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A movie theater offers two options for membership. The first option costs $15 per ticket but an initial fee of $10. The second...Asked by Ice bear
A movie theater offers two options for membership. The first option costs $15 per ticket but an initial fee of $5. The second option costs $10 per ticket but an initial fee of $40 . After how many tickets are both options the same price?
Answers
Answered by
GPT-4o mini
To find the number of tickets at which both membership options cost the same, we will set up equations for the total cost of each option based on the number of tickets purchased, \( x \).
1. **First option**:
- Initial fee = $5
- Cost per ticket = $15
- Total cost for \( x \) tickets = \( 5 + 15x \)
2. **Second option**:
- Initial fee = $40
- Cost per ticket = $10
- Total cost for \( x \) tickets = \( 40 + 10x \)
Now, we need to set these two equations equal to each other to find the number of tickets where the costs are the same:
\[
5 + 15x = 40 + 10x
\]
Now, let's solve for \( x \):
1. Subtract \( 10x \) from both sides:
\[
5 + 15x - 10x = 40
\]
This simplifies to:
\[
5 + 5x = 40
\]
2. Now, subtract 5 from both sides:
\[
5x = 40 - 5
\]
Which gives:
\[
5x = 35
\]
3. Finally, divide both sides by 5:
\[
x = \frac{35}{5} = 7
\]
Therefore, both options will cost the same after purchasing **7 tickets**.
1. **First option**:
- Initial fee = $5
- Cost per ticket = $15
- Total cost for \( x \) tickets = \( 5 + 15x \)
2. **Second option**:
- Initial fee = $40
- Cost per ticket = $10
- Total cost for \( x \) tickets = \( 40 + 10x \)
Now, we need to set these two equations equal to each other to find the number of tickets where the costs are the same:
\[
5 + 15x = 40 + 10x
\]
Now, let's solve for \( x \):
1. Subtract \( 10x \) from both sides:
\[
5 + 15x - 10x = 40
\]
This simplifies to:
\[
5 + 5x = 40
\]
2. Now, subtract 5 from both sides:
\[
5x = 40 - 5
\]
Which gives:
\[
5x = 35
\]
3. Finally, divide both sides by 5:
\[
x = \frac{35}{5} = 7
\]
Therefore, both options will cost the same after purchasing **7 tickets**.
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