First, divide each segment of the system of inequalities by #color(red)(2)# to eliminate the need for parenthesis while keeping the system balanced:
#3/color(red)(2) < (2(2x - 1))/color(red)(2) <= 16/color(red)(2)#
#3/2 < (color(red)(cancel(color(black)(2)))(2x - 1))/cancel(color(red)(2)) <= 8#
#3/2 < 2x - 1 <= 8#
Next, add #color(red)(1)# to each segment of the system to isolate the #x# term while keeping the equation balanced:
#3/2 + color(red)(1) < 2x - 1 + color(red)(1) <= 8 + color(red)(1)#
#3/2 + color(red)(2/2) < 2x - 0 <= 9#
#(3 + color(red)(2))/2 < 2x <= 9#
#5/2 < 2x <= 9#
Now, multiply each segment of the system by #color(red)(1/2)# to solve for #x# while keeping the system balanced:
#5/2 xx color(red)(1/2) < 2x xx color(red)(1/2) <= 9 xx color(red)(1/2)#
#(5 xx color(red)(1))/(2 xx color(red)(2)) < 2/color(red)(2)x <= 9/2#
#5/4 < 1x <= 9/2#
#5/4 < x <= 9/2#
Or
#x > 5/4# and #x <= 9/2#
Or, in interval notation:
#(5/4, 9/2]#
