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Homework Help: Modulus Functions Equation

  1. Sep 24, 2010 #1
    I have to solve the equation: |12x -9| = |3 -4x| +2x -1

    I was wondering if it's supposed to be like:

    12x -9 = 3 -4x +2x -1 and -12x +9 = 3 -4x +2x -1 ?

    Or am I supposed to consider the positive and negative of 3 -4x also?
    I didn't attend lessons for this and I know little about modulus functions. :/
  2. jcsd
  3. Sep 24, 2010 #2


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    Actually what you do is you find what values of x make one of the absolute values more than zero, so lets do that now:

    [tex]12x-9\geq 0[/tex]

    [tex]x\geq 3/4[/tex]

    Therefore for all [tex]x\geq 3/4[/tex] that inequality is positive.

    For the other inequality, at x=3/4 we have 3-4x=0 and for any [tex]x\geq 3/4[/tex] we have [tex]3-4x\leq 0[/tex] So this means for all [tex]x\geq 3/4[/tex] we have the first inequality is more than zero, and the second is less than zero (so we take the negative of it). And things are just reversed when we consider when [tex]x<3/4[/tex].
  4. Sep 24, 2010 #3
    So does the inequality formed from |12x -9| applies to |3 -4x|?

    Then do I end up with something like 12x -9 = 0 +2x -1?
  5. Sep 25, 2010 #4


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    Yes it does apply to the other inequality.

    If you have |a|=|a+1| then you need to look at one of the absolute values. For the first, when [itex]a\geq 0[/itex] then you leave that as a, when [itex]a<0[/itex] you change that to -a (since the negative of a negative number is positive). But for the other inequality, when [itex]a+1\geq 0[/itex] or in other words, [itex]a\geq -1[/itex] then that absolute value is positive and thus left as a+1, so when [itex]a<-1[/itex] it becomes -(a+1)=-a-1.

    So this means when a<-1 we have both absolute values are negative so we solve -(a)=-(a+1) and when a>0 we have both are positive, so we solve a=a+1. What about for [itex]-1< a< 0[/itex]? Well we have that the first is negative, and the second is positive so we solve -a=a+1

    The same idea applies to your problem, so no, you don't solve 12x-9=0+2x-1. Use the idea of cases when x>3/4 and x<3/4 (this problem is easier since you only need two cases). Also, you can add in the value of x=3/4 later or just include it into each inequality, it doesn't really matter.
  6. Sep 25, 2010 #5
    Does this means that because |12x -9| is x > 3/4 and |3 -4x| is x< 3/4, when 12x -9 is positive, 3 -4x is negative and vice versa?

    So when I consider x > 3/4, I get
    12x -9 = 4x -3 +2x +1
    x = 7/6

    Is that correct or wrong?
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