Question

During the 2010 baseball​ season, the number of wins for three teams was three consecutive integers. Of these three​ teams, the first team had the most wins. The last team had the least wins. The total number of wins by these three teams was 243. How many wins did each team have in the 2010​ season?

Answer 1

The number of wins for the three teams in the 2010 baseball season were 80, 81, and 82, respectively.

Step-by-step explanation:

Let's represent the number of wins for the three teams as consecutive integers. Let's call the number of wins for the first team x, the second team x+1, and the third team x+2. According to the given information, x is the largest number, so the first team had the most wins and the third team had the least wins. We can express the total number of wins as the sum of the wins for each team:

x + (x+1) + (x+2) = 243

Combining like terms, we get:

3x + 3 = 243

Subtracting 3 from both sides, we get:

3x = 240

Dividing both sides by 3, we get:

x = 80

Therefore, the wins for the three teams in the 2010 season were 80, 81, and 82, respectively.

Answer 2

Step-by-step explanation:

Let x represent the number of wins of the least team.

The number of wins for three teams was three consecutive integers. Of these three​ teams, if the first team had the most wins and the last team had the least wins. It means that the number of wins of the second team would be x - 1

The number if wins on the last team would be x - 2

The total number of wins by these three teams was 243. It means that

x + x - 1 + x - 2 = 243

3x - 3 = 243

3x = 243 + 3 = 246

x = 246/3 = 82

The first team had 82 wins

The second team had 82 - 1 = 81 wins

The last team had 82 - 2 = 80 wins