To solve the inequality \(2|x - 3| > |3x + 1|\), we consider the critical points where the expressions inside the absolute values are zero.

1. Solve \(x - 3 = 0\) to get \(x = 3\).

2. Solve \(3x + 1 = 0\) to get \(x = -\frac{1}{3}\).

These critical points divide the number line into intervals: \((-\infty, -\frac{1}{3})\), \((-\frac{1}{3}, 3)\), and \((3, \infty)\).

Test each interval:

- For \(x < -\frac{1}{3}\), choose \(x = -1\):

\(2|-1 - 3| = 2|4| = 8\) and \(|3(-1) + 1| = |-3 + 1| = 2\).

Since \(8 > 2\), this interval satisfies the inequality.

- For \(-\frac{1}{3} < x < 3\), choose \(x = 0\):

\(2|0 - 3| = 2|3| = 6\) and \(|3(0) + 1| = |1| = 1\).

Since \(6 > 1\), this interval satisfies the inequality.

- For \(x > 3\), choose \(x = 4\):

\(2|4 - 3| = 2|1| = 2\) and \(|3(4) + 1| = |12 + 1| = 13\).

Since \(2 < 13\), this interval does not satisfy the inequality.

Thus, the solution is \(-7 < x < 1\).