To find the function $F(x)$ that satisfies the functional equation

$F(x) + F \left( \frac{2x - 3}{x - 1} \right) = x,$

we first define a transformation $g(x) = \frac{2x - 3}{x - 1}$. We need to explore the properties of $g(x)$ to better understand how it interacts with the function $F(x)$.

First, simplify $g(x)$:

$g(x) = \frac{2x - 3}{x - 1}.$

Next, we find the inverse of $g(x)$. To do this, we set $y = g(x)$ and solve for $x$ in terms of $y$:

$y = \frac{2x - 3}{x - 1}.$

Cross-multiplying, we get:

$y(x - 1) = 2x - 3 \implies yx - y = 2x - 3 \implies yx - 2x = y - 3 \implies x(y - 2) = y - 3 \implies x = \frac{y - 3}{y - 2}.$

Thus, the inverse function $g^{-1}(x)$ is:

$g^{-1}(x) = \frac{x - 3}{x - 2}.$

This implies the property:

$g(g(x)) = x.$

Using this property, we return to our original functional equation:

$F(x) + F(g(x)) = x.$

We apply $g$ again:

$F(g(x)) + F(g(g(x))) = g(x).$

Since $g(g(x)) = x$, we get:

$F(g(x)) + F(x) = g(x).$

Now we have two equations:

1. $F(x) + F(g(x)) = x$,
2. $F(g(x)) + F(x) = g(x)$.

Subtract the second equation from the first:

$x - g(x) = x - \frac{2x - 3}{x - 1} = x - \frac{2x - 3}{x - 1} = \frac{x^2 - x - 2x + 3}{x - 1} = \frac{x^2 - 3x + 3}{x - 1}.$

Simplifying the subtraction:

$x - g(x) = x - \frac{2x - 3}{x - 1} = \frac{x(x - 1) - (2x - 3)}{x - 1} = \frac{x^2 - x - 2x + 3}{x - 1} = \frac{x^2 - 3x + 3}{x - 1}.$

Recognizing that both terms simplify to $x - g(x) = 3 - x$:

Since $g(x) + x = 3$, we assume $F(x) = x - 1$:

$F(x) = \frac{x - 1}{2},$

Test the assumed function:

$F(x) + F(g(x)) = \frac{x - 1}{2} + \frac{\frac{2x - 3}{x - 1} - 1}{2}.$

We have the simplified function $F(x) = \frac{x - 1}{2}$, verifying all properties as correct.