Fractions. 3. Reduce 288 to its lowest terms. Operation. Explanation. 160 16 10) Dividing each term of 288 by 10 we have 200 1 tenth as many parts, which are 10 times as 16 4 large. Dividing each term of jo by 4 we have 4 20 1 fourth as many parts, which are 4 times as large. Hence, 268 = . But 4 and 5 are prime to each other, and the fraction is in its lowest terms. 20 RULE. Divide each term of the fraction by any common divisor except 1, and divide each term of the fraction thus obtained by any common divisor except 1, and so continue until the terms are prime to each other. 4. Reduce to higher terms — to 120ths. Operation. Explanation. 120 +8=15. In 75 there are 15 times as many parts as 5 x 15 there are in s, and the parts are 1 fifteenth as 75 large. Hence, m=s. 8 x 15 120 * Divide each term by 12). This involves the reduction of a complex to a simple fraction; but it will lead to thoughtful work for the pupil to solve such problems in this manner. + Divide each term by . | If the pupil has not had sufficient practice in addition of fractions to do this, the finding of the sum may be omitted until the book is reviewed. Fractions. 5 16 Reduce to higher terms to 160ths. (1) ģ (2) il (3) 26 (4) 37 (8) 21 5. Two or more fractions whose denominators are the same, are said to have a common denominator. 6. Two or more fractions that do not have a common denominator may be changed to equivalent fractions having a common denominator. EXAMPLE. 3 = 24 1=24 j = 36 = 36 7. Two or more fractions that do not have a common denominator may be changed to equivalent fractions having their least common denominator. The l. c. d. of two or more fractions is the l. c. m. of the given denominators. EXAMPLE Change 10, 4o, and 87 to equivalent fractions having their least common denominator. OPERATION. 11 x 4 44 30 x 4 120 9 x 3 27 (3) 12040=3 40 x 3 120 37 x 2 74 60 x 2 120 = Fractions. Reduce to equivalent fractions having their l. c. d. (1) it and y (6) 15, , and (2) 1 and 23 (7) 25 }, and 7 (3) ;; and 17 (8) 3%, , and (4) 35 and 63 (9) 31, 1U, and (5) 4 and 87 (10) 1, , and 23 (a) Find the sum of the twenty-five fractions. * 179. To add common fractions. RULE.— Reduce the fractions if necessary to equivalent fractions having a common denominator, add their numerators, and write their sum over the common denominator. EXAMPLE. Add 15, 17, and 1%. NOTE. — If the work that precedes this article has been well done, no explanation of the foregoing will be necessary. Pupils have already learned (presumably before using this book) (1) that fractions may be reduced to higher terms, (2) that two or more fractions whose denominators are not alike may be reduced to higher terms with like denominators, (3) that a common denominator of two or more fractions with unlike denominators, is a common multiple of the given denominators, and (4) that in reducing a fraction to higher terms the numerator and denominator must be multiplied by the same number. The simple problem of adding 44 180ths, 102 180ths, and 159 180ths, is not unlike the problem of adding 44 apples, 102 apples, and 159 apples. (For a continuation of this work, see page 91.) * This work may be omitted until the subject of fractions is reviewed. elaxaxa). Algebraic Fractions. 180. The expressions a, , 6 are algebraic fractions. D4' cd' The above expressions are read, a divided by b; x divided by 4; 6 divided by cd. REDUCTION OF ALGEBRAIC FRACTIONS. 6 aa a x 6 = Let a = 2, b= 3, c=5, and d= 7, and verify. Observe that to reduce c. fraction to its lowest terms we have only to str out the factors that are common to its numerator denominator. alb 4. What factors are common to both numerator aʼc and denominator ? Reduce and verify. 5. x2y3 What factors are common to both numerator 3 x 2. Change to a fraction whose denominator is 2 ay 2 ay Give any values you please to a, x, and y, 2 ay X y 2 aya and verify the reduction. 3 x x y 3 xy 183. REDUCE TO EQUIVALENT FRACTIONS HAVING A COMMON DENOMINATOR. X 1. Since the common denominator must be у and ab ard exactly divisible by each of the given denomi nators, it must contain all the prime factors * found in either of the given denominators. The new denominator must therefore be a x a xbxd= abd; abd -- ab = ad ; abd -- a’d = b. X X ad adx by ab x ada?bd a d x 6 у хь Give any values you please to a, b, d, x, and y, and verify. 3 The common denominator must contain the and bc2 factors a, b, b, c, C. Reduce and verify. ab2 The common denominator is 5 a. Reduce and verify a * Since the numerical values of the letters are unknown, each must be regarded as prime to all the others. The prime factors, then, in the first denominator are a and b; in the second, a, a, and d. |