1. **Problem statement:** We will solve the first subproblems of 2.224 and 2.225 by adding or subtracting the given rational expressions using the main denominator and simplifying the result. --- ### Problem 2.224 a) \( \frac{3a^2 + 2}{5a + 5} + \frac{1 - a}{10} \) 2. **Formula and rules:** - To add fractions, find the least common denominator (LCD). - Factor denominators if possible. - Rewrite each fraction with the LCD as denominator. - Add numerators and simplify. 3. **Work:** - Factor denominator: \(5a + 5 = 5(a + 1)\) - LCD of denominators \(5(a+1)\) and \(10\) is \(10(a+1)\) because \(10 = 2 \times 5\). Rewrite fractions: \[ \frac{3a^2 + 2}{5(a+1)} = \frac{(3a^2 + 2) \times 2}{10(a+1)} = \frac{6a^2 + 4}{10(a+1)} \] \[ \frac{1 - a}{10} = \frac{(1 - a)(a+1)}{10(a+1)} \] Expand numerator of second fraction: \[ (1 - a)(a + 1) = 1 \times a + 1 \times 1 - a \times a - a \times 1 = a + 1 - a^2 - a = 1 - a^2 \] 4. **Add numerators:** \[ 6a^2 + 4 + 1 - a^2 = (6a^2 - a^2) + (4 + 1) = 5a^2 + 5 \] 5. **Final fraction:** \[ \frac{5a^2 + 5}{10(a+1)} = \frac{5(a^2 + 1)}{10(a+1)} \] Simplify numerator and denominator by 5: \[ \frac{\cancel{5}(a^2 + 1)}{2 \times \cancel{5} (a+1)} = \frac{a^2 + 1}{2(a+1)} \] --- ### Problem 2.225 a) \( \frac{5 - 3g}{8g^2 + 2g} + \frac{5g}{12g^2} \) 2. **Formula and rules:** - Factor denominators. - Find LCD. - Rewrite fractions with LCD. - Add numerators and simplify. 3. **Work:** - Factor denominator \(8g^2 + 2g = 2g(4g + 1)\) - Denominator of second fraction is \(12g^2 = 12g^2\) Find LCD of \(2g(4g + 1)\) and \(12g^2\): - Prime factors: - \(2g(4g + 1)\) includes factor 2, g, and \(4g+1\) - \(12g^2 = 2^2 \times 3 \times g^2\) LCD must include: - Highest power of 2: \(2^2 = 4\) - Factor 3 - Highest power of g: \(g^2\) - Factor \(4g + 1\) So LCD = \(4 \times 3 \times g^2 \times (4g + 1) = 12 g^2 (4g + 1)\) Rewrite fractions: First fraction numerator multiplied by \(\frac{6g}{6g}\) to get denominator \(12 g^2 (4g + 1)\): \[ \frac{5 - 3g}{2g(4g + 1)} = \frac{(5 - 3g) \times 6g}{12 g^2 (4g + 1)} = \frac{6g(5 - 3g)}{12 g^2 (4g + 1)} \] Second fraction numerator multiplied by \(\frac{4g + 1}{4g + 1}\): \[ \frac{5g}{12 g^2} = \frac{5g (4g + 1)}{12 g^2 (4g + 1)} \] 4. **Add numerators:** \[ 6g(5 - 3g) + 5g(4g + 1) = 30g - 18g^2 + 20g^2 + 5g = (30g + 5g) + (-18g^2 + 20g^2) = 35g + 2g^2 \] 5. **Final fraction:** \[ \frac{2g^2 + 35g}{12 g^2 (4g + 1)} = \frac{g(2g + 35)}{12 g^2 (4g + 1)} \] Simplify numerator and denominator by \(g\): \[ \frac{\cancel{g}(2g + 35)}{12 g \times \cancel{g} (4g + 1)} = \frac{2g + 35}{12 g (4g + 1)} \] --- **Final answers:** - 2.224 a) \( \frac{a^2 + 1}{2(a + 1)} \) - 2.225 a) \( \frac{2g + 35}{12 g (4g + 1)} \)