Step-by-Step Solution

1. Find the inverse of matrix A:

   Given $A = \begin{bmatrix} 3 & 2 \\ 5 & -7 \end{bmatrix}$

   The determinant of A, $|A| = (3 \times -7) - (2 \times 5) = -21 - 10 = -31$

   The inverse of A, $A^{-1} = \frac{1}{|A|} \begin{bmatrix} -7 & -2 \\ -5 & 3 \end{bmatrix} = \frac{1}{-31} \begin{bmatrix} -7 & -2 \\ -5 & 3 \end{bmatrix} = \begin{bmatrix} \frac{7}{31} & \frac{2}{31} \\ \frac{5}{31} & \frac{-3}{31} \end{bmatrix}$

2. Express the system of equations in matrix form:

   The system of equations is: $3x + 5y = 11$ and $2x - 7y = -3$

   In matrix form, this is $AX = B$, where $A = \begin{bmatrix} 3 & 5 \\ 2 & -7 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, and $B = \begin{bmatrix} 11 \\ -3 \end{bmatrix}$

3. Solve for X using $X = A^{-1}B$:

   First, note that the matrix A here is the transpose of the matrix A given in the question. So, we need to find the inverse of this new A.

   Given $A = \begin{bmatrix} 3 & 5 \\ 2 & -7 \end{bmatrix}$

   The determinant of A, $|A| = (3 \times -7) - (5 \times 2) = -21 - 10 = -31$

   The inverse of A, $A^{-1} = \frac{1}{|A|} \begin{bmatrix} -7 & -5 \\ -2 & 3 \end{bmatrix} = \frac{1}{-31} \begin{bmatrix} -7 & -5 \\ -2 & 3 \end{bmatrix} = \begin{bmatrix} \frac{7}{31} & \frac{5}{31} \\ \frac{2}{31} & \frac{-3}{31} \end{bmatrix}$

   Now, $X = A^{-1}B = \begin{bmatrix} \frac{7}{31} & \frac{5}{31} \\ \frac{2}{31} & \frac{-3}{31} \end{bmatrix} \begin{bmatrix} 11 \\ -3 \end{bmatrix} = \begin{bmatrix} \frac{7 \times 11 + 5 \times -3}{31} \\ \frac{2 \times 11 + -3 \times -3}{31} \end{bmatrix} = \begin{bmatrix} \frac{77 - 15}{31} \\ \frac{22 + 9}{31} \end{bmatrix} = \begin{bmatrix} \frac{62}{31} \\ \frac{31}{31} \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$

4. State the solution:

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