Question

the solution set of the following inequality modulus x minus 1 greater than or equal to modulus x minus 3

Answer 1

I think this is what you mean:
| x-1 | ≥ | x-3 |
you can solve this question in more than one way.
One of these is,
| x-1 | - | x-3 | ≥ 0
Find the critical points, here they are 1 and 3.
solve for x<1,
-(x-1)-(-(x-3))≥0
-x+1+x-3≥0
-2≥0 (not possible)
Thus, x should be greater than 1
For 1 ≤ x ≤ 3
(x-1)-(-(x-3))≥0
2x-4≥0
x≥2
So one solution set is 2≤x≤3
For x>3
(x-1)-(x-3)≥0
2≥0
Therefore from above two results,
x≥2
Hope you got it

Answer 2

Given  : | x - 1 |  ≥  | x  - 3 |  

To find : Find x

Solution:

|x| =x  if  x≥ 0

|x| = - x if  x < 0

| x - 1 |  ≥  | x  - 3 |

case 1  

x  ≥  3

=> x - 1 ≥ x - 3

=> - 1 ≥  - 3

Which is true

Hence  x   ≥  3  

x ∈ [3 , ∞)

Case 2

x < 1

-(x - 1)   ≥  -(x - 3)

=> -x + 1  ≥ -x + 3

=> 1 ≥ 3

Not possible hence

x can not be < 1

case 3

1 ≤ x < 3

x - 1 ≥ -(x - 3)

=> x -  1  ≥ -x  + 3

=> 2x  ≥ 4

=> x  ≥ 2

=> x ∈ [ 2 , 3)

x ∈ [ 2 , 3)  ∪ x ∈ [3 , ∞)

=> x ∈ [2 , ∞)   or  x ≥ 2

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