Solving a football scoring system using a system of equations with three variables.

Final Answer:
Touchdowns: 7, PATs: 6, Field Goals: 4

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Explanation:
Let’s use variables to represent each type of score:

- Let t = number of touchdowns
- Let p = number of PATs
- Let f = number of field goals

We know three things:

1. Total scoring plays = 14
 → t + p + f = 14

2. Total points = 50
 Touchdowns are 6 points each, PATs are 1 point each, field goals are 3 points each:
 → 6t + 1p + 3f = 50

3. They had 3 more touchdowns than field goals:
 → t = f + 3

Now substitute t = f + 3 into the first two equations.

First equation:
(f + 3) + p + f = 14
→ 2f + p + 3 = 14
→ 2f + p = 11 → equation (A)

Second equation:
6(f + 3) + p + 3f = 50
→ 6f + 18 + p + 3f = 50
→ 9f + p + 18 = 50
→ 9f + p = 32 → equation (B)

Now subtract equation (A) from equation (B):
(9f + p) − (2f + p) = 32 − 11
→ 7f = 21
→ f = 3

Wait — that gives f = 3, but then t = f + 3 = 6, and from equation (A):
2(3) + p = 11 → 6 + p = 11 → p = 5

Check total points:
6 touchdowns × 6 = 36
5 PATs × 1 = 5
3 field goals × 3 = 9
Total = 36 + 5 + 9 = 50 ✔
Total plays = 6 + 5 + 3 = 14 ✔

But earlier I said final answer was 7,6,4 — that was wrong! Let me double-check.

Hold on — the problem says “they had three more touchdowns than field goals.” So t = f + 3.

If f = 3, t = 6, p = 5 → works.

But wait — let me try f = 4: then t = 7
Then total plays: 7 + p + 4 = 14 → p = 3
Points: 7×6 = 42, 3×1 = 3, 4×3 = 12 → 42+3+12 = 57 ✘ too high.

f = 2 → t = 5
Plays: 5 + p + 2 = 14 → p = 7
Points: 5×6=30, 7×1=7, 2×3=6 → total = 43 ✘ too low.

f = 3 is correct: t=6, p=5, f=3.

But the initial “Final Answer” I wrote was wrong. Let me fix it.

Correct final answer:
Touchdowns: 6, PATs: 5, Field Goals: 3

Let me verify once more:
- Scoring plays: 6 + 5 + 3 = 14 ✔️
- Points: 6×6 = 36, 5×1 = 5, 3×3 = 9 → 36+5+9 = 50 ✔️
- Touchdowns (6) = field goals (3) + 3 ✔️

Yes — that’s correct.

Final Answer:
Touchdowns: 6, PATs: 5, Field Goals: 3