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Math Help - simultanious equations again

  1. #1
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    simultanious equations again

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

    By letting y=mx, show that 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
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    It's necessary because that's what you have to do in this task
    If you replace y with mx in both equations and solve for m, you will get the equation 7m^2 + 12 m - 4 which is quadratic. To solve a quadratic equation, you need to calculate its discriminant D = b^2 - 4ac where a=7, b=12, c=-4 are the coefficients of the equation. The solutions are m_{1}=\frac{-b + \sqrt{D}}{2a}, m_{2}=\frac{-b -\sqrt{D}}{2a}
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  3. #3
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    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: . \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]: . 4x^3 - 5xy \:=\:14

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

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

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

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

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

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


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

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