; as, v/ a, cfi , and 2 4^ 1 -\-a. When the root indicated can be exactly obtained, it is called a rational quantity, and wh^n it cannot be exactly obtained, it is called an irrational or surd quantity. Thus, kJ/ 2*7 c?, which can be expressed by 8 a, is called a ra- 2 tional quantity ; and 4/ <^> or cr , is called an irrational or surd quantity. An even root of a negative quantity cannot be obtained, even approximately, and is therefore called an imaginary quantity (Art. 209). 228i The Coefficient of a radical is the quantity , or factor prefixed to it. Thus, in 2\/2h(?, and a{c-\-dY, 2 dud a are the coejBScients. 229. The Degeee of a radical is denoted by the index of the radical sign, or by the " denominator of the frac^ tional exponent. Thus, . \/ a, V' y. (« * c)^, are radicals of the s'econd degree ; Define a Radical. When is a quantity called rational ? When irra- tional or surd 1 When imaginary ? Define the Coefficient of a radical. The Degree of a radical. RADICALS. 201 is/3?, h , (2a^3?yY, are radicals of the third degree; \/ac, 3 ^ OT, (a -j- b)n are radicals of the wth degree. 230. Similar Eadicals are those of the same degree, with the same quantity under the radical sign. Thus, h^/ ax, and *l /^ ax are similar radicals ; and also a y"" and c yn . EEDUCTION OF EADICALS. 231. Eeduction of Eadicals is the process of changing their forms without altering their values. 232. The reduction of radicals depends upon the gen- eral principle, that The root of any quantity is equal to the product of the like roots of its several factors. For, in obtaining the root of a monomial, we obtain the root of each of its factors, whether numerical or literal (Art. 224). CASE I. 233. To reduce radicals to their simplest form. A radical is in its simplest form, when it has under the sign no factor which is a perfect power. 1. Eeduce a^ 135 a^b* to its simplest form. OPERATION. ^® ^^^^ resolve the quan- tity under the radical sign ^135an* = ^2n^X5S_ ;„to two factors, one of = \/21a^l^ X -^6 5 -which, 27a''6^ is a perfect . q -2 i '/KT cube. Then, since the root of a quantity is equal to the product of the roots of its Define Similar Badicals. Redaction of Eadicals. Upon What principle does the reduction of radicals depend ? Explain the opevation. 202 ELEMENTARY ALGEBRA. several factors (Art. 232), we find the root, Sa'b, of the rational part, and multiply it by the indicated root of the surd factor, or, which is the same thing, write it as the coefficient of the surd factor placed under the sign ; and thus we obtain Sa^b \/5 b, the simplest form of the radical. Hence the RULE. Resolve the quantity under the radical sign into two factors, one of which shall contain all the perfect powers of the same degree as the radical. Extract the required root of this factor, and write it as a coeJUcient of the other factor, placed under the sign. Examples. Reduce the following radicals to their simplest forms. 2. ^9a*x. An8.^aW^- 3. /^32a^x. Ans. 4.a\/~2x. 4. TV 80 a;. . Ans. 28\/5a;. 5. a V 125 4^. Ans. 5ab\/Tb. 6. 4/~6l¥¥. Ans. iaS'^'^K 1. s^WoF?. 8. {a3?-\-h3^)^. Ans. x{a-\-hx)^. 9. 2 {a? — a' a^)*. Ans. 2x{x — a^)*. 10. ^"5^' + «*i). Ans. a^5(l + ai). 11. 6^54a«J'c. Ans. 18aJV'^"«ic. 12. 3>^"32^^F?. Ans. Qao^lH?. 13. (T2a; + 108y)*. Ans. 6(2x-\-Zy)^. 14. 5 (a — J) >/«^« + 2aSc + 52g Ans. 5 (a" — 52) a/7. Repeat the Rule. RADICALS. 203 234. When the giveu radical is in a fractional' form, it will often be convenient, before applying the rule, to mukiply loth terms of the fraction hy such a quantity as wiU make the denominator a perfect power of the degree indicated. Then the factor under the sign in the simplest form of the radical will be an entire quantity. 1. Reduce v^f to its simplest form. OPERATION. 2. Reduce 3 i/-^ to its simplest form. OPERATION. Y"5" ~ V ^^ V ^ X 5 = 3 X ^ V5 = -g- -v/5 Reduce the following radicals to their simplest forms. 3. 2v'f. Ans. ^V6. , 3 iYa¥ . X ,- *• J i/ — 3 — Ans. - v6 a x. \CT)' Ans. — (2af. /3^ . a , 7 ysj- Ans. gVeai. 7. 2(^)i Ans. ^(10«5c)i CASE n. 235. To reduce a rational quantity to the form of a radical. 1. Reduce 2 a' to the form of the cube root. When the radical is in a fractional form, how may we proceed? Ex- plain the first operation. The second operation. 5 6. 4 204 ELEMENTARY ALGEBRA. OPERATION. S'°<=^ *^® ^^"^"^^ required I J 1 is of the third degree, we 2 a^ = (2 a^) = (8 a ) ^^^^ ^^^j^ ^^ ^^^ ^^^^^^^ ^^ 2 a^, and obtain 8 a°, which, written under the sign of the root indicated, gives the required form, or ^8 a'. The value of this expression is evidently 2 a'' ; and in general, since evolution is the reverse of involution, powers and roots of the same degree cancel each other like the terms of fractions. Hence the RtTLE. liaise the quantity to the power indicated hy the given wot and write it under the corresponding radical sign. Examples. 2. Eeduce 3 a a; to the form of the square root. Ans. i\/9 «* »*. 3. Eeduce ■. — b a^h to the form of the cube root, Ans. ^— 126^«K 4. Reduce 2 a; — 3 to a radical of the second degree, Ans. (4 a:^ — 12 sf + 9)*- 5. Eeduce 2a^a?y to a radical of the fourth degree. 6. Eeduce ^-^ to a radical of the fifth degree. , 5 7243 a* a;'" 236. A coefficient, or a factor of a coefficient, of a radi- cal may be placed under the radical sign, by raising it to the power indicated hy the radical, and multiplying the quan- tity already under the sign hy the result. 1. Eeduce 3a\/1 to a radical without a coefficient. Explain the first operation under Case U. Repeat the Rule. How may a coefficient, or a factor of a coefficient, be placed under the radical sign ? EADICALS. 205 OPERATION. 2. Eeduce 5 \/x y — 1 to a radical without a coefficient. Ans. is/Vlhxy — Vlh. 3. In the expression 2 a^ 4^3 a b, place the factor 2 under the sign. Ans. a^ ^48 a h. L Eeduce (a -|- J) v'c to a radical without a coefficient. Ans. (a^c-\-2ahc-\-b^c)^. 5, Reduce ^ i/ to a radical without a coefficient. CASE ni. 237t To reduce radicals of diflFerent degrees to equivalent ones having a common index. 1. Reduce cr and 5* to a common index. OPERATION Since any quantity may be „J _ „l _ c„8 J oj. ^^ '•^'sed to any power indicat- 1 ^ 4 ' ed by a given root, and writ- o =0 = (h') , or V ten under a corresponding radical sign, without chang- ing the value of the expression (Art. 235), it is evident that cr 1 1 A . and 6° are equal to cr and 6', respectively. That is, we may re- duce the given exponents to equivalent ones having a common de- nominator. , Now, a' is equal to (a') », or S/a\ and 6* is equal to (J=)^, or S/^. Hence the RULE. Reduce the given exponents to a common denominator ; raise eojoh quantity to the power denoted hy the numerator of the re- ' duced exponent, and indicate the root denoted hy the denomi- nator. Explain the second operation under Case II. Explain the first opera- tion under Pa§e III. Repeat the Rule. 1§ 206 ELEMENTARY ALGEgRA. Note. Radicals may be reduced to a common index, without the use of fractional exponents, by multiplying the index and the exponents of each by such a quantity as will make its index the least common mul- tiple of the given indices. Examples. 2. Eeduce aS/ x and Jv'y to a common index. OPEEATION. 1 i. J- ai^x-=ax'':=a3f"' = a (af")™" = av' a;". I i J- 3. Eeduce \^2 and 3\/3 to a common index. Ans. ^2 and 3v'27. 4. Reduce ^a, (5 6)*, and {6?-\-¥y to a common in- dex. Ans., (a»)^, (25 5^)^, and [(a= + 6^)*]!^. 5. Reduce a/Ii, \^ a — h, and 4^ a-{^b to a common index. Ans. ^^, -^"(^^^=^*. and ^ {a^hf. ADDITION OF RADICALS. 238. When radicals to be added are similar, the com- mon radical part, with the sum of their coefficients, will constitute the sum of the radicals. 1. Find the sum of \/18 and s/i. OPERATION. W« '^^'^"?« *" g*^*" '■^- — — icals to their simplest forms, \/18 = y /Q X 2 = 3 s/2 ^^^ ^^^^ 4^^^^ which _are ^ 8 = v/4: X 2 = 2v'2 similar. Finding, then, v'2 to g 5/2^ ^ ^^ common radical part, we have 3 times and 2 times V^2, equal to 5 times v'2, or Explain the operation of Example 2. What may constitute the unit of addition when radicals are added) Explain the operation of Ex- ample 1. RADICALS. 207 2. Find the sum of a/^, ^ 4«», and V"^. OPERATION. Reducing the given rad- — icals to those which are sim- V'_^ = \/^x^2^ ~ ^^ ilar, of which ^Ic is the com- ^4 a:* = ^4 a;^ X a; = 2 a; y/a; mon radical part, we have x Ja^x =:: y/ «/' v t n i/Z times, 2 X times, and a times c< ~7Z ; :^ — F= V'^j or 3 a; 4- a times i/a;, or Sum= (3a:4-a)t/F ' ' ^ _'^ ' * (3a; + a)v'a;. Hence the following RULE. Sedtiee each radical, if necessary, to its simplest form. If, then, the radicals are similar, add their coefficients, and to the ■ sum annex the common radical; but if they are dissimilar, indicate the addition by the proper sign. Note. Since dissimilar radicals have no common radical part, it is evident that their addition can only be indicated. Examples. 3. Find the sum of 5 \/98lc and 10 \/2x. Ans. 4:5\/2a;. 4. Find the sum of ^Wa and v^l62 a. Ans. ,5 ^^6 a. 6. Find the sum of ^32 and 5 4^2". Ans. 1 ^2? 6. Find the sum of a/^H and y/TWb. Ans. (a 4" *) 'V'ST. 1. Find the sum of 5 nJWcfx and 3 aJWc^x. Ans. \%as/bx. 8. What is the sum of (Sa'^S)*- and (2T,a'i)^ ? Ans. 4 a (3 S)i 9. What is the sum of (45 (?f, (80 (?f, and (5 e^ c)^ ? Explain the operation. Repeat the Rule. Why can the addition of dissimilar radicals be only indicated! 208 ELEMKNTAIiY ALGEBEA. 1,0. What is the sum of 4^b*^ and \/by*? __ Ans. (b-^y)\^b^. 11. Find the sum of \/3 and \/t^. Ans. ^v^S. Note, yj = ^^ X 3 == J ^3 ; ^3 + 1.^/3=1^3. 12. Find the sum of 12 v^^^ and 3 ^^. _ Ans. ^-4^2. 13. Find the sum of 2 \^5, 5 \/6 a, and V'4: a;. Ans. 2 ^5"+ 5 V'6a'+ 2 \/k^ 14. Find the sum of \/a; .+ 1 and \/4, x -\- i, Ans. 3 V'a; -f- 1. 15. Find the sum of (a^y)^ (ocf)^, and (8a*y)^. Ans., xy^ -{- x^ y -\- 2 a (ay) . 16. Find the sum of ^\/a^ic and ^\/4Jca;*. Ans. (2 + !i^^ v'A^.- 11. Find the value of ^16e»J + VT^ + ,.^54 a" 6 -f- \/o^. Ans. 5 a v'2l + 3 a \/6; SUBTEACTION OF EADIOALS. 239. When the radicals are similar, the common radi- cal part, with the difference of. their coefficients, will constitute the difference of the radicals. 1. Find the difference between \^l25a and \/20a. OPERATION. Eeducing the -given radi. cals to their simplest forms, ' ' ' we have those which are v/20a = v'4X5ffl = 2 y/S a similar. Finding, then, v/Sa Difference = 3 v/5a *° ^ *''® common radical part, we take 2 times ^5 a from 5 times ^5 a, and have as their difference 3 times ^Va, or 8 y/S a. Hence the When the radicals are similar, what constitutes the unit of subtraction i Explain the operation. RADICALS. 209 RULE. Reduce each radical, if necessary, to its simplest form. If then the radicals are similar, subtract the coefficient of the sub- trahend from that of the minuend, and to the difference annex the common radical; hut if they are dissimilar, indicate the subtraction hy the proper sign. Examples. 2. From -v/46 a take y/Wa. Ans. 2 s/Ta. 3. From ^l92 take /^2i. Ans. 2 ^3. 4. From (9a*a;)i take (4a*a;)i Ans. a' ^x. 5. From b^^iaH take ai^W^. Ans. a'h/^l. 6. From 4 (2 + y)^ take 3 (3/ + 2)^ Ans. (2 -f y)*. 7. Find the difference between \/108 a 3? and \/48aP. 8. From Vl take ^/l. Ans. |-v/3. 9. From 2 VT^i^- t^j^g ^"5Vj3_ Ans. 2fl!i/\/3c — 6\/5a6. 10; From -^32 a take 2 -^40 a. Ans. 2 (v^Ta — 2 s/Ta) . 11. Find the value of 4/ S aH-\-16 a* — A^ b* -^2 a&'. Ans. (2 a — J) 4/ 2 a + S. MULTIPLICATION OF EADICALS. 240i The multiplication of radicals depends upon the principle, that The product of like roots of two quantities is equal to the same root of their product. Kepeat the Bale. Upon what principle does the multiplication of rad- icals depend? 18* 210 ELEMENTARY ALGEBRA. To prove the principle, let a and h be any two quantities. Then, by Art. 232, ^'ab = \/a X V'*! Whence, ^aX^b^ y'a 6. 1. Multiply 4 a V 2 6y by 3 J \/ 2 6 a;. OPERATION. daV'Tiy X 35v'26a; = 4aX 3J\/26yX 26a: = 12aJV'TFiy = 12a5x 26\/^ = 2^:aW is/ xy Since it is immaterial in what order the factors are taken (Art. 58), we multiply together the coefficients \a and 36, and obtain 12 a 6; and then the radical parts yjlhyexA ^ibx, and, in accord- ance with the above principle, obtain ^iVxy\ or, for the whole product, 12 a 6 ^iVxy, which, reduced to its simplest form (Art. 233), is 24: a¥ \[xy'. 2. Multiply ZsfYa by ^^/Ya OPEEATION. 3v'2aX2A/3a = 3V2'a'X2,^32a2=3X2^2Vx3''a* = 6v"72a« We reduce the given radicals to equivalent ones having a com- mon index (Art. 237), and then multiplying, in the same manner as in the last example, obtain 3X2 ■5^2' c? X 3" cf, which, reduced to its simplest form, is 6 ^72a^ Hence we deduce the following RULE. Reduce the radical parts, if necessary, to a common index; then multiply the coefficients together for the coefficient of the prod- uct, and the^arts under the radicals for the radical part. Explain the first operation. The second operation. Bepeat the Rule. RADICALS. 211 EXAUFLES. 3. Multiply 3-v/S by 5^/"^. Ans. 15^1^ 4. Multiply 6 A^~bi by 3 v'"2. 5. Multiply 1 s/axy by 3\/2ax. Ans. 21 axx/^y. 6. Multiply aSylt by h>!fy. Ans. ahl^l^. 1. Multiply 4.4^1ix by Sa^Ic}. Ans. 12^"^^/. 8. Multiply iV'e by t^^/v/"9. Ans. ,VV6. 9. Multiply 2 4j^| by S/^f. Ans. 2^T5. 10. Multiply 3 5* by 4 o^. Ans. 12 ,^^T«. 11. Multiply Sa^YI^ by 2J^T^^. Ans. 12a«i^2^. 12. Multiply (a + J)^ by (a + J)*- Ans. (a+J)*. 13. Required the product of Jl^hj J^ lb l~x' . / 3 a« 6 Ans. t / - c 14. Required the product of '^6 a* 6c-i by 4^B-^a-*be*. Ans. ^2 6^. 241 ( When either or both of the radicals are connected with other quantities by the sign -\- or — , each term of the mukiplicand must he multiplied by each term of the midti- plier. (Art. 64.) 1. Multiply o + 2 -v/ * by « — V^- When the radicals are connected with other quantities by + or — , hew do we maltiplj ? 212 ELEMENTARY ALGEBRA. riRST OPERATION. SECOND OPEBATION. a -{- 2 k/I a + 2 5^ / _ 1 a — \/ 5 a — 6^ — ax/l — 2A/y — ah^ — 2b^ a=+ a/s/h — 2b a' -\- ab^ — 2b It -will be seen that both operations give precisely the same re- sult. From this and previous examples, it may be inferred that, in miil- tiplicalion of radicals expressed hy fractional exponents, the same rules apply as when the exponents are integral. If fractional exponents having different denominators are to be added, they must of course be reduced to a common denominator, and this is precisely the same process as reducing radicals to a common index. When fractional exponents are used, it is often most convenient '' to allow each factor to take a separate one ; but when the radical sign is used, it is most convenient to employ a common one for all the irrational factors of any given term. 2. Multiply 4 + 2 v^2 by 2 — a^J. Ans. L 3. Multiply (a + a:)* by {a — «)*. Ans. {a' — x'f. 4. Multiply ^/a -\- b hj \/a + b. Ans. a-\-b. 6. Eequired the product of J + J v'S by I + i V^.. Ans. ,1 -|- ^ v'5. 