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A school play ran for two nights, with audiences totaling 1390 adults and students. They paid 4285 dollars for admission. One adult ticket cost 4 dollars and one student ticket cost 2.50 dollars. The ratio of adults to students was 3:5 for the first night and 2:3 on the second night. How many students attended each night?

I have determined that there was 850 students and 540 adults on both nights, but for the life of me couldn't figure out how to incoporate those ratios into finding how much students were per day.

This is marked as a homework question so please don't give me the full answer but how to go about it. I also have the answers if that is any help: for the first night, there were 400 students on the first night, and 450 on the second.

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I have verified that the adult/student totals you got is correct.

For the second part:

Suppose a total of $x$ people (i.e. including students and adults) attended the first day and total $y$ people attended the second day.

What is the total number of students in terms of $x$ and $y$?

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    One more hint please? It seems to me that I'm back where I started if I do that.2010-10-05
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    @high: Do the same with number of adults. You have 2 variables, 2 equations. I believe you already know how to solve that.2010-10-05
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    Still stuck. I figured out that on the first night, there would be 240 adults and then 300 on the second day, but I'm still unsure how it all fits together...2010-10-05
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    So if you have the number of adults each night (which I have not verified) the ratios give the number of students2010-10-05
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    @high: On the first day, for every 3 adults, there were 5 students. So for 240 adults there must be how many students?2010-10-05
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    @Moron- 400. However, the only reason I knew there was 240 adults was by 'reverse-engineering' the answer I was given. I'm still not sure how to reach 240 adults though!2010-10-05
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    @user1827- which is 3:5. How do I take this ratio and find the number of adults in the first place though?2010-10-05
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    @high: Did you understand my comment about two equations two variables (x and y)? If x is total number of people on day 1, how many of them are students, given the ratio 3:5?2010-10-05
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    @Moron- Unfortunately, no. EDIT: 0.6?2010-10-05
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    @high: Not 0.6. It is 5x/8. Similarly you can find the number of students on the second day in terms of y. Since you know the total number of students you get 5x/8 + something in terms of y = 850. You get a similar equation for adults. Is it clearer?2010-10-05
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    @Moron- Maybe I'm tired, but how did you get 5x/8 from 3:5?2010-10-05
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    @high: If there were 8 people, there would be 3 adults and 5 students, so if there were x people, how many adults and how many students?2010-10-05
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    @Moron- Would I be correct saying it is `5x/8 + 3y/5`?2010-10-05
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    @high: Correct. 5x/8 + 3y/5 = 850. Now form another equation for adults, and solve for x and y.2010-10-05
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    @Moron-So I formed 3x/8+2y/5. Is this still correct?2010-10-05
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    @high: Yes, looks right.2010-10-05
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    @Moron- So if I solve `5x/8+3y/5=850`, I get x=8, y=-25. Then `3x/8+2y/5=540` gives me x=16, y=-15. What now?2010-10-05
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    @high: Can you find _one_ value of x and _one_ value of y so that both the equations are true?2010-10-05
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    @Moron- Would x=640, y=750 be correct?2010-10-05
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    @high: Looks right (you can verify it yourself by substituting it back in those two equations). So now you know how many total people attended on day1 and how many on day2.2010-10-05
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    @Moron- So how do I incorporate the ratios into the numbers?2010-10-05
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    @high: You already did when you formed the equation 5x/8+3y/8 = 850 and also in the other one. Wasn't 5x/8 the number of students on day1?2010-10-05
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    @Moron- That was so obvious I can't believe I didn't see that. I keep thinking about the ratios.... Well, thanks, I got the answer! Thanks for wasting your time with me :)2010-10-05
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    @high: No worries. Glad to have helped.2010-10-05