\( 2 \quad 8 \frac { 3 } { 4 } - 3 \frac { 7 } { 8 } = \)

To solve \( 2 \quad 8 \frac { 3 } { 4 } - 3 \frac { 7 } { 8 } \), first convert the mixed numbers into improper fractions. The first part \( 2 \quad 8 \frac { 3 } { 4 } \) can be calculated by: \[ 2 \times 8 + 3 = 16 + 3 = 19 \] Thus, \( 2 \quad 8 \frac { 3 } { 4 } = \frac{19}{4} \). For the second part \( 3 \frac { 7 } { 8 } \): \[ 3 \times 8 + 7 = 24 + 7 = 31 \] Thus, \( 3 \frac { 7 } { 8 } = \frac{31}{8} \). Now, we can rewrite the expression as: \[ \frac{19}{4} - \frac{31}{8}. \] To subtract these fractions, we need a common denominator. The least common multiple of 4 and 8 is 8. So, we convert \( \frac{19}{4} \) to an equivalent fraction with a denominator of 8: \[ \frac{19}{4} = \frac{19 \times 2}{4 \times 2} = \frac{38}{8}. \] Now our expression is: \[ \frac{38}{8} - \frac{31}{8} = \frac{38 - 31}{8} = \frac{7}{8}. \] Thus, the final answer is: \[ \frac{7}{8}. \]