Negative Exponents in Complex Fractions

**kpatch60** · May 29th 2011, 03:41 PM

((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.

**Also sprach Zarathustra** · May 29th 2011, 03:45 PM

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.

$x^{-1}=\frac{1}{x} \\ \text{and} \\ x^{-2}=\frac{1}{x^2}$

**kpatch60** · May 29th 2011, 03:47 PM

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

**kpatch60** · May 29th 2011, 03:50 PM

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

**e^(i*pi)** · May 29th 2011, 03:51 PM

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 $|x| \neq |y|$

**Plato** · May 29th 2011, 03:54 PM

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.

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

**kpatch60** · May 29th 2011, 03:59 PM

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.

**Also sprach Zarathustra** · May 29th 2011, 04:00 PM

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...

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

**kpatch60** · May 29th 2011, 04:04 PM

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.

**Plato** · May 29th 2011, 04:08 PM

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.

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

**HallsofIvy** · May 30th 2011, 12:10 AM

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 $x^2y^2$ .

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