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?

If Pipe 1 (P1) takes x hours to fill the pool, Pipe 2 (P1) and pipe 2 (P2) takes (x-9) hours to fill the pool, and pipe 2 (P2) takes (x+7) hours to fill the pool.

That is,
P1 = x hrs
P1+P2 = (x-9) hrs
P3 = (x+7) hrs

In 1 hour, P1 fills 1/x of the pool, P1+P2 fills 1/(x-9) of the pool and P2 fills 1/(1+7) of the pool.
Therefore,
1/x+1/(1+7) = 1/(x-9) => ((x+7)+x)/(x)(x+7)=1/(x-9) => (2x+7)/x^2+7x = 1/(x-9) => (2x+7)(x-9)=x^2+7x => x^2-18x-63 =0 
Solving for x
x= (-b+/- sqrt (b^2-4ac)/2a, where a=1, b=18, and c=63
Substituting;
x1=21 and x2=-3 (the  negative x is ignored as it does not make sense).
Therefore, x = 21
This means,
P1 takes 21 hours to fill the pool
P1+P2 takes (21-9) hours = 12 hours to fill the pool while P3 takes (21+7) hours = 28 hours

good luck

I'm sure the answer is a