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Thread: System of equations

  1. #1
    Junior Member
    Oct 2005

    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.
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  2. #2
    Super Member

    May 2006
    Lexington, MA (USA)
    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: .$\displaystyle \boxed{(3,\text{-}4)}\;\;\boxed{(\text{-}2,1)}$

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  3. #3
    MHF Contributor Quick's Avatar
    May 2006
    New England
    For your reference (cause it's hard to picture what it looks like), here's the graph...
    Attached Thumbnails Attached Thumbnails System of equations-weird.jpg  
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