# Thread: [SOLVED] Tricky Fraction Problem

1. ## [SOLVED] Tricky Fraction Problem

Question:
The numerator of a fraction is two less than the denominator. The sum of the fraction and its reciprocal is 25/12. Find the numerator and denominator is a positive integer.

Where do I start on this problem? Should there only be one variable?
Here is what I tried...but it did not work:

((x-2)/x) + (x/(x-2)) = 25/2

2. ## Figured it out!

I just didn't take it far enough...my equation was set up right.

3. Originally Posted by polt
Question:
The numerator of a fraction is two less than the denominator. The sum of the fraction and its reciprocal is 25/12. Find the numerator and denominator is a positive integer.

Where do I start on this problem? Should there only be one variable?
Here is what I tried...but it did not work:

((x-2)/x) + (x/(x-2)) = 25/12
Hi polt,

$\displaystyle \frac{x-2}{x}+\frac{x}{x-2}=\frac{25}{12}$

GCD=$\displaystyle 12x(x-2)$

$\displaystyle 12(x-2)^2+12x^2=25x(x-2)$

$\displaystyle 12(x^2-4x+4)+12x^2=25x^2-50x$

$\displaystyle 12x^2-48x+48+12x^2=25x^2-50x$

$\displaystyle -x^2+2x+48=0$

$\displaystyle x^2-2x+48=0$

$\displaystyle (x-8)(x+6)=0$

$\displaystyle x=8 \ \ or \ \ x=-6$

Use x = 8 to make it positive

$\displaystyle \frac{x-2}{x}=\frac{6}{8}=\frac{3}{4}$

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# In a fraction 20% less than numerator. The of fraction and its reciprocal is

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