Решить систему уравнений: x+y=4; xy=3.

\[\left\{ \begin{matrix} x + y = 4 \\ xy = 3\ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = 4 - x\ \ \ \ \ \ \\ x(4 - x) = 3 \\ \end{matrix} \right.\ \]

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

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

\[D_{1} = 4 - 3 = 1\]

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

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

\[y_{1} = 4 - 3 = 1;\]

\[y_{2} = 4 - 1 = 3.\]

\[Ответ:(3;1);(1;3).\]