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Answer 1 · 2017-05-19T04:20:13

#{2n+1, 2n+3, 2n+5}={11, 13, 15}#

If you have three, consecutive odd integers, you could write it like this
#{2n+1, 2n+3, 2n+5}# where #n# is an integer

The #2n# term is guaranteed to be even. So adding an odd value (i.e., 1, 3, and 5) ensures it will be odd.

The smallest odd is #2n+1#. The largest is #2n+5#. The problem tells us that

#2(2n+1)=2n+5+7#
#4n+2=2n+12#
#2n+2=12# (subtract #2n# from both sides)
#2n=10# (subtract #2# from both sides)
#n=5#

Plugging #n=5# back into our three consecutive integers gives
#{2n+1, 2n+3, 2n+5}={11, 13, 15}#

Answer 2 · 2017-05-19T04:27:43

Let #x# be the smallest of the three consecutive odd integers. Then the three consecutive odd integers are:
#x, x+2, x+4#

#2x = 7+(x+4)#
Left side of the equation comes from "Twice the smallest"
Right side of the equation comes from "seven more than the largest"

#2x=x+11#

x=11

Since we have #x# now, the three consecutive odd integers are:
#11, 11+2, 11+4#

#11, 13, 15#