|x+5|-|2x-4|>2 inequation

Given the inequality:
$$\left|{x + 5}\right| - \left|{2 x - 4}\right| > 2$$
To solve this inequality, we must first solve the corresponding equation:
$$\left|{x + 5}\right| - \left|{2 x - 4}\right| = 2$$
Solve:
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.

1.
$$x + 5 \geq 0$$
$$2 x - 4 \geq 0$$
or
$$2 \leq x \wedge x < \infty$$
we get the equation
$$\left(x + 5\right) - \left(2 x - 4\right) - 2 = 0$$
after simplifying we get
$$7 - x = 0$$
the solution in this interval:
$$x_{1} = 7$$

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

3.
$$x + 5 < 0$$
$$2 x - 4 \geq 0$$
The inequality system has no solutions, see the next condition

4.
$$x + 5 < 0$$
$$2 x - 4 < 0$$
or
$$-\infty < x \wedge x < -5$$
we get the equation
$$- (4 - 2 x) + \left(- x - 5\right) - 2 = 0$$
after simplifying we get
$$x - 11 = 0$$
the solution in this interval:
$$x_{3} = 11$$
but x3 not in the inequality interval


$$x_{1} = 7$$
$$x_{2} = \frac{1}{3}$$
$$x_{1} = 7$$
$$x_{2} = \frac{1}{3}$$
This roots
$$x_{2} = \frac{1}{3}$$
$$x_{1} = 7$$
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
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{1}{3}$$
=
$$\frac{7}{30}$$
substitute to the expression
$$\left|{x + 5}\right| - \left|{2 x - 4}\right| > 2$$
$$- \left|{-4 + \frac{2 \cdot 7}{30}}\right| + \left|{\frac{7}{30} + 5}\right| > 2$$

17    
-- > 2
10    

Then
$$x < \frac{1}{3}$$
no execute
one of the solutions of our inequality is:
$$x > \frac{1}{3} \wedge x < 7$$

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       x2      x1