To solve \( 3 \frac{2}{4} + 3 \frac{3}{4} \), we start by simplifying the fractions:
- Convert \( \frac{2}{4} \) to \( \frac{1}{2} \). So, \( 3 \frac{2}{4} = 3 \frac{1}{2} \).
- The second fraction \( 3 \frac{3}{4} \) remains the same.
Now we can rewrite the problem with improper fractions:
- \( 3 \frac{1}{2} = 3 + \frac{1}{2} = \frac{6}{2} + \frac{1}{2} = \frac{7}{2} \)
- \( 3 \frac{3}{4} = 3 + \frac{3}{4} = \frac{12}{4} + \frac{3}{4} = \frac{15}{4} \)
Now we need to add the two improper fractions:
- Find a common denominator. The least common multiple of 2 and 4 is 4.
- Convert \( \frac{7}{2} \) to a fraction with a denominator of 4: \[ \frac{7}{2} = \frac{7 \times 2}{2 \times 2} = \frac{14}{4} \]
Now, add \( \frac{14}{4} + \frac{15}{4} \):
\[ \frac{14}{4} + \frac{15}{4} = \frac{29}{4} \]
Now, convert \( \frac{29}{4} \) back to a mixed number:
- \( 29 \div 4 = 7 \) remainder \( 1 \)
- Therefore, \( \frac{29}{4} = 7 \frac{1}{4} \)
Thus, the final result of \( 3 \frac{2}{4} + 3 \frac{3}{4} \) is
\[ \boxed{7 \frac{1}{4}} \]