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hi Bunuel,

I solved this but I am not getting the -2 case, can you advise where I am going wrong or which case I am missing out?

I solved inside out from the inside modulus to outside:
Case I: x>=1:
| x - 1 -2 | = 1
|x-3| = 1
Case I.I:
x > 3
x-3 = 1
=> x = 4

Case 1.2
1<x<3
3-x = 1
x = 2

Case 2:
|1-x-2| = 1
|1-x| = 1
x = 0

thank you!
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miag
hi Bunuel,

I solved this but I am not getting the -2 case, can you advise where I am going wrong or which case I am missing out?

I solved inside out from the inside modulus to outside:
Case I: x>=1:
| x - 1 -2 | = 1
|x-3| = 1
Case I.I:
x > 3
x-3 = 1
=> x = 4

Case 1.2
1<x<3
3-x = 1
x = 2

Case 2:
|1-x-2| = 1
|1-x| = 1
x = 0


thank you!

When x < 1, we get |-(x - 1) - 2| = 1, which gives |-x - 1| = 1. So, x = 0 or x = -2.
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thank you - got it! Somehow was just unable to see this even after going through it multiple times! :|
Bunuel
miag
hi Bunuel,

I solved this but I am not getting the -2 case, can you advise where I am going wrong or which case I am missing out?

I solved inside out from the inside modulus to outside:
Case I: x>=1:
| x - 1 -2 | = 1
|x-3| = 1
Case I.I:
x > 3
x-3 = 1
=> x = 4

Case 1.2
1<x<3
3-x = 1
x = 2

Case 2:
|1-x-2| = 1
|1-x| = 1
x = 0


thank you!

When x < 1, we get |-(x - 1) - 2| = 1, which gives |-x - 1| = 1. So, x = 0 or x = -2.
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