Let's solve the problem step by step: --- **Step 1: Set up the rate equations for each tank.** Let's denote: - \( Q_1(t) \): Amount of solute in Tank 1 at time \( t \) - \( Q_2(t) \): Amount of solute in Tank 2 at time \( t \) **Tank 1:** - Inflow: 10 L/min * 0.3 kg/L = **3 kg/min** (from outside source) - Outflow to Tank 2: 10 L/min, concentration in Tank 1 is \( Q_1/1000 \) kg/L, so outflow = \( 10 \times Q_1/1000 = Q_1/100 \) kg/min - Inflow from Tank 2: 10 L/min, concentration in Tank 2 is \( Q_2/1000 \), so inflow = \( 10 \times Q_2/1000 = Q_2/100 \) kg/min - Outflow to outside: 10 L/min, concentration in Tank 1 is \( Q_1/1000 \), so outflow = \( 10 \times Q_1/1000 = Q_1/100 \) kg/min So, **net rate for Tank 1:** \[ Q_1' = \text{inflow from outside} + \text{inflow from Tank 2} - \text{outflow to Tank 2} - \text{outflow to outside} \] \[ Q_1' = 3 + \frac{Q_2}{100} - \frac{Q_1}{100} - \frac{Q_1}{100} \] \[ Q_1' = 3 + \frac{Q_2}{100} - \frac{2Q_1}{100} \] \[ Q_1' = 3 - \frac{2Q_1}{100} + \frac{Q_2}{100} \] --- **Step 2: Write in matrix form and set up the system for both tanks** **Tank 2:** - Inflow from Tank 1: \( 10 \times Q_1/1000 = Q_1/100 \) kg/min - Outflow to Tank 1: \( 10 \times Q_2/1000 = Q_2/100 \) kg/min - Inflow from outside: 0 (concentration is 0 kg/L) - Outflow to outside: \( Q_2/100 \) kg/min So, **net rate for Tank 2:** \[ Q_2' = \text{inflow from Tank 1} - \text{outflow to Tank 1} - \text{outflow to outside} \] \[ Q_2' = \frac{Q_1}{100} - \frac{Q_2}{100} - \frac{Q_2}{100} \] \[ Q_2' = \frac{Q_1}{100} - \frac{2Q_2}{100} \] Write both equations in matrix form: \[ \begin{bmatrix} Q_1' \\ Q_2' \end{bmatrix} = \begin{bmatrix} -\frac{2}{100} & \frac{1}{100} \\ \frac{1}{100} & -\frac{2}{100} \end{bmatrix} \begin{bmatrix} Q_1 \\ Q_2 \end{bmatrix} + \begin{bmatrix} 3 \\ 0 \end{bmatrix} \] **Explanation:** - The matrix coefficients come from the inflow/outflow rates and concentrations as explained above. - The vector \(\begin{bmatrix} 3 \\ 0 \end{bmatrix}\) represents the external solute inflow to Tank 1 only. --- **Final Answer:** \[ \boxed{ \begin{bmatrix} Q_1' \\ Q_2' \end{bmatrix} = \begin{bmatrix} -\frac{1}{100} & \frac{1}{100} \\ \frac{1}{100} & -\frac{1}{100} \end{bmatrix} \begin{bmatrix} Q_1 \\ Q_2 \end{bmatrix} + \begin{bmatrix} 3 \\ 0 \end{bmatrix} } \] **(As shown in the image, the boxed entries are correct!)**

**Step 1: Derive the differential equations for each tank** - **Tank 1:** - Inflow from external source: \(10 \text{ L/min} \times 0.3 \text{ kg/L} = 3 \text{ kg/min}\) - Inflow from Tank 2: \(10 \text{ L/min} \times \frac{Q_2}{1000} \text{ kg/L} = \frac{Q_2}{100} \text{ kg/min}\) - Outflow to Tank 2: \(10 \text{ L/min} \times \frac{Q_1}{1000} \text{ kg/L} = \frac{Q_1}{100} \text{ kg/min}\) - Outflow to outside: \(10 \text{ L/min} \times \frac{Q_1}{1000} \text{ kg/L} = \frac{Q_1}{100} \text{ kg/min}\) The rate of change of solute in Tank 1: \[ \frac{dQ_1}{dt} = \text{Inflow} + \text{Inflow from Tank 2} - \text{Outflow to Tank 2} - \text{Outflow to outside} \] \[ \frac{dQ_1}{dt} = 3 + \frac{Q_2}{100} - \frac{Q_1}{100} - \frac{Q_1}{100} = 3 + \frac{Q_2}{100} - \frac{2Q_1}{100} \] - **Tank 2:** - Inflow from Tank 1: \(\frac{Q_1}{100}\) kg/min - Outflow to Tank 1: \(\frac{Q_2}{100}\) kg/min - Outflow to outside: \(\frac{Q_2}{100}\) kg/min - No external inflow (concentration 0) Rate of change: \[ \frac{dQ_2}{dt} = \frac{Q_1}{100} - \frac{Q_2}{100} - \frac{Q_2}{100} = \frac{Q_1}{100} - \frac{2Q_2}{100} \] --- **Step 2: Write the system in matrix form** \[ \boxed{ \begin{bmatrix} \displaystyle \frac{dQ_1}{dt} \\ \displaystyle \frac{dQ_2}{dt} \end{bmatrix} = \begin{bmatrix} -\frac{2}{100} & \frac{1}{100} \\ \frac{1}{100} & -\frac{2}{100} \end{bmatrix} \begin{bmatrix} Q_1 \\ Q_2 \end{bmatrix} + \begin{bmatrix} 3 \\ 0 \end{bmatrix} } \] **Final answer:** The system of first-order differential equations modeling the process is as shown above.