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Math Help - Simultaneous equations?

  1. #1
    Newbie SuumEorum's Avatar
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    Simultaneous equations?

    Hm... I've been asked to find x and y in terms of a, b, c, d, e, f. It's supposed to be 'simple', but I keep getting the wrong answer...I've tried equating the equations and multiplying out, but there're just too many variables, please help .

    y=a+\left( \frac{b+c-2d}{e+f-2d} \right)(x-d)

    y=e+\left( \frac{a+c-2e}{d+f-2e} \right)(x-e)

    The solution should be:

    x=\frac{d+e+f}{3}

    y=\frac{a+b+c}{3}

    Ah well, thanks in advance to whoever can be bothered to solve it .
    Last edited by SuumEorum; June 19th 2008 at 02:26 PM.
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by SuumEorum View Post
    Hm... I've been asked to find x and y in terms of a, b, c, d, e, f. It's supposed to be 'simple', but I keep getting the wrong answer...I've tried equating the equations and multiplying out, but there're just too many variables, please help .

    y=a+\left( \frac{b+c-2d}{e+f-2d} \right)(x-d)

    y=e+\left( \frac{a+c-2e}{d+f-2e} \right)(x-e)

    The solution should be:

    x=\frac{d+e+f}{3}

    y=\frac{a+b+c}{3}
    Just looking at the given solution, it seems that you have written the equations wrongly. I think they should be

    y=a+\left( \frac{b+c-2{\color{red}a}}{e+f-2d} \right)(x-d)

    y={\color{red}b}+\left( \frac{a+c-2{\color{red}b}}{d+f-2e} \right)(x-e)
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  3. #3
    Newbie SuumEorum's Avatar
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    England
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    Aah I don't believe it, I've been solving the wrong equations, no wonder they've been impossible. Thanks for spotting that for me .
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