Example 3

Modeling Real Life

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A comedy club earns $1088 from an opening night performance and $1183 from a second performance. On opening night, the club sells 68 adult tickets and 136 student tickets. For the second performance, the club sells 79 adult tickets and 140 student tickets. What is the price of each type of ticket?

Solution

Understand the Problem You know the amounts earned for each performance, and the total numbers of adult and student tickets sold for each. You are asked to find the price of each type of ticket.

Make a Plan Use a verbal model to write a system that represents the problem. Then solve the system.

Solve and Check

Verbal Model

Number
of adult
tickets

$\cdot$

Adult
ticket
price

Number
of student
tickets

$\cdot$

Student
ticket
price

Total
amount
earned

Variables

Let $x$ be the price $in dollars$ of an adult ticket and let $y$ be the price $in dollars$ of a student ticket.

System

$\begin{align} 68x + 136y& = 1088 && \qquad \qquad \qquad \quad \color{red}{\text{Equation 1 (first performance)}}\\[1ex] 79x + 140y& = 1183 && \qquad \qquad \qquad \quad \color{red}{\text{Equation 2 (second performance)}}\\[1ex] \end{align}$

Step 1

Solve for $y$ in Equation 1.

$\begin{align} 68x + 136y& = 1088 && \qquad \quad \color{red}{\text{Equation 1}}\\[1ex] y& = 8 -\frac{1}{2}x && \qquad \quad \color{red}{\text{Solve for y.}}\\[1ex] \end{align}$

Step 2

Substitute $8-\small{\frac{1}{2}}x$ for $y$ in Equation 2 and solve for $x$ .

$\begin{align} 79x + 140y& = 1183 && \color{red}{\text{Equation 2}}\\[1ex] 79x + 140\left(\color{red}{8 - \frac{1}{2}x}\right)& = 1183 && \color{red}{\text{Substitute 8 }-\frac{1}{2}x\text{ for }y.}\\[1ex] 79x + 1120 -70x& = 1183 && \color{red}{\text{Distributive Property}}\\[1ex] x& = 7 && \color{red}{\text{Solve for }x.}\\[1ex] \end{align}$

Step 3

Substitute $7$ for $x$ in Equation 1 and solve for $y$ .

$\begin{align} 68x + 136y& = 1088 && \qquad \qquad \color{red}{\text{Equation 1}}\\[1ex] 68(\color{blue}{7}\color{black}{)} + 136y& = 1088 && \qquad \qquad \color{red}{\text{Substitute 7 for $x$.}}\\[1ex] y& = 4.5 && \qquad \qquad \color{red}{\text{Solve for $y$.}}\\[1ex] \end{align}$

►

The solution is $(7, 4.5)$ . So, an adult ticket costs $7 and a student ticket costs $4.50.

Check

$\color{red}{\text{Equation 1}}$
$\begin{align} 68(\color{blue}{7}) + 136(\color{red}{4.5})& \stackrel{\color{red}{?}}{=} 1088 \\[1ex] 1088 & = 1088 \ \color{red}{\checkmark} \end{align}$

$\color{red}{\text{Equation 2}}$
$\begin{align} 79(\color{blue}{7}) + 140(\color{red}{4.5})& \stackrel{\color{red}{?}}{=} 1183 \\[1ex] 1183& = 1183 \ \color{red}{\checkmark} \end{align}$

Solving Real-Life Problems

Modeling Real Life

Solution