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)