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

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?

• Could you give reasons justifying the second line? – illysial Jul 5 '14 at 15:32
• I multiplied both 1's diagonally. – user160137 Jul 5 '14 at 15:34
• The step after that i meant: where is $x^2+4x+4$ coming from in the numerator – illysial Jul 5 '14 at 15:35
• @user160137 I have typeset your equations into LaTeX. Please double-check that I transcribed correctly. – Neal Jul 5 '14 at 15:39
• There are some as would say that it’s already simplified. – Lubin Jul 5 '14 at 15:48

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