set up and solve a system of equations to answer the questions asked

Question

Callie’s school is selling tickets to a choral performance. On the first day of ticket sales the school sold 2 adult tickets and 2 student tickets for a total of $31.40. The school took in $189.80 on the second day by selling 14 adult tickets and 8 student tickets. Find the price of each kind of ticket.

Answer 1

Cory

 

You can solve this by setting up two equations with two unknown values and then solving for those values.

 

We know that the number of each ticket sold on any particular day times the value of that ticket when added up for both tickets should equal the total amount of money taken in on that day.

 

Let us say that a student ticket costs x dollars and an adult ticket costs y dollars. 

 

Then for the first day:

 

(2)(x) + (2)(y) = $31.40

 

And for the second day: 

 

(8)(x) + (14)(y) = $189.80

 

We can solve for x in the first equation above and then plug it into the second equation in order to solve for y.

Once we know the value of y, we can then solve for x.

 

Rearranging the first equation, we get:  

2x = 31.40 - 2y

x = 15.70 - y

 

Plugging that value of x into the second equation, we get:

(8)(15.70 - y) + 14y = 189.80

125.6 - 8y + 14y = 189.80

6y = 189.80 - 125.6 = 64.2

y = 10.70

 

We know that x = 15.70 - y (from above) so:

x = 15.70 - 10.70

x = 5.00

 

So the student tickets sold for $5.00 each and the adult tickets sold for $10.70 each.