6. Eequired tbe product of k/x-{- vV+^ by V'a + a; Ans. \/fl; X -\- x'' -\- a -\- X. t. Eequired the product of a^ + J* by a^ 2 J^. Ans. a^ + a^ d^ _ 2 a^ 5^ — 2 b^ . What is said of multiplication by fractional exponents 1 RADICALS. 213 DIVISION OP EADICALS. 242. Division of radicals depends upon the principle that The quotient of like roots of two quantities is equal to the same root of their quotient. To prove this principle, let a and 6 represent two quantities. Then, by Art. 240, y'^'X »/&= \i'ab, Whence, \f^ -=- V'" = \/^ = \fb. a 1. Divide 6 v'54a by 3-v/2a- OPERATION. We divide the coefficient, 6, of the 6 i/54a 6 I bia dividend by the coefficient, 3, of the „ ,— r- ■" 3 V/ ~2^ divisor, and obtain 2, and the radi- . cal part of the dividend by that of J yJ7 ijjig divisor, and, in accordance with = 6 y/ 3 the above principle, obtain v'27 ; or, for the whole quotient, 2^21, which, reduced to its simplest form, is 6 y^S. 2. Divide 16 ^^ by 8 Voi OPERATION. We reduce the given rad- 16 i/1^ 16 £/a* ica\s to equivalent ones hav- ~ — T= = g. — ■:= 2 4/<^ '°S '■' common index (Art. 237), and, dividing in the same manner as in the preceding example, obtain 2 S/a. Hence the following RULE. Reduce the radical parts, if necessary, to a common index; then divide the coefficient of the dividend hy the coefficient of the divisor, and the radical part of the dividend hy that of the Upon what principle does division of radicals depend? Explain the first operation. The second. Repeat the Rule. 214 ELEMENTARY ALGEBRA. ■ divisor, and prefix the first quotient to the last written under the common index. EZAUFLES. 3. Divide \/40 by a/2. Ans. 2 a/S. 4. Divide a" by i"- Ans. ^^• 5. Divide -^135 by 4^5. 6. Divide 4 V^ l5y 2 ^'a- Ans. 2 a ^a. 4 r^ 7. Divide 4^aa; by ^s/xy. Ans. g^^- 8. Divide bc/f/lTb by h^/a. Ans. c^6., 9. Kequired the quotient of (1 — «">* divided by (l+x)^. Ans. (1— ^r- 10. Required the quotient of t/^ divided by U-^- ^''^- \l ro- ll. Required the quotient of i^| divided by ^/^i. Ans. f ^12. 12. Required the value of (^72 + \/32 — 4) -r- VS". Ans. 3 + ^ \/i4. 13. Required the quotient of m t / divided by 14. Required the quotient of \/a" — V divided by « — h. — 6 . m —, — r- Ans. — + o » A»- v/S 6' The remarks already made (Art. 241) respecting fractional ex- ponents will apply also to the division of radicals. 15. Divide a" + a i^ — 6 6 by a — 2 i^. Ans. a-\-Zh^. EADICALS. 215 16. Divide a» + 2 o^ 6* — 4 o^ 6^ — 8 6^ by a* — 4= h^- Ans. a^-V-2 6*. INVOLUTION OF EADICALS. 243i Involution of radicals depends upon the same general principles as involution of rational quantities. 1. Baise 2 \/a to the third power., OPERATION. 2y/aX ^isfa X 2 Va = 8 V'^= 8 « Va. By the definition of involution (Art. 186) we take the given quan- tity, 2 y'a, three times as a factor, and performing the multiplication (Art. 240), we obtain 8 y/a^ This, reduced to its simplest form (Art. 233), gives 8 a \/a. 2. Eaise3a-|-\/y to the second power. OPEEATION. Since the second power is required, _ we take the given quantity twice as ~r Vjf a factor, and, it being a polynomial, 3 a -\- ^y we .perform the multiplication as in 9a'-\-3ayfy ^"^^ 2"' Hence the following RTTLE. Raise the rational part of a monomial to the required power, and annex the required power of the radioed part, written under the given sign. If the radical part is connected with other quantities hy -\- or — , perform the involution hy mvMiplication of the several terms, as in the multiplication of polynomials. Explain the first operation. The second. Bepeat the Rule. 216 ELEMENTARY ALGEBRA. Note 1. When a quantity is affected by a fractional exponent, its involution may be performed by multiplying that exponent by the expo- nent of the required power (Art. 190). Note 2. The result should be reduced to its simplest form. Any factor common to the index of the given radical and the exponent of the required power should be canceled (Art. 235). Hence, when the given radical and the required power are of the same degree, the involu- tion may be perforrried by removing the radical sign. Examples. 3. Eequired the square of 5 cT. Ans. 25 a^. 4. Kequired the cube oibai^y. Ans. 125 a' y. 5. Raise a:^/^6 to the second power. Ans. a;* ,^36. 6. Eaise 4a^\/3c to the fourth power. T., Raise 3 .^25 a x to the second power. Ans. 45V'aa;. 8. Required the fourth power of />,/% — \/2. Ans. 49 — 20\/6. 9. Raise S^Ya to the wth power. Ans. ^2" a». 10. Required the square oi \/% -\- x f>/%. Ans. 3 + 6 a; + 3 a;''. 11. Required the square of a:" — y ^. Ans. X — 2 a;^y~5 4" ^ • EVOLUTION OP RADICALS. 244i Evolution of radicals depends upon the same gen- eral principles as evolution of rational quantities. 1. Find the cube root of S i\/c^. Bepeat Note 1. Note 2. RADICALS. 217 OPERATION Since the coefficient 8 is a _ perfect cube, we have for its n/S \/a* = /i/S X V^a' = 2 v'a cube root 2, and the quan- tity under the sign, a', being also a perfect cube, we have for its cube root a ; hence the entire root is 2 v'a. (Art. 232.) 2. Find the square root of 20 /^Ta. OPERATION. V20 \^5 a = a/4 X 5 v'a a = */i X Vs \^5 o = 2 14^125 a. The coefficient 20 is not a perfect square, but is composed of two factors, 4 and 5, of which 4 is a perfect square. The square root of 4 is 2, which is the coefficient of the required root. As we cannot take the square root of 5, we square it and introduce it as a factor under the sign. As the quantity under the radical sign is not a perfect square, we denote its root by multiplying the index of the sign by the 'index of the required root, and we then haye as the entire result, 2 ^125 a. Hence the following RULE. Extract the required root of the rational part of a monomial, if it is a complete power of the required degree, otherwise in- troduce it under the radical sign. Extract the required root of the quardity under the radical sign, if it is a complete power of the required degree, other- wise multiply the index of the radical hy the index of the re- quired root. Note. When a quantity is affected by a fractional exponent, its evo- Intiou may be performed by dividing this exponent by the index of the required root (Art. 224). Explain the first operation. The second operation. Repeat the Kule. The Note. 19 218 ELEMENTAEY ALGEBEA. Examples. 3. Find the cube root oi *>/ aW. Ans. /^ aW. 4. Find the square root of 4 \/ 5 c. Ans. 2/^ be. 1 -ill 5. Find the cube root of cr'^ixyy. Ans. a 'x^ f. 6. Find the square root of 25 v^4a^J. 1. Find the sixth root of a\/«- -^°s. a^ a. 8. Eequired the cube root of g l/g- -^°^- ^V^a. 9. Eequired the cirbe root of 125 x^. Ans. 5 a; . 10. Extract the fourth root of 6ia^b*\/2cd. Ans. 2an4^32cd. 11. Extract the square root of x'-^-Gx^ 3' + 9 y, by means of the rule found in Art. 225. ^^^ x^-\-3v^ EATIONALIZATION. 245. Eationalization is the process of removing the radical sign from a quantity by multiplication. It is often necessary to transform a fraction having an irrational denominator, into one whose denominator is rational. CASE I. 246. To rationalize any monomial surd. 1. Eationalize \/ a. Define Rationalization. RADICALS. 219 OPERATION. Y^ multiply the given _ _ 1 1 radical, ^a, by the same ya X V = o X <* = « ■ quantity, with such an index as will make the sum of the corresponding fractional exponents of the two quantities equal to unity. J! 2. Rationalize a^. OPEBATioN. -yyg multiply the given radical, J, f f by the same quantity with such a fractional exponent as will make the sum of the fractional exponents of the two quantities equal to unity. Hence the EULE. Mvkvply the given surd hy the same quantity with such a fractional ea^onent as, when added to the- fractional exponent of the given quantity, shall he equal to unity. Examples. 3. What factor will rationalize a:*? Ans. a:*. 4. What factor will rationalize 4^a6^? Ans. i^o?h. 5. What factor will rationalize a'*, and at the same » 8 time make its exponent positive ? Ans. (fl. CASE n. 247. To rationalize a binomial surd containing only the square root. 1. Rationalize %/« + V'*- Explain the first operation. The second. Eepeat the Rule. 220 ELEMENTARY ALGEBRA. OPERATION. Since the product of the sum and _ _ diflference of two quantities is equal ^a -\- \o ^Q tjjg difference of their squares yja — v'* (Theo. III. Art. 78), we multiply the , — r given binomial by the same terms, ' with one of the signs changed, and — V « — " obtain a — J, a rational quantity. a , h Hence the RULE. the given hinomial hy the same terms, with one of the signs changed. Examples. 2. Eationalize a -|- ^b, Ans. a° — h. 3. Eationalize V^ — V'l- AjiS. 5 — 1, or 4. 248. A trimmial surd may be reduced to a binomial surd by multiplying it by the same terms, with the sign of one of them changed, and then the binomial may be rationalized. Thus, to rationalize /\/t -(- v'S — V'2, we have (VT+ \/3 — s/2) (a/1 4- V3 4- \/2) = 8 + 2 ^21, and then (8 + 2^21) (—8 + 2 v^2T) = 84 — 64 = 20. CASE in. 249. To rationalize either of the terms of a frac- tional surd. a 1. Reduce "^ to a fraction whose denominator shall be rational. Explain the operation. Repeat the Rule. How may a trinomial surd be rationalized ? RADICALS. 221 OPEEATION. ^^ multiply both tenns of the o ]/b a i/b fraction by y/6, and it becomes "-^, \/b \/i "~* '"^ which the denominator is rational, and the value of the fraction is not changed. Hence the RULE. MuMply both numerator and denominator by a factor that wiU render either of them rational, as may be required. Examples. 2. Reduce '-rr to a fraction having a rational numerator. )/b Ans. -^=;- 3. Eeduce -= — - to a fraction having a rational de- nominator. . v^3 — 1 2 4. Eeduce ^ to a fraction having a rational denom- mator. 5. Eeduce — = = to a fraction having a rational de- ^b + }/c ^ _ nominator. . a {^b — y/c) J — c 6. Eeduce — - — = to a fraction having a rational de- 3 -^ v'2 _ nominator. ^^^ 3v/2-|-2 1. Eeduce _"* _ to a fraction having a rational de- ^x — ^y _ _ nominator. ^^^ (y/a; + V^y)' . x — y Explain the operation. Bepeat the Bnle. 19* 222 ELEMENTARY ALGEBRA. IMAGINAEY QUANTITIES. 250. An Imaginary Quantity is an indicated even root of a negativ6 quantity. Other quantities, whether ration- al or irrational, are called real. Although this root of a negative quantity is a symbol of an impossible operation (Art. 209), yet it is not with- out its use in mathematical analysis. 251 • Every imaginary quantity can he resolved into two factors, one of which is real, and the other the imaginary expression, s/ — 1> oi" V — !• For, let a denote any real quantity, and y/ — o any imaginaiy quantity of the second degree. Then, ^i—a —\/a (— 1) = ± y/a X V'— 1 ; also, v'^^ = V'a' (— O = ± v'a' X »/=^ = ± « V'^^. and so on. Hence, \/ — 1., or v' — 1, may be regarded as a universal factor of every imaginary quantity, and, consequently, may be used as the only sjTnbol of such a quantity. 252. In the addition and subtraction of imaginary quan- tities, the operations are the same as for other radicals; but with regard to their multiplication and division, the rules for common radicals require some modification. 253. The prodtLct of two imaginary terms is real, with the sign before the radical as hy the common rule reversed. For, if we take the product of two imaginary quantities in which the imaginary parts are equal, it is evident that the sign of the product is changed by removing the radical. Thus, h y/ — a X c y/ — a = he (—a) = — ahc. Define an Imaginary Quantity. Into what two factors may an imagi- nary quantity be resolved? Give the illustration and inference. What is said of the addition and subtraction of imaginary quantities ■? What u said of the product of two imaginary terms 1 RADICALS. 223 But , if we take two unequal imaginary quantities, y/ — a and ^—b,-iiy the common rule (Art. 240), we have Now, since the quantity whose root is to be extracted was not pro- duced by that root, but from two unequal factors, it does not im- mediately appear whether the result obtained is to be .taken posi- tively or negatively. We may, however, resolve the imaginary quantity into two factors, of which one is ^ — 1 (Art. 251). ^Then we have (+ V^^^) (+ V^^ = (+ V'0< t/^=^+ Vb X V''^^^) = +\'abx w^^y = +V'a6X(-l) ^ab. ■ Hence it appears that the result is properly negative. In like manner it may be shown, that (- ^^) (- v'^ft) \/^ and (+ \f^^) (— v/:Zj) = _i_ y/^ Whence, also, it appears that lake signs produce — , and unlike signs -j-. 234i The quotient of one imaginary quantity divided by another is real, with the sign hefore the radical as by the com- mon rule. For, +^Eg _ +yaft X y^ = + t/g, + y/—a +^a xy-1 and -^-ab _ -V °ft X y- 1 ^ j_ H-y— a +yo xy— 1 • Whence it appears that jAke signs produce -\-, and unlike signs — . 255i To show the application of the foregoing princi- ples, there are introduced the following What is said of the quotient of one imaginary quantity divided by another f 224 ELKMENTAKY ALGEBKA. Examples. 1 . Multiply ^/'^=^ by V^^. Ans. — ^'36'= — 6. 2. Multiply 2 \/—S by 3 V'— 2. Ans. — 6 \/"6. 3. Multiply 1 + V — 1 by 1—\^'^^1. Ans. 2. 4. Divide a V'— 1 by i \/ — 1. Ans. j. 5. • Divide 6 v' — 3 by 2 \/ — ^. Ans. f V^- 6. Divide 2 V — 10 by — 5 ^/^^1. Ans. — f \/ 5. EADICAL EQUATIONS LEADING TO SIMPLE EQUATIONS. 256i Eadical Equations are those containing radical quantities ; as, ^"3^ + 4= = 5, or (4 + a;)* = 5. 257. The solution of a radical equation consists in ra- 'tionalizing the terms containing the unknown quantity, and in determining its value. Only those which reduce to simph equations can be considered here. More will depend upon the ingenuity of the learner than upon any rules that can be given. 1. Given /s/ x — 2 = 3, to find the value of x. OPERATION. V'"^— 2= 3 Transposing and uniting, v' a; =5 Squaring (Art. 151), x = 25. 2. Given i^/ \\-\-^ 5x=^i, to find the value of x. Define Radical Equations. In -vrbat does the solution of a radical equation consist ? Explain the first operation.'' EADICALS. 225 OPERATION. ^Il4-V5a: = 3 Involving to the fourth power, ll-|-v'5^= 81 ■ Transposing and uniting, \/^ = 10 Squaring, 5 a; = 4900 Whence, x = 980. 3. Given \/ax = ^a-\- s/l:, to find the value of x. OPEBATIOlf. t>/ ax = v' a -\rA/ls • Transposing, »y~ax — is/lc =\/a^ Factoring, s/ x [»^~a — 1) s= \/^ Squaring, x (%/ a — 1)^ :^ a Whence, Prom the foregoing illustrations are deduced the follow- ing general directions : — 1. Transpose cdl the terms so that a radioed expression inay stand on one side of the equation, and the rest of the terms on the other side; then involve eaxih side to a power of the same degree as the radical. 2. ^ there is stiU a radical expression remaining, the pro- cess of involution must he r^eated. 