Данная система уравнений:
\( \begin{cases} 4x + 5y = 11 \\ 3x - 2y = 6 \end{cases} \)
Решим систему методом подстановки.
\[ 3x = 6 + 2y \]
\[ x = \frac{6 + 2y}{3} \]
\[ 4\left(\frac{6 + 2y}{3}\right) + 5y = 11 \]
\[ \frac{24 + 8y}{3} + 5y = 11 \]
\[ 24 + 8y + 15y = 33 \]
\[ 23y = 33 - 24 \]
\[ 23y = 9 \]
\[ y = \frac{9}{23} \]
\[ x = \frac{6 + 2\left(\frac{9}{23}\right)}{3} = \frac{6 + \frac{18}{23}}{3} = \frac{\frac{6 \cdot 23 + 18}{23}}{3} = \frac{\frac{138 + 18}{23}}{3} = \frac{\frac{156}{23}}{3} = \frac{156}{23 \cdot 3} = \frac{156}{69} = \frac{52}{23} \]
Проверка:
4x + 5y = \( 4\left(\frac{52}{23}\right) + 5\left(\frac{9}{23}\right) = \frac{208}{23} + \frac{45}{23} = \frac{253}{23} = 11 \)
3x - 2y = \( 3\left(\frac{52}{23}\right) - 2\left(\frac{9}{23}\right) = \frac{156}{23} - \frac{18}{23} = \frac{138}{23} = 6 \)
Ответ: \( x = \frac{52}{23}, y = \frac{9}{23} \).