Решите систему уравнений: x=2+y; y^2-2xy=3.

Ответ:

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

\[\left\{ \begin{matrix} x = 2 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 2 \cdot (2 + y)y = 3 \\ \end{matrix}\text{\ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} x = 2 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 4y - 2y^{2} - 3 = 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 2 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - y^{2} - 4y - 3 = 0 \\ \end{matrix} \right.\ \]

\[- y^{2} - 4y - 3 = 0\]

\[y_{1} + y_{2} = - 4\ \ \ \ \ \ \text{\ y}_{1} = - 1\]

\[y_{1}y_{2} = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y_{2} = - 3\]

\[\left\{ \begin{matrix} x = 1\ \ \ \\ y = - 1 \\ \end{matrix}\ \ \ \ или\ \ \right.\ \text{\ \ \ \ }\left\{ \begin{matrix} x = - 1 \\ y = - 3 \\ \end{matrix} \right.\ \]

\[Овет:(1;\ - 1),\ ( - 1;\ - 3)\text{.\ }\]