17
x + y = 4 ⇒ y = 4 - x (1)
x^2 + y^2 = 16 (2)
Sub (1) into (2) and we have that
x^2 + (4 - x)^2 = 16
x^2 + (x - 4)^2 = 16
x^2 + x^2 - 8x + 16 = 16 subtract 16 from both sides
2x^2 - 8x = 0 factor
2x ( x - 4) = 0
Setting each factor to 0 we have that x = 0 and x = 4
And when x = 0 , y = 4 - 0 = 4
And when x = 4, y = 4 - 4 = 0
So.....the solutions are (0,4) and (4,0)
18.....easy, NSS
Just change the signs on the constants in the parentheses and we get
3, -2 , 2