Cộng hai pt , ta có:
\(\left(2m+1\right)x=m+3\)
\(\Leftrightarrow x=\dfrac{m+3}{2m+1}\)
Thay vào pt thứ 2, ta có:
\(y=m-\dfrac{m^2+3m}{2m+1}=\dfrac{m^2-2m}{2m+1}\)
\(\Rightarrow x+y=\dfrac{m^2-m+3}{2m+1}>0\)
\(\Rightarrow m^2-m+3>2m+1\)
\(\Leftrightarrow m^2-3m+2>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x< 1\\x>2\end{matrix}\right.\)