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|x+3|+|x-1| = 5

  1. Dec 13, 2011 #1
    1. The problem statement, all variables and given/known data

    |x+3|+|x-1| = 5


    2. Relevant equations

    absolute value = distance from origin
    distance between 2 points on a line = |x1-x2|

    so |x+3| can be viewed as the distance from x to -3
    and |x-1| can be viewed as the distance from x to 1

    so we are trying to find a point x which has a distance from -3 and a distance from 1 that combine to equal 5...

    3. The attempt at a solution

    The options that add up to equal 5 are (0+5), (5+0), (2+3), (3+2), (1+4), (4+1)

    if |x+3| = 0, then x=-3.... that gives 4 instead of 5
    if |x+3| = 5, then x=2 or -8.... that gives 6 and 14 instead of 5
    if |x+3| = 3, then x=0 or -6... that gives 4 and 10 instead of 5
    if |x+3| = 2, then x=-1 or -5... that gives 4 and 8 instead of 5
    if |x+3| = 1, then x=-2 or -4... that gives 4 and 6 instead of 5
    if |x+3| = 4, then x=1 or -7... that gives 4 and 12 instead of 5


    ^ not only is that a really inefficient way to try to solve it. but I also couldn't get the answer! :(

    can someone please tell me a way to solve these problems?



    edit: I just realized I neglected all the options where x is not an integer...
     
    Last edited: Dec 13, 2011
  2. jcsd
  3. Dec 13, 2011 #2

    Mark44

    Staff: Mentor

    So the approach is not only ineffecient, but ineffective as well.

    Divide the number line into three intervals: (-∞, -3), (-3, 1), and (1, ∞)

    If x is in (-∞, -3), the equation can be written as -(x + 3) + -(x - 1) = 5. Can you solve this?

    Now look at the resulting equation in each of the other two intervals.

    The basic idea I am using is that |x - a| = -(x -a) if x < a, and |x - a| = x - a if x > a.
     
  4. Dec 13, 2011 #3

    SammyS

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    "not only is that a really inefficient way to try to solve it... " True !

    An instructive way to solve it (in my opinion) is to graph y = |x+3|+|x-1| .

    Alternatively,
    consider the piecewise definition of the absolute value.

    Consider separately the graphs: y = |x+3|, and y = |x-1| .​
     
  5. Dec 13, 2011 #4
    Ok great!

    so if x <= -3 ........... -3-x+1-x = 5 ; (ans: x=-7/2)

    if 1 >= x >= -3 ............ x+3+1-x = 5 (0=1 :frown: ) I realize that this indicates there is no solution for x in this interval but is there anything else to be learned from the result or should I just accept it as indicating no answer in the interval?

    if x > 1 ....... x+3+x-1 = 5 ; (ans: x=3/2)
     
  6. Dec 13, 2011 #5

    SammyS

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    Notice that everywhere on this interval, x+3+1-x = 4. So, it can's be 5.
     
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