1: Thay m=1 và n=5 vào hệ, ta được:
\(\left\{{}\begin{matrix}x+\left(1+1\right)y=5\\\left(5-2\right)x-2y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+2y=5\\3x-2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2y+3x-2y=5-1\\x+2y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x=4\\x+2y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\2y=5-x=5-1=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
2: Thay x=-1 và y=2 vào hệ, ta được:
\(\left\{{}\begin{matrix}-1+2\left(m+1\right)=5\\\left(n-2\right)\cdot\left(-1\right)-2\cdot2=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\left(m+1\right)=6\\-\left(n-2\right)=-1+4=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m+1=3\\n-2=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=2\\n=-1\end{matrix}\right.\)
