1. **Problem Statement:** We have a network of one-way streets with unknown flow rates $x_1, x_2, x_3, x_4, x_5, x_6, x_7$ representing vehicles per hour. Given known flows and directions, we want to: - (a) Set up a linear system for the unknown flows. - (b) Solve the system. - (c) Determine if closing the road from A to B (flow $x_1$) is possible while keeping traffic flowing. 2. **Set up the linear system using flow conservation at each node:** - At node A: Total inflow = Total outflow Inflow: 500 (left) + 300 (top) + $x_7$ (from bottom left, leftward) Outflow: $x_1$ (to B) + $x_3$ (downward) + 350 (bottom) Equation: $500 + 300 + x_7 = x_1 + x_3 + 350$ - At node B: Inflow: $x_1$ (from A) + 200 (top) + $x_4$ (downward) Outflow: $x_2$ (right) + 100 (right) + $x_5$ (downward) Equation: $x_1 + 200 + x_4 = x_2 + 100 + x_5$ - At the bottom left node (where $x_6$ and $x_7$ are): Inflow: $x_6$ (downward) + 400 (left) Outflow: $x_7$ (leftward) + 350 (upward) Equation: $x_6 + 400 = x_7 + 350$ - At the bottom middle node (where $x_4$ and $x_6$ meet): Inflow: $x_3$ (downward) + 450 (left) Outflow: $x_4$ (downward) + $x_6$ (downward) Equation: $x_3 + 450 = x_4 + x_6$ - At the bottom right node (where $x_5$ and $x_7$ meet): Inflow: $x_5$ (downward) + 600 (downward) Outflow: 400 (downward) + $x_7$ (leftward) Equation: $x_5 + 600 = 400 + x_7$ 3. **Rewrite equations:** (1) $500 + 300 + x_7 = x_1 + x_3 + 350 \implies 800 + x_7 = x_1 + x_3 + 350$ $\Rightarrow x_1 + x_3 - x_7 = 450$ (2) $x_1 + 200 + x_4 = x_2 + 100 + x_5 \implies x_1 + x_4 - x_2 - x_5 = -100$ (3) $x_6 + 400 = x_7 + 350 \implies x_6 - x_7 = -50$ (4) $x_3 + 450 = x_4 + x_6 \implies x_3 - x_4 - x_6 = -450$ (5) $x_5 + 600 = 400 + x_7 \implies x_5 - x_7 = -200$ 4. **System of equations:** $$\begin{cases} x_1 + x_3 - x_7 = 450 \\ x_1 + x_4 - x_2 - x_5 = -100 \\ x_6 - x_7 = -50 \\ x_3 - x_4 - x_6 = -450 \\ x_5 - x_7 = -200 \end{cases}$$ 5. **Solve the system:** From (3): $x_6 = x_7 - 50$ From (5): $x_5 = x_7 - 200$ Substitute $x_6$ and $x_5$ into (4): $x_3 - x_4 - (x_7 - 50) = -450 \implies x_3 - x_4 - x_7 + 50 = -450 \implies x_3 - x_4 - x_7 = -500$ From (1): $x_1 + x_3 - x_7 = 450$ From (2): $x_1 + x_4 - x_2 - (x_7 - 200) = -100 \implies x_1 + x_4 - x_2 - x_7 + 200 = -100 \implies x_1 + x_4 - x_2 - x_7 = -300$ Now we have: (1) $x_1 + x_3 - x_7 = 450$ (2) $x_1 + x_4 - x_2 - x_7 = -300$ (4 modified) $x_3 - x_4 - x_7 = -500$ Express $x_1$ from (1): $x_1 = 450 + x_7 - x_3$ Substitute into (2): $450 + x_7 - x_3 + x_4 - x_2 - x_7 = -300 \implies 450 - x_3 + x_4 - x_2 = -300$ Simplify: $- x_3 + x_4 - x_2 = -750$ Rewrite: $x_4 - x_3 - x_2 = -750$ From (4 modified): $x_3 - x_4 - x_7 = -500 \implies - x_4 + x_3 = -500 + x_7$ Add the two equations: $x_4 - x_3 - x_2 = -750$ $- x_4 + x_3 = -500 + x_7$ Add left sides and right sides: $(x_4 - x_3) + (- x_4 + x_3) - x_2 = -750 + (-500 + x_7)$ $0 - x_2 = -1250 + x_7 \implies x_2 = 1250 - x_7$ Now from $x_4 - x_3 - x_2 = -750$: Substitute $x_2$: $x_4 - x_3 - (1250 - x_7) = -750 \implies x_4 - x_3 - 1250 + x_7 = -750$ Simplify: $x_4 - x_3 + x_7 = 500$ From (4 modified): $x_3 - x_4 - x_7 = -500$ Add these two: $(x_4 - x_3 + x_7) + (x_3 - x_4 - x_7) = 500 + (-500)$ $0 = 0$ This confirms consistency but does not determine $x_3, x_4, x_7$ uniquely. Choose $x_7 = t$ (parameter). From (4 modified): $x_3 - x_4 = -500 + t$ From $x_4 - x_3 + t = 500$: $x_4 - x_3 = 500 - t$ Add: $(x_3 - x_4) + (x_4 - x_3) = (-500 + t) + (500 - t) \implies 0 = 0$ Again consistent. Set $x_3 = s$ (parameter), then: $x_4 = s + 500 - t$ Recall: $x_1 = 450 + t - s$ $x_2 = 1250 - t$ $x_5 = t - 200$ $x_6 = t - 50$ $x_7 = t$ 6. **Interpretation:** The system has infinitely many solutions depending on parameters $s$ and $t$. All flows depend on these two parameters. 7. **Closing road from A to B ($x_1=0$):** Set $x_1=0$: $0 = 450 + t - s \implies s = 450 + t$ All other flows become: $x_3 = s = 450 + t$ $x_4 = s + 500 - t = 450 + t + 500 - t = 950$ $x_2 = 1250 - t$ $x_5 = t - 200$ $x_6 = t - 50$ $x_7 = t$ For all flows to be nonnegative: $x_5 = t - 200 \geq 0 \implies t \geq 200$ $x_6 = t - 50 \geq 0 \implies t \geq 50$ $x_7 = t \geq 0$ $x_3 = 450 + t \geq 0$ always true for $t \geq 0$ $x_2 = 1250 - t \geq 0 \implies t \leq 1250$ So $200 \leq t \leq 1250$. Therefore, it is possible to close the road from A to B and keep traffic flowing by choosing $t$ in this range and $s = 450 + t$. **Final answer:** $$\boxed{\begin{cases} x_1 = 0 \\ x_2 = 1250 - t \\ x_3 = 450 + t \\ x_4 = 950 \\ x_5 = t - 200 \\ x_6 = t - 50 \\ x_7 = t \end{cases} \text{ with } 200 \leq t \leq 1250}$$