Question

student tickets for the football game cost $12 each and adult ticket cost $20 $1,720 was collected for the 120 ticket sold at last game which system of equations can be used to solve for the number of each kind of ticket sold​

Answer 1

Students=85

Adults=35

Explanation:

-12s-20a=-1720

20k+20a=2400

8k=680

k=85

85+a=120

a=35

Answer 2

85 student tickets and 35 adult tickets were sold.

Explanation:

Howdy!

We know that student tickets cost $12, adult ticket cost $20 and 120 tickets were sold.

So:

$12×A + $20×B = $1,720. Where A and B are the number of student tickets and adult tickets sold, respectively.

Given that 120 tickets were sold in total, we have that:

A + B = 120

So the system of equations to be solved is the following:

$12×A + $20×B = $1,720 (1)

A + B = 120 (2)

Solving for 'A' in equation (2) we get:

A = 120 - B

Substituting this value into equation (1) we get:

$12×(120 - B) + $20×B = $1,720

Solving for 'B' we have:

$12×(120 - B) + $20×B = $1,720

$1,440 - $12×B + $20×B = $1,720

$1,440 + $8×B = $1,720

$8×B = $1,720 - $1,440

$8×B = $280

B = 35 tickets.

Given that B = 35, then A = 120 - 35 = 85 tickets.

So 85 student tickets and 35 adult tickets were sold.