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Math Help - Polynomials simultaneous solution

  1. #1
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    Polynomials simultaneous solution

    Solve the following simultaneous quadratic equation

    2x^2 +xy-2y^2=-2
    x^2-3xy-5y^2=5

    I have tried solving the equation by substituting y=mx after simplifying both equations above.

    I would like to check my answers.... working with latex is troublesome.. it would be great if you could just share the answers out...

    My answers are as follow..
    x= +/- \frac{4}{\sqrt 13}
    y= +/- \frac{3}{\sqrt 13}

    and

    x= +/- \frac{5}{\sqrt 351}
    y= +/- \frac{4}{3 \sqrt 39 }
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  2. #2
    MHF Contributor Swlabr's Avatar
    Joined
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    Quote Originally Posted by ecogreen View Post
    Solve the following simultaneous quadratic equation

    2x^2 +xy-2y^2=-2
    x^2-3xy-5y^2=5

    I have tried solving the equation by substituting y=mx after simplifying both equations above.

    I would like to check my answers.... working with latex is troublesome.. it would be great if you could just share the answers out...

    My answers are as follow..
    x= +/- \frac{4}{\sqrt 13}
    y= +/- \frac{3}{\sqrt 13}

    and

    x= +/- \frac{5}{\sqrt 351}
    y= +/- \frac{4}{3 \sqrt 39 }
    No, that is not correct. Simply substitute x= \frac{4}{\sqrt 13} into both equations and your y will be different both times. You can do the same for your other three x-terms and the same thing will happen.

    To solve the problem, get rid of the ugly xy term in the middle;

    3(2x^2 +xy-2y^2) + (x^2-3xy-5y^2)=3(-2) + 5
    \Rightarrow 7x^2-11y^2 = -1
    \Rightarrow 11y^2 - 7x^2 = 1.

    Now things get interesting. Have you ever heard of the Euclidean algorithm?

    (With respect to LaTeX, \pm gives you \pm which is neater than +/-.)
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