ANSWER
[tex]X = \begin{bmatrix} 1 & 7/4 \\ 8&39/4\\ -4&23/2\end{bmatrix}[/tex]
EXPLANATION
Solve for x, noting that scalar multiplication on matrices is distributive:
[tex]\begin{aligned}
4X + 5A &= B \\
4X &= B + (-5A) \\
X &= \tfrac{1}{4}( B + (-5A) ) \\
X &= \tfrac{1}{4} B + (-\tfrac{5}{4}A)
\end{aligned}[/tex]
Note that
[tex]\begin{aligned} \tfrac{1}{4} B &= \frac{1}{4} \begin{bmatrix} 4 & -8 \\ 2 & -1 \\ 9 & 1 \end{bmatrix} \\ \\ &= \begin{bmatrix} (1/4) \cdot 4 & (1/4) \cdot-8 \\ (1/4) \cdot2 & (1/4)\cdot-1 \\ (1/4)\cdot9 & (1/4)\cdot1 \end{bmatrix} \\ \\ &= \begin{bmatrix} 1 & -2 \\ 1/2 & -1/4\\ 9/4 & 1/4\end{bmatrix} \end{aligned}[/tex]
and
[tex]\begin{aligned}
(-\tfrac{5}{4}A) &=
-\frac{5}{4}
\begin{bmatrix}
0 & -3 \\
-6 & -8 \\
5 & -9
\end{bmatrix}
\\ \\
&= \begin{bmatrix}
(-5/4)\cdot 0 &(-5/4)\cdot -3 \\
(-5/4)\cdot-6 &(-5/4)\cdot -8 \\
(-5/4)\cdot5 &(-5/4)\cdot -9
\end{bmatrix}
\\ \\
&= \begin{bmatrix}
0 & 15/4 \\
15/2 & 10 \\
-25/4 & 45/4
\end{bmatrix}
\end{aligned} [/tex]
Therefore, since we add corresponding entries with matrix addition:
[tex]\begin{aligned} X &= \tfrac{1}{4} B + (-\tfrac{5}{4}A) \\
&= \begin{bmatrix} 1 & -2 \\ 1/2 & -1/4\\ 9/4 & 1/4 \end{bmatrix} + \begin{bmatrix} 0 & 15/4 \\ 15/2 & 10 \\ -25/4 & 45/4 \end{bmatrix} \\
&= \begin{bmatrix} 1 & 7/4 \\ 8&39/4\\ -4&23/2\end{bmatrix}
\end{aligned}[/tex]