Решите систему уравнений: x^2-xy+y^2=14; x-3y=10.

Ответ:

\[7y^{2} + 50y + 86 = 0\ \ \ \ \]

\[D = 625 - 602 = 23\]

\[y_{1,2} = \frac{- 25 \pm \sqrt{23}}{7};\]

\[x_{1} = 3 \cdot \left( \frac{- 25 - \sqrt{23}}{7} \right) + 10 =\]

\[= \frac{- 75 - 3\sqrt{23} + 70}{7} =\]

\[= \frac{- 5 - 3\sqrt{23}}{7}.\]

\[x_{2} = 3 \cdot \left( \frac{- 25 + \sqrt{23}}{7} \right) + 10 =\]

\[= \frac{- 75 + 3\sqrt{23} + 70}{7} =\]

\[= \frac{- 5 + 3\sqrt{23}}{7}.\]