At a local Brownsville play production, 400 tickets were sold. The ticket prices varied on the seating arrangements and cost $8, $10, or $12. The total income from ticket sales reached $3700. If the combined number of $8 and $10 priced tickets sold was 7 times the number of $12 tickets sold, how many tickets of each type were sold

number of $8 sold= 200

Number of $10 tickets sold= $150

Number of $12 tickets sold = $50

Step-by-step explanation:

Let the quantity of $8 ticket be represented as  x

Let the quantity of $10 ticket be represented as  y

Let the quantity of $12 ticket be represented as  z

Such that total tikets sold = 400 can be represented as x+y+z= 400

Prices of$8, $10 and $12 gave a total income of $3,700  such that it can be represented as

8x +10y + 12z =$3,700

Also given that combined number of $8 and $10 priced tickets sold was 7 times the number of $12 tickets sold, we have that

x+y = 7z

Giving us the equations to solve to  be

x+y+z= 400 equation 1

8x +10y + 12z =$3,700 equation 2

x+y = 7zequation 3

Step 2- solving

Putting equation 3 i n equation 1, we have that

7z+z = 400

8z= 400

z= 400/8= 50

also puting the value of z= 50 in equation 1

x+y+z= 400

x+y+50= 400

x+y = 350 equation 4

putting z= 50 into equaton 2 and solving out

8x +10y + 12z =$3,700

8x +10y + 12x 50 =$3,700

8x +10y + 600 =$3,700

8x +10y =3,700 - 600

8x +10y=3,100equation 5

Multiplying equation 4 by 8  and subtracting from equation 5

8x +10y=3,100

-8x+8y= 2,800

2y=300

y= 300/2

y=150

to find x, when y= 150 using equation 4

x+y = 350

x= 350-150

x= 200

Therefore number of $8 sold= 200

Number of $10 tickets sold= $150

Number of $12 tickets sold = $50