3. Simplify the equation as much as possible' lefore per- forming the involution. Note. Biidicals may sotlictimes be removed by multiplying or divid- ing both members of an equation by a radical expression ; hence they sometimes disappear on clearing of fractions. It is also occasionally con- venient to rationalize the denominator of a fraction before removini; de- nominators or involving. Explain the second operation. The third. Bepeat the general direc- tions. The Note. 226 ELEMENTABY ALGEBEA. Examples. 4. Given V^+l — 2 = 3, to find the value of x. Ans. X = 24. 5. Given \/^+^ = V^ + 1, to find the value of x. Ans. oj = 9. 6. Given 2 -j- \/3 a; — 30 = 5, to find the value of x. Ans. a; = 13. 7. Given v'S — ^ -1-6 = 8 — 1, to find the value of a;. 8. Given x^ — T = — 3, to find the value of x. Ans. X =. 16. 9. Given /k/x -f- 6 = 3, to find the value of x. Ans. a; = 3. 10. Given ik/i -\- x = 4 — ^/x, to find the value of x. ' Ans. X = 2^. 11. Given y^ -\- 5 = 11, to find the value of y. Ans. y = 216. 12. Given ^^20 — v'2^ — 2 = 0, to find the value of x. Ans. a; = 8. 13. Given — = — = — , to find the value of x. yx X Ans. X = 1—a 14. Given a; (a:* -)- a; *) := a a; , to find the value of x. Ans. a: = o — 1. 15, Given )[l±l + 3 = -1^, to find the value of x. \fx — 3 Ans. X = 36. a* — 1 j/^j. J IG. Given , — , , =4 4- - — , to find the value y a a; -j- 1 ' 2 ' Fa;. .81 Ans. X = — a Note. Perform the division indicated in the first member (Art. 78). QUADRATIC EQUATIONS. 227 QUADEATIC EQUATIONS. 258. A Quadratic Equation is an equation of the second degree (Art. 146), or one in which the square is the highest power of the unknown quantity ; as, ax^=b, a:" + 8 a; = 20, or a' ar' — J* a: = c^ 259. A Cubic Equation is an equation of the third de- gree; as, aa?z=b, si?-\-x^=12, or aa? — bx' -{- c x = d. 260. A Biquadratic EQUATioff is an equation of the fourth degree; as, ax^=zb, x'^ -\-a? =.2Q, or ax* — ba?-\-cx'' — dx=.e. 261. A Puke Equation is one which contains only a single power of the unknown quantity ; as, aa? = J, a? =L i, or c^a^ = m. PUKE QUADEATIC EQUATIONS. 262. A PuBE Quadratic Equation is one which con- tains only the second power of the unknown quanti- fy ! *^' aa?z=b, 7?= 100, or ^ = c\ Note. Pure qnadratic eqaations are sometimes called incomplete equa- tions of the second degree. 263. Any numerical pure equation may always be re- duced to two terms, one containing the unknown quantity, and the other consisting of all the known terms united in one. Thus, the equation _ 5a^-4=— +2-, Define a Quadratic Equation. A Cubic Equation. A Biquadratic Equa^ tion. A Pure Equation. A Pure Quadratic Equation. To how many- terms may a pure equation be reduced ? 228 ELEMENTARY ALGEBRA. by clearing of fractions and transposing, reduces to 5 a;2 = 80. In a literal pure equation, all the terms which contain the unknown quantity may be united, since', expressing the same power of the same letter, they are similar : and as the remaining terras are all known quantities, they may be considered as one. Hence the equation may be thus expressed : (a _|_ 5 _ c) a^ = ((^2 _ OT^ -1- „2). Pure equations are therefore sometimes called KnomiaH equations. 1. Given 5 a;^ = 80, to find the values of x. By dividing equation (1) by 5 (Art. 151), we obtain (2), and by extract^ ing the square root of both members of the equation (Art. 151), we obtain x^ ±i. As «n even root of a posi- tive quantity if either positive or neg- ative (Art. 207), we obtain a: = 4 and a; =- — 4, as the roots of the equation (Art. 155). These we write in one expression, thus, a; = ± 4. Note. It may soem to the student that x should also have the double sign, thus, ± a; = ± 4. The results are the same, however, in either case; for ± a; = ±4 would include four expressions, -|-ar = -[-4, -f-a: = — 4, — X = -1- 4, and — x = — 4, of which the third reduces to the second, and the fourth to' the first, by a change of signs (Art. 152, Note). Hence we obtain all the possible values much more readily by prefixing the double sign to the second member of the equation. From the preceding solution it will be seen that A pure quadratic equation has two roots, which are equal in numerical value, hut differ in their signs. Explain the first operation. Why is not the double sign prefixed to both members 1 What is said of the number and relation of the roots of a pure quadratic equation ? OPERATION. bx^= 80 (1) x""— 16 (2) X = ± I (3) QUADRATIC EQUATIONS. 229 2. Given — \- - = =, to find the values of x. a ^^ X X d' FIRST OPEKATION. X j_ i C X a ' X X d (1) dx^-\-aid = acd — ax^ (2) ffla;'+ da? = acd — ahd (3) (a -{-d)x' = ad{c—b) (*) ._ad(c-b) (5) ladU Clearing equation (1) of fractions gives us (2), trans- posing gives us (3), and factoring gives us (4). Di- viding both members by a -\- d, the coeflBcient of k", gives us (5), the value of a;*; and extracting the square root of both members (Art> ^ /«\ 151), gives us (6), the value (6) ^d ^-> of». SECOND OPHRATION. i+ \ =i -3 m cr^x -f- 6a;~* =car^ — dr^x (2) a-^x^-\-h = c — dr^a? (3) a-ia;2 _[_ ct^a? =c~b (4) (0-1 -\-d-^)a?=c—h (5) — h + d- Jo-V) {1) (8) Equation (1) expressed by the aid of negative exponents changes to the form of (2), and the negative exponents of the unknown quantity are removed by multi{)lying all the terms by x (Art. 153, Note 3), producing equation (3). This equation is solved as in the previous 'operation, giving (7) as the value of a;. This is freed from negative exponents by multiplying both numerator and de- nominator by a d (Art. 142, Note). Explain the operations of Example 2. 20 230 ELEMENTAKY ALGEBRA. From the foregoing principles and illustrations we in- fer that any pure quadratic equation may be solved by the following RULE. Obtain the value of the square of the unknown quantity as in simple equations (Art. 169). Mctract the square root of both members of the equation thvts obtained. Note I. A similar application of Ax. 8 will serve to obtain one root of any pure equation of a higher degree. In treating of equations which take the form of affected quadratics, we shall have occasion to solve various cubic, biquadratic, and higher pure equations. Note 2. It will be observed that many equations, which do not at first appear to be pure quadratics, reduce to such equations after clearing of fractions or performing the operations indicated. Examples. 3. Given 3 k" — 2 = 2 a^ -f 2, to find the values of x. Ans. X = ±2. 4. Given — - -far' = 126, to find the values of x. a Ans. a; ::= ± 6. 5. Given fz^^a^, to find the values of y. 6. Given t (2 a^ — 6) + 5 (3 — a^) = 198, to find x. Ans. a; = ± 5. 7. Given {x-^Vf = 1x-\-Vl, to find x. K-a.^. x=z ±4:. 8. Given ar* -}- a J = 5 a^, to find x. Ans. a; = ±^\^ ab. „ „. 2a:«-fl0 . 50 -I- r" 9. Given — -~^— =7 ^, to find x. Ans. a; = ± 5. Bepeat the Rule. What is Note 1 ? Note 2 ? QUADRATIC EQUATIONS. 231 10. Given (aj + 2)" = 4 a: + 5, to find x.. Ans. a; = ± 1. 11. Given (2x— hf z=a? — id x -\-n, to find x. 12. Given Ix — 150a;-i = a; — 3a;-\ to find x. Ans. »:= ± 7. 13. Given ~-^ + ^^ = 8, to find a;. Ans. x=±^. 14. Given a^ — ^-\-3ab=:2a''-\-l^, to find x. Ans. a:== ± /-a — jV 1&. Giyen c(si^-{-4cab-\-4:bc)=a{a-\-2cy-{-dx' — a'b, to find a;. . , /- i o \ /« — * Ans. x=±(a-\-2c) i/^^^. 16. Given ^ + ^^-^ = _, to find x. Ans. x= ±10. 11. Given X = 3 ar* , to find x. 5x Sx Ans. a: = ± 2. 2 10 18. Givenr»-3^:^=g^3^^-p^. to find y. Ans. y = ± 3. 19. Given ^T 8^=n = -3"' *° ^^"^ *' Ans. « = ± 2. 20.- Given (o + a;)' + (a — a:)' = 2 i*, to find x. Ans. . /6» — a» 264i Eadical Equations sometimes reduce to the form of pure quadratics. The preliminary reductions should be efiected as in Art. 251. 1. Given 22^ + ^5 (4 a:^ — 1) = 25, to find x. Ans. a; = ± f . 2. Given A/ay' + Vas* — «* = «> to fii^ '*'• • Ans. x= ±a. 232 ELEMENTAEY ALGEBBA. 3. Given i>M — 1^ = ^' *" ^"^^ ^■ 4. Given V"' + a;^ = '^i* + ^■'' *" ^^^l *• ,^ 5. Given ^/^^^^^ = 2 V^, to find x. Ans. a; = ± 4. 6. Given ^a; + a = kJx + V*'' + a:^ to find x. Ans. a; = ± s/a? — J'. 7. Given x (10 + a:=)^ = 5 — a:^ to find x. Ans. a;= ± ^\/5. 8. Given a: + (a" + a;*")^ = 2 a^ (a" + a;'')"^, to find x. Ans. a: = ± ^• 9. Given ' = h, to find a;. Ans. a: = ± i^l 1+6 Note. Rationalize the denominator of the first member of the last eqaation, and then extract the square root of both members (Art. 257, Note, and Art. 249, Ex. 7). 10. Given y/|±-^ + y^^ = 5, to find x. Ans. x=L±\ v'ai. KOTB. Square the last equation as it stands, and thus remore all radicals. SIMULTANEOUS EQUATIONS. 265f Simultaneous equations (Art. 170) sometimes pro- duce pure equations after elimination. The methods of elimination are the same as in simple equations ; but sub- What methods of elimination are here employed ? QUADRATIC EQUATIONS. 233 stitution will be found best adapted to most of the ex- amples which follow. Some of the equations will be found to be of a higher degree than the second. (Art. 263, Kule, Note 1.) 1. Given ix^ — 2y^ = 40, and a; — 2y = 0, to find the values of x and y. OPERATION. Equation (2) gives the 3 a^ 2 «2 ^^ 40 (1) ^*^"^ °f ^> i'l terms of y. "„ ,n\ Substituting 2y, this value 3(2y)^_2/ = J (3) °'^"' t'"' lV^"f°^ ^'^' ^ "' " ^ -" we obtain (3), which re- 122,^—22,^ = 40 (4) dices to (6). Extracting the lO^T = 4:0 (5) square root of (6), we have y''= 4 (6) (7), the value of y. Sub- y =±2 (T) stituting the value of ^ in (2), a; = 2« = ±4 (8) we have the value of x. 2. Given ^x^ — %y^='n., and ix-\-2y = Q, to find X and y. Ans. a; = ± 12 ; y = T 3. 8. Given bxy — ^y^= 100, and bx — 4y = 0, to find X and y. Ans. a; == ± 8 ; y = ± 10. 4. Given a;y -|- y'' =: 126, and 5 y = 2 a;, to find x and y. 5. Given 4 a^ + 1 j," = 148, and 3 aji" — y" = 11, to find X and y. Ans. a;=±3; y=±4. 6. Given x-\-y=Zx — Zy^ and a:' — y' = 56, to find X and y. Ans. a; = 4 ; y = 2. 7. Given a? -\-f : x^ — y^ :: Vl : 9,, and a;y« = 45, to find X and y. Ans. a; = 5 ; ■y=±Z. 8. Given a:* -|- y* = 97, and 9 a;" = 4"'y'', to find x and y. Ans. a: = ± 2 ; y ^ ± 3. NoTB. There are also two imaginary values for each of the unk nown qnaptities in the lajt example, viz. a; = ± 2 ^ — 1, y = ± 3 y/ — I ; for a* = 4 or —4, and y» = 9 or —9. Explain the first operation. 20* 234 ELEMENTAKY ALGEBRA. PEOBLEMS LEADING TO PURE EQUATIONS. 266. The methods of stating these problems are the same as in the case of those leading to simple equations. (Arts. 160, 16Y.) Either one or two unknown quantities may be used in many cases. 1. Find two numbers, one of -which is three times a; great as the other, and the sum of whose squares is 90. SOLUTION. Let X = the iirst number, and 3x = the second number. Then, x^. -|- 9 a;" = 90 Uniting terms, 10 a;'' = 90 Dividing by 10, a^ =: 9 Evolving, a; = ± 3, the first number, 3x = ±9, the second number. The only arithmetical numbers which will answer the conditions are 3 and 9. 2. Find two numbers, one of which is five times as great as the other, and the difference of whose squares is 96. Ans. 2 and 10. 3. The length of a field is to its breadth as 3 to 2, and its area is 3 acres and 3 roods. What are its di- mensions ? Ans. Length, 30 rods ; breadth, 20 rods. 4. A merchant bought two pieces of cloth, which to- gether measured 36 yards. Each of them cost as many dimes a yard as there were yards in the piece, and their entire prices were as 4 to 1. How many yards were there in each piece? Ans. 24 yards in one ; 12 yards in the other. Explain the solution of Problem 1. QUADRATIC EQUATIONS. 235 5. The product of two numbers is 150, and the quo- tient when one is divided by the other is 3^ ; find the numbers. Ans. 50 and 15. 6. A detachment from an army was marching in reg- ular column, with 5 men more in depth than in front ; but upon the enemy being discovered, the front was in- creased by 845 men, and by this movement the detach- ment was drawn up in five equal lines. What was the number of men ? Ans. 4550. t. Find three numbers which shall be to each other as 6, 1, and 9, a;nd the sum of whose squares shall be 620. Let 5 X, 7 X, and 9 x represent the numbers. 8. A certain number of 'Tboys went out to gather nuts, each taking as many bags as there were boys in all, each bag being of a capacity to contain as many nuts as there were boys. Upon filling the bags, they found them to contain exactly 1000 nuts. How many boys were there ? Ans. 10. 9. There are two cubical blocks of stone, one of which contains ll'T cubic feet more than the other, and the side of the larger is 2^ times as long as that of the smaller. Required the dimensions of each. Ans. 5 feet, side of the larger ; 2 feet, side of the smaller. 10. Two persons, A and B, set out from difierent places to meet each other. They started at the same time, and traveled on the direct road between the two places. On meeting, it appeared that A had traveled 18 miles more than B ; and that A could have gone B's distance in 15J days, but B would have been 28 days in going A's distance. How ikr did each travel? Ans. A, 12 miles; B, 54 miles. 236 ELEMENTAKY ALGEBRA. AFFECTED QUADEATIC EQUATIONS. 267 • An Affected Quadratic Equation is one which contains both the second and first powers of the unknown . quantity ; as, a;" -|- a a; = J, 2 a;^ -|- 16 a; = 40, or a x' -\- h x = c. Note. Affected quadratic equations are sometimes called complete equations of the second ai'gree. 268. Any affected quadratic equation may always be reduced to three terms, one containing the second power of the unknown quantity, another its first power, and the remaining one the known terms of the equation. Thus, the equation (^ + l)»U^-x + 2i = ^ + 8, after performing the operations indicated and clearing of fractions, reduces to 4 a;2 — 3 a: = 2T ; and ax^ -\- hx — e = bx' — ax -\- d may be thus expressed i (a — J) a:'i 4- (a + 5) a; = (c + d). Affected quadratic equations are therefore sometimes classed among trinomial equatiorts. Note. If the first of the three terms is wanting, the equation is evi- dently of the first degree ; if the second is wanting, the equation is a pure quadratic ; and if the third is wanting, the equation may be at once reduced to the first degree, by dividing both terms by the unknown quantity. FIRST METHOD OF COMPLETING THE SQUARE. 269i If the second power of the unknown quantity has any coeflScient expressed, the equation may be still Define an Affected Quadratic Equation. To what terms may any af- fected quadratic equation be reduced ? How may it be still further re- duced 1 QUADRATIC EQUATIONS. 237 further reduced by dividing all its terms by that coefficient. Thus, 3 a;" — 6 ic = 9 reduces to x^ — 2x = 3, and 4 a;^ — 3x = 21 reduces to x^ — # « = ■¥■■ Hence any affected quadratic equation may be made to assume the form x^ -\- p X = q, in which p and q are understood to represent any num- bers whatever, whether positive or negative, integral or fractional. 