Problem: One year ago a man was four times as old as his son. After 6 years, his age exceeds twice his son's age by 9 years. Find their present ages?

Solution:

Let's assume the current ages of the man and his son to be M and S respectively.

Step 1: Formulating Equations

We are given two conditions in the problem:

- One year ago a man was four times as old as his son.
- After 6 years, his age exceeds twice his son's age by 9 years.

Using the above conditions, we can form two equations:

- M-1 = 4(S-1) (Equation 1)
- M+6 = 2(S+6) + 9 (Equation 2)

Step 2: Solving Equations

We can solve the equations simultaneously to find the values of M and S.

- M-1 = 4S-4 (From Equation 1)
- M = 4S-3 (Add 1 to both sides)
- 4S-3+6 = 2(S+6) + 9 (Substitute M+6 for 2(S+6) in Equation 2)
- 4S+3 = 2S+21
- 2S = 18
- S = 9

Now that we know S=9, we can substitute this value in Equation 1 to find M.

- M-1 = 4(9)-4
- M-1 = 32
- M = 33

Therefore, the current age of the man is 33 years and the current age of his son is 9 years.

Step 3: Verification

We can verify our answer by checking if it satisfies both the conditions given in the problem.

- One year ago, the man was 32 years old and his son was 8 years old. 32 is four times as old as 8, thus the first condition is satisfied.
- After 6 years, the man will be 39 years old and his son will be 15 years old. 39 exceeds twice 15 by 9 years, thus the second condition is also satisfied.

Therefore, our answer is correct.

Answer: The current age of the man is 33 years and the current age of his son is 9 years.