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.
- \( t \) 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.
