Why is $|x-2|+4 \ge 10$ equal to $-(\infty, -4]\cup[8,\infty)$

algebra-precalculus

I'm trying to solve a inequality with absolute value variable:

$|x-2|+4 \ge 10$

The solution is provided as $-(\infty, -4]\cup[8,\infty)$

However, I arrived at

$-(\infty, -12]\cup[8,\infty)$

The book says -4 and I cannot see how it's not -12.

$|x-2|+4 \ge -10$

$x+2 \ge -10$

$x \ge -12$

For the other part, $x-2+4 \ge 10$, I'm aligned with the book and arrive at $8>=\infty$

How can I arrive at the text book solution of -4?

Best Answer

Tl;dr is you are having an issue with absolute value. Let's do it step by step.

$$|x-2|+4 \geq 10 $$ will hold if and only if $|x-2|\geq 6$ holds. Remind that $|a|= a$ wherever $a$ is positive, and $|a|=-a$ wherever $a$ is negative (for $a$=0, it doesn't matter, it's zero either way). In your case, $a= x-2$.

So we actually have two inequations:

  • $x-2 \geq 6$, for $x\geq 2$, and
  • $-(x-2) \geq 6$, for $x<2$.

The first one yields that $x\geq 8$, so $x\in [8,\infty)$ shall verify the inequality. The second one can be rewritten as $x-2\leq-6$ (why?), so it yields $x\leq-4$. Thus, $x\in (-\infty,-4]$ shall verify the inequality.

We put together the two intervals to say that $x\in (-\infty,-4]\cup [8, \infty)$ is the solution of this inequality.

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