Simplifying $(1/x-1/5 )/( 1/x^2-1/25)$

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How do I get $\frac{5x}{x+5}$ from simplifying the following? $$\frac{(\frac{1}{x}-\frac{1}{5} )}{( \frac{1}{x^2}-\frac{1}{25})}$$

My work:

I multiplied the top and bottom by the LCD: $25x^2$ (to get the same denominators). Then I got: $\frac{25x-5x}{25-x^2}$ Then I got this for my answer: $\frac{20}{-(x-5)(x+5)}$ But the real answer is: $\frac{5x}{x+5}$

So my questions is how to I get from $\frac{20}{-(x-5)(x+5)}$ to $\frac{5x}{x+5}$?

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3 Answers

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Easier: $\frac{1}{x^2} - \frac{1}{25} = \big( \frac{1}{x} -\frac{1}{5} \big)\big(\frac{1}{x} + \frac{1}{5} \big)$ and the first term happily cancels out, now sum two fractions and you are done.

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$$\frac{\frac1x-\frac15}{\frac1{x^2}-\frac1{25}}=\frac{\frac{5-x}{5x}}{\frac{25-x^2}{25x^2}}$$

$$=\frac{25x^2(5-x)}{5x(25-x^2)}=\frac{5x}{5+x}$$ assuming $x(5-x)\ne0$

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You should have gotten to

$$\frac{25x-5x^2}{25-x^2} = \frac{5x(5 - x)}{(5+x)(5-x)}$$

Now cancel the common factor to get the desired result, assuming $x\neq 5$.

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