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Thread: 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.

    $\displaystyle 3e^x + 3xe^x - y^2 + 3y = 0$

    and

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

    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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    $\displaystyle 3e^x + 3xe^x = e^x(-2y + 3) + 3x$

    $\displaystyle 3e^x + 3xe^x = 3e^x - 2ye^x + 3x$

    $\displaystyle 3xe^x + 2ye^x = 3x$

    $\displaystyle e^x(3x + 2y) = 3x$

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

    Substitute back into the first equation:

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

    $\displaystyle (3x + 3)(3x) + (3x + 2y)(3y - y^2) = 0$

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

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

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

    is

    $\displaystyle The\ unstoppable\ object\ doesn't\ stop$

    and

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

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

    is

    $\displaystyle The\ unstoppable\ object\ doesn't\ stop$

    and

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