Question

What is the solution set to the following system?
X+y=4
X^2+y^2=16

Answer 1

x+y = 4, -----> y=4 - x
x² +(4-x)² =16
x² +16-2*4x+x² =16
2x² - 8x = 0
x(2x-8) = 0
2x-4=0, 2x=8, x=4
x=0 or x=4

If x=0, y=4-x=4-0=4.
If x=4, y= 4-x=4-4=0.

(0,4) or (4,0)

Check
x² +y² =16
For (0,4)
0²+4² =16
16 =16 True

For (4,0)
4² +0²=16
16=16 True

Answer (0, 4) and (4,0).