Let S be the rate of work of the small hose (= volume of water per 1 hour).

Let L be the rate of work of the large hose (= volume of water per 1 hour).


Then you have these two equation from the condition:

 3L +  8S = 0.5,    (1)
10L + 10S = 1       (2)

where 0.5 denotes half of pool volume and 1 denotes whole volume.


To solve the system, multiply eq(1) by 10 (both sides)  and  multiply eq(2) by 3 (both sides). You will get

30L + 80S = 5,      (3)
30L + 30S = 3.      (4)


Now subtract eq(4) from eq(3)  (both sides).


You will get

     50S = 2  ====>  S =  = .


Thus you found that the rate of work of the small hose is  of the pool volume per hour.

It means that the small hose can fill the pool in 25 hour, working alone.



Now from equation (2)

10L + 10*(1/25) = 1,   or   10L = 1 - 10/25 = 15/25 = 3/5  ====>  L = .


Thus you found that the rate of work of the large hose is  of the pool volume per hour.

It means that the small hose can fill the pool in  hours =  hours = 16 hours and 40 minutes, working alone.