Answer
The required solution is $\left\{ \left( -3,2 \right),\left( 3,-2 \right) \right\}$
Work Step by Step
The given equations are,
$\begin{align}
& 4{{x}^{2}}+xy=30\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\
& {{x}^{2}}+3xy=-9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right)
\end{align}$
By multiplying equation (I) by $-2$ and adding equation (I) and equation (II): we get,
$\begin{align}
& -12{{x}^{2}}-3xy=-90 \\
& \underline{{{x}^{2}}+3xy=-9} \\
& -11{{x}^{2}}=99 \\
& {{x}^{2}}=9 \\
\end{align}$
Now,
$ x=\pm 3$
Substitute $ x=-3$ in equation (II):
$\begin{align}
& {{\left( -3 \right)}^{2}}+3\left( -3 \right)y=-9 \\
& 9-9y=-9 \\
& -9y=-18 \\
& y=2
\end{align}$
Substitute $ x=3$ in equation (II):
$\begin{align}
& {{\left( 3 \right)}^{2}}+3\left( 3 \right)y=-9 \\
& 9+9y=-9 \\
& 9y=-18 \\
& y=-2
\end{align}$
Thus, the solution is $\left\{ \left( -3,2 \right),\left( 3,-2 \right) \right\}$.
