\(\Leftrightarrow\left\{{}\begin{matrix}14x+4y=50\\14x+7my=112\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}14x+4y=50\\\left(7m-4\right)y=62\end{matrix}\right.\)
Để hệ có nghiệm duy nhất \(\Rightarrow m\ne\frac{4}{7}\)
Khi đó: \(\left\{{}\begin{matrix}y=\frac{62}{7m-4}\\x=\frac{25-2y}{7}=\frac{25m-32}{7m-4}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\frac{25m-32}{7m-4}>0\\\frac{62}{7m-4}>0\end{matrix}\right.\) \(\Rightarrow m>\frac{32}{25}\)