First, multiply each side of the inequality by #color(red)(11)/color(blue)(4)# to isolate the absolute value function while keeping the inequality balanced:
#color(red)(11)/color(blue)(4) xx (4abs(-11/3m + 4/3))/11 <= color(red)(11)/color(blue)(4) xx 2/11#
#cancel(color(red)(11))/cancel(color(blue)(4)) xx (color(blue)(cancel(color(black)(4)))abs(-11/3m + 4/3))/color(red)(cancel(color(black)(11))) <= cancel(color(red)(11))/color(blue)(4) xx 2/color(red)(cancel(color(black)(11)))#
#abs(-11/3m + 4/3) <= 2/color(blue)(4)#
#abs(-11/3m + 4/3) <= 1/2#
The absolute value function takes any 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.
#-1/2 <= -11/3m + 4/3 <= 1/2#
Next, subtract #color(red)(4/3)# from each segment of the system of inequalities to isolate the #m# term while keeping the system balanced:
#-1/2 - color(red)(4/3) <= -11/3m + 4/3 - color(red)(4/3) <= 1/2 -
color(red)(4/3)#
#(3/3 xx -1/2) - (2/2 xx color(red)(4/3)) <= -11/3m + 0 <= (3/3 xx 1/2) - (2/2 xx color(red)(4/3))#
#-3/6 - 8/6 <= -11/3m <= 3/6 - 8/6#
#-11/6 <= -11/3m <= -5/6#
Now, multiply each segment by #color(red)(3)/color(blue)(-11)# to solve for #m# while keeping the system balanced:
#color(red)(3)/color(blue)(-11) xx (-11)/6 <= color(red)(3)/color(blue)(-11) xx (-11)/3m <= color(red)(3)/color(blue)(-11) xx (-5)/6#
#cancel(color(red)(3))/cancel(color(blue)(-11)) xx color(blue)(cancel(color(black)(-11)))/(color(red)(cancel(color(black)(6)))2) <= cancel(color(red)(3))/cancel(color(blue)(-11)) xx color(blue)(cancel(color(black)(-11)))/color(red)(cancel(color(black)(3)))m <= cancel(color(red)(3))/color(blue)(-11) xx (-5)/(color(red)(cancel(color(black)(6)))2)#
#1/2 <= m <= 5/22#
Or,
#m <= 5/22#; #m >= 1/2#
Or, in interval notation:
#(-oo, 5/22]#; #[1/2, +oo)#
