# How do you solve abs(2x + 3)>= -13?

Apr 9, 2015

The solution is any $x \in \mathbb{R}$.

The explanation is the following:

By definition, $| z | \ge 0 \forall z \in \mathbb{R}$, so, applying this definition to our question, we have that $| 2 x + 3 | \ge 0$, which is a stronger condition tan $| 2 x + 3 | \ge - 13$ ("stronger" means that $| 2 x + 3 | \ge 0$ is more restrictive than $| 2 x + 3 | \ge - 13$).

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

In order to solve $| 2 x + 3 | \ge 0$ we must again remember the definition of $| z |$, which is done by cases:
If $z \ge 0$, then $| z | = z$
If $z < 0$, then $| z | = - z$

Applying this to our problem, we have that:
If $\left(2 x + 3\right) \ge 0 \implies | 2 x + 3 | = 2 x + 3$ and then, $| 2 x + 3 | \ge 0 \implies 2 x + 3 \ge 0 \implies 2 x \ge - 3 \implies x \ge - \frac{3}{2}$
If $\left(2 x + 3\right) < 0 \implies | 2 x + 3 | = - \left(2 x + 3\right)$ and then, $| 2 x + 3 | \ge 0 \implies - \left(2 x + 3\right) \ge 0 \implies - 2 x - 3 \ge 0 \implies - 2 x \ge 3 \implies 2 x \le - 3$ (observe that the sign of the inequality has changed on changing the sign of both members) $\implies x \le - \frac{3}{2}$

Since the result obtained in the first case is $\forall x \ge - \frac{3}{2}$ and the result obtained in the second case is $\forall x \le - \frac{3}{2}$, both put together give us the final result that the inequation is satisfied $\forall x \in \mathbb{R}$.