How do you graph and solve #2x-1-abs[9-3x]<abs[3x]#? Algebra Linear Inequalities and Absolute Value Absolute Value Inequalities 1 Answer Cesareo R. Mar 21, 2017 See below. Explanation: #2x-1-3abs(x-3) < 3absx# now supposing #x ne 0# #2x/absx -1/absx-3abs(x-3)/absx < 3# but #x/abs x = pm 1# so #pm2-3<1/absx+3abs(x-3)/absx# or #abs x max(-5,-2) < 1+abs(x-3)# or #-2absx < 1 +abs(x-3)# or #2abs x + abs(x-3) + 1 > 0# So it is true for all #x# 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 1301 views around the world You can reuse this answer Creative Commons License