1. **State the problem:** Add the rational expressions $\frac{5}{x-7} + \frac{x-3}{x-4}$ and simplify the result. 2. **Find a common denominator:** The denominators are $x-7$ and $x-4$. The least common denominator (LCD) is $(x-7)(x-4)$. 3. **Rewrite each fraction with the LCD:** $$\frac{5}{x-7} = \frac{5(x-4)}{(x-7)(x-4)}$$ $$\frac{x-3}{x-4} = \frac{(x-3)(x-7)}{(x-4)(x-7)}$$ 4. **Add the numerators over the common denominator:** $$\frac{5(x-4) + (x-3)(x-7)}{(x-7)(x-4)}$$ 5. **Expand the numerators:** $$5(x-4) = 5x - 20$$ $$(x-3)(x-7) = x^2 - 7x - 3x + 21 = x^2 - 10x + 21$$ 6. **Combine the expanded numerators:** $$5x - 20 + x^2 - 10x + 21 = x^2 - 5x + 1$$ 7. **Write the final expression:** $$\frac{x^2 - 5x + 1}{(x-7)(x-4)}$$ 8. **Check for factorization:** The numerator $x^2 - 5x + 1$ does not factor nicely with integer roots, so this is the simplified form. **Final answer:** $$\frac{x^2 - 5x + 1}{(x-7)(x-4)}$$