First, subtract #color(red)(3)# from each segment of the system of inequalities to isolate the #x# term while keeping the system balanced:
#-color(red)(3) - 3 <= -color(red)(3) + 3 - 2x < -color(red)(3) + 11#
#-6 <= 0 - 2x < 8#
#-6 <= -2x < 8#
Now, divide each segment by #color(blue)(-2)# 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)(-2) color(red)(>=) (-2x)/color(blue)(-2) color(red)(>) 8/color(blue)(-2)#
#3 color(red)(>=) (color(blue)(cancel(color(black)(-2)))x)/cancel(color(blue)(-2)) color(red)(>) -4#
#3 color(red)(>=) x color(red)(>) -4#
Or
#x > -4# and #x <= 3#
Or, in interval notation:
#(-4, 3]#
