# [SOLVED] subtracting fractions

• Jan 23rd 2010, 09:49 AM
satis
[SOLVED] subtracting fractions
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.

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

I started by factoring the denominators to

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

then I multiplied the stuff out to

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

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

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

then cancelled the (x-2) out, leaving

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

next I distributed the numerator to

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

then finally simplified to

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

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

close, but close doesn't count in math. Can someone please point out my mistake?
• Jan 23rd 2010, 10:16 AM
skeeter
$\frac{1}{x^2-x-2} - \frac{x}{x^2-5x+6}$

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

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

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

finish up
• Jan 23rd 2010, 10:27 AM
satis
ok, so that distributes to

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

then simplifies to

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

which is what I got doing it the other way :p So how do I get from here to

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

Am I doing something wrong in distributing, or am I just missing some simple transform or something?
• Jan 23rd 2010, 10:50 AM
skeeter
the given "answer" is not correct.

the final result should be

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

you sure you copied it correctly?
• Jan 23rd 2010, 12:01 PM
satis
hah....sure enough, I didn't copy it right. The negative in the given answer is in front of the entire fraction. So, for all intents and purposes, my answer was correct. :p

thanks for the assistance.