To determine the break-even point between the two venues based on the number of attendees, let's define the costs for each venue as follows:
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Cost of the hotel (H): \[ H = 500 + 8x \] where \( x \) is the number of attendees.
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Cost of the restaurant (R): \[ R = 600 + 6x \]
Finding the Break-even Point
To find how many attendees it would take for both venues to cost the same, we can set the two equations equal to each other:
\[ 500 + 8x = 600 + 6x \]
Solving for \( x \)
- Rearranging the equation: \[ 8x - 6x = 600 - 500 \] \[ 2x = 100 \] \[ x = 50 \]
So, it takes 50 attendees for the costs of the hotel and restaurant to be the same.
Setting Up Inequalities
Next, to create a system of inequalities to determine which venue is more cost-effective for different numbers of attendees, we can express the inequalities based on comparative costs:
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For the hotel to be cheaper than the restaurant: \[ 500 + 8x < 600 + 6x \]
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For the restaurant to be cheaper than the hotel: \[ 600 + 6x < 500 + 8x \]
Graphing the Inequalities
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Hotel cheaper than Restaurant: Rearrange the first inequality: \[ 8x - 6x < 600 - 500 \] \[ 2x < 100 \implies x < 50 \]
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Restaurant cheaper than Hotel: Rearrange the second inequality: \[ 6x - 8x < 500 - 600 \] \[ -2x < -100 \implies x > 50 \]
Summary of the Inequalities
- The hotel is cheaper when \( x < 50 \).
- The restaurant is cheaper when \( x > 50 \).
Graphing the Inequalities
You can graph these two inequalities on a number line:
- From 0 to 50, you shade the region indicating the hotel is cheaper (below the line).
- After 50, you shade the region for the restaurant being cheaper (above the line).
Conclusion
- At \( x = 50 \), both venues will cost the same.
- For fewer than 50 attendees, the hotel is the more cost-effective choice.
- For more than 50 attendees, the restaurant becomes the cheaper option.