\(\hept{\begin{cases}2.\frac{1}{x}+5.\frac{1}{x+y}=2\\3.\frac{1}{x}+\frac{1}{x+y}=1,7\end{cases}}\)
Đặt \(\frac{1}{x}\)=a
\(\frac{1}{x+y}=b\)
ta có \(\hept{\begin{cases}2a+5b=2\\3a+b=1,7\end{cases}}\)
\(\hept{\begin{cases}a=\frac{1}{2}\\b=\frac{1}{5}\end{cases}}\)
=> \(\frac{1}{x}=\frac{1}{2}\Rightarrow x=2\)
\(\frac{1}{x+y}=\frac{1}{5}\)\(\Rightarrow x+y=5\)\(\Rightarrow y=3\)