SOLUTION: Find three consecutive even numbers such that the sum of the first, one half of the second, and one fourth of the third is twenty less than twice the first.

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Question 976787: Find three consecutive even numbers such that the sum of the first, one half of the second, and one fourth of the third is twenty less than twice the first.
Answer by FrankM(1040) About Me  (Show Source):
You can put this solution on YOUR website!
3 even numbers can be written as
%282n%29%2B%282n%2B2%29%2B%282n%2B4%29

2n%2B%281%2F2%29%282n%2B2%29%2B%281%2F4%29%282n%2B4%29=%284n-20%29

2n%2B%28n%2B1%29%2B%281%2F4%29%282n%2B4%29=%284n-20%29

%283n%2B1%29%2B%281%2F4%29%282n%2B4%29=%284n-20%29

multiply both sides by 2 to get rid of that 1/4

%286n%2B2%29%2B%28n%2B2%29=%288n-40%29
7n%2B4=8n-40
0=n-44
n=44
The numbers are
88,90,92
the sum of the first (88), one half of the second (45), and one fourth of the third (23) (total 156) is twenty less than twice the first (176-20)