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Absolute within an absolute

  1. Oct 18, 2012 #1
    Solve:

    | |x+1| +2| - | x-2 | = 3

    Relevant equations:

    if |x| = a, then x = a; x = -a

    My attempt:
    |x+1| +2 - (x-2) = 3 ; |x+1| + 2 - (x-2) = -3 (by theorem provided by teacher above)
    |x+1| = x- 1 ; |x+1| = x-7

    if |x+1| < 0:

    -(x+1) = x -1
    -x - 1 = x - 1
    -2x = 0
    x = 0
    ------------------------
    -(x+1) = x-7
    -x-1 = x-7
    -2x = -6
    x = 3


    If |x+1| => 0:

    x+1 = x-1
    1≠ -1
    -------------------------
    x+1 = x-7
    1≠ -7



    The answer is 1 and I'm not sure if the theorem my teacher provided applies...
     
    Last edited: Oct 18, 2012
  2. jcsd
  3. Oct 18, 2012 #2

    HallsofIvy

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    Are you giving this any thought or just trying to copy what you have seen before?

    x= -1 is clearly NOT the solution, it does not satisfy the equation:
    |x+ 1|= 0 so ||x+1|+ 2|= |2|= 2 while |x- 2|= |-1- 2|= |-3|= 3. 2- 3 is NOT equal to 3.

    I would start off doing this is a series of cases:

    case 1) x> 2
    Then x> -1 so |x+1|= x+ 1 and then |x+1|+ 2= x+ 1+ 2= x+ 3> 0 so ||x+1|+ 2|= x+ 3.

    case 2) [itex]-1\le x\le 2[/itex]
    |x+ 1| is still equal to x+ 1 so ||x+1|+ 2|= |x+ 3|= x+ 3 but now |x- 2|= 2- x. The equation becomes x+ 3- (2- x)= 2x+ 1= 3. 2x= 2, x= 1.
    If x= 1, |x+ 1|= 2 so ||x+1|+ 2|.

    case 3) x< -1.
    |x+ 1|= -x- 1 so ||x+ 1|+ 2|= |-x-1+ 2|= |-x+ 1|. With x< -1, that is positive so ||x+ 1|+ 2|= -x+ 1.
     
  4. Oct 18, 2012 #3
    Oh i meant to type 1, sorry for the typo :(
     
  5. Oct 18, 2012 #4

    SammyS

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    Supposing that the theorem provided by teacher is:
    If |x| = a, where a ≥ 0, then x = a, or x = -a .​
    That theorem cannot be directly applied to the equation
    | |x+1| +2 | - | x-2 | = 3​
    in the manner which you applied it.

    Another way to state that theorem is:
    If |x| = a, where a ≥ 0, then x = a, or -x = a .​
    If you carefully consider the first term in your equation, the one with the absolute value inside the absolute value, you can simplify the equation a bit.

    Is |x+1| + 2 ever negative? ... No !

    Then | |x+1| +2 | = |x+1| +2 .
     
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