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) - 12 < -color(red)(3) + 3 - 3x <= -color(red)(3) + 15#
#-15 < 0 - 3x <= 12#
#-15 < -3x <= 12#
Now, divide each segment by #color(blue)(-3)# 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:
#(-15)/color(blue)(-3) color(red)(>) (-3x)/color(blue)(-3) color(red)(>=) 12/color(blue)(-3)#
#5 color(red)(>) (color(blue)(cancel(color(black)(-3)))x)/cancel(color(blue)(-3)) color(red)(>=) -4#
#5 color(red)(>) x color(red)(>=) -4#
Or
#x < 5# and #x >= -4#
Or, in interval notation:
#[-4, 5)#
