Solve for x and y :

$\begin{bmatrix$}[r] x + y & x - 4 \end{bmatrix}\begin{bmatrix}[r] -1 & -2 \ 2 & 2 \end{bmatrix} = \begin{bmatrix}[r] -7 & -11 \end{bmatrix}

Given,

$\Rightarrow \begin{bmatrix$}[r] x + y & x - 4 \end{bmatrix}\begin{bmatrix}[r] -1 & -2 \ 2 & 2 \end{bmatrix} = \begin{bmatrix}[r] -7 & -11 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] (x + y) \times -1 + (x - 4) \times 2 & (x + y) \times -2 + (x - 4) \times 2 \end{bmatrix} = \begin{bmatrix}[r] -7 & -11 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] -x - y + 2x - 8 & -2x - 2y + 2x - 8 \end{bmatrix} = \begin{bmatrix}[r] -7 & -11 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] x - y - 8 & - 2y - 8 \end{bmatrix} = \begin{bmatrix}[r] -7 & -11 \end{bmatrix}

By definition of equality of matrices we get,

-2y - 8 = -11

⇒ -2y = -3

⇒ y = $\dfrac{3}{2}$.

x - y - 8 = -7

⇒ x - y = 1

⇒ x = 1 + y

⇒ x = $1 + \dfrac{3}{2} = \dfrac{5}{2}$.

Hence, $x = \dfrac{5}{2}, y = \dfrac{3}{2}.$