Answer
The required solution is $\left\{ \left( -2,1 \right),\left( 2,-1 \right) \right\}$
Work Step by Step
The given equations are,
$\begin{align}
& 2{{x}^{2}}+xy=6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\
& {{x}^{2}}+2xy=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \\
\end{align}$
By multiplying equation (I) by $-2$ and adding equation (I) and equation (II): we get,
$\begin{align}
& -4{{x}^{2}}-2xy=-12 \\
& \underline{{{x}^{2}}+2xy=0} \\
& -3{{x}^{2}}=12 \\
& x=\pm 2 \\
\end{align}$
Substitute $x=-2$ in equation (II):
$\begin{align}
& {{\left( -2 \right)}^{2}}+2\left( -2 \right)y=0 \\
& 4-4y=0 \\
& y=1
\end{align}$
Substitute $x=2$ in equation (II):
$\begin{align}
& {{\left( 2 \right)}^{2}}+2\left( 2 \right)y=0 \\
& 4+4y=0 \\
& y=-1
\end{align}$
Thus, the solution is $\left\{ \left( -2,1 \right),\left( 2,-1 \right) \right\}$.
