SOLUTION: Frida is five years older than Fredelyn. In three years the product of her age and Fredelyn's age five years ago will be 9o years. zFind their present ages.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Frida is five years older than Fredelyn. In three years the product of her age and Fredelyn's age five years ago will be 9o years. zFind their present ages.      Log On


   

Question 127362: Frida is five years older than Fredelyn. In three years the product of her age and Fredelyn's age five years ago will be 9o years. zFind their present ages.
Answer by solver91311(24713) About Me  (Show Source):
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Frida's current age is x and Fredelyn's current age is y.

The first thing we know is that x=y%2B5

Three years from now Frida will be x%2B3 and five years ago, Fredelyn was y-5 and we know that the product of these two values is 90, so:

%28x%2B3%29%28y-5%29=90.

Substitute the expression for x in terms of y from the first equation into the second equation:
%28%28y%2B5%29%2B3%29%28y-5%29=90

Simplify, expand, and put into standard form:
%28y%2B8%29%28y-5%29=90
y%5E2%2B3y-40=90
y%5E2%2B3y-130=0

-130=%28-10%29%2813%29 and 3=-10%2B13, so:

%28y-10%29%28y%2B13%29=0, hence Fredelyn is either 10 or -13 years old. Since a negative number for the age of a person is absurd, exclude this root. Fredelyn is therefore 10 years old.

From equation 1: Frida is 10%2B5=15 years old.

Check:
%2815%2B3%29%2810-5%29=18%2A5=90 answer checks.