# Precalc absolute value fuction

Detemine |X-16| - |X-2| = ? given X<7

anomaly
Boy, it has been ages since I have seen/solved one of these.

I am probably mistaken, but I think this might be the way this is solved:

Detemine |X-16| - |X-2| = ? given X<7

0 < |x-16| - |x-2| < 7
I think at this step, the absolute value symbols disappear.
Then solve for x.

Anyone else, please feel free to correct me. It has been far too long for me to recall if this is the correct path to the solution.

Homework Helper
Detemine |X-16| - |X-2| = ? given X<7

If X< 7 then X-16< 7-16= -9. Since this X-16 is negative,
|X-16|= -(X-16)= 16-X.
If X< 7, then X-2< 5. "< 5" may be EITHER positive or negative so this is not enough to tell us what |X-2| is. It should be clear that the "break" occurs at X= 2 (just as the "break" in |X-16| occurs at X= 16. If x< 7, then X must be less than 16).

If 2<= X< 7, then |X-16|= -(X-16)= 16- X (because X< 7< 16) and
|X-2|= X- 2 (X-2 is non-negative). |X-16|- |X-2|= 16-X- X+ 2= 18- 2X

If X<= 2, then |X-16|= 16- X as above and |X-2|= -(X-2)= 2-X (X- 2 is now negative). |X- 16|- |X-2|= 16-X-(2-X)= 16-X-2+X= 14.

|X-16|- |X-2|= 18- 2X if 2<= X< 7
= 14 if X< 2