Solve using substitution.

A local theater sells out for their show. They sell at 500 tickets for a total purse of $8,010
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?

1 answer

Let's assign variables to each type of ticket: let $a$ be the number of adult tickets, $s$ be the number of student tickets, and $c$ be the number of children's tickets.
We have two equations: \begin{align*}
a+s+c&=500\\
18a+15s+12c&=8010
\end{align*}Since we know that three times as many adult tickets as children's tickets were sold, we can write $a=3c$. Now substitute$ a$ with $3c$ in the equation $a+s+c=500$.

We have now two equations with two variables: \begin{align*}
a+s+c=3c+s+c&=500\\
\Rightarrow 2c+s&=500 \hspace{20mm}(1)\\
18a+15s+12c&=8010\hspace{20mm}(2)
\end{align*}
Substitute $a=3c$ in equation $(2)$ to get $$18(3c)+15s+12c=8010$$ $$\Rightarrow 54c+15s+12c=8010$$ $$\Rightarrow 66c+15s=8010$$ Now multiply both sides of equation $(1)$ by $-15$ to get $$-30c-15s=-7500.$$ Adding this to $(2)$, we get $$36c=510 \Rightarrow c=14.$$

Now we substitute $c=14$ in equation $(1)$ to get $$2(14)+s=500$$ $$\Rightarrow 28+s=500$$ $$\Rightarrow s=472$$ and in $a=3c$ we get $$a=3(14)=42.$$

So, they sold 42 adult tickets, 472 students tickets, and 14 children's tickets.
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