Solve the system of equations \[ \left\{\begin{array}{l} -8 x-8 y=-752 \\ -4 x-3 y=-296 \end{array}\right. \] by converting to a matrix equation and using the inverse of the coefficient matrix. \[ x= \] \( \square \) \[ y= \] \( \square \) Question Help: Video Submit Question

To solve the system of equations using matrices, we start by rewriting the equations in the form \(Ax = b\). Given the equations: 1. \(-8x - 8y = -752\) 2. \(-4x - 3y = -296\) We can express this in matrix form as: \[ \begin{bmatrix} -8 & -8 \\ -4 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -752 \\ -296 \end{bmatrix} \] Let \(A = \begin{bmatrix} -8 & -8 \\ -4 & -3 \end{bmatrix}\), \(x = \begin{bmatrix} x \\ y \end{bmatrix}\), and \(b = \begin{bmatrix} -752 \\ -296 \end{bmatrix}\). Next, we find the inverse of the matrix \(A\). To calculate the inverse of a 2x2 matrix, we use the formula: \[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] where \(a = -8\), \(b = -8\), \(c = -4\), and \(d = -3\). Calculating the determinant: \[ ad - bc = (-8)(-3) - (-8)(-4) = 24 - 32 = -8 \] Now, applying the inverse formula: \[ A^{-1} = \frac{1}{-8} \begin{bmatrix} -3 & 8 \\ 4 & -8 \end{bmatrix} = \begin{bmatrix} \frac{3}{8} & -1 \\ -\frac{1}{2} & 1 \end{bmatrix} \] Now, to find \(x\), we multiply the inverse of \(A\) with \(b\): \[ \begin{bmatrix} x \\ y \end{bmatrix} = A^{-1} b = \begin{bmatrix} \frac{3}{8} & -1 \\ -\frac{1}{2} & 1 \end{bmatrix} \begin{bmatrix} -752 \\ -296 \end{bmatrix} \] Calculating the first row: \[ x = \frac{3}{8} \cdot (-752) - 1 \cdot (-296) = -282 + 296 = 14 \] Calculating the second row: \[ y = -\frac{1}{2} \cdot (-752) + 1 \cdot (-296) = 376 - 296 = 80 \] Thus, the solution to the system of equations is: \[ x = 14, \] \[ y = 80. \] So, the final answer is: \[ x = 14, \quad y = 80. \]