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stira [4]
3 years ago
14

student tickets for the football game cost $12 each and adult ticket cost $20 $1,720 was collected for the 120 ticket sold at la

st game which system of equations can be used to solve for the number of each kind of ticket sold​
Mathematics
2 answers:
erik [133]3 years ago
4 0

Answer:

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.

pochemuha3 years ago
3 0

Answer:

Students=85

Adults=35

Step-by-step explanation:

-12s-20a=-1720

20k+20a=2400

8k=680

k=85

85+a=120

a=35

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