## College Algebra (10th Edition)

$(-2,1)$
This problem is connected to exercise $35$, as the coefficient matrix of the system $\left\{\begin{array}{l}{2 x+y=-3}\\{ax+ay=-a}\end{array}\right., a\neq 0$ is $A=\left[\begin{array}{ll}{2}&{1}\\{a}&{a}\end{array}\right]$, and we found $A^{-1}=\displaystyle \left[\begin{array}{rr}{1}&{-\displaystyle \frac{1}{a}}\\{-1}&{\displaystyle \frac{2}{a}}\end{array}\right]$ Writing the system in matrix form, $AX=B,$ the solution is $X=A^{-1}B$ $\left[\begin{array}{l} x\\ y \end{array}\right]=\left[\begin{array}{rr} {1}&{-\displaystyle \frac{1}{a}}\\ {-1}&{\displaystyle \frac{2}{a}}\end{array}\right]\left[\begin{array}{l} -3\\ -a \end{array}\right]$ $\left[\begin{array}{l} x\\ y \end{array}\right]=\left[\begin{array}{l} -3+1\\ 3-2 \end{array}\right]=\left[\begin{array}{l} -2\\ 1 \end{array}\right]$ Solution: $(x,y)=(-2,1)$