How do you solve #abs(3x-6) + 3<15#? Algebra Linear Inequalities and Absolute Value Absolute Value Inequalities 1 Answer A. S. Adikesavan Mar 10, 2016 #-2 < x < 6# Explanation: #abs a# < k can be expanded as #-k < a < k# #-12<3x-6<12#. Add 6 and divide by 3. 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 1278 views around the world You can reuse this answer Creative Commons License