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Math Help - Rational, absolute value inequality

  1. #1
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    Question Rational, absolute value equality

    Hello there!
    I have been struggling with the next equality for some time now. The problem is, that I can get to the correct answers, but they are in the wrong intervals, and I can not figure out, what is it, that I am missing.

    So:
    | (3x-2)/(x-1) | = 2

    I understand, that the root is 2/3, and that if x>= 2/3, the expression inside the absolute value sign is positive, and negative otherwise. After taking the different cases into consideration, I get that if x is the element of (-inf,2/3), the solution is 4/5, and if x is the element of [2/3,inf), the solution is 0. I simply don't know, what am I doing wrong, I have had no problems, solving absolute value equations before, but this rational one confuses me a bit.
    Last edited by szucslaszlo; March 14th 2011 at 05:07 AM.
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  2. #2
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    This is not an inequality...
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  3. #3
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    Yes, you are right, I have mistyped it. Any suggestions concerning my problem?
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  4. #4
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    Quote Originally Posted by szucslaszlo View Post
    ...
    So:
    \begin{vmatrix}<br />
\frac{3x-2}{x-1}<br />
\end{vmatrix} = 2

    I understand, that the root is 2/3, and that if x>= 2/3, the expression inside the absolute value sign is positive, and negative otherwise. ...
    First, the sign of the "inside" of the absolute value depends on the

    3x - 2 (making x = 2/3 an "interesting point") AND on
    x - 1 (making x = 1 an "interesting point").

    You might want to make a number line, and mark the points x = 2/3 and x = 1. ...
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  5. #5
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    Ah, yes, I understand. I forgot to inspect the two polinomials seperatly. I'll check back if it worked out his way. Thanks.
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  6. #6
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    Quote Originally Posted by szucslaszlo View Post
    Hello there!
    I have been struggling with the next equality for some time now. The problem is, that I can get to the correct answers, but they are in the wrong intervals, and I can not figure out, what is it, that I am missing.

    So:
    | (3x-2)/(x-1) | = 2

    I understand, that the root is 2/3, and that if x>= 2/3, the expression inside the absolute value sign is positive, and negative otherwise. After taking the different cases into consideration, I get that if x is the element of (-inf,2/3), the solution is 4/5, and if x is the element of [2/3,inf), the solution is 0. I simply don't know, what am I doing wrong, I have had no problems, solving absolute value equations before, but this rational one confuses me a bit.
    Clearly, (though maybe not) this is an inequality, since if it is an equality, the solutions are 0, 4/5.

    Assuming this is |\frac{3x-2}{x-1}| > 2

    you require the fraction to be >2 or <-2

    \frac{3x-2}{x-1}>2 when we have +/+ or -/-

    \frac{3x-2}{x-1}<-2 when we have +/- or -/+



    x>1 and 3x>2 for +/+

    3x-2>2x-2\Rightarrow\ x>0\Rightarrow\ x>1

    x<1 and 3x<2 for -/-

    3x-2<2x-2\Rightarrow\ x<0

    x>1 and 3x<2 for -/+

    3x-2<-2(x-1)\Rightarrow\ 5x<4 contradiction as x>1

    x<1 and 3x>2 for +/-

    3x-2>-2(x-1)\Rightarrow\ 5x>4\Rightarrow\ 0.8<x<1


    All in all, you have x>0.8 and x<0

    (sorry, my modem was jammed and saw no responses)
    Last edited by Archie Meade; March 14th 2011 at 05:36 AM.
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  7. #7
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    Thank you, it worked out flawlessly!
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  8. #8
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    Thank you, though it was in fact an equality, I just had some (embarrasing) mehtodical problems. Thanks for your reply though, I will check the problem as an enaquality and it might come in handy.
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  9. #9
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    Hello, szucslaszlo!

    . . \displaystyle \left|\frac{3x-2}{x-1}\right| \:=\: 2

    I blush to confess that I got tangled up in my algebra,
    . . so I tried a graphical approach.

    The graph of y \:=\:\dfrac{3x-2}{x-1} looks like this:

    Code:
    
            |
            |         :
            |         :*
            |         :
            |         : *
            |         :  *
            |         :    *
            |         :       *
           3|         :             *
      - - - + - - - - : - - - - - - - -
        *   |         :
            *         :
            |  *      :
    --------+----*----:------------------
            |      *  :1
            |         :
            |       * :
            |         :
            |         :
            |        *:
            |         :
            |         :
            |         :
            |        *:
            |



    In the graph of y \,=\,\left|\dfrac{3x-2}{x-1}\right|
    . . everything below the x-axis is reflected upward.

    Code:
    
            |
            |         :
            |         :*
            |        *:
            |         : *
            |         :  *
            |         :    *
            |        *:       *
           3|         :             *
      - - - + - - - - : - - - - - - - -
        *   |       * :
            *         :
            |  *   *  :
    --------+----*----:------------------
            |         :1
            |         :
            |


    And we want the points where y \,=\,2.

    Code:
    
            |
            |         :
            |         :*
            |         :
            |         : *
            |         :  *
            |         :    *
            |        *:       *
            |         :             *
      . . . + . . . . : . . . . . . . .
        *   |        *:
      - - - ♥ - - - ♥ : - - - - -  y = 2
            |  *   *  :
     -------+----*----:------------------
            |         :1
            |         :
            |

    Now that I can see the solution, I can do the algebra.

    The answers are: . x \:=\:0,\,\frac{4}{5}

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