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Thread: simultanious equations again

  1. #1
    Junior Member
    Sep 2008

    simultanious equations again

    Consider the equations $\displaystyle 12x^2-4xy+11y^2=64$ and $\displaystyle 16^2x-9xy+11y^2=78$

    By letting $\displaystyle y=mx$, show that $\displaystyle 7m^2+12m-4=0$ and thus solve the equation

    okay, can you help me with this equation and also can someone tell me why it is neccesary to put in the form of y=mx?
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  2. #2
    Apr 2009
    It's necessary because that's what you have to do in this task
    If you replace $\displaystyle y$ with $\displaystyle mx$ in both equations and solve for $\displaystyle m$, you will get the equation $\displaystyle 7m^2 + 12 m - 4$ which is quadratic. To solve a quadratic equation, you need to calculate its discriminant $\displaystyle D = b^2 - 4ac$ where $\displaystyle a=7, b=12, c=-4$ are the coefficients of the equation. The solutions are $\displaystyle m_{1}=\frac{-b + \sqrt{D}}{2a}, m_{2}=\frac{-b -\sqrt{D}}{2a}$
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  3. #3
    Super Member

    May 2006
    Lexington, MA (USA)
    Hello, requal!

    That substitution is a useful method for solving. .Did you try it?

    We can, of course, solve it head-on if you prefer . . .

    Solve: .$\displaystyle \begin{array}{cccc}12x^2-4xy+11y^2&=&64 & {\color{blue}[1]} \\ 16x^2-9xy+11y^2&=&78 & {\color{blue}[2]} \end{array}$

    Subtract [2] - [1]: .$\displaystyle 4x^3 - 5xy \:=\:14$

    Solve for $\displaystyle y\!:\;\;y \;=\;\frac{4x^2-14}{5}\;\;{\color{blue}[3]}$

    Substitute into [1]: .$\displaystyle 12x^2 - 4x\left(\frac{4x^2-14}{5x}\right) + 11\left(\frac{4x^2-14}{5x}\right)^2 \:=\:64$

    . . which simplifies to: .$\displaystyle 9x^4 - 58x^2 + 49 \:=\:0$

    . . which factors: .$\displaystyle (x^2-1)(9x^2 - 49) \:=\:0$

    . . and has roots: .$\displaystyle x \;=\;\pm1,\:\pm\tfrac{7}{3}$

    Substitute into [3]: .$\displaystyle y \;=\;\mp2,\;\pm\tfrac{2}{3}$

    Answers: .$\displaystyle (\pm1,\:\mp2),\;\left(\pm\frac{7}{3},\:\pm\frac{2} {3}\right) $

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