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Question 350304: find three consecutive odd integers such that the sum of the first and third is 33 more than the second
find three consective even integers such that twice the largerst is equal to the sum of thetwo smallest ones
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! Hi again,
As to odd integers, they differ by 2, let n,(n+2),(n+4)represent them
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the odd integers that the sum of the first and third is 33 more the second
n +(n + 4) = (n+2) +33
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Simplify and solve for n.
Then list n,(n+2), (n+4) as being the 3 odd consecutive integers
Hi
Note: consecutive interger differ by 1. Let x, 9x+1),x+2) represent the integers.
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Question states
n + (n+2) = (n+1) +33
n = 32
n =
Intergers are 32,33 and 34
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check your answer
32 + 34 = 33 + 33
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twice the largerst is equal to the sum of thetwo smallest ones
Write as you Read and Solve
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2(n+2) = n + (n+1)
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