Question 161939: A year ago, a mother was eight times as old as her daughter. Now, her age is the square of her daughter's age. How old are they now?
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! A year ago, a mother was eight times as old as her daughter. Now, her age is the square of her daughter's age. How old are they now?
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M-1 = 8*(D-1) One year ago
M = D^2 Now
Sub for M in 1st eqn
D^2 - 1 = 8*(D-1)
D^2 - 1 = 8D-8
D^2 - 8D + 7 = 0
(D-7)*(D-1) = 0
D = 7, D = 1
The D=1 solution isn't reasonable, as one year ago the Daughter would be zero.
So the daughter is 7, and the mother is 49.
One year ago, they were 6 and 48.
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