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.

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 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 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 x_4 - x_3 = 300 Step 2: Solve the system of linear equations. We have the system: x_1 + x_4 &= 475 &(1) \\ x_1 - x_2 &= 175 &(2) \\ x_2 - x_3 &= -550 &(3) \\ x_4 - x_3 &= 300 &(4) 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 &(5) Now we have a system with x_2 and x_3: x_2 - x_3 &= -550 &(3) \\ x_2 + x_3 &= 0 &(5) 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: 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