SOLUTION: The numerator of a fraction is 1 less than the denominator. If the numerator and the denominator are both increased by 4, the new fraction will be 1/8 more than the original fract

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: The numerator of a fraction is 1 less than the denominator. If the numerator and the denominator are both increased by 4, the new fraction will be 1/8 more than the original fract      Log On


   

Question 136648This question is from textbook Algebra Structure and Method
: The numerator of a fraction is 1 less than the denominator. If the numerator and the denominator are both increased by 4, the new fraction will be 1/8 more than the original fraction. Find the original fraction. This question is from textbook Algebra Structure and Method

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The numerator of a fraction is 1 less than the denominator. If the numerator and the denominator are both increased by 4, the new fraction will be 1/8 more than the original fraction. Find the original fraction.
:
Write an equation for what it says:
x%2F%28%28x%2B1%29%29 + 1%2F8 = %28%28x%2B4%29%29%2F%28%28x%2B1%2B4%29%29
:
x%2F%28%28x%2B1%29%29 + 1%2F8 = %28%28x%2B4%29%29%2F%28%28x%2B5%29%29
:
The common denominator will be 8(x+1)(x+5)
Multiply each term by this, cancel out the denominators, leaving us with:
x(8(x+5) + (x+1)(x+5) = 8(x+4)(x+1)
:
8x^2 + 40x + x^2 + 6x + 5 = 8(x^2 + 5x + 4)
:
8x^2 + 40x + x^2 + 6x + 5 = 8x^2 + 40x + 32
Combine on the left:
8x^2 + x^2 - 8x^2 + 40x + 6x - 40x + 5 - 32 = 0
A quadratic equation
x^2 + 6x - 27 = 0
Factors to:
(x+9)(x-3) = 0
We want the positive solution:
x = 3
:
Check solution in original equation
3%2F%28%284%29%29 + 1%2F8 = %28%287%29%29%2F%28%288%29%29
:
Actually the negative solution x = -9 will work also:
%28-9%29%2F%28%28-8%29%29 + 1%2F8 = %28%28-5%29%29%2F%28%28-5%29%29
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Did I make this understandable? Any questions?