Answer
$\left( x,y \right)=\left( 4,\frac{1}{3} \right)$.
Work Step by Step
According to Cramer’s rule
$x=\frac{{{D}_{x}}}{D}$, $y=\frac{{{D}_{y}}}{D}$
As
$\begin{align}
& D=\left| \begin{matrix}
2 & -9 \\
3 & -3 \\
\end{matrix} \right| \\
& {{D}_{x}}=\left| \begin{matrix}
5 & -9 \\
11 & -3 \\
\end{matrix} \right| \\
& {{D}_{y}}=\left| \begin{matrix}
2 & 5 \\
3 & 11 \\
\end{matrix} \right|
\end{align}$
Calculate the given determinants.
$\begin{align}
& D=\left| \begin{matrix}
2 & -9 \\
3 & -3 \\
\end{matrix} \right| \\
& =-6+27 \\
& =21
\end{align}$
$\begin{align}
& {{D}_{x}}=\left| \begin{matrix}
5 & -9 \\
11 & -3 \\
\end{matrix} \right| \\
& =-15+99 \\
& =84
\end{align}$
$\begin{align}
& {{D}_{y}}=\left| \begin{matrix}
2 & 5 \\
3 & 11 \\
\end{matrix} \right| \\
& =22-15 \\
& =7
\end{align}$
Substitute the given values
$\begin{align}
& x=\frac{{{D}_{x}}}{D} \\
& =\frac{84}{21} \\
& =4
\end{align}$
$\begin{align}
& y=\frac{{{D}_{y}}}{D} \\
& =\frac{7}{21} \\
& =\frac{1}{3}
\end{align}$
Therefore, $\left( x,y \right)=\left( 4,\frac{1}{3} \right)$
