How do you solve #abs(-2x -5) + 2 > 9#?

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Answer 1 · 2018-03-16T13:25:18

The solution set #S = {x in RR | (x< -6) or (x>2)}#

First let's clean up:

#abs(-2x-5) > 7#

We have to distinguish two cases:

(1) #(-2x-5) > 7 # which is obvious ;-)

and

(2) #(-2x-5) < -7 # which is true, because the absolute of a nuber #<-7# is #>7#.

Lets solve case (1) with some cleanup:

#-2x-5 > 7#

#-2x>12#

#x<-6#

and case (2) results as follows:

#-2x-5 < -7#

#-2x < -2#

#x > 2#

The solution set #S = {x in RR | (x< -6) or (x>2)}#