Solving an equation with absolute values inside absolute value

Question

How do I solve this? $$\big||2x-6|-|x-9|\big|<3$$

I know you can just square $[()^2]$ the whole inequation, but in this case it becomes a very ugly expression. I'm sure there's an easier way!

Answer 1

The first step would be to write it in an equivalent way: $$\big||2x-6|-|x-9|\big|<3\iff -3+|x-9|<|2x-6|<3+|x-9|.$$

Now, the absolute values have important points in $x=3$ and $x=9$. So one can write, for example, for $x\ge 9$: $$ -3+ x-9 < 2x-6 <3+ x-9 $$ or $$ x-12 < 2x-6 < x-6, $$ which gives us the system $$-6<x \quad\text{and}\quad x<0\quad\text{and}\quad x\ge 9.$$ Clearly, this system has no solutions.

Can you take it from here and write all remaining cases?

Answer 2

Assuming $x$ to be a complex number

Using this, $$||w| - |z|| ≤ |z - w| $$

$$\implies \big||2x-6|-|x-9|\big|\le |(2x-6)-(x-9)|=|x+3|$$

So we need $|x+3|<3$

If $x$ is purely real, $-3<x+3<3\iff -6<x<0$

If $x$ is purely imaginary $=ib$(say), where $b$ is real $|x+3|=\sqrt{9+b^2}\ge3$

else $x=a+ib $ where $a,b$ are real and $ab\ne0$

$\implies |x+3|=\sqrt{(a+3)^2+b^2}<3\iff (a+3)^2+b^2<9$

So, $(a,b)$ lies inside a circle with center $(-3,0)$ and radius $3$ unit