Решите методом подстановки систему

x^2-3y^2=12

x+y=6

X^2 - 3y^2 = 12

x + y = 6

x^2 - 3y^2 = 12

y = 6 - x

x^2 - 3 (6 - x) ^2 - 12 = 0

x^2 - 3 (x^2 - 12x + 36) - 12 = 0

x^2 - 3x^2 + 36x - 108 - 12 = 0

- 2x^2 + 36x - 120 = 0 / / : (-2)

x^2 - 18x + 60 = 0

D = 324 - 240 = 84

x₁ = (18 + 2√21) / 2 = 9 + √21

x₂ = (18 - 2√21) / 2 = 9 - √21

x₁ = 9 + √21

y₁ = 6 - (9 + √21) = - 3 - √21

x₂ = 9 - √21

y₂ = 6 - (9 - √21) = - 3 + √21