Chapter 10 - Systems of Equations and Inequalities - Chapter Review - Review Exercises - Page 815: 30

$(2,3)$

Based on the given equations, we have $D=\begin{vmatrix} 2 & 3\\ 3 & -2 \end{vmatrix}=2(-2)-(3)(3)=-13$, Use the Cramer’s Rule, we have $D_x=\begin{vmatrix} 13 & 3\\ 0 & -2 \end{vmatrix}=13(-2)-(3)(0)=-26$, $D_y=\begin{vmatrix} 2 & 13\\ 3 & 0 \end{vmatrix}=2(0)-(13)(3)=-39$, Thus $x=\frac{D_x}{D}=2, y=\frac{D_y}{D}=3$ or $(2,3)$