To solve the inequality \(3x + 5 < |2x + 1|\), we consider two cases for the absolute value.

Case 1: \(2x + 1 \\geq 0\)

In this case, \(|2x + 1| = 2x + 1\).

So, the inequality becomes \(3x + 5 < 2x + 1\).

Subtract \(2x\) from both sides: \(x + 5 < 1\).

Subtract 5 from both sides: \(x < -4\).

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

In this case, \(|2x + 1| = -(2x + 1)\).

So, the inequality becomes \(3x + 5 < -(2x + 1)\).

Which simplifies to \(3x + 5 < -2x - 1\).

Add \(2x\) to both sides: \(5x + 5 < -1\).

Subtract 5 from both sides: \(5x < -6\).

Divide by 5: \(x < -\frac{6}{5}\).

The critical value is \(x = -\frac{6}{5}\), and the solution is \(x < -\frac{6}{5}\).