$\displaystyle x^2+y^2=16$
$\displaystyle y+2x=-1$
Use substitution. I'd solve the bottom equation for y since it is linear in y, and thus simpler.
$\displaystyle y = -2x - 1$
Insert this into the top equation:
$\displaystyle x^2+(-2x - 1)^2=16$
$\displaystyle x^2 + 4x^2 + 4x + 1 = 16$
Solve this for x (there will be two of them), then find your y values. (And check to make sure that both pairs fit your original equations, just to be on the safe side.)
-Dan