Let a, b, c represent the numbers of $8, $10, and $12 tickets sold, respectively. The problem statement gives rise to three equations:
- a + b + c = 500
- 8a +10b +12c = 4700
- a + b - 4c = 0
Solving these equations by your favorite method gives ...
... (a, b, c) = (250, 150, 100)
250 $8 tickets, 150 $10 tickets, and 100 $12 tickets were sold.
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After you subtract the 3rd equation from the first to find 5c=500, you can substitute c=100 into the first two equations to get two equations in two unknowns. You know several ways to solve such equations, including elimination, substitution, and graphing, at least. Cramer's method is another viable choice.
