SOLUTION: Janet is three years younger than Rebecca. Eight years ago, Janet was half Rebecca's age. How old is each girl now?

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Question 1110164: Janet is three years younger than Rebecca. Eight years ago, Janet was half Rebecca's age. How old is each girl now?
Answer by ikleyn(52786)   (Show Source): You can put this solution on YOUR website!
.
J = R - 3          (1)    ("Janet is three years younger than Rebecca.".)
J-8 = 0.5*(R-8)    (2)    ("8 years ago . . . ")


Substitute eq(1) into eq(2). You will get

(R-3) - 8 = 0.5*(R-8)


R-11 = 0.5*(R-8)


Multiply it by 2 (both sides)

2R - 22 = R - 8

2R - R = 22 - 8 = 14  ====>  R = 14.


Answer.  Rebecca is 14 years old.  Janet is  R-3 = 14-3 = 11 years old.

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On age word problems,  see the lessons
    - Age problems and their solutions
    - A fresh formulation of a traditional age problem
    - Really intricate age word problem
    - Selected age word problems from the archive
    - Age problems for mental solution
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