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Math Help - Transformation using matrices (algebra help)

  1. #1
    RAz
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    Transformation using matrices (algebra help)

    I'm stuck on simplifying this:

    \frac{-1}{2}x' + \frac{\sqrt3}{2}y' = \frac{-\sqrt3}{2}x' - \frac{1}{2}y' + 4

    The negative is apart of the entire fraction, not just the numerator.

    I think I have to apply the distributive law and rationalise the denominator, but I've never been any good at fractions to be honest.

    (This is for matrix rotation, if you are curious)
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  2. #2
    Super Member Gamma's Avatar
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    Quote Originally Posted by RAz View Post
    I'm stuck on simplifying this:

    \frac{-1}{2}x' + \frac{\sqrt3}{2}y' = \frac{-\sqrt3}{2}x' - \frac{1}{2}y' + 4

    The negative is apart of the entire fraction, not just the numerator.

    I think I have to apply the distributive law and rationalise the denominator, but I've never been any good at fractions to be honest.

    (This is for matrix rotation, if you are curious)
    \frac{-1}{2}x' + \frac{\sqrt3}{2}y' = \frac{-\sqrt3}{2}x' - \frac{1}{2}y' + 4


    \frac{-1}{2}x' + \frac{\sqrt3}{2}x'  = - \frac{1}{2}y' -\frac{\sqrt3}{2}y' + 4


    \frac{\sqrt3-1}{2}x'  =  -\frac{\sqrt3+1}{2}y' + 4

    \frac{2}{\sqrt3-1}\frac{\sqrt3-1}{2}x'  =  \frac{2}{\sqrt3-1}\left(-\frac{\sqrt3+1}{2}y' + 4 \right)

    x'  =  -\frac{\sqrt3+1}{\sqrt3 - 1}y' + \frac{8}{\sqrt3 - 1}

    rationalize denominator:

    x'  =  -\frac{4+2\sqrt3}{4}y' + \frac{8\sqrt3+8}{4} = -(1 + \frac{\sqrt3}{2})y' + 2(\sqrt3 +1)
    Last edited by Gamma; May 8th 2009 at 11:35 PM. Reason: forgot to rationalize denominator for you
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  3. #3
    RAz
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    Thankyou so much!
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