To solve the inequality \(|x - 2| > 2x - 3\), we consider two cases based on the definition of absolute value.

Case 1: When \(x - 2 \geq 0\), i.e., \(x \geq 2\).

The inequality becomes \(x - 2 > 2x - 3\).

Simplifying, we get:

\(x - 2 > 2x - 3\)

\(-2 + 3 > 2x - x\)

\(1 > x\)

Since \(x \geq 2\) and \(1 > x\) cannot both be true, there is no solution in this case.

Case 2: When \(x - 2 < 0\), i.e., \(x < 2\).

The inequality becomes \(-(x - 2) > 2x - 3\).

Simplifying, we get:

\(-x + 2 > 2x - 3\)

\(2 + 3 > 2x + x\)

\(5 > 3x\)

\(x < \frac{5}{3}\)

Since \(x < 2\) and \(x < \frac{5}{3}\) are consistent, the solution is \(x < \frac{5}{3}\).