(4x−1)\9≥1 inequation

Given the inequality:
$$\frac{4 x - 1}{9} \geq 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{4 x - 1}{9} = 1$$
Solve:
Given the linear equation:

(4*x-1)/9 = 1

Expand brackets in the left part

4*x/9-1/9 = 1

Move free summands (without x)
from left part to right part, we given:
$$\frac{4 x}{9} = \frac{10}{9}$$
Divide both parts of the equation by 4/9

x = 10/9 / (4/9)

$$x_{1} = \frac{5}{2}$$
$$x_{1} = \frac{5}{2}$$
This roots
$$x_{1} = \frac{5}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{5}{2}$$
=
$$\frac{12}{5}$$
substitute to the expression
$$\frac{4 x - 1}{9} \geq 1$$
$$\frac{-1 + \frac{4 \cdot 12}{5}}{9} \geq 1$$

43     
-- >= 1
45     

but

43    
-- < 1
45    

Then
$$x \leq \frac{5}{2}$$
no execute
the solution of our inequality is:
$$x \geq \frac{5}{2}$$

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