Question 1207733
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Answers: <font color=red>x = 0, x = 1</font>



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. 
<font color=red>x = 0 is confirmed as a solution</font>.


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. 
<font color=red>x = 1 is confirmed as a solution</font>.


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 <font color=red>NOT</font> 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 <font color=red>x = 0, x = 1</font>



If we use Desmos to graph things out, then we get a V shape curve that represents y = |x + |3x-2||
<a href="https://www.desmos.com/calculator/aavxbn9myo">https://www.desmos.com/calculator/aavxbn9myo</a>
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.
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