\(\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}\Leftrightarrow4x\left(1+5y\right)=5x\left(1+7y\right)\\ \Leftrightarrow4x+20xy=5x+35xy\\ \Leftrightarrow x+15xy=0\\ \Leftrightarrow x\left(15y+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\15y+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\y=-\dfrac{1}{15}\end{matrix}\right.\)
\(y=-\dfrac{1}{15}\Leftrightarrow\dfrac{1-3\cdot\dfrac{1}{15}}{12}=\dfrac{1-5\cdot\dfrac{1}{15}}{5x}\\ \Leftrightarrow\dfrac{\dfrac{4}{5}}{12}=\dfrac{\dfrac{2}{3}}{5x}\\ \Leftrightarrow\dfrac{4}{5}\cdot5x=12\cdot\dfrac{2}{3}=8\\ \Leftrightarrow4x=8\Leftrightarrow x=2\)
Vậy \(\left(x;y\right)=\left(2;-\dfrac{1}{15}\right)\)
