1. **Stating the problem:** Add the mixed numbers $3 \frac{7}{8}$ and $2 \frac{1}{3}$. 2. Convert the mixed numbers to improper fractions: - $3 \frac{7}{8} = \frac{3 \times 8 + 7}{8} = \frac{24 + 7}{8} = \frac{31}{8}$ - $2 \frac{1}{3} = \frac{2 \times 3 + 1}{3} = \frac{6 + 1}{3} = \frac{7}{3}$ 3. Find the least common denominator (LCD) of $8$ and $3$, which is $24$. 4. Convert both fractions to have denominator $24$: - $\frac{31}{8} = \frac{31 \times 3}{8 \times 3} = \frac{93}{24}$ - $\frac{7}{3} = \frac{7 \times 8}{3 \times 8} = \frac{56}{24}$ 5. Add the fractions: $$\frac{93}{24} + \frac{56}{24} = \frac{93 + 56}{24} = \frac{149}{24}$$ 6. Convert back to a mixed number: - Divide $149$ by $24$: $149 ÷ 24 = 6$ remainder $5$ - So, $\frac{149}{24} = 6 \frac{5}{24}$ 7. **Final answer:** $3 \frac{7}{8} + 2 \frac{1}{3} = 6 \frac{5}{24}$