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].

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;

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

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