tickets sold in advance cost $3. Tickets sold at the door cost $5. If 100 tickets were sold for a total of $460, how many tickets were sold at the door?

80 tickets

Step-by-step explanation:

Make tickets sold in advance be 3x and tickets sold at the door 5y. We will make a system of equations.

Make the tickets sold in advance and sold at the door equal to the amount of 100 tickets.

x + y = 100

Now make the expressions for the amount of money the tickets cost equal to $460.

3x + 5y = 460

Solve for x in the first equation by subtracting y from both sides.

x = 100 - y

Substitute x into the second equation.

3 (100 - y) + 5y = 460

Distribute.

300 - 3y + 5y = 460

Simplify.

300 + 2y = 460

Subtract 300 from both sides.

2y = 160

Divide both sides by 2.

y = 80

Substitute 80 for y into the equation made for x (x = 100 - y).

100 - (80) = 20

Since the variable y represented the number of tickets sold at the door, then 80 tickets were sold at the door.

80 tickets

Step-by-step explanation:

Let x be the number of tickets that were sold at the door and y be the number of tickets that were tickets sold in advance. So we can write an equation for the total amount of money as:

5x + 3y = 460 - - - (1)

and equation for the number of tickets sold as:

x + y = 100 - - - (2)

From equation (2), we get y = 100 - x.

Substituting this value of x in equation (1) to get:

5x + 3 (100 - x) = 460

5x + 300 - 3x = 460

5x - 3x = 460 - 300

2x = 160

x = 80

Therefore, 80 tickets were sold at the door.