Solve using elimination, \[ \begin{array}{l}-8 x-6 y=12 \\ -6 x-6 y=18\end{array} \]

To solve the system of equations using elimination, we start with the two equations: 1) \(-8x - 6y = 12\) 2) \(-6x - 6y = 18\) We can eliminate \(y\) by subtracting the second equation from the first. First, let's multiply the second equation by 1 (to keep it the same) and align the terms: \[ \begin{array}{rcl} -8x - 6y & = & 12 \\ -6x - 6y & = & 18 \end{array} \] Subtract the second equation from the first: \[ (-8x - 6y) - (-6x - 6y) = 12 - 18 \] This simplifies to: \[ -8x + 6x - 6y + 6y = -6 \] which can be simplified further to: \[ -2x = -6 \] Dividing both sides by -2 gives: \[ x = 3 \] Now that we have the value for \(x\), we can substitute \(x = 3\) back into one of the original equations to find \(y\). Let’s use the second equation: \[ -6(3) - 6y = 18 \] This simplifies to: \[ -18 - 6y = 18 \] Add 18 to both sides: \[ -6y = 36 \] Dividing both sides by -6 gives: \[ y = -6 \] The solution to the system of equations is \(x = 3\) and \(y = -6\). Final answer: \( \boxed{(3, -6)} \)