Problema Solution
A basketball team sells tickets cost $10,$20,or , for VIP seats, $30. The team has sold 515 tickets overall. it has sold 171 more $20 tickets than $10 tickets. the total sales are $9060. How many of each kind have been sold?
Answer provided by our tutors
let
x = the number of $10 tickets sold
y = the number of $20 tickets sold
z = the number of $30 tickets sold
The team has sold 515 tickets overall:
x + y + z = 515
it has sold 171 more $20 tickets than $10 tickets:
y = 171 + x
the total sales are $9060:
10x + 20y + 30z = 9060 divide both sides by 10
x + 2y + 3z = 906
by solving the system of equations:
x + y + z = 515
y = 171 + x
x + 2y + 3z = 906
we find:
x = 156 tickets
y = 327 tickets
z = 32 tickets
click here to see the step by step solution of the system of equations:
there were 156 tickets of $10, 327 tickets of $20 and 32 tickets of $30 sold.