Step 1: Set up the system of linear equations based on the principle that the total traffic entering an intersection must equal the total traffic leaving it. Intersection A: Traffic in: $125 + 350 = 475$ Traffic out: $x_1 + x_4$ Equation 1: $x_1 + x_4 = 475$ Intersection B: Traffic in: $x_1 + 225$ Traffic out: $x_2 + 400$ Equation 2: $x_1 + 225 = x_2 + 400 \implies x_1 - x_2 = 175$ Intersection C: Traffic in: $x_2 + 800$ Traffic out: $x_3 + 250$ Equation 3: $x_2 + 800 = x_3 + 250 \implies x_2 - x_3 = -550$ Intersection D: Traffic in: $x_4 + 300$ Traffic out: $x_3 + 600$ Equation 4: $x_4 + 300 = x_3 + 600 \implies x_4 - x_3 = 300$ Step 2: Solve the system of linear equations. We have the system: \begin{align} x_1 + x_4 &= 475 \quad &(1) \\ x_1 - x_2 &= 175 \quad &(2) \\ x_2 - x_3 &= -550 \quad &(3) \\ x_4 - x_3 &= 300 \quad &(4)\end{align} From (2), $x_1 = x_2 + 175$. From (3), $x_2 = x_3 - 550$. From (4), $x_4 = x_3 + 300$. Substitute $x_1$ and $x_4$ in terms of $x_2$ and $x_3$ into (1): $(x_2 + 175) + (x_3 + 300) = 475$ $x_2 + x_3 + 475 = 475$ $x_2 + x_3 = 0 \quad &(5)$ Now we have a system with $x_2$ and $x_3$: \begin{align} x_2 - x_3 &= -550 \quad &(3) \\ x_2 + x_3 &= 0 \quad &(5)\end{align} Add (3) and (5): $(x_2 - x_3) + (x_2 + x_3) = -550 + 0$ $2x_2 = -550$ $x_2 = -275$ Substitute $x_2 = -275$ into (5): $-275 + x_3 = 0$ $x_3 = 275$ Substitute $x_2 = -275$ into $x_1 = x_2 + 175$: $x_1 = -275 + 175$ $x_1 = -100$ Substitute $x_3 = 275$ into $x_4 = x_3 + 300$: $x_4 = 275 + 300$ $x_4 = 575$ The average hourly traffic values are: $x_1 = -100$ $x_2 = -275$ $x_3 = 275$ $x_4 = 575$ The negative values for $x_1$ and $x_2$ indicate that the actual traffic flow is in the opposite direction to what was initially assumed for those specific segments. The values are: $\boxed{x_1 = -100, x_2 = -275, x_3 = 275, x_4 = 575}$ Step 1: Write down the balanced equation with coefficients $x_1, x_2, x_3, x_4