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Math Help - Simplifying Rationals

  1. #1
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    Simplifying Rationals

    \frac {5x^{3a} + 5x^{2a} y^a}{x^{2a} - y^{2a}} = \frac {5x^{2a}}{x^a-y^a} right?


    \frac {m^3 + n^3}{mp-mq-np+nq} \div \frac {mn-m^2-n^2}{mp-mq+np-nq} \Longrightarrow \frac {(m+n)(m^2-mn+n^2)}{(p-q)(m-n)} \cdot \frac{(p-q)(m+n)}{mn-[(m+n)(m-n)]} \Longrightarrow \frac {m^2 -mn + n^2}{m-n} \cdot \frac {m+n}{mn-(m-n)} = \frac {m^3+n^3}{-m^2+m^2n-mn^2+n^2} Did I do this right?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by chrozer View Post
    \frac {5x^{3a} + 5x^{2a} y^a}{x^{2a} - y^{2a}} = \frac {5x^{2a}}{x^a-y^a} right?


    \frac {m^3 + n^3}{mp-mq-np+nq} \div \frac {mn-m^2-n^2}{mp-mq+np-nq} \Longrightarrow \frac {(m+n)(m^2-mn+n^2)}{(p-q)(m-n)} \cdot \frac{(p-q)(m+n)}{mn-[(m+n)(m-n)]} \Longrightarrow \frac {m^2 -mn + n^2}{m-n} \cdot \frac {m+n}{mn-(m-n)} = \frac {m^3+n^3}{-m^2+m^2n-mn^2+n^2} Did I do this right?
    THe first one is correct you factored the difference of squares on the bottom and factored out the common factor on top and cancelled right?
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    THe first one is correct you factored the difference of squares on the bottom and factored out the common factor on top and cancelled right?
    Yep. How about the 2nd one?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by chrozer View Post
    \frac {5x^{3a} + 5x^{2a} y^a}{x^{2a} - y^{2a}} = \frac {5x^{2a}}{x^a-y^a} right?


    \frac {m^3 + n^3}{mp-mq-np+nq} \div \frac {mn-m^2-n^2}{mp-mq+np-nq} \Longrightarrow \frac {(m+n)(m^2-mn+n^2)}{(p-q)(m-n)} \cdot \frac{(p-q)(m+n)}{mn-[(m+n)(m-n)]} \Longrightarrow \frac {m^2 -mn + n^2}{m-n} \cdot \frac {m+n}{mn-(m-n)} = \frac {m^3+n^3}{-m^2+m^2n-mn^2+n^2} Did I do this right?
    No...I think you did the second one incorrectly you cancelled m+n in the second things denominator but it was a binomial you cannot cancel this way you...I know you know that you just might have missed it.......for example

    \frac{a}{b}\cdot{b}=a

    but \frac{a}{b+a}\cdot{b}\ne\frac{a}{a} you cant cancel when theyre are terms like that
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    No...I think you did the second one incorrectly you cancelled m+n in the second things denominator but it was a binomial you cannot cancel this way you...I know you know that you just might have missed it.......for example

    \frac{a}{b}\cdot{b}=a

    but \frac{a}{b+a}\cdot{b}\ne\frac{a}{a} you cant cancel when theyre are terms like that
    What else can i cancel out other than the (p-q) ?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by chrozer View Post
    What else can i cancel out other than the (p-q) ?
    Nothing that I can see of
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  7. #7
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    Quote Originally Posted by Mathstud28 View Post
    Nothing that I can see of
    So would the answer equal \frac {(m+n)^2(m^2-mn+n^2)}{(m-n)(mn-m^2-n^2)} ?
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by chrozer View Post
    So would the answer equal \frac {(m+n)^2(m^2-mn+n^2)}{(p-q)(m-n)(mn-m^2-n^2)} ?
    I believe so...I could be wrong though
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  9. #9
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    Quote Originally Posted by Mathstud28 View Post
    I believe so...I could be wrong though
    NVM. I found the right answer. It is -\frac {(m+n)^2}{(m-n)}.
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  10. #10
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    Quote Originally Posted by chrozer View Post
    NVM. I found the right answer. It is -\frac {(m+n)^2}{(m-n)}.
    Ya that is correct.

    Mathstud28: You didnt notice m^2 -mn + n^2 = -( mn - m^2 - n^2) ??
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