How to solve: A movie theater has a seating capacity of 355. The theater charges $5.00 for children, $7.00 for students, and $12.00 of adults.

Question

A movie theater has a seating capacity of 355. The theater charges $5.00 for children, $7.00 for students, and $12.00 of adults. There are half as many adults as there are children. If the total ticket sales was $ 2578, How many children, students, and adults attended?

Answer 1

Break this down - read through it fully, and find out from the last sentence what you are asked to find - the NUMBER of (C)hildren, (S)tudents, and (A)dults.

For 3 variables (C, S, A) you usually need 3 conditions to solve to a unique answer (this is a system of equations). I see the following:

1) A movie theater has a seating capacity of 355

2) There are half as many adults as there are children

3) The total ticket sales was $ 2578 - uses the cost per type of ticket

From (1) - seating capacity is a maximum for the total number of individuals, So C + S + A < 355

From (2) - you can see that there are MORE children than adults - actually twice as much. So 2A = C (which is easier to work with than the equivalent equation 1/2C = A). Use real numbers to make sure this equation makes sense - otherwise you are using nonsense moving forward.

From (3) and the cost per type of ticket: The theater charges $5.00 for children, $7.00 for students, and $12.00 of adults. NOTE THAT 5*C is the ticket sales collected for the C children tickets sold. EX- if 10 children tickets are sold, we collect $50. Putting together all sales, 5C + 7S + 12A = 2578

To solve the 3x3 system, Start with substituting C for 2A in the other 2 equations. This will reduce it down to the variables A and S in the equations

4) 3A + S < 355

5) 22A + 7S = 2578

Use elimination by multiplying equation (4) by -7 and add the two equations. The answer is:

A = 93

C = 186

S = 76

I usually find that students have the most trouble with converting the word problem into math, so that's what I concentrated on.

When it comes to 3x3 systems, there is not one method to attack these problems

a) Your first goal is to reduce it to 2x2 by eliminating 1 of the variables with 2 equations (they are now new equations that are a combination from 2 of the 3 original equations).

b) Stay focused on what variable you are eliminating (I eliminated C by substituting twice in the remaining 2 equations).

c) Keep practicing to get sharp on the skills you learned in 2x2 systems.