L = rate of the large hose
S = rate of the small hose
we know the sum of their rates deliver the swimming pool filled up in 35 minutes, thus L + S = 35.
now, the large hose can do it in 60 minutes, that means that in 1 minute, the large hose has only done 1/60 th of the work.
the small hose can do the whole thing in S minutes, that means in 1 minute it has only done 1/S th of the whole work.
and since both of them working together can do it in 35 minutes, then in 1 minute they both have done 1/35 th of the whole job.
[tex]\bf \stackrel{\textit{large's rate in 1 minute}}{\cfrac{1}{L}}+\stackrel{\textit{small's rate in 1 minute}}{\cfrac{1}{S}}=\stackrel{\textit{done in 1 minute}}{\cfrac{1}{35}} \\\\\\ \cfrac{1}{60}+\cfrac{1}{S}=\cfrac{1}{35}\implies \cfrac{S+60}{60S}=\cfrac{1}{35}\implies 35S+2100=60S\implies 2100=25S \\\\\\ \cfrac{2100}{25}=S\implies 84=S[/tex]
