solve the system of equations?

Question

y=x^2-4
y=x-2

Answer 1

y=x^2-4....................... (1)
y=x-2............................(2)

Using difference of two squares,
y = x^2 - 4 = (x + 2)(x - 2)......................... (3)
Equating the two y terms we have:
(x + 2)(x - 2) = (x - 2)........ subtracting (x - 2) from both sides:
(x + 2)(x - 2) - (x-2) = 0
(x - 2)(x + 2 - 1) = 0
(x - 2)(x + 1) = 0
x - 2 = 0 or x + 1 = 0
x = 2 or x = -1
when x = 2, y = x - 2 = 2 - 2 = 0.
when x = -1, y = x - 2 = -1 -2 = -3.
Therefore x = 2, y = 0 or x = -1, y = -3.

Answer 2

x-2 = x^2 -4
x^2 - x - 2 = 0
(x-2)(x+1)=0
x = 2 or -1
y = 0 or -3

Answer 3

y = (x²-4) = (x+2)(x-2)
y = (x-2)

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

x = {-1, 2}
y = {-3, 0}

(x, y) = {(-1,-3), (2,0)}