Linear ages equation: A father is twice his son's present age. If the son is now 20 years old, how many years ago was the father three times the son's age?

Introduction / Context:
Ages problems typically translate English statements into linear equations in one variable, then solve for the elapsed time. Here the relationship “twice now, thrice some years ago” becomes a simple one-unknown equation.

Given Data / Assumptions:

• Son's present age S = 20.
• Father's present age F = 2S = 40.
• Find t > 0 such that (F − t) = 3(S − t).

Concept / Approach:
Use one unknown (t) for “years ago” and set up the linear equality that matches the ratio condition at that time.

Step-by-Step Solution:
1) Equation: 40 − t = 3(20 − t).2) Expand: 40 − t = 60 − 3t.3) Rearrange: −t + 3t = 60 − 40 ⇒ 2t = 20 ⇒ t = 10.

Verification / Alternative check:
Ten years ago: father 30, son 10; indeed 30 = 3 × 10.

Why Other Options Are Wrong:
They do not satisfy the equation (plug them back to see the mismatch between 3× and the actual ages then).

Common Pitfalls:
Confusing “twice” and “three times,” or subtracting t from only one age. Both ages change with time.

Final Answer:
10 years