How do you solve #abs(x+a)<b#? Algebra Linear Inequalities and Absolute Value Absolute Value Inequalities 1 Answer Cesareo R. Jul 25, 2016 #a-b < x < b-a# Explanation: #abs(x+a) < b -> {(x+a < b->x < b-a), (-x+a < b->x > a-b) :}# so #a-b < x < b-a# 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