How to solve for unknown matrix? How can I solve this?
$$   \begin{bmatrix}
    1 & 1 \\
    1 & 2 \\
    \end{bmatrix}
X + 
\begin{bmatrix}
2 & -1\\
-1 & 1\\
\end{bmatrix}
X
\begin{bmatrix}
1 & 5 \\
1 & 2 \\
\end{bmatrix}
 = 
\begin{bmatrix}
1 & 1\\
1 & 1\\
\end{bmatrix}
$$
I know there's similar question like: 
Solve for unknown matrix.
But this one is much more complex as there are two separate terms with $X$.
Can I perform something similar like:
$Ax + Bx = C \implies (A+B)x = C$? But at the second term, the $X$ is at the middle and that order is important in matrix.
Any help would be appreciated!

If we can't see the trick suggested in the comments, by $X=\begin{bmatrix}
    a & b \\
    c & d \\
    \end{bmatrix}$ we obtain 
$$  \begin{bmatrix}
    1 & 1 \\
    1 & 2 \\
    \end{bmatrix}X= \begin{bmatrix}
    a+b & a+2b \\
    c+d & c+2d \\
    \end{bmatrix}$$ 
$$ 
\begin{bmatrix}
2 & -1\\
-1 & 1\\
\end{bmatrix}
X
\begin{bmatrix}
1 & 5 \\
1 & 2 \\
\end{bmatrix}
 =$$
$$=\begin{bmatrix}
2 & -1\\
-1 & 1\\
\end{bmatrix}
\begin{bmatrix}
a+b & 5a+2b\\
c+d & 5c+2d\\
\end{bmatrix}
=\begin{bmatrix}
2(a+b)-(c+d) & 2(5a+2b)-(5c+2d)\\
-(a+b)+(c+d) & -(5a+2b)+(5c+2d)\\
\end{bmatrix}$$
then the system
$$\begin{bmatrix}
3(a+b)-(c+d) & (11a+6b)-(5c+2d)\\
-(a+b)+2(c+d) & -(5a+2b)+(6c+4d)\\
\end{bmatrix}= \begin{bmatrix}
    1 & 1 \\
    1 & 1 \\
    \end{bmatrix}$$
