1. **State the problem:** Simplify the expression $$\frac{3c^2 - 7c + 4}{9c^2 - 16}$$. 2. **Factor the numerator:** We look for two numbers that multiply to $3 \times 4 = 12$ and add to $-7$. These numbers are $-3$ and $-4$. Rewrite the numerator: $$3c^2 - 7c + 4 = 3c^2 - 3c - 4c + 4$$ Group terms: $$= (3c^2 - 3c) - (4c - 4)$$ Factor each group: $$= 3c(c - 1) - 4(c - 1)$$ Factor out the common binomial: $$= (3c - 4)(c - 1)$$ 3. **Factor the denominator:** Recognize the difference of squares: $$9c^2 - 16 = (3c)^2 - 4^2 = (3c - 4)(3c + 4)$$ 4. **Rewrite the expression with factors:** $$\frac{(3c - 4)(c - 1)}{(3c - 4)(3c + 4)}$$ 5. **Cancel common factors:** $$\frac{\cancel{(3c - 4)}(c - 1)}{\cancel{(3c - 4)}(3c + 4)} = \frac{c - 1}{3c + 4}$$ 6. **Final answer:** $$\boxed{\frac{c - 1}{3c + 4}}$$ This simplification is valid as long as $3c - 4 \neq 0$ and $3c + 4 \neq 0$, so $c \neq \frac{4}{3}$ and $c \neq -\frac{4}{3}$ to avoid division by zero.