Answer
$X=\left[\begin{array}{rr}
-1 & -1/2\\
1 & 2
\end{array}\right]$
Work Step by Step
Observing the dimensions, B is a 2$\times$2 matrix, A is 2$\times$2 matrix.
For the LHS to be defined, X must be a 2$\times$2 matrix.
So, there is a solution.
$2X+A=B\qquad $subtract A from both sides
$2X=B-A\qquad $... muliply with the scalar $\displaystyle \frac{1}{2}$
$X=\displaystyle \frac{1}{2}(B-A)=\frac{1}{2}\left[\begin{array}{ll}
2-4 & 5-6\\
3-1 & 7-3
\end{array}\right]$
$X=\left[\begin{array}{ll}
-2/2 & -1/2\\
2/2 & 4/2
\end{array}\right]$
$X=\left[\begin{array}{ll}
-1 & -1/2\\
1 & 2
\end{array}\right]$
