algebra
posted 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.

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