Thanks.  In my head it said that, but, as you know, I'm getting old and doddery, so what my fingers type doesn't always square with what my brain thinks.  I have edited the post.

If you look carefully at the graph, you see the end point pixels are included. 

Bob

Hi;

- 1 / 2 is a solution, isn't it?

hi Sabbir,

Welcome to the forum.

For any 'linear' expression

The critical value for x is

So, for your equation you need to consider these cases:

(i) x < - 1/2

expression becomes

Although, strictly that value is outside the range that I'm checking, a numeric check shows it is a solution.

(ii) x = -1/2 Just done that.

(iii) -1/2 < x < 1/2

expression becomes

This is true for all values of x in the range.

(iv) x = +1/2

expression becomes

So this is a solution.

(v)

expression becomes

Putting (i) to (v) together, the solution set is {x : -1/2 ≤ x ≤ 1/2}  or [-1/2,1/2] as noelevans has already explained.

Hope that helps,

Bob

Hi!

The general approach to absolute value equalities is to consider where the quantities inside the
absolute values are positive, zero, or negative.  Try the following:

Where both of 2x+1 and 2x-1 are positive replace the absolute values with these and solve.
Where both of 2x+1 and 2x-1 are negative replace the absolute values with the opposite of these
    and solve.
Where 2x+1 is positive and 2x-1 is negative replace |2x+1| with 2x+1 and |2x-1| with -(2x-1)
    and solve.
You will also find solutions where each of these quantities individually are zero.

You will find the solutions in [-1/2, 1/2].