|2x-4|>=1 inequation

Given the inequality:
$$\left|{2 x - 4}\right| \geq 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\left|{2 x - 4}\right| = 1$$
Solve:
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.

1.
$$2 x - 4 \geq 0$$
or
$$2 \leq x \wedge x < \infty$$
we get the equation
$$\left(2 x - 4\right) - 1 = 0$$
after simplifying we get
$$2 x - 5 = 0$$
the solution in this interval:
$$x_{1} = \frac{5}{2}$$

2.
$$2 x - 4 < 0$$
or
$$-\infty < x \wedge x < 2$$
we get the equation
$$\left(4 - 2 x\right) - 1 = 0$$
after simplifying we get
$$3 - 2 x = 0$$
the solution in this interval:
$$x_{2} = \frac{3}{2}$$


$$x_{1} = \frac{5}{2}$$
$$x_{2} = \frac{3}{2}$$
$$x_{1} = \frac{5}{2}$$
$$x_{2} = \frac{3}{2}$$
This roots
$$x_{2} = \frac{3}{2}$$
$$x_{1} = \frac{5}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{3}{2}$$
=
$$\frac{7}{5}$$
substitute to the expression
$$\left|{2 x - 4}\right| \geq 1$$
$$\left|{-4 + \frac{2 \cdot 7}{5}}\right| \geq 1$$

6/5 >= 1

one of the solutions of our inequality is:
$$x \leq \frac{3}{2}$$

 _____           _____          
      \         /
-------•-------•-------
       x2      x1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq \frac{3}{2}$$
$$x \geq \frac{5}{2}$$