Answer
$(9,23)$
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 $\left\{\begin{aligned}3x-y&=4\\-2x+y&=5\end{aligned}\right.$
in matrix form, $AX=B, $ the solution is
$X=A^{-1}B=\left[\begin{array}{rr}
{1}&{1}\\
{2}&{3}\end{array}\right]\left[\begin{array}{l}
4\\
5
\end{array}\right]$
$\left[\begin{array}{l}
x\\
y
\end{array}\right]=\left[\begin{array}{l}
4+5\\
8+15
\end{array}\right]=\left[\begin{array}{l}
9\\
23
\end{array}\right]$
Solution: $(x,y)=(9,23)$