Question

|2x – 3| < |x + 5|, then x belongs to

• A. (– 3, 5)
• B. (5, 9)
• C. $\left( - \frac{2}{3},8 \right)$
• D. $\left( - 8,\frac{2}{3} \right)$

Answer 1

Answer: (C)

$\left( - \frac{2}{3},8 \right)$

Answer 2

|2x – 3| < |x + 5|

case (i) x < – 5 . . . (1)

3 – 2x < – (x + 5) Ю x > 8 . . . (2)

from (1) & (2) x = f

case (ii) – 5 Ј x < $\frac{3}{2}$ . . . (3)

3 – 2x < x + 5 Ю x > $\frac{- 2}{3}$ . . . (4)

from (3) & (4) xО$\left( \frac{- 2}{3},\frac{3}{2} \right)$

case (iii) x і $\frac{3}{2}$ . . . (5)

2x – 3 < x + 5 Ю x < 8 . . . (6)

from (5) & (6) xО$\left\lbrack \frac{3}{2},8 \right)$

\ combining all cases xО$\left( \frac{- 2}{3},8 \right)$