\(\left\{{}\begin{matrix}\left(x+1\right)\left(y-1\right)=xy-1\\\left(x-3\right)\left(y-3\right)=xy-3\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}xy-x+y-1=xy-1\\xy-3x-3y+9=xy-3\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-x+y=0\\-3x-3y=-12\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-x+y=0\\x+y=4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}2y=4\\x+y=4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}y=2\\x+2=4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}y=2\\x=2\end{matrix}\right.\)
Vậy (2;2) là nghiệm