If $2 \times \begin{bmatrix$}[r] x & 6 \ y & 8 \end{bmatrix} + 3 \times \begin{bmatrix}[r] 1 & -1 \ 0 & 2 \end{bmatrix} = 3 \times \begin{bmatrix}[r] 3 & 3 \ 4 & 6 \end{bmatrix}, find the values of x and y.

Solving,

$\Rightarrow 2 \times \begin{bmatrix$}[r] x & 6 \ y & 8 \end{bmatrix} + 3 \times \begin{bmatrix}[r] 1 & -1 \ 0 & 2 \end{bmatrix} = 3 \times \begin{bmatrix}[r] 3 & 3 \ 4 & 6 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2x & 12 \ 2y & 16 \end{bmatrix} + \begin{bmatrix}[r] 3 & -3 \ 0 & 6 \end{bmatrix} = \begin{bmatrix}[r] 9 & 9 \ 12 & 18 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2x + 3 & 12 + (-3) \ 2y + 0 & 16 + 6 \end{bmatrix} = \begin{bmatrix}[r] 9 & 9 \ 12 & 18 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2x + 3 & 9 \ 2y & 22 \end{bmatrix} = \begin{bmatrix}[r] 9 & 9 \ 12 & 18 \end{bmatrix} \\[1em]

From above equation we get,

⇒ 2x + 3 = 9 and 2y = 12

⇒ 2x = 6 and y = $\dfrac{12}{2}$

⇒ x = $\dfrac{6}{2}$ = 3 and y = 6.

Hence, x = 3 and y = 6.