How to solve a convoluted absolute value inequality? $$ \lvert \lvert x-2\rvert -3\rvert \lt 5 $$
How can I attack this the best way? I see that both sides are positive. 
Squaring yields:
$$ \lvert x-2\rvert ^2 -6 \lvert x-2\rvert +9\lt 25 $$
$$ x^2-4x -6\lvert x-2\rvert \lt 12 $$
This is where I'm stuck. How to proceed?
 A: $||x-2|-3|<5$
$-5<|x-2|-3<5$
$\implies -2<|x-2|<8$
now modulus of any value is always greater than or equal to zero so
$0\le |x-2|<8$
I think you can carryout from here
A: Here are two approaches:
The first builds on your initial attempt:  Since both sides of the inequality
$$||x - 2| - 3| < 5$$
are nonnegative, the inequality is preserved if we square both sides, which yields 
$$|x - 2|^2 - 6|x - 2| + 9 < 25$$
Subtracting $25$ from each side of the inequality yields
$$|x - 2|^2 - 6|x - 2| - 16 < 0$$
Factoring the quadratic in $|x - 2|$ yields
$$(|x - 2| - 8)(|x - 2| + 2) < 0$$
Since $|x - 2| \geq 0$ for each real number $x$, the factor $|x - 2| + 2$ is always positive.  Thus, the inequality is satisfied when 
$$|x - 2| - 8 < 0$$
which is equivalent to the statement 
$$|x - 2| < 8$$
The inequality $|x - 2| < 8$ is satisfied when
$$-8 < x - 2 < 8$$
Hence, 
$$-6 < x < 10$$
The second approach is simpler:
If $||x - 2| - 3| < 5$, then 
$$-5 < |x - 2| - 3 < 5$$
Adding $3$ to each side of the inequality yields
$$-2 < |x - 2| < 8$$
Since $|x - 2| \geq 0$ for each real number $x$, this inequality reduces to 
$$|x - 2| < 8$$
Proceed as above.
A: Hint:
By the definition of $|.|$, we have $-5<|x-2|-3<5$ $\Rightarrow$ $-2<|x-2|<8$. By using a property of absolute value (which one ?) you can say $0<|x-2|<8$.
Now again apply the definition of $|.|$ to see the range for $x$.
A: These inequalities can be done by first finding the $x$ making both sides equal. So first let $|x-2|=a,$ then $|a-3|=5$ has solution $a=8, a=-2$ but the latter is discarded since $a=|x-2|\ge 0.$ Thus we have it simplified to $|x-2|=8$ with solutions $x=-6,\ x=10.$
Once it is known the sides are equal at only two specific points, by continuity one can finish the problem by using any three "test points" in the intervals of the real line remaining after removing the two specific points.
A: A systematic approach is to use the rule
$$x\le0\implies |x|=-x,\\
x\ge0\implies |x|=x$$ to get rid of the absolute value.
So we have
$$x-2\le0\to|2-x-3|=|-x-1|<5,\\
x-2\ge0\to|x-2-3|=|x-5|<5$$
which we further subdivide
$$x-2\le0,-x-1\le0\to x+1<5,\\
x-2\le0,-x-1\ge0\to-x-1<5,\\
x-2\ge0,x\le5\to5-x<5\\
x-2\ge0,x\ge5\to x-5<5$$
Remains to cleanup a little.
