Решите систему уравнений: x^2-xy+y=16; 3y-x=14.

Ответ:

\[\left\{ \begin{matrix} x^{2} - xy + y = 16 \\ 3y - x = 14\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix}\text{\ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} 6y^{2} - 69y + 180 = 0\ \ |\ :3 \\ x = 3y - 14\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[2y^{2} - 23y + 60 = 0\]

\[D = 529 - 480 = 49\ \ \ \]

\[y_{1} = \frac{23 + 7}{4} = 7,5;\ \ \ \ \ \ \ \ \ \]

\[y_{2} = \frac{23 - 7}{4} = 4\]

\[\left\{ \begin{matrix} x = 8,5 \\ y = 7,5 \\ \end{matrix}\ \ \ \ \ \ или\ \ \ \right.\ \text{\ \ \ }\left\{ \begin{matrix} x = - 2 \\ y = 4\ \ \ \\ \end{matrix} \right.\ \]

\[Овет:(8,\ 5;\ \ 7,5),\ \ \ ( - 2;\ \ 4).\]