# Combine the fractions

• Feb 22nd 2009, 04:42 AM
bryang
Combine the fractions
I'm having trouble finding the answer to this problem. Here is how I have been trying to solve it:

First, here is the problem:

$\displaystyle \frac{x}{x+1} + \frac{4}{1-x} + \frac{x^2-5x-8}{x^2-1} =$

The denominator of the second fraction is arranged wrong (x should be first, like the others) so I multiply it by -1.

$\displaystyle \frac{x}{x+1} - \frac{4}{x-1} + \frac{x^2-5x-8}{x^2-1} =$

Next, I factor the third fraction and I see it's factors are (x+1)(x-1), so that must be the common denominator.

$\displaystyle \frac{x(x-1)-4(x+1)+x^2-5x-8}{(x+1)(x-1)}$

Next, I multiply:

$\displaystyle \frac{x^2-x-4x-4+x^2-5x-8}{(x+1)(x-1)}$

Last, I combine like terms and I have the answer:

$\displaystyle \frac{2x^2-10x-12}{(x+1)(x-1)}$

I think I did something wrong though, because the book says this fraction here is the correct answer:

$\displaystyle \frac{2(x-6)}{x-1}$

But it doesn't show how to get that answer, so I have no way to figure out what I'm doing wrong. I have done it three time and I still come up with the same incorrect answer.

What am I doing wrong? :confused:
• Feb 22nd 2009, 04:45 AM
Mush
Quote:

Originally Posted by bryang
I'm having trouble finding the answer to this problem. Here is how I have been trying to solve it:

First, here is the problem:

$\displaystyle \frac{x}{x+1} + \frac{4}{1-x} + \frac{x^2-5x-8}{x^2-1} =$

The denominator of the second fraction is arranged wrong (x should be first, like the others) so I multiply it by -1.

$\displaystyle \frac{x}{x+1} - \frac{4}{x-1} + \frac{x^2-5x-8}{x^2-1} =$

Next, I factor the third fraction and I see it's factors are (x+1)(x-1), so that must be the common denominator.

$\displaystyle \frac{x(x-1)-4(x+1)+x^2-5x-8}{(x+1)(x-1)}$

Next, I multiply:

$\displaystyle \frac{x^2-x-4x-4+x^2-5x-8}{(x+1)(x-1)}$

Last, I combine like terms and I have the answer:

$\displaystyle \frac{2x^2-10x-12}{(x+1)(x-1)}$

I think I did something wrong though, because the book says this fraction here is the correct answer:

$\displaystyle \frac{2(x-6)}{x-1}$

But it doesn't show how to get that answer, so I have no way to figure out what I'm doing wrong. I have done it three time and I still come up with the same incorrect answer.

What am I doing wrong? :confused:

$\displaystyle \frac{2x^2-10x-12}{(x+1)(x-1)}$

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

$\displaystyle \frac{2(x-6)(x+1)}{(x+1)(x-1)}$

$\displaystyle \frac{2(x-6)}{x-1}$
• Feb 22nd 2009, 04:50 AM
red_dog
You did nothing wrong. You must continue.

$\displaystyle 2x^2-10x-12=2(x^2-5x-6)=2(x+1)(x-6)$

Then $\displaystyle \frac{2x^2-10x-12}{(x+1)(x-1)}=\frac{2(x+1)(x-6)}{(x+1)(x-1)}=\frac{2(x-6)}{x-1}$