Janelle and her friends are selling bracelets and headbands ahead of the school pep rally to support the team during homecoming. They figured they would sell the headbands for $4 and the bracelets for $2. They’d like to make more than $200 and sell no more than 100 items.

To help Janelle and her friends with their fundraising project, we can set up some equations based on the information provided.

Let:
- \( x \) = the number of headbands sold
- \( y \) = the number of bracelets sold

The pricing is as follows:
- Headbands sell for $4 each
- Bracelets sell for $2 each

They have two main conditions:
1. They want to make more than $200.
2. They want to sell no more than 100 items.

We can write these conditions mathematically:

1. **Revenue Condition**:
\[
4x + 2y > 200
\]

2. **Quantity Condition**:
\[
x + y \leq 100
\]

### Solving the inequalities:

We can rearrange the revenue condition to find the minimum number of items they need to sell:

\[
4x + 2y > 200
\]

Dividing the entire inequality by 2 gives us:

\[
2x + y > 100
\]

So now we have two inequalities:

1. \( 2x + y > 100 \)
2. \( x + y \leq 100 \)

### Graphing the Constraints:

To visualize these inequalities, we can graph them on a coordinate plane with \( x \) (headbands) on the x-axis and \( y \) (bracelets) on the y-axis.

1. For \( x + y = 100 \):
- If \( x = 0 \), then \( y = 100 \) (the point (0, 100)).
- If \( y = 0 \), then \( x = 100 \) (the point (100, 0)).

This line creates a boundary, and the area below this line (including the line itself) satisfies the quantity condition.

2. For \( 2x + y = 100 \):
- If \( x = 0 \), then \( y = 100 \) (the point (0, 100)).
- If \( y = 0 \), then \( x = 50 \) (the point (50, 0)).

This line also creates a boundary, and the area above this line (not including the line itself) satisfies the revenue condition.

### Feasible Region:

The feasible region will be where the area that meets both conditions intersects:

- Above the line \( 2x + y = 100 \) (but not including the line).
- Below the line \( x + y = 100 \) (including the line).

### Finding Integer Solutions:

Now we need to consider integer values for \( x \) and \( y \) that fall within this feasible area. Assuming they want to maximize their earnings, they can try combinations of \( x \) and \( y \) that satisfy both conditions:

A few potential pairs that work:

1. \( x = 40 \), \( y = 60 \):
- \( x + y = 100 \) (satisfies quantity condition)
- \( 2(40) + 60 = 80 + 60 = 140 > 100\) (satisfies revenue condition)

2. \( x = 45 \), \( y = 55 \):
- \( x + y = 100 \) (satisfies quantity condition)
- \( 2(45) + 55 = 90 + 55 = 145 > 100 \) (satisfies revenue condition)

3. \( x = 50 \), \( y = 50 \):
- \( x + y = 100 \) (satisfies quantity condition)
- \( 2(50) + 50 = 100 + 50 = 150 > 100 \) (satisfies revenue condition)

By choosing combinations of \( x \) and \( y \) based on these equations, Janelle and her friends can find a suitable number of headbands and bracelets to sell to meet their fundraising goals!