The absolute value |x| is defined as

|x| = `{(x if x>=0),(-x if x<0):}`

The equation to be solved is |x - 1| = |3 - x|

If `x - 1 < 0` and `3 - x < 0` , the given equation is 1 - x = x - 3

=> x = 2

But this does not satisfy `x < 1` or `x > 3`

If `x - 1 >= 0` and `3 - x < 0` , the equation is x - 1 = x - 3

There is no solution in this case.

A similar case arises with `x - 1 < 0` and `3 - x >= 0`

For `x - 1 >= 0` and `3 - x >= 0` , the equation is x - 1 = 3 - x

=> x = 2

x = 2 satisfies x > 1 and x < 3

**The solution of the equation |x - 1| = |3 - x| is x = 2**

Whenever you have such equations divide the number line into, 1+ the number of constants in the equation. Here the total constants are two so the number line will be divided into 3 parts. The constants should be determined by individually equating the LHS and RHS to zero so here the constants will be 1 and 3. The three parts here should be

1.-infinity to first constant( the smaller one among the two) here it is 1

2. first constant to second constant here 1 to 3

3. Second constant to infinity

Case I.-infinity to 1.

Here x is less than 1 and also less than 3.

so |x-1|=-(x-1)( since |x|=-x for x<0)

and |3-x|=3-x(since |x|=x for x>0).

This implies 1=3 hence no solution of equation in this portion of number line

II.1 to 3

Here x is greater than 1 but less than 3

so |x-1|=(x-1)( since |x|=x for x>0)

and |3-x|=3-x(since |x|=x for x>0).

this implies x=2. Hence 1 solution in this region

III.3 to infinity.

Here x is greater than 1 and greater than 3

so |x-1|=(x-1)( since |x|=x for x>0)

and |3-x|=-(3-x)(since |x|=-x for x>0)

this implies -1=-3 which is not possible

Hence the only solution is x=2