Answer
$(2,1)$
Work Step by Step
We will write the system
$\left\{\begin{array}{r}{bx+3y=2b+3}\\{bx+2y=2b+2}\end{array}\right.$
in matrix form as: $AX=B$
where, $X=\left[\begin{array}{l}x_1\\x_2 \end{array}\right]$
We have:
$A=\left[\begin{array}{ll}{b}&{3}\\{b}&{2}\end{array}\right]$,
and its inverse is:
$A^{-1}=\left[\begin{array}{rr}{-2/b}&{3/b}\\{1}&{-1}\end{array}\right]$
Thus, the solution of the given matrix can be expressed as:
$X=A^{-1}B=\left[\begin{array}{rr}{-2/b}&{3/b}\\{1}&{-1}\end{array}\right]\left[\begin{array}{l}
2b+3\\2b+2\end{array}\right]$
$\left[\begin{array}{l}
x_1\\x_2 \end{array}\right]=\left[\begin{array}{l} -4-\dfrac{6}{b}+6+\dfrac{6}{b}\\
1 \end{array}\right]=\left[\begin{array}{l}
2\\1\end{array}\right]$
So, our solution is: $(x_1,x_2)=(2,1)$
