\(\left\{{}\begin{matrix}\sqrt{x+y+1}=\sqrt{2x+y+3}\\x\left(y+1\right)-4\left(x+y\right)+54=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+1=2x+y+3\\x\left(y+1\right)-4\left(x+y\right)+54=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\-2\left(y+1\right)-4\left(-2+y\right)+54=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=10\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(-2;10\right)\)