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Question

In a baseball league of ten teams, Joseph's team is currently in the first place with thirty-five wins. Benjamin's team is eight games behind Joseph's team. Sydney's team is three games behind Benjamin's team. In the next forty games, Benjamin's team won four less than the number of games won by Joseph's team. Sydney's team won fourteen less than the number of games won by Benjamin's team. Three times the number of games won by Joseph's team, during these forty games, was eighty-eight more than the number of games won by Sydney's. The sum of the games won by these three teams after the complete season is 169. How many games did each of these three teams win?

Answer 1

Joseph's team won\35+35 = 70 games
Benjamin's team won 27+31 = 58 games
Sydney's team won 24+17 = 41 games

good luck

Answer 2

Initially:
Let J0 be the number of games Joseph's team has won.
Let B0 be the number of games Benjamin's team has won.
Let S0 be the number of games Sydney's team has won.

Then
J0=35
B0=J0-8=27
S0=B0-3=24

During the next 40 games:
Let J be the number of games Joseph's team has won.
Let B be the number of games Benjamin's team has won.
Let S be the number of games Sydney's team has won.

We have these equations:
B=J-4
S=B-14
3J=S+88

And, after the complete season:
(B+B0)+(J+J0)+(S+S0)=169

First, subtract off the current number of wins each team has from 169, and get
B+J+S = 169-(B0+J0+S0) = 169-(35+27+24) = 169-86 = 83.

So now we must solve this system of equations:
B=J-4
S=B-14
3J=S+88
B+J+S=83

Replace B with J-4:
S=J-18
3J-S=88
2J+S=87

Add the last two equations:
5J=175, so J=35. That means S=J-18=17 and B=J-4=31.

For the complete season:
Joseph's team won J0+J = 35+35 = 70 games
Benjamin's team won B0+B = 27+31 = 58 games
Sydney's team won S0+S = 24+17 = 41 games