First, divide each side of the inequality by #color(red)(3)# to isolate the absolute value term while keeping the inequality balanced:
#(3abs(4 - 5x))/color(red)(3) <= 9/color(red)(3)#
#(color(red)(cancel(color(black)(3)))abs(4 - 5x))/cancel(color(red)(3)) <= 3#
#abs(4 - 5x) <= 3#
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.
#-3 <= 4 - 5x <= 3#
Next, subtract #color(red)(4)# from each segment of the system of inequalities to isolate the #x# term while keeping the system balanced:
#-3 - color(red)(4) <= 4 - color(red)(4) - 5x <= 3 - color(red)(4)#
#-7 <= 0 - 5x <= -1#
#-7 <= -5x <= -1#
Now, divide each segment by #color(blue)(-5)# 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:
#(-7)/color(blue)(-5) color(red)(>=) (-5x)/color(blue)(-5) color(red)(>=) (-1)/color(blue)(-5)#
#7/5 color(red)(>=) (color(blue)(cancel(color(black)(-5)))x)/cancel(color(blue)(-5)) color(red)(>=) 1/5#
#7/5 color(red)(>=) x color(red)(>=) 1/5#
Or
#x >= 1/5#; #x <= 7/5#
Or, in interval notation
#[1/5, 7/5]#
