# Thread: \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. ## Solve the following system of equations

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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

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$\displaystyle x^2+y^2+z^2=yz+\dfrac{8}{x}=2zx-\dfrac{2}{y}=3xy+\dfrac{18}{z}$

2. ## 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

Originally Posted by mathlinks
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$\displaystyle \begin{cases} & 6x^4-(x^2-x)y^2-(y+12)x^2=-6 \\ &5x^4-(x^2-1)y^2-11x^2=-5 \end{cases}$

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

3/

$\displaystyle x^2+y^2+z^2=yz+\dfrac{8}{x}=2zx-\dfrac{2}{y}=3xy+\dfrac{18}{z}$
Is this a question?

3. ## 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

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

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

the solution is (x,y) ?