SOLUTION: I don't know how to do this math problem. I need help fast because I have a high school Regents exam tomorrow morning and I don't know if a question similar to this one could be on

Question 637872: I don't know how to do this math problem. I need help fast because I have a high school Regents exam tomorrow morning and I don't know if a question similar to this one could be on the exam. The problem is: "The sum of three consecutive odd integers is 18 less than five times the middle number. Find the three integers. [Only alebraic solution can receive full credit.]" PLEASE HELP!
Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
My granddaughter may be taking the same regents, so I'm rushing the solution to you.
Let n = the first odd integer
Then the next higher odd integer is n+2 (n+1 is an even number)
And the third and largest odd integer is n+4 (or n+2+2)
The sum of them = 5*(middle odd integer) less 18
n + (n+2) + (n+4) = 5*(n+2) - 18
Simplify by adding like terms
3n + 6 = 5n + 10 -18
6 - 10 + 18 = 5n - 3n
2n = 14
n = 7 OK?
Answer: the three consecutive odd integers are 7, 9, 11
Let's check.
Does (7+9+11 = 5(9)-18)?
Does (27 = 45-18)?
Does (27=27)? Yes
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