SOLUTION: 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

Question 193710: 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 the youngest. Find their ages from youngest to oldest.
Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
from youngest to oldest, let ages be w,,x,,y,,z
,
1) w+x+y+z = 384
2) y-w=14, ,,y=(14+w)
3) x=(w+3)
4) z=( 20 + (y+w)/2 ),,,= (20+y/2+w/2) = 20+(14+w)/2 +w/2= 20+7+w/2+w/2= (27+w)
,
with 4 eqn and 4 unknown, we can solve
,
w+x+y+z = 384
(w)+(w+3)+(14+w)+(27+w) = 384
44+4w=384
4w=340
w=85 ********
substituting
x=(w+3)=88 ********
y= (14+w)=99******
z=(27+w)=112******
,
checking
85+88+99+112=384
384=384,,,,ok
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