Question

The father is 7 times older than his son. Two years ago, the father was 13 times as old as his son. What are their present ages?

Hint:

When solving age problems, translate the relationships into algebraic equations and solve step by step.

Answer 1

Let the present age of the son be \( x \) years and the present age of the father be \( y \) years.
We are given two conditions:
1. The father is 7 times older than his son: \[ y = 7x. \] 2. Two years ago, the father was 13 times as old as the son: \[ y - 2 = 13(x - 2). \] Step 1: Substitute \( y = 7x \) into the second equation: \[ 7x - 2 = 13(x - 2). \] Simplify the equation: \[ 7x - 2 = 13x - 26.
7x - 13x = -26 + 2.
-6x = -24.
x = 4. \] Step 2: Find \( y \): Now, substitute \( x = 4 \) into the first equation \( y = 7x \): \[ y = 7(4) = 28. \] Thus, the present age of the son is 4 years, and the present age of the father is 28 years.
Conclusion:
The present age of the son is 4 years, and the present age of the father is 28 years.