Question

4\ \frac{1}{6} + 5\ \frac{2}{3}

Answer 1

To add the mixed numbers \(4\ \frac{1}{6}\) and \(5\ \frac{2}{3}\), we first convert them into improper fractions.

1. Convert \(4\ \frac{1}{6}\) into an improper fraction:
\[
4\ \frac{1}{6} = 4 \times 6 + 1 = 24 + 1 = 25 \quad \text{(so it is } \frac{25}{6})
\]

2. Convert \(5\ \frac{2}{3}\) into an improper fraction:
\[
5\ \frac{2}{3} = 5 \times 3 + 2 = 15 + 2 = 17 \quad \text{(so it is } \frac{17}{3})
\]

Now we need to add \(\frac{25}{6}\) and \(\frac{17}{3}\). To do this, we need a common denominator. The least common multiple of 6 and 3 is 6.

3. Convert \(\frac{17}{3}\) to an equivalent fraction with a denominator of 6:
\[
\frac{17}{3} = \frac{17 \times 2}{3 \times 2} = \frac{34}{6}
\]

4. Now we can add the fractions:
\[
\frac{25}{6} + \frac{34}{6} = \frac{25 + 34}{6} = \frac{59}{6}
\]

5. Convert \(\frac{59}{6}\) back to a mixed number:
\[
59 \div 6 = 9 \quad \text{(with a remainder of } 5\text{)}
\]
So,
\[
\frac{59}{6} = 9\ \frac{5}{6}
\]

Thus, the answer is:
\[
4\ \frac{1}{6} + 5\ \frac{2}{3} = 9\ \frac{5}{6}
\]