(1)
(2)
re arrange (2)
y = 4 - x^2
putting this back into (1)
4 - x^2 = x - 8
x^2 + x - 12 = 0
(x+4)(x-3) = 0
x = -4 or 3
when x = -4 (1) y = -4 -8 = -12 (2) (-4)^2 + y = 4 16 + y = 4 = -12
when x = 3 (1) y = 3 - 8 = -5 (2) (3)^2 + y = 4 9 + y + 4 = -5
therefore 2 answers fulfil this sim equation
x = -4 y = -12
x = 3 y = -5
I did it this way as when I saw a x squared I automatically knew it could have a negative value as well as a positive.