((x^-1)-(y^-1))/((x^-2)-(y^-2))? I know the answer is xy/x+y but I do not know how to get it.

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- May 29th 2011, 03:41 PMkpatch60Negative Exponents in Complex Fractions
((x^-1)-(y^-1))/((x^-2)-(y^-2))? I know the answer is xy/x+y but I do not know how to get it.

- May 29th 2011, 03:45 PMAlso sprach Zarathustra
- May 29th 2011, 03:47 PMkpatch60
Its question #2 on this page if looking at it in the image format would help. Sample test for LAVC's Pre-Calculus Diagnostic

- May 29th 2011, 03:50 PMkpatch60
so i have (1/x-1/y)/(1/x^2-1/y^2)/ From here should I try using conjugates or something?

- May 29th 2011, 03:51 PMe^(i*pi)
Use the difference of two squares on the denominator nothing that $\displaystyle x^{-2} = (x^{-1})^2$. If you've done it correctly you'll get a common factor on the top and bottom which can be cancelled since we know that $\displaystyle |x| \neq |y|$

- May 29th 2011, 03:54 PMPlato
- May 29th 2011, 03:59 PMkpatch60
- May 29th 2011, 04:00 PMAlso sprach Zarathustra
Alright, I'll play your game with you...

$\displaystyle \frac{x^{-1}-y^{-1}}{x^{-2}-y^{-2}}=\frac{\frac{1}{x}-\frac{1}{y}}{\frac{1}{x^2}-\frac{1}{y^2}}=\frac{\frac{y-x}{xy}}{\frac{y^2-x^2}{(xy)^2}}=\frac{(xy)^2(y-x)}{(xy)(y^2-x^2)}=\frac{(xy)(y-x)}{(x+y)(x-y)}=\frac{xy}{x+y}$ - May 29th 2011, 04:04 PMkpatch60
- May 29th 2011, 04:08 PMPlato
- May 30th 2011, 12:10 AMHallsofIvy