Question 152228: The present age of a father exceeds 3 times the present age of his son by 2 years. Six years ago, the age of the father exceeded 4 times the age of the son by 4 years. Find the present ages of the son and the father.
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! The present age of a father exceeds 3 times the present age of his son by 2 years. Six years ago, the age of the father exceeded 4 times the age of the son by 4 years. Find the present ages of the son and the father.
.
Let F = present age of father
and S = present age of son
.
Since we have two unknowns we'll need to find two equations.
.
From:"The present age of a father exceeds 3 times the present age of his son by 2 years. " we get equation 1:
F = 3S + 2
.
From:"Six years ago, the age of the father exceeded 4 times the age of the son by 4 years." we get equation 2:
F-6 = 4(S-6) + 4
.
Using the value of F from equation 1, substitute it into equation 2 and solve for S:
F-6 = 4(S-6) + 4
(3S + 2)-6 = 4S - 24 + 4
3S + 2 -6 = 4S - 24 + 4
3S - 4 = 4S - 20
-4 = S - 20
16 years = S (age of son)
.
Use the above, and substitute it into equation 1 and solve for F:
F = 3S + 2 = 3(16) + 2 = 50 years (age of father)
|
|
|
