Answer
$X=\left[\begin{array}{rr}
-2 & -\dfrac{7}{3}\\
\dfrac{1}{3} & \dfrac{1}{3}
\end{array}\right]$
Work Step by Step
$2A$ on the LHS is a 2$\times$2 matrix.
If X is a 2$\times$2 matrix, then the RHS also produces a 2$\times$2 matrix,
by which the equation is defined, and a solution exists.
$ 2A=B-3X\qquad$ ... add -B to both sides
$-B+2A=-3X \qquad$ ... multiply with the scalar $-\displaystyle \frac{1}{3}$
$\displaystyle \frac{1}{3}B-\frac{2}{3}A=X$
$X=\displaystyle \frac{1}{3}\left[\begin{array}{ll}
2 & 5\\
3 & 7
\end{array}\right]-\displaystyle \frac{2}{3}\left[\begin{array}{ll}
4 & 6\\
1 & 3
\end{array}\right]$
$=\left[\begin{array}{ll}
\frac{2}{3}-\frac{8}{3} & \frac{5}{3}-\frac{12}{3}\\
\frac{3}{3}-\frac{2}{3} & \frac{7}{3}-\frac{6}{3}
\end{array}\right]$
$X=\left[\begin{array}{rr}
-2 & -\dfrac{7}{3}\\
\dfrac{1}{3} & \dfrac{1}{3}
\end{array}\right]$
