Question 668869: A man is three times as old as his son and six years ago the product of their age was 288. Find their present ages. Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=son's age; x-6=son's age 6 years ago
Then 3x=father's age: 3x-6=father's age 6 years ago
Now we are told the following:
(x-6)(3x-6)=288 expand left side using FOIL
3x^2-6x-18x+36=288
3x^2-24x+36=288 divide each term by 3
x^2-8x+12=96 subtract 96 from each side
x^2-8x-84=0 quadratic in standard form and it can be factored
(x+6)(x-14)=0
x=-6----------------No good--can't have negative ages
x=14------------------son's present age
3x=42---------------------father's present age
CK
(14-6)(42-6)=288
8*36=288
288=288
Hope this helps---ptaylor
