Answer
$X=\left[\begin{array}{ll}
4 & 7\\
7 & 4\\
2 & 2
\end{array}\right]$
Work Step by Step
Because D on the RHS is a 3$\times$2 matrix,
for the equation to be defined, the LHS must also be a a 3$\times$2 matrix.
If X is a 3$\times$2 matrix, then so is $X-C$, as is $5(X-C).$
A solution exists.
$ 5(X-C)=D \qquad$ ... multiply with the scalar $\displaystyle \frac{1}{5}$
$ X-C=\displaystyle \frac{1}{5}D \qquad$ ... add C to both sides
$X=\displaystyle \frac{1}{5}D+C=\frac{1}{5}\left[\begin{array}{ll}
10 & 20\\
30 & 20\\
10 & 0
\end{array}\right]+\displaystyle \left[\begin{array}{ll}
2 & 3\\
1 & 0\\
0 & 2
\end{array}\right]$
$=\left[\begin{array}{ll}
2+2 & 4+3\\
6+1 & 4+0\\
2+0 & 0+2
\end{array}\right]$
$X=\left[\begin{array}{ll}
4 & 7\\
7 & 4\\
2 & 2
\end{array}\right]$
