Question

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?

Answer 1

Answer:
Player A had 122, while player B had 110. 

Explanation:
This is solved by:
1- setting Player A = x and Player B = x -12.

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

3- solving.
x + (x - 12) = 232.
2x - 12 = 232.
2x = 244.
x = 122. 

Answer 2

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