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I'm getting stuck on the end result. I set it up

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

I then factor out the left side denominator to get $(x-2)(x+2)$

I multiply the right side by $x-2$ since it already has $x+2$ and I get on top:

$$\frac{x^2-x^2-2x+x-2}{x^2-4}$$

I then get $-x-2$ on top, I figure I should factor out the negative to make the top $-1(x+2)$

so cancel out the $x+2$ from the top and bottom and it should leave me with

$\frac{-1}{x-2}$ but according to my book the answer is $\frac{1}{x-2}$

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  • $\begingroup$ You can format the equation by putting it into dollar signs, so "dollarsign" equation "dollarsign". Will edit it for you, look at what I did and try to repeat in other questions. $\endgroup$
    – 5xum
    Jan 31 '14 at 14:52
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You have not handled the minus sign properly.

As $x^2-4=(x-2)(x+2),$

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

Which is $\displaystyle\frac1{x-2}$ if $x+2\ne0$

else undefined

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  • $\begingroup$ I forgot the sign, thank you. $\endgroup$
    – Joshhw
    Jan 31 '14 at 14:57

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