A town's population has been growing linearly. In 2003, the population was 65000, and the population has been growing by 700 people each year.

Write an equation for the population x years after 2003.

The town's population is growing linearly, so we can use the equation of a straight line in the form of:

\[
P(x) = mx + b
\]

Where:
- \( P(x) \) is the population \( x \) years after 2003,
- \( m \) is the rate of population growth per year,
- \( b \) is the initial population at \( x = 0 \), which corresponds to the year 2003.

### Given:
- In 2003, the population was 65,000, so \( b = 65,000 \).
- The population has been growing by 700 people per year, so \( m = 700 \).

Now, substitute these values into the equation:

\[
P(x) = 700x + 65,000
\]

Thus, the equation for the population \( x \) years after 2003 is:

\[
P(x) = 700x + 65,000
\]

Would you like more details or further clarification?

Here are five related questions to consider:
1. What will the population be in 2025?
2. How many years will it take for the population to reach 80,000?
3. How would the equation change if the growth rate doubled?
4. What if the town's population decreased by 200 people per year instead of increasing?
5. How does linear growth differ from exponential population growth?

**Tip:** In linear growth, the rate of change is constant, making it easier to predict future values compared to non-linear growth.

A town's population has been growing linearly. In 2003, the population was 65000, and the population has been growing by 700 people each year. Write an equation for the population x years after 2003.

Learn how to write the linear equation representing the population growth of a town starting in 2003, with an initial population of 65,000 and a growth rate of 700 people per year. This problem explores key algebraic concepts and is suitable for students in Grades 6-8.

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