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Math Help - Hard simultaneous equation.

  1. #1
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    Hard simultaneous equation.

    For the simultaneous equations ax+by=p and bx-ay=q, show that x=(ap+bq)/(a^2+b^2) and y=(bp-aq)/(a^2+b^2)
    I just can't figure out which method to use and how to do it. Because I've never done one without numbers before.
    Many thanks in advance.
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  2. #2
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    Do it the same way as you would with numbers...

    \displaystyle a\,x + b\,y = p
    \displaystyle b\,x - a\,y = q.

    Multiply equation 1 by \displaystyle b and multiply equation 2 by \displaystyle a to get

    \displaystyle ab\,x + b^2y = bp
    \displaystyle ab\,x - a^2y = aq.


    Now subtract equation 2 from equation 1 to eliminate the \displaystyle x term.
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  3. #3
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    Thanks for that. I ended up getting y=(aq-bp)/(a^2+b^2).
    But now how do I get the x value, I know I have to sub in the y but which equation do I sub it into?
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  4. #4
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    Either one, since the solution satisfies both equations at the same time.

    BTW you should have gotten \displaystyle y = \frac{bp-aq}{a^2 + b^2}.
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  5. #5
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    Ok, thanks so much.
    I was wondering why mine wasn't working.
    I got it all fixed up now.
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