of the quantities; and since a2 — b2 = (a + b) (a — b), we have (ab)÷ (a + b) = a−b. Hence we shall have (a3 + b3) (a — b) = a* — a3b + ab3 - b for the least common multiple. - Ex. 4. To find the least common multiple of a2 — b2, a+b3, and ao — bo. By Ex. 3, (a3 + b3) (a — b) is the least common multiple of a2 — b3, a3 + b3; and since ao — 6o is divisible by (a3 + b3) (a - b), we shall have a b for the sought least common multiple. Ex. 5.-To find the least common multiple of 3a2 -2a-1 and 4a3 2a2 Since a 3a + 1. 1 is the greatest common divisor of the quantities, and that, (3a2a-1) (a-1)=3a+1, we get (4a3 — 2a3 — 3a + 1) (3a + 1) = 12a 2a3 11a2 + 1 for the sought least common multiple. 1. Find the least common multiple of 13 and 19. Ans. 19 x 13 = 247. 2. Find the least common multiple of 6, 7, and 21. Ans. 42. 3. Find the least common multiple of a + b and x-y. Ans. (a + b) (x − y). 4. Find the least common multiple of x + y and x-y. Ans. x2-y3. 5. Find the least common multiple of a2 + 2ab + b3, a 2ab+b2, and a2- b3. 6. Find the least common multiple of a Ans. 7. Find the least common multiple of a a2 + b2. Ans. (a + b) (a1 — a3b + a2b3 — ab3 + b1) (a2 — ab + b3). 8. Find the least common multiple of a2 ab 262 and a2 3ab + 2b2. Ans. (a-ab- 263) (a - b). 9. Find the least common multiple of 6a+ Tax-3x2 and 6a2+11ax + 3x2. Ans. (6a2+7ax-3x2) (3a + x). (37.) Remark.-Many arithmetical and algebraic ques tions are easily solved by the use of the least common multiple. For examples, take the following questions. 1. A and B denote two individuals, such that A alone can do a piece of work in 12 days, and B alone can do the work in 14 days; then, in what time can A and B together do the work? The least common multiple of 12 and 14 is 84. And since 84127, and 84 ÷ 146, it follows that A alone in 84 days can do 7 times the work, and B alone in 84 days can do 6 times the work; consequently, both together in 84 days can do 13 times the work. Hence, if we divide 84 by 13, we shall have 6 84 = the number of days in which A and B together can do the work, as required. 2. Two pipes, A and B, are such that A will fill a certain cistern in an hour and a half, and B will fill it in two hours and a quarter. In what time will A and B together fill the cistern? Take a quarter of an hour for the unit of time, then 6 and 9 will respectively represent the times in which A and B can fill the cistern. Now 18 is the least common multiple of 6 and 9; and since 1863 and 1892, it follows that A alone will fill the cistern 3 times and B alone will fill the cistern 2 times in 18 units of time; consequently A and B together will fill the cistern 5 times in 18 units of time. Hence, if we 18 3 divide 18 by 5, we have = 3 units of time in which A and B together will fill the cistern; or, since our unit of time 3 equals 15 minutes, we have 3 x 15 54 the number of = = minutes in which A and B together will fill the cistern. 3. A man and his wife usually drank out a cask of beer in 12 days, but when the man was from home it lasted the woman 30 days. How many days would the man alone be in drinking it? The least common multiple of 12 and 30 is 60; and since 60125 and 60 ÷ 30 = 2, it follows that the man and woman would drink out the cask of beer 5 times in 60 days, and that the woman alone would drink out the cask of beer 2 times in 60 days. Hence, by subtraction, 5 times - 2 times = 3 times, is the number of times that the man alone would drink out the cask of beer in 60 days. Consequently, if we divide 60 by 3, we have 60 ÷ 3 = 20 = the number of days in which the man alone would drink out the cask of beer. 4. Let A, B, and C denote three individuals, such that A can do a certain piece of work in the time a, and B can do the same piece of work in the time b, and C can do it in the time c. Required, the time in which A, B, and C can jointly perform the work? Since a, b, and c are under the form of primes, or are not explicitly composed of factors, their product abc must be taken for their least common multiple. Hence, abc ÷ a = bc, abc÷b= ac, abc÷c= ab, and be + ac + ab = the number of times that A, B, and C together can do the work in abc abc days. Hence, we shall have = the time in which A, B, C, jointly, can perform the work. ab + ac + bc If a 6 days, b = 8 days, c = 24 days, then putting 6 abc for a, 8 for b, and 24 for c, we shall have ab + ac + be = 1152 = = 3 = the number of days in which 6 × 8 × 24 48 +144+ 192 384 A, B, and C can jointly perform the work. 5. If A and B can perform a piece of work in a days, A and C in b days, B and C in c days, then in what time can A, B, and C jointly perform the work? Here, abc expresses the least common multiple of a, b, and c. And abc ÷ a = be the number of times that A and B can do the work in abc days; also, abc ÷ b ac the number of times that A can do the work in abe days; and abc÷ cab the number of times that B and C can do the work in abc days. Hence, A and B, A and C, B and C, or 2A + 2B + 2C, can perform (ab + ac + be) times the work in abc days, or ab + ac + bc A+B+C can perform times the work in abe 2 ab+ac+be days. Hence we shall have abc ÷ = 2 = the time in which A, B, and C together can perform the 2abc at + ac+bc = work. If a 8, 69, c = 10, then the time in which A, B, and C can jointly perform the work will be found to be in which A alone can perform the work; and in like manner = is the time in which B alone can do the work, be is the time in which C alone can do the If a 8, b = 9, c = 10, it will be found that A can do 34 49 23 1741 the work in 14 days, and that B can do the work in 17 7 days, and C can do the work in 2331 days. SECTION VII. OF FRACTIONS. (1.) If we represent any quantity by 1, and wish to show that the quantity is to be divided into a equal parts, and that b of the parts are to be used, we proceed as follows, viz. : Conceive the quantity to consist of a equal parts; then, 1 since 1 stands for the quantity, it is clear that will stand for α one of the equal parts into which the quantity is to be 1 divided, since the quotient expresses one part out of the a parts which are supposed to compose the quantity. 1 Since b of the equal parts are to be used, we must multiply 1 a by b, and we shall have xb for the proper expression of the portion of the quantity that is to be used. Because 1 a x b is an expression for b equal parts out of the a equal parts into which the quantity (that is represented by 1) is supposed to be broken or divided, it is called a fraction; and since a denominates or shows into how many equal parts the quantity is to be divided, it is called the denominator of the fraction; and because enumerates or shows how many of the equal parts into which the quantity is divided are to be taken, it is called the numerator of the fraction; and since the value of the fraction depends on its numerator and denominator, we call the numerator and denominator the terms of the fraction. b 1 we shall have a' a 1 α + etc., to b terms (by addition) equals хъ = which shows that the fraction can be expressed by writing the denominator below the numerator, with a right line between them; and it is clear that the b α fraction expresses the quotient of the division of the numerator by the denominator. Hence, generally, any quotient which is indicated by writing the divisor below the dividend, with a right line between them, is called a fraction, particularly if the quotient expresses a portion of some quantity that is represented by 1; the dividend being called the numer ator and the divisor the denominator of the fraction; also the numerator and denominator are called the terms of the fraction. (2.) Any fraction whose numerator is less than its denom347 inator is called a proper fraction. Thus, Thus, 4' 5' 13 are proper fractions. |