Question

(4х+1)^2=12y
(x+4)^2=3y

решите систему уравнений ​

Answer 1

a² - b² = (a - b)(a + b)

(4х+1)^2=12y

(x+4)^2=3y |*4   4(x + 4)^2 = 12y

вычитаем второе из первого

(4x + 1)^2 - (2(x + 4))^2 = 12y - 12y

(4x + 1 - 2x - 8)(4x + 1 + 2x + 8) = 0

(2x - 7)(6x + 9) = 0

x = 7/2

(4*7/2 + 1)^2 = (14 + 1)^2 = 15^2 = 225 = 12y

y = 225/12

x = -9/6 = -3/2

(4*(-3/2) + 1)^2 = (-6 + 1)^2 = 25 = 12y

y = 25/12

(-3/2, 25/12) (7/2, 225/12)