Asked by tempest
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.
Answers
Answered by
Steve
let the integers be n-3, n-1, n+1, n+3
5(n-3 + n+1) = 4(n-1 + n+3)+14
5(2n-2) = 4(2n+2)+14
10n-10 = 8n+22
2n = 32
n=16
The numbers are 13,15,17,19
Check:
5(13+17) = 150
4(15+19)+14 = 150
5(n-3 + n+1) = 4(n-1 + n+3)+14
5(2n-2) = 4(2n+2)+14
10n-10 = 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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