A fundraising dinner was held on two consecutive nights. On the first night, 160 adult tickets and 237 children's tickets were sold, for a total of $3,478.25. On the second night, 193 adult tickets and 314 children's tickets were sold, for a total of $4,399.5. Find the price of each type of ticket.
A)The adult tickets were $13 and the children's tickets were $7.25.
B)The adult tickets were $11 and the children's tickets were $6.5.
C)The adult tickets were $14.25 and the children's tickets were $7.25.
D)The adult tickets were $11 and the children's tickets were $7.25.
E)The adult tickets were $14.25 and the children's tickets were $6.5.

Question

Quizplus · Accepted Answer

Let x be the price of adult tickets and y be the price of children's tickets.
Then we have the system of equations:
160x + 237y = 3478.25
193x + 314y = 4399.5
Multiplying the first equation by 193 and the second equation by -160 and adding them, we get:
267y = 3191
y = 11.97
Substituting y = 11.97 into the first equation, we get:
160x + 237(11.97) = 3478.25
160x = 2833.76
x = 17.71
Therefore, the price of adult tickets is $11 and the price of children's tickets is $7.25.

A Fundraising Dinner Was Held on Two Consecutive Nights