\[ \begin{aligned} 9 x-y & =-4 \\ -8 x-2 y & =7 \end{aligned} \] Answer Attempt 1 out of 2 Write the three relevant determinants in any order: \( \square \) , \( \square \) , \( \square \) Solution to the system: \( x= \) \( \square \) \( y= \) \( \square \)

To find the relevant determinants for the system of equations, we start with the coefficients from the original equations. The equations can be represented in a matrix form as follows: \[ \begin{pmatrix} 9 & -1 \\ -8 & -2 \end{pmatrix} \] The determinants are calculated as follows: 1. The first determinant (D) is the main determinant of the coefficient matrix: \(D = \begin{vmatrix} 9 & -1 \\ -8 & -2 \end{vmatrix} = (9)(-2) - (-1)(-8) = -18 - 8 = -26\). 2. The second determinant (D_x) is found by replacing the first column with the outputs: \(D_x = \begin{vmatrix} -4 & -1 \\ 7 & -2 \end{vmatrix} = (-4)(-2) - (-1)(7) = 8 + 7 = 15\). 3. The third determinant (D_y) is done similarly by replacing the second column with outputs: \(D_y = \begin{vmatrix} 9 & -4 \\ -8 & 7 \end{vmatrix} = (9)(7) - (-4)(-8) = 63 - 32 = 31\). So, the three relevant determinants are: \(D = -26 \), \(D_x = 15\), \(D_y = 31\). Now for the solution of the system using Cramer's rule: \[ x = \frac{D_x}{D} = \frac{15}{-26} \quad \text{and} \quad y = \frac{D_y}{D} = \frac{31}{-26} \] Thus, the solution to the system is: \( x = -\frac{15}{26} \), \( y = -\frac{31}{26} \).