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Question 209881: There are three consecutive odd integers such that twice their sum is 30 less than 8 times the third integer. Find these odd integers.
Answer by math problem solving(30) (Show Source):
You can put this solution on YOUR website! Let the middle number is n.
Then the three numbers can be expressed as n-2,n,n+2.
Their sum is n-2 + n + n + 2 = 3n.
Now use the given relationship to set up an equation:
twice their sum is 30 less than 8 times the third integer
The relationship can be written as:
2* their sum = 8*the third number - 30
2*3n =8(n+2)-30
Solving for n, we have
6n = 8n + 16 -30
0 = 2n - 14
2n = 14
n = 7
So the three numbers are 5, 7, 9.
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