1. **State the problem:** Simplify the expression $$\frac{1}{1 - 3y} + \frac{21y - 9}{1 - 9y^2}$$. 2. **Identify the formula and rules:** To add these fractions, we need a common denominator. Note that $$1 - 9y^2$$ is a difference of squares and can be factored as $$ (1 - 3y)(1 + 3y) $$. 3. **Rewrite the expression with factored denominators:** $$\frac{1}{1 - 3y} + \frac{21y - 9}{(1 - 3y)(1 + 3y)}$$ 4. **Find the common denominator:** The common denominator is $$ (1 - 3y)(1 + 3y) $$. 5. **Rewrite the first fraction with the common denominator:** $$\frac{1}{1 - 3y} = \frac{1 \times (1 + 3y)}{(1 - 3y)(1 + 3y)} = \frac{1 + 3y}{(1 - 3y)(1 + 3y)}$$ 6. **Add the fractions:** $$\frac{1 + 3y}{(1 - 3y)(1 + 3y)} + \frac{21y - 9}{(1 - 3y)(1 + 3y)} = \frac{1 + 3y + 21y - 9}{(1 - 3y)(1 + 3y)}$$ 7. **Simplify the numerator:** $$1 + 3y + 21y - 9 = (1 - 9) + (3y + 21y) = -8 + 24y$$ 8. **Final expression:** $$\frac{24y - 8}{(1 - 3y)(1 + 3y)}$$ 9. **Factor numerator:** $$24y - 8 = 8(3y - 1)$$ 10. **Rewrite denominator:** $$ (1 - 3y)(1 + 3y) $$ can be rewritten as $$ -(3y - 1)(1 + 3y) $$ because $$1 - 3y = -(3y - 1)$$. 11. **Substitute and simplify:** $$\frac{8(3y - 1)}{-(3y - 1)(1 + 3y)} = \frac{8 \cancel{(3y - 1)}}{-\cancel{(3y - 1)}(1 + 3y)} = \frac{8}{-(1 + 3y)} = -\frac{8}{1 + 3y}$$ **Final answer:** $$-\frac{8}{1 + 3y}$$