Вопрос:

Решите систему уравнений: 5/x-6/y=2; 10/x-9/y=13.

Ответ:

\[\left\{ \begin{matrix} \frac{5}{x} - \frac{6}{y} = 2\ \ \ \ \\ \frac{10}{x} - \frac{9}{y} = 13 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[Пусть\ \ \frac{5}{x} = a;\ \ \frac{3}{y} = b:\]

\[\left\{ \begin{matrix} a - 2b = 2\ \ \ \ \ \\ 2a - 3b = 13 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} a = 2b + 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2 \cdot (2b + 2) - 3b = 13 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} a = 2b + 2\ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4b + 4 - 3b = 13 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} a = 2b + 2 \\ b = 9\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} b = 9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ a = 2 \cdot 9 + 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} b = 9\ \ \\ a = 20 \\ \end{matrix} \right.\ \]

\[Подставим:\]

\[\left\{ \begin{matrix} \frac{5}{x} = 20\ \ \ | \cdot x \\ \frac{3}{y} = 9\ \ \ \ \ | \cdot y \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 5 = 20x \\ 3 = 9y\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = \frac{5}{20} = \frac{1}{4} \\ y = \frac{3}{9} = \frac{1}{3}\text{\ \ } \\ \end{matrix} \right.\ \]

\[Ответ:\left( \frac{1}{4};\frac{1}{3} \right).\]


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