Найдите все значения параметра a, при каждом из которых система уравнений x+(a+2)y=3; (a+3)x+6y=a+9 имеет бесконечно много решений.

\[\left\{ \begin{matrix} x + (a + 2)y = 3\ \ \ \ \ \ \ \ \ \\ (a + 3)x + 6y = a + 9 \\ \end{matrix} \right.\ \]

\[\frac{1}{a + 3} = \frac{a + 2}{6} = \frac{3}{a + 9}\]

\[\frac{1}{a + 3} = \frac{a + 2}{6}\]

\[6 = (a + 2)(a + 3)\]

\[6 = a^{2} + 5a + 6\]

\[a^{2} + 5a = 0\]

\[a(a + 5) = 0\]

\[a_{1} = 0;\ \ \ a_{2} = - 5.\]

\[\frac{a + 2}{6} = \frac{3}{a + 9}\text{\ \ \ }\]

\[a = 0:\ \ \ \]

\[\frac{0 + 2}{6} = \frac{3}{0 + 9}\text{\ \ \ }\]

\[\frac{2}{6} = \frac{3}{9}\text{\ \ \ }\]

\[\frac{1}{3} = \frac{1}{3}.\]

\[a = - 5:\ \ \]

\[\frac{- 5 + 2}{6} = \frac{3}{- 5 + 9}\text{\ \ \ \ \ \ \ }\]

\[\frac{- 3}{6} = \frac{3}{4}\text{\ \ \ \ }\]

\[- \frac{1}{2} \neq \frac{3}{4}.\]

\[Ответ:a = 0.\]