Finding the Present Ages of Man and Son

Video Explanation

Question

Six years hence, a man’s age will be three times the age of his son. Three years ago, he was nine times as old as his son. Find their present ages.

Solution

Step 1: Let the Variables

Let present age of the man = \(x\) years

Let present age of the son = \(y\) years

Step 2: Form the Equations

Six years hence:

\[ x + 6 = 3(y + 6) \]

\[ x + 6 = 3y + 18 \]

\[ x – 3y = 12 \quad (1) \]

Three years ago:

\[ x – 3 = 9(y – 3) \]

\[ x – 3 = 9y – 27 \]

\[ x – 9y = -24 \quad (2) \]

Step 3: Solve by Elimination Method

Subtract equation (2) from equation (1):

\[ (x – 3y) – (x – 9y) = 12 – (-24) \]

\[ x – 3y – x + 9y = 36 \]

\[ 6y = 36 \]

\[ y = 6 \]

Step 4: Find the Value of x

Substitute \(y = 6\) in equation (1):

\[ x – 3(6) = 12 \]

\[ x – 18 = 12 \]

\[ x = 30 \]

Conclusion

Present age of the man:

\[ \boxed{30 \text{ years}} \]

Present age of the son:

\[ \boxed{6 \text{ years}} \]