## Algebra and Trigonometry 10th Edition

$x=10, y=-12$
Add $4R_2$ to $R_1$, and $R_1$ to $R_2$ . We have: $\begin{bmatrix} 1 &8 & : -86 \\ 0 & 9 & : -108 \end{bmatrix}$ The operation $\dfrac{R_2}{9}$ gives: $\begin{bmatrix} 1 &8 & : -86 \\ 0 & 1 & : -12 \end{bmatrix}$ Write the system of equations corresponding to the augmented matrix as follows: $x+8y=-86 \\ y =-12$ and $x+8(-12)=-86 \implies x=10$ Therefore, $x=10, y=-12$