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Math Help - simultaneous equation

  1. #1
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    simultaneous equation

    Solve the simultaneous equation:

    x-2y+3=0

    x^2+xy-27=0
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  2. #2
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    Quote Originally Posted by mastermin346 View Post
    Solve the simultaneous equation:

    x-2y+3=0

    x^2+xy-27=0
    Dear mastermin346,

    Subject "y" in the first equation and substitute it in the second equation. Then you would have a quadratic equation.

    Hope this helps.
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  3. #3
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    Quote Originally Posted by mastermin346 View Post
    Solve the simultaneous equation:

    x-2y+3=0

    x^2+xy-27=0
    From the first equation, we can see that x = 2y - 3.

    Substituting into the second equation:

    (2y - 3)^2 + (2y - 3)y - 27 = 0

    4y^2 - 12y + 9 + 2y^2 - 3y - 27 = 0

    6y^2 - 15y - 18 = 0

    2y^2 - 5y - 6 = 0

    y^2 - \frac{5}{2}y - 3 = 0

    y^2 - \frac{5}{2}y + \left(-\frac{5}{4}\right)^2 - \left(-\frac{5}{4}\right)^2 - 3 = 0

    \left(y - \frac{5}{4}\right)^2 - \frac{25}{16} - \frac{48}{16} = 0

    \left(y - \frac{5}{4}\right)^2 - \frac{73}{16} = 0

    \left(y - \frac{5}{4}\right)^2 = \frac{73}{16}

    y - \frac{5}{4} = \pm \frac{\sqrt{73}}{4}

    y = \frac{5 \pm \sqrt{73}}{4}.


    So substitute y = \frac{5 - \sqrt{73}}{4} and y = \frac{5 + \sqrt{73}}{4}

    into x = 2y - 3

    to find the two values of x.
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