Treat the matrices on the right side of each equation like you would a constant.
Let 2X + Y = A and 3X - 4Y = B.
Then you can eliminate Y by taking the sum
4A + B = 4 (2X + Y) + (3X - 4Y) = 11X
==> X = (4A + B)/11
Similarly, you can eliminate X by using
-3A + 2B = -3 (2X + Y) + 2 (3X - 4Y) = -11Y
==> Y = (3A - 2B)/11
It follows that
[tex]X=\dfrac4{11}\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\dfrac1{11}\begin{bmatrix}7&-10\\-7&11\end{bmatrix} \\\\ X=\dfrac1{11}\left(4\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\left(\begin{bmatrix}48&-12\\40&88\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\begin{bmatrix}55&-22\\33&99\end{bmatrix} \\\\ X=\begin{bmatrix}5&-2\\3&9\end{bmatrix}[/tex]
Similarly, you would find
[tex]Y=\begin{bmatrix}2&1\\4&4\end{bmatrix}[/tex]
You can solve the second system in the same fashion. You would end up with
[tex]P=\begin{bmatrix}2&-3\\0&1\end{bmatrix} \text{ and } Q=\begin{bmatrix}1&2\\3&-1\end{bmatrix}[/tex]