1. Given a;*" -j- 4 a; -j- 4 = 9, to find the values of x. OPERATION. I' 'S evident that ^i!_|_4a: + 4 = 9 (1) _ x' + Ax + i (x -i- 2Y = 9 (2) ^^ ^ perfect square of a bi- _i_ „ , „ ,„•. nomial; for a? and 4 are n o /A\ positive squares, while 4a; is ^ \. / twice the product of their '"' ^^ ■'■' '"' ** square roots. Equation (1) may therefore take the form VERIFICATION. ^ ^^y (Art. 90.) This may l°-[-4X 1-1-4 = 9 (1) be regarded as a pure quad- ( 5)2_1_4 ( — 5) 4-4 = 9) ratio, in which the unknown 25 ■ 20 -I- 4 ::= 9 j ' ' quantity is not x, but x-\-2. Extracting the square root of both members, we have ±3 as the value of x -\- 2, and the equation is now reduced to a simple one. Taking the upper of the two signs, and transposing 2, we have x ^ — 2-|-3 = l; but taking the lower, we have x = — 2 — 3= — 5; and these values are found to satisfy the equation. We thus obtain two roots of the equation, which differ both in sign aftd in numerical value. Note. The reason for prefixing the double sign to only the second member of the equation, in extracting the square root, has already been given. (Art. 263, Ex. 1, Note.) 2. Given a? — Qx-{-l2 — 3, to find the values of x. Explain the first operation. OPEKATION. a;a_6x+ 12 = 3 (1) a^—.Qx+ 9 = (2) a;— 3= ±0 (3) a; = 3 ±0 (^) x:=B, or 3 238 ELEMENTARY ALGEBRA. Subtracting 3 from both members of equation (1), we obtain (2), whose first member happens to be a perfect square ; for a? and 9 are the squares of x and 3, while 6 a; is twice their prod- net. Extracting the square root of each member, we obtain (3), which reduces by transpo- sition to a; = 3. It will be seen that the two roots of the above equation are alike, both in sign and in numerical value. Such an equation is said to have equal roots. Note. The two roots of a pure quadratic equation are alike in m- mericai value, bat differ in their signs (Art. 263), and hence are not equal, in an algebraic sense. No qnadratic equation can have equal roots, unless its second member is when its first member is a perfect square, that is, unless its three terms make a perfect square when collected in one mem- ber. 3. Given x' — Sx = 20, to find the values of x. It is evident that the first member is not a perfect square, as -in the first exam- ple, neither can it be made such by the transposition of ■the known term, as in the second example. Such a term must therefore be added to z" — 8 a; as will make it the square of some binomial. As a" is the first term of the equation, x, its square root, must be the first term of the binomial sought. The next term of the equation, 8 a;, must be twice the product of the two terms of the binomial ; and one half of 8 a;, or 4 x, must be their product. But 4 a; is the pro- duct of 4 and x ; hence 4 is the second term of the binomial sought, and its square, or 16, must be added to the first member of the equation to make it a perfect square, and also to the second mem- Explain the second operation. What is said of the roots of the equa- tion t Explain the third operation. OPERATION. a;2— 8a; = 20 (1) ar". — 8a; -f 16 = 36 (2) a;— 4= ±6 (3) x — l ±6 W X = 10, or- -2 QUADRATIC EQUATIONS. 239 ber to preserve the equality, thus producing equation (2). The square root can now be extracted, thus producing equation (3). Taking 6, the positive root of 36, and transposing and uniting terms, we find X = 10 ; but taking — 6, the negative root, we find 1= —2. 4g->f m _ I X Did OCT ■ <^i^^ii 1 — -r = 5 + -r- to find the values of x. We first multiply by 5, to free the unknown quantity of fractions, and, after trans- posing and uniting terms, ob- tain equation (3). As the square of the unknown quan- tity must be positive, we then divide all the terms by — 3, and obtain (4), the equation in its reduced form. If the first member of equation (4) is to be made a perfect square, — x must be twice the product of the two terms of the root. As X is one of those terms, one 7 half its coefficient, or — , must D be the other term, and the 7 49 1. square of -, or — , must oe added to both members of the equation. Extracting the square root of equation (5), we obtain (6). Taking the positive root of gg, and transposing and uniting terms, we obtain x = — - = — gj but taking the nega- , . 10 5 tive root, we obtam a;"= — — = — g-. It will be seen that the two roots of the above equation have the same sign, but differ in numerical value. 6. Given si^-{-px = q, to find the values of a;. OPERATION. 7x 5 , Sar" 5 ~3+ 6 (1) 5 — 1x = ~-\-3x' (2) -3.-V.= ¥ (3) ^^ + ^=-¥ (4) *'+3'"+36 = 36 (5) '+I=±5 (6) 7 , 3 a) 4 a; = --,or- 10 ~ 6 2 a; = -3,or- 5 ~3 Explain the fourth operation. 240 ELEMENTAEY ALGEBRA. • OPERATION. x^-\-px^q (1) P x = -- ^ + f=±y/y+f (3) As in the other examples, p x being twice the product of the two terms of the root of the ^completed square, and x being one of those terms, ^ must be the other, and 3? -\- px can be made a perfect square only by adding to it j-. After adding the same quantity to the second member of the equation, we extract the square root of both members. The root of the second member, howeyer, can only be expressed. By transposing ^, we find the two val- ues of X to be — l + i/g+l^and— | — 4/}+£^. From the foregoing principles and illustrations we in- fer that any affected quadratic equation may be solved by the following RULE. Reduce the equation to three terms, placing the two which contain the unknown quantity on the first side, the higher power first, and the known quantity on the second side. Divide each side by the coefficient of the first term, and the equation wiU h reduced to the form 3? -\- p x ^ q. Add the square of half the coefficient of x to both members of the equation, and the first member wiU be a complete square. Extract the square root of both members, and solve the sim- . pie equation thus produced. Note 1. If the coefficient of the square of the unknown quantity happens to be negative, all the signs must be changed. This may be effected by using the negative coefBcieut as a divisor. Explain the fifth operation. Repeat the Rule. Note 1. QUADRATIC EQUATIONS. 241 Note 2. After conTpleting the square of the first member of the equa- tion, its first and third terms must be positive squares ; but its second term may be either positive or negative. If the second member is then positive and » perfect square, both roots will be real and rational; if it is (lositive, but not a perfect square, both roots will be real, but irrational (Art. 227) ; and if it is negative, both roots will be imaginary (Art. 250). Note 3. The square root of the first member of the equation is com- posed of the square roots of its first and third terms, connected by the sign of the second term. . The above rule may be applied in the solution of the following Examples. 6. Given si^-\-2x = 8, to find the values of a;. Ans. X = 2, or — i. ■?. Given a? — ix^ — 4, to find the values of a:. Ans. X ^2, or 2. 8. Given x' — 6 a; = 55, to find the values of x. 9. Given a^ -|- 12 a; -f- 35 = 0, to find the values of x. Ans. x = — 5, or — T. 10. Given 3 2:^ -f- 48 = 30 s, to find the values of e. Ans. 2 = 8, or 2. 11. Given x' — 2 ax = J, to find the values of x. Ans. x = a ± \/ a''-\-b. 12. Given a^ = 3 a; -|- 10, to find the values of x. Ans. a; = 5, or — 2. 13. Given 2 a; + 60 = 2 a;^ to find the values of x. Ans. x=G, or — 5. 14. Given ii^ -{-81/ = 5, to find the values of y. Ans. y = i, or — |-. 15. Given 5 a;^ + 20 = 25 x, to find the values of x. Ans. a; = 4, or 1. Kepeat Note 2. Note 3. 21 242 ELEMENTAEY ALGEBRA. 16. Given 3 x -\-4:=39x~^, to find the values of a;. Ans. X ^5, or — 4^. It. Given 5x^ — 40 a; = 70, to find the values of a;. Ans. 03 = 4 ± \/~s6l 18. Given 3 a; = 10 -j- ^^ a^, to find the values of x. Ans. x = Q ± 2^/— 1. 19. Given a? — 6 a; = 0, to find the values of x. Ans. a; = 6, or 0. Note. Sach an eqnation may be solved as an affected quadratic ;^at one of its roots will be found to be 0, as it evidently should be, since the equation can be at once reduced to a simple one (Art. 268, Note). 20. Given a~^ x-\-a ar^ = 2 a~^, to find the values of x. Ans. a; = 1 ± \/ 1 — a!'. 270. The equation aP -\-px =: q may be regarded as the general expression of any afiected quadratic equation re- duced to that form. (Art. 167, Prob. 34.) As this equation has already been solved (Art. 269, Ex. 6), we may use its roots, — 1^+1/ y+f and — | — t/ y + l", as the general formulas for the roots of any afiected quadratic equation. Instead, then, of going through the full pro- cess of solving each equation by itself, according to the foregoing rule, we may write out its roots at once, by substituting the particular values of ja and q in the above formulas. Hence, The roots of any equation reduced to the form a?-\- px-=zq may be found by taking one half the coefficient of x, with a contrary sign, plus or minus the square root of the sum of the second member and the square of half the coefficient of x. How may any affected quadratic eqnation be solved without going through the full process of completing the square ? QUADRATIC EQUATIONS. 243 This process is to be employed in the solution of the following Examples. 1. Given x''—8x=9, to find the values of x. Ans. a; = 4 ± \/9 + 16 = 9, or —1. • 2. Given x= + 16 a: = —55, to find the values of x. Ans. a; = — 8 ± Ay/^n>5~^M = — 5, or — 11. 3. Given a;^ — 20 a; = 800, to find the values of a;. 4. Given a;^ + 5 a; = 14, to find the values of x. Ans. a: = — I ± ^/14 + -^^ = 2, or — T. sc^ 3 X 5. Given - + — = 21, to find the values of x. Ans. a; = 6, or — 10 J. 6. Given ^a;« — ^ a; + 7| ±= 8, to find the values of x. Ans. x = ^, or — f. SECOND METHOD OE COMPLETING THE SQUARE. 271 • Any afiected quadratic equation whatever may be solved by the method employed in Art. 269. It will be seen, however, that a fraction must be added to complete the square, unless the coefficient of the first power of the unknown quantity in the reduced equation becomes an even whole number ; and even then the second member may sometimes be fractional. But by employing another mode of completing the square, sometimes called the "Hin- doo method," all fractions can be avoided till the roots are obtained. 272. Any equation may be reduced to three terms, as Why do we introduce a second method of completing the square ? 244 ELEMENTARY ALGEBRA. before, cleared of all fractions, and divided by the great- est common measure of its terms. It will thus be reduced to the form in which a, h, and c represent any whole numbers whatever which have.no common measure greater than unity. 1. Given a? — 5 a; -(- 4 = 0, to find the values of x. Transposing the known term to tlie second member, we have (2). If we wish to complete the square of the first member without intro- ducing fractions, it is evident that the second term should be divisible by 2, as it is twice the product of the two terms of the root. But the first term must be a perfect square; hence, we multiply- all the terms of the equatiou by 4, the smallest even square number, and obtain (3). The square root of the first term is 2x, which must be the first term of the binomial root, and as 20 a; is twice the product of the two terms of the root, 10 a; must be their product, and the other term must be ^^ = 5. Hence 5^ or 25, must be added to the first 2a; member to render it a perfect square, and to the second member to preserve the equality, thus producing equation (4). Extracting the square root, we obtain (5), which, by transposing and uniting terms, and dividing by 2, the coefficient of a;, gives 4 and 1, as the values of x ; and these values satisfy the equation. It will be observed that we have thus avoided the fractions which must be employed in solving this example by the previous rule. (Art. 269, Ex. 15.) To what form is the equation here reduced ? Explain the first op- eration. OPERATION. a:2_5a; + 4 = (1) a:= — 5a;=:— 4 (2) 4a;'^ — 20a; = — 16 (3) lar' — 20a; + 25 = 9 (4) 2 a; — 5 = .± 3 (5) 2a;=5 ± 3 (6) 2 a; = 8, or 2 a) a; = 4, or 1 (8) VEKIPICATION. 4^ —5X4 + 4 = (1) p_5xl+4=0 (2) QUADRATIC EQUATIONS. 245 It will also be seen that the quantity added to complete the square is the square of the coefficient of x in equation (2). 2. Given --^ — 8 a:-i = 24 + i-, to find the values of a;. OPERATION. ii^-8a^i = 24 + ^ (1) 18 «= — 40 = 120 a; + 2 (2) 18 a;" — 120 a; = 42 (3) 3 ar" — 20 a; = Y (4) 9 ar" — 60 a; =: 21 (5) 9 a:» — 60 a; + 100 = 121 (6) 3 a; — 10 = ± 11 (\) 8a;=:10 ± 11 (8) 3 a: = 21, or — 1 (9) x= 1, or — i (10) We first remove the denominators and the negative exponent, by multiplying both members by 5 a; ; then, after transposing and uniting terms, and dividing by 6, the greatest common measure of the three terms of equation (3), we obtain (4) as the reduced equation. The coefficient of the second term, — 20 a;, is an even num- ber; hence there is no necessity for multiplying by 4, as in the last example. But 3 s" is not • a perfect square, and we must there- fore multiply by 3, to render the first term a square, producing equation (5). 3 x, the square root of 9 ^, must be the first term of the binomial root, and 30 x, one half of 60 x, must be the pro- duct of the two terms of the root; hence — — , or 10, must be the second term, and its square, 100, must be added to both members. We then extract the square root of both members, and reduce Ets in the previous example. The number added to Complete the square in the above example is the square of one half the coefficient of x in equation (4). 3. Given a a:^ + J a; = c, to find the values of x. Explain the second operation, 21* 246 ELEMENTARY ALGEBRA. OPERATION. ax'^bx = c (1) 4a''ar' + 4aJa; = 4ac (2) ia!'x'-{-4.abx-\-b^= 4:ac-\-¥ (3) 2ax-\-b= ± \J4.ae + ¥ (4) 2ax= —b± v'4ac-j-62 (5) (6) — b± ^iac-{-¥ 2a To make the first term a square, and the second term divisible by 2, we multiply both members by 4 a, producing equation (2). 2 ax, the square root of ia's^, must be the first term of the root, and — - — , or 2 a 6a;, must be the product of the two terms; hence -^ , or 6, must be the second term of the root, and J" 2ax . ' must be added to complete the square, producing equation (3). We next extract the square root of the first member, and express the square root of the second ; then, by transposing and dividing, we ob- tain the values of x in equaticm (6). The quantity added to complete the square is the square of the coe£Sci^nt of x in equation (1). As a, b, and c in the last example may have any value whatever, we derive from the solution of that equation the following RULE. Seduce the equation to three integral terms, placing the two which contain the unknown quantity on the first side, the higher power first, and the known quantity on the second side. Divide the three terms by their greatest common measure, and the equa- tion will be reduced to the form aa? -\-bx = c. Multiply both members of the equation by four times the co- efficient of a?, and add to each the square of the coefficient of X. Extract the square root of both members, and solve the sim- ple equation thus produced. Explain th^ third operation. Bepeat the Bule. QUADRATIC EQUATIONS. 