3x-(x-2)>=2 inequation

Given the inequality:
$$3 x + \left(2 - x\right) \geq 2$$
To solve this inequality, we must first solve the corresponding equation:
$$3 x + \left(2 - x\right) = 2$$
Solve:
Given the linear equation:

3*x-(x-2) = 2

Expand brackets in the left part

3*x-x+2 = 2

Looking for similar summands in the left part:

2 + 2*x = 2

Move free summands (without x)
from left part to right part, we given:
$$2 x = 0$$
Divide both parts of the equation by 2

x = 0 / (2)

$$x_{1} = 0$$
$$x_{1} = 0$$
This roots
$$x_{1} = 0$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10}$$
=
$$- \frac{1}{10}$$
substitute to the expression
$$3 x + \left(2 - x\right) \geq 2$$
$$\frac{\left(-1\right) 3}{10} + \left(2 - - \frac{1}{10}\right) \geq 2$$

9/5 >= 2

but

9/5 < 2

Then
$$x \leq 0$$
no execute
the solution of our inequality is:
$$x \geq 0$$

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