Solving an absolute value inequality with fractions I'm having a hard time figuring out this inequality: $\bigg|\dfrac{x-4}{x+5}\bigg| \le 4$
I use one of the absolute value properties, which results in: $-4 \leq \dfrac{x-4}{x+5} \leq 4$.
From there, I get $x \geq -\frac{16}{5}$ and $x \geq -8$, yet when I look at Wolfram alpha, the answer is $[-\frac{16}{5}, \infty)$ and $(-\infty, -8]$, which makes sense. What am I missing?
These are my steps to get $x \ge -8$:
\begin{align*}
&\frac{x - 4}{x + 5} \le 4 \\
&x - 4 \le 4(x + 5) \\
&x - 4 \le 4x + 20 \\
&x \le 4x + 24 \\
&-3x \le 24 \\
&x \ge -8.
\end{align*}

From your steps, a couple of red flags jump out at me. First, as I predicted in my comment, it appears you have multiplied both sides by $x + 5$ without worrying whether $x + 5$ is positive or negative (it can't be zero, for obvious reasons!). Second, you are ignoring the other inequality $-4 \le \frac{x - 4}{x + 5}$, presumably because you thought that this simply resulted in $x \ge -\frac{16}{5}$, though I'm guessing that a similar error was committed there.
First, let me say that the best way to do this is to follow Macavity's suggestion. Start by squaring both sides. If you know that $a$ and $b$ are both non-negative, then $a \le b$ is equivalent to $a^2 \le b^2$ (Why? Because $x^2$ is a strictly increasing function on $[0, \infty)$, so it respects the order. Also, $\sqrt{x}$ is strictly increasing too.) Therefore,
$$\left|\dfrac{x-4}{x+5}\right| \le 4 \iff \left|\dfrac{x-4}{x+5}\right|^2 \le 4^2 \iff \frac{(x - 4)^2}{(x + 5)^2} \le 16.$$
The advantage now is that we can multiply both sides by $(x + 5)^2$, safe in the knowledge that, assuming $x \neq -5$ so that the expression is well-defined, we are multiplying both sides by a positive number, so we need not change the sign. This leads to a quadratic that can be factored (as Macavity does in the comment), and can be solved with cases from there (if the product is positive, then either both factors are positive, or both factors are negative).
However, one can proceed as you have, but we just need to be more cautious.
It's a good idea to consider $\frac{x - 4}{x + 5} \le 4$ and $\frac{x - 4}{x + 5} \ge -4$ separately, taking the intersections at the end. Let's consider the former first.
We wish to multiply both sides by $x + 5$. Let us first consider the case that $x + 5 > 0$. Then, under this assumption,
$$\frac{x - 4}{x + 5} \le 4 \iff x - 4 \le 4(x + 5) \iff x \ge -8,$$
as your working showed. But, this is all under the overarching assumption that $x > -5$. So, the only valid solutions in this case satisfy both $x > -5$ and $x \ge -8$, which is to say $x > -5$.
Next, consider the case $x + 5 < 0$. Then
$$\frac{x - 4}{x + 5} \le 4 \iff x - 4 \ge 4(x + 5) \iff x \le -8.$$
So, our valid solutions in this case satisfy both $x < -5$ and $x \le -8$, which is true precisely when $x \le -8$. Thus, the entire solution to $\frac{x - 4}{x + 5} \le 4$ is the union of these intervals, i.e.
$$(-\infty, -8] \cup (-5, \infty).$$
I'll leave it to you, but the next step would be to repeat this analysis (taking into account cases as we have done) for $-4 \le \frac{x - 4}{x + 5}$. You should get another union of intervals, as above. The full solution to $\left|\frac{x-4}{x+5}\right| \le 4$ will be the intersection of the two sets.
