Algebra Absolute Value

Question

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.

Answer 1

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 $|a-d|=3$.

Answer 2

\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}