(4+x)/(4-x)<3 inequation

Given the inequality:
$$\frac{x + 4}{4 - x} < 3$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{x + 4}{4 - x} = 3$$
Solve:
Given the equation:
$$\frac{x + 4}{4 - x} = 3$$
Multiply the equation sides by the denominator 4 - x
we get:
$$- \frac{\left(4 - x\right) \left(x + 4\right)}{x - 4} = 12 - 3 x$$
Expand brackets in the left part

-4-x4+x-4+x = 12 - 3*x

Looking for similar summands in the left part:

-(4 + x)*(4 - x)/(-4 + x) = 12 - 3*x

Move free summands (without x)
from left part to right part, we given:
$$- \frac{\left(4 - x\right) \left(x + 4\right)}{x - 4} + 4 = 16 - 3 x$$
Move the summands with the unknown x
from the right part to the left part:
$$3 x - \frac{\left(4 - x\right) \left(x + 4\right)}{x - 4} + 4 = 16$$
Divide both parts of the equation by (4 + 3*x - (4 + x)*(4 - x)/(-4 + x))/x

x = 16 / ((4 + 3*x - (4 + x)*(4 - x)/(-4 + x))/x)

$$x_{1} = 2$$
$$x_{1} = 2$$
This roots
$$x_{1} = 2$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 2$$
=
$$\frac{19}{10}$$
substitute to the expression
$$\frac{x + 4}{4 - x} < 3$$
$$\frac{\frac{19}{10} + 4}{4 - \frac{19}{10}} < 3$$

59    
-- < 3
21    

the solution of our inequality is:
$$x < 2$$

 _____          
      \    
-------ο-------
       x1