Page 20 - MarceAlgebra Demystified

CHAPTER 1 Fractions 7

Solutions

14 2 7 ð2 7Þ 1 2 7 1 14 1 1
1: ¼ ¼ ¼ ¼ ¼
42 2 3 7 ð2 7Þ 3 2 7 3 14 3 3
5 5 5 1 5 1 1
2: ¼ ¼ ¼ ¼
35 5 7 5 7 5 7 7

48 2 2 2 2 3 ð2 3Þ 2 2 2 2 3 2 2 2 6 8 8
3: ¼ ¼ ¼ ¼ ¼
30 2 3 5 ð2 3Þ 5 2 3 5 6 5 5
22 2 11 11 2 2
4: ¼ ¼ ¼
121 11 11 11 11 11

39 3 13 3 13 13
5: ¼ ¼ ¼
123 3 41 3 41 41
18 2 3 3 2 3 3 9
6: ¼ ¼ ¼
4 2 2 2 2 2
7 7 7 1 7 1 1
7: ¼ ¼ ¼ ¼
210 2 3 5 7 7 2 3 5 7 2 3 5 30
240 2 2 2 2 3 5 ð3 5Þ 2 2 2 2 3 5 2 2 2 2
8: ¼ ¼ ¼
165 3 5 11 ð3 5Þ 11 3 5 11
15 16 16
¼ ¼
15 11 11
55 5 11 11 5 11 5 5
9: ¼ ¼ ¼ ¼
33 3 11 11 3 11 3 3

150 2 3 5 5 ð2 3 5Þ 5 2 3 5 5 30
10: ¼ ¼ ¼ ¼ 5 ¼ 5
30 2 3 5 ð2 3 5Þ 1 2 3 5 1 30

Fortunately there is a less tedious method for reducing fractions to their
lowest terms. Find the largest number that divides both the numerator
and the denominator. This number is called the greatest common divisor
(GCD) . Factor the GCD from the numerator and denominator and rewrite
the fraction. In the previous examples and practice problems, the product of
the common primes was the GCD.