Question

Решите систему уравнений:
{(x+2)√y-3=0
{x+y=4

Answer 1

$left { {{(x+2) sqrt{y-3}=0 } atop {x+y=4}} right. == textgreater x=4-y \ \ (4-y+2) sqrt{y-3} =0 \ (6-y) sqrt{y-3}=0 \ \ 6-y=0 \ y_1=6 \ \ sqrt{y-3}=0 \ y-3=0 \ y_2=3 \ \ x+y=4 \ x+6=4 \ x_1=-2 \ \ x+3=4 \ x_2=1$