A minor league baseball team plays 89 games in a season . If the team won 17 more than twice as many Games as they lost. How many wins and loses did the team have?
Games lost ---- x
games wond = 89-x
89-x = 2x + 17
solve for
Let's break down the problem step-by-step:
Step 1: Let's assume the number of games the team lost is x.
Step 2: Since the team won 17 more than twice as many games as they lost, the number of games they won is (2x + 17).
Step 3: The total number of games played in a season is 89, so we can set up an equation: x + (2x + 17) = 89.
Step 4: Simplify the equation: 3x + 17 = 89.
Step 5: Subtract 17 from both sides: 3x = 72.
Step 6: Divide by 3: x = 24.
Therefore, the team lost 24 games (x) and won (2x + 17) = (2*24 + 17) = 65 games.
The team had 65 wins and 24 losses.
To solve this problem, we can set up an equation based on the given information. Let's assume the number of games the team lost is "x".
According to the given information, the team won 17 more than twice the number of games they lost. This can be expressed as "(2x + 17)".
Since the team played a total of 89 games, the sum of the wins and losses is equal to the total number of games played. Therefore, we can write the equation as:
Wins + Losses = Total games
(2x + 17) + x = 89
Now, let's solve this equation to find the value of "x" (number of games lost).
Combine like terms:
2x + x + 17 = 89
3x + 17 = 89
Subtract 17 from both sides:
3x = 89 - 17
3x = 72
Divide both sides by 3:
x = 72 / 3
x = 24
So, the team lost 24 games.
To find the number of wins, we substitute the value of "x" back into the equation:
Wins = 2x + 17
Wins = 2 * 24 + 17
Wins = 48 + 17
Wins = 65
Therefore, the team had 65 wins and 24 losses.