University of Georgia and Florida State University scored a total of 39 points during the 2003 sugar bowl. The points cameo from a total of 11 different scoring plays which were a combination of touchdowns extra point kick some field goals worth 6, 1, and 3 points respectively. The same number of touchdowns and field goals were scored. How many touchdowns, extra-point kicks, and field goals were scored during the game?

Let's define variables to represent the unknown quantities:
- Let \( t \) be the number of touchdowns scored.
- Let \( e \) be the number of extra-point kicks.
- Let \( f \) be the number of field goals scored.

### Step 1: Understanding the problem

We know that:
1. Each touchdown is worth 6 points.
2. Each extra-point kick is worth 1 point.
3. Each field goal is worth 3 points.

Also, we are given:
- The total number of scoring plays is 11, so:
  \[
  t + e + f = 11
  \]
- The total score is 39 points, so:
  \[
  6t + e + 3f = 39
  \]
- The same number of touchdowns and field goals were scored, so:
  \[
  t = f
  \]

### Step 2: Substituting \( t = f \)

Using \( t = f \), we can reduce the number of variables:
- From \( t + e + f = 11 \), substituting \( f \) for \( t \), we get:
  \[
  f + e + f = 11 \quad \Rightarrow \quad 2f + e = 11 \quad \text{(Equation 1)}
  \]
- From \( 6t + e + 3f = 39 \), substituting \( f \) for \( t \), we get:
  \[
  6f + e + 3f = 39 \quad \Rightarrow \quad 9f + e = 39 \quad \text{(Equation 2)}
  \]

### Step 3: Solving the system of equations

Now, we have the following system:
1. \( 2f + e = 11 \)
2. \( 9f + e = 39 \)

We can subtract Equation 1 from Equation 2 to eliminate \( e \):
\[
(9f + e) - (2f + e) = 39 - 11
\]
\[
7f = 28
\]
\[
f = 4
\]

### Step 4: Solving for \( e \) and \( t \)

Now that \( f = 4 \), we can substitute this value into Equation 1:
\[
2(4) + e = 11
\]
\[
8 + e = 11 \quad \Rightarrow \quad e = 3
\]
Since \( t = f \), we have \( t = 4 \).

### Step 5: Conclusion

- \( t = 4 \) touchdowns were scored.
- \( e = 3 \) extra-point kicks were made.
- \( f = 4 \) field goals were scored.

Thus, the game included:
- 4 touchdowns,
- 3 extra-point kicks, and
- 4 field goals.

Let me know if you need further details or have any questions!

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Here are 5 related questions to expand on this:

1. How would the solution change if the total score was 42 points?
2. What if the same number of touchdowns and extra-point kicks were scored instead of touchdowns and field goals?
3. How can we calculate the total score if the number of extra-point kicks is unknown?
4. What would happen if a safety (worth 2 points) was included in the scoring plays?
5. How does the equation change if the points per play differ in another sport?

**Tip**: For problems involving equations, always substitute known relationships early to simplify your calculations.

University of Georgia and Florida State University scored a total of 39 points during the 2003 Sugar Bowl. The points came from a total of 11 different scoring plays, which were a combination of touchdowns, extra point kicks, and field goals worth 6, 1, and 3 points, respectively. The same number of touchdowns and field goals were scored. How many touchdowns, extra-point kicks, and field goals were scored during the game?

This math problem involves solving a system of equations based on the scoring breakdown in the 2003 Sugar Bowl. Learn how to calculate the number of touchdowns, extra-point kicks, and field goals scored using algebraic methods. Suitable for students in Grades 8-10.

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