To solve the inequality \(2|3x - 1| < |x + 1|\), we first consider the critical points where the expressions inside the absolute values are zero: \(3x - 1 = 0\) and \(x + 1 = 0\).

Solving these gives \(x = \frac{1}{3}\) and \(x = -1\). These points divide the number line into intervals.

Consider the intervals: \((-\infty, -1)\), \((-1, \frac{1}{3})\), and \((\frac{1}{3}, \infty)\).

For each interval, remove the absolute values by considering the sign of the expressions:

1. For \(x < -1\), both \(3x - 1\) and \(x + 1\) are negative, so the inequality becomes \(2(-(3x - 1)) < -(x + 1)\).

2. For \(-1 < x < \frac{1}{3}\), \(3x - 1\) is negative and \(x + 1\) is positive, so the inequality becomes \(2(-(3x - 1)) < x + 1\).

3. For \(x > \frac{1}{3}\), both \(3x - 1\) and \(x + 1\) are positive, so the inequality becomes \(2(3x - 1) < x + 1\).

Solving these inequalities gives the critical values \(x = \frac{3}{5}\) and \(x = \frac{1}{7}\).

The solution is \(\frac{1}{7} < x < \frac{3}{5}\).