Question

divide 105 two parts such that one third of one part exceeds one seventh of the other part by 8 find the two parts​

Answer 1

This is the final answer hope this helps you.

Mark as brainliest

Answer 2

Solution:

let the two part be x and (105-x)

then,

Or, 1/3x-1/7(105-x)=8

Or, 1/3s-105/7+x/7=8

Or, 1/3x+1/7x=105/7+8

Or, 7x+3x/21=8+105/7

Or, 10x/21=56+105/7

Or, 10x/21=161/7

Or, x=161*21/7*10

Therefore,  x=48.3

Now,

second part =105-48.3 =56.7

.'. the two part are 48.3 & 56.7