Let [tex]A_1[/tex] be the amount of salt in tank 1 at time [tex]t[/tex], and [tex]A_2[/tex] the amount of salt in the tank 2.
The volume of solution in either tank stays constant. In tank 1, at time [tex]t[/tex] (min) we have
[tex]90\,\mathrm L+\underbrace{\left(3.5\dfrac{\rm L}{\rm min}-6\dfrac{\rm L}{\rm min}+2.5\dfrac{\rm L}{\rm min}\right)}_0t=90\,\mathrm L[/tex]
In tank 2,
[tex]60\,\mathrm L+\underbrace{\left(4\dfrac{\rm L}{\rm min}+6\dfrac{\rm L}{\rm min}-2.5\frac{\rm L}{\rm min}-7.5\frac{\rm L}{\rm min}\right)}_0t=60\,\mathrm L[/tex]
Then the concentration of salt in tanks 1 and 2 at any given time is [tex]\dfrac{A_1\,\rm g}{90\,\rm L}[/tex] and [tex]\dfrac{A_2\,\rm g}{60\,\rm L}[/tex].
The net rate of change of the amount of salt in tanks 1 and 2 follows a simple rule:
[tex]\dfrac{\mathrm dA_i}{\mathrm dt}=(\text{rate in})-(\text{rate out})[/tex]
Each rate is in units of g/min. Each L coming in or going out contributes or removes some salt depending on the flow rate (L/min) and concentration (g/L) of the solution in either tank. For tank 1, we have
[tex]\text{rate in}=\left(50\dfrac{\rm g}{\rm L}\right)\left(3.5\dfrac{\rm L}{\rm min}\right)+\left(\dfrac{A_2}{60}\dfrac{\rm g}{\rm L}\right)\left(2.5\dfrac{\rm L}{\rm min}\right)[/tex]
[tex]\text{rate out}=\left(\dfrac{A_1}{90}\dfrac{\rm g}{\rm L}\right)\left(6\dfrac{\rm L}{\rm min}\right)[/tex]
Then the amount of salt in tank 1 has rate of change (ignoring units)
[tex]\dfrac{\mathrm dA_1}{\mathrm dt}=-\dfrac{A_1}{15}+\dfrac{A_2}{24}+175[/tex]
A similar breakdown for tank 2 shows a rate of change of
[tex]\dfrac{\mathrm dA_2}{\mathrm dt}=\dfrac{A_1}{15}-\dfrac{A_2}6+160[/tex]
In matrix form, the system is described by
[tex]\begin{pmatrix}A_1\\A_2\end{pmatrix}'=\begin{pmatrix}-\frac1{15}&\frac1{24}\\\frac1{15}&-\frac16\end{pmatrix}\begin{pmatrix}A_1\\A_2\end{pmatrix}+\begin{pmatrix}175\\160\end{pmatrix}[/tex]
You can solve this with the usual eigenvalue method and method of undetermined coefficients. You should get a general solution of
[tex]\begin{pmatrix}A_1\\A_2\end{pmatrix}=C_1\begin{pmatrix}5\\-6-2\sqrt{19}\end{pmatrix}e^{\frac{-7+\sqrt{19}}{60}t}+C_2\begin{pmatrix}5\\-6+2\sqrt{19}\end{pmatrix}e^{\frac{-7-\sqrt{19}}{60}t}+\begin{pmatrix}4300\\2680\end{pmatrix}[/tex]
Then use the initial values [tex]A_1(0)=290[/tex] and [tex]A_2(0)=245[/tex] to solve for [tex]C_1,C_2[/tex] and find the particular solution.
