# Negative Exponents in Complex Fractions

• May 29th 2011, 04:41 PM
kpatch60
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.
• May 29th 2011, 04:45 PM
Also sprach Zarathustra
Quote:

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.

$x^{-1}=\frac{1}{x} \\ \text{and} \\ x^{-2}=\frac{1}{x^2}$
• May 29th 2011, 04:47 PM
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
• May 29th 2011, 04:50 PM
kpatch60
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, 04:51 PM
e^(i*pi)
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 $|x| \neq |y|$
• May 29th 2011, 04:54 PM
Plato
Quote:

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.

$\frac{{x^{ - 1} - y^{ - 1} }}{{x^{ - 2} - y^{ - 2} }} = \frac{{\frac{1}{x} - \frac{1}{y}}}{{\frac{1}{{x^2 }} - \frac{1}{{y^2 }}}} = \frac{{xy^2 - x^2 y}}{{y^2 - x^2 }}$
• May 29th 2011, 04:59 PM
kpatch60
Quote:

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.
• May 29th 2011, 05:00 PM
Also sprach Zarathustra
Quote:

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...

$\frac{x^{-1}-y^{-1}}{x^{-2}-y^{-2}}=\frac{\frac{1}{x}-\frac{1}{y}}{\frac{1}{x^2}-\frac{1}{y^2}}=\frac{\frac{y-x}{xy}}{\frac{y^2-x^2}{(xy)^2}}=\frac{(xy)^2(y-x)}{(xy)(y^2-x^2)}=\frac{(xy)(y-x)}{(x+y)(x-y)}=\frac{xy}{x+y}$
• May 29th 2011, 05:04 PM
kpatch60
Quote:

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.
• May 29th 2011, 05:08 PM
Plato
Quote:

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.

$\frac{{xy^2 - x^2 y}}{{y^2 - x^2 }} = \frac{{xy\left( {y - x} \right)}}{{\left( {y + x} \right)\left( {y - x} \right)}}$
• May 30th 2011, 01:10 AM
HallsofIvy
Quote:

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 $x^2y^2$.