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Math Help - Finding x and y

  1. #1
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    Finding x and y

    Hi,

    I have two equations where I need to find x and y but I have no idea how.

    3e^x + 3xe^x - y^2 + 3y = 0

    and

    e^x(-2y + 3) + 3x - y^2 + 3y = 0

    I am unsure whether logs would be much use or if they would make it worse.
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  2. #2
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    3e^x + 3xe^x = e^x(-2y + 3) + 3x

    3e^x + 3xe^x = 3e^x - 2ye^x + 3x

    3xe^x + 2ye^x = 3x

    e^x(3x + 2y) = 3x

    e^x = \frac{3x}{3x + 2y}

    Substitute back into the first equation:

    (3x + 3)\left(\frac{3x}{3x + 2y}\right) - y^2 + 3y = 0

    (3x + 3)(3x) + (3x + 2y)(3y - y^2) = 0

    9x^2 + 9x + 9xy + 6y^2 - 3xy^2 - 2y^3 = 0

    Not pretty, but it gets rid of e^x quite nicely. All solutions to this equation such that \frac{3x}{3x + 2y} > 0 are the solutions to the original system of equations.
    Last edited by icemanfan; March 3rd 2010 at 12:55 PM.
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  3. #3
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    I'd imagine the answer to

    What\ happens\ when\ an\ unstoppable\ object\ meets\ an\ immovable\ target

    is

    The\ unstoppable\ object\ doesn't\ stop

    and

    The\ immoveable\ target\ doesn't\ move
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  4. #4
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    Quote Originally Posted by Archie Meade View Post
    I'd imagine the answer to

    What\ happens\ when\ an\ unstoppable\ object\ meets\ an\ immovable\ target

    is

    The\ unstoppable\ object\ doesn't\ stop

    and

    The\ immoveable\ target\ doesn't\ move
    its a paradox, i think it is said in The Dark Knight
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