For the high school's production of "grease," 490 tickets were sold. the cost of a student
ticket was $8, a pre-sale non-student ticket was $10, and a ticket at the door cost $12. the
total income from ticket sales was $4600. the combined number of student and pre-sale
tickets was 6 times more than the tickets sold at the door. which system of equations could
be used to determine the number of each type of ticket sold? (teks a2.3a-s)

The system of equations could  be used to determine the number of each type of ticket sold is:

s + p + t = 490

8s + 10p + 12t = 4600

s + p = 6t

Number of each type of ticket sold is:

number of student tickets sold = 220

number of pre-sale non student ticket sold = 200

number of ticket sold at door = 70

Solution:

Let "s" be the number of student tickets sold

Let "p" be the number of pre-sale non student ticket sold

Let "t" be the number of ticket sold at door

Cost of 1 student ticket = $ 8

Cost of 1 pre-sale non student ticket = $ 10

Cost of 1 ticket sold at door = $ 12

From given information,

490 tickets were sold. So we can frame a equation as:

number of student tickets sold + number of pre-sale non student ticket sold + number of ticket sold at door = 490

s + p + t = 490 ---------- eqn 1

Also given that, The  total income from ticket sales was $4600

So we can frame a equation as:

number of student tickets sold x Cost of 1 student ticket + number of pre-sale non student ticket sold x Cost of 1 pre-sale non student ticket + number of ticket sold at door x Cost of 1 ticket sold at door = $ 4600

8s + 10p + 12t = 4600 ------ eqn 2

The combined number of student and pre-sale  tickets was 6 times more than the tickets sold at the door

s + p = 6t --------- eqn 3

So eqn 1 eqn 2 and eqn 3 could  be used to determine the number of each type of ticket sold

Let us solve eqn 1, eqn 2, eqn 3 to find values of "s" "p" and "t"

Substitute eqn 3 in eqn 1

6t + t = 490

7t = 490

Substitute t = 70 in eqn 2

8s + 10p + 12(70) = 4600

8s + 10p = 3760 ----- eqn 4

Substitute t = 70 in eqn 3

s + p = 6(70)

s + p = 420  ----- eqn 5

Multiply eqn 5 by 8

8s + 8p = 3360  ----- eqn 6

Subtract eqn 6 from eqn 4

8s + 10p = 3760

8s + 8p = 3360

(-) ---------------------

2p = 400

Substitute p = 200 in eqn 5

s + 200 = 420

Summarizing the results:

number of student tickets sold = 220

number of pre-sale non student ticket sold = 200

number of ticket sold at door = 70

total amount of money that he should pay if he has not used his duty free exemption is $98

step-by-step explanation:

given a united states citizen returning to the states declares the following items at the customs office: 3 shirts at $8.50 each 2 dresses at $27.50 each 1 pair of gold cuff links at $17.50 per pair. we have to find how much he pay

cost of 3 shirts if one cost at $8.50 is

cost of 2 dresses if one cost at $27.50 is  

cost of 1 pair of gold cuff at $17.50

so, total amount of money that he should pay if he has not used his duty free exemption is   $25.5+$55+$17.50=$98

answer: 80,000

step-by-step explanation:

the population doubles every day.

10,000*2=20,000 day 4

20,000*2=40,000 day 5

40,000*2=80,000 day 6