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

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

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

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

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

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

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

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

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

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

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

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

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

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