1. ## System of equations

Well then a last question before I go out for some more beer.

We have the system of these two equations:
$\displaystyle (y+2)(x+y+1)=0$
$\displaystyle x^2+y^2+4y=9$

I am supposed to solve 'em for a couple of different x but I can only find the one that gives y+2=0.

2. Hello, a4swe!

$\displaystyle \begin{array}{cc}(1)\\(2)\end{array}\;\begin {array} {cc}(y+2)(x+y+1)\,=\,0 \\ x^2+y^2+4y\,=\,9\end{array}$

Equation (1) is a pair of intersecting lines; equation (2) is a circle.
. . They can intersect in (at most) four points.

Equation (1) give us two equations: .$\displaystyle \begin{array}{cc}(a)\\(b)\end{array}$ $\displaystyle \;\begin{array}{cc}y + 2\,=\,0\quad\Rightarrow\quad y \,=\,\text{-}2 \\ x + y + 1\,=\,0\quad\Rightarrow\quad y\,=\,\text{-}x-1\end{array}$

Substitute (a) into (2): .$\displaystyle x^2 + (\text{-}2)^2 + 4(\text{-}2) \:=\:9\quad\Rightarrow\quad x^2 =$ $\displaystyle 13\quad\Rightarrow\quad x = \pm\sqrt{13}$

. . and we have two intersections: .$\displaystyle \boxed{\left(\sqrt{13},\text{-}2\right)}\;\;\boxed{\left(\text{-}\sqrt{13},\;\text{-}2\right)}$

Substitute (b) into (2): .$\displaystyle x^2 + (-x-1)^2 + 4(-x-1)\:=\:9$

. . which simplifies to the quadratic: .$\displaystyle x^2 - x - 6\:=\:0$

. . which factors: .$\displaystyle (x - 3)(x + 2)\:=\:0$

. . and has roots: .$\displaystyle x\,=\,3,\;\text{-}2$

Substitute into (b) and we get: .$\displaystyle y \,=\,\text{-}4,\;1$

. . and we have two more intersections: .$\displaystyle \boxed{(3,\text{-}4)}\;\;\boxed{(\text{-}2,1)}$

3. For your reference (cause it's hard to picture what it looks like), here's the graph...