Question

Система x^2-xy+y^2=7 x+y=5 решите методом постановки

Answer 1

$\left\{\begin{matrix} x^2-xy+y^2 &= & 7\\ x+y &= &5 \end{matrix}\right.\ \ \left\{\begin{matrix} x^2-xy+y^2 &= &7 \\ x &= &5-y \end{matrix}\right.\\\\\\ (5-y)^2-y(5-y)+y^2=7\\ 25-10y+y^2-5y+y^2+y^2=7\\ 3y^2-15y+18=0\ \ |:(3)\\ y^2-5y+6=0\\ D=25-24=1; \ \sqrt D=1\\\\ y_{1/2}=\frac{5\pm1}{2}\\\\ y_1=3;y_2=2\\\\ x_1=5-3=2\\ x_2=5-2=3$

ответ: $(2;3)(3;2)$