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This problem describes a linear growth in population. Let's break it down:

### Step 1: Writing the equation for population

- The population in 2003 was **46,000**.
- Each year, the population increases by **1,700** people.
  
The general form of a linear equation is:

\[
P = P_0 + r \cdot x
\]

Where:
- \( P_0 \) is the initial population (in 2003), which is **46,000**.
- \( r \) is the rate of change (increase per year), which is **1,700**.
- \( x \) is the number of years after 2003.

Thus, the equation becomes:

\[
P = 46000 + 1700 \cdot x
\]

### Step 2: Finding the population in 2009

To find the population in 2009, calculate the number of years after 2003:

\[
x = 2009 - 2003 = 6
\]

Substitute \( x = 6 \) into the equation:

\[
P = 46000 + 1700 \cdot 6
\]

Now, calculate the result:

\[
P = 46000 + 10200 = 56,200
\]

Thus, the population in 2009 is **56,200**.

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Would you like a detailed explanation or further clarification on any part of this? 

Here are five related questions for practice:
1. What would the population be in 2015?
2. How many years will it take for the population to reach 80,000?
3. If the population growth rate changed to 2,000 per year, how would the equation change?
4. What is the population growth rate per month based on the yearly increase?
5. If the initial population was 50,000 in 2003, how would the equation change?

**Tip**: In linear models, the slope (rate of change) shows how quickly values increase or decrease per unit of time.

A town's population has been growing linearly. In 2003 the population was 46,000. The population has been growing by 1700 people each year. Write an equation for the population, P, x years after 2003. Use the formula to find the population in 2009.

Learn how to calculate the linear population growth for a town, starting from 46,000 people in 2003 with an increase of 1,700 per year. This problem includes creating a linear equation for population and finding the population in 2009. Suitable for Grades 7-9.

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