How do you solve #abs(x)>=5#? Algebra Linear Inequalities and Absolute Value Absolute Value Inequalities 1 Answer Ratnaker Mehta Sep 3, 2016 #|x|>=5 hArr x in RR-(-5,5)=(-oo,5]uu[5,oo)#. Explanation: Recall that, #|y|<,a, (a>0) hArr y in (-a,a)#. #:. not (|y|<,a, (a>0)) hArr not (y in (-a,a))#. #:. |y|>=a, (a>0) hArr y !in (-a,a) hArr y in RR-(-a,a), i.e.,# #y in (-oo,-a]uu[a,oo)# Clearly, #|x|>=5 hArr x in RR-(-5,5)=(-oo,5]uu[5,oo)#. 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 1805 views around the world You can reuse this answer Creative Commons License