# Algebra Absolute Value

Let $a,b,c$, and $d$ be real numbers with $$|a-b|=2, \hspace{.2in} |b-c|=3, \hspace{.2in} |c-d|=4$$ What is the sum of all possible values of $|a-d|$? I am completely clueless on how to begin! It's due tomorrow and I need help.

• Why did you wait until today? – TBrendle Nov 1 '13 at 21:38
• Let's be constructive, @TBrendle. – vadim123 Nov 1 '13 at 21:39
• Put $s either side of mathematics in future. – Shaun Nov 1 '13 at 21:39 ## 2 Answers There are eight cases. 1.$b>a$or$b<a$2.$c>b$or$c<b$3.$d>c$or$d<c$For all$2\times 2\times 2=8$possibilities, work out$|a-d|$. Since the problem is invariant under translation, you may as well assume$a$is something simple, like$0$. Here's one of the eight to get you started. Suppose$b>a, c<b, d>c$. We start with$a=0$. Then$b=2$since$b>a$. Then$c=-1$since$c<b$. Lastly,$d=3$since$d>c$. Hence in this case$|a-d|=3\$.

\begin{align} |a-b|=2 &\implies a-b = \pm 2\\ |b-c|=3 &\implies b-c = \pm 3\\\ |c-d|=4 &\implies c-d = \pm 4 \\ &\implies a-d = \begin{cases} 2 + 3 + 4 &= 9 \\ 2 + 3 - 4 &= 1 \\ 2 - 3 + 4 &= 3 \\ 2 - 3 - 4 &= -5 \\ -2 + 3 + 4 &= 5 \\ -2 + 3 - 4 &= -3 \\ -2 - 3 + 4 &= -1 \\ -2 - 3 - 4 &= -9 \end{cases} \end{align}