SOLUTION: A mother is twice as old
as her daughter now, but ten
years ago, she was 3 times as old
as her daughter. Fins their ages
at present.
Algebra.Com
Question 1054890: A mother is twice as old
as her daughter now, but ten
years ago, she was 3 times as old
as her daughter. Fins their ages
at present.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
let x equal the mother's age now.
let y equal her daughter's age now.
the mother is twice as old as the daughter now.
this means that x = 2y.
the mother was 3 times as old as her daughter 10 years ago.
the mother was x - 10 years old ten years ago.
the daughter was y - 10 years old ten years ago.
this means that x - 10 = 3 * (y - 10).
you have two equations that need to be solved simultaneously.
they are:
x = 2y
x - 10 = 3 * (y - 10)
since x = 2y from the first equation, replace x with 2y in the second equation to get:
x - 10 = 3 * (y - 10) becomes 2y - 10 = 3 * (y - 10)
simplify to get:
2y - 10 = 3y - 30
subtract 2y from both sides of the equation and add 30 to both sides of the eqution to get:
30 - 10 = 3y - 2y
combine like terms to get:
20 = y
since x = 2y, this means that x = 40.
you have:
x = 40
y = 20
currently:
the mother is x years old = 40.
the daughter is y years old = 20.
the mother is 2 times as old as the daughter.
10 years ago:
the mother was x - 10 = 30 years old.
the daughter was y - 10 = 10 years old.
the mother was 3 times as old as the daughter.
solution looks good.
solution is:
mother is currently 40 years old.
daughter is currently 20 years old.
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