complex absolute value equations

Question

How on earth does one solve this problem?

I know that to solve abs value equations we have to consider both the negative and postive possibilities, but the constant (6) makes this a little more challenging

$|x-8|=|2-x|-6$

Answer 1

Split it into three parts:

• $x<2$
• $2\leq x\leq 8$
• $8<x$

Note that the three parts follow straight forward from the equation, $x=2$ and $x=8$ are the points where $|2-x|$ and $|x-8|$ flip sign. Now you can inspect each part seperately, such that you can drop the absolute values:

• If $x<2$ then $|x-8|=8-x$ and $|2-x|=2-x$, so the equation becomes: $$8-x=2-x-6 \iff 8-x=-4-x,$$ which has no solution.
• If $2\leq x\leq 8$ then $|x-8|=8-x$ and $|2-x|=x-2$, so the equation becomes: $$8-x=x-2-6 \iff 8-x=x-8,$$ which has solution $x=8$.
• If $8<x$ then $|x-8|=x-8$ and $|2-x|=x-2$, so the equation becomes: $$x-8=x-2-6 \iff x-8=x-8,$$ which is true for every $x$.

Concluding, $|x-8|=|2-x|-6$ for $x\geq8$.

Edit to answer Johns comment:

By definition $|y|=\begin{cases}y\ \mbox{ if }\ y\geq0\\ -y\ \mbox{ if }\ y<0\end{cases}$. So when we look at $|x-8|$ for $x<2$, we can use this definition as follows: $$x<2 \Rightarrow x-8<-6 \Rightarrow |x-8|=-(x-8)=8-x.$$ The same reasoning is used in the rest of the answer.