\(\left\{{}\begin{matrix}\left(x-1\right)\left(y+3\right)=xy+27\\\left(x-2\right)\left(y+1\right)-xy=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy+3x-y-3=xy+27\\xy+x-2y-2-xy=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-y=30\\x-2y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-30\\x-2y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-30\\x-2\left(3x-30\right)=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-30\\x-6x+60=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-30\\-5x=-50\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=3.10-30\\x=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=10\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm là: \(\left\{{}\begin{matrix}y=0\\x=10\end{matrix}\right.\)