|4*x+3|>5 inequation

Given the inequality:
$$\left|{4 x + 3}\right| > 5$$
To solve this inequality, we must first solve the corresponding equation:
$$\left|{4 x + 3}\right| = 5$$
Solve:
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.

1.
$$4 x + 3 \geq 0$$
or
$$- \frac{3}{4} \leq x \wedge x < \infty$$
we get the equation
$$\left(4 x + 3\right) - 5 = 0$$
after simplifying we get
$$4 x - 2 = 0$$
the solution in this interval:
$$x_{1} = \frac{1}{2}$$

2.
$$4 x + 3 < 0$$
or
$$-\infty < x \wedge x < - \frac{3}{4}$$
we get the equation
$$\left(- 4 x - 3\right) - 5 = 0$$
after simplifying we get
$$- 4 x - 8 = 0$$
the solution in this interval:
$$x_{2} = -2$$


$$x_{1} = \frac{1}{2}$$
$$x_{2} = -2$$
$$x_{1} = \frac{1}{2}$$
$$x_{2} = -2$$
This roots
$$x_{2} = -2$$
$$x_{1} = \frac{1}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-2 + - \frac{1}{10}$$
=
$$- \frac{21}{10}$$
substitute to the expression
$$\left|{4 x + 3}\right| > 5$$
$$\left|{\frac{\left(-21\right) 4}{10} + 3}\right| > 5$$

27/5 > 5

one of the solutions of our inequality is:
$$x < -2$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < -2$$
$$x > \frac{1}{2}$$