247 Note 1. It must be observed, that in this rule we take the square of the coefficient which * has bejbre it is multiplied ; but in the previous one we take the square of one half the coefficient which it has after it is divided. Note 2. Any quadratic equation may also be solved by using the co- efficient of x^ as a, multiplier, instead of four times that coefficient, and adding the square of one half the coefficient of x, instead of the square of that coefficient. If the coefficient of x is an even number, this method will avoid fraptions, and at the same time make each term only one fourth as great as it would be by the rule given above. Note 3. If the coefficient of i^ is negative in the reduced form of the equation, all the signs must be changed. This may be effected by including the negative sign in the multiplier. Note 4. Tl^e formula x = ~ ^V <"=+ — ^ obtained by the solution of Example 3, may be used for the solution of any quadratic equa- tion of the for m ax^ + hx = c, in the same manner as the formula x = — S.± ^Jq + ^ is used in Art. 270. The use of fractions is to be avoided in the solution of the following Examples. 4. Given ar* — '7a;-|-6 = 0, to find the values of x. Ans. a; =: 6, or 1. 5. Given a^ -]- - = 3, to find the values of x. Ans. X = 1 J, or — 2. 6. Given 10a: = 6a:* + 4:, to find the values of x. 1. Given 5 3^=51 — 4 a;, to find the values of a;. Ans. X = S, or — 3f . 8. Given °°~^'' = 2 a a; — ca^, to find the values of x. " . a±h Ans. X = . c 9 Given - -\-bx-^ = c, to find the values of x. a ac±s/ a'c' — iab Ans. X = s • Repeat Note I. Note 2. Note 3. Note 4. 248 ELEMENTARY ALGEBRA. 10. Given (a;+l) (2x + 3) = 4a;= — 22, to find the values of a;. Ans. x=5, or — |. 11. Given f (/ — 3)=:i(y— 3), to find the values of y. Ans. y = |, or — |. 12. Given ar^ + 1 a;-^ + 1 = 0, to find the values of x. Ans. x = — 1, or — §. Note. "When aU the exponents of the unknown quantity are negative, as in the last equation, the negative exponents nmy be retained until the value of ar-i is found. The value of x will be its reciprocal. If it is preferred, however, the negative exponents may be removed at once, as in previous examples. THIRD METHOD OF COMPLETING THE SQUARE. 273. The following statement includes all the methods of completing the square already given, for it is founded directly upon the nature of the square of a binomial. It ■wUl be seen that it is substantially the method employed in the solutions on which the rules have been founded, the main difierence being that we here multiply the divi- sor by two, instead of dividing the dividend by that number. The terms of the equation being arranged in the same manner as before, Make the coefficient of the first term a positive square, either by multiplication or hy division. Divide the second term by twice the square root of the first, and add the square of the quotient to both sides. The character -of the solution will depend upon the multiplier or divisor used to render the coefficient of the first term a perfect square. If the smallest factor or divisor be used, this method will, of course, frequently What statement will , include all the rules for completing the square 1 On what does the character of the solution depend? QUADRATIC EQUATIONS. 249 require the use of fractions ; but it may occasionally be applied to advantage, and give a solution preferable to that obtained by either of the rules before given, espe- cially when the coeflScient of the first term is either a square, or contains a square factor, as in the following EXAUFLES. 1. Given 9 ar" — 6 a: = 8, to find the values of x. Ans, x=.\, or — |. Note. ^%:c^ = ^x, and g^ = 1- Hence the addition of 1 will com- plete the square. 2. Given 4 a;'' -|- 4 x = 3, to find the values of x. W 3. Given 2T ax" + 6 5a; = -, to find the values of x. Ans. X = — , or — — • i) a o a Note. Multiply by 3 a, or multiply by a and divide by 3. The for- mer course will avoid fractions, and [-^^ ) = 6^ will complete the square. 4. Given 50 x'^ -\- 4= x =: ^^, to find the values of x. Ans. x = xV. or — /w Note. Either divide by 2, or multiply by 2. Fractions must be used in either case. 5. Given 5 x"^ (5 ar^ — 12) = — 36, to find the values of X. Ans. X = f . 6. Given |-4x=lT, to find the values of a;. Ans. X = 4, or — 68. 1. Given [- 3 x = 21, to find the values of x. Ans. X = 6, or — 42. When may this method be used in pieferenoe to the rules before given? 250 . ELEMENTARY ALGEBRA. 274, The following equations are to be solved by ei- ther of the methods which have been explained, care being taken to select the method best adapted to the example under consideration. The radical equations introduced in this connection are such as reduce to quadratics by the methods already ex- plained and applied. (Art. 257.) Examples. 1. Given x^ — 14 a; = 120. Ans. x = 20, or —6. 2. Given a? ^ = 2*1. Ans. a; = 6, or —4^. 3. Given 2 a:= — 10 a; = 100. 4. Given 16 a;-' — 4 = 12 ar*. Ans. a; = 3, or 1. 5. Given a;" — A a; = xV- Ans. x = ^, or — |. 6. Given ^^+3^ = 1 + 8. !^ 20 T. Given — + y = 2 ». Ans. « = 20, or4. ^ 8. Given 2 a;" + 15 = 3 a;. Ans. x — ^ =^/^"" "^ 9. Given a;" — 6 a; -|- 19 = 13, to find the approximate values of x. Ans. x = 4.V32, or 1.268. 10. Given 4aa;=' — 25a: = c. Ans. a. = ^±4^1+?. 11. Given a;^ — 4 = 16 — (a; — 2)=. Ans. a; = 4, or — 2. 12. Given (3 a; — 5) (2 a; — 5) = (a: + 3) (x — 1). Ans. X = 4, or f. 13. Given (2 a: — 3)" = 8 x. Ans. a: = |, or ^. 14. Given ^ (a; — 3)'' + f = a;. Ans. a; = 6, or 6. 15. Given a;" + (a; + 1)^ = ^ a; (a: + 1). Ans. a; = 2, or — 3. How are the equations of Art. 274 to be solved ? QUADRATIC EQUATIONS. 251 16. Given 3 (2 — a;) + 2 (3 — «) = 2 (4 + 3 ar"). Ans. x=:^, or — ^. 11. -Given 4 (x — 1) — ZUl — qs x—l Ans. a: = 2, or -ji^. 18. Given ^-p- + j = 3. Ans. x = 2, or — f 7 2 19. Given ^-^ + j — 5 = 0, to find the approximate values of a;. Ans. a; = 1,148, or —0.348. 2<>- ^^^^"^ ^+60 = 3^—5- ^°«- »= = 1^. or - 10.- 21. Given 8 a; + 11 + Y a:"* = 3 + — . Ans. a;= T, or — J. 21 a; 22. Given r ^ = 3f Ans. a; = — 2, or — 16. 23. Given ^^^^ = a; — 3 + a:-\ a; -J- 3 ' 24. Given - ^ a; Ans. a; = 1, or f. ^6 + 8--^ = ^,- Ans. a; = — 4, or — 4. _£. ~. a:+3 , s — 3 2a; — 3 25. Given — f-^ -J r = ,-• x-\- 2 ' X — 2 X — 1 Ans. a; = 4, or 0. Note. If the second member of the reduced equation becomes before completing the square, one of the roots will be (Ex. 19, Note, Art. 269)-; but if the second member becomes after completing the square, the roots will be equal (Ex. 2 and Note, Art. 269). „. 3a: — 2 , 2a; — 5 _ 10 26. Given -z: z •+■ z a = "S"' 2a; — 5 ' 3a; — 2 3 Ans. a; = 4^, or }. 21. Given ar'4-aa; + &a; + aJ = 0. Ans. X = — a, or — b. 28. Given adx — aca^ = bcx — hd. d h Ans. xz=-, or — -. -252 ELEMENTARY ALGEBRA. ac 29. Given (a 4- b) x^ — cx=z. — r—j-- Ans.x = '-^-!^4l^ 30. Given \/x^ + \/a;= = 6 a/x. Ans. a; = 2, 'or — 3. ;, 31. Given (4 a; + 5)^ (T a; + 1)^ = 30. Ans. a; = 5, or — -y,,*. 32. Given \/2 x — 1x = — 52. Ans. a; = 8, or \^-^. 33. Given /s/x -\- S -\- \^x -\- 8 = 5 \/x. Ans. x=l, or ^\. 34. Given \/x -\- i — a^x = a/x -\- ^ Ans. a: = ^, or — ^^, 35. Given J^^ = o^ + g)*- Ans. a. = |, or §, 36. Given a/x -{- \/a — a; =: \/b. . a ± 1/2 a 6 — i" Ans. a; = . 2 4 EQUATIONS IN THE QUADEATIO FORM. 275. The rules already given for the solution of qusJ- ratic equations will apply to any equation which can te made to assume the quadratic form. An equation takes the quadratic form when it is ex- pressed in three terms, and of the two terms which con- tain the unknown quantity, one has an eayxment twice as great as the other. The quadratic form, then, is ax^-\-baf = c, or x^" -{- p af ^ q, in which n may have any value whatever, positirs or negative, integra,l or fractional, x may also represent To what other equations may the rules for quadratics bs sxtended? What is the quadratic form? QUADKATIC EQUATIONS. 253 «ither the unknown quantity itself, or some expression containing the unknown quantity. 276i Higher Equations in the quadratic form usually reduce to pure equations of some higher degree than the first, after the completion of the square and extrac- tion of the square root. The solution must, therefore, be completed by the rule for pure equations. (Art. 263, Kule, Note 1.) 1. Given si^-\- 3 a? =810, to find the values of a;. This equation evidently has the quadratic form, since a* is the square of a?. As the coefficient of the second term is an odd number, we avoid fractions by using the second method of completing the square ; that is, we multiply by 4, and add the square of 3 to both members. Extract- ing the square root of (3), we obtain (4), a pure cubic equa- tion, which reduces to (7) by transposing, uniting, and di- viding. Extracting the cube root of both members, we ob- tain (8). to find the values of x. OPERATION. a;« -f 3 £C» = 810 4a;«-l-12a^ = 3240 4a;'+12a:» + 9 = 3249 2a^ + 3= ±5T 2a^ = — 3 ± 51 2 a^ = 54, or a^ = 21, or —30 as = 3, or v'- 60 30 VERIFICATION. T29 + 3 X ST = 810 900 + 3 (—30) = 810 2. Given t'-\-Ax-^ = 5, (1) (2) (3) (4) (5) (6) 0) (8) (1) (2) FIRST OPERATION. a^ + ^ = 5 (1) a;*-f 4 = 5a;^ (2) ^— 5a;^ = — 4 (3) -5a:" + V- = l W ^-^=±§ (5) 3? = 4, or 1 (6) ar= ±2, or ±1 (7). We first remove the de- nominator by multiplying both members of the equar tion by ^, and then transpose the terms, producing equa- tion (3), which is evidently in the quadratic form. Com- pleting the square by the first method, extracting the 22 254 ELEMENTAKY ALGEBRA. square root, transposing and uniting terms, we obtain the value of a? in equation (6). Extracting the square root again, we have the value of X in equation (7). SECOND OPERATION. As ^ and 4ar" are both a;2 _(- 4 a;~" ^^ 5 (1) positive squares, and the let- ^2 I 4 I 4 3.-2 __ 9 /'2') *s''s cancel each other when _. „ _i , „ ,„, those terms or their roots are ^ ~L^ -Z 2 — + 3 m ™'>l'^y'^8d together, we may I ± ^ V*^ complete the square by sup- a; T 3 a; = 2 (5) plying the middle term, which «' =F 3 a; -|- f ^ J (6) must be twice the product of a; T f = ± J C?) ^^ square roots of ^ and a;= ±f ± ^ (8) 4a;-», or 4. Extracting the a; = ± 2, or ± 1 (9) square root, we have (3). Multiplying by x to remove the negative exponent (Art. 153, Note 3), we find that the equation becomes an affected quadratic, instead of a pure quadratic. Solving equation (5), we obtain the same results as by the other process. 3. Given a;* — 9 a;^ + 20 = 0, to find the values of a;. Ans. a; = ± \/ 5, or ±2. 4. Given a;« — 35 a^ + 216 = 0, to find the values of x. Ans. a; = 3, or 2. 5. Given 5a:«— 90a;' — 2'70 = 945, to find the values of a;. Ans. a; = 3, or ^ — 9. 6. Given a;*" + 31 a;* = 32, to find the values of x. Ans. a; = 1, or — 2. Y. Given ^ — ia?" =: 10, to find the values of x. 1 Ans. a;= (2 ± v^M)"- 8. Given ^ + 1225 a;-^ = T4, to find the values of x. Ans. a; =: ± Y, or ± '5. 277 1 Kadical Equations sometimes take the quadratic form, and reduce to pure equations. Explain the first and second operations of Example a. QUADRATIC EQUATIONS. 255 NoTB. Some of the following equations may be changed to trne quad- ratics by removing the radicals, as has heretofore been done (Art. 274). It is intended, however, that they should be solved by the quadratic form, and not as true quadratics. 1. Given x-\-2s/x= 15, to find the values of a;. OPERATION. The exponent of the first term is 1, or §, and that of the second is \, for 2 y'a; ^ 2 ai ; hence the equation has the quadratic form. Completing the square by the first meth- od, and extracting the square root, we obtain (3) ; trans- posing and uniting, we have (4) ; and,squaring both mem- bers, we have (5) . In verifying these values, we find that 9 is limited to the positive square root, while 25 is limited to the negative square root, as those roots only will satisfy the equa- tion. It will be seen that (-)- 3)* and ( — 5)' are, then, the real roots of the equation, as we might infer from the origin of 9 and 25, equan tion (4). 2. Given 3 a:^ -f- a:^ = 3104 x*, to find the values of x, OPERATION. Dividing both members of J 1 the equation by «», we ob- 3a:« + a:Tf = 3104a:* (1) tain (2), which is in the 3 a;^ -f a:^ ; a; + 2 Va; = 15 (1) a; -f 2 Va: + 1 = 16 (2) Va: + 1 = ± 4= (3) ^/x = 3, or — 5 (4) a: = 9, or 25 (5) VERIFICATION. 9 + 2 X 3 = 15 (1) 25 + 2 (—5) = 15 (2) 3104 36 x^ + 12 X' ,l_ 3T248 36 a:^ + 12 a;^ + 1 = 3t249 6a;*+ 1 = ±193 6a;^=192, or —194 a;* = 32, or — -»/ x^= 2, or (—■?/-)* a; = 64, or (— V)* (2) (3) (4) (5) (6) 0) (8) (9) quadratic form, because the exponent f is twice as great as the exponent -f, and a;* is therefore the square of a*. Multiplying by 4 X 3, or 12, adding 1, the square of the coefficient of a*, extracting the square root, transposing, uniting, and dividing by 6, we obtain the value of a* Explain the operation of Example 1. Of Example 2. 256 ELEMENTARY ALGEBRA. in equation (7). Extracting the fifth root of both sides of the equa- tion, -we obtain (8) ; and raising both sides to the sixth power, we obtain the values of x in equation (9). As we extract the corresponding root to remove the numerator of the exponent of x, and raise to a corresponding power to remove its denominator,, the effect is the same as if we at once transfer the exponent of x to the second member hy inverting it. Note. It may be well to state in connection the three ways in which a quantity may be transferred from one member of an equation to the other, corresponding with the three changes of addition or subtraction (Art. 38, Ax. 1, 2), multiplication or division (Ax. 3, 4), and involution or evolution (Ax. 8). 1. Any term may be transposed from one member of an equation to the other by changing its sign, that is, the sign of its coefficient. 2. A factor of either member of an equation may be transposed to the other member by changing the sign of its exponent. 3. An exponent of either member of an equation may be transferred to the other member by inverting it. It will be se£n that the factor and exponent must belong to the whok member, not to a single term only, nor, in the case of the exponent, to a single &ctor only. 3. Given a; -f- 4= \/ a: = 21, to find the values of x. Ans. X = 9, or 49. 4. Given x~^ -j- x'^ = 6, to find the values of x. Ans. x=: ^, or ^. 