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Solving absolute value equations

  1. Apr 25, 2012 #1
    Is this a correct a way of thinking for solving absolute value equations? Say I have |2x+6|-|x+3|=|x| and want to solve for x, then I have:

    For |2x+6|
    2x + 6 if x ≥ -3
    -2x - 6 if x < -3

    For |x+3|
    x+3 if x ≥ -3
    -x-3 if x<-3

    For |x|
    x if x ≥ 0
    -x if x < 0

    Am I supposed to look at the cases where x is in a valid interval? e.g. I can't have 2x+6-(x+3) = x because x can't be equal to or greater than both 0 and 3.

    If this is the case, then why can't I have (-2x-6) - (-x+3) = -x? This is where x<-3 and x<0, isn't this valid? If x is less than -3 then it's also less than 0
     
  2. jcsd
  3. Apr 25, 2012 #2
    Yes, that is also a valid equation for x. You just have to look at each case, evaluating one absolute value at a time.
    If |2x+6|-|x+3|=|x|, then 2x + 6 - |x + 3| = |x| when x >= -3 and -2x - 6 - |x + 3| = |x| when x < -3.
    Now we evaluate |x + 3| in this pair of equations to get four possibilities:
    2x + 6 - x - 3 = |x| when x >=-3 and x >= - 3
    2x + 6 + x + 3 = |x| when x >=-3 and x < - 3
    -2x - 6 - x - 3 = |x| when x < -3 and x >= - 3
    -2x - 6 + x + 3 = |x| when x < -3 and x < -3
    Out of the four, only the first and last equations correspond to real values of x. Now we evaluate the |x| in each of those two equations to get four possibilities:
    2x + 6 - x - 3 = x when x >= -3 and x >= 0
    2x + 6 - x - 3 = -x when x >= -3 and x < 0
    -2x - 6 + x + 3 = x when x < -3 and x >= 0
    -2x - 6 + x + 3 = -x when x < -3 and x < 0
    The third equation does not correspond to any values of x, so we now have 3 equations without absolute values whose solutions are the same as those of the original equation with absolute values.
    x + 3 = x when x >= 0
    x + 3 = -x when x >= -3 and x < 0
    -x - 3 = -x when x < -3
    Since the first and last equation have no solutions, the middle equation contains the only valid solution for the original equation.
     
  4. Apr 25, 2012 #3

    mathman

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    The left side can be simplified. |2x+6| - |x+3| = |x+3|.

    So you have |x+3| = |x|.

    You have 3 cases.
    x < -3, -x -3 = -x, (no solution)
    -3 < x < 0, x + 3 = -x or x = -3/2
    0 < x, x+3 = x, (no solution)
     
  5. Apr 25, 2012 #4
    Thank you for the replies, I don't want to sound arrogant or anything but at the moment I'm kind of more interested in whether the way I currently think of it is accepted and if it is, why the last part of my initial post doesn't make sense (aside from the fact that you just can't get a solution when you solve it, I'm more interested in the intervals)
     
  6. Apr 25, 2012 #5

    HallsofIvy

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    Yes, that's all true.

    That's perfectly valid. If x< -3, then it is also less than 0 so all three of |2x+6|= -2x- 6, |x+3|= -x-3, and |x|= -x. The equation becomes -2x- 6- (-x- 3)= -x. That gives -x- 9= -x, which, since the two "-x" terms cancel, reduces to -9= 0 which is false for all x. Therefore, there is NO x<-3 satisfying the equation.

    If -3< x< 0, |2x+ 6|= 2x+ 6 and |x+3|= x+ 3 but |x| is still -x. The equation becomes 2x+ 6- (x+3)= -x. That gives x+ 3= -x which reduces to 2x= -3= or x= -3/2. Since that is between -3 and 0, that is a valid solution: |2x+6|-|x+3|= |-3+ 6|-|-3/2|= 3- 3/2= 3/2= |-3/2|.

    If 0< x, |2x+ 6|= 2x+ 6, |x+3|= x+ 3, and |x|= x. The equation becomes 2x+ 6- (x+3)= x. That gives x+ 3= x or 3= 0. Again, that is not true so there is no x larger than 0 that satisfies the equation.

    It is not a matter of "x< -3" not being valid- it is simply that, in that case, the equation reduces to one that has no solution.
     
  7. Apr 26, 2012 #6
    ^ Ah ok, I was thinking all valid intervals yielded valid equations...thank you
     
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