2) \(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}\)
Ta có: \(\frac{1+5y}{5x}=\frac{1+7y}{4x}=\frac{\left(1+5y-1+7y\right)}{5x-4x}=\frac{-2y}{x}\)
\(\Rightarrow\frac{1+5y}{5}=-2y\)
\(\Rightarrow y=-\frac{1}{15}\)
\(\Rightarrow x=2\)
\(\Rightarrow x+y=\frac{-1}{15}+2=\frac{29}{15}\)