2 men and 5 boys can finish a piece of work in 4 days,while 3 men and 6 boys can finish it in 3 days. find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work.

let 1 men can finish the work in x days and 1 boy can finish the work in y days.

then 1 man's 1 day work = 1/x and 1 boy's one day work = 1/y

according to the question---

2/x + 5/y = 1/4 ------ (1)

3/x + 6/y = 1/3 --------(2)

put 1/x = u and 1/y = v in eq, (1) and (2)

2u + 5v = 1/4

= 8u + 20 v = 1 ----------- (3)

3u+ 6v =1/3

= 9u + 18 v = 1 ------------(4)

solving equation 3 and 4

multiply equation 3 with 9 and 4 with 8

9 (8u +20v =1)

8( 9u + 18v =1)

by multiplying we get

72u + 180v =9 ------ (5)

72u + 144v =8 -------(6)

now subtract eq. 5 from 6

72u + 180v =9

72u + 144v =8

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36V = 1

so v= 1/36

put v = 1/36 in eq. 3

8u + 20 X 1/36 = 1

8u =4/9

72u = 4

u = 1/18

u= 1/x = 1/18

x = 18 days

v = 1/y = 1/36

y = 36 days

so man can finish work in 18 days and boy can finsh work in 36 days