Question

Решите систему уравнений x-2=y y2-2xy=3

Answer 1

y(y-2x)=3
x=2+y

y(y-4-2y)=3
y(-4-y)=3

-4y-y²-3=0
y²+4y+3=0
D=16-12=4
y1=-4+2/2=-2/2=-1
y2=-4-2/2=-6/2=-3
x1=2-1=1
x2=2-3=-1