Answer
$(x,y)=(2,1)$
Work Step by Step
This problem is connected to exercise $36$, as the coefficient matrix of the system
$\left\{\begin{array}{l}{b x+3 y=14 b+3}\\{bx+2y=2b+2}\end{array}\right., b\neq 0$
is
$A=\left[\begin{array}{ll}{b}&{3}\\{b}&{2}\end{array}\right]$,
and we found
$A^{-1}=\displaystyle \left[\begin{array}{rr}{-\dfrac{2}{b}}&{\displaystyle \frac{3}{b}}\\{1}&{-1}\end{array}\right]$
Writing the system in matrix form, $AX=B,$ the solution is $X=A^{-1}B$
$\left[\begin{array}{l}
x\\
y
\end{array}\right]=\left[\begin{array}{rr}
{-\displaystyle \frac{2}{b}}&{\displaystyle \frac{3}{b}}\\
{1}&{-1}\end{array}\right] \left[\begin{array}{l}
2b+3\\
2b+2
\end{array}\right]$
$\left[\begin{array}{l}
x\\
y
\end{array}\right]=\left[\begin{array}{l}
-\frac{2}{b}(2b+3)+\frac{3}{b}(2b+2)\\
2b+3-(2b+2)
\end{array}\right]=\left[\begin{array}{l}
-4-\frac{6}{b}+6+\frac{6}{b}\\
1
\end{array}\right]=\left[\begin{array}{l}
2\\
1
\end{array}\right]$
Solution: $(x,y)=(2,1)$
