(3x-2):(3-x)=<2 inequation

Given the inequality:
$$\frac{3 x - 2}{3 - x} \leq 2$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{3 x - 2}{3 - x} = 2$$
Solve:
Given the equation:
$$\frac{3 x - 2}{3 - x} = 2$$
Multiply the equation sides by the denominator 3 - x
we get:
$$\frac{\left(2 - 3 x\right) \left(3 - x\right)}{x - 3} = 6 - 2 x$$
Expand brackets in the left part

2+3*x3+x-3+x = 6 - 2*x

Looking for similar summands in the left part:

(2 - 3*x)*(3 - x)/(-3 + x) = 6 - 2*x

Move free summands (without x)
from left part to right part, we given:
$$\frac{\left(2 - 3 x\right) \left(3 - x\right)}{x - 3} + 3 = 9 - 2 x$$
Move the summands with the unknown x
from the right part to the left part:
$$2 x + \frac{\left(2 - 3 x\right) \left(3 - x\right)}{x - 3} + 3 = 9$$
Divide both parts of the equation by (3 + 2*x + (2 - 3*x)*(3 - x)/(-3 + x))/x

x = 9 / ((3 + 2*x + (2 - 3*x)*(3 - x)/(-3 + x))/x)

$$x_{1} = \frac{8}{5}$$
$$x_{1} = \frac{8}{5}$$
This roots
$$x_{1} = \frac{8}{5}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{8}{5}$$
=
$$\frac{3}{2}$$
substitute to the expression
$$\frac{3 x - 2}{3 - x} \leq 2$$
$$\frac{-2 + \frac{3 \cdot 3}{2}}{3 - \frac{3}{2}} \leq 2$$

5/3 <= 2

the solution of our inequality is:
$$x \leq \frac{8}{5}$$

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