[matticus0123456]

so I got this problem for school and I'm working on it but I think I set it up wrong..

In Super Bowl I, on January 15, 1967, the Green Bay Packers defeated the Kansas City Chiefs by a score of 35 to 10. The total points scored came from 13 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 extra point kicks were scored. How many touchdowns, extra-point kicks, and field goals were scored during the game?

that is the problem and this is what I did so far...

45= 6a+3b+c

because 45 is the total amount of points scored between the teams, 6 is the amount awarded for a touchdown, 3 is the amount awarded for a field goal and 1 is the amount awarded for a field goal.

the second equation was 2=a+b+c. I got that because there are two teams total and they are both contributing an unknown amount of each one.

and the third equation is a=b because the amounts have to be even...
am I on the right track here.. I'm kind of confused :-(

[engee182]

You've got your first equation correct, 45 is the total number of points. However, the question never asked anything about the number of teams. They said the number of plays.

Therefore your second equation would have to be 13 = a + b + c.

After your second equation, I think you mixed up your third equation. You said that in your question the number of touchdowns and extra-point kicks were even; not field-goals.

So your third equation would have to be a = c; not a = b.

After working on the equation, my answer came out to be:

a = 6, b = 1, c = 6

Hope this helps.