two groups of volunteers are cleaning up the football stadium after the homecoming game. Volunteers from the band booster club have already cleaned six rows of bleachers and will continue to clean at a rate of one row per minute. The leadership class has completed three rows and will continue working at two rows per minute. Once the two groups get to the point where they have cleaned the same number of Rose they will take a break and decide how to split up the remaining work. How many rows will each group have cleaned by them?

Let's define the problem using variables and then set up a system of equations.

Let:
- \( t \) = time in minutes after the homecoming game
- \( B(t) \) = number of rows cleaned by the band booster club after \( t \) minutes
- \( L(t) \) = number of rows cleaned by the leadership class after \( t \) minutes

From the information provided:
1. The band booster club has already cleaned 6 rows and continues at a rate of 1 row per minute:
\[
B(t) = 6 + t
\]

2. The leadership class has already cleaned 3 rows and continues at a rate of 2 rows per minute:
\[
L(t) = 3 + 2t
\]

We want to find when the number of rows cleaned by both groups is the same, which can be represented by the equation:
\[
B(t) = L(t)
\]

Substituting in the equations for \( B(t) \) and \( L(t) \):
\[
6 + t = 3 + 2t
\]

Now, let's solve this equation for \( t \).

### Solve for \( t \):
1. Rearranging the equation gives:
\[
6 - 3 = 2t - t \\
3 = t
\]

### Find the number of rows cleaned by each group at \( t = 3 \) minutes:
- For the band booster club:
\[
B(3) = 6 + 3 = 9 \text{ rows}
\]

- For the leadership class:
\[
L(3) = 3 + 2(3) = 3 + 6 = 9 \text{ rows}
\]

Thus, at \( t = 3 \) minutes, both groups will have cleaned 9 rows.

### Summary:
- The system of equations is:
\[
\begin{align*}
B(t) & = 6 + t \\
L(t) & = 3 + 2t
\end{align*}
\]
- The solution is:
- Number of rows cleaned by each group: **9 rows** each.

### Graphing:
You can graph both equations by plotting several points and drawing lines through them:

1. For \( B(t) = 6 + t \):
- At \( t = 0, B(0) = 6 \) (point: (0, 6))
- At \( t = 3, B(3) = 9 \) (point: (3, 9))
- At \( t = 6, B(6) = 12 \) (point: (6, 12))

2. For \( L(t) = 3 + 2t \):
- At \( t = 0, L(0) = 3 \) (point: (0, 3))
- At \( t = 3, L(3) = 9 \) (point: (3, 9))
- At \( t = 6, L(6) = 15 \) (point: (6, 15))

You would see the two lines intersecting at the point (3, 9), confirming the solution.