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Let's define variables for the populations of Numberton and Decimalville over time, and set up equations to describe their growth patterns.

1. **Numberton**:
   - Numberton’s population increases linearly by 1,000 people each year, starting from an initial population of 25,000.
   - Let \( N(t) \) represent the population of Numberton after \( t \) years.
   - The growth equation for Numberton is:
     \[
     N(t) = 25000 + 1000 \cdot t
     \]

2. **Decimalville**:
   - Decimalville’s population grows exponentially by 10% of its previous year's population, starting from an initial population of 15,000.
   - Let \( D(t) \) represent the population of Decimalville after \( t \) years.
   - The growth equation for Decimalville is:
     \[
     D(t) = 15000 \cdot (1.10)^t
     \]

### Solving for When Populations Are Equal
To find the year when both populations will be the same, we can set \( N(t) = D(t) \) and solve for \( t \):
\[
25000 + 1000 \cdot t = 15000 \cdot (1.10)^t
\]

This equation involves an exponential term and a linear term, so it doesn’t have a straightforward algebraic solution. We could solve this using numerical methods or approximation techniques to find when the populations will be approximately equal.

Would you like me to calculate an approximate solution for \( t \)?

Consider the population growth patterns of two towns since the turn of the century. Numberton has been growing steadily by 1000 every year, from an initial population of 25,000. Decimalville has been growing by 10% of its previous year’s population every year, from an initial population of 15,000.

Explore the population growth patterns of two towns, Numberton and Decimalville, with one town growing linearly by 1,000 people per year and the other growing exponentially by 10% annually. Learn how to set up and solve a system of equations to find when the populations will be equal. Suitable for high school students.

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