SOLUTION: find the two number that differ by seven such that one fifth of the smaller exceeds one ninth of the larger by 5

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Question 987561: find the two number that differ by seven such that one fifth of the smaller exceeds one ninth of the larger by 5
Answer by macston(5194) About Me  (Show Source):
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S=smaller number; L=larger number=S+7
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(1/5)S-5=(1/9)L
(S/5)-5=L/9
(S/5)-5=(S+7)/9
9S/5-45=S+7
9S/5-5S/5=52
4S/5=52
4S=260
S=65 ANSWER 1: The smaller number is 65
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L=S+7=65+7=72 ANSWER 2: The larger number is 72.
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CHECK:
(1/5)S=(1/9)L+5
(1/5)(65)=(1/9)72+5
13=8+5
13=13.