First, subtract #color(red)(8)# from each segment of the system of inequalities to isolate the #x# term while keeping the system balanced:

#2 - color(red)(8) > 8 - color(red)(8) - 3x > 1/2 - color(red)(8)#

#-6 > 0 - 3x > 1/2 - (2/2 xx color(red)(8))#

#-6 > -3x > 1/2 - color(red)(16)/2#

#-6 > -3x > (1 - color(red)(16))/2#

#-6 > -3x > -15/2#

Next, divide each segment by #color(blue)(-3)# to solve for #x# while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative number we must reverse the inequality operators:

#(-6)/color(blue)(-3) color(red)(<) (-3x)/color(blue)(-3) color(red)(<) (-15/2)/color(blue)(-3)#

#2 color(red)(<) (color(blue)(cancel(color(black)(-3)))x)/cancel(color(blue)(-3)) color(red)(<) 15/6#

#2 color(red)(<) x color(red)(<) 5/2#

Or

#x > 2#; #x < 5/2#

Or, in interval notation:

#(2, 5/2)#