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Thread: simultaneous equation

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    simultaneous equation

    Solve the simultaneous equation:

    $\displaystyle x-2y+3=0$

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

    $\displaystyle x-2y+3=0$

    $\displaystyle x^2+xy-27=0$
    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:

    $\displaystyle x-2y+3=0$

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

    Substituting into the second equation:

    $\displaystyle (2y - 3)^2 + (2y - 3)y - 27 = 0$

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

    $\displaystyle 6y^2 - 15y - 18 = 0$

    $\displaystyle 2y^2 - 5y - 6 = 0$

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

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

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

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

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

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

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


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

    into $\displaystyle x = 2y - 3$

    to find the two values of $\displaystyle x$.
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