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Bunuel
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Graphical approach is much better
We can easily draw the graph of y=||x-3|-2| by the following steps:-
Step 1:We will first draw a line which has slope 1 and y intercept -3 (y=x-3)
Step 2:Then we would flip the line over x axis. This would result in a 'V' shaped graph with vertex(of 'V') on x=3. This is the graph of y=|x-3|
Step 3:Then we will shift the V shaped Line 2 units down which would result in the graph of y=|x-3|-2
Step 4: Further we will again flip the graph over x axis to get a 'W' shaped graph. This is finally the graph of y=||x-3|-2|

We would find the number of intersection points of this graph with the line y=1 (line || x axis) to finally get the answer

NOTE- In question if it is asked 'no. of solutions', then graphical method is the best and hassle-free approach. But if it would ask the values of those soultions then we need to solve by cases
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We need to find how many solutions does the equation ||x - 3| - 2| = 1 have?

Now, | |x - 3| - 2 | = 1
=> |x - 3| - 2 = 1 or -1

Now we have |x - 3| = 3 or 1
-Case 1: |x - 3| = 3
=> x - 3 = 3 or -3
=> x = 6, 0 is a SOLUTION
-Case 2: |x - 3| = 1
=> x - 3 = 1 or -1
=> x = 4, 2 is a SOLUTION

So, there are four solutions for x = 0, 2, 4, 6

So, Answer will be E
Hope it helps!

Watch the following video to MASTER Absolute Values

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I'm making this response because its a topic that I really struggle with and It will help me to breakdown the logic behind the question:

||x-3|-2|=1

take |x-3| and label it as y. y>=0

|y-2|=1

when looking at this equation we can see that y can take two values: 3 and 1

if y=3 then |3-2|=1 - this checks out
if y=1 then |1-2|=1 - this checks out

Therefore y which is actually |x-3| can assume two values - 3 and 1

Thus |x-3|=3 or 1

|x-3|=3 - x can be 0 or 6
|x-3|=1 - x can be 2 or 4

Hence the solutions has 4 values



Bunuel
How many solutions does the equation ||x - 3| - 2| = 1 have?

A. 0
B. 1
C. 2
D. 3
E. 4
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