Select the correct answer. What is the solution to this system of equations? \( \begin{array}{l}5 x-6 y=12 \\ 2 x+3 y=-33 \\ \text { A. } \quad(-6,-7) \\ \text { B. }(-7,-6) \\ \text { C. }(-7,6) \\ \text { D. }(-6,7)\end{array} \)

To solve the system of equations, we'll use substitution or elimination. First, let's express both equations: 1) \( 5x - 6y = 12 \) 2) \( 2x + 3y = -33 \) We can solve for \(x\) or \(y\) in one equation and substitute it into the other. If we multiply the second equation by 2 (to make elimination easier), we can align the \(y\) terms for subtraction: \[ 2(2x + 3y) = 2(-33) \implies 4x + 6y = -66 \] Now, we have: 1) \( 5x - 6y = 12 \) 2) \( 4x + 6y = -66 \) Adding these two equations to eliminate \(y\): \[ (5x - 6y) + (4x + 6y) = 12 - 66 \implies 9x = -54 \implies x = -6 \] Now, we substitute \(x = -6\) back into one of the original equations to solve for \(y\): \[ 5(-6) - 6y = 12 \implies -30 - 6y = 12 \implies -6y = 42 \implies y = -7 \] So the solution is \((-6, -7)\). The correct answer is **A. (-6, -7)**. Isn't it fascinating how algebra works? By systematically eliminating variables, you can unveil the hidden solutions in just a few steps! Plus, it's like solving a puzzle where every piece fits into place perfectly. Always double-check your calculations, and you’ll avoid common mistakes like misplacing signs or losing track of operations. Happy solving!