Negative Exponents in Complex Fractions

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May 29th 2011, 03:41 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, 03:45 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, 03:47 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, 03:50 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, 03:51 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, 03:54 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, 03:59 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, 04:00 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, 04:04 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, 04:08 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, 12:10 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 $x^2y^2$ .