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Math Help - System solution

  1. #1
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    System solution

     x=zx+zy
     y=zx+2zy
     x^2+2xy+y^2=100
    Is this system solvable? I can solve for z in terms of x and y, but I can't separate x and y to get a solution. Its three unkowns and three equations, but I don't know if all three unkowns have to be in all three equations. Anyway, if I solve for z I get x^2+xy=y^2 but that's as far as I can get. Any suggestions?
    Thanks
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  2. #2
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    Re: System solution

    I'm showing no solutions.

    The 3rd equation gets you

    $$(x+y)^2=100 \Rightarrow \pm (x+y)=10$$

    $$(x=10-y) \vee (x=-(10+y))$$

    when you try substituting those into equations 1 and 2 the result is no solution.
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  3. #3
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    Re: System solution

    Hey, thanks for your time. I wrote it down wrong. It should be
    x=zx+zy

    y=zx+2zy
    x^2+2xy+2y^2=100
    Last edited by polarbear73; February 25th 2014 at 01:23 PM.
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  4. #4
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    Re: System solution

    Quote Originally Posted by polarbear73 View Post
     x=zx+zy
     y=zx+2zy
     x^2+2xy+y^2=100
    Is this system solvable? I can solve for z in terms of x and y, but I can't separate x and y to get a solution. Its three unkowns and three equations, but I don't know if all three unkowns have to be in all three equations. Anyway, if I solve for z I get x^2+xy=y^2 but that's as far as I can get. Any suggestions?
    Thanks
    $Case\ I:\ x \ne - y\ and\ x \ne - 2y$

    $x = zx + zy = z(x + y) \implies z = \dfrac{x}{x + y}.$

    $y = zx + 2zy = z(x + 2y) \implies z = \dfrac{y}{x + 2y}.$

    So $\dfrac{x}{x + y} = \dfrac{y}{x + 2y} \implies x^2 + 2xy = xy + y^2 \implies x^2 + xy - y^2 = 0 \implies$

    $x = \dfrac{- y \pm \sqrt{y^2 - 4 * 1 * y^2}}{2} = \dfrac{- y \pm y\sqrt{-3}}{2}.$ No real solution.

    $Case\ II:\ x = - y.$

    $x = - y \implies x^2 + 2xy + y^2 = x^2 + 2x(-x) + (-x)^2 = x^2 - 2x^2 + x^2 = 0 \ne 100 \implies x \ne - y.$ No solution.

    $Case\ III:\ x = - 2y.$

    $x = - 2y \implies y = zx + 2zy = z(x + 2y) = z * 0 = 0 \implies x = - 2 * 0 = 0 = y = - y,\ which\ is\ case\ II.$ No solution.

    It is an inconsistent system (at least in the reals).
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