Help with math problem 10pts

At a college theater 400 tickets were sold. The ticket prices were $8, $10, $12. The total income 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?

You need to set up a system of equations.

x + y + z = 400
8x + 10y + 12z = 3700
x + y = 7z

Now you just have to solve for x, y and z. See if you can do it with the equations I set up for you.

No. of $12-tickets sold—x:
= 1/8(400)
= 400/8 or 50

No. of $10-tickets sold—y:
$10y + $8(400 - 50 - y) = $3,700 - $12(50)
10y + 8(350 - y) = 3,700 - 12(50)
5y + 4(350 - y) = 1,850 - 6(50)
5y + 1,400 - 4y = 1,850 - 300
y = 150

No. of $8-tickets sold:
= 400 - 50 - 150
= 200

Answer: sold for each type: $8-tickets, 200; $10-tickets, 150; $12-tickets, 50

Check—$3,700 was the total income:
= $8(200) + $10(150) + $12(50)
= $1,600 + $1,500 + $600
= $3,700

Let a = number of $8 tickets, b = no of $10 tickets, c = no of $12 tickets.
a + b + c = 400
a + b = 7c
c = 400 - 7c
8c = 400
c = 50 (tickets sold at $12)
8a + 10b + 12c = 3700
8a + 10b + 600 = 3700
a + b + 50 = 400
8a + 8b + 400 = 3200
2b + 200 = 400
2b = 200
b = 200
a + 200 + 50 = 400
a = 150

8(150) + 10(200) + 12(50) = 1200 + 2000 + 500 = 3700

150 $8 Tickets, 200 $10 tickets, 50 $12 tickets.

12$- 50 tickets
10$- 150 tickets
8$- 200 tickets