Solution of $|1-2|x||\le3$ What is the set of all solutions of $|1-2|x||\le3$? Is there a general approach for problems like this with nested absolute functions?
 A: Hint.
In this case, begin by noting that $|A| \leq 3$ is equivalent to $-3 \leq A \leq 3$. Next manipulate the inequality you obtain to one about $|x|$. Then use the same idea again. 
A: $|1-2|x||\le  3$  
Squaring both sides gives
$1 + 4x^2 - 4|x| \le 9$
$|x|^2 - |x| -2 \le 0$  
$|x| \le 2  ~\wedge  ~|x| \ge  -1$
$-2 \le x \le 2  $  
A: The solution set is $[-2,2]$. To see this, each of the following lines is equivalent:
$$\left|1-2|x|\right|\leq3$$
$$-3\leq1-2|x|\leq3$$
$$-2\leq2|x|\leq4$$
$$-1\leq|x|\leq2$$
$$|x|\leq2$$
$$-2\leq x\leq2$$
The general approach is to replace the absolute value bars with a compound inequality, working from the outside in. You may have "and" or "or" statements introduced in the process of doing this.
A: $$-3 \le 1-2|x| \le 3 \implies -4 \le -2|x| \le 2 \implies 2 \ge |x| \ge -1$$
A: For realize this you can plot this:first you plot |x| and then -2*|x| then -2*|x|+1
and then get abs from this your Plot should be have W shape and then you solve 1-2|x|=3
and obtain max x and we'll have -2
A: $|1-2|x|| \leq 3 \implies -3 \leq 1-2|x| \leq 3 \implies -4 \leq -2|x| \leq 2 \implies 4 \geq 2|x| \geq -1 \implies 2 \geq |x| \geq -\frac 12$
In the last inequality, the right-hand half ($|x| \geq -\frac 12$) is trivially satisfied, so the only solution that matters is $|x| \leq 2$, giving $-2 \leq x \leq 2$.
