First, subtract #color(red)(2)# from each side of the inequality to isolate the absolute value TERM while keeping the inequality balanced:
#-color(red)(2) + 2 - 3abs(2x + 7) <= -color(red)(2) - 16#
#0 - 3abs(2x + 7) <= -18#
#-3abs(2x + 7) <= -18#
Next, divide each side of the inequality by #color(blue)(-3)# 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 also reverse the inequality operator:
#(-3abs(2x + 7))/color(blue)(-3) color(red)(>=) (-18)/color(blue)(-3)#
#(color(blue)(cancel(color(black)(-3)))abs(2x + 7))/cancel(color(blue)(-3)) color(red)(>=) 6#
#abs(2x + 7) color(red)(>=) 6#
The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.
#-6 >= 2x + 7 >= 6#
Then, subtract #color(red)(7)# from each segment of the inequality system to isolate the #x# term while keeping the system balanced:
#-6 - color(red)(7) >= 2x + 7 - color(red)(7) >= 6 - color(red)(7)#
#-13 >= 2x + 0 >= -1#
#-13 >= 2x >= -1#
Now, divide each segment of the system by #color(red)(2)# to solve for #x# while keeping the system balanced:
#-13/color(red)(2) >= (2x)/color(red)(2) >= -1/color(red)(2)#
#-13/2 >= (color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) >= -1/2#
#-13/2 >= x >= -1/2#
Or
#x <= -13/2# and #x >= -1/2#
Or. in interval notation:
#(-oo, -13/2]# and #[-1/2, oo)#
