Question: $$3$\frac{1}{6}$ - 1$\frac{1}{2}$ + $\frac{1}{3}$
1. **State the problem:** Calculate the value of $$3\frac{1}{6} - 1\frac{1}{2} + \frac{1}{3}$$.
2. **Convert mixed numbers to improper fractions:**
- $$3\frac{1}{6} = \frac{3 \times 6 + 1}{6} = \frac{19}{6}$$
- $$1\frac{1}{2} = \frac{1 \times 2 + 1}{2} = \frac{3}{2}$$
3. **Rewrite the expression with improper fractions:**
$$\frac{19}{6} - \frac{3}{2} + \frac{1}{3}$$
4. **Find a common denominator:**
The denominators are 6, 2, and 3. The least common denominator (LCD) is 6.
5. **Convert all fractions to have denominator 6:**
- $$\frac{19}{6}$$ stays the same.
- $$\frac{3}{2} = \frac{3 \times 3}{2 \times 3} = \frac{9}{6}$$
- $$\frac{1}{3} = \frac{1 \times 2}{3 \times 2} = \frac{2}{6}$$
6. **Perform the operations:**
$$\frac{19}{6} - \frac{9}{6} + \frac{2}{6} = \frac{19 - 9 + 2}{6} = \frac{12}{6}$$
7. **Simplify the fraction:**
$$\frac{12}{6} = \cancel{\frac{12}{6}} = 2$$
**Final answer:** $$2$$