Решите систему уравнений

$\left \{ {{x-y=4} \atop {xy+y^2=6}} \right. \left \{ {{x=4+y} \atop {(4+y)y+y^2=6}} \right. \left \{ {{x=4+y} \atop {(4y+y^2+y^2=6}} \right. \left \{ {{x=4+y} \atop {(4y+2y^2=6}} \right. \left \{ {{x=4+y} \atop {(2y^2+4y-6=0}} \right. \\\\ 2y^2+4y-6=0|/2\\ y^2+2y-3=0\\ D=b^2-4ac=2^2-4*1*(-3)=4-(-12)=16\\ x_1= \frac{-b+ \sqrt{D} }{2a} = \frac{-2+4}{2}=1\\ x_2= \frac{-b- \sqrt{D} }{2a} =-3\\\\ x-1=4\\ x=4+1=5\\ 5*1+1^2=6\\ 5+1=6\\6=6\\\\ x-(-3)=4\\ x=4-3\\ x=1\\ 1*(-3)+(-3)^2=6\\$
$-3+9=6\\ 6=6$

Ответ: 1, 5; 1, (-3)