10

One way to solve this problem is to set up an equation using one variable. Let n represent the number of 3 dollars tickets sold; then 50 minus n represents the number of 5 dollars tickets sold.
(3 times n) + (5 times (50 minus n)) = 230
(3 times n) +250 minus (5 times n) = 230
20 = 2 times n
10 = n
There were 10 3 dollar tickets sold.
Simultaneous equations can also be set up to solve the problem. Let n represent the number of 3 dollar tickets sold and p represent the number of 5 dollar tickets sold.
n + p = 50
(3 times n) + (5 times p) = 230
Another way to solve this problem is to realize that each of the tickets costs at least 3 dollars. At 3 dollars each, a total of 150 dollars would be collected from selling all 50 of the tickets. Since the remaining 80 dollars was the extra 2 dollars collected from each of the 5 dollar tickets, 40 of the 5 dollar tickets were sold. This implies that 10 of the 3 dollar tickets were sold.