Let a(t) and b(t) denote the amounts of salt in tanks A and B, respectively.
The volume of liquid in tanks A and B after t minutes are
A: 50 L + (5 L/min + 3 L/min - 8 L/min)t = 50 L
B: 40 L + (7 L/min + 8 L/min - 3 L/min - 12 L/min)t = 40 L
so the amount of solution in the tanks stays constant.
Salt flows into tank A at a rate of
(2 g/L)*(5 L/min) + (b(t)/40 g/L)*(3 L/min) = (10 + 3/40 b(t)) g/min
and out at a rate of
(a(t)/50 g/L)*(8 L/min) = 4/25 a(t) g/min
so the net flow rate is given by the differential equation
[tex]\dfrac{\mathrm da(t)}{\mathrm dt}=10+\dfrac{3b(t)}{40}-\dfrac{4a(t)}{25}[/tex]
We do the same for tank B: salt flows in at a rate of
(3 g/L)*(7 L/min) + (a(t)/50 g/L)*(8 L/min) = (21 + 4/25 a(t)) g/min
and out at a rate of
(b(t)/40 g/L)*(3 L/min + 12 L/min) = 3/8 b(t) g/min
and hence with a net rate of
[tex]\dfrac{\mathrm db(t)}{\mathrm dt}=21+\dfrac{4a(t)}{25}-\dfrac{3b(t)}8[/tex]
Replace a(t) and b(t) with x and y. Then the system is (in matrix form)
[tex]\dfrac{\mathrm d}{\mathrm dt}\begin{bmatrix}x\\y\end{bmatrix}=\dfrac1{200}\begin{bmatrix}-32&15\\32&-75\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}+\begin{bmatrix}10\\21\end{bmatrix}[/tex]
with initial conditions x(0) = 60 g and y(0) = 80 g.
