Solve (|x|-1)/(|x|-2)≥0, x ∈ R, x ≠ ± 2.

Question

Solve \(\frac{|x|-1}{|x|-2}\) ≥ 0, \(x\) ∈ R, \(x\) ≠  ± 2.

Answer 1

Let |x| = y. Then

 \(\frac{|x|-1}{|x|-2}\) ≥ 0 ⇒ \(\frac{y-1}{y-2}≥0\)

On equating (y – 1) and (y – 2) equal to zero, we have the critical points as y = 1, 2. Now using the real number line, we see that the expression \(\frac{y-1}{y-2}\) is greater than equal to zero (positive) only when, y < 1 or y > 2.

⇒ |x| < 1 or |x| > 2 

⇒ –1 < x < 1 or (x < –2 or x > 2) 

⇒ \(x\)∈ [–1, 1] ∪ (–∞, –2) ∪ (2, ∞)

Comments

Hello.  To answer your question about why use y =  |x|
The answer is that it was absolutely not necessary to do so and it just made it confusing for you.  The answer is correct as stated though:  |x| <= 1 or |x| > 2
On other words, you could have solved this problem without making that substitution and just used |x| instead of using y.