Write the system of equations in matrix form AX = B.

The given system of equations is:

$$4x + 3y = 18$$ $$5x - 2y = 11$$

This can be written as:

$$\begin{bmatrix} 4 & 3 \\ 5 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 18 \\ 11 \end{bmatrix}$$

Here, $A = \begin{bmatrix} 4 & 3 \\ 5 & -2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, and $B = \begin{bmatrix} 18 \\ 11 \end{bmatrix}$.

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

$$\det(A) = (4)(-2) - (3)(5)$$ $$\det(A) = -8 - 15$$ $$\det(A) = -23$$

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

$$A^{-1} = \frac{1}{-23} \begin{bmatrix} -2 & -3 \\ -5 & 4 \end{bmatrix}$$

$$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{-23} \begin{bmatrix} -2 & -3 \\ -5 & 4 \end{bmatrix} \begin{bmatrix} 18 \\ 11 \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{-23} \begin{bmatrix} (-2)(18) + (-3)(11) \\ (-5)(18) + (4)(11) \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{-23} \begin{bmatrix} -36 - 33 \\ -90 + 44 \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{-23} \begin{bmatrix} -69 \\ -46 \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{-69}{-23} \\ \frac{-46}{-23} \end{bmatrix}$$ $$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}$$

$$x = 3$$ $$y = 2$$

The final answer is $x=3, y=2$.