\(\Leftrightarrow\left\{{}\begin{matrix}\left(2m+2\right)x+2y=8\\mx+2y=2m\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(m+2\right)x=-2m+8\\y=4-\left(m+1\right)x\end{matrix}\right.\)
Để pt có nghiệm \(\Leftrightarrow m\ne-2\)
Khi đó: \(\left\{{}\begin{matrix}x=\frac{-2m+8}{m+2}\\y=\frac{2m^2-2m}{m+2}\end{matrix}\right.\)
\(x+y=2\Leftrightarrow\frac{-2m+8}{m+2}+\frac{2m^2-2m}{m+2}=2\)
\(\Leftrightarrow2m^2-4m+8=2m+4\)
\(\Leftrightarrow2m^2-6m+4=0\Rightarrow\left[{}\begin{matrix}m=1\\m=2\end{matrix}\right.\)