\( x+y=12 \) \( x-y=10 \) What is the value of the x-determinant for the system shown? -22 -12 -2

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+y=12\\x-y=10\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=12-y\\x-y=10\end{array}\right.\) - step2: Substitute the value of \(x:\) \(12-y-y=10\) - step3: Subtract the terms: \(12-2y=10\) - step4: Move the constant to the right side: \(-2y=10-12\) - step5: Subtract the numbers: \(-2y=-2\) - step6: Change the signs: \(2y=2\) - step7: Divide both sides: \(\frac{2y}{2}=\frac{2}{2}\) - step8: Divide the numbers: \(y=1\) - step9: Substitute the value of \(y:\) \(x=12-1\) - step10: Calculate: \(x=11\) - step11: Calculate: \(\left\{ \begin{array}{l}x=11\\y=1\end{array}\right.\) - step12: Check the solution: \(\left\{ \begin{array}{l}x=11\\y=1\end{array}\right.\) - step13: Rewrite: \(\left(x,y\right) = \left(11,1\right)\) To find the value of the x-determinant for the given system of equations, we first need to solve the system: 1. The equations are: \[ x + y = 12 \quad (1) \] \[ x - y = 10 \quad (2) \] 2. We can solve these equations simultaneously. From equation (1), we can express \( y \) in terms of \( x \): \[ y = 12 - x \quad (3) \] 3. Now, substitute equation (3) into equation (2): \[ x - (12 - x) = 10 \] Simplifying this gives: \[ x - 12 + x = 10 \] \[ 2x - 12 = 10 \] \[ 2x = 22 \] \[ x = 11 \] 4. Now, substitute \( x = 11 \) back into equation (3) to find \( y \): \[ y = 12 - 11 = 1 \] Thus, the solution to the system is \( (x, y) = (11, 1) \). Now, to find the x-determinant, we can use the coefficients of the equations. The determinant \( D \) for the system of equations in the form \( ax + by = c \) is given by: \[ D = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} \] Calculating this determinant: \[ D = (1)(-1) - (1)(1) = -1 - 1 = -2 \] Therefore, the value of the x-determinant for the system is \(-2\).