Question

Player A led a baseball league in runs batted in for the 2006 regular season. Player B, who came in second to player A, had 16 fewer runs batted in for the 2006 regular season. Together, these two players brought home 228 runs during the 2006 regular season. how many runs batted in did player A and player B each account for?

Answer 1

Player A led the league in runs batted in, and Player B came in second with 16 fewer runs batted in. Let's denote the number of runs batted in by Player A as "A" and by Player B as "B."

According to the information given, we can write two equations:

1. A = B + 16 (Player A had 16 more runs batted in than Player B).

2. A + B = 228 (Together, they brought home 228 runs).

Now, we can solve this system of equations simultaneously. Substitute the value of A from the first equation into the second equation:

(B + 16) + B = 228

Combine like terms:

2B + 16 = 228

Subtract 16 from both sides:

2B = 228 - 16

2B = 212

Now, divide by 2:

B = 212 / 2

B = 106

So, Player B accounted for 106 runs batted in during the 2006 regular season.

Now that we know B, we can find A using the first equation:

A = B + 16

A = 106 + 16

A = 122

Player A accounted for 122 runs batted in during the 2006 regular season.

To summarize:

- Player A accounted for 122 runs batted in.

- Player B accounted for 106 runs batted in.