Solving Simple Mixing Problem

The Question: Assume there are two tanks, Tank A and Tank B. Assume Tank A contains 6 gallons of liquid and Tank B contains 9 gallons of liquid. Initially, Tank A contains $A_0$ pounds of salt and Tank B contains $B_0$ pounds of sale and $A_0+B_0=20$. Starting at time $t=0$, the liquid in the system begins to circulate, with liquid flowing from Tank A to Tank B at a rate of 1 gallon per minute, and from Tank B back to Tank A at the same rate. Fnd the initial value problem satisfied by $A(t)$ and $B(t)$. DO NOT SOLVE THE INITIAL VALUE PROBLEM.
My Work: The solution to any mixing problem involves finding the rate of change in the tanks, and in this case we have a closed system. Thus, $$\frac{dA}{dt}=\text{Coming In-Going Out}= \frac{B(t)\cdot 1gal/min}{9}-\frac{A(t)\cdot 1gal/min}{6}$$ and $$\frac{dB}{dt}=\text{Coming In-Going Out}= \frac{A(t)\cdot 1gal/min}{9}-\frac{B(t)\cdot 1gal/min}{6}$$We also know that $A(0)=A_0$ and $B(0)=B_0$. My Question: I am not sure if this is the right coming and going of liquid within the system. It makes sense to me but I just want to make sure. Any help is appreciated. Thanks

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\cdot to get $\cdot$, \text{ } to type regular text in-between \$\$ –  Git Gud Jan 26 '13 at 17:59

I assume that $A(t)$, $B(t)$ are the amounts of salt in A and B repectively at time $t$.
The second equation is not right. The right-hand side should be the negative of the right-hand side of the first! For note that $A+B$ is constant, so its derivative is $0$.
Or else do in-out analysis. In $1$ minute, we get a gallon transferred from A to B, which carries $\frac{A(t)}{6}$ of salt, not $\frac{A(t)}{9}$.