How do you solve #|x+2|>2#? Algebra Linear Inequalities and Absolute Value Absolute Value Inequalities 1 Answer A. S. Adikesavan Apr 8, 2016 x > 0 and x < #-4#. The graduation x on the real line, under this condition, is outside the closed interval #[-4, 0]#. Explanation: #|x+2|>2# is equivalent to the pair of inequalities #x+2 > 2 and -(x+2) > 2#. So, x > 0 and (rearranging for +x) #x < -4#. 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 1259 views around the world You can reuse this answer Creative Commons License