Posted by tempest on Tuesday, July 17, 2012 at 1:51pm.
Find four consecutive odd integers so that five times the sum of the first and third integers excceeds four times the sum of the second and last integers by 14.

algebra  Steve, Tuesday, July 17, 2012 at 2:43pm
let the integers be n3, n1, n+1, n+3
5(n3 + n+1) = 4(n1 + n+3)+14
5(2n2) = 4(2n+2)+14
10n10 = 8n+22
2n = 32
n=16
The numbers are 13,15,17,19
Check:
5(13+17) = 150
4(15+19)+14 = 150
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