Решите систему уравнений: 5/(3x-2y)+2/(2x+y)=21; 9/(3x-2y)+8/(2x+y)=40.

Ответ:

\[\left\{ \begin{matrix} \frac{5}{3x - 2y} + \frac{2}{2x + y} = 21 \\ \frac{9}{3x - 2y} + \frac{8}{2x + y} = 40 \\ \end{matrix} \right.\ \]

\[Пусть\ \ \frac{1}{3x - 2y} = t;\ \frac{1}{2x + y} = c:\]

\[\left\{ \begin{matrix} 5t + 2c = 21\ \ | \cdot 4 \\ 9t + 8c = 40\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} 20t + 8c = 84 \\ 9t + 8c = 40\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} 11t = 44\ \ \ \ \\ c = \frac{40 - 9t}{8} \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} t = 4 \\ c = \frac{1}{2} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \frac{1}{3x - 2y} = 4 \\ \frac{1}{2x + y} = \frac{1}{2}\text{\ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 12x - 8y - 1 = 0\ \ \ \ \ \ \ \\ 2x + y - 2 = 0\ \ \ \ \ | \cdot 8 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 12x - 8y - 1 = 0\ \ \\ 16x + 8y - 16 = 0 \\ \end{matrix} \right.\ ( + )\]

\[\left\{ \begin{matrix} 28x = 17\ \ \\ y = 2 - 2x \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = \frac{17}{28\ }\text{\ \ \ \ \ \ \ } \\ y = 2 - \frac{17}{14} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = \frac{17}{28} \\ y = \frac{11}{14} \\ \end{matrix} \right.\ \]

\[Ответ:\left( \frac{17}{28};\frac{11}{14} \right).\]