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Thread: absolute inequality

  1. #1
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    [solved]absolute inequality

    Hi, good day

    Can anyone helps me in solving this absolute inequality?
    $\displaystyle
    \frac{|x+1|}{|3x-5|}<1
    $

    Part 1:
    $\displaystyle \frac{x+1}{3x-5}<1$

    $\displaystyle \frac{x+1}{3x-5}-1<0$

    $\displaystyle \frac{-2x+6}{3x-5}<0$

    Part 2:
    $\displaystyle \frac{x+1}{3x-5}>-1$

    $\displaystyle \frac{x+1}{3x-5}+1>0$

    $\displaystyle \frac{4x-4}{3x-5}>0$

    Next?

    p/s:I know there is other method, something like squaring both sides, but I find it untrue for certain questions.
    Last edited by imforumer; Mar 21st 2009 at 12:24 AM. Reason: Problem solved.
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  2. #2
    MHF Contributor
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    I don't get why you are leaving these as fractions both to work with and as your final answer for each part. You haven't established a domain for x for which this inequality is true, which is what you want to do.
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  3. #3
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    Post

    Ok, I've got the answer already. TQ
    Last edited by imforumer; Mar 21st 2009 at 12:23 AM.
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