SOLUTION: 1) The sum of the ages of Luis and Shannon is 108 years. 9 years ago, Luis's age was 4 times Shannon's age. How old is Luis now?
2) Jason is 80 years old. Erin is 35 years old.
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-> SOLUTION: 1) The sum of the ages of Luis and Shannon is 108 years. 9 years ago, Luis's age was 4 times Shannon's age. How old is Luis now?
2) Jason is 80 years old. Erin is 35 years old.
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Question 1016704: 1) The sum of the ages of Luis and Shannon is 108 years. 9 years ago, Luis's age was 4 times Shannon's age. How old is Luis now?
2) Jason is 80 years old. Erin is 35 years old. How many years until Jason's age is 2 times Erin's age? years Answer by macston(5194) (Show Source):
You can put this solution on YOUR website! .
L=Luis's age; S=Shannon's age
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L+S=108
L=108-S
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L-9=4(S-9)
L-9=4S-36
L=4S-27
108-S=4S-27
135=5S
45=27
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L=108-S=108-27=81
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ANSWER: Luis is 81 and Shannon is 27.
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CHECK:
L-9=4(S-9)
81-9=4(27-9)
72=4(18)
72=72
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