\(\dfrac{1+7y}{4x}=\dfrac{1+5y}{5x}=\dfrac{1+3y}{12}\\ \Rightarrow\dfrac{2\left(1+5y\right)}{2\cdot5x}=\dfrac{1+7y}{6x}\\ \Rightarrow\dfrac{2+10y}{10x}=\dfrac{1+7y}{4x}=\dfrac{1+3y}{6x}\)
Vậy \(\dfrac{1+3y}{12}=\dfrac{1+3y}{6x}\Rightarrow12=6x\Rightarrow x=2\)