To solve the inequality \(|2x - 1| < 3|x + 1|\), we first consider the non-modular inequality \((2x - 1)^2 < 3^2(x + 1)^2\).

Expanding both sides, we get:

\((2x - 1)^2 = 4x^2 - 4x + 1\)

\(3^2(x + 1)^2 = 9(x^2 + 2x + 1) = 9x^2 + 18x + 9\)

Setting up the inequality:

\(4x^2 - 4x + 1 < 9x^2 + 18x + 9\)

Rearranging terms gives:

\(4x^2 - 4x + 1 - 9x^2 - 18x - 9 < 0\)

\(-5x^2 - 22x - 8 < 0\)

Solving the quadratic equation \(-5x^2 - 22x - 8 = 0\) gives the critical values \(x = -4\) and \(x = -\frac{2}{5}\).

The solution to the inequality is \(x < -4\) or \(x > -\frac{2}{5}\).