Add. Write your answer as a mixed number in simplest form. \( 2 \frac{1}{6}+4 \frac{1}{3} \)

To solve \( 2 \frac{1}{6} + 4 \frac{1}{3} \), first convert both mixed numbers to improper fractions. For \( 2 \frac{1}{6} \): \[ 2 \times 6 + 1 = 12 + 1 = 13 \] So, \( 2 \frac{1}{6} = \frac{13}{6} \). For \( 4 \frac{1}{3} \): \[ 4 \times 3 + 1 = 12 + 1 = 13 \] So, \( 4 \frac{1}{3} = \frac{13}{3} \). Now, to add these improper fractions, find a common denominator. The least common multiple of 6 and 3 is 6. Convert \( \frac{13}{3} \) to have a denominator of 6: \[ \frac{13}{3} = \frac{13 \times 2}{3 \times 2} = \frac{26}{6} \] Now add: \[ \frac{13}{6} + \frac{26}{6} = \frac{39}{6} \] Next, simplify \( \frac{39}{6} \) by dividing both the numerator and the denominator by 3: \[ \frac{39 \div 3}{6 \div 3} = \frac{13}{2} \] Finally, convert \( \frac{13}{2} \) back to a mixed number: 13 divided by 2 gives 6 with a remainder of 1, so: \[ 6 \frac{1}{2} \] Thus, the final answer is: \[ 6 \frac{1}{2} \]