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.
#-4 > (3x - 5)/6 > 4#
First, multiply each segment of the system of inequalities by #color(red)(6)# to eliminate the fraction while keeping the system balanced:
#color(red)(6) xx -4 > color(red)(6) xx (3x - 5)/6 > color(red)(6) xx 4#
#-24 > cancel(color(red)(6)) xx (3x - 5)/color(red)(cancel(color(black)(6))) > 24#
#-24 > 3x - 5 > 24#
Next, add #color(red)(5)# to each segment to isolate the #x# term while keeping the system balanced:
#-24 + color(red)(5) > 3x - 5 + color(red)(5) > 24 + color(red)(5)#
#-19 > 3x - 0 > 29#
#-19 > 3x > 29#
Now, divide each segment by #color(red)(3)# to solve for #x# while keeping the system balanced:
#-19/color(red)(3) > (3x)/color(red)(3) > 29/color(red)(3)#
#-19/3 > (color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) > 29/3#
#-19/3 > x > 29/3#
Or
#x < -19/3# and #x > 29/3#
Or, in interval notation:
#(-oo, -19/3)# and #(29/3. +oo)#
