First, subtract #color(red)(4)# from each side of the inequality to isolate the absolute value term while keeping the inequality balanced:
#-1/2abs(5 - 9x) + 4 - color(red)(4) <= -10 - color(red)(4)#
#-1/2abs(5 - 9x) + 0 <= -14#
#-1/2abs(5 - 9x) <= -14#
Next, multiply each side of the inequality by #color(blue)(-2)# to isolate the absolute value function while keeping the inequality balanced. However, because we are multiplying or dividing an inequality by a negative term we must reverse the inequality operator:
#color(blue)(-2) xx 1/-2abs(5 - 9x) color(red)(>=) color(blue)(-2) xx -14#
#cancel(color(blue)(-2)) xx 1/color(blue)(cancel(color(black)(-2)))abs(5 - 9x) color(red)(>=) 28#
#1abs(5 - 9x) color(red)(>=) 28#
#abs(5 - 9x) color(red)(>=) 28#
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.
#-28 >= 5 - 9x >= 28#
First, subtract #color(red)(5)# from each segment of the system of inequalities to isolate the #x# term while keeping the system balanced:
#-color(red)(5) - 28 >= -color(red)(5) + 5 - 9x >= -color(red)(5) + 28#
#-33 >= 0 - 9x >= 23#
#-33 >= -9x >= 23#
Now, divide each segment of the system by #color(blue)(-9)# to solve for #x# while keeping the system balanced. However, because we are multiplying or dividing an inequalities by a negative term we must reverse the inequality operator:
#(-33)/color(blue)(-9) color(red)(<=) (-9x)/color(blue)(-9) color(red)(<=) 23/color(blue)(-9)#
#33/9 color(red)(<=) (color(blue)(cancel(color(black)(-9)))x)/cancel(color(blue)(-9)) color(red)(<=) -23/9#
#(3 xx 11)/(3 xx 3) color(red)(<=) x color(red)(<=) -23/9#
#(color(blue)(cancel(color(black)(3))) xx 11)/(color(blue)(cancel(color(black)(3))) xx 3) color(red)(<=) x color(red)(<=) -23/9#
#11/3 color(red)(<=) x color(red)(<=) -23/9#
Or
#x >= 11/3# and #x <= -23/9#
Or, in interval notation:
#[11/3, +oo)# and #(-oo, -23/9]#
