\(\left\{{}\begin{matrix}x-y=m\\2x+y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x-m\\2x+x-m=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=x-m\\3x=4+m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x-m\\x=\frac{4+m}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{4-2m}{3}\\x=\frac{4+m}{3}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm (x,y) = \(\left(\frac{4+m}{3};\frac{4-2m}{3}\right)\)