\[\left\{ \begin{matrix} \frac{1}{x} + \frac{1}{y} = 1\ \ \\ \frac{1}{2x} + \frac{2}{y} = 8 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[Пусть\ \ \ \frac{1}{2x} = a;\ \ \frac{1}{y} = b:\]
\[\left\{ \begin{matrix} 2a + b = 1\ \ | \cdot ( - 2) \\ a + 2b = 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} - 4a - 2b = - 2\ \ (1) \\ a + 2b = 8\ \ \ \ \ \ \ \ \ \ \ (2) \\ \end{matrix} \right.\ \]
\[\Longrightarrow (1) + (2):\]
\[- 3a = 6\]
\[a = - 2.\]
\[\left\{ \begin{matrix} a = - 2\ \ \ \ \ \ \ \\ 2a + b = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} a = - 2\ \ \ \ \ \ \ \\ b = 1 - 2a \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} a = - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ b = 1 - 2 \cdot ( - 2) \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\left\{ \begin{matrix} a = - 2 \\ b = 5\ \ \ \ \\ \end{matrix} \right.\ \]
\[Подставим:\ \]
\[\left\{ \begin{matrix} \frac{1}{2x} = - 2\ \ \ | \cdot 2x \\ \frac{1}{y} = 5\ \ \ \ \ \ \ | \cdot y\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} 1 = - 4x \\ 1 = 5y\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x = - \frac{1}{4} \\ y = \frac{1}{5}\text{\ \ \ } \\ \end{matrix} \right.\ \]
\[Ответ:\left( - \frac{1}{4};\frac{1}{5} \right).\]