# Thread: solve the system

1. ## solve the system

$x^2+y^2=16$

$y+2x=-1$

2. Originally Posted by uglygreencouch
$x^2+y^2=16$

$y+2x=-1$
Use substitution. I'd solve the bottom equation for y since it is linear in y, and thus simpler.

$y = -2x - 1$

Insert this into the top equation:
$x^2+(-2x - 1)^2=16$

$x^2 + 4x^2 + 4x + 1 = 16$

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