(2-x)/(2x-8)>0 inequation

Given the inequality:
$$\frac{2 - x}{2 x - 8} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{2 - x}{2 x - 8} = 0$$
Solve:
Given the equation:
$$\frac{2 - x}{2 x - 8} = 0$$
Multiply the equation sides by the denominator -8 + 2*x
we get:
$$\frac{\left(2 - x\right) \left(2 x - 8\right)}{2 \left(x - 4\right)} = 0$$
Expand brackets in the left part

-8+2*x2+x2*-4+x) = 0

Looking for similar summands in the left part:

(-8 + 2*x)*(2 - x)/(2*(-4 + x)) = 0

Move free summands (without x)
from left part to right part, we given:
$$\frac{\left(2 - x\right) \left(2 x - 8\right)}{2 \left(x - 4\right)} + 8 = 8$$
Divide both parts of the equation by (8 + (-8 + 2*x)*(2 - x)/(2*(-4 + x)))/x

x = 8 / ((8 + (-8 + 2*x)*(2 - x)/(2*(-4 + x)))/x)

$$x_{1} = 2$$
$$x_{1} = 2$$
This roots
$$x_{1} = 2$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 2$$
=
$$\frac{19}{10}$$
substitute to the expression
$$\frac{2 - x}{2 x - 8} > 0$$
$$\frac{2 - \frac{19}{10}}{-8 + \frac{2 \cdot 19}{10}} > 0$$

-1/42 > 0

Then
$$x < 2$$
no execute
the solution of our inequality is:
$$x > 2$$

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