Question 1005937
From
x^2+y^2=16
y=3-x
we substitute the second into the first equation and get
x^2 + (3-x)^2 = 16
x^2 + 9 - 6x + x^2 = 16
2x^2 - 6x - 7 = 0
This doesn't factor...you'll have to use the quadratic formula...not sure what you wanted to do here...
It appears to be a circle and a line...