algebra help!

jvignacio · #1 $Old$ 11-19-2008, 07:37 PM

$\frac{x}{x-y} - \frac{y}{x+y} - \frac{2xy}{x^2-y^2}$

$\frac{x}{x-y} - \frac{y}{x+y} - \frac{2xy}{(x-y)(x+y)}$

$\frac{(x-y)-2xy}{(x-y)(x+y)}$

$\frac{-2xy}{(x+y)}$

correcT?

Shyam · #2 $Old$ 11-19-2008, 07:53 PM

Quote:

Originally Posted by jvignacio $View Post$

$\frac{x}{x-y} - \frac{y}{x+y} - \frac{2xy}{x^2-y^2}$

$\frac{x}{x-y} - \frac{y}{x+y} - \frac{2xy}{(x-y)(x+y)}$

$\frac{(x-y)-2xy}{(x-y)(x+y)}$

$\frac{-2xy}{(x+y)}$

correcT?

NO, see this:

$\frac{x}{x-y} - \frac{y}{x+y} - \frac{2xy}{x^2-y^2}$

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

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

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

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

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

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

$=\frac{x-y}{x+y}$

jvignacio · #3 $Old$ 11-19-2008, 10:26 PM

Quote:

Originally Posted by Shyam $View Post$

NO, see this:

$\frac{x}{x-y} - \frac{y}{x+y} - \frac{2xy}{x^2-y^2}$

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

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

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

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

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

$=\frac{(x-y)^2}{(x-y)(x+y)}$ <<------ isnt the numerator ment to be (x+y)^2 ?

$=\frac{x-y}{x+y}$

so the last fraction you dont cross multiply that one? -2xy/(x-y)(x+y)