simplify the expression lx+1l - lx-1l. (X is 75% between 0 and 1 on the numberline)
|x+1| - |x-1|
If x>1, |x+1| = x+1 and |x-1| = x-1, so
|x+1| - |x-1| = (x+1)-(x-1) = 2
Similarly, if x < -1,
|x+1| - |x-1| = -(x+1) - -(x-1) = -2
If -1 < x < 1,
|x+1| > 0
|x-1| < 0, so
|x+1| - |x-1| = (x+1)- -(x-1) = 2x