The inequality $|11x + 3| \leq 0$ has no solutions.

To understand why, we first recognize that the absolute value of any real number is always non-negative. Specifically, for any real number $a$, it holds that $|a| \geq 0$. Consequently, this implies that $|11x + 3| \geq 0$ for all $x$ in $\mathbb{R}$.

Now, let’s consider the possibility of there being some $x$ in $\mathbb{R}$ such that $|11x + 3| \leq 0$. According to the definition of absolute value, this condition can only be satisfied if $11x + 3 = 0$. Solving this equation for $x$, we find:

However, we need to check whether this value of $x$ satisfies the original inequality. Substituting $x = -\frac{3}{11}$ into the expression, we get:

While this shows that $|11x + 3|$ equals $0$, it does not satisfy the condition $|11x + 3| < 0$ because $0$ is not less than $0$.

Thus, we conclude that there are no solutions to the inequality $|11x + 3| \leq 0$. In summary, since the absolute value of any real number cannot be negative, the inequality $|11x + 3| \leq 0$ has no solutions.