How do you solve this absolute-value inequality #6abs(2x + 5 )> 66#?

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Answer 1 · 2015-04-03T22:57:12

Answer: #x < -8 uu x > 3#

Solution:
#6|2x + 5|> 66#

#=> |2x + 5|> 11#

By definition, #|x| > k# where #k# is a constant, means:
# x< -k uu x > k#
This implies,

#2x + 5 < -11 uu 2x + 5 > 11#

Case1: 2x + 5 < -11

#=> 2x < -16 => x< -8#

Case2: 2x + 5 > 11

#=> 2x > 6 => x > 3#