Answer
$$x=1, \quad y=\frac{3}{2}, \quad z=-\frac{1}{4}, \quad w=-1.$$
Work Step by Step
We have
\begin{align*}
&\left[\begin{array}{rrr}{w} & {x} \\ {y} & {x} \end{array}\right]=\left[\begin{array}{rrr}{-4} & {3} \\ {2} & {-1} \end{array}\right]+2\left[\begin{array}{rrr}{y} & {w} \\ {z} & {x} \end{array}\right]\\
\Longrightarrow &
\left[\begin{array}{rrr}{w} & {x} \\ {y} & {x} \end{array}\right]=\left[\begin{array}{rrr}{-4} & {3} \\ {2} & {-1} \end{array}\right]+\left[\begin{array}{rrr}{2y} & {2w} \\ {2z} & {2x} \end{array}\right]\\
\Longrightarrow &
\left[\begin{array}{rrr}{w} & {x} \\ {y} & {x} \end{array}\right]=\left[\begin{array}{rrr}{-4+2y} & {3+2w} \\ {2+2z} & {-1+2x} \end{array}\right]\\
\end{align*}
By comparing the corresponding entries of the matrices in both sides of the above equation, we get the following system of equations
\begin{align*}
w&=-4+2y\\
x&=3+2w\\
y&=2+2z\\
x&=-1+2x.
\end{align*}
From which, we have the solution
$$x=1, \quad y=\frac{3}{2}, \quad z=-\frac{1}{4}, \quad w=-1.$$
