Posted by tempest on .
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,
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