Question

It takes the first pipe 9 more hours to fill the pool than the first and the second pipes together and 7 less hours than it would take the second pipe if it was working alone. How long would it take to fill up the pool if both pipes were working together?

Answer 1

If the two together take z hours, then
1/x + 1/y = 1/z
x = z+9
x = y-7

now just crank it out

Answer 2

Let T1 be the time for pipe1 to fill the pool, T2...

so the combined time for both pipes is
1/T12 = 1/T1 + 1/T2 or T12=T1*T2/(T1+T2)

given T1=9+T12
and T1=T2-7

t1=9 + T1*T2/(T1+T2) replacing T1 with T2-7
T2-7=9 + (T2-7)(T2)/(2T2-7) lets replace t2 with x
(x-7)(2x-7)=9(2x-7)+x^2-7x

you can do all that, it looks to be a quadratic.
once x (T2) is found, then
x-7 is t1

so the combined time is (T1*T2/(T1+T2))

Answer 3

first pipe fills in f hours so 1 pool/f hours rate
second pipe fill in s hours so 1 pool /s hours rate

rate with both working = (1/f+1/s)pools/hour
so (1/f + 1/s) T = 1

f = 9 + T
f = s - 7 so s = f+7 = 9+T +7 = 16 + T

1/(T+9) + 1/(T+16) ] = 1/T
(T+16)+ (T+9) =(T^2+ 25 T + 144)/T
2 T^2 + 25 T = T^2 + 25 T +144
T^2 = 144
T = 12