\(\begin{cases}3x-2y=5\\3x+y=-1\end{cases}\\↔\begin{cases}3x+y-3y=5\\3x=-1-y\end{cases}\\↔\begin{cases}-1-3y=5\\x=\dfrac{-1-y}{3}\end{cases}\\↔\begin{cases}3y=-6\\x=\dfrac{-1-y}{3}\end{cases}\\↔\begin{cases}y=-2\\x=\dfrac{1}{3}\end{cases}\\Vậy\,\,(x;y)=(\dfrac{1}{3};-2)\)