Number of integral values of 'x' satisfying the equation 3 |x + 1| – 2.3 x = 2.|3 x – 1| + 1 are
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3 |x + 1| – 2.3 x = 2.|3 x – 1| + 1
critical points are x = –1, 3 x – 1 = 0 ⇒ x = 0
Case-I : x < –1
3 –(x + 1) – 2.3 x = –2(3 x –1) + 1 ⇒ 
= –2.3 x + 2 + 1
3 –x = 9 ⇒ x = – 2
Case-II –1 ≤ x < 0
3 (x+1) – 2.3 x = –2(3 x – 1) + 1 ⇒ 3 x = 1 ⇒ x = 0
Not in solution.
Case-III : x ≥ 0
3 x+1 – 2.3 x = 2.3 x – 2 + 1 ⇒ x = 0
so x = 0, 2 satisfy the equation.
3 |x + 1| – 2.3 x = 2.|3 x – 1| + 1
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