SOLUTION: three times the first of three consecutive odd integers is 3 more than twice the third. find the third integer?

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Question 640968: three times the first of three consecutive odd integers is 3 more than twice the third. find the third integer?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Let the last/greatest of the three consecutive odd integers be n .
(I like n. You could use x if you like that better).
We state that n has to be odd.
(If the n we find turns to be something that is not an odd integer, then the problem has no solution).
The odd integer before n is n-2, and the odd integer before that is n-4.
The three consecutive odd integers, in order, are:
n-4 , n-2 , and n .
Three times the first is 3%28n-4%29.
Twice the third one is 2n, and 3 more than that is 2n%2B3 .
Our equation is
3%28n-4%29=2n%2B3 .
To solve, we first apply the distributive property to do the multiplication on the left side of the equal sign.
3%28n-4%29=2n%2B3 --> 3n-3%2A4=2n%2B3 --> 3n-12=2n%2B3
Next, we subtract 2n from both sides of the equal sign:
3n-12=2n%2B3 --> 3n-12-2n=2n%2B3-2n --> n-12=3
Then we add 12 to both sides of the equal sign:
n-12=3 --> n-12%2B12=3%2B12 --> highlight%28n=15%29
The third of the three consecutive odd integers is highlight%2815%29 .
The three consecutive odd integers are 11, 13, and 15, and 3 times the first, 3%2A11=33 , is 3 more than twice the third (2%2A15=30 and 30%2B3=33).