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How do you solve #abs(x+2)-5>=8#?

1 Answer

Apr 13, 2017

Solution: # x <= -15 or x >= 11 #. In interval notation: #(-oo,-15]uu[11,oo)#

Explanation:

#| x+2| -5 >= 8 or | x+2| >= 8 +5 or x+2 >=13 or x >= 11# OR

#| x+2| -5 >= 8 or | x+2| >= 8 +5 or x+2 <= -13 or x <= -15#

Solution: # x <= -15 or x >= 11 #. In interval notation: #(-oo,-15]uu[11,oo)#[Ans]

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