SOLUTION: ten years ago the age of a father was four times of his son . Ten years hence the age of the father will be twice that of his son . The present ages of the father and the son are.

Algebra ->  Customizable Word Problem Solvers  -> Age -> SOLUTION: ten years ago the age of a father was four times of his son . Ten years hence the age of the father will be twice that of his son . The present ages of the father and the son are.      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 507474: ten years ago the age of a father was four times of his son . Ten years hence the age of the father will be twice that of his son . The present ages of the father and the son are.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let D = Dad's present age and S = Son's present age.
1) D-10 = 4(S-10) "Ten years ago the age of a father (D-10) was four times the age of his son (4(S-10))."
2) D+10 = 2(S+10) "Ten years hence, the age of the father (D+10) will be twice that of his son ((2(S+10))."
So now you have two equations with two unknowns (D & S). Simplify the two equations:
1a) D-10 = 4S-40
2a) D+10 = 2S+20 Subtract equation 2a) from equation 1a) to get:
3) -20 = 2S-60 Now add 60 to both sides.
3a) 40 = 2S Divide both sides by 2.
3b) 20 = S or S = 20 Now substitute this value of S into 1a) or 2a) and solve for D.
2a) D+10 = 2S+20 Substitute S = 20
2b) D+10 = 2(20)+20 Simplify.
3c) D+10 = 60 Subtract 10 from both sides.
3d) D = 50
The father is 50 years old and the son is 20 years old.