Let's break down the solution into 2 steps as requested: --- **Step 1: Write the differential equations for each tank.** - **Tank 1:** - **Inflow:** - From external source: \(10 \text{ L/min} \times 0.3 \text{ kg/L} = 3 \text{ kg/min}\) - From Tank 2: \(10 \text{ L/min} \times \frac{Q_2}{1000} \text{ kg/L}\) - **Outflow:** - To Tank 2: \(10 \text{ L/min} \times \frac{Q_1}{1000} \text{ kg/L}\) - **Rate of change:** \[ Q_1' = 3 + 10\frac{Q_2}{1000} - 10\frac{Q_1}{1000} \] \[ Q_1' = 3 + \frac{Q_2}{100} - \frac{Q_1}{100} \] - **Tank 2:** - **Inflow:** - From Tank 1: \(10 \text{ L/min} \times \frac{Q_1}{1000} \text{ kg/L}\) - **Outflow:** - To Tank 1: \(10 \text{ L/min} \times \frac{Q_2}{1000} \text{ kg/L}\) - **Rate of change:** \[ Q_2' = 10\frac{Q_1}{1000} - 10\frac{Q_2}{1000} \] \[ Q_2' = \frac{Q_1}{100} - \frac{Q_2}{100} \] **Explanation:** Here, the rates of change for the solute in each tank are determined by the net inflow and outflow of solute, considering the concentration and flow rates. --- **Step 2: Write the system in matrix form.** \[ \begin{bmatrix} Q_1' \\ Q_2' \end{bmatrix} = \begin{bmatrix} -\frac{1}{100} & \frac{1}{100} \\ \frac{1}{100} & -\frac{1}{100} \end{bmatrix} \begin{bmatrix} Q_1 \\ Q_2 \end{bmatrix} + \begin{bmatrix} 3 \\ 0 \end{bmatrix} \] **Explanation:** This matrix system captures the coupled nature of the tanks: each tank's solute changes depend both on its own current solute and the solute in the other tank, as well as any external input. --- **Final Answer:** \[ \boxed{ \begin{bmatrix} Q_1' \\ Q_2' \end{bmatrix} = \begin{bmatrix} -\frac{1}{100} & \frac{1}{100} \\ \frac{1}{100} & -\frac{1}{100} \end{bmatrix} \begin{bmatrix} Q_1 \\ Q_2 \end{bmatrix} + \begin{bmatrix} 3 \\ 0 \end{bmatrix} } \]