5. Given x^-\-10x^ = HI, to find the values of a;. Ans. x = 21, or (—19)^ 6. Given 5 ^^ -j- ^ — 22, to find the values of y. Ans. y=16, or (— "g)- 1. Given 4^ x -{-*(/ 3^ = 6, to find the values of k. Ans. x= 32, or —243. How may an exponent be transferred from one member of an equation to the other? QUADRATIC EQUATIONS. 257 1 2 8. Given a;» — ac» -\-2 = 0, to find the values of x. Ans. a; = 2", or (—1)". n 9. Given x" -\- px^ z= q, to find the values of x. Ans. =o=:{—ip±^q-^ip')''. 10. Given y/ x^ — 3 a; = 40 a; *, to find the values of x. Ans. a: =i 4, or ( — 5)^. 278i Polynomials may sometimes take the place of the unknown quantity, as the basis of the quadratic form. These polynomials may have the exponents 2 and 1, or they may have higher or fractional exponents, bearing the same ratio. Note. Most of the equations which belong' to this class must also be considered either higher or radical e(]aa,tions. The first Note found in the last Article will apply to the latter. 1. Given x — \/a;-|- 5 = 1, to find the values of x. OPERATION. a: — Va: -f 5 = 1 (1) a: + 5 — v'S^^^ = 6 (2) (a; + 5) -(a; + 5)^ = 6 (3) (a,+ 5)-(a;+5)* + i=:^^ (4) (aj+5)*-J=±f (5) (a; + 5)* = 3, or —2 (6) a: + 5 = 9, or 4 (Y) a; = 4, or — 1 (8) VEEIFICATION. 4-3 = 1 (1) — 1 — (— 2) = 1 (2) We first add 5 to both members of the equation, in order that we may make the quantity without the radical the same as that within. The equatJMi then assumes the quadratic form, (a; -|- 5) 22* 258 ELEMENTARY ALGEBRA* being its basis, instead of x. The coefficient of (x -\- 5)* is 1, jmd we therefore add (^)'' to both members to complete the square. Extracting the square root, transposing, and uniting, we find the value of (x -\- 5)* in equation (6). Squaring, and transposing 5, we have the value of x. In verifying these values of x, we are obliged to take the positive root of a; -|- 5 when x = 4, but the negative root when x = — 1, as these only will satisfy the equation. 2. Given (x — 5)' — 3 (at — 5)' = 40, to find the values of X. OPERATION. (x _ 5)» — 3 (a; — 5)^ = 40 (1) y" =(x — 5)», and y = (x — 5)^ (2) f — 3y = A0 (3) y=8, or— 5 (4) (« — 5)^=8, or —5 . (5) a — 5=4, or (—5)* (6) a;=9, or5 + (— 5)* (T) a; =9, or 5 4- -4^25" (8) This equation is of the quadratic form, because the exponent 3 is twice as great as |. We may carry through the solution without any change of letters, as in the last example ; or we may substitute y* for (x — 5)', and ff for (a; — 5) t, when equation (1) becomes (3). This last equation, solved by either of the rules for quadratics, gives (4), or, replacing the value of y, (5), which readily reduces to (7) or (8). ■ 3. Given (x — 1 )* — x = — J, to find the values of x. Ans. X = 2^, or J. Note. The above equation, by adding 1 to both members, assumes the form {x — lf — {x — 1) = |, and may thus bo solved. It will be seen, however, that the given equation can be readily reduced to a com- mon quadratic by expanding (x — 1)^ and uniting terms, as in the Ex- amples of Art. 274. ' Explain the first operation. The second. QUADRATIC EQUATIONS. 259 4. Given (/ — 4 j^)^ — 6 (jr'' — 4 y) + 5 = 0, to find_the values of y. Ans. y = 5, or — 1, or 2 ± v'^. 5. Given a:* — 2 a: + 6 Va^^ — 2a; + 5 = 11, to find the values of x. Ans. a; = 1, or 1 ± 2\/^- 6. Given v'a:4-2 + 2,^ar + 2 = 8, to find the values of X. Ans. X = 14, or 254. 1. Given (a:^ -fT)^ -|- 2 (a:^ + T)* = 80, to find the real values of x. Ans. a; = ±5. 8. Given (x — 3)* + 4 (a; — 3)= = IIY, to find the v al- ues of X. Ans. a; = 6, or 0, or 3 ± V — 13. 9. Given V« + 12 +^« + 12 = 6, to find the values of «. Ans. « = 4, or 69. SIMULTANEOUS EQUATIONS INVOLVING QUAD- KATICS. 279. The Degree of an equation containing more than one unknown quantity is indicated by the highest sum of the exponents of the unknown quantities contained in any one of its terms. (Art. 145.) Thus, &xy-\-2x-{-3y = i3 is of the second degree, and asi^y-\-l^xi/ = c* is of the third degree. 280i A Homogeneous Equation is one whose terms, ex- cept those which contain only known quantities, are ho- mogeneous with respect to the unknown quantities. (Art. 30r) Thus, the equations 5a:y4-2a;2-j-3/=66, and ax' y-\-bx f = e*, are homogeneous, for in each equation the sum of the ex- ponents of the unknown quantities is the same in every term which contains an unknown quantity. How is the degree of an equation containing more than one unknown quantity indicated 1 Define a Homogeneous Equation. 260 ELEMENTARY ALGEBRA. 281. A Symmetrical Equation is one in which the un- known quantities are similarly involved. Thus, the equar tions are symmetrical ; for in each of the equations x and y are affected by the same coefficients and exponents, and per- form the same office. Note. Many equations are both homogeneons and symmetrical; as 3 a:2 _|- 3^! = 39^ or !x^ + xy + y^^ S^. 282. In general, two quadratic equations containing two unknown quantities will produce an equation of the fourth degree after elimination. The rules for quadratics are not, therefore, sufficient to solve a?^ simultaneous equations of the second, degree. Most of those which are capable of' solution by means of rules already given may be included in three cases : — I. When one equation is of the first degree and the other of the second. *■ II. When both equations are homogeneous and of the second degree. III. When the equations are symmetrical. CASE I. . 283. When one equation is of the first degree and the other of the second. Equations belonging to this class can always be solved. It is usually most convenient to find an expression for Define » Symmetrical Equation. Are the rules for quadratics sufB- cient to solve all simultaneous equations of the second degree? What ones can be solved ? • How are equations belonging to Case 1 nsnally solved t QUADBATIC EQUATIONS. 261 the value of one of the unknown quantities in the simple equation, and eliminate by substitution. Examples have already been given in which a pure equation is thus ob- tained. (Art. 265.) Examples. to find the values of x and y. OPERATION. 5(a? — x)-\-3xy—2f = 10 (1) 2x + y= 1 (2) Prom (2), " "^ y=T — 2 x (3) Subs, in (1), 5(x'—x)-\-3x(1 — 2x)—2(1—2xy=10(4:) Expanding, 5a^— 5a: -|-21a;— 63:^—984-56 a;— 8a?=10 (5) Uniting terms, — 9 a:^ + T2 a; = 108 (6) Dividing by— 9, 3:^—8 a: = —12 (Y) Completing square, a? — 8a;+16^:4 (8) Evolving, x — i=±2 (9) Whence, a;_6 (,j.2 Substituting in (3), y = 7 — 12, or 1 — 4 Whence, . y= — 5, or 3 VEEIPICATION. First set of ( 150 — 90 — 50 = 10 values, I 12 — 5 = T (1) (2) Second set oft 10 + 18 — 18=10 values, I 4 + 3 = T fl) (2) It will te observed that the values of x and y must be taken in the same order; that is, wheii x ^ 6, y = — 5; and when Explain the operation. 262 ELEMENTARY ALGEBRA. 2. Given a; + y=7, and a^4-2/ = 34, to find the values of x and y. Ans. X = 4, or y. V = 3, or -. «. a; — » , < x-i- Sy ,./.■■ 3. Given x —^^ = 4, and y V-^ = 1, to find the values of x and y. . ( a: = 2, or 5. ■ -ly =6, or 3. 4. Given a:-l-4y:=23, and as^ -[- 3 a; y = 54, to find the values of x and ^. v I a; = 3, 6r — 'r2. 4 y = 5, or s^. 5. Given 49 a;? = 36 f, and a: (2 a; + J) -|- 3 a; y — y (6y~l~^) 4" 128 = 0, to find the values of a; and y. Ans. i^ = ^' °'^-^- I y = 7, or — ^, Note. It is evident that one of the equations can be readily reduced to a simple one. CASE n. 284. When both equations are homogeneous and of the second degree. Equations belonging to this class can always be solved. It is usually most convenient to substitute for one of the unknown jquantities the product of the other by a third unknown quantity. EXAUFLES. 1. Given 2y'^ — 4.xy -\-Za?=Vl, and f — x^ = U, to find the values of x and y. r~ ' 11 — ■ I . .1 ■- ■ ■ ■ ' r- i .1 ' ■'■'■■■■■■'■ III ■ ■ --. J I II . ■ ...ii - V ' How are equations belonging to Case II. usually solved ? QUADRATIC EQUATIONS. 263 OPERATION. Sy — 4a;3^4.3a;«=lt (1) i^ — a?= 16 (2) Let y = vx (3) Subs.in(l), 2v''a:'' — i:va?-\-%x^ = Vl (4) Subs, in (2), v^x^ — a?=:lQ (5) From (5), ^ = ^ (^) ^"'^*="' 2^-I„+3 = i;^ (8) Cleaning of fractions, 1*1 v^ — 1T==32«* — 64^4-48 (9) Tr. and uniting, — 15 1)^ -[^ 64 « = 65 (10) Dividing by —15, ir" — f | r = — f | (11) Whence, " r = 3^, or ^ 1 fi 1 fl Substituting in (7), a:« = ^ _ j^ , or ^-— ^ Reducing, a;^ ^ fy, or 9 Evolving, «= ±f, or ±3 Substituting in (3), 3'=±^X-^, or±3Xf Reducing, y= ±.^, or ±5 2. Given y^ — x^ = Z, and y" — 2a:y + 2a;» = 2, to find the values .of x and y. ^jjg ( x= ±1, or ±1^/5. b=±2, or ±|V5. 3. Given ar" + 3a!y — y* = 2t, and 3*'' + 2a:y = 63, to find the values of x and y. Ans \^=±^' 0^^ ±1*^23 |y=±6,. or ±AV23. Note. If either x or y be directly eliminated from sach equations as the above, the result will be a biquadratic equation in the quadratic form. (Art. 275.) Two homogeneous quadratic equations, containing two un- known quantities, may therefore be solved in that manner, without the aid of a third unknown quantity. Explain the operation. What method of solving homogeneous equa- tions is mentioned in the Note? 264 ELEMENTARY ALGEBRA. It will be seen that the square root must be taken twice, whatever the method used, and that each unknown quantity must have four values, two of which differ only in their signs. (Art. 263.) CASE m. 285i When the equations are symmetrical. It is often convenient to combine and simplify quad- ratic and higher equations belonging to this class, before attempting to eliminate either unknown quantity. The proper application of the various expedients employed must be learned by experience, as the details vary with each new class of examples. , The student must be thor- oughly conversant with the forms of the powers of bino- mials, the principles of factoring, and especially with the relations existing between the sum, difference, and prod- uct of two quantities. Examples. 1. Given a; -|- y = T, and a; y = 12, to find the values of X and y. OPERATION. ■x^y= 1 (1) xy=12 (2) Squaring (1), x'-\-2xy-^f = i9 (3) Multiplying (2) by 4, 4:xy =48 . (4) Subtracting (4) from (3) , x'-2xy-^f= 1 (5) Evolving, x — y=±l (6) Equation (1), x + y= 1 Adding (6) and (1), 2a; = 8,or 6 Whence, X = 4, or 3 Subtracting (6) from (1) 2y = 6,or8 Whence, y=3,or4 What is said of the number of roots of homogeneous equations ? What is said of the methods of solving equations Under Case III. 1 Explain the operation. QUADRATIC EQUATIONS. 265 Many complicated equations reduce to the sum and product of the two unknown quantities. After reaching that point, the work may conform to the above. It is evident, however, that such eqna. tions as the above belong under Case I. as well as Case III., and may therefore.be solved by eliminating one of the unknown quan- tities from the original equations by substitution. It must not be inferred that x and y are equal to each other} for when a; = 4, y = 3, and when a; = 3, j, = 4. Whenever two simultaneous equations are symmetrical in their dgns, as weU as in other rpspects, it is evident that the letters may be exchanged without affecting the equation; hence the values of the letters must be interchangeable, and when the two values of one letter are found, the same values may be assigned to the other letter, the order being reversed. 2. Given :i?-\-f=.2h, and xy= 12, to find the val- ues of X and y. OPEKATION. 3!= -f j^ = 25 (1) xyi=\2 (2) Multiplying (2) by 2, 2 a; y = 24 (3) Adding (1) and (3), a^-f 2a;y-|-/ = 49 (4) Subtracting (3) from (1), x^ — 1xy\-f^ 1 (5) Extracting square root of (4), a; -|-y = ±T (6) Extracting square root of (5), x — y = ± 1 (f) Adding (6) and (7), ' 2a;= ±8, or ±6 Subtracting (1) from (6), 2 y = ±6, or ±8 Whence, a: = ±4, or ±3 Also, y=±3, or±4 It is evident that the above Example might be classed under Case II. £is well as under Case lU. ; but the method here adopted gives a simpler solution. 3. Given oj^ — ^'^ = 19> ^i^d ^ y — xy' ^=6, to find the values of x and y. By what other method might the equations be solved? What is said of the relative values of x and y in such equations ? Explain the second operation. By what other method might Example 2 be solved! 23 266 ELEMENTARY ALGEBEA. OPERATION. a^ — / = 19 (1) o,?y — xf= 6 (2) Multiplying (2) by 3, 3a^y— 3a:/=18 (3) Subtr. (3) from (1), ^ — Zx'y-^-Zxf—f^ 1 W Extr. cube root of (4), x — y= 1 (5) Dividing (2) by (5), xy— 6 (6) Squaring (5), a?—2xy + f= 1 (^) Multiplying (6) by 4, 4a;y =24 (8) Adding (7) and (8), a:2_|.2a:y + / = 25 (9) Extr. square root of (9), x-\-y — ±5 (10) Equation (5), x — y=z 1 Adding (5) and (10), 2cB= 6, or- — 4 Whence, al = 3,or • — 2 Subtracting (5) from (10), 2y = 4,or- — 6 Whence, y = 2,or- -8 As the original equations are not symmetrical in their signs, ihe ■values of x and y are not interchangeable. 4. Given a; -f- y = 4, and x~^ -\- y~^ = 1, to find the vaU ues of X and y. ^ ( a; = 2. Note. Bemove negative exponents, apply Axiom 7, and solve like Example 1. 5. Given a^-j-^ =: 65, and a;-|-y = 5, to find the val- ues of X and y. . ( a; = 4, or 1. ' I y = 1, or 4. Kot;b. Divide one equation by the other (Art. 87), and square the second. a? 11^ 6. Given — f- - z= 9, and a; + y = 6, to find the val- y ^ ues of X and y. . ( x = 4, or 2. = 2, or 4. Explain the third operation. Why are not the values of the two un known quantities interehangeahle ? QUADRATIC KQUATIONS. 267 r. Given ot^-^x" f-j-f = QSl, and a:^+ a; y -f y" = 49, to find the values of x and y. Ans. |«'=±5,or±3. (y = ±3, or ±6, Note. Divide one equation by the other. 8. Given a: + y=61, a.nA x^ -{- y^ z= 11, to find the values of x and y. . ( xz= 36, or 25. ' ■ I y = 26, or 36.. Substitute » for a', and a for y*, and the equations will become w"'-!" a* = fit and v -\- z =^ 11; from which the values of v and z, and oonseqaently of x Mid g., are readily found. Or,. Subtract x-{^y= 61 from, the, square of x^ -\- y^= 11, and square th^ result. 286. Sometimes one of the given equations, or some combination of the two given equations, takes the- quai^ ratio form, an expression containing both unknown qnani" tiles feeing^ the basis. (Arts. 216-218.) 1. Given a^-j-^-["^y — ^"^ — 2y=9, and x y = 6, to find the values of x and y. . j ar = 3, or 2. {x = 3, or 3. After adding the second equation to the first, their sum may be put in the quadratic form, thus: (x-{-yy—2(x + y} = 15. Completing the square, evolving, and reducing, we obtain x + y = h or — 3 ; but as the latter value, produces imaginary results, we use only the former. 2. Given 4a;y = 96 — n^i^, and x-\-i/ = 6, to find the values of x and y. ^^g t a; = 4, or 2, or 3 ±\/21. = 2, or 4, or 3 TV'Sl. How may an equation containing two unknown quantities take the. quadratic form 1 268 ELEMENTARY ALGEBRA. Note. The Bigns ± and q:, if used independently, would have the same signification; but when taken in connection, one is the reverse of the other. When x takes the upper sign, or +, y must also take the upper sign, or — ; and when x takes the lower sign, or — , g must also take the lower sign, or -{- ; that is, x and y must always take opposite signs. In the course of the operation, the sign ± is changed to :p whenever + would be changed to — . 