|3*x+5|=>20 inequation

Given the inequality:
$$\left|{3 x + 5}\right| \geq 20$$
To solve this inequality, we must first solve the corresponding equation:
$$\left|{3 x + 5}\right| = 20$$
Solve:
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.

1.
$$3 x + 5 \geq 0$$
or
$$- \frac{5}{3} \leq x \wedge x < \infty$$
we get the equation
$$\left(3 x + 5\right) - 20 = 0$$
after simplifying we get
$$3 x - 15 = 0$$
the solution in this interval:
$$x_{1} = 5$$

2.
$$3 x + 5 < 0$$
or
$$-\infty < x \wedge x < - \frac{5}{3}$$
we get the equation
$$\left(- 3 x - 5\right) - 20 = 0$$
after simplifying we get
$$- 3 x - 25 = 0$$
the solution in this interval:
$$x_{2} = - \frac{25}{3}$$


$$x_{1} = 5$$
$$x_{2} = - \frac{25}{3}$$
$$x_{1} = 5$$
$$x_{2} = - \frac{25}{3}$$
This roots
$$x_{2} = - \frac{25}{3}$$
$$x_{1} = 5$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{25}{3} + - \frac{1}{10}$$
=
$$- \frac{253}{30}$$
substitute to the expression
$$\left|{3 x + 5}\right| \geq 20$$
$$\left|{\frac{\left(-253\right) 3}{30} + 5}\right| \geq 20$$

203      
--- >= 20
 10      

one of the solutions of our inequality is:
$$x \leq - \frac{25}{3}$$

 _____           _____          
      \         /
-------•-------•-------
       x2      x1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq - \frac{25}{3}$$
$$x \geq 5$$