Question

Player Aled a baseball league in runs batted in for the 2008 regular season Player B, who came in second to player A, had 18 fewer runs batted in for the 2008 regular season Together, these two players

brought home 230 runs during the 2008 regular season. How many runs bafted in did player A and player Beach account for?
Player A had _______runs and player 6 had ______runs batted in for the 2008 regular season

Answer 1

Answer: Let's call the number of runs batted in by Player A "RA" and the number of runs batted in by Player B "RB". We can set up two equations based on the given information:

RA = RB + 18 (Player A had 18 more runs batted in than Player B)

RA + RB = 230 (Together, the two players brought home 230 runs)

We can use the first equation to substitute for RA in the second equation:

(RB + 18) + RB = 230

Simplifying the equation:

2RB + 18 = 230

Subtracting 18 from both sides:

2RB = 212

Dividing both sides by 2:

RB = 106

So Player B had 106 runs batted in. To find out how many runs Player A had, we can substitute into the first equation:

RA = RB + 18

RA = 106 + 18

RA = 124

Therefore, Player A had 124 runs batted in and Player B had 106 runs batted in, and together they accounted for 230 runs batted in during the 2008 regular season.

Explanation: