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Math Help - I would lkike a problem done and the steps explained (reducing rational expressions)

  1. #1
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    I would lkike a problem done and the steps explained (reducing rational expressions)

    (18u^6v^5+24u^3v^3)/42u^2v^5

    help??!
    I don't quite understand how to decide which elements to pull/ factor out etc.
    step by step explanation would be lovely!
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  2. #2
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    Quote Originally Posted by lindseyb View Post
    (18u^6v^5+24u^3v^3)/42u^2v^5

    help??!
    I don't quite understand how to decide which elements to pull/ factor out etc.
    step by step explanation would be lovely!
    Numerator:
    18=6\cdot 3

    24=6\cdot 4

    u^6=u^3\cdot u^3

    v^5=v^3\cdot v^2

    Denominator:
    42=6\cdot 7

    What do you think you should factor out?
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  3. #3
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    The highest common factor of the numerator is \displaystyle 6u^3v^3. After you have factorised the numerator, is there anything that cancels with the denominator?
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  4. #4
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    so I pull the 6 out front of them all, then the ^3 in the numerator?
    which would make it

    6(3u^3v^2)^3 6(4uv)3/ 6(7u^2v^5)?
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  5. #5
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    Quote Originally Posted by lindseyb View Post
    so I pull the 6 out front of them all, then the ^3 in the numerator?
    which would make it

    6(3u^3v^2)^3 6(4uv)3/ 6(7u^2v^5)?
    \displaystyle\frac{6u^3v^3(3u^3v^2+4)}{6u^2v^3(7)}
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  6. #6
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    Alrighty! That makes sooo much more sense!
    from here can I cancel the 6u^3 to become u?

    so the final answer would be

    u(3u^3v^2+4v^2)/ v^3(7)
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  7. #7
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    Quote Originally Posted by lindseyb View Post
    Alrighty! That makes sooo much more sense!
    from here can I cancel the 6u^3 to become u?

    so the final answer would be

    u(3u^3v^2+4v^2)/ v^3(7)
    Yes, what about the v?
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  8. #8
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    oh! Can I combine them??

    3v^2+4v^2?

    so
    u(3u^7v^2)/ v^3(7)
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  9. #9
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    Quote Originally Posted by lindseyb View Post
    oh! Can I combine them??

    3v^2+4v^2?

    so
    u(3u^7v^2)/ v^3(7)
    Quote Originally Posted by dwsmith View Post
    \displaystyle\frac{uv^3(3u^3v^2+4)}{v^3(7)}
    How about cancel them?
    Last edited by dwsmith; January 22nd 2011 at 07:26 PM. Reason: forgot u
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  10. #10
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    AH! I'm sorry, I forgot to rewrite the v^3 in the denominator....
    that makes sense
    so the final FINAL result is u(3u^3v^2+4)/ 7
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  11. #11
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    Quote Originally Posted by lindseyb View Post
    AH! I'm sorry, I forgot to rewrite the v^3 in the denominator....
    that makes sense
    so the final FINAL result is u(3u^3v^2+4)/ 7
    Correct.
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  12. #12
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    yay!! Thank you!
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  13. #13
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    one more??!

    if (x^2-4)/(x^2+4)
    am I allowed to cancel the x^2's?
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  14. #14
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    Quote Originally Posted by lindseyb View Post
    one more??!

    if (x^2-4)/(x^2+4)
    am I allowed to cancel the x^2's?
    No.

    It doesn't simplify.
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  15. #15
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    Alrighty, thank you.
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