8 men and 12 boys can finish a piece of work in 5 days while 6 men and 8 boys can finish it in 7 days find the time taken by a man and a boy to do the same work

Let W be the work
No. of days taken by a man alone to complete the work be x days &
no. of days taken by a boy alone to complete the work be y days
Therefore one days work by a man = W/x
and one days work by a boy = W/y
 
CASE 1
one days work of 8 men and 12 boys = 8 W/x+12 W/y
and ten days of work by 8 men and 12 boys = W
therefore, 10 [8 W/x +12 W/y]=W
80 W/x+120 W/y = w
=80/x + 120/y = 1                     (1)

CASE 2
following the same steps according to the given condition we get the 2 n d equation as
84/x+112/y=1                          (2)

take 1/x =u and 1/y=v
(1) will change to 80 u+ 120 v=1          (3)
(2) will change to 84 u+ 112 v=1          (4)

(3)*21 = 1680 u+2520 v=21     (5)
(4)*20 = 1680 u+2240 v=20     (6)
              -          -             -
solving we will get v = 1/280
substituting value of v in (3)
80 u +120 *1/280 =1
80 u= 1-3/7
80 u= 7-3/7 {l c m}
80 u =4/7
u= 4/7*80
u= 1/140

when v = 1/280
1/y=1/280
therefore y = 280

when u = 1/140
1/x  = 1/140
therefore x = 40

therefore no.of days taken by a man alone to complete the work = x days = 140 days &
no. of days taken by a boy alone to complete the work = y days = 280 days

:)