x^2+xy=6 ; x-y=4 решите методом подстановки

Решите задачу:

$\displaystyle \left \{ {{x^2+xy=6} \atop {x-y=4}} \right.\\\\ \left \{ {{x^2+xy=6} \atop {x=4+y}} \right. \\\\ \left \{ {{(4+y)^2+(4+y)y=6} \atop {x=4+y}} \right.\\\\ \left \{ {{16+8y+y^2+4y+y^2-6=0} \atop {x=4+y}} \right.\\\\ \left \{ {{2y^2+12y+10=0} \atop {x=4+y}} \right.\\\\2(y^2+6y+5)=0\\y^2+6y+5=0\\D=36-4*5=16=4^2\\y_1=(-6+4)/2=-1\\y_2=(-6-4)/2=-5\\\\y_1=-1; x_1=4-1=3\\y_2=-5; x_2=4-5=-1$