Find the value of $m$:
$$m-\frac{2}{3}=\frac{5}{6}+\frac{1}{2}$$

Given linear equation:
$$m-\frac{2}{3}=\frac{5}{6}+\frac{1}{2}$$

First, solve the right-hand side (RHS) by taking the LCM of $6$ and $2$, which is $6$:

$$\frac{5}{6}+\frac{1}{2}=\frac{5+3}{6}=\frac{8}{6}=\frac{4}{3}$$

Now, substitute this value back into the equation:

$$m-\frac{2}{3}=\frac{4}{3}$$

Move $-\frac{2}{3}$ to the RHS:

$$m=\frac{4}{3}+\frac{2}{3}$$

$$m=\frac{4+2}{3}=\frac{6}{3}=2$$

Thus, the value of $m$ is $2$.