Problem 75
Question
Answer with an algebraic expression. See Example 9. Tickets to a circus cost \(\$ 25\) each. What will tickets cost for a family of \(x\) people if they also pay for two of their neighbors?
Step-by-Step Solution
Verified Answer
The cost expression is \(25(x + 2)\).
1Step 1: Identify variables
Let the variable \(x\) represent the number of people in the family.
2Step 2: Determine total number of people
The family consists of \(x\) people, plus they are paying for 2 neighbors. Therefore, the total number of tickets required is \(x + 2\).
3Step 3: Calculate cost per ticket
Each ticket costs \$25. So the cost for one ticket is \(25\). This will be multiplied by the total number of tickets to find the total cost.
4Step 4: Write the algebraic expression
The total cost can be written as an expression: \(25(x + 2)\), where \(25\) is multiplied by the total number of tickets \(x + 2\).
Key Concepts
Ticket PricesVariables in AlgebraMultiplication in Algebra
Ticket Prices
When we talk about ticket prices, it's essential to understand that each ticket has a specific cost. In our example, one ticket to the circus costs \(25. This is a fixed price, meaning no matter how many tickets you buy, each one will still cost \)25.
When buying tickets for multiple people, we need to know two things:
Calculating the overall cost involves multiplying the number of tickets by the cost of each ticket.
When buying tickets for multiple people, we need to know two things:
- The individual ticket price (in this case, $25)
- The total number of people who need tickets
Calculating the overall cost involves multiplying the number of tickets by the cost of each ticket.
Variables in Algebra
Variables are symbols that represent numbers or quantities in mathematics. They are incredibly useful in algebra because they allow us to write expressions that can stand for many similar situations.
In this exercise, \(x\) is the variable representing the number of people in the family. It could be any number. By using \(x\), we create a flexible way to calculate the total ticket cost for different family sizes, without having to write out long equations or calculations each time.This adaptability makes variables powerful; they allow expressions to change based on different inputs—like different family sizes—while keeping the solution process consistent. Understanding how to set and use variables is a foundational skill in algebra, enabling easier problem-solving across various scenarios.
In this exercise, \(x\) is the variable representing the number of people in the family. It could be any number. By using \(x\), we create a flexible way to calculate the total ticket cost for different family sizes, without having to write out long equations or calculations each time.This adaptability makes variables powerful; they allow expressions to change based on different inputs—like different family sizes—while keeping the solution process consistent. Understanding how to set and use variables is a foundational skill in algebra, enabling easier problem-solving across various scenarios.
Multiplication in Algebra
In algebra, we often use multiplication to combine values and variables. When calculating the cost of tickets, multiplication becomes crucial.
The algebraic expression from the exercise is \(25(x + 2)\). Here’s what’s happening:
Using multiplication this way simplifies calculations and applies the same rule to any number of tickets. It’s efficient and reduces errors, making it easier to solve real-world problems consistently.
The algebraic expression from the exercise is \(25(x + 2)\). Here’s what’s happening:
- Inside the parentheses: \((x + 2)\). This shows the total number of people. \(x\) represents family members, and \(+ 2\) accounts for the neighbors.
- Outside the parentheses: The multiplication factor, which is \(25\), stands for the cost of each ticket.
Using multiplication this way simplifies calculations and applies the same rule to any number of tickets. It’s efficient and reduces errors, making it easier to solve real-world problems consistently.