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Math Help - Negative Exponents in Complex Fractions

  1. #1
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    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.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by kpatch60 View Post
    ((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}
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    Its question #2 on this page if looking at it in the image format would help. Sample test for LAVC's Pre-Calculus Diagnostic
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    so i have (1/x-1/y)/(1/x^2-1/y^2)/ From here should I try using conjugates or something?
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    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|
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    Quote Originally Posted by kpatch60 View Post
    ((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 }}
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    Quote Originally Posted by Plato View Post
    \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.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by kpatch60 View Post
    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}
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    Quote Originally Posted by Also sprach Zarathustra View Post
    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.
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    Quote Originally Posted by kpatch60 View Post
    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)}}
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  11. #11
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    Quote Originally Posted by kpatch60 View Post
    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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