Solve |2x + 3| > 4x + 1: Absolute Value Inequality Challenge

To solve $\left|2x + 3\right| > 4x + 1$, we need to consider two cases based on the definition of absolute value:

1. $2x + 3 > 4x + 1$

2. $2x + 3 < -(4x + 1)$

Solving the first inequality:

$2x + 3 > 4x + 1 \Rightarrow 3 > 4x - 2x + 1 \Rightarrow 3 > 2x + 1 \Rightarrow 2 > 2x \Rightarrow x < 1$

Solving the second inequality:

$2x + 3 < -4x - 1 \Rightarrow 6x < -4 \Rightarrow x < -\frac{2}{3}$

These inequalities indicate that $x < 1$ is the range that satisfies the original inequality.