Absolutely Outrageous

solving via graph is too easy and comfortable :-)

How would you present a simpler solution?

You'll have to plot the graphs of those functions using transformations. Let's take the LHS .

We have $|||x-1|+1|-1|$ .

Now its obvious that this function is a composite function.

Now lets look at the simplest function in this composite function.

$\rightarrow |x-1|$

Let it be f(x)

Now Its easy to plot f(x)

<http://www.wolframalpha.com/input/?i=plot+%7Cx-1%7C>

now the function becomes $||f(x)+1|-1|$ as $|x-1|$ was replaced by f(x)

And now we will be doing operations on f(x) instead of x

So now the next simplest function we can see is $|f(x)+1|$

Now you already have the plot of f(x) All you gotta do to make the graph of $|f(x)+1|$ is Push f(x) 1 unit up on the Y axis this will give f(x)+1

here it is

<http://www.wolframalpha.com/input/?i=plot+%7Cx-1%7C+%2B1>

And this leads to our next step, Taking absolute of this function.

i.e, we have to make $|f(x)+1|$ now.

As you can know, the absolute function converts all the -ve values to +ve values.

So the best method to make the graph of $|g(x)|$ is to flip the negative portion of the graph of g(x) into the positive part taking X axis as the hinge.

here g(x) is a random function

you can try it on your own, let g(x)=x , now make $|g(x)| = |x|$

you'll see that the -ve portion of $y=x$ was flipped on the +ve y axis side

So the function we had was $|f(x)+1|$ , you already have f(x)+1 and using the method i told you above, make $|f(x)+1|$ .

Its obvious that the graph of $|f(x)+1|$ will be the same as of $f(x)+1$

as there is no -ve part in $f(x)+1$

here it is

<http://www.wolframalpha.com/input/?i=plot+%7C%7Cx-1%7C+%2B1%7C>

OK now we have the graph of $||x-1|+1|$

let $||x-1|+1| = h(x)$ ,

so the objective now is to make the graph of $|h(x)-1|$

so again, its simple, push down h(x) 1 unit on y axis

<http://www.wolframalpha.com/input/?i=plot+%7C%7Cx-1%7C+%2B1%7C+-1+>

and take the absolute of this, i.e. flip the -ve part of the plot to +ve part.

obviously if there's no -ve part then the graph will not change

you'll get this $|h(x)-1|$ or $|||x-1|+1|-1|$

Similarly plot the graph of the function on the RHS

and where ever they will intersect will give us the soution

if you need help with this or how to make the graph of the function on rhs, let me know, i'll comment the solution to that