Answer
$(\displaystyle \frac{1}{2},2)$
Work Step by Step
Solving $AX=B,\ \qquad X=A^{-1}B$.
This problem is connected to exercise $33$, as the coefficient matrix is
$A=\left[\begin{array}{ll}{6}&{5}\\{2}&{2}\end{array}\right]$, and $A^{-1}=\displaystyle \left[\begin{array}{rr}{1}&{-\displaystyle \frac{5}{2}}\\{-1}&{3}\end{array}\right]$
Writing the system $\left\{\begin{array}{l}{6 x+5 y=13}\\{2x+2y=5}\end{array}\right.$
in matrix form, $AX=B, $ the solution is
$X=A^{-1}B=\left[\begin{array}{rr}
{1}&{-5/2}\\
{-1}&{3}\end{array}\right]\left[\begin{array}{l}
13\\
5
\end{array}\right]$
$\left[\begin{array}{l}
x\\
y
\end{array}\right]=\left[\begin{array}{l}
13-25/2\\
-13+15
\end{array}\right]=\left[\begin{array}{l}
1/2\\
2
\end{array}\right]$
Solution: $(x,y)=(\displaystyle \frac{1}{2},2)$
