|x+2|-|2x-6|≤4 inequation

Given the inequality:
$$\left|{x + 2}\right| - \left|{2 x - 6}\right| \leq 4$$
To solve this inequality, we must first solve the corresponding equation:
$$\left|{x + 2}\right| - \left|{2 x - 6}\right| = 4$$
Solve:
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.

1.
$$x + 2 \geq 0$$
$$2 x - 6 \geq 0$$
or
$$3 \leq x \wedge x < \infty$$
we get the equation
$$\left(x + 2\right) - \left(2 x - 6\right) - 4 = 0$$
after simplifying we get
$$4 - x = 0$$
the solution in this interval:
$$x_{1} = 4$$

2.
$$x + 2 \geq 0$$
$$2 x - 6 < 0$$
or
$$-2 \leq x \wedge x < 3$$
we get the equation
$$- (6 - 2 x) + \left(x + 2\right) - 4 = 0$$
after simplifying we get
$$3 x - 8 = 0$$
the solution in this interval:
$$x_{2} = \frac{8}{3}$$

3.
$$x + 2 < 0$$
$$2 x - 6 \geq 0$$
The inequality system has no solutions, see the next condition

4.
$$x + 2 < 0$$
$$2 x - 6 < 0$$
or
$$-\infty < x \wedge x < -2$$
we get the equation
$$- (6 - 2 x) + \left(- x - 2\right) - 4 = 0$$
after simplifying we get
$$x - 12 = 0$$
the solution in this interval:
$$x_{3} = 12$$
but x3 not in the inequality interval


$$x_{1} = 4$$
$$x_{2} = \frac{8}{3}$$
$$x_{1} = 4$$
$$x_{2} = \frac{8}{3}$$
This roots
$$x_{2} = \frac{8}{3}$$
$$x_{1} = 4$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{8}{3}$$
=
$$\frac{77}{30}$$
substitute to the expression
$$\left|{x + 2}\right| - \left|{2 x - 6}\right| \leq 4$$
$$- \left|{-6 + \frac{2 \cdot 77}{30}}\right| + \left|{2 + \frac{77}{30}}\right| \leq 4$$

37     
-- <= 4
10     

one of the solutions of our inequality is:
$$x \leq \frac{8}{3}$$

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

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq \frac{8}{3}$$
$$x \geq 4$$