Q. 129 - Intermediate Algebra

Question

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

Tickets for a Minnesota Twins baseball game are \($69$ for Main Level seats and \)$39$ for Terrace Level seats. A group of sixteen friends went to the game and spent a total of $$804$ for the tickets. How many of Main Level and how many Terrace Level tickets did they buy? 

Step-by-Step Solution

Answer

The group of friends bought $6$ Main Level Seats and $10$ Terrace Level Seats

Step 1. Given Information

The price for one Main Level seat is $$69$ and the price of one Terrace Level seat is $$39$.

The price of a total of sixteen tickets is $$804$

Step 2. Identify and name what we are looking for

The objective is to find the number of Main Level and Terrace Level tickets bought.

Let the number of Main Level tickets bought be $x$ and the number of Terrace Level tickets bought be $y$

Step 3. Form the equations

The total number of tickets sold is $16$. So an equation can be formed as

$x+y=16 ...\left(1\right)$

Now the total cost of $x$ Main Level tickets each at $$69$ and $y$ Terrace Level tickets each at $$39$ is $$804$. So an equation can be written as

$69x+39y=804 ...\left(2\right)$

Step 4. Solve using substitution

The first equation can be solved for $y$ as

$\begin{array}{rcl}x+y& =& 16\\ x+y-x& =& 16-x\\ y& =& 16-x\end{array}$

Substitute $16-x$ for $y$ in the second equation and solve for $x$

$\begin{array}{rcl}69x+39y& =& 804\\ 69x+39(16-x)& =& 804\\ 69x+624-39x& =& 804\\ 69x-39x+624-624& =& 804-624\\ 30x& =& 180\\ \frac{30x}{30}& =& \frac{180}{30}\\ x& =& 6\end{array}$

Step 5. Find the value of y

Substitute $6$ for $x$ in the third equation and solve for $y$

$$\begin{gathered} y=16-x \\ y=16-6 \\ y=10 \end{gathered}$$

So they bought $6$ Main Level tickets and $10$ Terrace Level tickets.

Step 6. Check the solution.

Substitute $6$ for $x$ and $10$ for $y$ in the first equation formed.

$\begin{array}{rcl}x+y& =& 16\\ 6+10& =& 16\\ 16& =& 16\end{array}$

It is a true statement.

Again, substitute the values in the second equation formed.

$\begin{array}{rcl}69x+39y& =& 804\\ 69·6+39·10& =& 804\\ 414+390& =& 804\\ 804& =& 804\end{array}$

This is also a true statement.

So the point satisfies both the equations.