SOLUTION: 3 consecutive odd integers that four times the smallest is 56 more than the sum of the other two.
Algebra.Com
Question 780350: 3 consecutive odd integers that four times the smallest is 56 more than the sum of the other two.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
First integer 
, so next consecutive odd integer is 
and the one after that is 
So
\ +\ 56)
Solve for , then calcualte and 
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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