(3-x)/(x+2)>0 inequation

Given the inequality:
$$\frac{3 - x}{x + 2} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{3 - x}{x + 2} = 0$$
Solve:
Given the equation:
$$\frac{3 - x}{x + 2} = 0$$
Multiply the equation sides by the denominator 2 + x
we get:
$$3 - x = 0$$
Move free summands (without x)
from left part to right part, we given:
$$- x = -3$$
Divide both parts of the equation by -1

x = -3 / (-1)

$$x_{1} = 3$$
$$x_{1} = 3$$
This roots
$$x_{1} = 3$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 3$$
=
$$\frac{29}{10}$$
substitute to the expression
$$\frac{3 - x}{x + 2} > 0$$
$$\frac{3 - \frac{29}{10}}{2 + \frac{29}{10}} > 0$$

1/49 > 0

the solution of our inequality is:
$$x < 3$$

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