Question 1077455: A basketball team sells tickets that cost $10, $20, and VIP seats for $30. The team has sold 3408 tickets overall. It has sold 129 more $20 tickets than $10 tickets. The total sales are $64,460. How many $10 tickets, $20 tickets, and $30 tickets have been sold?
Answer by ikleyn(52787)   (Show Source):
You can put this solution on YOUR website! .
Let x be the number of $10 tickets, and let y be the number of $30 tickets.
Then the number of $20 tickets is (x+129).
The system of two equations in two unknowns is
x + (x+129) + y = 3408. (1) (The total number of tickets)
10x + 20(x+129) + 30y = 64460 (2) (The total sales)
Simplify and solve.
The lesson to learn from his solution:
This problem reduces to two unknowns.
It is not about 3 unknowns.
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