Simplifying Algebra

Kaynight · #1 $Old$ November 7th, 2008, 11:24 PM

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

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

$\frac{1}{x} - \frac{1}{x}$

This just doesn't look right to me especially the third step as I'm unsure whether I can do what I did by canceling out on either side. Can someone check to make sure I’m doing it right and if I'm not tell me what I’ve done wrong. Thanks!

Rapha · #2 $Old$ November 7th, 2008, 11:36 PM

Quote:

Originally Posted by Kaynight $View Post$

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

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

$\frac{1}{x} - \frac{1}{x}$

This just doesn't look right to me especially the third step as I'm unsure whether I can do what I did by canceling out on either side. Can someone check to make sure I’m doing it right and if I'm not tell me what I’ve done wrong. Thanks!

The third step doesnt make sense

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

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

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

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

$\frac{-4x}{x(x+1)(x-1)}$

$\frac{-4}{(x+1)(x-1)}$

$\frac{4}{(x+1)(1-x)}$

$\frac{4}{1-x^2}$

Kaynight · #3 $Old$ November 8th, 2008, 12:28 AM

I didn't think so

Thanks for the help