Tickets for the homecoming dance cost $20 for a single ticket or $35 for a couple. Ticket sales totaled $2280, and 128 people attended. How many tickets of each type were sold?

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Tickets for the homecoming dance cost $20 for a single ticket or $35 for a couple. Ticket sales totaled $2280, and 128 people attended. How many tickets of each type were sold?

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Answer 1 · 2018-03-20T21:29:26

$16$ $16$ singles, $56$ $56$ couples

Explanation:

There's two linear equations that we can make: one for money and one for people.

Let the number of single tickets be $s$ $s$ and the number of couple tickets be $c$ $c$ .

We know that the amount of money we make is
$$ = 20 s + 35 c = 2280$ $$ = 20 s + 35 c = 2280$

We also how many people can come
$P = 1 s + 2 c = 128$ $P = 1 s + 2 c = 128$

We know that both $s$ $s$ are the same and both $c$ $c$ are the same. We have two unknowns and two equations, so we can do some algebra to solve for each.

Take the first minus twenty times the second:
$20 s + 35 c = 2280$ $20 s + 35 c = 2280$
$- 20 s - 40 c = - 2560$ $-20 s - 40 c = -2560$
$- 5 c = - 280 \Rightarrow c = 56$ $-5c = -280 implies c = 56$

Plugging this back into the second equation,
$s + 2 c = s + 2 \cdot 56 = s + 112 = 128 \Rightarrow s = 16$ $s + 2c = s + 2 * 56 = s + 112 = 128 implies s = 16$

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Tickets for the homecoming dance cost $20 for a single ticket or $35 for a couple. Ticket sales totaled $2280, and 128 people attended. How many tickets of each type were sold?

1 Answer

Explanation:

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