Problem 40 - Beginning and Intermediate Algebra

Question

Write a system of equations and solve. A movie theater charges \(\$ 9.00\) for an adult's ticket. S7. OO for a ticket for scriors 60 and over, and \(\$ 6,00\) for a child's ticket. For a particular movie, the theater sold a total of 290 tickets. which brought in \(\$ 2400\). The number of semors' 'tickets sold was twice the number of children's tickets sold. Determine the number of ndults", scniors', and children's tickets sold.

Step-by-Step Solution

Answer

The number of tickets sold in each category is: 200 adult tickets, 60 senior tickets, and 30 child tickets.

Step 1: Define Variables

Let's define the variables for the number of adult, senior, and children's tickets sold: - Let \(x\) be the number of adult tickets. - Let \(y\) be the number of senior tickets. - Let \(z\) be the number of child tickets.

Step 2: Write the Equations

We will set up three linear equations: 1. Total number of tickets sold: \[ x + y + z = 290 \] 2. Total sales from the tickets: \[ 9x + 7y + 6z = 2400 \] 3. Number of senior tickets sold is twice the number of children tickets sold: \[ y = 2z \]

Step 3: Solve the System of Equations

We will solve the system of equations using substitution or elimination method. In this case, let's use substitution. From the third equation, we have \(y = 2z\), which we will substitute into the first two equations: 1. Total number of tickets sold (substitute \(y=2z\)): \[ x + (2z) + z = 290 ⇒ x + 3z = 290 \] 2. Total sales from the tickets (substitute \(y=2z\)): \[ 9x + 7(2z) + 6z = 2400 ⇒ 9x + 20z = 2400 \] Now we have two linear equations with two variables, \(x\) and \(z\). Let's solve for \(x\) from the first equation: \[ x = 290 - 3z \] Now, substitute this expression for \(x\) into the second equation: \[ 9(290 - 3z) + 20z = 2400 \]

Step 4: Solve for x and z

We now have a linear equation in terms of z only, so let's solve for z: \[ 2610 - 27z + 20z = 2400 ⇒ -7z = -210 \] Divide by -7 on both sides: \[ z = 30 \] Now, we can find the value of x using the expression for \(x\): \[ x = 290 - 3(30) = 290 - 90 = 200 \]

Step 5: Find the Number of Senior Tickets Sold

We can now find y using the third equation: \[ y = 2z = 2(30) = 60 \]

Step 6: Present the Results

The number of tickets sold in each category is as follows: - Adult tickets: \(x = 200\) - Senior tickets: \(y = 60\) - Child tickets: \(z = 30\)

Key Concepts

Linear Equations

Linear equations are mathematical expressions that create a straight line when graphed on a coordinate plane. They typically have variables that are first degree, meaning the variable is not raised to any power other than one. In our problem, we have three linear equations:\ \

• The total number of tickets sold is represented by: \(x + y + z = 290\).
• The total amount in dollars collected from the ticket sales is modeled by the equation \(9x + 7y + 6z = 2400\).
• The relationship between senior and children's tickets is \(y = 2z\).

\ These equations together form a system of linear equations. Understanding how to set up and interrelate these equations is crucial for solving problems involving multiple variables.

Substitution Method

The substitution method is a technique used to solve systems of equations where you solve one equation for one variable and substitute that expression into another equation. This replaces one of the variables, making it easier to solve.
In our particular exercise, we used the equation \(y = 2z\) from step 2. By expressing \(y\) in terms of \(z\), we can substitute into other equations to reduce the number of variables.
For example, the substitution in the first equation, \(x + (2z) + z = 290\), simplifies to \(x + 3z = 290\). Substituting this into the equation for total ticket sales and simplifying leads to more straightforward calculations.
Substitution is especially useful when one equation is simple and lets you reduce the complexity of the system. This technique is valuable in breaking down complex word problems like the one here.

Equation Solving

Equation solving is about finding values for variables that satisfy a given set of equations. With a system of equations like those from our exercise, this process involves both understanding the relationships between variables and performing algebraic manipulations.
In our exercise, after substitutions, we've reduced the equations to two simpler linear equations: \(x + 3z = 290\) and \(9x + 20z = 2400\).

• Using substitutions, we found \(x = 290 - 3z\).
• Replaced \(x\) in the other equation to solve for \(z\).
• Once \(z\) was known, substitute it back to find \(x\), followed by calculating \(y\) using initial given relationships.

This step-by-step reduction and substitution are key for effective equation solving in multi-variable scenarios.

Word Problems

Word problems involve real-world scenarios translated into mathematical language. They test your ability to comprehend and extract mathematical equations from a story.
The movie theater problem we're solving is an excellent example of transforming worded information into a system of equations: deciding on appropriate variables and creating equations that accurately represent the given scenario.
It's essential to:

• Identify what's known and what needs to be found.
• Translate relationships into equations, like knowing seniors' tickets are twice children's tickets.
• Ensure the arithmetic expressions accurately map to these relationships.

Mastering word problems involves practice. With every solved problem, you're learning to bridge the gap between words and numbers.