First, multiply each segment of the system of inequalities by #color(red)(3)# to eliminate the fraction and keep the system balanced:
#color(red)(3) xx -5 <= color(red)(3) xx (-2x + 5)/3 <= color(red)(3) xx 3#
#-15 <= cancel(color(red)(3)) xx (-2x + 5)/color(red)cancel(color(black)(3))) <= 9#
#-15 <= -2x + 5 <= 9#
Next, subtract #color(red)(5)# from each segment of the system to isolate the #x# term while keeping the system in balance:
#-15 - color(red)(5) <= -2x + 5 - color(red)(5) <= 9 - color(red)(5)#
#-20 <= -2x + 0 <= 4#
#-20 <= -2x <= 4#
Now, divide each segment of the system by #color(blue)(-2)# to solve for #x# while keeping the system balanced. However, because these are inequalities and we are multiplying or dividing the system by a negative term we must reverse the inequality signs:
#(-20)/color(blue)(-2) color(red)(>=) (-2x)/color(blue)(-2) color(red)(>=) 4/color(blue)(-2)#
#10 color(red)(>=) (color(blue)(cancel(color(black)(-2)))x)/cancel(color(blue)(-2)) color(red)(>=) -2#
#10 >= x >= -2#
