How do you solve #abs(4x+10)<20#?

1 Answer
Apr 7, 2018

Answer:

#-15/2< x < 5/2#

Explanation:

#"inequalities of the type "|x| < a#

#"always have solutions of the form"#

#-a < x < a#

#rArr-20<4x+10<20#

#"subtract 10 from each interval"#

#-20color(red)(-10)>4xcancel(+10)cancel(color(red)(-10))20color(red)(-10)#

#rArr-30<4x <10#

#"divide each interval by 4"#

#-30/4 < x<10/4#

#rArr-15/2 < x < 5/2#

#x in(-15/2,5/2)larrcolor(blue)"in interval notation"#