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How to solve absolute value equation with two absolute values

  1. Sep 16, 2015 #1
    How does one solve an equation with two absolute value functions as below
    My algebra book does not show how to solve with two abs functions.

    2|4x-1| = 3|4x+2|

    I thought this might work..

    |4x-1|/|4x+2| = 3/2 then

    |(4x-1)/(4x+2)| = 3/2 and solve the normal way..
     
  2. jcsd
  3. Sep 16, 2015 #2

    symbolipoint

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    Reforming as rational equation will not help.
    Each expression inside absolute value function maybe be non-negative or negative, giving you four possible cases, but you really only need (placing stress on my logic), two cases.
    You could be able to do this using : Both expressions are nonnegative; or one expression is nonnegative while the other is negative.

    4x-1 maybe be nonnegative or it may be negative;
    4x+2 may be nonnegative or it may be negative.
    IF you want to check for the four possible cases, which is probably redundant,

    2(4x-1)=3(4x+2)
    ----------------------------------------
    2(1-4x)=3(4x+2)
    ---------------------------------------
    2(4x-1)=3(-1)(4x+2)
    ---------------------------------------
    2(1-4x)=3(-1)(4x+2)
    ---------------------------------------

    You can soon determine which of these are the two redundant equations.
     
  4. Sep 17, 2015 #3
    There's even a method to simply check for two cases only, by dividing both sides by ##|4x+2|##, $$\begin{align*}2|4x-1| = 3|4x+2|&\overset{\ ?}\iff \frac{2|4x-1|}{|4x+2|}=\frac{3|4x+2|}{|4x+2|} \end{align*}$$ Since we're dividing by ##|4x+2|## we must have ##x\neq -1/2##, so for the equivalence to hold we must check that either: ##x=-1/2## isn't a solution, and if it's a solution then we must include it. (As it turns out, it isn't, so we wont include it) $$\begin{align}2|4x-1| = 3|4x+2|&\iff \frac{2|4x-1|}{|4x+2|}=\frac{3|4x+2|}{|4x+2|}\\ &\iff2\frac{|4x-1|}{|4x+2|}= 3\\ &\iff\frac{|4x-1|}{|4x+2|}=\frac32\\ &\iff\left|\dfrac{4x-1}{4x+2}\right|=\frac32\\ &\iff\dfrac{4x-1}{4x+2}=\frac32\ \text{or}\ \dfrac{4x-1}{4x+2}=-\frac32\\ \end{align}$$ I'm sure you could take it from here.
     
    Last edited by a moderator: Sep 17, 2015
  5. Sep 17, 2015 #4
    Anama, I see why you should check to see if x = -1/2 is a solution. You say it is, I checked and it isn't , I don't think???
     
  6. Sep 17, 2015 #5
    My mistake, thanks, corrected
     
  7. Sep 17, 2015 #6

    symbolipoint

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    I would avoid trying to form the rational form of the equation, mostly because it creates, at least temporarily, a possible undefined x value. The rational form also should not be necessary.
     
  8. Sep 19, 2015 #7

    HallsofIvy

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    You could do this by looking at the three cases. 4x- 1= 0 when x= 1/4. 4x+ 2= 0 when x= -1/2.
    So the three cases are x< -1/2, -1/2< x< 1/4, and x> 1/4.
    1) If x< -1/2, it is also less than 1/4 so both 4x- 1 and 4x+ 2 are negative: |4x- 1|= -(4x- 1)= 1- 4x and |4x+ 2|= -(4x+ 2)= -4x- 2.
    2|4x- 1|= 2(1- 4x)= 2- 8x= 3|4x+ 2|= 3(-4x- 2)= -12x- 6.
    Solve 2- 8x= -12x- 6. (And check that any solution is less than -1/2.)

    2) If -1/2< x< 1/4 then 4x+ 2 is positive but 4x- 1 is still negative: |4x- 1|= 1- 4x.
    2|4x- 1|= 2(1- 4x)= 2- 8x= 3|4x+ 2|= 3(4x+ 2)= 12x+ 6.
    Solve 2- 8x= 12x+ 6. (And check that any solution is between -1/2 and 1/4.)

    3) If x> 1/4 then both 4x+ 2 and 4x- 1 are positive.
    2|4x- 1|= 2(4x- 1)= 8x- 2= 3|4x+ 2|= 3(4x+ 2)= 12x+ 6
    Solve 8x- 2= 12x+ 6. (And check that any solution is larger than 1/4.)
     
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