How do you solve #abs(2x + 2) - 7 <= -5#? Algebra Linear Inequalities and Absolute Value Absolute Value Inequalities 1 Answer Alan P. Jun 14, 2015 #-2 <= x <= 0# Explanation: #abs(2x+2) -7 <= -5# #color(white)("XXXX")##rArr abs(2x+2) <= 2# If #(2x+2) < 0##color(white)("XXXX")##rarr x<-1# #abs(2x+2) <=2# means #(-2x-2) <=2# #color(white)("XXXX")##-2x <=4# #color(white)("XXXX")##x>= -2# If #(2x+2) >=0##color(white)("XXXX")##rarr x>=-1# #abs(2x+2) <= 2# means #(2x+2) <=2# #color(white)("XXXX")##x<=0# for #x<-1#, #rarr x>=-2# for #x>=-1#, #rarr x<=0# So #-2<=x<=0# 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 2359 views around the world You can reuse this answer Creative Commons License