Question 116119
What you have done is listed the integers as being two numbers apart ... consecutive 
even integers (such as 10, 12, 14, ...) are two numbers apart. So are consecutive odd integers
(such as 17, 19, 21 ...). But consecutive integers ... such as 11, 12, and 13 ... are just one 
apart. Therefore, you represent three consecutive integers as:
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1st integer: n
2nd integer: n + 1
3rd integer: n + 2
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The sum of the first and second is n + n + 1 = 2n + 1 and four times that is:
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4(2n + 1)
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Seven times the third is 7(n + 2)
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The problem says that 4(2n + 1) is 17 more than 7(n + 2). So if you take 17 away from 4(2n + 1)
it will equal 7(n + 2). In equation form this is:
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4(2n + 1) - 17 = 7(n + 2)
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Multiply out the two distributed multiplications. First multiply 4 times each of the two terms
in its related parentheses. Then multiply 7 times each of the two terms in its related 
parentheses. When you do those multiplications the equation becomes:
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8n + 4 - 17 = 7n + 14
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Combine the +4 and -17 on the left side and you have:
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8n - 13 = 7n + 14
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Get rid of the -13 on the left side by adding 13 to both sides to reduce the equation to:
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8n = 7n + 27
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Then get rid of the 7n on the right side by subtracting 7n from both sides. When you do that
subtraction you get:
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n = 27
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If n is 27, then the next two consecutive integers are 28 and 29. So the three consecutive 
numbers that you are looking for are 27, 28, and 29.
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Check: 4 times (27 + 28) = 4(55) = 220. 7 times 29 = 203

Is the difference between 220 and 203 equal to 17. Yes it is, so the answer checks.
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Hope this helps you to see where you made your mistake and what you can do to solve the problem.
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