What is $\frac{9}{3} - \frac{1}{2}$?

$$\frac93-\frac12=3-\frac12=\frac62-\frac{1}{2}=\frac{5}{2}$$

$$\text{or}$$

$$\frac93-\frac12=\frac{9\times2-3\times1}{6}=\frac{15}{6}=\frac{5}{2}$$

So we are trying to compute $\frac{9}{3} - \frac{1}{2}$. To add or subtract fractions, we need to find a common denominator. I think you already understand that the common denominator here is $6$.

To get a $6$ in the denominator of $\frac{9}{3}$, we need to multiply this by $\frac{2}{2}$, and so this becomes $\frac{18}{6}$. To get a $6$ in the denominator of $\frac{1}{2}$, we need to multiply this by $\frac{3}{3}$, and we get $\frac{3}{6}$.

So, we get

$\frac{9}{3} - \frac{1}{2}$

$ = \frac{9}{3} \cdot \frac{2}{2} - \frac{1}{2} \cdot \frac{3}{3}$

$ = \frac{18}{6} - \frac{3}{6}$

$= \frac{18 - 3}{6} $

$ = \frac{15}{6}$

But the numerator and denominator have a common factor of $3$, since $15 = 5 \cdot 3$ and $6 = 2 \cdot 3$, so we can cancel the $3$s, and we get:

$\frac{15}{6}$

$= \frac{5 \cdot 3}{2 \cdot 3}$

$= \frac{5}{2}$.