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Math Help - `\begin{cases}&6x^4-(x^2-x)y^2-(y+12)x^2=-6 \\ &5x^4-(x^2-1)y^2-11x^2=-5 \end{cases}`

  1. #1
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    Solve the following system of equations

    1/

    6x^4-(x^2-x)y^2-(y+12)x^2=-6
    5x^4-(x^2-1)y^2-11x^2=-5

    2/
    ( x-2011 )( 2011+2012\sqrt[3]{y-2013} )=1
    ( \sqrt[3]{x-2010}) ( y-4024 )= 2012

    3/

    x^2+y^2+z^2=yz+\dfrac{8}{x}=2zx-\dfrac{2}{y}=3xy+\dfrac{18}{z}
    Last edited by mathlinks; November 18th 2011 at 10:10 PM.
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  2. #2
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    Re: `\begin{cases}&6x^4-(x^2-x)y^2-(y+12)x^2=-6 \\ &5x^4-(x^2-1)y^2-11x^2=-5 \end{cas

    Quote Originally Posted by mathlinks View Post
    1/

    \begin{cases} & 6x^4-(x^2-x)y^2-(y+12)x^2=-6 \\ <br />
  &5x^4-(x^2-1)y^2-11x^2=-5 \end{cases}

    2/
    \begin{cases}  ( x-2011 )( 2011+2012\sqrt[3]{y-2013}  )=1 \\ <br />
   ( \sqrt[3]{x-2010}) ( y-4024 )= 2012\end{cases}

    3/

    x^2+y^2+z^2=yz+\dfrac{8}{x}=2zx-\dfrac{2}{y}=3xy+\dfrac{18}{z}
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  3. #3
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    Re: `\begin{cases}&6x^4-(x^2-x)y^2-(y+12)x^2=-6 \\ &5x^4-(x^2-1)y^2-11x^2=-5 \end{cas

    Quote Originally Posted by Prove It View Post
    Is this a question?
    Solve the following system of equations:

    6x^4-(x^2-x)y^2-(y+12)x^2=-6
    and  5x^4-(x^2-1)y^2-11x^2=-5

    the solution is (x,y) ?
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