At a college production of a streetcar named desire, 400 tickets were sold. the ticket prices were $8, $10, and $12, and the total income from ticket sales was $3700. how many tickets of each type were sold if the combined number of $8 and $10 tickets sold was 7 times the number of $12 tickets sold?

There are 200 $8 tickets

There are 150 $10 tickets

There are 50 $12 tickets

Step-by-step explanation:

Let a = number of $8 tickets

Let b = number of $10 tickets

Let +c = number of $12 tickets

given:

(1) a+b+c = 400

(1) 8a+10b+12c = 3700

(1) a+b = 7c

This is 3 equations and 3 unknowns, so

it should be solvable

Substitute (3) into (1)

(1) 7c+c = 400

(1) 8c= 400

(1) c = 50

Plug this value into (2)

(2) 8a+10b+12*50= 3700

(2) 8a+10b+600= 3700

(2) 8a+10b= 3100

(2) 4a+5b= 1550

and

(1) a+b+c= 400

(1) a+b+50= 400

(1) a+b= 350

Multiply both sides of (1) by +4+

(1) 4a+4b= 1400

Subtract (1) from (2)

(2) 4a+5b+=+1550+

(1) -4a-4b= -1400

b= 150

and, since

(3) a+b= 7c

(3) a+150= 7*50

(3) a= 350-150

(3) a= 200

There are 200 $8 tickets

There are 150 $10 tickets

There are 50 $12 tickets

check:

(2) 8*200+10*150+12*50= 3700

(2) 1600+1500+600= 3700

(2) 3700= 3700

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -