Question

In the following exercise, translate to a system of equations and solve.

Tickets to a Broadway show cost \($35$for adults and \)$15$for children. The total receipts for $1650$tickets at one performance were $$47,150$. How many adult and how many child tickets were sold?

Short answer

The number of adult tickets sold is $1120$

and the number of child tickets sold is $530$.

Step-by-step solution

Step 1. Given Information

Given that the price of one adult ticket is $$35$and the price of one child ticket is $$15$.

The total number of tickets sold is $1650$and the amount collected is $$47,150$.

Step 2. Identify and name what we are looking for

The objective is to find the number of adult and child tickets sold.

Let the number of adult tickets sold be $x$and the number of child tickets sold be$y$

Step 3. Form the equations

The total number of tickets sold is $1650$. So an equation can be formed as

$x+y=1650...\left(1\right)$

Now the total cost of $x$adult tickets each at $$35$and $y$child tickets each at $$15$is $$47,150$. So another equation can be written as

$35x+15y=47150...\left(2\right)$

Step 4. Solve using substitution

The first equation can be solved for $y$as

$\begin{array}{rcl}x+y& =& 1650\\ x+y-x& =& 1650-x\\ y& =& 1650-x...\left(3\right)\end{array}$

So using the third equation substitute $1650-x$for $y$in the second equation and solve for data-custom-editor="chemistry" $\mathrm{x}$

$\begin{array}{rcl}35x+15y& =& 47150\\ 35x+15(1650-x)& =& 47150\\ 35x+24750-15x& =& 47150\\ 35x-15x+24750-24750& =& 47150-24750\\ 20x& =& 22400\\ \frac{20x}{20}& =& \frac{22400}{20}\\ x& =& 1120\end{array}$

Step 5. Find the value of y

Substitute $1120$for $x$in the third equation and find the value of $y$

$$\begin{gathered} y=1650-x \\ y=1650-1120 \\ y=530 \end{gathered}$$

So the number of adult tickets sold is$1120$

and the number of child tickets sold is $530$.

Step 6. Check the solution.

Substitut$1120$for $x$and $530$for $y$in the first equation formed.

$\begin{array}{rcl}x+y& =& 1650\\ 1120+530& =& 1650\\ 1650& =& 1650\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}35x+15y& =& 47150\\ 35·1120+15·530& =& 47150\\ 39200+7950& =& 47150\\ 47150& =& 47150\end{array}$

This is also a true statement.

So the point satisfies both the equations.