Answer
$X=\left[ \begin{array}{*{35}{l}}
\frac{27}{2} & \frac{31}{2} \\
-4 & -18 \\
-\frac{29}{2} & 6 \\
\end{array} \right]$
Work Step by Step
Consider the matrix equation, $4A+3B=-2X$.
This implies that, $X=-\frac{1}{2}\left[ 4A+3B \right]$
Now we will consider,
$\begin{align}
& X=-\frac{1}{2}\left[ 4A+3B \right] \\
& =-\frac{1}{2}\left( 4\left[ \begin{matrix}
-3 & -7 \\
2 & -9 \\
5 & 0 \\
\end{matrix} \right]+3\left[ \begin{matrix}
-5 & -1 \\
0 & 0 \\
3 & -4 \\
\end{matrix} \right] \right) \\
& =-\frac{1}{2}\left( \left[ \begin{matrix}
-12 & -28 \\
8 & -36 \\
20 & 0 \\
\end{matrix} \right]+\left[ \begin{matrix}
-15 & -3 \\
0 & 0 \\
9 & -12 \\
\end{matrix} \right] \right) \\
& =-\frac{1}{2}\left[ \begin{array}{*{35}{l}}
-27 & -31 \\
8 & -36 \\
29 & -12 \\
\end{array} \right] \\
& =\left[ \begin{array}{*{35}{l}}
\frac{27}{2} & \frac{31}{2} \\
-4 & -18 \\
-\frac{29}{2} & 6 \\
\end{array} \right]
\end{align}$
