Equivalent Fractions: Simplifying And Building With Answers Page 10
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16c.
First find the prime factorizations of 168 and 224:
(
)
(
)
(
)
(
)
168 = 4 • 42 = 2 • 2
= 2 • 2
= 2 • 2 • 2 • 3 • 7
• 2 • 21
• 2 • 3 • 7
(
)
(
)
(
)
(
)
224 = 4 • 56 = 2 • 2
= 2 • 2
= 2 • 2 • 2 • 2 • 2 • 7
• 4 • 14
• 2 • 2 • 2 • 7
Now rewrite the fraction using prime numbers, and simplify:
168
2 • 2 • 2 • 3 • 7
!
= !
prime factorizations
224
2 • 2 • 2 • 2 • 2 • 7
/ 2 • / 2 • / 2 • 3 • / 7
= !
cancelling common factors
/ 2 • / 2 • / 2 • 2 • 2 • / 7
3
= !
writing the remaining factors
2 • 2
3
= !
multiplying
4
d.
This may not seem to “fit”, but look carefully at the fraction. Treating the
variables as prime numbers, we simplify:
2
3
5x
y
5 • x • x • y • y • y
=
prime factorizations
4
4x
y
2 • 2 • x • x • x • x • y
5 • / x • / x • y • y • / y
=
cancelling common factors
2 • 2 • / x • / x • x • x • / y
5 • y • y
=
writing the remaining factors
2 • 2 • x • x
2
5y
=
multiplying
2
4x
This example is a typical algebra problem, and it is included here to
illustrate that the ideas we are developing extend directly to algebra.
In the last section, we found that some fractions result in a terminating decimal, while others
result in a repeating decimal. Is there a way to tell in advance which will occur? Consider the
two fractions and their decimal forms:
17
= 0.425
40
19
= 0.63
30
154
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