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

To find the relevant determinants and the solution to the system of equations, we first express the given equations in standard form and identify the coefficients. The system is: 1) \( 9x - y = -4 \) 2) \( -8x - 2y = 7 \) The determinant \(D\) for the coefficients of \(x\) and \(y\) is calculated as: \[ D = \begin{vmatrix} 9 & -1 \\ -8 & -2 \end{vmatrix} = 9 \cdot (-2) - (-1) \cdot (-8) = -18 - 8 = -26 \] Next, we calculate \(D_x\) (which replaces the first column with the constant terms): \[ D_x = \begin{vmatrix} -4 & -1 \\ 7 & -2 \end{vmatrix} = -4 \cdot (-2) - (-1) \cdot 7 = 8 + 7 = 15 \] Lastly, we calculate \(D_y\) (which replaces the second column with the constant terms): \[ D_y = \begin{vmatrix} 9 & -4 \\ -8 & 7 \end{vmatrix} = 9 \cdot 7 - (-4) \cdot (-8) = 63 - 32 = 31 \] So, the relevant determinants are \(D = -26\), \(D_x = 15\), and \(D_y = 31\). For the solution to the system: \[ x = \frac{D_x}{D} = \frac{15}{-26} \] Thus, the solution is: \[ x = -\frac{15}{26} \] Now, substituting back to find \(y\): Using the first equation: \[ 9\left(-\frac{15}{26}\right) - y = -4 \] \[ -\frac{135}{26} - y = -4 \] \[ -y = -4 + \frac{135}{26} \] \[ -y = -\frac{104}{26} + \frac{135}{26} \] \[ -y = \frac{31}{26} \] Thus, \[ y = -\frac{31}{26} \] In conclusion: Relevant determinants: \( -26, 15, 31 \) Solution to the system: \( x = -\frac{15}{26} \) \( y = -\frac{31}{26} \)