$||x-2|-3| >1$, then $x$ belongs to which interval 
$||x-2|-3| >1$, then $x$ belongs to: (a) $(-\infty, -2) \cup (0,4) \cup(6,\infty)$ (b) $(-1,1)$ (c) $(-\infty, 1)\cup (1,
> \infty)$  (d) $(-2,2)$

Answer:(a)
My solution:
let $|x-2| = p,$
$|p-3|>1$ and finally i got four inequalities after more calculations :
$x>6, x<-2,x>0, x<4$
I think I am doing this question wrong but i don't know how to do this question either way. How do I do this?
 A: Just do it with two simultaneous inequalities:
$$
||x-2| - 3| > 1 \Leftrightarrow |x-2|-3 > 1 \text{ or } |x-2|-3 < -1
$$
$$
\Leftrightarrow |x-2| > 4 \text{ or } |x-2| < 2
$$
$$
\Leftrightarrow x-2 > 4 \text{ or } x-2 < -4 \text{ or } -2 < x-2 < 2
$$
$$
\Leftrightarrow x>6 \text{ or } x < -2 \text{ or } 0 < x < 4
$$
So your answer is
$$
(6,\infty)\cup (-\infty,-2)\cup (0,4)
$$
A: $||x−2|−3|>1$ then either $|x-2| - 3 > 1$ or $|x-2| - 3 < -1$; respectively
$$|x-2| > 4 \ \text{ or } |x-2| < 2$$
The first of these is equivalent to $x - 2 > 4$ or $x - 2 < -4$; i.e., $$x > 6 \text{ or } x < -2$$
The second is equivalent to $-2 < x - 2 < 2$; i.e., $$0 < x < 4$$
Now put that all together.
A: Ok, so $|x - 2| = p$, and so $|p - 3| > 1$.
Then that means $p - 3 > 1$ or $p - 3 < -1$.
Then that means $p > 4$ or $p < 2$.
But $p = |x - 2|$, so $|x - 2| > 4$ or $|x - 2| < 2$.
That means either $x - 2 > 4$ or $x - 2 < -4$ or $-2 < x - 2 < 2$.
That means either $x > 6$ or $x < -2$ or $0 < x < 4$.
Notice that the last set of inequalities (the $0 < x < 4$ one) is an and.  Your answer isn't represented exactly like this.  You wrote "$x > 0$ or $x < 4$" but that's different than writing $0 < x < 4$.
