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?

3 answers

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
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))
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