algebra word problem

Question

A movie theater charges $5 for adults ticket and $2 for childs ticket. One Saturday the theater sold 785 tickets for a total of $3280.00. How many children's tickets were sold for the movie that Saturday?

Answer 1

In order to answer this question, you must set up two different equations since you are dealing with a two part question.

The first equation deals with how many tickets are being sold. For this, you have two variables children tickets (x) and adult tickets (y). According to the information, there was 785 tickets sold. Therefore:

1. x (children tickets) + y (adult tickets) = 785 tickets sold
2. x + y = 785

The second equation deals with how much money is made. According to the data, the theater made a total of 3280 that Saturday. We know that a children's ticket (x) costs 2 dollars and an adult ticket (y) costs 5 dollars. Using this data, we need to set up an equation to determine cost. Therefore:

1. 2x (cost of children ticket) + 5y (cost of adult ticket) = 3280 total dollars
2. 2x + 5y = 3280

Since we now have two equations, we need to use the substitution method to solve this. In order to do that, we need solve the first equation for a variable. I tend to go with y= for my equations so that is what I am going to solve for:

1. x + y = 785 (subtract x from both sides)
2. y = 785 - x

Now I can insert this into my second equation

1. 2x + 5y = 3280 (substitute y)
2. 2x + 5 (785 - x) = 3280 (distribute 5 across the parenthesis)
3. 2x + 3925 - 5x = 3280 (combine x's)
4. -3x + 3925 = 3280 (subtract 3925 from both sides)
5. -3x = -645 (divide each side by -3 )
6. x = 215

Therefore we know that there were 215 children tickets sold. We can know put this in our equation y = 785 - x to determine the number of adult tickets.

1. y = 785 - 215
2. y = 570

Therefore, 570 adult tickets were sold. In order to check our equation we need to make sure that both numbers satisfy our initial equation

1. x + y = 785
2. 215 + 570 = 785
3. 785 = 785 *correct!

1. 2x + 5y = 3280
2. 2(215) + 5(570) = 3280
3. 430 + 2850 = 3280
4. 3280 = 3280 *correct!

In conclusion, there was a total of 215 children tickets and 570 adults tickets sold on Saturday for a total sale of $3280.

Answer 2

Good morning Gin. Here is your solution...

Since there are two variables and two conditions stated in this problem, we have a system of equations. In algebra, since we can only solve for one variable at a time, we're going to use a process called "elimination".

Setup

Let c = # of children's tix

Let a = # of adult tix

Therefore, we can set up two different equations to describe whats taking place.

First, the total number of tickets sold. A + C = 785

Second, the total amount of sales. 5A + 2C = 3280

In these two equations the variables represent the same number of students,

Solution

I'm going to multiply the top line by 5[A + C = 785]. the goal is to get the coefficients of the adult variables to match, so that we may "eliminate" them from our system. So we get...

5A + 5C = 3925 (the new first equation)

5A + 2C = 3280 (the rewritten second equation)

_____________ (Subtract the second line from the first line)

3C = 645 Divide both sides by 3)

C = 215

Therefore, 215 child's tickets were sold.

Hope this helps!

Answer 3

5A + 2C = 3280

A +C = 785

multiply 2nd by 2

and subtract from 1st equation

3A = 3280 - 2(785) = 3280-1570 = 1710

A = 1710/3 = 570 adult tickets

C = 785 - 570 = 215 child tickets