(-6, 8)

That means this interval:

-------(=========================================)------
-8 -7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8  9 10

locate the mid point of this interval by averaging -6 and 8



So 1 in the exact middle of that interval. I'll mark it M

-------(====================M====================)------
-8 -7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7  8  9 10

The interval (-6,8) contains all the points up to 8-1 or 7 units
to the right of M=1 and all the points up to 1-(-6) or 7 units to
the left of M=1.

Let x be any point on the interval.  

If x is to the right of 1, if you subtract 1 from it, x-1, you
will always get a number between 0 and 6.

If x is to the left of 1, if you subtract 1 from it, x-1, you
will always get a number between -6 and 0.

So you combine the two preceding statements by saying:

If x is on the interval, regarless of whether it's to the right or left
of -1, if you subtract 1 from it, x-1 AND TAKE THE ABSOLUTE VALUE of it,
|x-1|, it will always be between 0 and 6, so

     |x-1|<6 is the answer.

If you want a formula,  

(a,b) is equivalent to  

If you want to go the other way,

|x-h| < k (if k > 0) is equivalent to h-k < x < h+k

--------------------------------------------------

Using "oo" for "infinity",

(-oo,a) U (b,oo) (if a < b) is equivalent to 

|x-h| > k (if k > 0), is equivalent to " x < h-k OR x > h+k ".

Edwin