# \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}

• Nov 18th 2011, 09:49 PM
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/

$x^2+y^2+z^2=yz+\dfrac{8}{x}=2zx-\dfrac{2}{y}=3xy+\dfrac{18}{z}$
• Nov 18th 2011, 10:21 PM
Prove It
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:

1/

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

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

3/

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

Is this a question?
• Nov 18th 2011, 10:49 PM
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
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) ?