SOLUTION: This is my 3rd and last problem that has me stumped. I just want to say that yall are wonderful people to help me to work out and understand these problems. Thank you so much! T

Question 465917: This is my 3rd and last problem that has me stumped. I just want to say that yall are wonderful people to help me to work out and understand these problems. Thank you so much!
Transulate the situation into an equation and show work.
The sum of three consecutive odd integers is 85 less than four times the first integer?
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Let's begin by calling the first odd integer N. The next higher integer has to be an even number. (Think of counting odd numbers by skipping every other number. For example 3, 5, 7, 9, etc.)
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So if the first odd integer is N, the next consecutive odd integer is two numbers higher and can be represented as N + 2. The third consecutive odd integer is 2 more than the second so it is N+2 plus 2 more or N + 4.
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The sum of these three odd integers is:
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N + (N+2) + (N+4)
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And if you add these terms together the result is:
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3N + 6
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This sum is to equal 85 less than 4 times the first odd integer. This is:
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4N - 85
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Setting this equal to the sum of the three consecutive odd integers results in:
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3N + 6 = 4N - 85
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Subtract 4N from both sides to get:
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-N + 6 = -85
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Subtracting 6 from both sides simplifies this equation to:
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-N = -91
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Solve for N by multiplying both sides by -1 to get:
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N = 91
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So the three consecutive odd integers are 91, 93, and 95.
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The sum of these three is 91 + 93 + 95 = 279.
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4 times the first of these is 4*91 = 364 and 85 less than that is 364 - 85 = 279. Since this is equal to the sum of the three consecutive odd integers, the answer checks out.
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Hope this helps you to understand the problem.
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