First, divide each segment of the system of equation by #color(red)(5)# to eliminate the term outside the parenthesis while keeping the system balanced:
#30/color(red)(5) < (5(x + 6))/color(red)(5) < 85/color(red)(5)#
#6 < (color(red)(cancel(color(black)(5)))(x + 6))/cancel(color(red)(5)) < 17#
#6 < x + 6 < 17#
Now, subtract #color(red)(6)# from each segment to solve for #x# while keeping the system balanced:
#6 - color(red)(6) < x + 6 - color(red)(6) < 17 - color(red)(6)#
#0 < x + 0 < 11#
#0 < x < 11#
Or
#x > 0# and #x < 11#
Or, in interval notation:
#(0, 11)#