Question

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

Answer 1

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