this should probably be fairly easy, but my answer ended up being slightly different than what the answer should be. I'm hoping someone can be so kind as to point out my mistake.

$\displaystyle \frac{1}{x^2-x-2} - \frac{x}{x^2-5x+6}$

I started by factoring the denominators to

$\displaystyle \frac{1}{(x+1)(x-2)} - \frac{x}{(x-3)(x-2)}$

then I multiplied the stuff out to

$\displaystyle \frac{(x-3)(x-2)-x(x+1)(x-2)}{(x+1)(x-2)(x-3)(x-2)}$

I then factored out the (x-2) in the numerator to

$\displaystyle \frac{(x-2)[(x-3)-x(x+1)]}{(x+1)(x-2)(x-3)(x-2)}$

then cancelled the (x-2) out, leaving

$\displaystyle \frac{(x-3)-x(x+1)}{(x+1)(x-2)(x-3)}$

next I distributed the numerator to

$\displaystyle \frac{x-3-x^2-x}{(x+1)(x-2)(x-3)}$

then finally simplified to

$\displaystyle \frac{-3-x^2}{(x+1)(x-2)(x-3)}$

unfortunately, the answer is apparently

$\displaystyle \frac{-x^2+3}{(x+1)(x-2)(x-3)}$

close, but close doesn't count in math. Can someone please point out my mistake?