Решите систему уравнений:x-y=4xy+y в квадрате=6

$\left \{ {{x-y=4} \atop {xy+ y^{2} =6}} \right.$
$\left \{ {{x=4+y} \atop {xy+ y^{2} =6}} \right.$
$\left \{ {{x=4+y} \atop {(4+y)y+ y^{2} =6}} \right.$
$\left \{ {{x=4+y} \atop {4y+ y^{2} + y^{2} =6}} \right.$
$\left \{ {{x=4+y} \atop {2 y^{2}+4y-6 =0}} \right.$
$\left \{ {{x=4+y} \atop {y=1}} \right.$ $\left \{ {{x=4+y} \atop {y=-3}} \right.$
$\left \{ {{x=1} \atop {y=-3}} \right.$
$\left \{ {{x=5} \atop {y=1}} \right.$

Ответ:(-3;1);(1;5)