Solve.
A local theater sells out for their show. They sell all 500 tickets for a total purse of $8,010.00.
The tickets were priced at $15.00 for students, $12.00 for children, and $18.00 for adults. If the band sold three times as many adult tickets as children's tickets, how many of each type were sold?
Let $x$ be the number of student tickets, $y$ be the number of children tickets, and $z$ be the number of adult tickets. We have the following system of equations:
\begin{align*}
x+y+z &= 500 \\
15x + 12y + 18z &= 8010 \\
z &= 3y
\end{align*}
Substituting the third equation into the first equation to eliminate $z$, we have $x+y+3y = x+4y = 500$. Now, substituting the third equation into the second equation to eliminate $z$, we have $15x + 12y + 18(3y) = 15x + 12y + 54y = 8010$. So, we get the following system of equations:
\begin{align*}
x + 4y &= 500 \\
15x + 66y &= 8010
\end{align*}
We can solve for $x$ in terms of $y$ from the first equation: $x = 500-4y$. Substituting this expression into the second equation to eliminate $x$, we get $15(500-4y) + 66y = 8010$.
Expanding and simplifying this equation, we get $7500-60y+66y=8010$, so $6y=510$, which gives $y=85$. Thus, there were 85 children tickets sold.
Substituting this value into the equation $z=3y$, we get $z=3(85)=255$. Thus, there were 255 adult tickets sold.
Finally, substituting this value into the equation $x+4y=500$, we get $x+4(85)=500$, so $x=500-340=160$. Thus, there were 160 student tickets sold.
So, the theater sold 160 student tickets, 85 children tickets, and 255 adult tickets.
# Answer
160, 85, 255