SOLUTION: Find three consecutive numbers such that the sum of one-fourth the first and one-fifth the second is five less than one-seventh the third.

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Question 972802: Find three consecutive numbers such that the sum of one-fourth the first and one-fifth the second is five less than one-seventh the third.
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
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n, n+1, n+2= three consecutive numbers
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%281%2F4%29n%2B%281%2F5%29%28n%2B1%29=%281%2F7%29%28n%2B2%29-5

%2863%2F140%29n%2B%2828%2F140%29=%2820%2F140%29n-%28660%2F140%29 Multiply by 140.
63n%2B28=20n-660 Subtract (20n+28) from each side.
43n=-668 Divide each side by 43.
n=-16 The first number is -16.
n%2B1=-16%2B1=-15 The second number is -15.
n%2B2=-16%2B2=-14 The third number is -14.
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CHECK:
%281%2F4%29-16%2B%281%2F5%29-15=%281%2F7%29-14-5
-4-3=-2-5
-7=-7