First, multiply each segment of the system of inequalities by #color(red)(2)# to eliminate the parenthesis while keeping the system balanced:
#color(red)(2) xx -5 < color(red)(2) xx 1/2(2m + 8) <= color(red)(2) xx 11#
#-10 < cancel(color(red)(2)) xx 1/color(red)(cancel(color(black)(2)))(2m + 8) <= 22#
#-10 < 2m + 8 <= 22#
Next, subtract #color(red)(8)# from each segment to isolate the #m# term while keeping the system balanced:
#-10 - color(red)(8) < 2m + 8 - color(red)(8) <= 22 - color(red)(8)#
#-18 < 2m + 0 <= 14#
#-18 < 2m <= 14#
Now, divide each segment by #color(red)(2)# to solve for #m# while keeping the equation balanced:
#-18/color(red)(2) < (2m)/color(red)(2) <= 14/color(red)(2)#
#-9 < (color(red)(cancel(color(black)(2)))m)/cancel(color(red)(2)) <= 7#
#-9 < m <= 7#
Or
#m > -9# and #m <= 7#
Or, in interval notation:
#(-9, 7]#
To graph this we will draw vertical lines at #-9# and #7# on the horizontal axis.
The line at #-9# will be a dashed line because the inequality operator does not contain and "or equal to" clause. The line a #7# will be a solid line because the inequality operator does contain an "or equal to" clause.
We will shade between the two lines to show the interval for the system of inequalities:
