Question

Решить систему уравнений!

Answer 1

x² + 3xy = 3y + x (1)

y² - xy = 3x + y (2)

(1) решаем

x(3y + x) - (3y + x ) = 0

(x - 1)(3y + x) = 0

1/ x = 1

во (2)

y² - y = 3 + y

y² - 2y - 3 = 0

D = 4 + 12 = 16

y12 = (2 +- 4)/2 = -1  3

2. x = -3y

во (2)

y² - y*(-3y) = 3*(-3y) + y

y² + 3y² = -9y + y

4y² + 8y = 0

4y(y + 2) = 0

y = 0   x = 0

y = -2  x = 6

ответ (1, -1) (1, 3) (0, 0) (6, -2)