Question 881039: a father is twice as old as his son. 12 years ago the father was 4 times as old as the son was then.
Determine their present age.
Answer by algebrahouse.com(1659)   (Show Source):
You can put this solution on YOUR website! x = son's age
2x = father's age {he is twice as old as his son}
x - 12 = son's age 12 years ago
2x - 12 = father's age 12 years ago
2x - 12 = 4(x - 12) {twelve years ago the father was 4 times his son}
2x - 12 = 4x - 48 {used distributive property}
-12 = 2x - 48 {subtracted 2x from each side}
36 = 2x {added 48 to each side}
x = 18 {divided each side by 2}
2x = 36 {substituted 18, in for x, into 2x}
son is 18
father is 36
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