# Equation with double absolute value

I have problem with solving the following equation.

$$2x - |5-|x-2|| = 1$$

How to handle absolute value in absolute value? I have tried multiple times to solve it but I get no solution.

## 2 Answers

No "easy way out". Solve the first absolute value $$|x-2|=\begin{cases}-x+2, & x<2\\\phantom{-}x-2, & x\ge 2 \end{cases}$$ to take cases

• Case 1: $x<2$. Then $$2x-|5-|x-2||=2x-|5-(-x+2)|=2x-|x+3|$$ And now take subcases (but do not forget that you are in $x<2$) depending on the second absolute value $$|x+3|=\begin{cases}-x-3, & \hspace{26.5pt}x<-3\\ \phantom{-}x+3, & -3\le x <2 \\\end{cases}$$ So, $$2x-|x+3|=\begin{cases}2x-(-x-3)=3x+3, & \hspace{26.5pt}x<-3 \\2x-(x+3)=x-3, & -3\le x < 2 \end{cases}$$ And now treat each subcase separately: $$3x+3=1\iff x=-\frac23\not <-3$$ so, no solution here. Second subcase: $$x-3=1\iff x=4 \notin-3\le x<2$$ so, nothing here either.
• Case 2: $x\ge 2$. Then $$2x-|5-|x-2||=2x-|5-(x-2)|=2x-|7-x|$$ And now take subcases (but do not forget that you are in $x\ge 2$) depending on the second absolute value $$|7-x|=\begin{cases} \phantom{-}7-x, & 2\le x<7\\-7+x, & 7\le x \end{cases}$$ So, $$2x-|7-x|=\begin{cases}2x-(7-x)=3x-7, & 2\le x<7 \\2x-(-7+x)=x+7, & 7\le x \end{cases}$$ And now treat each subcase separately: $$3x-7=1\iff x=\frac83\approx2.67\in [2,7)$$ so, bingo! first solution here. Second subcase: $$x+7=1\iff x=-6 \not \ge 7$$ so, nothing here.

The only solution is $x_0=\frac83=2\frac23$.

• Yes, but when I do it I get that subcases( x=-6 range[2<x<7] and x=8/3 range[x>7] don't fit in these ranges? @Jimmy R. – Adnan Selimovic Oct 5 '16 at 15:26
• No, this is not correct. See my edit. – Jimmy R. Oct 5 '16 at 15:26
• Sorry all clear now. Thank you for the help very much Sir! – Adnan Selimovic Oct 5 '16 at 15:39
• I do not understand your question, sorry. – Jimmy R. Oct 5 '16 at 15:41

$$2x - |5-|x-2|| = 1$$ $$|5-|x-2||=2x-1$$ Hence, $x\ge\frac12$

$5-|x-2|=2x-1$ or $5-|x-2|=1-2x$

$|x-2|=6-2x$ or $|x-2|=4+2x$