Найдите все значения параметра a, при каждом из которых система уравнений ax+2y=a+4; 4x+(a+2)y=12 имеет бесконечно много решений.

\[\left\{ \begin{matrix}
ax + 2y = a + 4\ \ \ \ \ \ \\
4x + (a + 2)y = 12 \\
\end{matrix} \right.\ \]

\[\frac{a}{4} = \frac{2}{a + 2} = \frac{a + 4}{12}\]

\[\frac{a}{4} = \frac{2}{a + 2}\]

\[a(a + 2) = 8\]

\[a^{2} + 2a - 8 = 0\]

\[D = 2^{2} - 4 \cdot 1 \cdot ( - 8) = 4 + 32 =\]

\[= 36\]

\[a_{1} = \frac{- 2 + \sqrt{36}}{2} = \frac{- 2 + 6}{2} = \frac{4}{2} =\]

\[= 2\]

\[a_{2} = \ \frac{- 2 - \sqrt{36}}{2} = \frac{- 2 - 6}{2} =\]

\[= \frac{- 8}{2} = - 4\]

\[\frac{2}{a + 2} = \frac{a + 4}{12}\]

\[24 = (a + 4)(a + 2)\]

\[24 = a^{2} + 6a + 8\]

\[a^{2} + 6a - 16 = 0\]

\[D = 6^{2} - 4 \cdot 1 \cdot ( - 16) =\]

\[= 36 + 64 = 100\]

\[a_{1} = \frac{- 6 + \sqrt{100}}{2} = \frac{- 6 + 10}{2} =\]

\[= \frac{4}{2} = 2\]

\[a_{2} = \frac{- 6 - \sqrt{100}}{2} = \frac{- 6 - 10}{2} =\]

\[= \frac{- 16}{2} = - 8\]

\[Ответ:при\ a = 2.\]