Ta có : \(\left\{{}\begin{matrix}\left(m+1\right)x+y=4\\mx+y=2m\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left(m+1\right)x+2m-mx=4\\y=2m-mx\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}mx+x+2m-mx=4\\y=2m-mx\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=4-2m\\y=2m-m\left(4-2m\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=4-2m\\y=2m-4m+2m^2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=4-2m\\y=2m^2-2m\end{matrix}\right.\)
- Ta có : \(x+y=2\)
=> \(4-2m+2m^2-2m=2\)
=> \(2m^2-4m+2=0\)
=> \(\left(2m-2\right)\left(m-1\right)=0\)
=> \(m-1=0\)
=> \(m=1\)
Vậy với m = 1 thì thỏa mãn điều kiện trên .