Question

$x - y = 1 \ xy = 6$
решите методом подстановки

Answer 1

$left { {{x-y=1} atop {xy=6}} right.Leftrightarrowleft { {{x=1+y} atop {(1+y)y=6}} right.\(1+y)y=6\y+y^{2}=6\y{2}+y-6=0\D=1-4times1times(-6)=1+4times6=25=5^{2}\y_{1,2}=frac{-1pm5}{2}\y_{1}=-3\y_{2}=2\left { {{y_{1}=-3} atop {x_{1}=1+y_{1}}} right. Leftrightarrowleft { {{y_{1}=-3} atop {x_{1}=-2}} right. \left { {{y_{2}=2} atop {x_{2}=1+y_{2}}} right. Leftrightarrowleft { {{y_{2}=2} atop {x_{2}=3}} right.$

Ответ:$(-2;-3);(3;2).$