Adult tickets for a play cost $17 and child tickets cost $2.?

Question

Adult tickets for a play cost $17 and child tickets cost $2. If there were 32 people at a performance and the theater collected $364 from ticket sales, how many adults and how many children attended the play?

Answer 1

Let a = adults and c = children.
a + c = 32, so a = 32-c

17a + 2c = 364
Substitue 32-c for a:
17(32-c) + 2c = 364
Use the distributive property for the binomial in parentheses:
544-17c+2c = 364
Combine the c's:
544-15c = 364
Subtract 544 from both sides:
-15c = -180
Divide both sides by -15:
c = 12 children
Plug this into the first equation to find the # of adults:
a + c = 32
a + 12 = 32
Subtract 12 from both sides:
a = 20 adults

So, there were 12 children and 20 adults.

Answer 2

There were 20 adults and 12 children. I got this answer by setting up the system of equations:
{17a + 2c = 364
a + c = 32
Then I used the substitution method to solve for a adults, and c children.

Answer 3

a= number of adult tickets sold
c= number of child tickets sold

$17a + $2c = $364
a + c = 32

Now, in order to solve for one variable, you'll need to multiply the second equation by two. That way, when you subtract the two equations, the c will equal to zero.
So:

17a + 2c = 364 (nothing changed here)
2a + 2c = 64 (everything multiplied by two)

(subtract)

15a = 300
a = 20

Plug it back into one of the equations, and.....

20 + c = 32
c = 12

To double-check, plug the two numbers into the other equation:
17 (20) + 2 (12) =364
340 + 24 =364
364=364

(Yay! it's right)

Therefore, 20 adults and 12 children attended the play.

Answer 4

x+y=32
17x+2y=364

y=32-x

17x+2(32-x)=364
17x+64-2x=364
15x+64-64=364-64
15x=300
x=20

20+y=32
y=12

20 adults and 12 children

Answer 5

possibly 20 adults and 12 kids

Answer 6

It's something like 20 adults and12 children.