Решить систему уравнения

{ (2x+4) ^2=3y

{ (4x+2) ^2=3y

{ (2x+4) ^2=3y

{ (4x+2) ^2=3y

{2x+4) ^2 = (4x+2) ^2

{3y=3y

4x^2+16x+16=16x^2+16x+4

-12x^2+12=0

x^2-12=0

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

x+1=0 = > x ₁ = - 1

x-1=0 = > x ₂ = 1

3y = (2 * (-1) + 4) ^2 = (4 * (-1) + 2) ^2

3y=4 = > y ₁=4/3=1 1/3

3y = (2*1+4) ^2 = (4*1+2) ^2

3y=36 = > y₂=12

x₁=-1; y₁=1 1/3

x₂=1; y₂=12