SOLUTION: a minor league baseball team plays 98 games in a season. if the team won 14 more than three times as many games they lost, how many wins and losses did the team have?

Question 462174: a minor league baseball team plays 98 games in a season. if the team won 14 more than three times as many games they lost, how many wins and losses did the team have?
Found 2 solutions by algebrahouse.com, graphmatics:
Answer by algebrahouse.com(1659) (Show Source): You can put this solution on YOUR website!
a minor league baseball team plays 98 games in a season. if the team won 14 more than three times as many games they lost, how many wins and losses did the team have?

x = number of losses
3x + 14 = number of wins

x + 3x + 14 = 98 {number of wins plus number of losses equals total of 98}
4x + 14 = 98 {combined like terms}
4x = 84 {subtracted 14 from both sides}
x = 21 {divided both sides by 4}
3x + 14 = 77 {substituted 21, in for x, into 3x + 14}

21 losses and 77 wins
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Answer by graphmatics(170) (Show Source): You can put this solution on YOUR website!
Let x be the number of games the team won. Let y be the number of games the team lost. Then x+y = 98
Because the team won 14 more than three times as many games they lost
x = 14 + 3*y
Substitute the value of x from the second equation into the first equation to get

(14 + 3*y) + y = 98
4*y = 98 -14
4*y = 84
y = 84/4
y = 21
so to get x we substitute y = 21 into the first equation to get
x + 21 = 98
x = 98 -21
x = 77
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