1. The problem is to solve the inequality $\left| \frac{x-4}{3x+1} \right| \geq 0$. 2. Recall that the absolute value of any real number or expression is always greater than or equal to zero. That is, for any expression $A$, $|A| \geq 0$. 3. Therefore, $\left| \frac{x-4}{3x+1} \right| \geq 0$ is true for all values of $x$ where the expression inside the absolute value is defined. 4. The expression $\frac{x-4}{3x+1}$ is undefined when the denominator is zero, so we find where $3x+1=0$. 5. Solve for $x$: $$3x+1=0 \implies 3x = -1 \implies x = -\frac{1}{3}.$$ 6. So the expression is undefined at $x = -\frac{1}{3}$. 7. Therefore, the solution to the inequality is all real numbers except $x = -\frac{1}{3}$. 8. In interval notation, the solution is $$(-\infty, -\frac{1}{3}) \cup (-\frac{1}{3}, \infty).$$