« ΠροηγούμενηΣυνέχεια »
3. Reduce joy to its lowest terms. Operation.
Explanation. 160 16 Dividing each term of 188 by 10 we have 10);
200 20 1 tenth as many parts, which are 10 times as
16 4 large. Dividing each term of 18 by 4 we have 4); 20 5
1 fourth as many parts, which are 4 times as
large. Hence, 188 = 4. But 4 and 5 are prime to each other, and the fraction is in its lowest terms.
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.
(a) Find the sum of the ten results. I 4. Reduce & to higher terms to 120ths.
Operation. 120 - 8 = 15.
E.rplanation. In 73 there are 15 times as many parts as there are in 3, and the parts are 1 fifteenth as large. Hence, 1 = g.
5 x 15 8 x 15
* 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 2.
# 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.
Reduce to higher terms to 160ths.
(7) 1 (8) 13 (9)
(10) 88 (a) Find the sum of the ten results.
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.
* = 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 1. c. d. of two or more fractions is the 1. c. m. of the given denominators.
EXAMPLE Change it, do, and 7 to equivalent fractions having their least common denominator.
11 X 4 44
30 X 4 120
9 x 3 27 (3) 120 – 40 : 3
40 X 3
37 x 2 74 (4) 120 x 60 2
60 X 2 120 3t = 120, 10 87
Reduce to equivalent fractions having their 1. c. d. (1) 1 and
(6) 15, %, and 11 (2) and 13
(7) , , and 2% (3) and 13
(8) 36, %, and is (4) 3 and
(9) 8 %, and 2 (5) and
(10) 13, 2, and x3 (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.
Add 11, 37, and 3.
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.
180. The expressions z I cd are algebraic fractions.
The above expressions are read, a divided by b; x divided by 4; 6 divided by cd.
5, and d = 7, and verify. Observe that to reduce a fraction to its lowest terms we have only to strike out the factors that are common to its numerator and denom
a b 4.
What factors are common to both numerator and a'c denominator ? Reduce and verify.
What factors are common to both numerator and
182. REDUCE TO HIGHER TERMS.
to a fraction whose denominator is ahc.
bc 2a x a
2 a bc xa
Let a =
= 2, b = 3, and c= 5, and verify the reduction.
2. Change to a fraction whose denominator is 2 ay.
2 ay 3 x x y 3 xy Give any values you please to a, x, and y 2 ay x y " 2 ay and verify the reduction.
183. REDUCE TO EQUIVALENT FRACTIONS HAVING A
Since the common denominator must be 1.
у and ab a2d exactly divisible by each of the given denomin.
ators, it must contain all the prime factors* found in either of the given denominators. The new denominator must therefore be a X a X 6 Xd= a?bd ; a2bd ; ab=ad; a2bd a?d=b.
x x ad
adx ab x ad abd
by ad x7 abd
Give any values you please to a, b, d, x, and y, and verify.
The common denominator is 5 a. Reduce and verify.
* 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.