Age Word Problem: Find the Father's and Son's Ages

A classic algebra problem, worked out step by step by Dr. E — with the full reasoning behind every move, a video walkthrough, and a free practice worksheet.

➗ Algebra🎬 Video solution📄 Free worksheet
The Problem
x son 3x father (now) = 3×
Two snapshots in time: now, and 12 years later

A father is 3 times as old as his son. In 12 years, the father will be twice as old as the son. Find their present ages.

Video solution coming soon. Dr. E walks through this exact problem on the whiteboard, step by step.

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Step-by-Step Solution
1
Let the variable be the younger age.
Let the son's age now be x. The father is 3 times as old, so the father is 3x. Anchoring the variable to the smaller quantity keeps the numbers positive.
2
Write everyone's age in 12 years.
Add 12 to each person. A small table keeps it organized:
NowIn 12 years
Sonxx + 12
Father3x3x + 12
3
Turn the future condition into an equation.
"In 12 years the father is twice the son" means 3x + 12 = 2(x + 12).
4
Solve.
3x + 12 = 2x + 24, so subtracting 2x and 12 gives x = 12.
5
Find both ages.
Son = 12, father = 3 × 12 = 36.
Check in 12 years: son 24, father 48, and 48 = 2 × 24. ✓
Answer the son is 12 and the father is 36 The trick to every age problem is the time table: write each age now and later, then translate the later condition into one equation.

💡 Why this works — the habit to remember

Age problems feel wordy, but they're tame once you build the Now / Later table. Everybody ages by the same amount, so each "later" cell is just the "now" cell plus 12. The single equation always comes from the sentence about the future relationship.

Extension question: change "twice as old" to "the father is 30 years older than the son" and see which clue actually controls the answer.

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