3. Given -j -| = -;r. and x — y ^ 2, to find the val- ues of X and y. Ans. x=5,ov -. y = 3,or--- 287. Two equations, neither of which is strictly sym- metrical in itself, may sometimes produce a symmetrical equation when properly combined. ■ Two equations which are not symmetrical in respect to the unknown quantities themselves, may be symmetrical in respect to some multiple or power of those unknown quantities ; that is, the same multiple or power is the basis of the forms found in both equations. Sometimes it is convenient to obtain one simple equa- tion by means of the expedients used in Case III., and then complete the solution as in Case I. 1. Given a^-\-x^ = 60, and ^-j-a;y=84, to find the ralues of x and y. . I x= ±5. Add the two equations, and the result is symmetrical. Extract Ae square root of the sum, and divide each equation by the result. 2. Given x' + 9f=z52, and a; + 3y= 10, to find the values of x and y. . ( a; = 6, or 4. ''^' |y = ior2. These equations are symmetrical in respect to x and Sy. By substituting z for 3 y, each equation will become strictly symmetri- cal, and the values of x and z will be interchangeable. How may equations not strictly symmetrical be brought under Case III. ' QUADRATIC EQUATIONS. 269 3. Given a:* + y* = >j, and a:*y = 144, to find the_yal- ues of X and ^. ^^g | a; = ±8, or ±3 V 3. 1 y = 9, or 16, Substituting v for a:% and z for 2^^, the equations become » + 2 = 7. and ««!!?= 144, from which the values of » and z are readily obtained. Keplacing a;' and y^, the values of a; and y are found. After going through the operation as above, the student may take precisely the same steps, and find the values of x^ and y^, without the use of v and z. Note, x = (J-^^^ and y = (L5j^y may also be obtained from the equations last given. 4. Given a^-|-3a;y = 54, and xi/-\-iy^ =: 115; to find the values of a: and y. ' {x= ±8, or ±36 (y = ±5, or Ty' Add the two equations together, and the result is a perfect square. 288t The foUpwing equations are to be solved by either of the itethods already explained. As has already been shown, many of those which come under Case III. may also be classed under one of the first two Casesi Several solutions of the same set of equations are often possible, and the student should therefore seek to obtain the best. Examples. 1. Given i-=-^^ = 2M. \x -\-y = 6 J 2. Given P + ^ = :!. Ans. ( a:y= 6 J 23* -. I;: = 5. = 1. a ± v'a" - -46 2 -46 — 2 270 ELEMENTARY ALGEUEA. X -\-y =0! ^ I ^■ 3. Gwen ]" T ^ = " } . Ans. J "^ ~ J * [y = S 4. Given i"_+^_, = ^i. Abs. jf=;- 5. Given f^"; + 3^; = *|. Ans. |^=^' ""^ 3. 6. Given |«''-2'^ = 8]. ^ns. 1^=^, or 0. «— y=2) (3^=0, or— 2. 7. Given j*^+^' = 82 1 ^^^ j .= ±3, JToms. There «re also .the 'Same jiumber -of imaginary roots, = ±3, or±l. or ±3. 8. Given i ="-11=8 (^^->^y)L ( Vxy = 15 j ,Ana. |»^=25, or 9. ( y = 9, or 25. 9. -Given j \ + ^^ = '^^ | ^^^ C ^ = 64, or 8. ix^ -\-y«z= %) J y = 8, or 64, x-\-y ix — y::13:5 y* + a: = 25 Ans. 1=^=9, or -14tV. (y = 4, or — 6J. 11. Given I a: + 4y=14) 4ar — 2y + y^=ll I Ans. ^*=2, or -46. 10. Given y= 3, or 15. 12. Given J ^ »; + 3y = n (a!» — 3a:y + 3y2 = ? (• Ans. }*=1. or4. y = 2, or 1. QUADRATIC EQUATIONS. 271 13. Given \ =^-^y'= 9) . ^ {x=±b. 14. Given \ ^' + «=}/ + ^f=U \^x' + 2xy-\-f^>l% Ans. |»' = ±3, or ^8. 15. Given \ ^^+23^ = 9) \xy-\-2a?=z4.\ Ans. |^=±1' o'^ ±^V2. |y=±2, or q:|V2. 16. Given \^ "^ f + '^ -^y=l^ \ \ 2xy=12\' .. Ans. \^—^' OJ" 2, or — 3 ± V3. ( y = 2, or 3, or — 3 :f ^3. THE0EY OF QUADRATIC EQUATIONS. 289. Every complete quadratic equation may be re- duced to the form a?-\-px = g, whose roots are — | + \/? + — » and ~ I — v/s" + i- i-^^- 269, Ex. 5. ) It is evident that the sum of these roots is — p, and their product is ^ — (q^^\z= — q. Hence, When the coefficient of the first term is unity, 1. The algebraic sum of the two roots of a quadratic equa- tion is equal to the coefficient of the second term, with its sign changed. 2. The product of the two roots is equal to the second mem- her, with its sign changed. What relation exists between the roots of a quadratic eqaation and the coefficient of the second term? Between the roots and the second member t 272 ELKMENTAEY ALGEBRA. 290i Using r and r' for the two roots of the quadratic equation 3? -^ fx — §'=0, we have a; = r, and x = r', or, X — r :^ 0, and x — r' = 0. Multiplying the last two expressions together, (a; — r) (x — r') = 0, or, s? — (r -\-r') x-\-r r' = 0. But, by Art. 289, r-\-r' = — p, and?-r' = — q; hence, a?-\-px — q z= {x — r) [x — ?•') = 0. That is. If all the terms of a quadratic equation he transposed to the jirst member, it may he resolved into the two hinomial factors formed hy subtracting each of the two roots of Ike equation from the unknown quantity. Examples. 1. Eesolve x' — 4a;-f-3 = into binomial factors. Ans. (x ~3) {x—l)= 0. A solution of the equation gives 3 and 1 as the roots. 2. Eesolve a^ — - — -s- := into binomial factors. Ans. (x — §)(a; + J)=0. 3. Eesolve a^ — Ta;-j-12 = into binomial factors. 4. Eesolve x^-\-6x-\-8 = into binomial factors. 291. The principle established in the last Article fur- nishes a method of resolving into factors any trinomial which contains the first and second powers of a letter or quantity. (Art. 90,) Such a trinomial is called a quad- ratic expression. Examples. 1. Eesolve xF — 5 a; + 6 into binomial factors. Ans. (a;— 3)(a: — 2.) How may a quadratic equation be factored? What is a quadratic ex- pression ? ' QtTADEATIC EQUATIONS. 273 Although this trinomial may have any value whatever, yet we find its factors by supposing it equal to 0, and obtaining the roots of the equation thus produced. The faciors of the trinomial will re- main the same, whatever the values of x, and of the trinomial. 2. Kesolve a;^ _ -1 _ ^ into binomial factors. Ans. (a. — |)(a; + |). 3. Eesolve x^-f 3 a; — 28 into binomial factors. 4 Eesolve a=+18« + 80 into binomial factors. Note. The' factors will be irrational or imaginary whenever the roots of the assumed equation are of that nature. FOKMATION 01" EQUATIONS. 292. The principle established in Art. 290 also fur» nishes a method of forming a quadratic equation which shall have any two given roots. RULE. Subtract each of the given roots from the unknown quantity, 'and place the product of the two binomial factors equal to 0. Examples. 1. What is the equation whose roots are 1 and — 2 ? OPEEATION. (a; — l)(ar-f2)=a^ + a; — 2 = Or, sc'-\-x = 2 2. What is the equation whose roots are 4 and 5 ? Ans. a^— 9a; = — 20. 3. What is the equation whose roots are 6 and 7 ? How may a quadratic expression be factored ? - When will the factors be irrational or imaginary ? Repeat the Rule for forming an equation . having any two given roots. Explain the operation. : 274 ELEMENTAEY ALGEBRA. 4. Form an equation whose roots shall Ire — 1 and — 2. Ans. a^ + 3x = —2. 5. Form an equation whose roots shall be 20 and — 30. Ans. 3^=+ 10 a: =600. 6. Form an equaition whose roots shall be IJ- and — 2f. Ans. 10 a;2 + 13 a; = 42. 'I. Required the equation whose roots are a and b. Ans. -a^ — (a -\-b) x^^ — ab. 8. Required the equation whose ^roots are w -j- « and •m — n. Ans. a^ — 2 m a: = n^ — »»?. Note. Most of the foregoing principles might be extended, with suit- able modifications, to equations of a higher degree than the second. In general, the number of roots, and consequently of binomial factors, will correspond with the degree Of the equation. DISCUSSION OF THE GENERAL EQUATION. 293. If y = 0, in the general equation sd' -{- p x=:q, the roots will become — ■^ ± ^', or — p and 0, and the equation may be considered a simple one. (Art. 269, Ex. 19, Note.) If ^ = 0, the term p x disa,ppears, the roots become 4- ■>/ S" and — \/ q, and the equation is found to be . a pure quadratic. "Hence the principles 'Stated in Arts. 289, 290, and 292 may be applied to both pure and affected quadratic equations. 294. The values of p and q in the general equation may be either positive or negative. If those letters be consid- ered essentially positive, and the signs be expressed, we shall have four forms. si?-\-px = q, a: = — |± y/y+l; .(1) If 9' = 0, what will he the effect on the general equation? What is (Said of pure quadratic eq^uations 1 What are the four foitns ? QUADRATIC EQUATIONS. 275 ^-px==q, a:=f±y/y+f; (2) x^^px = -q, a. = __£±^_^4.j'; (3) :^-px = -q, a!=|±y/_<^+J. (4), 295. As q is positive in the first ana second forms, and negative in the third and fourth, the roots must have different signs in the first two forms, and the same sign in the last two. (Art. 289.) Moreover, since - must be numerically greater than ^ — qJ^^, the signs of the roots in the last two forms must be controlled by the sign of 5. Hence, lb 1. In the first and second forms, one root is positive and the other negaJtive. 2. In the third form, both roots are negative. 3. £1 the fourth form, both roots are positive. 296. It is evident that the quantities under the radical sign of the roots in the first two forms can never be neg- ative, and that they can be negative in the last two forms only when j- is numerically less than q. Hence; 1. In the first and second forms, both roots are always real. 2. In the third and fourth formiS, both roots are imaginary when the square of half the coefficient of x is numerically less than the second member; otherwise they are real. 297. It is evident' that the radical portion of the roots can never become in the first two forms ; but if — = g'. What will be the signs of the roots in each form? When will the «roots be real, and when imaginary i 216 ELEMENTAKY ALGEBRA. the radical portion will become- in the last two forms, and both roots will be — ^, or ^. Hence, 1. In the first and second forms, the two roots are always numerinally unequal. 2. In the third and fourth forms, the two roots are equal when the square of half the coefficient of x is numerically equal to the second member; otherwise they are unequal. Note. Examples to illustrate the foregoing principles, as well as a statement of some of the principles themselves, may be found in Art. 269. PROBLEMS LEADING TO AFFECTED QUADRATIC EQUATIONS. 298 1 The general principles involved in stating prob- lems leading to quadratic equations are the same as those which have already been given in connection with simple equations. The principles established in the Discussion of Prob- lems (Arts. 179-184) are also equally applicable here; but we must note certain peculiarities, arising from the facts that every quadratic -equation has two roots, and that those roots are sometimes imaginary. 1. The positive root of the equation is usually the true answer to the given problem. If there are two positive roots, there may be two answers to the problem, either of which conforms to the given conditions. 2. A negative result is sometimes the answer to an- other analogous problem, formed by attributing to the unknown quantity a quality directly opposite to that which has been attributed to it. As the algebraic mode When will the roots be unequal, and when equal? What is said of the statement of problems leading to quadratic equations ? What of the interpretation of results ? QUADRATIC EQUATIONS. 277 of expression is more general than ordinary language, the same equation often represents two analogous prob- lems in this manner. 3. An imaginary result shows that the problem is an impossible one. 299. Some of the following problems require the use of but a single unknown quantity, others Tequire the use of two, and others still may be solved by either method. Some problems may also lead to either pure or affected quadratic equations, according to the notation assumed. PROBLEMS. 1. A man buys a watch, which he sells again for $24, and jfinds that he loses as much per cent as the watch cost ; required the price of the watch. SOLUTION. Let X = the price in dollars. Then x =: the loss per cent, and 100 X a; = Jq^ = his whole loss. Therefore, -^ = a; — 24 Or, a:» — 100 a; = — 2400 Completing square, a^— 100 a; + 2500 = 100 Whence, a; — 50 = ± 10 And, X = 60, or 40 The price was either $ 60 or $ 40, for each of these values satisfies all the conditions of the problem. 2. What number is that which exceeds the square of its fourth part by 3 ? Ans. 12, or 4. What is said of the namber of unknown quantities? Explain the solution of Problem 1. 24 278 ELEMENTARY ALGEBRA. 3. Divide the number 10 into two parts whose product shall be 24. soLrraioN. Let X = one part, and 10 — X = the other part. Then, x (10 — a;) = 24 Or, or"— 10 a: = —24 Completing the square, a;" — 10 aj -(- 25 = 1 Whence, x — 5 = ±1 And a; = 4, or 6 Also, 10 — x^ 6, or 4 One part must be 4, and the other 6, and there is only one mode of dividing 10 so that the product of the two parts shall be 24; but it is immaterial which part is 4, and which is 6. The same results may be obtained by tbe use of two unknown quantitifs, producing the symmetrical equations X -\- y = 10, and xy=2i. 4. A person bought a certain number of sheep for $ 80 ; if he had bought 4 more for the same sum, each sheep would have cost $ 1 less ; required the number of sheep, and the price of each. SOLUTION. Let X == the number of shee,p. Then 80 ^, ... I. — = the price of each. and 80 — r—r = the price of each if he a; + 4 ^ had bought 4 more. Tbprpfnrp 80 _ 80 X xicx cxvX ^f X-\-A X Or, 80a: = 80 (a; + 4) — ar" — 4a; Hence, a:'' + 4a: = 320 Completing square, x^ + 4a: + 4 = 324 Explain the solution of Problem 3. Problem 4. QUADEATIC EQUATIONS. 27.9 Whence, x -\- 2 = ±18 And x= 16, or —20 Also, — = 5, or —4 The negative results are not admissible, as answers to the above problem in its present form. (Art. 298.) The nvunber of sheep was therefore 16, and the price of each $ 5. If, in the above problem, " bought " be changed to sold, " 4 more " to 4 fewer, and " $ 1 less " to S 1 ■more, 20 and 4 will be the true answers. It will be well for the pupil to interpret the negative results in the problems which follow, whenever an obvious interpretation occurs. 5. Having sold a piece of goods for $56, I gained as much per cent as the whole cost me. How much did it cost? Ans. $40. 6. A person bought a lot of chickens for 96 cents, which he sold again at 13J cents a piece, and gained as much as one chicken cost him. What number did he buy? Ans. 8. 1. A printer, reckoning the cost of printing a book at so much per page, made the whole book come to $ 80. It turned out^ however, that the book contained 5 pages more than he reckoned, and an abatement also was made of 50 cents per^page. He received $6T.50. How many- pages did the book contain ? Ans. 45 pages. 