Решите систему уравнений: x=5-y; y^2+4xy=33.

Ответ:

\[\left\{ \begin{matrix} x = 5 - y\ \ \ \ \ \ \ \ \ \\ y^{2} + 4xy = 33 \\ \end{matrix}\text{\ \ \ } \right.\ \text{\ \ }\]

\[\left\{ \begin{matrix} x = 5 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + 4 \cdot (5 - y)y = 33 \\ \end{matrix}\text{\ \ \ } \right.\ \]

\[\left\{ \begin{matrix} x = 5 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + 20y - 4y^{2} - 33 = 0 \\ \end{matrix} \right.\ \]

\[- 3y^{2} + 20y - 33 = 0\]

\[D = 400 - 396 = 4\]

\[y = \frac{- 20 + 2}{- 6} = 3,\ \ \ \ \ \ \ \]

\[y = \frac{- 20 - 2}{- 6} = 3\frac{2}{3}\]

\[\left\{ \begin{matrix} x = 2 \\ y = 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 1\frac{1}{3} \\ y = 3\frac{2}{3} \\ \end{matrix} \right.\ \]

\[Ответ:(2;3);\ \left( 1\frac{1}{3};3\frac{2}{3} \right).\]