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Math Help - Another Algebraic Fraction

  1. #1
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    Another Algebraic Fraction

    Simplify:

    <br />
\frac{1}{x - y} + \frac{2 x - y}{x^2 - y^2}
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  2. #2
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    Quote Originally Posted by stellina_91 View Post
    Simplify:

    <br />
\frac{1}{x - y} + \frac{2 x - y}{x^2 - y^2}
    Note that x^2 - y^2 = (x-y)(x+y) ....
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  3. #3
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    Same with this one.
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  4. #4
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    Quote Originally Posted by stellina_91 View Post
    Same with this one.
    Ditto.
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  5. #5
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    <br />
\frac{1}{x - y} + \frac{2 x - y}{x^2 - y^2}<br />

    = <br />
\frac{1}{x - y} + \frac{2 x - y}{(x + y)(x - y)}<br />

    = <br />
\frac{x^2 - y^2}{x - y} + \frac{(2 x - y)(x - y)}{(x + y)(x - y)}<br />

    = <br />
\frac{x^2 - y^2 + 2 x^2 - 2 x y - x y - y^2}{(x - y)(x + y)(x - y)} <br />

    = <br />
\frac{x^2 - y^2 + 2 x^2 - 3 x y - y^2}{(x - y)(x + y)(x - y)} <br />

    = <br />
\frac{3 x^2 - 2 y^2 - 3 x y}{(x - y)(x + y)(x - y)} <br />
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  6. #6
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    Quote Originally Posted by stellina_91 View Post
    <br />
\frac{1}{x - y} + \frac{2 x - y}{x^2 - y^2}<br />

    = <br />
\frac{1}{x - y} + \frac{2 x - y}{(x + y)(x - y)}<br />

    = <br />
\frac{x^2 - y^2}{x - y} + \frac{(2 x - y)(x - y)}{(x + y)(x - y)}<br />
Mr F says: No. The common denominator is (x + y)(x - y). The next line is:

    {\color{red}\frac{x + y}{(x - y)(x+y)} + \frac{2x - y}{(x + y)(x-y)}}


    [snip]
    ..
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  7. #7
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    Why didn't you multiply 2x + y by x - y??

    Thanks.
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  8. #8
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    Quote Originally Posted by stellina_91 View Post
    Why didn't you multiply 2x + y by x - y??

    Thanks.
    I'm not sure what you're asking.

    You get each fraction over a common denominator. The common denominator is (x + y)(x - y). Only one fraction needs to be re-written so that it is has the common denominator. I did that. The other fraction already has the common denominator - so nothing gets done to it.
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  9. #9
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    It's ok, I understand now, thank for your help
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