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Math Help - Absolute Value EQ w/ Fraction

  1. #1
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    Absolute Value EQ w/ Fraction

    It doesn't seem especially tough but there's not an example in the book and it was on a quiz today. Please note that I did consult the Inequality sticky at the top of the forum but I didn't see one quite like this. I apologize in advance if I missed it.


    3 \mid \frac{\2x-1}{3}\mid -1 \geq2

    My first thought was to multiply everything by the denominator 3 so that:

    3\mid \2x-1\mid -3 \geq6 Then: add 3 to 6 so that:

     3\mid \2x-1\mid \geq9 Then: divide by 3 so that:

    2x-1\geq3 Then: add 1 to both sides so that:

    2x\geq4 So that...

    x is equal to +2 and x is equal to -2

    It seemed tidy enough at the time but I have a feeling it's wrong; that maybe I should multiplied 2x-1 by 3 at some point or that I'm remiss in some way or another. Any and all help is appreciated. Thanks.
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  2. #2
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    Re: Absolute Value EQ w/ Fraction

    Quote Originally Posted by Trypanosoma View Post
    3 \mid \frac{\2x-1}{3}\mid -1 \geq2
    Simply note that
    3\left|\dfrac{2x-1}{3}\right|=\left|{2x-1}\right|
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    Re: Absolute Value EQ w/ Fraction

    Quote Originally Posted by Plato View Post
    Simply note that
    3\left|\dfrac{2x-1}{3}\right|=\left|{2x-1}\right|
    A Picard facepalm moment for me... I've also noticed that not multiplying -1 or 2 that my answer still comes out to same. Will that stand up for precisely similar problems?

    But what of something like

    A. 3\mid \frac{\2x-1}{<b>4</b>}\mid -1\geq 2

    would I then have to go with the LCD of 12 and then have...

    9\mid2x-1\mid -3 \geq6

    Again, thank you so much.

    Or does that 3 just get left alone, so that:

    3\mid2x-1\mid -3 \geq6

    And (sorry), would I then divide the right side by the 9 or 3 (depending on which is correct--if correct at all).
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    Re: Absolute Value EQ w/ Fraction

    Quote Originally Posted by Trypanosoma View Post
    A Picard facepalm moment for me... I've also noticed that not multiplying -1 or 2 that my answer still comes out to same. Will that stand up for precisely similar problems?
    But what of something like.
    I in no way suggested that your solution is wrong.
    I just wanted to point that it was awkward.
    You should know that |ab|=|a||b|.

    So your problem quickly simplifies to |2x-1|\ge 3.
    Thus 2x-1\ge 3\text{ or }2x-1\le -3~.
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