[tex]\begin{gathered} 2x+3y=-6 \\ 3x-4y=-12 \\ 2x=-6-3y \\ x=-\frac{6}{2}-\frac{3y}{2} \\ x=-3-\frac{3}{2}y \\ 3(-3-\frac{3}{2}y)-4y=-12 \\ -9-\frac{9y}{2}-4y=-12 \\ -\frac{9y}{2}-4y=-12+9 \\ \frac{-9y-8y}{2}=-3 \\ -\frac{17y}{2}=-3 \\ -17y=-6 \\ y=\frac{6}{17} \\ \\ 2x+3(\frac{6}{17})=-6 \\ 2x+\frac{18}{17}=-6 \\ 2x=-6-\frac{18}{17} \\ 2x=\frac{-102-18}{17} \\ 2x=\frac{-120}{17} \\ 34x=-120 \\ x=-\frac{120}{34} \\ x=-\frac{60}{17} \end{gathered}[/tex]
Therefore, it has exactly one solution.
