# Thread: Help with algebra - Fractions in simplest form

1. ## Help with algebra - Fractions in simplest form

Hi All,

Been about ten years since I was in school and am now doing a maths refresher course externally. I am really really stuck on a problem:

$\displaystyle \frac{x}{x^2-1} - \frac{x+1}{x-1}$
I would assume that the common denominator is x(x-1) therefore

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

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

That's where I get stuck. I'm sure it has something to do with the x-x(x+1)

$\displaystyle x-x^2+x$ maybe???

Thanks guys (and gals),

CHeers,

J

2. Originally Posted by perjac
Hi All,

Been about ten years since I was in school and am now doing a maths refresher course externally. I am really really stuck on a problem:

$\displaystyle \frac{x}{x^2-1} - \frac{x+1}{x-1}$
I would assume that the common denominator is x(x-1) therefore

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

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

That's where I get stuck. I'm sure it has something to do with the x-x(x+1)

$\displaystyle x-x^2+x$ maybe???

Thanks guys (and gals),

CHeers,

J
Hello .
$\displaystyle \frac{x}{x^2-1} - \frac{x+1}{x-1}=\frac{x}{{\color{red}(x-1)}(x+1)} - \frac{x+1}{\color{red}{x-1}}$

Did you catch it ? =)

3. Originally Posted by General
Hello .
$\displaystyle \frac{x}{x^2-1} - \frac{x+1}{x-1}=\frac{x}{{\color{red}(x-1)}(x+1)} - \frac{x+1}{\color{red}{x-1}}$

Did you catch it ? =)
Damnit of course.

Thank you so much.