1. **Problem statement:** We need to set up and solve the system of linear equations representing the traffic flow in the city network to find the values of $x_1, x_2, x_3, x_4, x_5$. 2. **Understanding the problem:** Each node (intersection) in the network must satisfy the flow conservation principle: the total inflow equals the total outflow. 3. **Write the flow balance equations for each node:** - Node A: inflow = outflow Inflow: $x_2$ (from B to A) Outflow: 600 (to left), 200 (upwards) Equation: $x_2 = 600 + 200 = 800$ - Node B: Inflow: $x_1$ (from D to B) Outflow: $x_2$ (to A), $x_4$ (to C), 300 (to right) Equation: $x_1 = x_2 + x_4 + 300$ - Node C: Inflow: $x_4$ (from B to C), $x_5$ (from D to C) Outflow: 150 (downward), 450 (to right) Equation: $x_4 + x_5 = 150 + 450 = 600$ - Node D: Inflow: None given explicitly Outflow: $x_1$ (to B), $x_3$ (to right), $x_5$ (to C) Equation: $x_1 + x_3 + x_5 = $ total inflow at D (unknown, so we use other equations) 4. **From the above, we have the following system:** $$ \begin{cases} x_2 = 800 \\ x_1 = x_2 + x_4 + 300 \\ x_4 + x_5 = 600 \end{cases} $$ 5. **We need one more equation to solve for all variables. Using node D's balance:** Assuming total inflow at D equals total outflow, and since no inflow is given, we consider the network is closed and consistent with the above equations. 6. **Substitute $x_2 = 800$ into the second equation:** $$x_1 = 800 + x_4 + 300 = x_4 + 1100$$ 7. **From the third equation:** $$x_5 = 600 - x_4$$ 8. **Express $x_1$ and $x_5$ in terms of $x_4$ and use the flow at node D:** At node D, outflow is $x_1 + x_3 + x_5$. Since no inflow is given, we assume the total outflow equals the sum of inflows at other nodes, which is consistent with the network. 9. **Choose $x_3$ as a free variable (since no equation fixes it), then express $x_1$ and $x_5$ in terms of $x_4$ and $x_3$.** 10. **Final solution:** $$ \begin{cases} x_2 = 800 \\ x_1 = x_4 + 1100 \\ x_5 = 600 - x_4 \\ x_3 = \text{free parameter} \end{cases} $$ This means $x_3$ and $x_4$ can be chosen freely, and $x_1, x_2, x_5$ depend on these choices. **Summary:** The system has infinitely many solutions parameterized by $x_3$ and $x_4$.