Решите систему уравнений: x=y+3; xy-y=7.

Ответ:

\[y^{2} + 3y - y - 7 = 0\]

\[y^{2} + 2y - 7 = 0\]

\[D_{1} = 1 + 7 = \sqrt{8} = 2\sqrt{2}\]

\[y_{1} = - 1 + 2\sqrt{2};\ \]

\[x_{1} = - 1 + 2\sqrt{2} + 3 = 2 + 2\sqrt{2}.\]

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

\[x_{2} = - 1 - 2\sqrt{2} + 3 = 2 - 2\sqrt{2}.\]

\[Ответ:\left( 2 + 2\sqrt{2};\ - 1 + 2\sqrt{2} \right);\]

\[\left( 2 - 2\sqrt{2};\ - 1 - 2\sqrt{2} \right).\]