7. \( \left\{\begin{array}{l}3 x+2 y=6 \\ 2 x+y=0\end{array}\right. \)

To solve the system of equations, we can use the substitution or elimination method. Let's go with substitution! From the second equation \(2x + y = 0\), we can express \(y\) in terms of \(x\): \[ y = -2x \] Next, we substitute \(y = -2x\) into the first equation: \[ 3x + 2(-2x) = 6 \] This simplifies to: \[ 3x - 4x = 6 \] \[ -x = 6 \] Now, we solve for \(x\): \[ x = -6 \] Next, we substitute \(x = -6\) back into the expression for \(y\): \[ y = -2(-6) = 12 \] Thus, the solution to the system of equations is: \[ (x, y) = (-6, 12) \]