Solving Simultaneous Equations: A Real-World Example with Age Problems

When dealing with age problems that involve relationships and sums, using simple algebraic equations can help us find solutions. One such problem is: Eunice is 3 times as old as Mavis. The sum of their ages is 60. How old is each?

Formulating the Problem

First, let's denote Mavis's age as ( m ). Given that Eunice is 3 times as old as Mavis, we denote Eunice's age as ( 3m ). Now, we have two key pieces of information: The sum of their ages is 60. Eunice's age is 3 times Mavis's age. This problem can be formulated into the following equation:

( m 3m 60 )

Step-by-Step Solving Process

The above equation simplifies to a more straightforward form, where we combine like terms:

( 4m 60 )

Next, solve for ( m ) by dividing both sides of the equation by 4:

( m frac{60}{4} 15 )

Therefore, Mavis is 15 years old. Since Eunice is 3 times as old as Mavis, we can find Eunice's age by multiplying 15 by 3:

( e 3m 3 times 15 45 )

Thus, Eunice is 45 years old and Mavis is 15 years old.

Verification

To verify, we simply add their ages together:

( 15 45 60 )

This confirms our solution is correct and satisfies the original problem statement.

Conclusion

In this example, we used a straightforward algebraic approach to solve a real-world age problem involving two individuals. Understanding and solving these types of equations are crucial skills in various applications, from basic arithmetic to more complex algebraic problems. This problem also serves as a practical example of how algebra can be applied to solve everyday problems.

Further Reading

There are many resources available to help you practice and improve your skills in solving similar kinds of algebraic word problems. Look for textbooks on algebra, online courses, or interactive problem sets on websites like Khan Academy and Brilli.enci