[m]\left\{\begin{matrix}x\cdot (x+y)=12 \\ x+y=2\end{matrix}\right.[/m]
Заменим [m] x+y[/m] на 2 согласно второму уравнению
[m]\left\{\begin{matrix}x\cdot 2=12 \\ y=2-x\end{matrix}\right.[/m]
[m]\left\{\begin{matrix}x=6 \\ y=2-6\end{matrix}\right.[/m]
[m]\left\{\begin{matrix}x=6 \\ y=-4\end{matrix}\right.[/m]
О т в е т. (6;-4)
или
так:
Решаем способом подстановки:
[m]\left\{\begin{matrix}x^2+x\cdot (2-x)=12 \\ y=2-x\end{matrix}\right.[/m]
[m]\left\{\begin{matrix}x^2+2x-x^2=12 \\ y=2-x\end{matrix}\right.[/m]
[m]\left\{\begin{matrix}2x=12 \\ y=2-x\end{matrix}\right.[/m]
[m]\left\{\begin{matrix}x=6 \\ y=2-6\end{matrix}\right.[/m]
[m]\left\{\begin{matrix}x=6 \\ y=-4\end{matrix}\right.[/m]
О т в е т. (6;-4)