Solve the given inequality below in the body.

Question

$$\frac{1}{|x|-3} \le \frac 12$$

Let’s consider $|x|=y$

So $$\frac{1}{y-3}-\frac 12 \le 0$$

$$\frac{2-y+3}{y-3} \le 0$$ $$\frac{y-5}{y-3} \ge 0$$ $$y \in (-\infty , 3)\cup [5, \infty)$$ Now this is where the problem starts. I cannot figure out on how to break the modulus function, so that’s the part where I need help in. Thanks a lot!

score 4 · Accepted Answer · answered Sep 04 '19 at 16:25

Anyway, in your case, $y\ge 0$, so you should transform all this to $$|x|<3\;\textit{ or }\; |x|\ge 5\iff -3<x<3\;\textit{ or }\; x\ge 5\;\textit{ or }\; x\le -5$$ or, as a set, the union of three intervals: $$(-\infty,-5]\cup(-3,3)\cup[5,+\infty). $$

score 2 · Answer 2 · 2019-09-04T16:37:53.930

2

The equivalences $$\lvert A\rvert< B\iff\begin{cases}B\ge 0\\ A> -B\\ A< B\end{cases}\\ \lvert A\rvert\ge B\iff B\le 0\lor \begin{cases}B> 0\\ A\le -B\end{cases}\lor\begin{cases}B>0\\ A\ge B\end{cases}$$

are a good place to start from.

edited Sep 04 '19 at 16:37

answered Sep 04 '19 at 16:19

In your first equivalence, the inequalities should be strict. – Bernard Sep 04 '19 at 16:34
@Bernard Yes, thank you. – Sep 04 '19 at 16:38
Even $B$ should be positive, since $|A|\ge 0$. – Bernard Sep 04 '19 at 16:41
@Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$). – Sep 04 '19 at 16:45
There remains that $|A|<B\Rightarrow B>0$, from a strictly logical point of view. – Bernard Sep 04 '19 at 16:47
Anyone may squeeze as he best sees fit. – Sep 04 '19 at 16:52
As you wish, mylord… – Bernard Sep 04 '19 at 16:56

score 1 · Answer 3 · 2019-09-04T18:20:47.447

1

For $|x|<3$, the LHS is negative and the inequation certainly holds. Now assuming $|x|>3$, we can solve the equation

$$\frac1{|x|-3}=\frac12,$$ which gives $|x|=5$. The inequation holds for $|x|\ge5$. Your resolution was right.

So the solution set is

$$(-3,3)\cup(-\infty,-5]\cup[5,\infty).$$

edited Sep 04 '19 at 18:20

answered Sep 04 '19 at 16:20

Solve the given inequality below in the body.

3 Answers3