How do you solve #abs(3x + 2) - 1 >=10#?

1 Answer
Jul 11, 2016

Answer:

#x <= -13/3#
#x >= 3#

Explanation:

#I3x + 2I >= 11#
Separate solving into 2 parts:
a.#(3x + 2) >= 11#
#3x >= 9@ #x >= 3# b.# -(3x + 2) >= 11# # - 3x - 2 >= 11 # #- 3x >= 13# #x <= -13/3# Note. The 2 end points #(-13/3)# and (3) are included in the solution set.
Graph:

=============== -13/3 ---------0---------- 3 =============