Question

x^2-3xy+2y^2=0
x^2+y^2=20
это система
НА ЗАВТРО НАДО

Answer 1

$x^2-3xy+2y^2=0\ x^2+y^2=20\ \ (y-x)(2y-x)=0\ x^2+y^2=0\ \ left { {{y=x} atop {x^2+y^2=20}} right. \ left { {{2y=x} atop {x^2+y^2=20}} right. \ 1)\ 2x^2=20\ x=+-sqrt{10}\ y=+-sqrt{10}\ 2)\ 5y^2=20\ y=+-2\ x=+-4$