\[\left\{ \begin{matrix} (a - 1)x + 3y = 5\ \ \\ 4x + (a + 3)y = 10 \\ \end{matrix} \right.\ \]
\[1)\ \ a \neq 0:\ \ \ \frac{a - 1}{4} \neq \frac{3}{a + 3}\]
\[\frac{a - 1}{4} = \frac{3}{a + 3}\]
\[(a - 1)(a + 3) = 3 \cdot 4\]
\[a^{2} + 2a - 3 = 12\]
\[a^{2} + 2a - 15 = 0\]
\[D = 2^{2} - 4 \cdot 1( - 15) = 4 + 60 =\]
\[= 64\]
\[a_{1} = \frac{- 2 + \sqrt{64}}{2} = \frac{- 2 + 8}{2} = \frac{6}{2} =\]
\[= 3\]
\[a_{2} = \frac{- 2 - \sqrt{64}}{2} = \frac{- 2 - 8}{2} =\]
\[= \frac{- 10}{2} = - 5\]
\[Единственное\ решение\ при\ \]
\[любых\ a \neq 0\ и\ \ a \neq - 5.\]
\[2)\ a = 0:\ \ \]
\[\left\{ \begin{matrix} - x + 3y = 5\ \ \ \ \ \ (1)\ \\ 4x + 3y = 10\ \ \ \ (2)\ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]
\[(2) - (1):\ \ \ 5x = 5\]
\[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }x = 1\]
\[\left\{ \begin{matrix} x = 1\ \ \ \ \ \ \ \ \ \\ 3y = x + 5 \\ \end{matrix} \right.\ \]
\[3y = 1 + 5\]
\[3y = 6\]
\[y = 2\]
\[Единственное\ решение \Longrightarrow\]
\[\Longrightarrow (1;2).\]