SOLUTION: find the sum of 3 consecutive odd integers such that 4 times the sum of the 1st and second is 17 more than 7 times the 3rd

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Question 144830: find the sum of 3 consecutive odd integers such that 4 times the sum of the 1st and second is 17 more than 7 times the 3rd
Answer by solver91311(24713) About Me  (Show Source):
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First odd integer: x
Second consecutive odd integer: x%2B2
Third consecutive odd integer: x%2B4

The sum of the first and second: x+%2B+%28x+%2B+2%29=2x+%2B2
4 times the sum of the 1st and 2nd: 4%282x%2B2%29=8x%2B8

7 times the third: 7%28x%2B4%29=7x%2B28
17 more than 7 times the 3rd: 7x%2B28%2B17=7x%2B45

So: 8x%2B8=7x%2B45. Solve this for x and verify that x is odd.

The sum of the three integers is: x+%2B+%28x%2B2%29+%2B+%28x%2B4%29=3x%2B6. Take the solution you found above, multiply by 3 and then add 6 to get your final answer.