Q. 131 - Intermediate Algebra

Question

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

Tickets for the community fair cost \($12$ for adults and \)$5$ dollars for children. On the first day of the fair, $312$ tickets were sold for a total of $$2204$. How many adult tickets and how many child tickets were sold? 

Step-by-Step Solution

Answer

The number of adult tickets sold is $92$ and the number of child tickets sold is $220$.

Step 1. Given Information

Given that the price of one adult ticket is $$12$ and the price of one child ticket is $$5$ .

The total number of tickets sold is $312$ and the amount collected is $$2204$

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 $312$. So an equation can be formed as

$x+y=312 ...\left(1\right)$

Now the total cost of $x$ adult tickets each at $$12$ and $y$ child tickets each at $$5$ is $$2204$. So another equation can be written as

$12x+5y=2204 ...\left(2\right)$

Step 4. Solve using substitution

The first equation can be solved for $y$ as

$\begin{array}{rcl}x+y& =& 312\\ x+y-x& =& 312-x\\ y& =& 312-x ...\left(3\right)\end{array}$

Now substitute $312-x$ for $y$ in the second equation and solve for $x$

$\begin{array}{rcl}12x+5y& =& 2204\\ 12x+5(312-x)& =& 2204\\ 12x+1560-5x& =& 2204\\ 12x-5x+1560-1560& =& 2204-1560\\ 7x& =& 644\\ \frac{7x}{7}& =& \frac{644}{7}\\ x& =& 92\end{array}$

Step 5. Find the value of y

Substitute $92$ for $x$ in the third equation and solve for $y$

$$\begin{gathered} y=312-x \\ y=312-92 \\ y=220 \end{gathered}$$

So $92$ adult tickets and $220$ children tickets were sold. 

Step 6. Check the solution

Substitute $92$ for $x$ and $220$ for $y$ in the first equation formed.

$\begin{array}{rcl}x+y& =& 312\\ 92+220& =& 312\\ 312& =& 312\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}12x+5y& =& 2204\\ 12·92+5·220& =& 2204\\ 1104+1100& =& 2204\\ 2204& =& 2204\end{array}$

This is also a true statement.

So the point satisfies both the equations.