<PRE><B>Question 894700</B>
Im trying to figure out how to write the equation for the following problem: find three consecutive even intergers such that one-forth of the first and one-seventh of the second plus one -half of the third is 13.
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Note:: Even numbers are always a multiple of 2.
1st:: 2(x-1)
2nd:: 2(x)
3rd:: 2(x+1)
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Equation:
(1/4)2(x-1) + (1/7)(2x) + (1/2)2(x+1) = 13
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Multiply thru by 28 to get:
14(x-1) + 8(x) + 28(x+1) = 28*13
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14x - 14 + 8x + 28x + 28 = 28*13
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50x + 14 = 28*13
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50x = 350
x = 7
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1st:: 2(x-1) = 12
2nd:: 14
3rd:: 16
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Cheers,
Stan H.
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