- 9 ⅐ + 8 ⁷/⁷

To solve the expression \( 9 \frac{1}{7} + 8 \frac{7}{7} \), we first convert the mixed numbers to improper fractions. 1. Convert \( 9 \frac{1}{7} \): \[ 9 \frac{1}{7} = 9 + \frac{1}{7} = \frac{9 \times 7}{7} + \frac{1}{7} = \frac{63}{7} + \frac{1}{7} = \frac{64}{7} \] 2. Convert \( 8 \frac{7}{7} \): Since \( \frac{7}{7} = 1 \), it simplifies to: \[ 8 \frac{7}{7} = 8 + 1 = 9 \] or as a fraction: \[ 8 \frac{7}{7} = 8 + \frac{7}{7} = 8 + 1 = 9 = \frac{9}{1} \] Now we can add the two improper fractions: \[ \frac{64}{7} + 9 = \frac{64}{7} + \frac{9 \times 7}{7} = \frac{64}{7} + \frac{63}{7} \] \[ = \frac{64 + 63}{7} = \frac{127}{7} \] Now, we can convert \( \frac{127}{7} \) back to a mixed number: \[ 127 \div 7 = 18 \quad \text{with a remainder of } 1 \] So, we have: \[ \frac{127}{7} = 18 \frac{1}{7} \] Thus, the final answer is: \[ 9 \frac{1}{7} + 8 \frac{7}{7} = 18 \frac{1}{7} \]