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Math Help - Absolute value equation

  1. #1
    Senior Member sakonpure6's Avatar
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    Absolute value equation

    |x+3| = |2x+1|

    How do we go about solving this? I know there are two cases when x>0 and when x<0, but how do we set it up afterwards?

    Edit: I may have solved this:

    Case 1: If x > 0, then

     x+3 = 2x +1
    x=2

    Case 2: If x < 0. then
     -(x+3) = - (2x+1)
    x=2

    So we can conclude that the solution to the system is at x=2?

    I checked Wolfram but they have a solution of \frac{-4}{3} http://www.wolframalpha.com/input/?i...3D%7C2x%2B1%7C

    Any Help is appreciated.
    Last edited by sakonpure6; August 22nd 2014 at 08:15 AM.
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  2. #2
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    Re: Absolute value equation

    x being greater or less than zero isn't of issue here.

    there are 2 ways that $|a|=|b|$

    $a=b$ (this is the same as $-a=-b$)

    $-a=b$ (this is the same as $a=-b$)


    $x+3=2x+1 \Rightarrow x=2$

    $-(x+3)=2x+1 \Rightarrow x=-\dfrac 4 3$
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  3. #3
    Senior Member sakonpure6's Avatar
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    Re: Absolute value equation

    So we have two solutions at x= 2 and -4/3 ?
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  4. #4
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    Re: Absolute value equation

    Quote Originally Posted by sakonpure6 View Post
    |x+3| = |2x+1|

    How do we go about solving this? I know there are two cases when x>0 and when x<0, but how do we set it up afterwards?

    Edit: I may have solved this:

    Case 1: If x > 0, then
    No, that's not true. the cases for |a| are a> 0 and a< 0. Here, you have |x+ 3| and |2x+ 1| so the cases, for |x+ 3| are x+ 3> 0 and x+ 3< 0 which give x>-3 and x< -3. The "cut point" for x is -3, not 0. For |2x+ 1|, the cases are 2x+1> 0 and 2x+ 1< 0 which give x> -1/2 and x< -1/2. The "cut point" for x is -1/2.

    So there are really three cases: x< -3, -3< x< -1/2, and x> -1/2.

     x+3 = 2x +1
    x=2

    Case 2: If x < 0. then
     -(x+3) = - (2x+1)
    x=2

    So we can conclude that the solution to the system is at x=2?

    I checked Wolfram but they have a solution of \frac{-4}{3} |x+3|=|2x+1| - Wolfram|Alpha

    Any Help is appreciated.
    Personally, I would not look a those cases. |x+ 3|= |2x+ 1| immediately gives x+ 3= 2x+ 1 and x+ 3= -(2x+ 1).
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  5. #5
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    Re: Absolute value equation

    Quote Originally Posted by sakonpure6 View Post
    So we have two solutions at x= 2 and -4/3 ?
    yes
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  6. #6
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    Re: Absolute value equation

    Quote Originally Posted by sakonpure6 View Post
    |x+3| = |2x+1|

    How do we go about solving this?
    Here is a different way to do these.
    a^2\le\ b^2~\iff~|a|\le|b|~. So solve (x+3)^2=(2x+1)^2
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