Our aim is to solve inequalities such as $|x|<5$ and $|x| \geq 3$. 

Let us first do a quick recap of how we solve quadratic inequalities like $(x+2)(x-3) < 0$. We make a quick sketch of the graph of $y=(x+2)(x-3)$ and observe where the graph lies below the $x$-axis ($y=0$). This gives us the solution $-2 < x < 3$.

We can approach the modulus the same way. To solve $|x|<5$, we draw the graphs of $y=|x|$ and $y=5$.

From the graph, we can see that the solution, where the graph of $y=|x|$ is below the graph of $y=5$, is $-5 < x < 5$.

The graphs to draw for $|x| \geq 3$ are similar. This will lead us to the answer $x \leq -3$ or $x \geq 3$.

In fact, this technique can be summarized as follows: given a positive real number $k$,

It is important that $k$ is positive for the above formulas to be valid! The solution for $|x| < -1$ is that "there is no solution". Meanwhile, the solution for $|x| > -2$ is "all real values of $x$" (understand why?).

As an additional exercise, try to solve $1 < |x| \leq 2$ $ -2 \leq x < -1 \quad \textrm{or} \quad 1 < x \leq 2$
Hover over or click on the inequality for the answer.