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Solving an inequality

  1. Jan 8, 2008 #1
    1) [tex]|2x+1|<4x-2[/tex]

    2) [tex]|2x-1|>x+2[/tex]

    3) [tex]|\frac {x-2}{x+1}|<3[/tex]

    4) [tex]|2x-1|>\frac {1}{x}[/tex]


    (Show me as many methods as possible. Thanks)
     
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  3. Jan 8, 2008 #2

    symbolipoint

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    Each of those shows only a single absolute value in the left member. Ask yourself what happens when the expression inside the absolute value function is positive or zero; and what happens when the expression inside is less than zero. Continue with each condition to find a solution set for each exercise.
     
  4. Jan 8, 2008 #3

    arildno

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    Take 1:

    Split up in the two cases:
    A) 2x+1<4x-2 AND 2x+1>0
    B) -(2x+1)<4x-2 AND 2x+1<0

    Take A:
    The second inequality requires x>-1/2
    The first requires 3<2x, that is x>3/2

    Thus, to fulfill both of these inequalities, we must have x>3/2 as the solution to A.

    Now, let us tackle B:
    The second inequality requires x<-1/2

    The first requires:
    1<6x, implying x>1/6

    But these two inequalities cannot be fulfilled simultaneously, i.e, there are no solutions to case B

    Thus, the entire solution to 1) is x>3/2
     
  5. Jan 8, 2008 #4

    epenguin

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    Since you ask as many methods as possible, sketch or drwa a graph of both functions. Eg the | | part for the first two has a V-shape. It will help give you a sense of what is happening and, also as a habit, pick out mistakes sometimes.
     
  6. Jan 8, 2008 #5
    1) [tex]|2x+1|<4x-2[/tex]

    [tex]2x+1<4x-2[/tex]
    [tex]3<2x[/tex]
    [tex]3/2<x[/tex] ...... i


    [tex]2x+1>2-4x[/tex]
    [tex]6x>1[/tex]
    [tex]x>1/6[/tex] ...... ii

    To fulfill both i and ii, therefore [tex]3/2<x[/tex].

    Can I do like that?
     
    Last edited: Jan 9, 2008
  7. Jan 8, 2008 #6

    symbolipoint

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    You could always take the three different regions around the two critical points in case you feel confused.
     
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