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Simplify:

$$\frac{1}{x-2}-\frac{1}{x+2}$$

What I did was multiply both sides to get the denominator equal:

$$\frac{x+2}{(x-2)(x+2)}-\frac{x-2}{(x-2)(x+2)}=\frac{x^2+4x+4}{x^2-4} =\frac{4x+4}{ -4}=\frac{4 (x+1)}{-1}$$

Apparently this is not correct. Can anyone show me what I did wrong in steps?

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  • $\begingroup$ Could you give reasons justifying the second line? $\endgroup$
    – recmath
    Jul 5, 2014 at 15:32
  • $\begingroup$ I multiplied both 1's diagonally. $\endgroup$
    – user160137
    Jul 5, 2014 at 15:34
  • $\begingroup$ The step after that i meant: where is $x^2+4x+4$ coming from in the numerator $\endgroup$
    – recmath
    Jul 5, 2014 at 15:35
  • $\begingroup$ @user160137 I have typeset your equations into LaTeX. Please double-check that I transcribed correctly. $\endgroup$
    – Neal
    Jul 5, 2014 at 15:39
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    $\begingroup$ There are some as would say that it’s already simplified. $\endgroup$
    – Lubin
    Jul 5, 2014 at 15:48

2 Answers 2

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$\frac{1}{x-2}-\frac{1}{x+2} = \frac{x+2}{(x-2)(x+2)}- \frac{x-2}{(x+2)(x-2)} = \frac{x+2 - (x-2)}{(x+2)(x-2)} = \frac{x+2 - x + 2 }{(x+2)(x-2)} = \frac{4}{(x+2)(x-2)} = \frac{4}{x^2-4}$.

First step is getting common denominator, second step is combining fractions with common denominator, third step is distributing the $-$, 4th step is combining like terms and the final step is expanding out the denominator using difference of squares.

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  • $\begingroup$ I see it now. I multiplied instead of combining like terms. Thank you! $\endgroup$
    – user160137
    Jul 5, 2014 at 15:39
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$$\frac{x+2}{(x-2)(x+2)}-\frac{x-2}{(x-2)(x+2)} $$ $$ = \frac{x+2 - (x-2)}{(x-2)(x+2)} $$ $$= \frac{x+2 - x+2}{(x-2)(x+2)}$$ $$ = \frac{x - x+2 +2}{(x-2)(x+2)} $$ $$ = \frac{4}{(x-2)(x+2)}$$

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