## Precalculus (6th Edition) Blitzer

$\left( x,y \right)=\left( 1,-2 \right)$.
According to Cramer’s rule $x=\frac{{{D}_{x}}}{D}$, $y=\frac{{{D}_{y}}}{D}$ As \begin{align} & D=\left| \begin{matrix} 4 & 1 \\ 2 & -3 \\ \end{matrix} \right| \\ & {{D}_{x}}=\left| \begin{matrix} 2 & 1 \\ 8 & -3 \\ \end{matrix} \right| \\ & {{D}_{y}}=\left| \begin{matrix} 4 & 2 \\ 2 & 8 \\ \end{matrix} \right| \end{align} Calculate the given determinants. \begin{align} & D=\left| \begin{matrix} 4 & 1 \\ 2 & -3 \\ \end{matrix} \right| \\ & =-12-2 \\ & =-14 \end{align} \begin{align} & {{D}_{x}}=\left| \begin{matrix} 2 & 1 \\ 8 & -3 \\ \end{matrix} \right| \\ & =-6-8 \\ & =-14 \end{align} \begin{align} & {{D}_{y}}=\left| \begin{matrix} 4 & 2 \\ 2 & 8 \\ \end{matrix} \right| \\ & =32-4 \\ & =28 \end{align} Substitute the given values \begin{align} & x=\frac{{{D}_{x}}}{D} \\ & =\frac{-14}{-14} \\ & =1 \end{align} \begin{align} & y=\frac{{{D}_{y}}}{D} \\ & =\frac{28}{-14} \\ & =-2 \end{align} Therefore, $\left( x,y \right)=\left( 1,-2 \right)$