(i) |x| + 2|x – 6| = 12

Case-I : x ≥ 6 3x = 24 ⇒ x = 8

Case-II : 0 ≤ x < 6

x + 12 – 2x = 12 ⇒ x = 0

Case-III : x < 0

–x + 12 – 2x = 12 ⇒ x = 0

so solution is x = 0, 8

(ii) ||x + 3| – 5| = 2

⇒ |x + 3| – 5 = 2, –2 ⇒ |x + 3| = 7 or |x + 3| = 3

⇒ x + 3 = 7, –7 or x + 3 = 3, –3

⇒ x = 4, –10 or x = 0, –6

so , x = –10, –6, 0, 4

(iii) |||x – 2| – 2 | – 2| = 2

⇒ ||x – 2| – 2 | – 2 = ±2

either ||x – 2| – 2 | = 4 or 0

case-I : ||x – 2| – 2| = 4

|x – 2| – 2 = ±4 ⇒ |x – 2| = 6 or –2 ⇒ x – 2 = ±6 ⇒ x = 8 or –4

case-II : ||x – 2| – 2| = 0

|x – 2| – 2 = 0 ⇒ |x – 2| = 2 ⇒ x – 2 = ±2 ⇒ x = 4 or 0

hence four solutions 0, –4, 4 & 83

(iv) |4x + 3| + |3x – 4| = 12

case-1 : x < –

–4x – 3 – 3x + 4 = 12 ⇒ –7x = 11 ⇒ x = –

case-2 : – ≤ x ≤ ⇒ 4x + 3 – 3x + 4 = 12

x = 5, not acceptable.

case-3 : x ≥

4x + 3 + 3x – 4 = 12 ⇒ 7x = 13 ⇒ x =

∴ x = , .

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