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Thread: help finding the common denominator with this rational expression

  1. #1
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    help finding the common denominator with this rational expression

    4xy/x^2-y^2 + x-y/x+y

    the answer is x+y/x-y

    im completely lost on this particular problem.

    if someone can point me in the correct direction here that would be great, somehow i got 4+x-y/(x-y)^2
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  2. #2
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    Quote Originally Posted by NecroWinter View Post
    4xy/x^2-y^2 + x-y/x+y

    the answer is x+y/x-y

    im completely lost on this particular problem.

    if someone can point me in the correct direction here that would be great, somehow i got 4+x-y/(x-y)^2
    note that the common denominator is $\displaystyle x^2-y^2 = (x+y)(x-y)$
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  3. #3
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    $\displaystyle \dfrac{4xy}{x^2-y^2} + \dfrac{x-y}{x+y}$

    $\displaystyle = \dfrac{4xy}{(x+y)(x-y)}+ \dfrac{x-y}{x+y}$

    $\displaystyle =\dfrac{4xy+(x-y)(x-y)}{(x+y)(x-y)}$

    $\displaystyle =\dfrac{4xy+(x-y)^2}{(x+y)(x-y)}$

    simplify the numerator..
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  4. #4
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    Quote Originally Posted by harish21 View Post
    $\displaystyle \dfrac{4xy}{x^2-y^2} + \dfrac{x-y}{x+y}$

    $\displaystyle = \dfrac{4xy}{(x+y)(x-y)}+ \dfrac{x-y}{x+y}$

    $\displaystyle =\dfrac{4xy+(x-y)(x-y)}{(x+y)(x-y)}$

    $\displaystyle =\dfrac{4xy+(x-y)^2}{(x+y)(x-y)}$

    simplify the numerator..
    im a little confused

    the only thing i would know to do is to start cancelling things out

    when i see (x-y)^2 i assume (x+y)(x-y) and i can cancel out the denom and be left with 4xy.
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  5. #5
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    Quote Originally Posted by NecroWinter View Post
    im a little confused

    the only thing i would know to do is to start cancelling things out

    when i see (x-y)^2 i assume (x+y)(x-y) and i can cancel out the denom and be left with 4xy.

    numerator ... $\displaystyle 4xy + (x-y)^2 = 4xy + (x^2 - 2xy + y^2) = x^2 + 2xy + y^2 = (x+y)^2$
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  6. #6
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    Quote Originally Posted by skeeter View Post
    numerator ... $\displaystyle 4xy + (x-y)^2 = 4xy + (x^2 - 2xy + y^2) = x^2 + 2xy + y^2 = (x+y)^2$
    thanks i get it now

    but i have one more small question

    when you have (x-y)^2 how do you know that it isnt a difference of two squares, but rather a perfect square trinomial?
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  7. #7
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    Quote Originally Posted by NecroWinter View Post
    thanks i get it now

    but i have one more small question

    when you have (x-y)^2 how do you know that it isnt a difference of two squares, but rather a perfect square trinomial?
    a difference of two squares is $\displaystyle x^2 - y^2$ , not $\displaystyle (x-y)^2$ , which is the square of a difference.
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