Question

найдите произведение х и у где (х;у) решение системы х2+ху+у2=36-ху, х-у=8

Answer 1

$left { {{x^{2}+xy+y^{2}=36-xy} atop {x-y=8}} right. \ left { {{x^{2}+2xy+y^{2}=36} atop {x-y=8}} right. \ left { {{(x+y)^{2}=36} atop {x=8+y}} right. \ (8+y+y)^{2} = 36 \ (8+2y)^{2} = 36 \ 64+32y+4y^{2}-36=0 \ 4y^{2}+32y+28=0 \ y^{2}+8y+7=0 \ y_{1} = -1, y_{2}=-7; \ x_{1}=7, x_{2}=1$ 
Тогда x*y = -7