Answer
$(12,28)$
Work Step by Step
Solving $AX=B,\ \qquad X=A^{-1}B$.
This problem is connected to exercise $32$, as the coefficient matrix is
$A=\left[\begin{array}{rr}{3}&{-1}\\{-2}&{1}\end{array}\right]$, and $A^{-1}=\left[\begin{array}{ll}{1}&{1}\\{2}&{3}\end{array}\right]$
Writing the system $\left\{\begin{aligned}3x-y&=8\\-2x+y&=4\end{aligned}\right.$
in matrix form, $AX=B, $ the solution is
$X=A^{-1}B=\left[\begin{array}{ll}
{1}&{1}\\
{2}&{3}\end{array}\right]\left[\begin{array}{l}
8\\
4
\end{array}\right]$
$\left[\begin{array}{l}
x\\
y
\end{array}\right]=\left[\begin{array}{l}
8+4\\
16+12
\end{array}\right]=\left[\begin{array}{l}
12\\
28
\end{array}\right]$
Solution: $(x,y)=(12,28)$