# How to solve absolute value equation with two absolute values

1. Sep 16, 2015

### barryj

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. Sep 16, 2015

### symbolipoint

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)
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2(1-4x)=3(4x+2)
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2(4x-1)=3(-1)(4x+2)
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2(1-4x)=3(-1)(4x+2)
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You can soon determine which of these are the two redundant equations.

3. Sep 17, 2015

### Anama Skout

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
4. Sep 17, 2015

### barryj

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???

5. Sep 17, 2015

### Anama Skout

My mistake, thanks, corrected

6. Sep 17, 2015

### symbolipoint

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.

7. Sep 19, 2015

### HallsofIvy

Staff Emeritus
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.)