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Math Help - Stumped

  1. #1
    Newbie
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    Aug 2007
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    Stumped

    Could someone pls show me why:

    (m/(n+m))*(V+mK) - mK

    is equivalent to:

    (m/(n+m))*(V-nK)

    Thx for the help
    Ben
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
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    Quote Originally Posted by NYKarl View Post
    Could someone pls show me why:

    (m/(n+m))*(V+mK) - mK

    is equivalent to:

    (m/(n+m))*(V-nK)

    Thx for the help
    Ben
    Let's play algebra.

    {(m/(n+m))*(V+mK) - mK} =? {(m/(n+m))*(V-nK)}

    Notice that [m / (m+n)] is common to both expressions. So we eliminate that to simplify the two expressions. Divide both sides by [m / (m+n)],

    (V +mK) -(mK / [m /(m+n)]) =? V -nK

    Then simplify. Expand the Lefthand Side,

    (V +mK) -(mK * [(m+n) / m]) =? V -nK
    (V +mK) -[(mK)(m+n) / m] =? V -nK
    Clear the fraction, multiply both sides by m,
    m(V +mK) -mK(m+n) =? mV -mnK
    mV +(m^2)K -(m^2)K -mnK =? mV -mnK
    mV -mnK =? mV -mnK
    Yes.

    So, it is shown.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by NYKarl View Post
    Could someone pls show me why:

    (m/(n+m))*(V+mK) - mK

    is equivalent to:

    (m/(n+m))*(V-nK)

    Thx for the help
    Ben
    <br />
\frac{m}{n+m}(V+mK) - mK = \frac{mV+m^2K-nmK-m^2K}{n+m} =\frac{mV-nmK}{n+m}=\frac{m}{n+m}(V+nK)<br />

    RonL
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