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Sonja [21]
2 years ago
5

Player a led a baseball league in runs batted in for the 2008 regular season. player​ b, who came in second to player​ a, had 12

fewer runs batted in for the 2008 regular season.​ together, these two players brought home 232 runs during the 2008 regular season. how many runs batted in did player a and player b each account​ for?
Mathematics
2 answers:
Ksenya-84 [330]2 years ago
7 0

Answer:

Runs batted in year 2008 by player A = 122

Runs batted in year 2008 by player B = 110

Step-by-step explanation:

Runs batted in year 2008 by player A = X

Runs batted in year 2008 by player B = X – 12

The total runs batted by the player A and B in all together = 232  

Thus, X + (X -12) = 232

2X = 232 + 12  

X = 244/2 = 122

Runs batted in year 2008 by player A = 122

Runs batted in year 2008 by player B = 122 – 12 = 110

riadik2000 [5.3K]2 years ago
6 0
<span><u><em>Answer:</em></u>
Player A had 122, while player B had 110. 

<u><em>Explanation:</em></u>
<u>This is solved by:</u>
<u>1- setting Player A = x and Player B = x -12. </u>

<u>2- adding both of those, setting equal to 232:</u>
x + x - 12 = 232

<u>3- solving. </u>
x + (x - 12) = 232.
2x - 12 = 232.
2x = 244.
x = 122. </span>
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