How do you graph and solve #2x-abs[x+4]> 8#? Algebra Linear Inequalities and Absolute Value Absolute Value Inequalities 1 Answer Cesareo R. Oct 27, 2017 #x > 12# Explanation: #2x-abs(x+4)=8+epsilon^2# with #epsilon ne 0# or #2x-8-epsilon^2= abs(x+4)# or #(2x-8-epsilon^2)^2-(x+4)^2=0# or #(-epsilon^2 - 12 + x) (-epsilon^2 - 4 + 3 x) = 0# or #{(-epsilon^2 - 12 + x=0),(-epsilon^2 - 4 + 3 x=0):}# or #{(x > 12),(x > 4/3):}# and the solution is #x > 12# 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 1234 views around the world You can reuse this answer Creative Commons License