Abigail sold $3\frac{3}{6}$ cakes and Cody sold $7\frac{2}{3}$ cakes. We need to find how many more cakes Cody sold than Abigail. This requires subtracting the number of cakes Abigail sold from the number of cakes Cody sold, and giving the answer as a mixed number in its simplest form.

1. Problem Description

Abigail sold $3\frac{3}{6}$ cakes and Cody sold $7\frac{2}{3}$ cakes. We need to find how many more cakes Cody sold than Abigail. This requires subtracting the number of cakes Abigail sold from the number of cakes Cody sold, and giving the answer as a mixed number in its simplest form.

2. Solution Steps

First, we need to subtract the number of cakes Abigail sold from the number of cakes Cody sold.

$7\frac{2}{3} - 3\frac{3}{6}$

We can simplify $3\frac{3}{6}$ to $3\frac{1}{2}$.

$7\frac{2}{3} - 3\frac{1}{2}$

Now, we convert the mixed numbers to improper fractions.

$7\frac{2}{3} = \frac{7 \times 3 + 2}{3} = \frac{21+2}{3} = \frac{23}{3}$

$3\frac{1}{2} = \frac{3 \times 2 + 1}{2} = \frac{6+1}{2} = \frac{7}{2}$

So, the subtraction becomes:

$\frac{23}{3} - \frac{7}{2}$

To subtract the fractions, we need a common denominator. The least common multiple of 3 and 2 is

6. $\frac{23}{3} \times \frac{2}{2} = \frac{46}{6}$

$\frac{7}{2} \times \frac{3}{3} = \frac{21}{6}$

Now we can subtract:

$\frac{46}{6} - \frac{21}{6} = \frac{46-21}{6} = \frac{25}{6}$

Now, convert the improper fraction $\frac{25}{6}$ to a mixed number:

$25 \div 6 = 4$ with a remainder of $1$.

So, $\frac{25}{6} = 4\frac{1}{6}$

3. Final Answer

$4\frac{1}{6}$