Answer:
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 = 7z------equation 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=3,100---------equation 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
