Solve each system of equations by ELIMINATION. Clearly identify your solution. \( \begin{array}{ll}\text { 7. }\left\{\begin{array}{l}x-y=-10 \\ x-6 y=-25\end{array}\right. & \text { 8. } \\ 2 x+2 y=28 \\ 8 x-2 y=22\end{array} \)

To solve these systems of equations by elimination, let's tackle each system one by one. **For system 7:** \[ \begin{cases} x - y = -10 \quad (1) \\ x - 6y = -25 \quad (2) \end{cases} \] We'll eliminate \(x\). First, we can multiply equation (1) by 6: \[ 6(x - y) = 6(-10) \\ 6x - 6y = -60 \quad \text{(3)} \] Now we subtract equation (2) from equation (3): \[ (6x - 6y) - (x - 6y) = -60 - (-25) \\ 6x - x = -60 + 25 \\ 5x = -35 \\ x = -7 \] Substituting \(x = -7\) back into equation (1) to find \(y\): \[ -7 - y = -10 \\ -y = -3 \\ y = 3 \] Thus, the solution for system 7 is \( (x, y) = (-7, 3) \). --- **For system 8:** \[ \begin{cases} 2x + 2y = 28 \quad (1) \\ 8x - 2y = 22 \quad (2) \end{cases} \] Let’s simplify equation (1) by dividing everything by 2: \[ x + y = 14 \quad (3) \] Now, we can express \(y\) in terms of \(x\): \[ y = 14 - x \quad (4) \] Next, substitute equation (4) into equation (2): \[ 8x - 2(14 - x) = 22 \\ 8x - 28 + 2x = 22 \\ 10x - 28 = 22 \\ 10x = 50 \\ x = 5 \] Now substitute \(x = 5\) back into equation (4): \[ y = 14 - 5 = 9 \] Thus, the solution for system 8 is \( (x, y) = (5, 9) \). **Final Answers:** - System 7: \( (x, y) = (-7, 3) \) - System 8: \( (x, y) = (5, 9) \)