Answer:
The number A(t) of pounds of salt in the tank at time 't' is;

Step-by-step explanation:
In the question, we have;
The volume of pure water initially in the tank = 700 gal
The concentration of brine pumped into the tank = 3 pounds per gallon
The rate at which the brine is pumped into the tank, = 7 gal/min
The rate at which the well mixed solution is pumped out = The same 7 gal/min
The number of pounds of salt in the tank at time 't' is found as follows;
The rate of change in A(t) with time = The rate of salt input - The rate of salt output

The rate of salt input = 7 gal/min × 3 lbs/gal = 21 lbs/min
The rate of salt output = (A(t)/700) lb/gal × 7 gal/min = (A(t)/100) lb/min
Therefore, we have;

Therefore;

The integrating factor is 

![\dfrac{d}{dt} \left[ e^{\dfrac{t}{100} } \times{A(t)}{} \right]= e^{\dfrac{t}{100} } \times21](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%7D%7Bdt%7D%20%5Cleft%5B%20e%5E%7B%5Cdfrac%7Bt%7D%7B100%7D%20%7D%20%5Ctimes%7BA%28t%29%7D%7B%7D%20%5Cright%5D%3D%20e%5E%7B%5Cdfrac%7Bt%7D%7B100%7D%20%7D%20%5Ctimes21)
Using an online tool, we get;

At time t = 0, A(t) = 0
We get;

c₁ = -2,100
Therefore, the number A(t) of pounds of salt in the tank at time 't' is given as follows;
