The linear system is: \[ 2x + 5y = 1 \quad \text{(1)} \] \[ 3x + 2y = 7 \quad \text{(2)} \] This system can be represented in matrix form: \[ \begin{pmatrix} 2 & 5 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \end{pmatrix} \] Let \( A = \begin{pmatrix} 2 & 5 \\ 3 & 2 \end{pmatrix} \), \( \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix} \), and \( \mathbf{b} = \begin{pmatrix} 1 \\ 7 \end{pmatrix} \). The matrix equation is \( A\mathbf{x} = \mathbf{b} \). To find \( \mathbf{x} \), we compute \( A^{-1} \). The inverse of a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is \( \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \). For matrix \( A \), the determinant is \( \text{det}(A) = (2 \times 2) - (5 \times 3) = 4 - 15 = -11 \). Thus, \[ A^{-1} = \frac{1}{-11} \begin{pmatrix} 2 & -5 \\ -3 & 2 \end{pmatrix} = \begin{pmatrix} -\frac{2}{11} & \frac{5}{11} \\ \frac{3}{11} & -\frac{2}{11} \end{pmatrix} \] Now, we solve for \( \mathbf{x} \) by multiplying \( A^{-1} \) with \( \mathbf{b} \): \[ \mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix} -\frac{2}{11} & \frac{5}{11} \\ \frac{3}{11} & -\frac{2}{11} \end{pmatrix} \begin{pmatrix} 1 \\ 7 \end{pmatrix} \] Performing the multiplication: \[ \mathbf{x} = \begin{pmatrix} -\frac{2}{11}(1) + \frac{5}{11}(7) \\ \frac{3}{11}(1) - \frac{2}{11}(7) \end{pmatrix} = \begin{pmatrix} \frac{-2 + 35}{11} \\ \frac{3 - 14}{11} \end{pmatrix} = \begin{pmatrix} \frac{33}{11} \\ -\frac{11}{11} \end{pmatrix} = \begin{pmatrix} 3 \\ -1 \end{pmatrix} \] Therefore, the solution is \( (x, y) = (3, -1) \).