A minor league baseball team plays 11 games in a season. If the team won 15 more than three times as many games as they lost. How many wins and losses did the team have?

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Answer 1 · 2017-09-18T06:59:49

5 wins and 10 losses

Let the number of wins be $x$ and the number of losses be $y$ .

Assuming there were no ties,the total number of games played by team would be equal to sum of number of losses and number of wins.

Therefore, (1) $x+y=15$
Also,the team won 15 more than three times as many games as they lost.
Thus,(2) $3x=15+3y$

Equations (1) and (2) are linear equations in variables $x$ and $y$
and can be solved by method of substitution.

From (1) , $x=15-y$

Substituting $x=15-y$ in (2),
$3(15-y)=15+3y$
$45-3y=15+3y$
$45-15=3y+3y$
$30=6y$
Therefore, $y=5$

From (1), $x=15-y$
Substituting $y=5$ in above equation,
$x=15-5$
Thus, $x=10$

Thus, the team had 10 wins and 5 losses.