Solve ((∣x−2∣+1)/(∣x−2∣−1))<3


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Question Number 108961 by ZiYangLee last updated on 20/Aug/20

$Solve \frac{∣ x - 2 ∣ + 1}{∣ x - 2 ∣ - 1} < 3$
Commented byRasheed.Sindhi last updated on 20/Aug/20

$\frac{∣ x - 2 ∣ + 1}{∣ x - 2 ∣ - 1} < 3$ $\frac{∣ x - 2 ∣ + 1}{∣ x - 2 ∣ - 1} < \frac{2 + 1}{2 - 1}$ $\frac{∣ x - 2 ∣}{∣ x - 2 ∣} < \frac{2}{2}$ $\frac{∣ x - 2 ∣}{∣ x - 2 ∣} < 1$ $? ¿ ? ¿ ? I s t h i s c o r r e c t ? ¿ ? ¿ ?$
Commented bybemath last updated on 20/Aug/20

$h a h a . . n o t c o r r e c t o r f a l s e$
	Answered by 1549442205PVT last updated on 20/Aug/20

	$Solve \frac{∣ x - 2 ∣ + 1}{∣ x - 2 ∣ - 1} < 3 (1)$ $Set x - 2 = y (y ⩾ 0) we have$ $(1) \Leftrightarrow \frac{y + 1}{y - 1} < 3 \Leftrightarrow \frac{y + 1}{y - 1} - 3 < 0$ $\Leftrightarrow \frac{- 2 y + 4}{y - 1} < 0 \Leftrightarrow \frac{- y + 2}{y - 1} < 0$ $\Leftrightarrow y \in (- \infty; 1) \cup (2; + \infty)$ $combining to the condition y ⩾ 0 we$ $get y \in [0; 1) \cup (2; + \infty)$ $i) y \in [0; 1) \Leftrightarrow ∣ x - 2 ∣< 1$ $\Leftrightarrow - 1 < x - 2 < 1 \Leftrightarrow 1 < x < 3$ $ii) y \in (2; + \infty) \Leftrightarrow∣ x - 2 ∣> 2 \Leftrightarrow [\begin{matrix} x - 2 > 2 \Leftrightarrow x > 4 \\ x - 2 < - 2 \Leftrightarrow x < 0 \end{matrix}]$ $Combining (i) and (ii) we get$ $x \in (1; 3) \cup (- \infty; 0) \cup (4; + \infty)$
	Answered by john santu last updated on 20/Aug/20
	$__J_⊸ S_⊸ __ (1) ∣x−2∣ ≠ 1 → { ((x ≠ 3 )),((x ≠ 1)) :} (2) let ∣x−2∣ = z →((z+1)/(z−1)) < 3 ((z−1+2)/(z−1)) < 3 ⇒ 1+(2/(z−1)) < 3 (2/(z−1)) < 2 ⇒ (1/(z−1)) < ((z−1)/(z−1)) ((2−z)/(z−1)) < 0 ⇒ ((z−2)/(z−1)) > 0 we got z < 1 or z > 2 now ∣x−2∣ < 1 or ∣x−2∣ > 2 ⇒−1< x−2< 1 or { x > 4 ∪ x < 0 } ⇒1 < x < 3 or { x < 0 ∪ x > 4 } the solution set is x ∈ (−∞,0) ∪(1,3) ∪ (4,∞)$
	$__\underset{⊸}{J} \underset{⊸}{S__}$ $(1) ∣ x - 2 ∣ \neq 1 \to {\begin{cases} x \neq 3 \\ x \neq 1 \end{cases}$ $(2) l e t ∣ x - 2 ∣ = z \to \frac{z + 1}{z - 1} < 3$ $\frac{z - 1 + 2}{z - 1} < 3 \Rightarrow 1 + \frac{2}{z - 1} < 3$ $\frac{2}{z - 1} < 2 \Rightarrow \frac{1}{z - 1} < \frac{z - 1}{z - 1}$ $\frac{2 - z}{z - 1} < 0 \Rightarrow \frac{z - 2}{z - 1} > 0$ $w e g o t z < 1 o r z > 2$ $n o w ∣ x - 2 ∣ < 1 o r ∣ x - 2 ∣ > 2$ $\Rightarrow - 1 < x - 2 < 1 o r {x > 4 \cup x < 0}$ $\Rightarrow 1 < x < 3 o r {x < 0 \cup x > 4}$ $t h e s o l u t i o n s e t i s$ $x \in (- \infty, 0) \cup (1, 3) \cup (4, \infty)$
	Commented bybemath last updated on 20/Aug/20

	$w a w . . . c r e a t i v e a n s w e r$
	Answered by Rasheed.Sindhi last updated on 21/Aug/20

	$\frac{∣ x - 2 ∣ + 1}{∣ x - 2 ∣ - 1} - 1 < 3 - 1$ $\frac{2}{∣ x - 2 ∣} < 2$ $\frac{1}{∣ x - 2 ∣} < 1$ $∣ x - 2 ∣> 1 {\begin{cases} x - 2 > 1 \Rightarrow x > 3 \\ - x + 2 > 1 \Rightarrow x < 1 \end{cases}$
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