Решите систему способом подстановки x=3-y y^2-x=39 x-y=3 xy=-2

Решите задачу:

$1) \left \{ {{x=3-y} \atop { y^{2} }-x=39} \right. \\ y^{2}+y-42=0 \\ D=1^{2}-4*(-42)=1+168=169 \\ y_{1} =(-1+13)/2=12/2=6 \\ y_{2}=(-1-13)/2=-14/2=-7 \\ x_1=3-6=-3 \\ x_2=3+7=10$
$2) \\ \left \{ {{x-y=3} \atop {xy=-2} \right. \\ \left \{ {{x=y+3} \atop {(y+3)*y=-2} \right. \\ y^{2}+3y+2=0 \\ D=9-8=1 \\ y_1=(-3+1)/2=-1 \\ y_2=(-3-1)/2=-2 \\ x_1=-1+3=2 \\ x_2=-2+3=1$