Система уравнений

2x^2-3xy+y^2=0

y^2-x^2=12

{x² - xy + (x-y) ² = 0

{ (y-x) (x+y) = 12

{-x (y-x) + (y-x) ² = 0

{ (y-x) (x+y) = 12

{ (y-x-x) (y-x) = 0

{ (y-x) (y+x) = 12

y≠x

y = 2x

(2x-x) (2x+x) = 12

3x² = 12

x² = 4

x1 = - 2

x2 = 2

y1 = - 4

y2 = 4

Ответ: (-2; - 4); (2; 4)