Решите систему уравнений

х2+y2=13

xy=6

X^2 + y^2 = 13

(x + y) ^2 - 2xy = 13

xy = 6

- 2xy = - 12

(x + y) ^2 - 12 = 13

(x + y) ^2 = 25

(x + y) ^2 = 5^2

x + y = 5

y = 5 - x

x (5 - x) = 6

5x - x^2 = 6

x^2 - 5x + 6 = 0

D = 25 - 24 = 1

x1 = (5 + 1) : 2 = 3

x2 = (5 - 1) : 2 = 2

y1 = 5 - 3 = 2

y2 = 5 - 2 = 3

Ответ (3; 2); (2; 3)