1
$\begingroup$

How would one proceed in solving this difficult inequation with multiple absolute values? is there a way one should proceed ?

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

thanks

$\endgroup$
3
$\begingroup$

Hint

Discuss several cases: $$x\in(-\infty,-2)\cup(-2 ,0]\; ; \; x\in(0 ,2)\; ;\; x\in(2 ,3)\; ;\; x\in(3 ,+\infty)$$

$\endgroup$
  • 1
    $\begingroup$ @franck It might also help to cross multiply the fractions as well. $\endgroup$ – Ataraxia Aug 14 '13 at 20:03
  • $\begingroup$ @Ataraxia how come don't we consider the case from x ∈ (1,2) ? $\endgroup$ – franck Aug 14 '13 at 20:06
  • $\begingroup$ @franck The $x-1$ doesn't form its own separate case because it's not absolute valued in the equation. $\endgroup$ – Ataraxia Aug 14 '13 at 20:24
  • $\begingroup$ Thanks for the comment @AndréNicolas $\endgroup$ – user63181 Aug 14 '13 at 20:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.