## Precalculus (6th Edition) Blitzer

$(-2,3)$
Step 1. From the given system, we have $D=\begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix}=2+3=5$ Step 2. From the given system, we have $D_x=\begin{vmatrix} -5 & -1 \\ 0 & 2 \end{vmatrix}=-10+0=-10$ Step 3. From the given system, we have $D_y=\begin{vmatrix} 1 & -5 \\ 3 & 0 \end{vmatrix}=0+15=15$ Step 4. Using the Cramer’s Rule, we have $x=\frac{D_x}{D}=-2$ and $y=\frac{D_y}{D}=3$. Thus the solution to the system is $(-2,3)$