|3x+5|>7 inequation

Given the inequality:
$$\left|{3 x + 5}\right| > 7$$
To solve this inequality, we must first solve the corresponding equation:
$$\left|{3 x + 5}\right| = 7$$
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
$$\left(3 x + 5\right) - 7 = 0$$
after simplifying we get
$$3 x - 2 = 0$$
the solution in this interval:
$$x_{1} = \frac{2}{3}$$

2.
$$3 x + 5 < 0$$
or
$$-\infty < x \wedge x < - \frac{5}{3}$$
we get the equation
$$\left(- 3 x - 5\right) - 7 = 0$$
after simplifying we get
$$- 3 x - 12 = 0$$
the solution in this interval:
$$x_{2} = -4$$


$$x_{1} = \frac{2}{3}$$
$$x_{2} = -4$$
$$x_{1} = \frac{2}{3}$$
$$x_{2} = -4$$
This roots
$$x_{2} = -4$$
$$x_{1} = \frac{2}{3}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-4 + - \frac{1}{10}$$
=
$$- \frac{41}{10}$$
substitute to the expression
$$\left|{3 x + 5}\right| > 7$$
$$\left|{\frac{\left(-41\right) 3}{10} + 5}\right| > 7$$

73    
-- > 7
10    

one of the solutions of our inequality is:
$$x < -4$$

 _____           _____          
      \         /
-------ο-------ο-------
       x2      x1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < -4$$
$$x > \frac{2}{3}$$