Системи уравнения

[Гост]

решете системата [tex]\begin{tabular}{|l}x+xy=3\\xy^2+xy^3=12 \end{tabular}[/tex]

[Добромир Глухаров]

[tex]\begin{tabular}{|lr}x+xy=3&(1)\\xy^2+xy^3=12&(2) \end{tabular}[/tex]

[tex](2):\ y^2(x+xy)=12\ \stackrel{(1)}{\Rightarrow}\ y^2.3=12\ \Rightarrow\ y^2=4[/tex]

[tex]y_1=2\ \stackrel{(1)}{\Rightarrow}\ x+x.2=3\Rightarrow x_1=1\\
y_2=-2\ \stackrel{(1)}{\Rightarrow}\ x+x.(-2)=3\Rightarrow x_2=-3[/tex]

[tex](x;y)\in\{(1;2),(-3;-2)\}[/tex]