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Thread: Equation system

  1. #1
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    Equation system

    Solve the following system of equation for $\displaystyle x,y\in\mathbb{R}$:
    $\displaystyle x^3+4y=y^3+16x$
    $\displaystyle \frac {1+y^2}{1+x^2}=5$
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  2. #2
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    Quote Originally Posted by james_bond View Post
    Solve the following system of equation for $\displaystyle x,y\in\mathbb{R}$:
    $\displaystyle x^3+4y=y^3+16x$
    $\displaystyle \frac {1+y^2}{1+x^2}=5$
    $\displaystyle x^3-16x = y^3-4y$

    $\displaystyle x(x^2-16) = y(y^2-4)
    $

    $\displaystyle \frac {1+y^2}{1+x^2}=5$

    $\displaystyle 1+y^2 = 5+5x^2$

    $\displaystyle y^2 = 4+5x^2$


    $\displaystyle x(x^2-16) = y(5x^2)$

    $\displaystyle y = \frac{x^2-16}{5x}$


    $\displaystyle \frac{(x^2-16)^2}{25x^2} = 4 + 5x^2$

    $\displaystyle x^4 - 32x^2 + 256 = 100x^2 + 125x^4$

    $\displaystyle 0 = 124x^4 + 132x^2 - 256$

    $\displaystyle 0 = 31x^4 + 33x^2 - 64$

    $\displaystyle 0 = (x^2-1)(31x^2 + 64)$

    $\displaystyle 0 = (x-1)(x+1)(31x^2+64)$

    can you finish?
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