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
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.
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.
