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To solve this problem, let's define the variables and set up the equations based on the given information:

### Given Information:
- **Budget**: $97,600 per month
- **Costs**:
  - Television ad: $2,000
  - Radio ad: $200
  - Newspaper ad: $600
- **Total ads**: 80 ads per month
- **Condition**: The number of television ads is equal to the combined number of radio and newspaper ads.

### Variables:
- Let $ x $ be the number of television ads.
- Let $ y $ be the number of radio ads.
- Let $ z $ be the number of newspaper ads.

### Equations:

1. **Total number of ads**:
   \[
   x + y + z = 80
   \]

2. **Budget constraint**:
   \[
   2000x + 200y + 600z = 97,600
   \]

3. **Television ads equals the combined number of radio and newspaper ads**:
   \[
   x = y + z
   \]

### Solving the System of Equations:

Substitute $ x = y + z $ into the first and second equations.

1. Substitute into the total number of ads equation:
   \[
   (y + z) + y + z = 80
   \]
   \[
   2y + 2z = 80
   \]
   \[
   y + z = 40
   \]
   
2. Substitute $ x = y + z $ into the budget equation:
   \[
   2000(y + z) + 200y + 600z = 97,600
   \]
   Simplify this equation:
   \[
   2000y + 2000z + 200y + 600z = 97,600
   \]
   \[
   2200y + 2600z = 97,600
   \]
   Divide by 200:
   \[
   11y + 13z = 488
   \]

Now we have the following two equations:
1. $ y + z = 40 $
2. $ 11y + 13z = 488 $

Let's solve this system:

1. From $ y + z = 40 $:
   \[
   y = 40 - z
   \]

2. Substitute $ y = 40 - z $ into $ 11y + 13z = 488 $:
   \[
   11(40 - z) + 13z = 488
   \]
   \[
   440 - 11z + 13z = 488
   \]
   \[
   2z = 48
   \]
   \[
   z = 24
   \]

3. Substitute $ z = 24 $ back into $ y + z = 40 $:
   \[
   y = 40 - 24 = 16
   \]

4. Substitute $ y = 16 $ and $ z = 24 $ into $ x = y + z $:
   \[
   x = 16 + 24 = 40
   \]

### Solution:
- **Television ads (x)**: 40
- **Radio ads (y)**: 16
- **Newspaper ads (z)**: 24

The department can run **40 television ads**, **16 radio ads**, and **24 newspaper ads** per month.

Do you have any questions or would you like more details?

Here are some related questions that could further enhance your understanding:
1. How would the solution change if the budget were different?
2. What if the costs of each type of ad were different?
3. How can this method be used to solve other advertising budget problems?
4. What happens if there is no restriction on the total number of ads?
5. How does the system of equations approach help in solving real-world problems?

**Tip:** When solving systems of equations, try to reduce the system by substitution or elimination to simplify the equations progressively.

A health insurance company advertises on television, on radio, and in the local newspaper. The marketing department has an advertising budget of $97,600 per month. A television ad costs $2,000, a radio ad costs $200, and a newspaper ad costs $600. The department wants to run 80 ads per month and have as many television ads as the combined number of radio and newspaper ads. How many of each type of ad can the department run each month?

Learn how to solve a real-world advertising budget allocation problem using systems of equations. The problem involves determining the number of television, radio, and newspaper ads for a health insurance company with a budget of $97,600. Suitable for Grades 9-12 students studying algebra and linear equations.

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