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Solve for x |x+1| = 2x-1

  1. Apr 1, 2006 #1
    Is this right:

    x+1 = -2x+1
    2x+x = 1-1
    3x = 0

    No solution?
     
  2. jcsd
  3. Apr 1, 2006 #2

    VietDao29

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    No, that's not correct.
    I'll give you some hints to start of the problems.
    First, you should notice that, can |A| ever be negative?
    So if you have an equation: |A| = B, can B be negative?
    And you will have:
    [tex]|A| = \left\{ \begin{array}{l} A, \quad \mbox{if } A \geq 0 \\ -A , \quad \mbox{if } A < 0 \end{array} \right.[/tex]
    So to solve something like |A| = B, you will have to solve the system of equations:
    [tex]\left\{ \begin{array}{l} B \geq 0 \\ \left[ \begin{array}{l} A = B \\ A = -B \end{array} \right. \end{array} \right.[/tex]
    Can you get it? Can you go from here? :)
     
  4. Apr 1, 2006 #3
    ok so
    -x+1 = 2x-1
    -3x = -2
    x=-2/3

    ?
     
  5. Apr 1, 2006 #4

    VietDao29

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    Again, it's wrong... :frown:
    First -(x + 1) is -x - 1, not -x + 1.
    Now, just do it step by step, and see if you can get the answer.
    First, in the equation you have:
    |x + 1| = 2x - 1.
    Just compare it with |A| = B, so what's A, and what's B, you think?
    And can you solve this system
    [tex]\left\{ \begin{array}{l} B \geq 0 \\ \left[ \begin{array}{l} A = B \\ A = -B \end{array} \right. \end{array} \right.[/tex]?
    But anyway, you should read my previous post again to understand the concept more deeply.
    Can you go from here? :)
     
  6. Apr 1, 2006 #5
    so |A| is |x+1| and B is 2x-1
    |A| can be negative
    B cannot
    -x-1 = 2x-1
    -2x-x = 1-1
    -3x = 0 ???

    I'm sorry if I'm just not seeing this
     
  7. Apr 1, 2006 #6

    VietDao29

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    The absolute value of A is never negative.
    Hence |A| >= 0
    If |A| = B, and A >= 0, it's true that B >= 0, right? Now, just look at my first post in this thread to see if you can get it.
    Now can you solve for the value of x that makes B non-negative?
    By the way, there should be some examples in your books, let's give it a glance to see if you find any.
    I'll give you an example, though:
    ------------------
    Example:
    Solve the equation:
    |2x + 1| = x - 1
    [tex]\Leftrightarrow \left\{ \begin{array}{l} x - 1 \geq 0 \\ \left[ \begin{array}{l} 2x + 1 = x - 1 \\ 2x + 1 = -(x - 1) \end{array} \right. \end{array} \right.[/tex]
    [tex]\Leftrightarrow \left\{ \begin{array}{l} x \geq 1 \\ \left[ \begin{array}{l} x = - 2 \\ 2x + 1 = -x + 1 \end{array} \right. \end{array} \right.[/tex]
    [tex]\Leftrightarrow \left\{ \begin{array}{l} x \geq 1 \\ \left[ \begin{array}{l} x = - 2 \\ 3x = 0 \end{array} \right. \end{array} \right.[/tex]
    [tex]\Leftrightarrow \left\{ \begin{array}{l} x \geq 1 \\ \left[ \begin{array}{l} x = - 2 \\ x = 0 \end{array} \right. \end{array} \right.[/tex]
    x = -2 is not a valid solution, right? Because in order for x - 1 >= 0, x must be greater than or equal to 1, and -2 is obviously less than 1.
    And neither is x = 0, hence there's no x such that:
    |2x + 1| = x - 1
    ------------------
    Can you go from here? :)
     
    Last edited: Apr 1, 2006
  8. Apr 1, 2006 #7
    alright, I'll work on it, thanks for your help!
     
  9. Apr 1, 2006 #8

    krab

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    Science Advisor

    |x+1| is always >0. Now look at the right side. It means that x MUST be larger than 1/2. So x+1 is >0 and we drop the absolute value sign: x+1=2x-1. It's pretty easy to solve.
     
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