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Solving inequality with absolute values |3x-2|<=x+1

  1. Feb 14, 2005 #1
    1) │3x-2│<= x+1 ; x>=-1
    Case1: 3x-2>=0
    x>= 2/3
    3x-2<=x+1
    x<=3/2

    what is case 2???

    2) │2-3x│ < 3x-4

    3) │x-3│=x-2
    How do u solve these?
     
  2. jcsd
  3. Feb 14, 2005 #2

    dextercioby

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    Apply the definition of the absolute value correctly.
    Case 1.3x-2>=0 (therefore x>=2/3) -------->3x-2<=x+1------------>x<=3/2.The solution:[tex] x\in [\frac{2}{3},\frac{3}{2}] [/tex]

    Case 2.3x-2<0 (therefore x<2/3)---------->-(3x-2)<=x+1------------>x>=1/4.The solution is:[tex] x\in [\frac{1}{4},\frac{2}{3}) [/tex]

    The solution of the problem is found by reuniting the partial solutions
    [tex] x\in [\frac{1}{4},\frac{2}{3}] [/tex]

    Do the same for the other 2...

    Daniel.
     
    Last edited: Feb 15, 2005
  4. Feb 14, 2005 #3
    how do u know case 2 would be -(3x-2)<=x+1 with a negative sign?
     
  5. Feb 14, 2005 #4

    dextercioby

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    Because that's the definition of the absolute value
    |x|=x,for x>=0 and -x for x<0...

    Daniel.
     
  6. Feb 14, 2005 #5
    hm.....can u do one more for me?
     
  7. Feb 14, 2005 #6

    dextercioby

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    Nope.What is the result of applying the definition of an absolute value to point b)...?

    Daniel.
     
  8. Feb 14, 2005 #7
    What point b?
     
  9. Feb 14, 2005 #8

    dextercioby

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    Exercise 2),sorry...

    Daniel.
     
  10. Feb 14, 2005 #9
    hm...ok...let me try
    case 1: 2-3x>=0
    x<=2/3
     
  11. Feb 14, 2005 #10

    learningphysics

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    I believe the above should be:
    [tex] x\in [\frac{1}{4},\frac{2}{3}) [/tex]

    And the final solution, union:
    [tex] x\in [\frac{1}{4},\frac{3}{2}] [/tex]
     
  12. Feb 15, 2005 #11

    learningphysics

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    I don't think you need to consider the value inside the || separately for positive and negative. (Unless your teacher wants you to do it that way to fully understand the steps.)

    For the first question, I'd just say:
    │3x-2│<= x+1 so
    -(x+1)<=3x-2<=x+1
    or in other words
    -(x+1)<=3x-2 AND 3x-2<=x+1
    The first inequality gives x>=1/4, the second gives x<=3/2

    The solution is the intersection of x>=1/4 and x<=3/2, so the solution is:
    1/4<=x<=3/2

    Although it is instructive to consider separately the positive and negative values inside ||, it isn't necessary to solve the inequality.
     
  13. Feb 15, 2005 #12
    Do you mean consider the value of x is bigger or equal to 2/3 first? or Less than or equal ?
    In the past, I have tried inequalities that with several absolute sign inside. It is extremely important to define the value first.
    However, I haven't learnt this in my lessons. Maybe later. Therefore, I don't know whether in this question this distinction is needed.
     
  14. Feb 15, 2005 #13

    learningphysics

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    I don't see why it is necessary...

    if |a|<b, then

    we know that a<b and a>-b

    This is true whether or not a is positive or negative or 0.

    If |a|>b, then

    we know that a>b or a<-b. This statement is also true whether or not a is positive or negative or zero.
     
  15. Feb 15, 2005 #14

    dextercioby

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    Yes,thank you for noticing.I edited my post and now it's "dandy"... :wink:

    Daniel.
     
  16. Feb 15, 2005 #15
    Yes. It is not necessary for this case.
     
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