Решить систему равенст

x+y=4

xy=3

X = 4 - y

(4 - y) * y = 3 *

*4y - y^2 - 3 = 0 / * (-1)

y^2 - 4y + 3 = 0

D = 16 - 12 = 4

y1 = (4 + 2) / 2 = 3

y2 = (4 - 2) / 2 = 1

y1 = 3

x1 = 4 - y1 = 4 - 3 = 1

y2 = 1

x2 = 4 - y2 = 4 - 1 = 3

Ответ

(1; 3); (3; 1)