In a basketball game, team a defeated team b with a score of 97 to 63. team a won by scoring a combination of two-point baskets, three-point baskets, and one-point free throws. the number of two-point baskets was 11 more than the number of free throws. the number of free throws was three less than the number of three-point baskets. what combination of scoring accounted for the team a's 97 points?

14 free throw baskets , 25 two point baskets and 11 three point baskets

Step-by-step explanation:

Let n₁ represent the number of free-throw baskets, n₂ represent the number of two point baskets and n₃ represent the number of three point baskets.

Now, from the question, the number of two point baskets, n₂ is greater than the free throw baskets by 11. This is written as n₂ = n₁ + 11. Also, the number of three point baskets n₃ is three less than the number of free point baskets. This is written as n₃ = n₂ - 3. Since our total number of points equals 97, it follows that, sum of number of points multiplied by each point equals 97. So, ∑(number of points × each point) = 97. Thus,

n₁ + 2n₂ + 3n₃ = 97. Substituting n₂ and n₃ from above, we have n₁ +2(n₁ + 11) + 3(n₁ - 3) = 97.

Expanding the brackets, we have, n₁ + 2n₁ + 22 + 3n₁ - 9 = 97

collecting like terms, we have 6n₁ + 13 = 97

6n₁ = 97 - 13

6n₁ = 84

dividing through by n₁ we have, n₁ = 84/6 =14

so n₁ our free throw baskets equals 14. Substituting this into n₂ our number of two point baskets equals n₂ = n₁ + 11 = 14 + 11 = 25. Our number of three point baskets n₃ = n₁ - 3. So, n₃ = 14 -3 = 11.

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