Answer
$X=\left[ \begin{matrix}
\frac{1}{3} & \frac{13}{3} \\
-\frac{4}{3} & 6 \\
-\frac{7}{3} & -\frac{4}{3} \\
\end{matrix} \right]$
Work Step by Step
Consider the matrix equation, $3X+2A=B$
This implies that, $3X=B-2A$ or $X=\frac{B-2A}{3}$.
Now we will consider,
$\begin{align}
& X=\frac{B-2A}{3} \\
& =\frac{1}{3}\left( \left[ \begin{matrix}
-5 & -1 \\
0 & 0 \\
3 & -4 \\
\end{matrix} \right]-2\left[ \begin{matrix}
-3 & -7 \\
2 & -9 \\
5 & 0 \\
\end{matrix} \right] \right) \\
& =\frac{1}{3}\left( \left[ \begin{matrix}
-5 & -1 \\
0 & 0 \\
3 & -4 \\
\end{matrix} \right]-\left[ \begin{matrix}
-6 & -14 \\
4 & -18 \\
10 & 0 \\
\end{matrix} \right] \right) \\
& =\frac{1}{3}\left( \left[ \begin{matrix}
1 & 13 \\
-4 & 18 \\
-7 & -4 \\
\end{matrix} \right] \right) \\
& =\left[ \begin{matrix}
\frac{1}{3} & \frac{13}{3} \\
-\frac{4}{3} & 6 \\
-\frac{7}{3} & -\frac{4}{3} \\
\end{matrix} \right]
\end{align}$
