First multiply the middle term of the system of inequalities by #(-1)/-1# which is the same a multiplying by #1# while does not change the value of the expression:
#-4 <= (-1)/-1 xx (-5x + 1)/-3 <= 2#
#-4 <= (-1(-5x + 1))/(-1 xx -3) <= 2#
#-4 <= (5x - 1)/3 <= 2#
Next, multiply each segment of the system of inequalities by #color(red)(3)# to eliminate the fraction while keeping the system balanced:
#color(red)(3) xx -4 <= color(red)(3) xx (5x - 1)/3 <= color(red)(3) xx 2#
#-12 <= cancel(color(red)(3)) xx (5x - 1)/color(red)(cancel(color(black)(3))) <= 6#
#-12 <= 5x - 1 <= 6#
Then, add #color(red)(1)# to each segment of the system of inequalities to isolate the #x# term while keeping the system balanced:
#-12 + color(red)(1) <= 5x - 1 + color(red)(1) <= 6 + color(red)(1)#
#-11 <= 5x - 0 <= 7#
#-11 <= 5x <= 7#
Now, divide each segment by #color(red)(5)# to solve for #x# while keeping the system balanced:
#-11/color(red)(5) <= (5x)/color(red)(5) <= 7/color(red)(5)#
#-11/5 <= (color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5)) <= 7/5#
#-11/5 <= x <= 7/5#
Or
#x >= -11/5#; #x <= 7/5#
Or, in interval notation:
#[-11/5, 7/5]#
