Solving $|1-\frac{3}{x}|\le 2$

Question

I did:

$$|1-\frac{3}{x}|\le 2 \iff \\ 1-\frac{3}{x}\le2 \lor -1+\frac{3}{x}\le2\iff \\ -\frac{3}{x}\le2-1 \lor \frac{3}{x}\le 2+1 \iff \\ \frac{3}{x}\ge-1 \lor \frac{3}{x}\le3 \iff \\ x \ge -3 \lor x\le1$$

Apparently this is correct only the symbols are the other way around on the last step. Why? Was it because when you raise numbers to -1 you have to invert the signs? Is this true for all negative powers?

Answer 1

We can write $$|x-3|\le 2|x|,$$ three cases arise and this equation is to be re-written each time.

1- $-\infty< x \le 0,$ then $$-(x-3)\le -2x \implies x \le -3 \implies x \in (-\infty, -3].$$

2- $ 0 <x \le 3,$ then $$-(x-3) \le 2x \implies -3x \le -3 \implies x\ge 1 \implies x \in [1,3].$$

3- $3<x <\infty,$ then $$(x-3) \le 2x \implies -x \le 3 \implies x \ge -3 \implies x \in [3,\infty).$$

So the net solution is $$x \in (-\infty,-3] \cup [1,\infty)$$

Answer 2

A clean procedure is to set

$$|1-\frac{3}{x}|=2$$

for the two break points at $x=-3$ and $x=1$. For each of the resulting intervals, just pick any point within to check the validity of the inequality.

$$x<-3,\>\>\> |1-\frac{3}{-5}|\le 2, \>\>\>true$$ $$-3<x<1,\>\>\> |1-\frac{3}{-1}|\le 2, \>\>\>false$$ $$x>1,\>\>\> |1-\frac{3}{2}|\le 2, \>\>\>true$$

Thus, the solutions are

$$x\le -3\>\>\>\text{and}\>\>\>x\ge 1$$