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Thread: Solving linear equations

  1. November 6th 2008, 12:06 PM #1
    natali.ferno
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    Solving linear equations

    Hello,

    hoping some maths genius on here can help me out!

    \frac{2}{x-3} + \frac{1}{x+2} = \frac{7}{x^2-x-6}

    i know that i have to factor the bottom \frac{2}{x-3} + \frac{1}{x+2} = \frac{7}{(x+2) (x-3)} were do i go from here?
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  2. November 6th 2008, 12:17 PM #2
    clic-clac
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    Hi,
    first you can say that -2,3 can't be chosen as solutions.
    Then multiply your equation by (x+2)(x-3), so you get a second order equation you can solve.

    Edit: first order equation not second...
    Last edited by clic-clac; November 6th 2008 at 12:57 PM. Reason: oups
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  3. November 6th 2008, 12:32 PM #3
    natali.ferno
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    Quote Originally Posted by clic-clac View Post
    Hi,
    first you can say that -2,3 can't be chosen as solutions.
    Then multiply your equation by (x+2)(x-3), so you get a second order equation you can solve.
    Okay so I multiplied each term by (x+2)(x+3) which gave me 2x + 4 + x -3 =7 than I just solve for x (right?)
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  4. November 6th 2008, 12:40 PM #4
    natali.ferno
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    is my equation right? cause i keep getting the wrong answer when i solve it!!
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  5. November 6th 2008, 12:50 PM #5
    masters
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    Quote Originally Posted by natali.ferno View Post
    is my equation right? cause i keep getting the wrong answer when i solve it!!
    Your equation looks fine. What answer did you get? I got x=2
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  6. November 6th 2008, 12:55 PM #6
    clic-clac
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    You found 2x+4+x-3=7, that's correct and becomes x=2.
    So 2 is your solution (because it's different from -2 and 3).
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