$$9-3\frac{1}{3}$$
Transform the mixed numbers into improper fraction. Use this formula (can you see why it's true):
$$a\frac{b}{c}=\frac{ac + b}{c}$$
We get:
$$9 - \frac{3*3+1}{3}=$$
$$9 - \frac{10}{3}=$$
Transform the integer into a fraction with a common denominator (bottom), and calculate the answer:
$$9 - \frac{10}{3}=$$
$$\frac{27}{3}- \frac{10}{3}= \frac{17}{3}$$
.Really nice presentation Tetration.
I would do it slightly differently. There is no need to change the mixed numeral into an improper fraction.
I have spread it out so that it is easy to follow but most of these steps are not normally written.
This really is a good method to use.
$$\\9-3\frac{1}{3}\\\\
\mbox{you are taking away more than 3 so it must be }-3-\frac{1}{3}\\\\
=9-3-\frac{1}{3}\\\\
=6-\frac{1}{3}\\\\
$Now 6 minus a bit must equal 5 and a bit, so what do we do next$\\
$I'll change the 6 into 5+1$\\\\
=5+1-\frac{1}{3}\\\\
=5+\frac{3}{3}-\frac{1}{3}\\\\
=5+\frac{2}{3}\\\\
=5\frac{2}{3}$$