Question

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

Answer 1

let 1st part be x
so second part is 184-x
$\frac{1}{3} x = \frac{1}{7} \times (184 - x) + 8 \\ \frac{x}{3} + \frac{x}{7} = \frac{184}{7} + 8 \\ \frac{10x}{21} = \frac{240}{7} \\ x = 72$
1st part = 72
2nd part = 112