(x-1)/((x+2)(x-3))≤0 inequation

Given the inequality:
$$\frac{x - 1}{\left(x - 3\right) \left(x + 2\right)} \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{x - 1}{\left(x - 3\right) \left(x + 2\right)} = 0$$
Solve:
Given the equation:
$$\frac{x - 1}{\left(x - 3\right) \left(x + 2\right)} = 0$$
the denominator
$$x - 3$$
then

x is not equal to 3

the denominator
$$x + 2$$
then

x is not equal to -2

Because the right side of the equation is zero, then the solution of the equation is exists if at least one of the multipliers in the left side of the equation equal to zero.
We get the equations
$$x - 1 = 0$$
solve the resulting equation:
1.
$$x - 1 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = 1$$
We get the answer: x1 = 1
but

x is not equal to 3

x is not equal to -2

$$x_{1} = 1$$
$$x_{1} = 1$$
This roots
$$x_{1} = 1$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 1$$
=
$$\frac{9}{10}$$
substitute to the expression
$$\frac{x - 1}{\left(x - 3\right) \left(x + 2\right)} \leq 0$$
$$\frac{-1 + \frac{9}{10}}{\left(-3 + \frac{9}{10}\right) \left(\frac{9}{10} + 2\right)} \leq 0$$

 10     
--- <= 0
609     

but

 10     
--- >= 0
609     

Then
$$x \leq 1$$
no execute
the solution of our inequality is:
$$x \geq 1$$

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       x1