8. A company at a tavern had $ 8..'r5 to ,pay ; but, be- fore the bill was paid, two of them went away, when those who remained had, in consequence, 50 cents more to pay. How many persons were in the company at first? Ans. 1. 9. .A sum of $1000 has to be divided equally among a number of persons ; but two new claimants appearing, it is found that each person will receive $25 less than he expected. Eequired the original number of persons. Ans. .8. Interpret the negative results. 280 ELEMENTARY ALGEBRA. 10. The plate of a looking-glass ia 18 inches by 12, and it is to be surrounded by a plain frame of uniform width, having a surface equal to that of the glass. Re- quired the width of the frame. Ans. 3 inches. 11. Twenty persons contribute to send a donation of $ 48 to a benevolent society, one half of the whole being furnished in equal portions by the women, and the other half by the men ; but each man gives a dollar more than each woman. How many are there of each sex, and what does each person contribute? SOLUTION. Let X = number of women, and y = contribution of each in dollars. Also, 20 — X = number of men, and y -J- 1 = contribution of each in dollars. Then «y = whole contrib. by the women, and (20 — jc) (y -f- 1) = whole contrib. by the men. Therefore, a;^ = 24 (1) Also, (20 — a;) (y + 1) = 24 (2) rrom(l), y=^ (3) From (2), 20 y + 20 — a; y — x = 24 (4) Subst. (3) in (4), 20 (^\ + 20 — 24 — a; = 24 (5) Or, l£2_a; = 28 (6) Clearing of fractions, a:^ -f- 28 a; = 480 (T) Completing the square, ' a:* + 28 a; -j- 196 = 616 (8) TEvolving, a; -4-14 =±26 (9) Whence, ' , ' a; =12, or —40 And y= 2,or — | Also, 20 — ?;= 8, or 60 And 2^ + 1= 3,orf Explain the solution of Problem II. QUADRATIC EQUATIONS. 281 The negative values of x and y, although furnishing a solution of the equations, evidently do not belong to the problem, and no obvious interpretation occurs for them; consequently, the positive values of x and y are the only admissible results. Therefore, the number of women is 12, contributing 2 dollars each, and the num- ber of men is 8, contributing 3 doUars each. 12. It is required to divide the number 40 into two such parts, that the sum of their squares shall be 818. Ans. 23 and It. 13. Divide the number 60 into two such parts, that their product shall be to the sum of their squares in the ratio of 2 to 5. Ans. 20 and 40. The last two Problems, as well as some others, lead to pure quadratic equations, when we let x—y and x + y represent the numbers. 14. The fore wheel of a carriage makes 6 revolutions more than the hind wheel, in going 120 yards ; but it is found that, if the circumference of each wheel be in- creased one yard, it will make only 4 revolutions more than the hind wheel, in the same distance; required the circumference of each wheel. SOLUTION. Let and Then X = circumference of hind wheel in yards y = circumference of fore wheel in yards. 120 — - = numljer of revolutions of hind wheel. X and 120 — = number of revolutions of fore wheel. Therefore, , ,, T, , , 120 120 by the Problem, — = 6 ^ y (1) Or, xy = 2Qx — 2Qy (2) AIro by th" P-'-W^.r, ^2** 120 ■ (3) ^ , ^+1— y_,_l - Explain the solution of Problem 14. 24* 283 ELEMENTARY ALGEBRA. Therefore, 30 (y + 1) = (a; -f 1 ) ■ (29 — y) (4) Or, 30y-[-30 = 29a; + 29 — a;y— y (5) Transposing and uniting, a; y = 29 a; — Sly — 1 (6) From (2) and (6), 29 a: — 31 y — 1 = 20 a; — 20 y (\) Or, 9a;=lly+l (8) Therefore, x = ^~^ <9) Substituting (9) in (2), ii£+i? = 220^+!? _ gOy (lo) Eeducing, 11 y^ + y = 40y + 20 (U) Or, 11/ — 39y==20 (12) Whence, y = 4,or — ^ And a;=5,or — f The negative values of x and y being inadmissible, the circum- ference of the hind wheel is 5 yards, and that of the fore wheel i^ 4 yards. If we take ^j for the fore wheel, and ^ for the hind wheel, the hind wheel must make 6 j-evolutions more than the fore wheel; and if "each circumference .be made equsil to the difference between itself and unity, the fore wheel will make 4 revolutions more than the hind -wheeL 15. A merchant buys two bailes tff clofe, -each contain- ing 80 yards, for $ 60. By selling the first at a gain of as much per cent as the second cost him per yard, in cents, and the second at a loss of as much per cent, he finds he has made a pro^t of 1 5 on the whole. Eequired the cost of each bale per yard. Ans. First, 50 cents ; second, 25 cents : or, first, 62J cts. ; second, 12J cents. 16. There are two numbers whose sum multiplied by the greater gives 144, and whose difierence multiplied by the less gives 14 ; what are the numbers ? Ans. .9 and 1. IT. A merchant bought as many bushels &f corn as cost him $ 60, and, after reserving for his own use 13 QUADRATIC EQUATIONS. 283 bushels, sold the remainder for $ 54, and gained 10 cents a bushel ; how many bushels did he buy ? Ans. 75 bushels. 18. The sum of the digits of a certain number is 15 ; and if 31 be added to their -product, the sum will be " equal to the number with its digits transposed. What is the number? ^ns_ fg. 19. In a purse containing 8 coins of gold and silver, each gold coin is worth as many dollars as there are sil- ver coins, and each silver coin is worth as many cents as there are gold coins ; and the whole is worth $ 15.15., How many are there of each ? Ans. 3 ^Id coins and 5 silver coins ; or, 5 gold and 3 silver. 20. What are eggs a dozen, when two more for twelve cents lowers the price one cent per dozen-?. Ans. 9 i!ents. 21. A farmer has inclosed a rectangular piece of lani, containing 1 acre and 32 square rods, with 88 panels of fence, each 4 yards long ; how many panels has he placed in each side of the rectangle ? Ans. 33 on one side, and 11 on the other. 22. There are four consecutive numbers, of which if the first two be taken for the digits of a number, that number is the product of the other two. What are the four nnmbers? ' Ans. 5, 6, 1, 8; or 1, 2, 3, 4. 23. A student traveled on a coach 6 miles into the country, and -walked bacik at a rate 5 miles less per hour than that of the coach. He found that he was 60 min- utes more in returning than in going. What was the speed of the coach ? Ans. 9 miles per hour. 24. A gentleman sent a lad into the market to buy 12 cents' worth of peaches. The lad having eaten a couple, the gentleman paid at the rate of a cent for fif- 284 ELEMENTARY ALGEBRA. teen more than the market price. How many did the gentleman receive ? Ans. 18. 25. A and B run a race. B, who runs slower than A by a mile in 5 hours, starts first by 2^ minutes, and they get to the 5 mile stone together. Eequired their rate's of running. Ans. A, 5 miles an hour ; B, 4f miles. 26. A room is 40 feet long, and twice as broad as it is high. The cost of papering its walls, at S'T^ cents per yard, is $ Yl.SO. Eequired the height of the room, no allowance being made for doors or windows. Ans. 13 feet. 2'7. A mirror is in the shape of a double square. The cost of the glass, at $ 1.25 a square foot, exceeds the cost of the frame, at 75 cents a linear foot, measured on the inside of the frame, by $ 22. Eequired the di- mensions of the glass. Ans. 8 feet by 4 feet. 28. Two detachments of soldiers being ordered to a station at the distance of 39 miles frorh their present quarters, begin their march at the same time ; but one party, by traveling |- of a mile an hour faster than the other, arrives there an hour sooner. Eequired their rates of marching. Ans. 3^ and 3 miles per hour. 29. Find two numbers whose sum is 6 and whose pro- duct is 10. Ans. Impossible. The imaginary expressions 3 -}- V — i and 3 — V — 1 ajpne an- swer the conditions, and these are readily obtained. 30. There are two lots,, each of which is an exact square ; it requires 200 rods of fence to inclose both, and their contents are 1800 square rods. What is the value of each at $ 2.25 per square rod ? Ans. The smaller, $ 900 ; the larger, $ 2025. • 31. A grocer sold 80 pounds of tea, and 100 pounds of coffee, for $ 65 ; but he sold 60 pounds more of coffee EATIO AND rEOPOETION. 285 for $20 than he did of tea for $ 10. What was the price of 1 pound of. each ? Ans. Tea, 50 cents ; coflfee, 25 cents. 32. Two farmers drove to market 100 sheep between them, and returned with equal sums. If each of them had sold his sheep at the same price that the other ac- tually did, the one would have returned with $ 180, and the other with $80. At what price per sheep did they sell, respectively, and how many sheep had each? Ans. At $ 2 and $ 3 per sheep ; the one had 60 sheep, and the other 40. * EATIO AND PROPOETION. 300i The Eatio of one quantity to another of the same kind is the quotient arising from dividing the first quan- tity by the second. (Art. 162.) Thus, the ratio of a to J is -r, or a : b. soli The Terms of a ratio are the two quantities re- quired to form it. The first term is called the antecedent of the ratio, and the second, the conseqiient. Thus, in the ratio of a to i, or a : i, a and h are the ■ terms, of which a is the antecedent and h the consequent. 302t A Direct Ratio is one in which the antecedent is divided by the consequent. An Inverse Ratio is one in which the consequent is divided by the antecedent. Thus, the . direct ratio of ' 6 to 3, or 6 : 3, is f , or 2, and the inverse ratio is %, or ^. Define Eatio. The Terms of a ratio. A Direct Eatio. An Inverse Batio. 286 ELEMENTAEY ALQEBEA. Note. When the kind of ratio is not mentioned, the interpretation is nnderstood to be that of the direct ratio. This method has the almost universal sanction of mathematicians in all countries. The so-called French interpretation is not that of the modern French mathematicians. 303. A Peopoetion is an equality of ratios. Four quantities are in proportion wlien the ratio of the Jirst to the second is the same as that of the third to the fourth. Thus, the ratios a : h and c : d, if equal to each other, forjn a proportion, when written o : 6 = c : In every proportion, the product of the extremes is equd to the product of the means. Let a : h : : c : d, a c -b = d Clearing of fractions, ad=bc. Hence, if three terms of a proportion be given, the fourth may he found. Let a : h :: c : X, Then, ax =^hc; whence, x= — • ' a TECEOREM H. 316. If the product of two quantities he equal to the product of two others, two of them may he made the extreme and the other two the means of a proportion. Let ad-=. he. a c Dividing hj id and reducing, j = 5 ' or, a : b : : c : d. THEOREM in. 317. ^ three ' quantities be in continued proportion, the product of the two extremes is equal to the square of the mean. Let a : b :: b : e. Then, by Theo. I., ac = bb = V. Demonstrate Theorem I. Show that the fourth term of a proportion may be found when three are giren. Demonstrate Theorem II. The- orem III. RATIO AND PKOPOBTION. 289 THEOREM IV. 318f If four quantities be in proportion, they wiU be in proportion by alteknation. Let a : b :: c : d, or, ° 1. h d ~, and reducing, - = ^, c c a or, a : e :: b : d. Multiplying by -, and reducing, - = -, c c d THEOREM V. 319i I^ four quantities be in proportion, they wiU be in proportion by inversion. ' Let a : b :: e : d. Then, by Theo. L, a rf = J c, or 5 c = a rf ; whence, by Theo. II., b : a :: d : c, THEOREM VI. 320t J^ four quantities be in proportion, they will be in . proportion by composition. Let a : b :: c : d; then, a-\-b : b :: c -\- d : d. For, by equality of ratios, j ^^ 3" Adding 1 to each side, r + 1 = ^ + 1, or ° + ^ _ g + <^ . whence, a-\-b : b :: c -\- d : d. THEOREM VII. 321 f If four quantities be in proportion, they will be in proportion by division. Demonstrate Theorem IV. Theorem V. Theorem VI. Theorem VII. 25 290 ELEMENTAEY ALGEBKA. Let a : h : : e : d; then, « — i '• i '• '• c — did, ' a c For, by equality of ratios, h^^ d' ct c Subtracting 1 from each side, j — ^ = ;? — 1» or, b d a — h c — d h d ' whence, a — b : h : : c — d : d. THEOKEM Tin. 322 • ^ four quantities be in proportion, the sum of the first and second is to their difference as the sum of the third and fourth is to their difference. Let a : b :■ : e : d; Then, a-\-b : a — b : : c-\-d : c — d ByTheo.VL, ^* = ^^. and by Theo. VII., ^ = "^ ; ^, - a-\-h a — J c-\-d c — d therefore, -ir_^__^^^ ___.^ a-\-l c-)-rf whence, a-\-b : a — 6: : c-\-d : c — d. THEOREM IX. 323 • Quantifies which are proportional to the same quan- tities are proportional to each other. Let a : b and c : d then a : b ■e:f •e:f; : e : d. For, by equality of ratios, .^ = -, Demonstrate Theorem YIII. Theorem IXi RATIO AND PEOPOETION. 291 and £ =1 d r f hereforiB, by Art, 38, Ax. Y, - = -, or> a:h::c'. A THEOREM X. 324i ^ any number of quantities are proportional,, any an^ tecedent is to its consequent as the sum of aU the antecedents is to the sum of aU the consequents. Let a:h::c:d::e:f; then a : & : ; g-f-c-f-e : l-\.d-\ - f. For, by Tlieo. I., ad=bc, and afz= h e ; also, ah=zha. Adding, ah-\-ad-^f-a f=ha\-lc^he, or, a{h-\-d-\-fy =h{a-\-c + e); whence, by Theo. II., a ih : : a -\- c ^^^ e : h -\- d -\- f. THBOBEM XI. 325> When four quantities are in proportion, if the jirsi and second he mitUipked or -divided hy the same quantity, as cdso the third and fourth, the resulting quantities will he pro- portional. Let a : h : : c ; d; then, « ma : mh : : nc :n d. For, by equality of ratios, t- = ^i' , ma n c and -^ = -— i, mo nd or, m a ', mh : : n c ', n d. d h c d m ' m ' ' n ' n' In like manner. Either m or n may be made equal to unity ; that is, either couplet maybe multiplied or divided ■without multiplying or dividing the other. Demonstrate Theorem' X. Theorem XT. 292 ELEMENTARY ALGEBEA. THEOREM Xn. 326i When four quantities, are in proportion, if the fint and third be multiplied or divided by the same quantity, as also the second and fourth, the resisting quantities wiU be proportioned. Let then, For, by equality of ratios, therefore, and, whence, ma : nh ; ; mc : nd, T Ti a h c d In like manner, —:-:: — :-. m n m n Either m or n may be made equal to unity. THEOREM Xni. 327. J^ there be two sets of proportioned quantities, the pro* ducts of the corresponding terms will be proportioned. a : b I I c : d 1 ma : nb : : mc : n d. a h c d' ma T mc d ' ma nb m G Td' Let a :i : : c : d, and e\f: :g:h; then. ae : bf: : eg : dh. For, by equality of ratios. a c b ~d' and By multiplication, ae eg bf—Jh' or. ae : b f : : eg : dh. THEOREM XIV. 328 1 If four quantities be in proportion, like powers or roots of these quantities wiU be proportioned. Demonstrate Theorem XII. Theorem XIII. Theorem XIV. EATIO AND PEOPOETION. 293 Let ^ a lb : :c :d; then a" : ^ : : c" : d", 11 11 and a" : J» : : c" : rf». For, by equality of ratios, raising to «th power. a c _ b ~d' a" e» 1 1 ^ c» _ 1 — ~ » extracting the nth root, whence, a" : J" : : C» : yir'y.|;J