Вопрос:

Решите систему уравнений: 1/x-1/y=1; 2/x-1/2y=5.

Ответ:

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

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

\[\left\{ \begin{matrix} a - 2b = 1 \\ 2a - b = 5 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} a = 2b + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2 \cdot (2b + 1) - b = 5 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} a = 2b + 1\ \ \ \ \ \ \ \\ 4b + 2 - b = 5 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} a = 2b + 1 \\ 3b = 3\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} b = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ a = 2 \cdot 1 + 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} b = 1 \\ a = 3 \\ \end{matrix} \right.\ \]

\[Подставим:\]

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

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

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

\[Ответ:\left( \frac{1}{3};\frac{1}{2} \right).\]


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