System of equations

**a4swe** · August 25th 2006, 06:58 AM

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

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

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

**Soroban** · August 25th 2006, 08:35 AM

Hello, a4swe!

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

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

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

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

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

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

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

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

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

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

**Quick** · August 25th 2006, 08:59 AM

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

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