SOLUTION: The denominator of a fraction is five more than twice the numerator. If both numerator and denominator are decreased by seven, the simplified result is 1/6. . Find the original fr

Question 1148882: The denominator of a fraction is five more than twice the numerator. If both numerator and denominator are decreased by seven, the simplified result is 1/6.
. Find the original fraction. (Do NOT simplify.)
Found 2 solutions by josgarithmetic, jim_thompson5910:
Answer by josgarithmetic(39625) (Show Source): You can put this solution on YOUR website!
The fraction

The second sentence

The fraction:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Answer: 10/25
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Work Shown:

Let
n = original numerator
d = original denominator

We know that d = 2n+5 since "The denominator of a fraction is five more than twice the numerator"

"If both numerator and denominator are decreased by seven, the simplified result is 1/6" meaning that (n-7)/(d-7) = 1/6

Let's solve for n after doing a substitution
(n-7)/(d-7) = 1/6
(n-7)/(2n+5-7) = 1/6 ... replace d with 2n+5
(n-7)/(2n-2) = 1/6
6(n-7) = 1(2n-2) .... cross multiply
6n-42 = 2n-2
6n-42-2n = 2n-2-2n ... subtract 2n from both sides
4n-42 = -2
4n-42+42 = -2+42 ... add 42 to both sides
4n = 40
4n/4 = 40/4 ... divide both sides by 4
n = 40/4
n = 10

If n = 10, then d is
d = 2n+5
d = 2*10+5
d = 20+5
d = 25

So n/d = 10/25 is the original fraction (we arent reducing it). After decreasing the numerator and denominator by 7, we end up with 3/18 = 1/6, which confirms our answer.

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