What is the solution set to the following system?

Question

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

Answers:
A. {(-4, 0)(0, 4)}

B. {(4, 0)(0, 4)}

C. {(4, 0)(0, -4)}

D. {(-4, 0)(0, -4)}

Answer 1

y = 4 - x

x^2 + (16 - 8 x + x^2) = 16

2 x^2 - 8 x = 0
x( 2x-8 ) = 0
x = 0 or x = 4
y = 0 or y = 4 of course, symmetry :)
(0,4) or (4,0)