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Thread: What is the shape of the function: x^2+xy+x+3y=6

  1. #1
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    What is the shape of the function: x^2+xy+x+3y=6

    Please explain how.

    Already tried graphing it.
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    Re: What is the shape of the function: x^2+xy+x+3y=6

    $x^2 + x y + x + 3y = 6$

    $y(3+x)=6-x-x^2$

    $y(x+3) = -(x^2+x-6) = -(x+3)(x-2)$

    $y = -(x-2) = 2-x$

    $y = -x + 2$

    it's a line
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    Re: What is the shape of the function: x^2+xy+x+3y=6

    Quote Originally Posted by romsek View Post
    $x^2 + x y + x + 3y = 6$
    it's a line
    Actually it is the graph of two lines. SEE HERE
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    Re: What is the shape of the function: x^2+xy+x+3y=6

    actually, two lines ...

    $ x^2+xy+x+3y=6$

    $x^2 + (y+1)x + (3y-6) = 0$

    $x = \dfrac{-(y+1) \pm \sqrt{(y+1)^2 - 4(3y-6)}}{2}$

    $x = \dfrac{-(y+1) \pm (y-5))}{2}$

    $x = \dfrac{-2y+4}{2} = -y+2$

    $x = \dfrac{-6}{2} = -3$
    Attached Thumbnails Attached Thumbnails What is the shape of the function: x^2+xy+x+3y=6-impgraph1.png  
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    Re: What is the shape of the function: x^2+xy+x+3y=6

    Quote Originally Posted by romsek View Post
    $x^2 + x y + x + 3y = 6$

    $y(3+x)=6-x-x^2$

    $y(x+3) = -(x^2+x-6) = -(x+3)(x-2)$
    ... it's a line
    If x=-3 then y(x+3)=-(x+3)(x-2) is true for every value of y, hence the second line.

    The best way to solve this is

    \begin{align*} y(x+3) &= -(x+3)(x-2) \\ (x+3)(y+x-2) &= 0 \end{align*}

    and so we have x+3=0 or y+x-2 = 0 (or both).
    Last edited by Archie; Feb 16th 2017 at 08:37 PM.
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