How do you solve and graph #abs((3h+1)/2)<8#?

The absolute value function takes any term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

#-8 < (3h + 1)/2 < 8#

First, multiply each segment of the system of inequalities by #color(red)(2)# to eliminate the fraction while keeping the system balanced:

#color(red)(2) xx -8 < color(red)(2) xx (3h + 1)/2 < color(red)(2) xx 8#

#-16 < cancel(color(red)(2)) xx (3h + 1)/color(red)(cancel(color(black)(2))) < 16#

#-16 < 3h + 1 < 16#

Next, subtract #color(red)(1)# from each segment to isolate the #h# term while keeping the system balanced:

#-16 - color(red)(1) < 3h + 1 - color(red)(1) < 16 - color(red)(1)#

#-17 < 3h + 0 < 15#

#-17 < 3h < 15#

Now, divide each segment by #color(red)(3)# to solve for #h# while keeping the system balanced:

#-17/color(red)(3) < (3h)/color(red)(3) < 15/color(red)(3)#

#-17/3 < (color(red)(cancel(color(black)(3)))h)/cancel(color(red)(3)) < 5#

#-17/3 < h < 5#

Or

#h > -17/3# and #h < 5#

Or, in interval notation:

#(-17/3, 5)#