Solving an equation like $|x-a| = 3$, is simple, where one just splits this into two equations for values of $x$ less than $a$ and values equal to or greater than $a:$ $$|x-a| = \left\{ \begin{array}{rcr} x-a & , & \text{if} \ \ x \geq a \\ -(x-a) & , & \text{if} \ \ x < a \\ \end{array} \right.$$

But how do you split it up if you have, for example, the equation $$||x-a|-a|=3\ ?$$


2 Answers 2


I would recommend just looking at different cases:

  • $x>a  \Rightarrow \left|\left|x-a \right|-a \right| = \left| x-2a \right|$
  • $x<a \Rightarrow \left|\left| x-a \right|-a \right| = \left|-x+a -a \right|= \left|x \right|$

So as you see, it gets a little bit more complicated and there are four cases in total. You just need to keep your head calm and do the math.

  • $\begingroup$ Oh I see, so for the RHS's I need to do what I did in my post but twice. So if we'd have 3 levels of nested absolute values, I'd get 8 cases? $\endgroup$
    – Parseval
    May 8, 2017 at 6:10
  • $\begingroup$ Well, in general it's difficult to say how many cases you will get. Anyway, I think these kind of questions are quite straightforward - you just need to advance logically, opening up the expressions one after the another. $\endgroup$
    – Matti P.
    May 8, 2017 at 6:14

The same process will find the "possible solutions", namely for each absolute value $|u|$ in the equation, replace it by the two choices $u,-u.$ The more absolute values originally there, the more such equations will come this way. [some power of 2 "possible equations".]

However once each possible equation is solved one must put the solution(s) back into the original absolute value equation and calculate the sides to see if equal.


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