1. **Problem statement:** Simplify the expressions by adding or subtracting the given algebraic fractions in exercise 2.224. 2. **Recall the formula for addition/subtraction of fractions:** $$\frac{a}{b} \pm \frac{c}{d} = \frac{ad \pm bc}{bd}$$ where $b$ and $d$ are denominators. 3. **Important rules:** - Find a common denominator. - Rewrite each fraction with the common denominator. - Add or subtract the numerators. - Simplify the resulting fraction if possible. --- ### a) $\frac{3a^2 + 2}{5a + 5} + \frac{1 - a}{10}$ 4. Factor denominators if possible: $$5a + 5 = 5(a + 1)$$ 5. The denominators are $5(a+1)$ and $10$. The least common denominator (LCD) is $10(a+1)$. 6. Rewrite each fraction with denominator $10(a+1)$: $$\frac{3a^2 + 2}{5(a+1)} = \frac{(3a^2 + 2) \cdot 2}{10(a+1)} = \frac{2(3a^2 + 2)}{10(a+1)}$$ $$\frac{1 - a}{10} = \frac{(1 - a)(a+1)}{10(a+1)}$$ 7. Add the numerators: $$\frac{2(3a^2 + 2) + (1 - a)(a+1)}{10(a+1)}$$ 8. Expand numerator: $$2(3a^2 + 2) = 6a^2 + 4$$ $$(1 - a)(a + 1) = 1 \cdot a + 1 \cdot 1 - a \cdot a - a \cdot 1 = a + 1 - a^2 - a = 1 - a^2$$ 9. Sum numerator: $$6a^2 + 4 + 1 - a^2 = (6a^2 - a^2) + (4 + 1) = 5a^2 + 5$$ 10. Final expression: $$\frac{5a^2 + 5}{10(a+1)}$$ 11. Factor numerator: $$5(a^2 + 1)$$ 12. Simplify fraction: $$\frac{5(a^2 + 1)}{10(a+1)} = \frac{\cancel{5}(a^2 + 1)}{\cancel{10}2(a+1)} = \frac{a^2 + 1}{2(a+1)}$$ --- ### b) $\frac{1 + 2b}{8b + 4} - \frac{2b + 9}{8b}$ 13. Factor denominator $8b + 4$: $$8b + 4 = 4(2b + 1)$$ 14. Denominators are $4(2b + 1)$ and $8b$. 15. Find LCD: $$\text{LCD} = 8b(2b + 1)$$ 16. Rewrite fractions with LCD: $$\frac{1 + 2b}{4(2b + 1)} = \frac{(1 + 2b) \cdot 2b}{8b(2b + 1)}$$ $$\frac{2b + 9}{8b} = \frac{(2b + 9)(2b + 1)}{8b(2b + 1)}$$ 17. Subtract numerators: $$\frac{(1 + 2b)2b - (2b + 9)(2b + 1)}{8b(2b + 1)}$$ 18. Expand numerator: $$(1 + 2b)2b = 2b + 4b^2$$ $$(2b + 9)(2b + 1) = 4b^2 + 2b + 18b + 9 = 4b^2 + 20b + 9$$ 19. Subtract: $$(2b + 4b^2) - (4b^2 + 20b + 9) = 2b + 4b^2 - 4b^2 - 20b - 9 = (2b - 20b) - 9 = -18b - 9$$ 20. Final expression: $$\frac{-18b - 9}{8b(2b + 1)}$$ 21. Factor numerator: $$-9(2b + 1)$$ 22. Simplify fraction: $$\frac{-9(2b + 1)}{8b(2b + 1)} = \frac{\cancel{-9}(\cancel{2b + 1})}{8b(\cancel{2b + 1})} = \frac{-9}{8b}$$ --- ### c) $\frac{12c^2 + 48}{3} + \frac{96 - 12c^3}{3c - 6}$ 23. Factor denominators and numerators: $$3c - 6 = 3(c - 2)$$ $$12c^2 + 48 = 12(c^2 + 4)$$ $$96 - 12c^3 = 12(8 - c^3)$$ 24. Denominators are $3$ and $3(c - 2)$. 25. LCD is $3(c - 2)$. 26. Rewrite first fraction: $$\frac{12(c^2 + 4)}{3} = \frac{12(c^2 + 4)(c - 2)}{3(c - 2)}$$ 27. Second fraction is already over $3(c - 2)$: $$\frac{12(8 - c^3)}{3(c - 2)}$$ 28. Add numerators: $$12(c^2 + 4)(c - 2) + 12(8 - c^3)$$ 29. Factor out 12: $$12[(c^2 + 4)(c - 2) + 8 - c^3]$$ 30. Expand $(c^2 + 4)(c - 2)$: $$c^3 - 2c^2 + 4c - 8$$ 31. Sum inside brackets: $$c^3 - 2c^2 + 4c - 8 + 8 - c^3 = (c^3 - c^3) + (-2c^2) + 4c + (-8 + 8) = -2c^2 + 4c$$ 32. Numerator becomes: $$12(-2c^2 + 4c) = -24c^2 + 48c$$ 33. Final expression: $$\frac{-24c^2 + 48c}{3(c - 2)}$$ 34. Factor numerator: $$-24c(c - 2)$$ 35. Simplify fraction: $$\frac{-24c(c - 2)}{3(c - 2)} = \frac{\cancel{-24}c(\cancel{c - 2})}{\cancel{3}(\cancel{c - 2})} = -8c$$ --- **Final answers:** - a) $\frac{a^2 + 1}{2(a + 1)}$ - b) $\frac{-9}{8b}$ - c) $-8c$