1. **State the problem:** Solve the inequality $$9|x+1| \geq \frac{1}{81}$$. 2. **Recall the properties of absolute values:** For any real number $a$, $|a| \geq 0$ and $|a| = \frac{b}{c}$ means $a = \pm \frac{b}{c}$. 3. **Isolate the absolute value:** Divide both sides by 9: $$\cancel{9}|x+1| \geq \frac{1}{81 \times \cancel{9}}$$ $$|x+1| \geq \frac{1}{729}$$ 4. **Rewrite the inequality without absolute value:** $$x+1 \geq \frac{1}{729} \quad \text{or} \quad x+1 \leq -\frac{1}{729}$$ 5. **Solve each inequality:** $$x \geq \frac{1}{729} - 1 = \frac{1}{729} - \frac{729}{729} = -\frac{728}{729}$$ $$x \leq -1 - \frac{1}{729} = -\frac{730}{729}$$ 6. **Final solution:** $$x \leq -\frac{730}{729} \quad \text{or} \quad x \geq -\frac{728}{729}$$ This means $x$ is either less than or equal to $-\frac{730}{729}$ or greater than or equal to $-\frac{728}{729}$.