\(\left\{{}\begin{matrix}\left(m-1\right)x+y=3m-4\\x+\left(m-1\right)y=m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m-1\right)x+y=3m-4\\\left(m-1\right)x+\left(m^2-2m+1\right)y=m^2-m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2-2m\right)y=m^2-4m-4\\x+\left(m-1\right)y=m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{\left(m-2\right)}{m}\\x=\frac{3m-2}{m}\end{matrix}\right.\)
Để \(x+y=2\Leftrightarrow\frac{4m-4}{m}=2\Leftrightarrow m=2\)