To solve the equation \(2|x - 1| = 3|x|\), consider the cases for the absolute values.

Case 1: \(x \geq 1\)

In this case, \(|x - 1| = x - 1\) and \(|x| = x\).

The equation becomes \(2(x - 1) = 3x\).

Simplifying gives \(2x - 2 = 3x\).

Rearranging, we get \(-2 = x\), which is not valid for \(x \geq 1\).

Case 2: \(0 \leq x < 1\)

Here, \(|x - 1| = 1 - x\) and \(|x| = x\).

The equation becomes \(2(1 - x) = 3x\).

Simplifying gives \(2 - 2x = 3x\).

Rearranging, we get \(2 = 5x\).

Thus, \(x = \frac{2}{5}\), which is valid for \(0 \leq x < 1\).

Case 3: \(x < 0\)

In this case, \(|x - 1| = 1 - x\) and \(|x| = -x\).

The equation becomes \(2(1 - x) = 3(-x)\).

Simplifying gives \(2 - 2x = -3x\).

Rearranging, we get \(2 = -x\).

Thus, \(x = -2\), which is valid for \(x < 0\).

Therefore, the solutions are \(x = -2\) and \(x = \frac{2}{5}\).