## Precalculus (6th Edition) Blitzer

$(9,13)$
Rewrite as $|11-x| \lt 2$ ... also, $|a-b|=|b-a|$ $|x-11| \lt 2$ ... $|u| \lt c$ is equivalent to $-c \lt u \lt c.$ $-2 \lt x-11 \lt 2 \qquad$ ... add $11$ $9 \lt x \lt 13$ Solution set = $(9,13)$