Решите систему уравнений: 5x+y=-7; (x+4)(y-5)=-4.

\[\left\{ \begin{matrix} 5x + y = - 7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x + 4)(y - 5) = - 4 \\ \end{matrix}\text{\ \ } \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = - 7 - 5x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x + 4)( - 7 - 5x - 5) = - 4 \\ \end{matrix}\text{\ \ } \right.\ \]

\[\left\{ \begin{matrix} y = - 7 - 5x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x + 4)( - 5x - 12) = - 4 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = - 7 - 5x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 5x^{2} - 32x - 44 = 0 \\ \end{matrix} \right.\ \]

\[- 5x^{2} - 32x - 44 = 0\]

\[D = 1024 - 880 = 144\ \ \]

\[x_{1} = \frac{32 - 12}{- 10} = - 2,\ \ \]

\[x_{2} = \frac{32 + 12}{- 10} = - 4,4\]

\[\left\{ \begin{matrix} x = - 2 \\ y = 3\ \ \ \\ \end{matrix} \right.\ \ \ \ \ \ \ \ \ \ \ \ или\ \ \ \ \ \ \ \left\{ \begin{matrix} x = - 4,4 \\ y = 15\ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:( - 2;3),\ ( - 4,4;15).\]