Question 1207733: Solve the absolute value equation.
Solve |x + |3x-2|| = 2
Note: Here we have an absolute value inside another absolute value.
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answers: x = 0, x = 1
Explanation
Rule: If |x| = k then x = -k or x = k
where k is nonnegative
Using that rule we go from this
|x + |3x-2|| = 2
to this
x + |3x-2| = 2 or x + |3x-2| = -2
Let's solve each separately.
x + |3x-2| = 2
|3x-2| = 2-x
3x-2 = 2-x or 3x-2 = -(2-x)
3x-2 = 2-x or 3x-2 = -2+x
3x+x = 2+2 or 3x-x = -2+2
4x = 4 or 2x = 0
x = 4/4 or x = 0/2
x = 1 or x = 0 are two potential solutions.
Now solve the other equation
x + |3x-2| = -2
|3x-2| = -2-x
3x-2 = -2-x or 3x-2 = -(-2-x)
3x-2 = -2-x or 3x-2 = 2+x
3x+x = -2+2 or 3x-x = 2+2
4x = 0 or 2x = 4
x = 0/4 or x = 4/2
x = 0 or x = 2
We have x = 0 show up again, but x = 2 is a new potential solution.
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Summary so far:
The 3 potential solutions are
x = 0, x = 1, or x = 2
Like with any equation, we need to check said solutions.
This step is very important.
Let's start with x = 0.
Replace each x with 0 and then apply PEMDAS.
|x + |3x-2|| = 2
|0 + |3*0-2|| = 2
|0 + |0-2|| = 2
|0 + |-2|| = 2
|0 + 2| = 2
|2| = 2
2 = 2
That checks out.
x = 0 is confirmed as a solution.
Now let's try x = 1
|x + |3x-2|| = 2
|1 + |3*1-2|| = 2
|1 + |3-2|| = 2
|1 + |1|| = 2
|1 + 1| = 2
|2| = 2
2 = 2
That works as well.
x = 1 is confirmed as a solution.
Lastly x = 2
|x + |3x-2|| = 2
|2 + |3*2-2|| = 2
|2 + |6-2|| = 2
|2 + |4|| = 2
|2 + 4| = 2
|6| = 2
6 = 2
The last statement is false since the numbers don't match up.
A false equation at the end means the first equation of this sub-block is false when x = 2.
It proves that x = 2 is NOT a solution. It's considered extraneous.
You can see why we needed to check each potential solution.
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Ultimately there are 2 solutions and they are x = 0, x = 1
If we use Desmos to graph things out, then we get a V shape curve that represents y = |x + |3x-2||
https://www.desmos.com/calculator/aavxbn9myo
I used the "abs" function to indicate absolute value in Desmos.
Example: abs(3x-2) = |3x-2|
The V shape and the horizontal line y = 2 intersect at the two points (0,2) and (1,2) shown in green and purple respectively.
The x coordinates of these points are the solutions mentioned above.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
If we think of this as a function, we know the behavior of the graph changes when (3x-2) is 0 -- i.e., at x=2/3. So let's solve two separate absolute value equations, considering only values of x greater than or equal to 2/3 in one equation and considering only values of x less than 2/3 in the other.
(1) If x is greater than or equal to 2/3, then |3x-2| is equal to 3x-2:
(1a) 
x=1 satisfies our condition that x is greater than or equal to 2/3, so x=1 is a solution.
OR
(1b) 
x=0 does NOT satisfy our condition that x is greater than or equal to 2/3, so x=0 is not a solution.
(2) If x is less than 2/3, then |3x-2| is equal to -(3x-2) = -3x+2:
(2a) 
x=0 satisfies our condition that x is less than 2/3, so x=0 is a solution.
OR
(2b) 
x=2 does NOT satisfy our condition that x is less than 2/3, so x=2 is not a solution.
We found two solutions above, in cases (1a) and (2a).
ANSWER: x=1 or x=0
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