Solve the system of linear equations using matrices: x − 2y + 3 = 0 and 3y + 2x − 8 = 0.

The given equations are:

1. $x - 2y + 3 = 0 \implies x - 2y = -3$
2. $3y + 2x - 8 = 0 \implies 2x + 3y = 8$

$$\begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -3 \\ 8 \end{bmatrix}$$

Here, $A = \begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, and $B = \begin{bmatrix} -3 \\ 8 \end{bmatrix}$.

For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the determinant is $ad - bc$.

$$\det(A) = (1)(3) - (-2)(2) = 3 - (-4) = 3 + 4 = 7$$

The formula for the inverse of a $2 \times 2$ matrix is $A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

$$A^{-1} = \frac{1}{7} \begin{bmatrix} 3 & -(-2) \\ -2 & 1 \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 3 & 2 \\ -2 & 1 \end{bmatrix}$$

$$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 3 & 2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} -3 \\ 8 \end{bmatrix}$$

Perform the matrix multiplication:

$$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{7} \begin{bmatrix} (3)(-3) + (2)(8) \\ (-2)(-3) + (1)(8) \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{7} \begin{bmatrix} -9 + 16 \\ 6 + 8 \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 7 \\ 14 \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{7}{7} \\ \frac{14}{7} \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$$

Therefore, $x = 1$ and $y = 2$.

The solution to the simultaneous equations is:
$x = \boxed{1}$
$y = \boxed{2}$