Problem 39
Question
Write each English phrase as an algebraic expression. Let the variable \(x\) represent the number. the sum of 10 divided by a number and that number divided by 10
Step-by-Step Solution
Verified Answer
The English phrase can be represented as the algebraic expression \(\frac{10}{x} + \frac{x}{10}\).
1Step 1: Identify the Operations
The phrase mentions the operations 'sum' and 'divided by'. 'Sum' refers to addition and 'divided by' refers to division. The operations are performed on the same number, represented by \(x\).
2Step 2: Break Down the Phrase and Formulate the Expressions
The phrase can be broken down into two parts for ease of translation into expressions: '10 divided by a number' and 'that number divided by 10'. '10 divided by a number' translates to \(\frac{10}{x}\) and 'that number divided by 10' translates to \(\frac{x}{10}\).
3Step 3: Combine Expressions
The term 'sum' in the phrase indicates addition of the two expressions formulated in Step 2. Hence, adding \(\frac{10}{x}\) and \(\frac{x}{10}\) gives the final algebraic expression: \(\frac{10}{x} + \frac{x}{10}\).
Key Concepts
VariablesDivision in AlgebraAddition in Algebra
Variables
In algebra, the concept of variables is fundamental. A variable is a symbol, often a letter like \(x\), that represents an unknown value or quantity. Variables allow us to write general expressions and equations that can apply to many situations. For example, rather than calculate a specific sum, we might use \(x\) in a formula to determine a sum for any number. Using variables makes it easier to work with changing values. In the algebraic expression from the exercise, let \(x\) represent the unknown number. This gives us flexibility: \(x\) can stand for any possible number that satisfies the conditions of the problem. Understanding variables is essential because they are the placeholders that can take on different values depending on the situation. Recognizing when to use a variable is key to setting up an expression or an equation effectively.
Division in Algebra
Division in algebra works similarly to division in arithmetic, but with variables involved, it becomes slightly more abstract. In the exercise, we see division expressed in the phrases '10 divided by a number' and 'that number divided by 10'. This translates algebraically to \(\frac{10}{x}\) and \(\frac{x}{10}\), respectively. Here, \(x\) can represent any non-zero number since division by zero is undefined.
It's important to understand how division affects expressions. When you divide a number by a variable, you're asking how many times that variable can "fit" into the number or vice versa. For instance:
It's important to understand how division affects expressions. When you divide a number by a variable, you're asking how many times that variable can "fit" into the number or vice versa. For instance:
- \(\frac{10}{x}\) means splitting 10 into \(x\) parts.
- \(\frac{x}{10}\) means splitting \(x\) into 10 parts.
Addition in Algebra
Addition in algebra combines different terms to form a single expression. In our context, we have two distinct terms: \(\frac{10}{x}\) and \(\frac{x}{10}\). The word "sum" in the problem indicates that these two expressions should be added together. When adding fractions, remember that having a common denominator is often necessary. However, in this problem, it's not explicitly required since we're simply demonstrating that these two terms form a sum.
When adding algebraic fractions, rely on basic fraction operations:
When adding algebraic fractions, rely on basic fraction operations:
- If the denominators are the same, combine the numerators directly.
- If the denominators differ, find a common denominator first.
Other exercises in this chapter
Problem 39
Perform the indicated subtraction. $$9.8-2.2$$
View solution Problem 39
Give an example of a rational number that is not an integer.
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Simplify each fraction by reducing it to its lowest terms. $$\frac{120}{813}$$
View solution Problem 40
Use the order of operations to simplify each expression. $$5(-3)^{2}-2(-4)^{2}$$
View solution