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