\[\left\{ \begin{matrix} ax + (a + 1)y = 3 \\ 3x + (a + 5)y = 9 \\ \end{matrix} \right.\ \]
\[1)\ a \neq 0:\ \]
\[\frac{a}{3} \neq \frac{a + 1}{a + 5}\]
\[\frac{a}{3} = \frac{a + 1}{a + 5}\]
\[a(a + 5) = 3 \bullet (a + 1)\]
\[a^{2} + 5a = 3a + 3\]
\[a^{2} + 2a - 3 = 0\]
\[D = 2^{2} - 4 \cdot 1 \cdot ( - 3) = 4 + 12 =\]
\[= 16\]
\[a_{1} = \frac{- 2 + \sqrt{16}}{2} = \frac{- 2 + 4}{2} = \frac{2}{2} =\]
\[= 1\]
\[a_{2} = \frac{- 2 - \sqrt{16}}{2} = \frac{- 2 - 4}{2} =\]
\[= \frac{- 6}{2} = - 3\]
\[Единственное\ решение\ при\ \]
\[a \neq 0,\ a \neq 1,\ a \neq - 3.\]
\[2)\ a = 0:\ \left\{ \begin{matrix} y = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3x + 5y = 9 \\ \end{matrix} \right.\ \]
\[3x + 5 \cdot 3 = 9\]
\[3x + 15 = 9\]
\[3x = - 6\]
\[x = - 2.\]
\[Есть\ решение \Longrightarrow ( - 2;3).\]
\[Ответ:\ \ a - любое,\ кроме\ \]
\[a = 1\ и\ a = - 3.\]