Question-175706

Question Number 175706 by daus last updated on 05/Sep/22

Commented by daus last updated on 05/Sep/22

j u s t h e l p m e i n j, k, l

Answered by mahdipoor last updated on 05/Sep/22

j :∣ \frac{2 x + 1}{x - 3} ∣= 1 \Rightarrow \frac{2 x + 1}{x - 3} = \pm 1 \Rightarrow

x = \frac{\mp 3 - 1}{2 \mp 1} = - 4 a n d \frac{2}{3}

(- \infty, - 4) \Rightarrow ∣ \frac{2 x + 1}{x - 3} ∣> 1

(- 4, \frac{2}{3}) \Rightarrow ∣ \frac{2 x + 1}{x - 3} ∣< 1

(\frac{2}{3}, + \infty) \Rightarrow ∣ \frac{2 x + 1}{x - 3} ∣> 1

y o u c a n a n s w e r k & l l i k e t h i s \dots

Commented by Tawa11 last updated on 05/Sep/22

Great sir

Answered by cortano1 last updated on 05/Sep/22

(l) ∣ \frac{x - 3}{x + 1} ∣< 2, x \neq - 1

∣ x - 3 ∣<∣ 2 x + 2 ∣

(3 x - 1) (- x - 5) < 0

(3 x - 1) (x + 5) > 0

x < - 5 \lor x > \frac{1}{3}

(- \infty, - 5) \cup (\frac{1}{3}, \infty)

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