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Thread: Absolute Value Inequality

  1. #1
    Super Member harpazo's Avatar
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    Absolute Value Inequality

    Solve |2x - 1| + 5 > 2

    My Work:

    This is an absolute value inequality. Thus, we are not solving for x.

    |2x - 1| + 5 > 2

    |2x - 1| > 2 - 5

    |2x - 1| > -3

    At this point, I created two cases.

    2x - 1 < 3

    2x < 3 + 1

    2x < 4

    x < 4/2

    x < 2

    or

    2x - 1 > -3

    2x > -3 + 1

    2x > -2

    x > -2/2

    x > -1

    The solution to |2x - 1| + 5 > 2 is found here -1 < x < 2.

    Correct?
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  2. #2
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    Re: Absolute Value Inequality

    |2x-1|>-3

    First Case:
    (2x-1) is +ve
    2x-1>=0
    x>=1/2 -----------(i)

    since supposed term to be +ve
    just remove modulus.

    (2x-1)>-3
    x>-1 ---------------(ii)

    On solving (i) and (ii):
    x belongs to [1/2,infinity]


    Second Case:

    (2x-1) is -ve

    Removing modulus:

    (2x-1)<-3
    x<-1

    x belongs to (-infinity,-1)

    so overall value (-infinity,-1)U[1/2,infinity)
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  3. #3
    MHF Contributor
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    Re: Absolute Value Inequality

    There's no work to do: $|2x-1| > -3$ for all $x$.
    Thanks from topsquark, mak29 and harpazo
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  4. #4
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    Re: Absolute Value Inequality

    Quote Originally Posted by mak29 View Post
    |2x-1|>-3

    First Case:
    (2x-1) is +ve
    2x-1>=0
    x>=1/2 -----------(i)

    since supposed term to be +ve
    just remove modulus.

    (2x-1)>-3
    x>-1 ---------------(ii)

    On solving (i) and (ii):
    x belongs to [1/2,infinity]


    Second Case:

    (2x-1) is -ve

    Removing modulus:

    (2x-1)<-3
    x<-1

    x belongs to (-infinity,-1)

    so overall value (-infinity,-1)U[1/2,infinity)
    Sorry for the solution.
    As stated by Archie , no need to do this.
    Thanks from harpazo
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  5. #5
    Super Member harpazo's Avatar
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    Re: Absolute Value Inequality

    Thank you everyone.
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