SOLUTION: Find three consecutive odd intergers such that twice the sum of the smaller two integers is 21 less than triple the third integer.

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Question 1203764: Find three consecutive odd intergers such that twice the sum of the smaller two integers is 21 less than triple the third integer.
Found 3 solutions by math_tutor2020, josgarithmetic, ikleyn:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Hint:

One equation to solve is
2(x+x+2) = 3(x+4)-21

where,
x = smallest odd integer
x+2 = odd integer just after x
x+4 = odd integer just after x+2

Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Odd consecutive integers----------system%28n%2Cn%2B2%2Cn%2B4%29

2%28n%2B%28n%2B2%29%29=3%28n%2B4%29-21
-
2%282n%2B2%29=3n%2B12-21
4n%2B4=3n-9-----------*MISTAKE MADE AFTER THIS STEP
cross%2813=n%29

13, 15, 17

Answer by ikleyn(52794) About Me  (Show Source):
You can put this solution on YOUR website!
.

The solution by @josgarithmetic, giving the answer 13, 15, 17, is INCORRECT.

The correct answer is -13, -11, -9. 

                     Solution


Let the numbers be  n, n+2, n+4.


Write an equation as you read the problem

    {2(n+(n+2)) = 3(n+4)-21


Simplify and find n


    2(2n+2) = 3n+12-21

    4n + 4 = 3n - 9

    4n - 3n = -9 - 4

       n    =   -13.

Solved (correctly)