divide $184$ into two parts such that one-third of one part may exceed one-seventh of the other part by $8$.

Given: A number $184$.

To do: To divide $184$ into two parts such that one-third of one part may exceed one-seventh of the other part by $8$.

Solution:

Let the two parts be  $x$ and $184-x$.

$\therefore \frac{1}{3} x-\frac{1}{7}(184-x)=8$

$\Rightarrow \frac{7x-552+3x}{21}=8$

$\Rightarrow 10x-552=168$

$\Rightarrow 10x=552+168$

$\Rightarrow 10x=720$

$\Rightarrow x=\frac{720}{10}$

$\Rightarrow x=72$

Thus, the numbers are: $72$ and $184-72=112$.