Answer: \((x, y) = (3, -2)\)
In the form \(a_1x + b_1y = c_1\), \(a_2x + b_2y = c_2\):\[x + 2y = -1\]\[2x - 3y = 12\]
\(D = \begin{vmatrix} 1 & 2 \\ 2 & -3 \end{vmatrix}\)\[D = 1\times (-3) - 2\times 2 = -7\]
\(D_x = \begin{vmatrix} -1 & 2 \\ 12 & -3 \end{vmatrix}\)\[D_x = (-1)\times (-3) - 2\times 12 = -21\]
\(D_y = \begin{vmatrix} 1 & -1 \\ 2 & 12 \end{vmatrix}\)\[D_y = 1\times 12 - (-1)\times 2 = 14\]
By Cramer's rule:\[x = \frac{D_x}{D} = \frac{-21}{-7} = 3\]\[y = \frac{D_y}{D} = \frac{14}{-7} = -2\]