\(\frac{2x+1}{7}=\frac{1}{y}\)\(\Leftrightarrow\left(2x+1\right)\times y=7\)
Ta có: \(7=1\times7=\left(-1\right)\times\left(-7\right)\)
*Nếu \(\left(2x+1\right)\times y=1\times7\Rightarrow\hept{\begin{cases}x=1\\y=7\end{cases}hoac}\hept{\begin{cases}x=4\\y=1\end{cases}}\)
*Nếu \(\left(2x+1\right)\times y=\left(-1\right)\times\left(-7\right)\Rightarrow\hept{\begin{cases}x=0\\y=-7\end{cases}hoac\hept{\begin{cases}x=-3\\y=-1\end{cases}}}\)