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

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