Question 85074: Find three consecutive even integers such that twice the sum of the first and third is twelve more than twice the second.
Thanks,
Tonya
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! To start with, recognize that even integers are 2 units apart. If the first even integer
is x, then the next even integer is x+2 and the third in the sequence is x + 4.
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Twice the sum of the first and third is, therefore, 2 times [x +(x+4)]. Inside the brackets
simplifies to x + x + 4 which is equivalent to 2x + 4 and 2 times that quantity is 2(2x + 4)
which multiplies out to 4x + 8.
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Twice the second is 2*(x+2) and this multiplies out to 2x +4. 12 more than that means add
12 to it to get (2x + 4) + 12 = 2x + 4 + 12 = 2x + 16.
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Now we can set the two results equal. The first result is 4x + 8 and the second is 2x + 16.
Setting them equal results in:
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4x + 8 = 2x + 16
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Get rid of the 8 on the left side by subtracting 8 from both sides to get:
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4x = 2x + 8
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Next get rid of the 2x on the right side by subtracting 2x from both sides. The result is:
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2x = 8
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Solve for x by dividing both sides by 2 and the result is:
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x = 8/2 = 4.
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So the first even number in the sequence is 4. That means the next even integer is 4 + 2
which is 6 and the next even integer is 4 + 4 = 8.
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The three even, consecutive integers that satisfy this problem are 4, 6, and 8.
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Check:
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Twice the sum of the first and third is 2 times (4 + 8) is 2 times 12 which is 24.
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12 more than 2 times the second is 12 + 2*6 = 12 + 12 = 24.
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These two answers are equal and so the problem is satisfied.
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Hope this helps you to understand the problem a little better.
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