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(52890)   (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?
-----------------------------------------------------
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 
|
|
|
