Another fraction simplifying.

**icesta1** · March 7th 2010, 04:52 AM

Sorry I'm crap at this..

simplify
1/x^2-1 - 2/x+1

I expanded x^2-1 then cross multiplied

which i got -2x+3 / (x+1)(x-1)

Would that be right? I'm not quite sure on this >< the negative signs screw me up alot.

I think it might be -2x-1 / (x+1)(x-1)

Please correct me

Thanks

EDIT: can i cancel here? 4+k-1 / (k+3)(k-1)

**Prove It** · March 7th 2010, 04:58 AM

Originally Posted by icesta1

simplify
1/x^2-1 - 2/x+1

I can't read this.

Is it $\frac{1}{x^2} - 1 - \frac{2}{x + 1}$

or

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

Please use LaTeX or else put brackets where they're needed.

**icesta1** · March 7th 2010, 05:03 AM

Originally Posted by Prove It

I can't read this.

Is it $\frac{1}{x^2} - 1 - \frac{2}{x + 1}$

or

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

Please use LaTeX or else put brackets where they're needed.

Sorry its the second one.

**Prove It** · March 7th 2010, 06:50 AM

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

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

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

$= \frac{3 - 2x}{x^2 - 1}$ .

**Prove It** · March 7th 2010, 06:51 AM

Originally Posted by icesta1

Can i cancel here? 4+k-1 / (k+3)(k-1)

If it's $\frac{4 + k - 1}{(k + 3)(k - 1)}$ then yes, you can.

$\frac{4 + k - 1}{(k + 3)(k - 1)} = \frac{k + 3}{(k + 3)(k - 1)}$

$= \frac{1}{k - 1}$ .

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