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: $x_4 + 300 = x_3 + 600 \implies x_4 - x_3 = 300$

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:
$\boxed{x_1 = -100, x_2 = -275, x_3 = 275, x_4 = 575}$