Решите систему уравнений: 1/x+2/y=11; 1/x-2/y=-1.

\[\left\{ \begin{matrix} \frac{1}{x} + \frac{2}{y} = 11 \\ \frac{1}{x} - \frac{2}{y} = - 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[Пусть\ \ \ \frac{1}{x} = a;\ \ \frac{2}{y} = b:\]

\[\left\{ \begin{matrix} a + b = 11\ \ \ (1) \\ a - b = - 1\ \ \ (2) \\ \end{matrix} \right.\ \]

\[(1) + (2) \Longrightarrow 2a = 10 \Longrightarrow a = 5\]

\[\left\{ \begin{matrix} a = 5\ \ \ \ \ \ \ \ \\ a - b = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} a = 5\ \ \ \ \ \ \ \ \\ b = a + 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} a = 5\ \ \ \ \ \ \ \ \\ b = 5 + 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} a = 5 \\ b = 6 \\ \end{matrix} \right.\ \]

\[Подставим:\]

\[\left\{ \begin{matrix} \frac{1}{x} = 5\ \ \ \ | \cdot x \\ \frac{2}{y} = 6\ \ \ \ | \cdot y \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 1 = 5x \\ 2 = 6y \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = \frac{1}{5}\text{\ \ \ \ \ \ \ \ \ } \\ y = \frac{2}{6} = \frac{1}{3} \\ \end{matrix} \right.\ \]

\[Ответ:\left( \frac{1}{5};\frac{1}{3} \right).\]