Solve each system of equations by ELIMINATION. Clearly identify your solution \( \begin{array}{ll}x-y=-10 & \text { 8. } \\ 2 x+2 y=28 \\ x-6 y=-2 y=22\end{array} \)

To solve the given system of equations by elimination, we first rewrite them clearly: 1) \( x - y = -10 \) 2) \( 2x + 2y = 28 \) 3) \( x - 6y = -2 \) 4) \( y = 22 \) Now, we simplify the second equation by dividing everything by 2: 2) \( x + y = 14 \) Now we have: 1) \( x - y = -10 \) 2) \( x + y = 14 \) Next, we can eliminate \(y\) by adding these two equations together: \[ (x - y) + (x + y) = -10 + 14 \\ 2x = 4 \] Dividing by 2 gives us: \[ x = 2 \] Now we can substitute \(x = 2\) back into one of the original equations to find \(y\). Using equation 1: \[ 2 - y = -10 \\ -y = -12 \\ y = 12 \] Thus, the solution to the system is: \[ (x, y) = (2, 12) \] Now let's check if this solution satisfies all the equations: For the first equation: \(2 - 12 = -10\) (True) For the second equation: \(2 + 12 = 14\) (True) For the third equation: \(2 - 6(12) = 2 - 72 = -70\) (False, this contradicts) For the fourth equation: \(y = 22\) doesn't match as \(y = 12\) This indicates a possible error in how the equations were listed or accessed. However, the elimination process performed on the first three relevant equations was correct. Based on what was analyzed, the elimination yielded \(x = 2\) and \(y = 12\) for the first set, but \(y\) contradicts \(y = 22\). The system does not have a consistent solution as presented, and further examination of the equations would be needed to clarify. If you need more assistance on this topic, just let me know!