A minor league baseball team plays 78 games in a season. If the team won 15 more than twice as many games as they lost, how many wins and losses did the team have?
If they lost x, then
x + 2x+15 = 78
Let's denote the number of losses as L.
According to the information given, the team won 15 more than twice as many games as they lost. This can be translated into the equation: W = 2L + 15, where W represents the number of wins.
We also know that the team played a total of 78 games. The total number of games played is the sum of wins and losses: W + L = 78.
Substitute the value of W from the first equation to the second equation:
(2L + 15) + L = 78
Combine like terms:
3L + 15 = 78
Subtract 15 from both sides:
3L = 63
Divide both sides by 3:
L = 21
Therefore, the team had 21 losses.
Substitute the value of L into the equation W = 2L + 15:
W = 2(21) + 15
Simplify:
W = 42 + 15
W = 57
Therefore, the team had 57 wins and 21 losses.
To solve this problem, let's set up an equation to represent the given information.
Let's assume the number of losses the team had is 'x,' and the number of wins can be represented by 'y.'
From the given information, we can gather two things:
1. The team played 78 games in a season, so the total number of wins and losses should add up to 78. We can represent this as:
x + y = 78
2. The team won 15 more than twice as many games as they lost. Mathematically, this can be represented as:
y = 2x + 15
Now, we have a system of equations, and we can solve them simultaneously to find the values of x and y.
Substitute the value of y from the second equation into the first equation:
x + (2x + 15) = 78
Combine like terms:
3x + 15 = 78
Subtract 15 from both sides of the equation:
3x = 78 - 15
3x = 63
Divide both sides of the equation by 3:
x = 63 / 3
x = 21
Now plug the value of x back into one of the earlier equations to find the value of y:
y = 2x + 15
y = 2(21) + 15
y = 42 + 15
y = 57
Therefore, the team had 21 losses and 57 wins in the season.