what is the value of $3\\frac{1}{3} + \\frac{2}{3}$?

Explanation:

Step1: Convert mixed number to improper fraction

The mixed number ( 3\frac{1}{3} ) can be converted to an improper fraction. The formula for converting a mixed number ( a\frac{b}{c} ) to an improper fraction is ( \frac{a\times c + b}{c} ). So for ( 3\frac{1}{3} ), we have ( a = 3 ), ( b = 1 ), ( c = 3 ). Then ( 3\times3+1 = 10 ), so ( 3\frac{1}{3}=\frac{10}{3} ).

Step2: Add the two fractions

Now we need to add ( \frac{10}{3} ) and ( \frac{9}{3} ). Since the denominators are the same, we add the numerators: ( \frac{10 + 9}{3}=\frac{19}{3} ). We can also convert this back to a mixed number: ( \frac{19}{3}=6\frac{1}{3} ), but if we consider the addition of the whole number and the fraction part separately, ( 3\frac{1}{3}+\frac{9}{3}=3+\frac{1 + 9}{3}=3+\frac{10}{3}=3 + 3\frac{1}{3}=6\frac{1}{3} ) or as an improper fraction ( \frac{19}{3} ). Wait, maybe I made a mistake. Wait, ( 3\frac{1}{3}+\frac{9}{3} ): ( \frac{9}{3}=3 ), so ( 3\frac{1}{3}+3 = 6\frac{1}{3} ), and ( 6\frac{1}{3}=\frac{19}{3}\approx6.333 ). But let's check again. ( 3\frac{1}{3} ) is ( 3 + \frac{1}{3} ), and ( \frac{9}{3}=3 ), so adding them together: ( 3+\frac{1}{3}+3 = 6+\frac{1}{3}=6\frac{1}{3} ), which is ( \frac{19}{3} ). Wait, maybe the original problem is ( 3\frac{1}{3}+\frac{9}{3} ), let's compute the numerators: ( 3\frac{1}{3}=\frac{10}{3} ), ( \frac{10}{3}+\frac{9}{3}=\frac{19}{3}=6\frac{1}{3} ). But if we do the addition of the fraction part first: ( \frac{1}{3}+\frac{9}{3}=\frac{10}{3}=3\frac{1}{3} ), then add the whole number 3: ( 3 + 3\frac{1}{3}=6\frac{1}{3} ).

Answer:

( 6\frac{1}{3} ) (or ( \frac{19}{3} ))