Algebra
posted by Shepp on .
The four oldest people in Golden City have lived a total of 384 years put together. The difference in ages for the youngest and the second oldest is 14. The second youngest is 3 years older than the youngest. The oldest is 20 years older than the average of the second oldest and youngest find their ages and enter them from youngest to oldest.

In order youngest to oldest, we have a,b,c,d where
a+b+c+d = 384
ca = 14
b = a+3
d = 20+(c+a)/2
85,88,99,112