\(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
\(\Rightarrow\frac{1+5y}{5x}=\frac{1+7y}{4x}=\frac{1+5y-1+7y}{\left(5x-4x\right)}=-\frac{2y}{x}\)
\(\frac{1+5y}{5}=-2y\)
Giải ra ta có : \(y=-\frac{1}{15}\)
\(\Leftrightarrow x=2\)
\(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
Ta có : \(\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
\(\Rightarrow\frac{\left(1+5y-1+7y\right)}{\left(5x-4x\right)}=\frac{-2y}{x}\)
\(\Rightarrow\frac{\left(1+5y\right)}{5}=-2y\)
Ta được \(y=\frac{-1}{15}\)
\(\Rightarrow x=2\)
\(\dfrac{1+5y}{5x}=\dfrac{1+7y}{4x}=\dfrac{1+5y-\left(1+7y\right)}{5x-4x}=\dfrac{1+5y-1-7y}{x}=\dfrac{-2y}{x}\Rightarrow y=\dfrac{-1}{15}\Rightarrow x=2\)
