How do you solve #abs(2x+9)-2x>=0#? Precalculus Solving Rational Inequalities Polynomial Inequalities 1 Answer Cesareo R. Feb 24, 2017 #abs(2x+9)-2x>=0# is true for all #x in RR# Explanation: #abs(2x+9)-2x>=0 -> abs(2x+9)-(2x+9)+9 ge 0# then, for #x ne -9/2# #1-(2x+9)/abs(2x+9)+9/abs(2x+9) ge 0# Calling #9/abs(2x+9) = epsilon ge 0# we have #1 pm 1 + epsilon ge 0#. This is always true so #abs(2x+9)-2x>=0# is true for all #x in RR# Answer link Related questions What are common mistakes students make when solving polynomial inequalities? How do I solve a polynomial inequality? How do I solve the polynomial inequality #-2(m-3)<5(m+1)-12#? How do I solve the polynomial inequality #-6<=2(x-5)<7#? How do I solve the polynomial inequality #1<2x+3<11#? How do I solve the polynomial inequality #-12<-2(x+1)<=18#? How do you solve the inequality #6x^2-5x>6#? How do you solve #x^2 - 4x - 21<=0# A) [-3, 7] B) (-∞, -3] C) (-∞, -3] [7, ∞) D) [7, ∞)? How do you solve quadratic inequality, graph, and write in interval notation #x^2 - 8x + 15 >0#? How do you solve #-x^2 - x + 6 < 0#? See all questions in Polynomial Inequalities Impact of this question 1087 views around the world You can reuse this answer Creative Commons License