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*}

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    $\begingroup$ The error is likely in the missing steps, but I would have to guess that you multiplied both sides by some quantity (likely $x + 5$) without regard for whether it is positive or negative. Remember, when multiplying/dividing both sides by a negative number, you must reverse the inequality. This means you must investigate whether $x + 5 > 0$ or $x - 5 < 0$, considering both cases separately if necessary. $\endgroup$ Nov 8, 2021 at 2:48
  • $\begingroup$ You should be getting $x \leq -8$ and that will do it. If you're still confused after the above hint, type in how you solved the part that led to $x \geq -8$, and we can find out where stuff has gone wrong. $\endgroup$ Nov 8, 2021 at 3:03
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    $\begingroup$ A way to cross check without much casework is to observe for $x\neq -5$, this is the same as $(x-4)^2\leqslant 16(x+5)^2 \iff 3(x+8)(5x+16) \geqslant 0$. Now when does this hold? $\endgroup$
    – Macavity
    Nov 8, 2021 at 6:38
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    $\begingroup$ Thank you guys for the help! I am still a little confused on how to get there though. Sorry about not including my steps, here is what I did to get $x \geq -8$ $ \dfrac{x-4}{x+5} \leq 4 \\ x-4 \leq 4(x+5) \\ x-4 \leq 4x + 20 \\ x \leq 4x + 24 \\ -3x \leq 24 \\ x \geq -8 $ Sorry about the formatting. $\endgroup$ Nov 8, 2021 at 15:12
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    $\begingroup$ Frederick, could you please take a look at my answer, which is just a graph of $y = \frac{x-4}{x+5}. $ I would also encourage you to get some graph paper and do the same graph, by plotting points when $x$ is an integer for example. $\endgroup$
    – Will Jagy
    Nov 8, 2021 at 18:32

2 Answers 2


This is

$$y = \frac{x-4}{x+5} $$

$$ $$

The graph is a hyperbola, one horizontal asymptote and one vertical. I encourage you to get some graph paper and draw the same thing, maybe by plotting points when $x$ is an integer. You can improve the picture by plotting points when $y$ is an integer, using $$ x = \frac{5y+4}{-y+1} $$

The educational bit: inequalities can be treacherous. A graph displays a good deal of firm information, while the process of plotting one and drawing in the curve by hand cements some concepts that are otherwise a bit uncertain.

Your mapping is called a Mobius Transformation. There is a simple rule for finding the inverse function, it is another Mobius transformation. Let me type in the rule, constants $a,b,c,d,$

$$ y = \frac{ax+b}{cx+d} \Longrightarrow x=\frac{dy-b}{-cy+a} $$

I figured out that Desmos will let me plot points. Here are enough points on the upper left arc of the hyperbola, then just a few on the lower right. Next I'll put more points on the lower right... enter image description here enter image description here


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.


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