What happens to $\frac{|x-3|}{x}<2$ for $x\geq 3$?

Question

I need to find real numbers $x$ that satisfy the inequality $\frac{|x-3|}{x}<2$

My analysis goes like this:

Case 1) $x<0$

Then, $\frac{|x-3|}{x}<0<2$, so all negative $x$ satisfy the inequality.

Case 2) $0<x<3$

Then, $\frac{3-x}{x}<2$, and as $x>0$, we can say $3-x<2x\implies x>1$

$\therefore x\in(1,3)$ satisfies the inequality, but $x\in(0,1]$ does not.

This is all fine, but my issue is in the 3rd case.

Case 3) $x\geq 3$

Then, $\frac{x-3}{x}<2$

$\iff x-3<2x$

$\iff x>-3$

What does the final line tell us about the $x\geq 3$ case?

When I plot $\frac{|x-3|}{x}$, I can see that all $x\geq 3$ satisfy the inequality, but why does the $x>-3$ result tell us this?

Answer 1

The logic might be more clear if you reverse the steps in the $x\ge3$ case, keeping only the reverse direction of the if-and-only-if double arrow:

$$x\ge3\implies x\gt-3\implies x-3\lt2x\implies{x-3\over x}\lt2\implies{|x-3|\over x}\lt2$$

where the first $\implies$ is trivial, the second is easy, and the final two depend on the assumption $x\ge3$ (which implies $x\gt0$ and $x-3=|x-3|$ for those two steps, respectively).

Answer 2

Any way, you don't have to consider 3 cases. If $x>0$ you get $$|x-3|<2x$$ so $$-2x<x-3<2x$$...

Answer 3

You get two cases: At first:$$x>0$$ and $$x\geq 3$$ so we get $$x-3<2x$$ Case two: $$x<0$$ and $$x<3$$ so we get $$-(x-3)>2x$$ And we get $$x<0$$ or $$x>1$$