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 252. How many wins did each team have in the 2010 season?

83,84 and 85

Step-by-step explanation:

Let the number of win for the teams be x, x+1 and x+2 respectively. From the question, we are told that the total number of wins by these three teams was 252 for the 2010 season. Therefore,

x+x+1+x+2 = 252

3x+3 = 252

3x = 252 - 3

3x = 249

x = 249/3

x = 83

This means that the team with least wins has 83 while the second team has 84 and the team with most wins has 85.

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1

c

step-by-step explanation: