Система уравнений:
\( \begin{cases} \frac{15x - 3y}{3x + 2y} = 3 \\ \frac{6}{3x + y} - \frac{x - 3y}{3x + y} = 9 \end{cases} \)
Из первого уравнения:
\( 15x - 3y = 3(3x + 2y) \)
\( 15x - 3y = 9x + 6y \)
\( 15x - 9x = 6y + 3y \)
\( 6x = 9y \)
\( x = \frac{9y}{6} = \frac{3y}{2} \)
Подставим во второе уравнение:
\( \frac{6}{3(\frac{3y}{2}) + y} - \frac{\frac{3y}{2} - 3y}{3(\frac{3y}{2}) + y} = 9 \)
\( \frac{6}{\frac{9y}{2} + y} - \frac{\frac{3y - 6y}{2}}{\frac{9y}{2} + y} = 9 \)
\( \frac{6}{\frac{9y + 2y}{2}} - \frac{\frac{-3y}{2}}{\frac{9y + 2y}{2}} = 9 \)
\( \frac{6}{\frac{11y}{2}} - \frac{\frac{-3y}{2}}{\frac{11y}{2}} = 9 \)
\( \frac{12}{11y} - \frac{-3y}{11y} = 9 \)
\( \frac{12 + 3y}{11y} = 9 \)
\( 12 + 3y = 9(11y) \)
\( 12 + 3y = 99y \)
\( 12 = 99y - 3y \)
\( 12 = 96y \)
\( y = \frac{12}{96} = \frac{1}{8} \)
Найдем x:
\( x = \frac{3y}{2} = \frac{3(\frac{1}{8})}{2} = \frac{\frac{3}{8}}{2} = \frac{3}{16} \)
Проверка:
Первое уравнение:
\( \frac{15(\frac{3}{16}) - 3(\frac{1}{8})}{3(\frac{3}{16}) + 2(\frac{1}{8})} = \frac{\frac{45}{16} - \frac{3}{8}}{\frac{9}{16} + \frac{2}{8}} = \frac{\frac{45 - 6}{16}}{\frac{9 + 4}{16}} = \frac{\frac{39}{16}}{\frac{13}{16}} = \frac{39}{13} = 3 \)
Второе уравнение:
\( \frac{6}{3(\frac{3}{16}) + \frac{1}{8}} - \frac{\frac{3}{16} - 3(\frac{1}{8})}{3(\frac{3}{16}) + \frac{1}{8}} = \frac{6}{\frac{9}{16} + \frac{2}{16}} - \frac{\frac{3}{16} - \frac{6}{16}}{\frac{9}{16} + \frac{2}{16}} = \frac{6}{\frac{11}{16}} - \frac{\frac{-3}{16}}{\frac{11}{16}} = \frac{96}{11} - \frac{-3}{11} = \frac{96 + 3}{11} = \frac{99}{11} = 9 \)
Ответ: x = 3/16, y = 1/8.