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Interesting Absolute Value problem

  1. Jul 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Find all x in

    |5x-2| = 6x-12





    3. The attempt at a solution

    http://www.wolframalpha.com/input/?i=Plot[|5x-2|+%3D+6x-12%2C{x%2C-1.5%2C1.5}]

    Notice how only x = 10 is a solution?

    So my question is, why isn't x = -14/11 a solution? I mean sure they don't intersect on the graph, but is there anyway of inspection on the equation to know that x = -14/11 is not a solution?
     
  2. jcsd
  3. Jul 3, 2011 #2
    Hi, the solutions are:

    5x-2=6x-12 x=10

    2-5x=6x-12 x=14/11

    why do you say -14/11?
     
  4. Jul 3, 2011 #3

    Mentallic

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    Because while we are solving [itex]5x-2=6x-12[/itex] and [itex]5x-2=-(6x-12)[/itex] we also have to place restrictions of the values of x in each of these cases.

    For example, in the first case we are assuming [itex]5x-2\geq 0[/itex] which means [itex]x\geq 2/5[/itex] and now if we get a value of [itex]x<2/5[/itex] then we have a contradiction and that solution is not valid.
    For the second case where we assume [itex]5x-2<0[/itex] thus [itex]x<2/5[/itex] but we instead get an answer of [itex]x=14/11>2/5[/itex] this answer is invalid, so we scrap it.
     
  5. Jul 4, 2011 #4
    But, if instead, we make fewer assumptions and take just the original formula at face value
    |5x-2| = 6x-12 obviously means [itex]0\leq 6x-12[/itex] by the definition of absolute values. This means [itex]x\geq 2[/itex]. This eliminates the same false answer for the OP.

    Is this incorrect? And what relation do the values involving 2/5 have?
     
  6. Jul 4, 2011 #5

    SammyS

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    Look at this graph from http://www.wolframalpha.com/input/?i=plot+|5x-2|+,+6x-12,+x=-4..12".

    A purely algebraic approach:
    [tex]|5x-2|=\left\{\begin{array} , 5x-2&,&\text{if }\ 5x-2\ge0\\ -5x+2&,&\text{if }\ 5x-2<0\end{array}\right.\quad=\left\{\begin{array} , 5x-2&,&\text{if }\ x\ge\frac{2}{5}\\ -5x+2&,&\text{if }\ x<\frac{2}{5}\end{array}\right.[/tex]
    Case1: x ≥ 2/5, therefore |5x-2| = 5x-2

    Solving |5x-2| = 6x-12 gives 5x-2 = 6x-12 which gives x = 10 .

    Case2: x<2/5, therefore |5x-2| = -5x+2

    Solving |5x-2| = 6x-12 gives -5x+2 = 6x-12 which gives x = 14/11 .
    but 14/11 > 2/5, so it is inconsistent with this case.

    Therefore, there is only one solution, x = 10 .
     
    Last edited by a moderator: Apr 26, 2017
  7. Jul 5, 2011 #6

    Mentallic

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    No, it's correct :smile:
    The method SammyS and I used by breaking it into cases and testing each one has a correlation with solving inequalities involving fractions where the unknown variable is in the denominator.

    Yours is much simpler though :wink:
     
  8. Jul 12, 2011 #7

    You are my hero.
     
    Last edited by a moderator: Apr 26, 2017
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