SOLUTION: Divide 140 into three parts such that the second is fifteen less than the first and the third exceeds twice the first by thirty five

Algebra ->  Customizable Word Problem Solvers  -> Numbers -> SOLUTION: Divide 140 into three parts such that the second is fifteen less than the first and the third exceeds twice the first by thirty five      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 898631: Divide 140 into three parts such that the second is fifteen less than the first and the third exceeds twice the first by thirty five
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
f + (f-15) + (2f+35) = 140 (f representing the first part)
4f = 120
f = 30
30, 15, 95 are the three parts