The solution is any #x in RR#.

The explanation is the following:

By definition, #|z| >= 0 AA z in RR#, so, applying this definition to our question, we have that #|2x+3| >=0#, which is a stronger condition tan #|2x+3| >= - 13# ("stronger" means that #|2x+3| >= 0# is more restrictive than #|2x+3| >= - 13#).

So now, instead of reading the problem as "solve #|2x+3| >= - 13#", we are going to read it as "solve #|2x+3| >= 0#" which, in fact, is easier to solve.

In order to solve #|2x+3|>= 0# we must again remember the definition of #|z|#, which is done by cases:

* If #z >= 0#, then #|z| = z#*

If #z < 0#, then #|z| = - z#

Applying this to our problem, we have that:

* If #(2x+3) >= 0 => |2x+3| = 2x+3# and then, #|2x+3| >= 0 => 2x+3 >= 0 => 2x >= - 3 => x >= - 3/2#*

If #(2x+3) < 0 => |2x+3| = - (2x+3)# and then, #|2x+3| >= 0 => -(2x+3) >= 0 => - 2x - 3 >= 0 => - 2x >= 3 => 2x <= -3# (observe that the sign of the inequality has changed on changing the sign of both members) #=> x <= - 3/2#

Since the result obtained in the first case is #AA x >= - 3/2# and the result obtained in the second case is #AA x <= - 3/2#, both put together give us the final result that the inequation is satisfied #AA x in RR#.