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

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

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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) (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
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