Question

Решите систему уравнений
(x+y)^2-2(x+y)=15
x+xy+y=11

Answer 1

$left { {{(x+y)^2-2(x+y)=15} atop {x+xy+y=11}} right. \ (x+y)^2-2(x+y)=15\ t=x+y\ t^2-2t-15=0\ (t-5)(t+3)=0\ t_1=5 t_2=-3\ left { {{x+y=5} atop {x+xy+y=11}} right. \ left { {{y=5-x} atop {xy+)5=11}} right. \ left { {{y=5-x} atop {xy=6}} right. \ left { {{y=5-x} atop {x(5-x)=6}} right.$
$left { {{y=5-x} atop {x^2-5x+6=0}} right. \ left { {{y=5-x} atop {(x-2)(x-3)=0}} right. \ left { {{y_1=3} atop {x_1=2}} right. left { {{y_2=2} atop {x_2=3}} right.$
$left { {{x+y=-3} atop {x+xy+y=11}} right. \ left { {{x+y=-3} atop {xy-3=11}} right. \ left { {{x+y=-3} atop {xy=14}} right. \ left { {{y=-3-x} atop {x(-3-x)=14}} right. \ left { {{y=-3-x} atop {x^2+3x+14=0}} right. \ x^2+3x+14=0\ a=1 b=3 c=14\ D=b^2-4ac=3^2-4*1*14=9-56=-47 textless 0$
нет решений
Ответ: (2;3), (3;2).