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Math Help - Help with simultaneous equation

  1. #1
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    Help with simultaneous equation

    2y-3x=2 & 4y2-4xy=18x2=5
    Last edited by mr fantastic; January 23rd 2011 at 04:19 PM. Reason: Title.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by jimbo28 View Post
    2y-3x=2 & 4y2-4xy=18x2=5
    It is not very clear. Did you mean?:

    \begin{Bmatrix} 2y-3x=2\\4y^2-4xy-18x^2=5\end{matrix}


    Fernando Revilla
    Last edited by FernandoRevilla; January 23rd 2011 at 03:40 AM.
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  3. #3
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    Yes apart from 2y - 3x = 2. Thankyou ever so much if you can help me.

    Kind regards
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by jimbo28 View Post
    Yes apart from 2y - 3x = 2. Thankyou ever so much if you can help me.
    All right, I've corrected it. Then from the first equation:

    y=\dfrac{2+3x}{2}

    Now, substitute y in the second one and you'll obtain a quadratic equation on x .

    Fernando Revilla
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  5. #5
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    Thankyou Fernando, any chance I can have the answers with working outs?

    Kind regards in advance,
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  6. #6
    Super Member Quacky's Avatar
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    Quote Originally Posted by jimbo28 View Post
    Thankyou Fernando, any chance I can have the answers with working outs?

    Kind regards in advance,
    What have you tried? The next stage is to substitute y=\dfrac{2+3x}{2} into the second equation.
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  7. #7
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    Here is a suggestion.
    Note that 2y=3x+2.
    Write the other equation as (2y)^2-2x(2y)-18x^2=5.
    Now substitute for (2y).
    Last edited by Plato; January 23rd 2011 at 10:40 AM.
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  8. #8
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    Quote Originally Posted by jimbo28 View Post
    Thankyou Fernando, any chance I can have the answers with working outs?

    Kind regards in advance,
    Please show some effort. See rule #11: http://www.mathhelpforum.com/math-he...ng-151418.html.

    Post what you can do and make it clear where you get stuck.
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