SOLUTION: Find 3 consecutive odd integers such that four times the sum of the first and second is 1 less than 7 times the third.

Question 184348: Find 3 consecutive odd integers such that four times the sum of the first and second is 1 less than 7 times the third.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Let the first odd integer be x.

Then the second odd integer must be x + 2.

Then the third odd integer must be x + 2 + 2 or x + 4.

The sum of the first and the second: (x) + (x + 2) or 2x + 2.

Four times the sum of the first and the second: 4(2x + 2) or 8x + 8.

"is" means equals.

Seven times the third: 7(x + 4) or 7x + 28.

One less than seven times the third: 7x + 28 - 1 or 7x + 27

Putting it all together:

Add -7x and -8 to both sides:

So the first odd integer is 19, the second is 2 more than that or 21, and the third is 2 more than that or 23.

Check:

The sum of the first and the second: 19 + 21 = 40.

Four times the sum of the first and the second: 4 X 40 = 160.

Seven times the third: 7 X 23 = 161, less 1 = 160.

Answer checks.

John

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