How do you solve #abs(5x+2)>=abs(3x-4)#? Algebra Linear Inequalities and Absolute Value Absolute Value Inequalities 1 Answer Cesareo R. Mar 21, 2018 #x le -3# and #x ge 1/4# Explanation: This inequality is equivalent to #sqrt((5x+2)^2) ge sqrt((3x-4)^2)# now squaring both sides #(5x+2)^2 ge (3x-4)^2# or #(5x+2)^2 - (3x-4)^2 ge 0# or #(5x+2-3x+4)(5x+2+3x-4) ge 0# or #(2x+6)(8x-2) ge 0# or #(x+3)(x-1/4) ge 0# or #x le -3# and #x ge 1/4# 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 1663 views around the world You can reuse this answer Creative Commons License