Answer
$X=\left[ \begin{matrix}
5 & 17 \\
-5 & \frac{45}{2} \\
-11 & -2 \\
\end{matrix} \right]$
Work Step by Step
Consider the matrix equation, $2X+5A=B$.
This implies that, $2X=B-5A$ or $X=\frac{1}{2}\left[ B-5A \right]$.
Now we will consider,
$\begin{align}
& X=\frac{1}{2}\left[ B-5A \right] \\
& =\frac{1}{2}\left( \left[ \begin{matrix}
-5 & -1 \\
0 & 0 \\
3 & -4 \\
\end{matrix} \right]-5\left[ \begin{matrix}
-3 & -7 \\
2 & -9 \\
5 & 0 \\
\end{matrix} \right] \right) \\
& =\frac{1}{2}\left( \left[ \begin{matrix}
-5 & -1 \\
0 & 0 \\
3 & -4 \\
\end{matrix} \right]-\left[ \begin{matrix}
-15 & -35 \\
10 & -45 \\
25 & 0 \\
\end{matrix} \right] \right) \\
& =\frac{1}{2}\left[ \begin{matrix}
10 & 34 \\
-10 & 45 \\
-22 & -4 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
5 & 17 \\
-5 & \frac{45}{2} \\
-11 & -2 \\
\end{matrix} \right]
\end{align}$
