Tickets at a show were $4 for students and $7 for adults. On the first night, the show sold 578 tickets and collected $3365 in sales. How many student tickets were sold? How many adult tickets?

Answer 1

To tackle this problem, we first begin with what we know:

Student tickets are $4 and adults are $7.

We also know 578 tickets were sold and $3365 was made in sales.

Now, we have to isolate what we want to find: the number of adult ticket sales and the number of student ticket sales. Let the number of adult ticket sales be x and the number of student ticket sales be y.

Now, we go back to what we know! Since 578 tickets were sold, that means the sum of the adult ticket sales (x) and the student ticket sales (y) equal 578. In algebraic form, we have:

Equation 1: x+ y = 578

Additionally, we know that $3365 was made in sales. We know the amount made from adult ticket sales must be the cost per ticket ($7) multiplied by the number of adult tickets sold (x). Similarly, the amount made from student ticket sales must be the cost per ticket ($4) times the number of student tickets sold (y). Translating this into an equation, we get that:

Equations 2: 7x + 4y = 3365

Now we have a system of equations! Since we have two variables (x and y) and two equations (1 and 2), we can solve this problem!

We will do this by solving for one variable at a time. Our goal is to combine both equations so that we end up with one equation with one variable. We will do this by using substitution.

First, let us solve for y (the number of student tickets sold). To do this, we must write y in terms of x in one equation and then plug that into the second equation.

Since we know that x+ y = 578, we can manipulate equation 1 by subtracting x from both sides.

Doing this, we get y = 578 - x.

Now, we can plug this into equation 2, which is our substitution step!

7(578-y) + 4y = 3365.

We then multiply out the 7(578-y) to get:

7*578-7y + 4y = 3365

Evaluating that product, we get:

4046 -7y + 4y = 3365

Now we combine like terms:

4046 -3y = 3365

Now we subtract 4046 from both sides:

-3y = -681

Dividing both sides by -3, we get:

y = 227!

This means that 227 student tickets were sold and we have solved half the problem!

What we have left to do is find the number of adult tickets sold! Since we have solved for y, we only need to use one equation to solve for x. Here, we will pick equation 1, as it is the simpler of the two, but it doesn't matter which one you pick.

Equation 1: x+ y = 578

Plugging in y, we get

x+ 227= 578

Subtracting 227 from both sides, we get

x = 351

Now we know 351 adult tickets were sold and 227 student tickets were sold, which means we have solved the problem!

The algebraic steps without the explanations are provided below:

7(578-y) + 4y = 3365.

7*578-7y + 4y = 3365

4046 -7y + 4y = 3365

4046 -3y = 3365

-3y = -681

y = 227!

Answer 2

S - the number of students' tickets sold

A - the number of adult tickets sold

Writing the system of equations:

4S + 7A = 3365 → equation (1)

S + A + 578 → equation (2)

From equation (2):

S = 578 - A → equation (3)

Substitute the expression for "s" into equation (1):

4(578 - A) + 7A = 3365

2312 - 4A + 7A = 3365

3A = 1053

A = 351

Calculating "S" from equation (3):

S = 578 - 351 = 227