Divide 184 into two parts such that one-third of one part may exceed one-seventh of the other part by 8 . Hint. Let the two parts be $x$ and $184-x$ . Then, $\frac{1}{3} x-\frac{1}{7}(184-x)=8 $

Given: Given number is 184

To do: Divide 184 in two parts such that one-tird exceeds one-seventh of other by 8

Solution:

Lets divide 184  into  two parts , such as    $x$   and   $184 - x  (x  +  184 - x = 184 )$

One third of one part  =  $\frac{1}{3} x$

One seventh of other part  =  $\frac{1}{7} (184 - x)$

One third of one part  exceeds One seventh of other part

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

multiply $\frac{1}{7}$  inside the brackets,

$\frac{1}{3}x- \frac{184}{7}+\frac{1}{7}x = 8$

$\frac{1}{3}x+\frac{1}{7}x = 8+ \frac{184}{7}$

Take LCM on  to add,

LCM of  3 , 7  is 21

$\frac{7}{21}x+\frac{3}{21}x = \frac{8 \times7}{7}+ \frac{184}{7}$

$\frac{10}{21}x = \frac{56}{7}+ \frac{184}{7}$

$\frac{10}{21}x = \frac{240}{7}$

$x =  \frac{240}{7} \times \frac{21}{10}$

            [$(\frac{240}{10}  =  24  ;  \frac{21}{7}  =  3)$]

$x$  =  24  $\times$  3

$x$   =  72

184  -  $x$  =  184 - 72  =  112

So, the parts are  72 and 112