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Math Help - Simultaneous Equation trouble

  1. #1
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    Sep 2009
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    Simultaneous Equation trouble

    Had a quick search but can't find what I need, so I'll make a new thread instead.
    I've got a quick problem here that I can't get my head around, and it's the only simultaneous equation out of 10 that I can't solve, I keep getting answers that satisfy one or the other!

    3x+2y=8 (1)
    4x-5y=6 (2)

    (solving x from 1)
    3x = 8-2y (subtract 2y from both sides)
    x = \frac{8-2y}{3} (divide by 3 to find x)

    (substituting x into into 2)
    4(\frac{8-2y}{3})-5y=6
    \frac{32-8y}{3}-5y=6 (expanded brackets)
    32-8y-15y=18 (multiply both sides by 3 to remove denominator)
    -23y=-14 (subtract 32 from both sides, -8y-15y=-23y)

    Which leaves me with y=\frac{14}{23} which of course then does not satisfy the equation very well... I've gone over my working and isolated the problem to this point, I hope it's not something glaring, but what have I done wrong?
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  2. #2
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    Joined
    Sep 2009
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    You did not do anything wrong.

    Using the first equation and your value y = \frac{14}{23}:

    3x + 2y = 8
    3x + 2(\frac{14}{23}) = 8
    3x + \frac{28}{23} = \frac{184}{23} (Multiplying and putting 8 in a different form)
    3x = \frac{156}{23}
    x = \frac{156}{69}
    x = \frac{52}{23} (Putting the factor into lowest terms)

    Substituting x = \frac{52}{23} and y = \frac{14}{23} into your equations gives the correct answers.

    So your problem was not something glaring, as there was no problem at all. If there is something not making sense to you still, let me know.
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