Question 1203174: Six years ago, Lydia was three times as old as Keiko. In three years time, Lydia will be twice as old as Keiko. What are their present ages?
Found 3 solutions by josgarithmetic, greenestamps, MathTherapy:
Answer by josgarithmetic(39618)   (Show Source):
You can put this solution on YOUR website!  , and if use some care the system could be formed using a single variable.
Solve the system.
If you see the way the form is written,
 , which is in a single variable.
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Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Let x be Keiko's age 6 years ago
then 3x = Lydia'a age 6 years ago
Then x+9 = Keiko's age 3 years from now
and 3x+9 = Lydia's age 3 years from now
3 years from now, Lydia's age will be 2 times Keiko's age:
3x+9 = 2(x+9)
3x+9 = 2x+18
x = 9
ANSWERS:
Keiko's age now = x+6 = 15
Lydia's age now = 3x+6 = 33
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
Six years ago, Lydia was three times as old as Keiko. In three years time, Lydia will be twice as old as Keiko. What are their present ages?
Let Lydia and Keiko's ages be L and K, respectively
Then we get L - 6 = 3(K - 6)____L - 6 = 3K - 18___L - 3K = - 12 ----- eq (i)
Also, L + 3 = 2(K + 3)_____L + 3 = 2K + 6_____L - 2K = 3 ----- eq(ii)
Subtracting eq (i) from eq (ii), we find that Keiko, or K = 12 + 3 = 15 years-old
Substituting 15 for K in eq (ii), we get: L - 2(15) = 3
L - 30 = 3
Lydia, or L = 3 + 30 = 33 years-old
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