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.

#-2 <= (2x - 3)/4 <= 2#

First, multiply each segment of the system of inequalities by #color(red)(4)# to eliminate the fraction while keeping the system balanced:

#color(red)(4) xx -2 <= color(red)(4) xx (2x - 3)/4 <= color(red)(4) xx 2#

#-8 <= cancel(color(red)(4)) xx (2x - 3)/color(red)(cancel(color(black)(4))) <= 8#

#-8 <= 2x - 3 <= 8#

Next, add #color(red)(3)# to each segment to isolate the #x# term while keeping the system balanced:

#-8 + color(red)(3) <= 2x - 3 + color(red)(3) <= 8 + color(red)(3)#

#-5 <= 2x - 0 <= 11#

#-5 <= 2x <= 11#

Now, divide each segment by #color(red)(2)# to solve for #x# while keeping the system balanced:

#-5/color(red)(2) <= (2x)/color(red)(2) <= 11/color(red)(2)#

#-5/2 <= (color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) <= 11/2#

#-5/2 <= x <= 11/2#

Or

#x >= -5/2# and #x <== 11/2#

Or, in interval notation:

#[-5/2, 11/2]#