SOLUTION: Determine three consecutive odd integers such that the sum of the smallest and three times the 2nd is 44 more than twice the 3rd. Is the question im asking. I get to like {{{ n

Question 1165456: Determine three consecutive odd integers such that the sum of the smallest and three times the 2nd is 44 more than twice the 3rd. Is the question im asking.
I get to like but im not sure if that is correct.
n = 1st odd z
n+2 = 2nd odd z
N+3 = 3rd odd z

Answer by ikleyn(52776) (Show Source): You can put this solution on YOUR website!
.


The correct starting equation is


    n + 3*(n+2) = 2*(n+4) + 44.

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Comment from student: Hey! Thanks for your input! I was wondering in the (n+4) part of your answer isn't it n+3 because
I was taught odd is n=3 and even is n+4.

My response. Thanks for asking.

The problem talks about three consecutive odd numbers.

If we denote the first (smallest of them) as "n", then the next odd number is (n+2), and the next after that is (n+4).

See the lesson
- Problems with consecutive integer numbers; odd even consecutive integer numbers
in this site.

Come again to this forum soon to learn something new (!)

Stay safe (!)

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