## Precalculus (6th Edition) Blitzer

$\left( x,y \right)=\left( 4,\frac{1}{3} \right)$.
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, $\left( x,y \right)=\left( 4,\frac{1}{3} \right)$