How do you solve #|p + 8| + 2\leq 29#?

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Answer 1 · 2017-09-24T08:19:56

see below

#abs(p+8)+2<=29#

Subtract #2# from LHS and RHS,

#abs(p+8)+2-2<=29-2#

#abs(p+8)<=27#

#-27<=p+8<=27# [this pattern for "less than or equal"]

#-27-8<=p+8-8<=27-8#

#-35<=p<=19#

The interval notation is #[-35,19]#