How do you solve #|-6x + 3| ≥ 39#? Algebra Linear Inequalities and Absolute Value Absolute Value Inequalities 1 Answer 256 Jan 17, 2017 #|-6x+3|>=39 <=> x in (-oo,-6]cup[7,oo)# Explanation: #(|x|=x <=> x>=0) and (|x|=(-x) <=> x<0)# So, if #(-6x+3)<0# then #|-6x+3|=6x-3# Then, #6x-3>=39# #<=># #6x>=42# #<=># #x>=7# Or, if #(-6x+3)>=0# then #-6x+3>=39# #<=># #-6x>=36# #<=># #x<=-6# Then #x<=-6# or #x>=7# #<=># #x in (-oo,-6] cup [7,oo) # Answer link Related questions How do you solve absolute value inequalities? When is a solution "all real numbers" when solving absolute value inequalities? How do you solve #|a+1|\le 4#? How do you solve #|-6t+3|+9 \ge 18#? How do you graph #|7x| \ge 21#? Are all absolute value inequalities going to turn into compound inequalities? How do you solve for x given #|\frac{2x}{7}+9 | > frac{5}{7}#? How do you solve #abs(2x-3)<=4#? How do you solve #abs(2-x)>abs(x+1)#? How do you solve this absolute-value inequality #6abs(2x + 5 )> 66#? See all questions in Absolute Value Inequalities Impact of this question 2040 views around the world You can reuse this answer Creative Commons License