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

85 student tickets and 35 adult tickets were sold.

Step-by-step 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.

Students=85

Adults=35

Step-by-step explanation:

-12s-20a=-1720

20k+20a=2400

8k=680

k=85

85+a=120

a=35