Question

система уравнений x^2+y^2=16,x-y=4

Answer 1

$left { {{ x^{2} +y^2=16} atop {x-y=4}} right.$
$left { {{ (x-y)^2+2xy=16} atop {x-y=4}} right.$
$left { {{ 4^2+2xy=16} atop {x-y=4}} right.$
$left { {{16+2xy=16} atop {x-y=4}} right.$
$left { {{2xy=0} atop {x-y=4}} right.$
$left { {{xy=0} atop {x=4+y}} right.$
$left { {{y(4+y)=0} atop {x=4+y}} right.$
$y=0$   или $y+4=0$
                   $y=-4$
$left { {{y=0} atop {x=4}} right.$   или   $left { {{y=-4} atop {x=0}} right.$

Ответ: (0;-4);  (4;0)