Compare the process of solving |x – 1| + 1 < 15 to that of solving |x – 1| + 1 > 15.

In the first case we'd subtract 1 from both sides, obtaining |x-1|<14.

In the second case we'd also subtract 1 from both sides, and would obtain
 |x-1|>14.

What would the graphs look like?

In the first case, the graph would be on the x-axis with "center" at x=1.  From this center count 14 units to the right, and then place a circle around that location (which would be at x=15).  Next, count 14 units to the left of this center, and place a circle around that location (which would be -13).  Draw a line segment connecting the two circles.  Notice that all of the solutions are between -13 and +15, not including these endpoints.

In the second case, x has to be greater than 15 or less than -13.  Draw an arrow from x=1 to the left, and then draw a separate arrow from 15 to the right.  None of the values in between are solutions.

Ithink if im not mistaking b is -3 but im not a 100% sude