Solve the following system of inequalities: $\frac{x}{2x+1} \geq \frac{1}{4}$ and $\frac{6x}{4x-1} < \frac{1}{2}$.

(D) For the first inequality: $\frac{x}{2x+1} - \frac{1}{4} \geq 0 \implies \frac{4x - (2x+1)}{4(2x+1)} \geq 0 \implies \frac{2x-1}{4(2x+1)} \geq 0$. The critical points are $x = -\frac{1}{2}$ and $x = \frac{1}{2}$. Testing intervals,we get $x \in (-\infty, -\frac{1}{2}) \cup [\frac{1}{2}, \infty)$.
For the second inequality: $\frac{6x}{4x-1} - \frac{1}{2} < 0 \implies \frac{12x - (4x-1)}{2(4x-1)} < 0 \implies \frac{8x+1}{2(4x-1)} < 0$. The critical points are $x = -\frac{1}{8}$ and $x = \frac{1}{4}$. Testing intervals,we get $x \in (-\frac{1}{8}, \frac{1}{4})$.
Since there is no common interval between $(-\infty, -\frac{1}{2}) \cup [\frac{1}{2}, \infty)$ and $(-\frac{1}{8}, \frac{1}{4})$,the solution set is $\emptyset$ (empty set).