Expressing age in terms of a variable, often denoted as $x$, is a common exercise in algebra. This concept is particularly useful in solving word problems where relationships between ages need to be described mathematically. Let’s break down the process with clear explanations and examples.
Basic Concept
The idea is to use a variable (usually $x$) to represent an unknown quantity, such as age. For example, if we say that John is $x$ years old, then $x$ is the variable representing John’s age. From this basic representation, we can describe other ages in relation to $x$
Example 1: Simple Age Relationship
Let’s say we have two people: John and his sister Mary. If John is $x$ years old and Mary is 5 years younger than John, we can express Mary’s age in terms of $x$ as follows:
- John’s age: $x$
- Mary’s age: $x – 5$
So, if John is 10 years old ($x = 10$), then Mary would be $10 – 5 = 5$ years old.
Example 2: Future Age
We can also use $x$ to represent future ages. Suppose John is currently $x$ years old. In 3 years, John’s age can be expressed as:
- John’s age in 3 years: $x + 3$
If John is currently 10 years old ($x = 10$), then in 3 years, he will be $10 + 3 = 13$ years old.
Example 3: Combined Ages
Sometimes, problems require us to find the sum of ages. Let’s consider John and Mary again. If John is $x$ years old and Mary is $x – 5$ years old, the combined age of John and Mary can be expressed as:
- Combined age: $x + (x – 5) = 2x – 5$
If $x = 10$, then the combined age is $2(10) – 5 = 20 – 5 = 15$ years.
Example 4: Age Difference
Age difference problems are another common scenario. If John is $x$ years old and Mary is $x – 5$ years old, the age difference between John and Mary is:
- Age difference: $x – (x – 5) = 5$
No matter what $x$ is, the age difference remains constant at 5 years.
Example 5: Multiple Variables
In some cases, you might have more than one variable. For instance, if John is $x$ years old, and his friend Alex is twice as old as John, then Alex’s age can be expressed as:
- Alex’s age: $2x$
If John is 10 years old ($x = 10$), then Alex is $2(10) = 20$ years old.
Solving Age Word Problems
Let’s apply these concepts to solve a typical word problem:
Problem: John is 4 years older than his sister Mary. In 5 years, John will be twice as old as Mary. How old are they now?
Solution:
Let $x$ be Mary’s current age.
Then, John’s current age is $x + 4$
In 5 years, Mary will be $x + 5$ years old, and John will be $(x + 4) + 5 = x + 9$ years old.
According to the problem, in 5 years, John will be twice as old as Mary:
$x + 9 = 2(x + 5)$
Solve the equation:
$x + 9 = 2x + 10$
$9 – 10 = 2x – x$
$-1 = x$
Therefore, Mary is currently -1 years old, which doesn’t make sense in real life. This indicates a mistake in the problem setup or interpretation. Let’s recheck the problem or try a different approach.
Revised Problem: John is 4 years older than his sister Mary. In 5 years, John will be 1.5 times as old as Mary. How old are they now?
Revised Solution:
Let $x$ be Mary’s current age.
Then, John’s current age is $x + 4$
In 5 years, Mary will be $x + 5$ years old, and John will be $(x + 4) + 5 = x + 9$ years old.
According to the revised problem, in 5 years, John will be 1.5 times as old as Mary:
$x + 9 = 1.5(x + 5)$
Solve the equation:
$x + 9 = 1.5x + 7.5$
$9 – 7.5 = 1.5x – x$
$1.5 = 0.5x$
$x = 3$
Therefore, Mary is currently 3 years old, and John is $3 + 4 = 7$ years old.
Conclusion
Expressing age in terms of $x$ is a powerful tool in algebra that helps us solve various age-related problems. By understanding the relationships between different ages and using algebraic equations, we can find unknown ages and solve complex word problems. Practice with different scenarios to become more comfortable with this concept.