6 5/8-(3 3/8 + 1 7/8 )

To solve the expression \( 6 \frac{5}{8} - (3 \frac{3}{8} + 1 \frac{7}{8}) \), we first need to simplify the expression inside the parentheses. 1. **Convert the mixed numbers to improper fractions:** - For \( 3 \frac{3}{8} \): \[ 3 \frac{3}{8} = \frac{(3 \times 8) + 3}{8} = \frac{24 + 3}{8} = \frac{27}{8} \] - For \( 1 \frac{7}{8} \): \[ 1 \frac{7}{8} = \frac{(1 \times 8) + 7}{8} = \frac{8 + 7}{8} = \frac{15}{8} \] 2. **Add the fractions:** \[ 3 \frac{3}{8} + 1 \frac{7}{8} = \frac{27}{8} + \frac{15}{8} = \frac{27 + 15}{8} = \frac{42}{8} \] We can simplify \(\frac{42}{8}\): \[ \frac{42}{8} = \frac{21}{4} \] 3. **Now we rewrite the initial expression:** \[ 6 \frac{5}{8} - (3 \frac{3}{8} + 1 \frac{7}{8}) = 6 \frac{5}{8} - \frac{21}{4} \] 4. **Convert \(6 \frac{5}{8}\) to an improper fraction:** \[ 6 \frac{5}{8} = \frac{(6 \times 8) + 5}{8} = \frac{48 + 5}{8} = \frac{53}{8} \] 5. **Now we perform the subtraction:** We need a common denominator to subtract \(\frac{53}{8}\) and \(\frac{21}{4}\). The common denominator is 8. Convert \(\frac{21}{4}\) to eighths: \[ \frac{21}{4} = \frac{21 \times 2}{4 \times 2} = \frac{42}{8} \] 6. **Now subtract the fractions:** \[ \frac{53}{8} - \frac{42}{8} = \frac{53 - 42}{8} = \frac{11}{8} \] 7. **Convert \(\frac{11}{8}\) back to a mixed number:** \[ \frac{11}{8} = 1 \frac{3}{8} \] Thus, the final answer is: \[ 1 \frac{3}{8} \]