Question 218847
Rate of flow into the tank at spout A=2 gal/hr
Let x=rate of flow out at Spout B (gal/hr)


Net rate of flow into the tank=2-x

Now we know that net rate of flow into the tank times 96 hrs=6 gal
So, our equation to solve is:

(2-x)*96=6
192-96x=6 subtract 192 from each side
-96x=6-192
-96x=-186 divide each side by -96
x=1.9375 gal/hr----rate of flow out at Spout B (gal/hr)

CK
(2-1.9375)*96=6
0.0625*96=6
6=6

Hope this helps----ptaylor