|6+x|+|x-6|<12 inequation

Given the inequality:
$$\left|{x - 6}\right| + \left|{x + 6}\right| < 12$$
To solve this inequality, we must first solve the corresponding equation:
$$\left|{x - 6}\right| + \left|{x + 6}\right| = 12$$
Solve:
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.

1.
$$x - 6 \geq 0$$
$$x + 6 \geq 0$$
or
$$6 \leq x \wedge x < \infty$$
we get the equation
$$\left(x - 6\right) + \left(x + 6\right) - 12 = 0$$
after simplifying we get
$$2 x - 12 = 0$$
the solution in this interval:
$$x_{1} = 6$$

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

3.
$$x - 6 < 0$$
$$x + 6 \geq 0$$
or
$$-6 \leq x \wedge x < 6$$
we get the equation
$$\left(6 - x\right) + \left(x + 6\right) - 12 = 0$$
after simplifying we get
the identity
the solution in this interval:

4.
$$x - 6 < 0$$
$$x + 6 < 0$$
or
$$-\infty < x \wedge x < -6$$
we get the equation
$$\left(6 - x\right) + \left(- x - 6\right) - 12 = 0$$
after simplifying we get
$$- 2 x - 12 = 0$$
the solution in this interval:
$$x_{2} = -6$$
but x2 not in the inequality interval


$$x_{1} = 6$$
$$x_{1} = 6$$
This roots
$$x_{1} = 6$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 6$$
=
$$\frac{59}{10}$$
substitute to the expression
$$\left|{x - 6}\right| + \left|{x + 6}\right| < 12$$
$$\left|{-6 + \frac{59}{10}}\right| + \left|{\frac{59}{10} + 6}\right| < 12$$

12 < 12

Then
$$x < 6$$
no execute
the solution of our inequality is:
$$x > 6$$

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