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

Bob

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

]]>The solution is correct, but the interval is wrong. It should be an open interval: (-1/2,1/2)=]-1/2,1/2[.

]]>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

]]>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].

]]>