# Solving an equation with absolute values inside absolute value

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!

• You could tackle it by breaking down into cases, since $\big||2x-6|-|x-9|\big|<3$ if and only if $|2x-6|-|x-9|<3$ and $|2x-6|-|x-9|>-3$. Then go to work on each inequality separately. – Old John Sep 15 '13 at 11:01

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?

• what if $x$ is complex? – lab bhattacharjee Sep 15 '13 at 11:04
• @labbhattacharjee judging by the tag (algebra-precalculus), I think it's safe to assume that we talk about real $x$. – TZakrevskiy Sep 15 '13 at 11:06
• Doesn't your first step mean we basically ignore the outer absolute value? I mean that's what I would do if the equation didn't contain it. – pingwin Sep 15 '13 at 12:19
• And for your questions, x is real. Thanks! – pingwin Sep 15 '13 at 12:25
• No, we don't ignore the outer absolute value, we "open the brackets": we rewrite it using $|a-b|<c\iff -c+b<a<b+c$. – TZakrevskiy Sep 15 '13 at 19:09

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