Asked by David lewis Washington jr
The denominator of a fraction is three more than twice the numerator. If both numerator and denominator are decreased by seven, the simplified result is 7 divided by19. Find the original fraction. (Do NOT simplify.)
Answers
Answered by
Reiny
In original fraction:
numerator ---- x
denominator ---- 2x+3
so the fraction was x/(2x+3)
after change:
fraction is (x-7)/(2x+3 - 7) = 7/19
(x-7)/(2x-4) = 7/19
19x - 133 = 14x - 28
5x = 105
x = 21
the original fraction is 21/45
numerator ---- x
denominator ---- 2x+3
so the fraction was x/(2x+3)
after change:
fraction is (x-7)/(2x+3 - 7) = 7/19
(x-7)/(2x-4) = 7/19
19x - 133 = 14x - 28
5x = 105
x = 21
the original fraction is 21/45
Answered by
oobleck
original fraction: n/(n+3)
(n-7)/(n+3-7) = 7/19
So solve for n
(n-7)/(n+3-7) = 7/19
So solve for n
Answered by
oobleck
my bad. I missed the "twice"
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