Question

помогите,пожалуйста $left{ {{x+y+xy=11} atop {x^2+xy+y^2=19}} right.$

Answer 1

$x+y+xy=11\ x^2+y^2+xy=19\ \ (x+y)^2=19+xy\ (11-xy)^2=19+xy\ 121-22xy+x^2y^2=19+xy\ xy=t\ t^2-23t+102=0\ t=6\\ xy=6\ y=frac{6}{x}\ frac{6}{x}+x+6=11\ x^2-5x+6=0\ x=2;3\ y=3;2$