Sanai Potter

2023-03-06

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

trackrunner92yuy

Beginner2023-03-07Added 3 answers

By definition, $\left|z\right|\ge 0\forall z\in \mathbb{R}$, 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 $\left|z\right|$, which is done by cases:

If $z\ge 0$, then $\left|z\right|=z$

If $z<0$, then $\left|z\right|=-z$

Taking this into account for our problem, we can conclude that:

If $(2x+3)\ge 0\Rightarrow |2x+3|=2x+3$ and then, $|2x+3|\ge 0\Rightarrow 2x+3\ge 0\Rightarrow 2x\ge -3\Rightarrow x\ge -\frac{3}{2}$

If $(2x+3)<0\Rightarrow |2x+3|=-(2x+3)$ and then, $|2x+3|\ge 0\Rightarrow -(2x+3)\ge 0\Rightarrow -2x-3\ge 0\Rightarrow -2x\ge 3\Rightarrow 2x\le -3$ (observe that the sign of the inequality has changed on changing the sign of both members) $\Rightarrow x\le -\frac{3}{2}$

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 $\forall x\in \mathbb{R}$.

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 $\left|z\right|$, which is done by cases:

If $z\ge 0$, then $\left|z\right|=z$

If $z<0$, then $\left|z\right|=-z$

Taking this into account for our problem, we can conclude that:

If $(2x+3)\ge 0\Rightarrow |2x+3|=2x+3$ and then, $|2x+3|\ge 0\Rightarrow 2x+3\ge 0\Rightarrow 2x\ge -3\Rightarrow x\ge -\frac{3}{2}$

If $(2x+3)<0\Rightarrow |2x+3|=-(2x+3)$ and then, $|2x+3|\ge 0\Rightarrow -(2x+3)\ge 0\Rightarrow -2x-3\ge 0\Rightarrow -2x\ge 3\Rightarrow 2x\le -3$ (observe that the sign of the inequality has changed on changing the sign of both members) $\Rightarrow x\le -\frac{3}{2}$

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 $\forall x\in \mathbb{R}$.