Question

Пусть (x0;y0)- решение системы {x+2y=4 y^+2x=5

Answer 1

$left { {{x+2y=4,} atop {y^2+2x=5,}} right. left { {{2x+4y=8,} atop {y^2+2x=5,}} right. left { {{2x=8-4y,} atop {y^2+8-4y=5,}} right. \ y^2-4y+4=5-4, \ (y-2)^2=1, \ y-2=pm1, \ left [ {{y=1,} atop {y=3;}} right. \ x=4-2y, \ left [ {{x=2,} atop {x=-2;}} right. \ 2cdot2-1=3, \ 2cdot(-2)-3=-7.$