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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 $184-72=112$.
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