The absolute value function takes any negative or positive term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.
#-2/3 >= 2 - 3x >= 2/3#
First, subtract #color(red)(2)# from each segment of the system of inequalities to isolate the #x# term while keeping the system balanced:
#-2/3 - color(red)(2) >= 2 - color(red)(2) - 3x >= 2/3 - color(red)(2)#
#-2/3 - (3/3 xx color(red)(2)) >= 0 - 3x >= 2/3 - (3/3 xx color(red)(2))#
#-2/3 - 6/3 >= -3x >= 2/3 - 6/3#
#-8/3 >= -3x >= -4/3#
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:
#(-8/3)/color(blue)(-3) color(red)(<=) (-3x)/color(blue)(-3) color(red)(<=) (-4/3)/color(blue)(-3)#
#((-8)/3)/(color(blue)(-3)/1) color(red)(<=) (color(red)(cancel(color(black)(-3)))x)/cancel(color(blue)(-3)) color(red)(<=) ((-4)/3)/(color(blue)(-3)/1)#
#(-8 xx 1)/(3 xx color(blue)(-3)) color(red)(<=) x color(red)(<=) (-4 xx 1)/(3 xx color(blue)(-3))#
#(-8)/(-9) color(red)(<=) x color(red)(<=) (-4)/(-9)#
#8/9 color(red)(<=) x color(red)(<=) 4/9#
Or
#x <= 4/9#; #x >= 8/9#
Or, in interval notation:
#(-oo, 4/9]#; #[8/9, +oo)#
