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A high school football team scored 39 points in the game last night, with a combination of touchdowns (6 points each), extra points (1 point each), and field goals (3 points each).

The team made 1 more extra point than field goals, represented by the equation \( e = f + 1 \), where \( e \) is the number of extra points and \( f \) is the number of field goals scored.

They also scored five times as many points on touchdowns as field goals, represented by the equation \( 6t = 15f \), where \( t \) is the number of touchdowns and \( f \) is the number of field goals scored.

Write a system of equations representing this situation and use it to complete the statement.

The team scored [tex]\(\quad \)[/tex] touchdowns, [tex]\(\quad \)[/tex] extra points, and [tex]\(\quad \)[/tex] field goals.



Answer :

Let's define the variables:
- \( t \) represents the number of touchdowns.
- \( e \) represents the number of extra points.
- \( f \) represents the number of field goals.

We know from the problem statement:
1. The total points scored is 39. The points from touchdowns, extra points, and field goals adds up to this total:
[tex]\[ 6t + e + 3f = 39 \][/tex]

2. The team made 1 more extra point than field goals:
[tex]\[ e = f + 1 \][/tex]

3. They scored five times as many points on touchdowns as field goals. Since each touchdown is worth 6 points and each field goal is worth 3 points:
[tex]\[ 6t = 15f \][/tex]

We use these equations to solve for \( t \), \( e \), and \( f \):

Rewriting the equations:
1. The total points equation:
[tex]\[ 6t + e + 3f = 39 \][/tex]

2. The extra points to field goals relation:
[tex]\[ e = f + 1 \][/tex]

3. The touchdowns to field goals relation:
[tex]\[ 6t = 15f \][/tex]
Simplifying this relation:
[tex]\[ 2t = 5f \][/tex]
[tex]\[ t = \frac{5}{2}f \][/tex]

Using these equations, we find the values of \( t \), \( e \), and \( f \):

- From the simplified third equation:
[tex]\[ t = \frac{5}{2} f \][/tex]

Next, substitute \( t \) and \( e \) into the first equation:

[tex]\[ 6 \left(\frac{5}{2} f\right) + (f + 1) + 3 f = 39 \][/tex]

This simplifies to:

[tex]\[ 15f + f + 1 + 3f = 39 \][/tex]
[tex]\[ 19f + 1 = 39\][/tex]
[tex]\[ 19f = 38 \][/tex]
[tex]\[ f = 2 \][/tex]

Using \( f = 2 \):

[tex]\[ t = \frac{5}{2} \cdot 2 = 5 \][/tex]

Using \( f = 2 \):

[tex]\[ e = f + 1 \][/tex]
[tex]\[ e = 2 + 1 = 3 \][/tex]

Thus, the team scored 5 touchdowns, 3 extra points, and 2 field goals.

So, the completed statement is:
The team scored [tex]\( 5 \)[/tex] touchdowns, [tex]\( 3 \)[/tex] extra points, and [tex]\( 2 \)[/tex] field goals.