# Math Help - Negative Exponents in Complex Fractions

1. ## Negative 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.

2. Originally Posted by kpatch60
((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.
$x^{-1}=\frac{1}{x} \\ \text{and} \\ x^{-2}=\frac{1}{x^2}$

3. Its question #2 on this page if looking at it in the image format would help. Sample test for LAVC's Pre-Calculus Diagnostic

4. so i have (1/x-1/y)/(1/x^2-1/y^2)/ From here should I try using conjugates or something?

5. Use the difference of two squares on the denominator nothing that $x^{-2} = (x^{-1})^2$. 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 $|x| \neq |y|$

6. Originally Posted by kpatch60
((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.
$\frac{{x^{ - 1} - y^{ - 1} }}{{x^{ - 2} - y^{ - 2} }} = \frac{{\frac{1}{x} - \frac{1}{y}}}{{\frac{1}{{x^2 }} - \frac{1}{{y^2 }}}} = \frac{{xy^2 - x^2 y}}{{y^2 - x^2 }}$

7. Originally Posted by Plato
$\frac{{x^{ - 1} - y^{ - 1} }}{{x^{ - 2} - y^{ - 2} }} = \frac{{\frac{1}{x} - \frac{1}{y}}}{{\frac{1}{{x^2 }} - \frac{1}{{y^2 }}}} = \frac{{xy^2 - x^2 y}}{{y^2 - x^2 }}$
What did you do to get from the second to third part? This is how I was shown to do the problem originally, but I forget how the 1/x's were eliminated.

8. Originally Posted by kpatch60
Its question #2 on this page if looking at it in the image format would help. Sample test for LAVC's Pre-Calculus Diagnostic
Alright, I'll play your game with you...

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

9. Originally Posted by Also sprach Zarathustra
Alright, I'll play your game with you...

$\frac{x^{-1}-y^{-1}}{x^{-2}-y^{-2}}=\frac{\frac{1}{x}-\frac{1}{y}}{\frac{1}{x^2}-\frac{1}{y^2}}=\frac{\frac{y-x}{xy}}{\frac{y^2-x^2}{(xy)^2}}=\frac{(xy)^2(y-x)}{(xy)(y^2-x^2)}=\frac{(xy)(y-x)}{(x+y)(x-y)}=\frac{xy}{x+y}$
Thank you very much. I never thought to do the top and bottom as seperate problems with their own common denominators but looking at it this way made it so much more simple.

10. Originally Posted by kpatch60
What did you do to get from the second to third part? This is how I was shown to do the problem originally, but I forget how the 1/x's were eliminated.
$\frac{{xy^2 - x^2 y}}{{y^2 - x^2 }} = \frac{{xy\left( {y - x} \right)}}{{\left( {y + x} \right)\left( {y - x} \right)}}$

11. Originally Posted by kpatch60
What did you do to get from the second to third part? This is how I was shown to do the problem originally, but I forget how the 1/x's were eliminated.
Multiply both numerator and denominator by $x^2y^2$.