Решите систему уравнений x+y=1; x^2-3y=1.

Ответ:

\[\left\{ \begin{matrix} x + y = 1\ \ \ \ \ \\ x^{2} - 3y = 1 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y = 1 - x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 3(1 - x) = 1 \\ \end{matrix} \right.\ \]

\[x^{2} - 3 + 3x - 1 = 0\]

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

\[x_{1} + x_{2} = - 3;\ \ x_{1} \cdot x_{2} = - 4\]

\[x_{1} = - 4;\ \ \ x_{2} = 1.\]

\[\left\{ \begin{matrix} x = - 4\ \ \ \ \\ y = 1 + 4 \\ \end{matrix} \right.\ \Leftrightarrow \left\{ \begin{matrix} x = 1\ \ \ \ \ \ \ \ \\ y = 1 - 1 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = - 4 \\ y = 5\ \ \ \\ \end{matrix} \right.\ \Leftrightarrow \left\{ \begin{matrix} x = 1 \\ y = 0 \\ \end{matrix} \right.\ \]

\[Ответ:( - 4;5)\ или\ (1;0).\]