SOLUTION: 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?

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Question 1016696: 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?

Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52897)   (Show Source): You can put this solution on YOUR website!
.
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?
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L + S = 108,       (1)
L - 9 = 4*(S-9).   (2)

Express L via S from (1): L = 108 - S, and substitute it into (2).
You will get a single equation for S. 

108 - S - 9 = 4*(S-9).

Solve it:

108 -S - 9 = 4S - 36,

108 - 9 + 36 = 4S + S,

135 = 5S,

S =  = 27.

Shannon is 27 years old.

Now calculate Luis' age. It is 108 - S = 108 - 27 = 81.

Check. 27 + 81 = 108.  
       27 - 9 = 18.  81 - 9 = 72.   = 4.  OK.

Answer. Luis is 81 years old. Shannon age is 27 years.


Answer by MathTherapy(10557)   (Show Source): You can put this solution on YOUR website!
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?
Let Luis' age, be L
Then Shannon's is: 108 - L
We then get: L - 9 = 4(108 - L - 9)
L - 9 = 4(99 - L)
L - 9 = 396 - 4L
L + 4L = 396 + 9
5L = 405
L, or Luis' current age = , or
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