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

Quote:

Originally Posted by Kaynight

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

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

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

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

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

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

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

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

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

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

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

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

The following users thank Rapha for this useful post:
Kaynight
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