Задание 4630

$$\left\{\begin{matrix}3x-y=2,\\x^{2}-4x+8=y\end{matrix}\right.\Leftrightarrow$$$$\left\{\begin{matrix}3x-2=y,\\x^{2}-4x+8=3x-2\end{matrix}\right.$$ $$x^{2}-4x+8=3x-2\Leftrightarrow$$$$x^{2}-7x+10=0\Leftrightarrow$$$$\left[\begin{matrix}x_{1}+x_{2}=7\\x_{1}*x_{2}=10\end{matrix}\right.\Leftrightarrow$$$$ \left[\begin{matrix}x_{1}=5\\x_{2}=2\end{matrix}\right.\Leftrightarrow$$$$ \left[\begin{matrix}y_{1}=3*5-2=13\\y_{2}=3*2-2=4\end{matrix}\right.$$ В итоге получаем две точки: (2; 4), (5; 13)