## Algebra and Trigonometry 10th Edition

$x=2; y=-3$
The system can be written as: $AX=B$ where, $B= \begin{bmatrix} x \\ y \end{bmatrix}$ Thus, $X=A^{-1} B = \begin{bmatrix} 2 & 1 \\ 9 & 5 \end{bmatrix} \begin{bmatrix} 13 \\ -24 \end{bmatrix} = \begin{bmatrix} (2) (13)+(1)(-24) \\ (9) (13) +(5)(-24) \end{bmatrix}$ So, $X= \begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 2 \\ -3 \end{bmatrix}$ Therefore, $x=2; y=-3$