Решите систему уравнений: x+y=3; x^2+y^2=29.

Ответ:

\[\left\{ \begin{matrix} x + y = 3\ \ \ \ \ \ \ \\ x^{2} + y^{2} = 29 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 3 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (3 - y)^{2} + y^{2} = 29 \\ \end{matrix} \right.\ \]

\[9 - 6y + y^{2} + y^{2} - 29 = 0\]

\[2y^{2} - 6y - 20 = 0\ \ |\ :3\]

\[y^{2} - 3y - 10 = 0\]

\[y_{1} + y_{2} = 3;\ \ y_{1} \cdot y_{2} = - 10;\]

\[y_{1} = - 2;\ \ y_{2} = 5;\]

\[x_{1} = 3 + 2 = 5;\]

\[x_{2} = 3 - 5 = - 2.\]

\[Ответ:( - 2;5);(5; - 2).\]