 Sanai Potter

2023-03-06

How to solve $|2x+3|\ge -13$? trackrunner92yuy

By definition, $|z|\ge 0\forall z\in ℝ$, so, applying this definition to our question, we know that $|2x+3|\ge 0$, which is a stronger condition tan $|2x+3|\ge -13$ ("stronger" means that $|2x+3|\ge 0$ is more restrictive than $|2x+3|\ge -13$).
Instead of interpreting the issue as "solve $|2x+3|\ge -13$", we are going to read it as "solve $|2x+3|\ge 0$" which, in fact, is easier to solve.
In order to solve $|2x+3|\ge 0$ we must again remember the definition of $|z|$, which is done by cases:
If $z\ge 0$, then $|z|=z$
If $z<0$, then $|z|=-z$
Taking this into account for our problem, we can conclude that:
If $\left(2x+3\right)\ge 0⇒|2x+3|=2x+3$ and then, $|2x+3|\ge 0⇒2x+3\ge 0⇒2x\ge -3⇒x\ge -\frac{3}{2}$
If $\left(2x+3\right)<0⇒|2x+3|=-\left(2x+3\right)$ and then, $|2x+3|\ge 0⇒-\left(2x+3\right)\ge 0⇒-2x-3\ge 0⇒-2x\ge 3⇒2x\le -3$ (observe that the sign of the inequality has changed on changing the sign of both members) $⇒x\le -\frac{3}{2}$
As 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 $\forall x\in ℝ$.

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