First, subtract #color(red)(1)# from each segment of the system of inequalities to isolate the #n# term while keeping the system balanced:

#-2 - color(red)(1) < -2n + 1 - color(red)(1) <= 7 - color(red)(1)#

#-3 < -2n + 0 <= 6#

#-3 < -2n <= 6#

Now, divide each segment of the system by #color(blue)(-2)# to solve for #n# while keeping the system balanced. However, because we are multiplying or dividing inequalities by a negative number we must reverse the inequality operators:

#(-3)/color(blue)(-2) color(red)(>) (-2n)/color(blue)(-2) color(red)(>=) 6/color(blue)(-2)#

#3/2 color(red)(>) (color(blue)(cancel(color(black)(-2)))n)/cancel(color(blue)(-2)) color(red)(>=) -3#

#3/2 color(red)(>) n color(red)(>=) -3#

Or

#n < 3/2# and #n >= -3#

Or, in interval notation:

#[-3, 3/2]

To graph this we will draw a vertical lines at #-3# and #3/2# on the horizontal axis.

The line at #-3# will be a solid line because the inequality operator contains an **"or equal to"** clause.

The line at #3/2# will be a dashed line because the inequality operator does contains an **"or equal to"** clause and therefore the line is not part of the solution set.

We will shade between the two lines: