Finding the Present Ages of Father and Son

Video Explanation

Question

Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. Find their present ages.

Solution

Step 1: Let the Variables

Let present age of father = \(x\) years

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

Step 2: Form the Equations

Two years ago:

\[ x – 2 = 5(y – 2) \]

\[ x – 2 = 5y – 10 \]

\[ x – 5y = -8 \quad (1) \]

Two years later:

\[ x + 2 = 3(y + 2) + 8 \]

\[ x + 2 = 3y + 6 + 8 \]

\[ x + 2 = 3y + 14 \]

\[ x – 3y = 12 \quad (2) \]

Step 3: Solve by Elimination Method

Subtract equation (2) from equation (1):

\[ (x – 5y) – (x – 3y) = -8 – 12 \]

\[ x – 5y – x + 3y = -20 \]

\[ -2y = -20 \]

\[ y = 10 \]

Step 4: Find the Value of x

Substitute \(y = 10\) in equation (2):

\[ x – 3(10) = 12 \]

\[ x – 30 = 12 \]

\[ x = 42 \]

Conclusion

Present age of father:

\[ \boxed{42 \text{ years}} \]

Present age of son:

\[ \boxed{10 \text{ years}} \]