Adult tickets to a basketball game cost $5. Student tickets cost $1. A total of $3,077 was collected on the
sale of 1,201 tickets. How many of each type of ticket were sold?
The basketball game sold
adult tickets and
student tickets.

Answers

Answer 1

Answer:

Your problem will require you setting up 2 algebraic equations. First, you should.

Let A represent the number of adult tickets

Let S represent the number of student tickets

The first equation will be:

A+S = 1,215 tickets

The number of adult tickets plus the number of student tickets equals 1,215 tickets

The second equation will be:

5.00A+1.00S=$2,591

The price and number of adult tickets plus the price and number of student tickets equals $2,591.

I hope this has made the problem easier to understand. I hope you are able to solve the problem from here. If you can not solve the problem from here, I will be more than happy to further assist you.

Step-by-step explanation:

You have two unknowns, two variables, therefore you need to have at least two equations in this word problem in order to find a solution. Let's review the info you are given:

The price of an Adult ticket: $5

The price of a Student ticket: $1

The two variables are: the number of Adult tickets sold (you can use variable "A"), and the number of Student tickets sold (use variable "S").

The other info you are given is the total number of tickets sold: 1,215 tickets. Use this to make your first equation: the number of adult tickets plus the number of student tickets equals 1,215.

You are also told how much money was collected: $2,591. That means, $5 from every Adult ticket and $1 from every student ticket. We don't know how many of each ticket, but we assigned variables A and S to stand in for this unknown info. The money made from all the adult tickets would be the price of the adult ticket times how many adult tickets sold (5 times A), and the money made from all the student tickets would be the price of the student ticket times how many student tickets sold (1 times S). Use this to make your second equation, which says that all the money collected from all the tickets (5 times A in addition to 1 times S) is (=) 2,591.

Now you have two equations, each equation has two variables. To solve, go back to the easiest equation (the first one) and isolate one of the variables. By isolate I mean get it on one side of the equation by itself. Since in this case these variables are being added together, you have to do the opposite of addition, which is of course subtraction. To do this, you will have to subtract the other variable ON BOTH SIDES of the equation. So, say we want to isolate the variable A, we will subtract S from both sides. Well, S minus S is zero, so now A is isolated, meaning the left side of the equation is just A = ...  Now you want to subtract the S on the other side. Remember that whatever you do to one side you must do to the other.

Now that you have your equation re-worked to show A = ..., you need to "plug it into" the second equation. What I mean is, you need to substitute instead of A, put in this equation. Then you will have one equation with one variable which you can solve! Remember the rules of equations, to do work in parenthesis first, the multiply or divide, then add or subtract. You will be able to solve for S.

Once you have found S (the number of student tickets sold), you can easily find A (the number of adult tickets sold) by going back to that first equation and subtracting your value of S from 1215.


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