a There are T touchdowns, and each one is awarded 6 points. After each touchdown, they can try for p extra points.
B
b Set the expression equal to the total points scored by the winning team.
C
c There is only one possible amount of touchdowns and extra points possible.
A
a 6T+p
B
b 6T+p=27
C
c In total, 4 touchdowns and 4 points for both teams.
Practice makes perfect
a If one touchdown is awarded with 6 points, the points from T touchdowns can be expressed as 6T. The team can also score p points worth 1 point each which gives the team p extra points. Adding these, we get an expression for the number of points scored during a game.
6T+p
b 6T+p is the expression that describes the number of points a team can score during a game. We know that the winning team scored 27 points, so the number of touchdowns and the number of extra points has to satisfy the following equation.
6T+p=27
c The number of touchdowns and the number of extra points that both teams scored in total has to satisfy the following equation.
6T+p=21+7
We can solve for T and p using a table of values. Remember that p cannot be greater than T, because the teams can only score the extra points after scoring touchdowns. Both T and p must be whole numbers, and neither T nor p can be negative numbers.
T
6T+p=28
Simplify
p=?
1
6( 1)+p=28
6+p=28
p=22
2
6( 2)+p=28
12+p=28
p=16
3
6( 3)+p=28
18+p=28
p=10
4
6( 4)+p=28
24+p=28
p=4
5
6( 5)+p=28
30+p=28
p=- 2
We can see that T must be 4, because it is the only value of T that is both greater than or equal to p while p is still positive.
