Solve for x
(i) |x + 1| = 4x + 3
(ii) |x + 1| = |x + 3|
(iii) 7|x – 2| – |x – 7| = 5
(iv) ||x – 1| – 2| = 6x + 8
(v) |2x 2 – 3x + 1| = |x 2 + x – 3|
Text Solution
Verified by Experts(i) (ii); (iii) (iv)
(i) 
(ii) – 2 (iii) 
(iv) 
(v) 
Sol. (i) Case -I : x ≥ 
⇒ x + 1 = 4x + 3 or x + 1 = – 4x – 3
⇒ x = 
or x = 
⇒ x = 
(ii) x + 1 = x + 3 or x + 1 = – x – 3
⇒ x = – 2
(iii)
Case – I : x ∈ (– ∞ ,2)
–7x + 14 + x – 7 = 5 ⇒ x = 1/3
Case – II : x ∈ [2,7)
7x – 14 + x – 7 = 5 ⇒ x = 
Case-III – x ∈ [7, ∞ )
7x – 14 – x + 7 = 5 ⇒ x = 2 Reject)
(iv) ||x – 1| – 2| = 6x + 8
Case – I : x ∈ (– ∞ ,–1)
–x – 1 =6x + 8 ⇒ x = 
Case – II : x ∈ [–1,1)
x + 1 = 6x + 8 ⇒ x = 
(reject)
Case - III : x ∈ [1,3)
–x + 3 = 6x + 8 ⇒ x = 
(reject)
Case -IV : x ∈ [3, ∞)
x – 3 = 6x + 8 ⇒ x = 
(Reject)
(v) |2x 2 – 3x + 1| = |x 2 + x – 3|
⇒ 2x 2 – 3x + 1 = x 2 + x – 3 or 2x 2 – 3x + 1 = – x 2 – x + 3
⇒ x 2 – 4x + 4 = 0 or 3x 2 – 2x – 2 = 0
x = 2 or x = 
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