Two pipes are connected to the same tank. When working​ together, they can fill the tank in 12 hrs. The larger​ pipe, working​ alone, can fill the tank in 18 hrs less time than the smaller one. How long would the smaller one​ take, working​ alone, to fill the​ tank?

Answer 1

The small pipe will fill the tank in 18 hours.

Answer 2

rate of smaller pipe --- 1/x
rate of larger pipe ---- 1/(x-18) , where x > 18

combined rate = (x-18 + x)/(x(x-18)
= (2x-18)/(x^2 - 18x)

1/[(2x-18)/(x^2 - 18x)] = 12

x^2 - 18x = 24x - 216
x^2 - 42x + 216 = 0
(x - 36)(x - 6)= 0
x = 36 or x = 6
but x = 6 does not fall into my defined domain

so x = 36 <----- It would take the smaller pipe 36 hours by itself

check:
rate #1 = 1/36
rate # 2 = 1/18
combined rate = 1/36 + 1/18 = 1/12

time at combined rate = 1/(1/12) = 12