System of Equations

Question

A high school sold out their 600 seat auditorium for their play. The prices for children were $2, students $3, and adults $7 for a total revenue of $2765. Next year they plan to raise the prices to $2.50 per child, $3.50 per student, and $8 per adult. If they sell the same number of each type of ticket, then the revenue will be $3201. How many of each type of ticket did they sell?

Answer 1

First, identify the unknowns.

-Set x as the number of tickets sold for children;

Set y as the number of tickets sold for students;

Set z as the number of tickets sold for adults.

Second, derive a equation for the problem statement.

Revenue=Number of tickets sold*Price, which is the sum of all three groups

There are 2 situations, for the first situation, the revenue equation is

Revenue 1=$2x+$3y+$7z=$2765

For the second situation, it is

Revenue 2=$2.50x+$3.50y+$8z=$3201

For both situations, they sell the same number of each type of ticket for both years, so

x+y+z=600 (This is the total number of tickets sold, with unknown number of tickets sold for children, students, and adults)

Now, we have 3 sets of equation with 3 unknowns. We are able to solve the equations.

Eq. 1: Revenue 1=$2x+$3y+$7z=$2765

Eq. 2: Revenue 2=$2.50x+$3.50y+$8z=$3201

Eq. 3: x+y+z=600

Eq. 3 can be arranged into x=600-y-z

Plug this equation into Eq. 1, then Eq. 1 becomes

Revenue 1=$2(600-y-z)+3y+7z=$2765

Solve this equation and we get

y=1565-5z

Plug this equation and x=600-y-z into Eq. 2, then Eq. 2 becomes

$2.50(600-(1565-5z)-z)+$3.50(1565-5z)+8z=$3201

Expand and solve for z

z=272

From y=1565-5z=1565-5(272)=205

From x=600-y-z=600-205-272=123

So our final answer is x=123, y=205, and z=272

Have a great night! :)