A minor league baseball team plays 74 games in a season. If the team won 14 more than twice as many games as they lost, how many wins and losses did the team have?
Let x stand for the losses.
x + (2x + 14) = 74
20
Right. The team lost 20 games. How many did it win?
54 wins?
Right. :-)
If a minor league baseball team plays 81 games in a season. If they won 15 more than twice as many games as they lost, how many wins and losses did the team have.
To solve this problem, we can use algebra. Let's assume the number of games the team lost is "x".
If the team won 14 more than twice as many games as they lost, then the number of games they won would be (2x + 14).
The total number of games played in a season is 74. So we can write the equation:
Number of games won + Number of games lost = Total number of games
(2x + 14) + x = 74
Simplifying the equation:
3x + 14 = 74
Subtracting 14 from both sides:
3x = 60
Dividing both sides by 3:
x = 20
So the team lost 20 games.
To find the number of wins, we substitute the value of x back into the equation:
Number of games won = 2x + 14
Number of games won = 2(20) + 14
Number of games won = 40 + 14
Number of games won = 54
Therefore, the team had 54 wins and 20 losses.