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solve the inequality |x+6-|3x+6||+|x+2|-x-4 less then or equal to 0

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LammettHash

It looks like the inequality is

[tex]\bigg|x + 6 - |3x + 6|\bigg| + |x + 2| - x - 4 \le 0[/tex]

or equivalently,

[tex]\bigg|x + 6 - 3 |x + 2|\bigg| + |x + 2| - x - 4 \le 0[/tex]

Recall the definition of absolute value:

[tex]|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}[/tex]

By this definition, we have

[tex]|x + 2| = \begin{cases} x + 2 & \text{if } x + 2 \ge 0 \\ -(x + 2) & \text{if } x + 2 < 0 \end{cases} = \begin{cases} x + 2 & \text{if } x \ge -2 \\ -x - 2 & \text{if } x < -2 \end{cases}[/tex]

Suppose [tex]x < -2[/tex]. Then

[tex]\bigg| x + 6 - |3x + 6| \bigg| + |x + 2| - x - 4 \\\\ = |x + 6 - 3 (-x - 2)| + (-x - 2) - x - 4 \\\\ = |4x + 12| - 2x - 6[/tex]

By definition of absolute value,

[tex]|4x + 12| = 4 |x + 3| = \begin{cases} 4x + 12 & \text{if } x \ge -3 \\ -4x - 12 & \text{if } x < -3 \end{cases}[/tex]

so we further suppose that [tex]x < -3[/tex]. Then

[tex]\bigg| x + 6 - |3x + 6| \bigg| + |x + 2| - x - 4 \\\\ = |4x + 12| - 2x - 6 \\\\ = (-4x - 12) - 2x - 6 \\\\ = -6x - 18 \le 0 \\\\ \implies x \ge -3[/tex]

but this is a contradiction, so this case gives no solution.

If instead we have [tex]-3 \le x < -2[/tex], then

[tex]\bigg| x + 6 - |3x + 6| \bigg| + |x + 2| - x - 4 \\\\ = |4x + 12| - 2x - 6 \\\\ = (4x + 12) - 2x - 6 \\\\ = 2x + 6 \le 0 \\\\ \implies x \le -3[/tex]

and [tex]-3 \le x[/tex] and [tex]x \le -3[/tex] means that [tex]\boxed{x = -3}[/tex].

Now suppose [tex]x \ge -2[/tex]. Then

[tex]\bigg| x + 6 - |3x + 6| \bigg| + |x + 2| - x - 4 \\\\ = |x + 6 - 3 (x + 2)| + (x + 2) - x - 4 \\\\ = |-2x| - 2 \\\\ = 2 |x| - 2[/tex]

If we further suppose that [tex]-2 \le x < 0[/tex], then

[tex]\bigg| x + 6 - |3x + 6| \bigg| + |x + 2| - x - 4 \\\\ = 2 |x| - 2 \\\\ = -2x - 2 \le 0 \\\\ \implies x \ge -1[/tex]

Otherwise, if [tex]x \ge 0[/tex], then

[tex]\bigg| x + 6 - |3x + 6| \bigg| + |x + 2| - x - 4 \\\\ = 2 |x| - 2 \\\\ = 2x - 2 \le 0 \\\\ \implies x \le 1[/tex]

Taken together, this case gives the solution [tex]\boxed{-1 \le x \le 1}[/tex]

So, the overall solution to the inequality is the set

[tex]\boxed{\left\{ x \in \mathbb R : x = -3 \text{ or } -1 \le x \le -1 \right\}}[/tex]

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