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Consider:

$$ \left|\left|\left|x-1\right|-2\right|-4\right|=4 $$

What is the number of solutions for this equation? This one was particularly easy to me. If first observed that if this inequality were to hold, then $ | x - 1 | $ should be 9. Because $ | x -1 |$ will always be positive, and once we remove $ | x- 1 |$, the only positive number that can lead to a possible solution is 9. So solving for $ | x - 1 | $ gives me $ 10 $ or $ -8 $, and I'm done. Number of solutions is 2.

But I was curious on what I'd do if I had different numbers. Say:

$$ \left|\left|\left|x-1\right|-2\right|-5\right|=4 $$

Now, Desmos says there are five possible solutions. How should one go about finding the number of solutions (or finding the solutions themselves) for something like this by hand? A general method would be appreciated.

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  • $\begingroup$ Each absolute value will double the maximum number of solutions. (Maximum number of solutions because zero may be an intermediate solution or two branches may give the same solution.) $\endgroup$ – Ertxiem Mar 24 at 12:35
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You can solve this by "going from the outside to the inside" of the equation and analyzing the different cases.

For your first example, it has to be true that

$$||x-1|-2|\in\{8,0\},$$ for the $0$ case, it follows that $$|x-1|=2\iff x\in\{3,-1\}.$$

For the $8$ case, it follows that $$|x-1|-2=\pm 8\implies |x-1|\in\{10,-6\},$$

which is only possible for $$|x-1|=10\implies x-1=\pm 10,$$

which leads to $$x\in\{11,-9\}.$$

So, there are $4$ solutions in total, namely

$$x\in\{-9,-1,3,11\}.$$

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Open one-by-one (remember, absolute value is always nonnegative): $$\left|\left|\left|x-1\right|-2\right|-4\right|=4 \iff \\ \left|\left|x-1\right|-2\right|-4=\pm 4 \iff \left|\left|x-1\right|-2\right|=0;8 \iff \\ |x-1|-2=\pm(0;8) \iff |x-1|=2;10 \iff \\ x-1=\pm(2;10) \iff x=-9;-1;3;11.$$

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