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Chapter 1: Problem 60

Write each sentence as an equation. Let the variable \(x\) represent the number. Five times a number is 35 .

Short Answer

Expert verified
The equation that represents the sentence is: \(5x = 35\)

Step by step solution

01

Understand the Sentence

From the sentence 'Five times a number is 35', the unknown number is referred to as 'a number' and will be represented by the variable \(x\). 'Five times a number' means that \(x\) is multiplied by 5, and 'is 35' means equal to 35.
02

Construct the Equation

By substituting the expressions in the sentence with mathematical operations, we obtain the equation: 5\(x\) = 35.
03

Check the Equation

Now, check if the equation correctly represents the sentence. In our case, 5\(x\) = 35 correctly represents 'Five times a number is 35'.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Translating Word Problems
One of the most common challenges in algebra is translating word problems into algebraic equations.
The key is to identify key phrases and their corresponding mathematical operations. For example, 'times' is a clear indicator of multiplication. The word 'is' often translates to the equals sign (\(=\) in an equation.

Let's consider the problem: 'Five times a number is 35'. The phrase 'five times' hints at a multiplication operation, while 'a number' is an unknown that we need to assign a variable to, usually represented by letters like \(x\) or \(y\).

To digest this process:
  • Identify the unknown and give it a name, such as \(x\).
  • Look for keywords that indicate mathematical operations: 'five times' suggests multiplication.
  • Understand that 'is' indicates equivalence, leading us towards forming an equation.
By practicing these steps, you'll become adept at forming algebraic expressions from word problems.
Variables in Algebra
In algebra, a variable is a symbol for a number we don't know yet. It is a placeholder for any value.

In our example, \(x\) is the variable that stands for 'a number' we're trying to find. Variables allow us to solve problems without knowing the exact values they represent.

Here are some tips to handle variables effectively:
  • Choose variables that make sense for the context (e.g., \(t\) for time, \(d\) for distance).
  • Be consistent. If you start with \(x\) as your variable, keep using \(x\) throughout the problem.
  • Remember that variables can represent any number but will have a specific value once the equation is solved.
Understanding the role of variables is crucial to mastering algebra.
Writing Equations
Writing equations is a critical skill that transforms real-world scenarios into solvable mathematical problems.
The equation 5\(x\) = 35 was derived from text. Writing an equation involves substituting words with mathematical symbols and ensuring they convey the same meaning.

Steps to writing an equation from a word problem include:
  • Determine what the problem is asking for and what is given.
  • Translate phrases into algebraic expressions using the right operations.
  • Link the expressions with an equal sign to create an equation.
Once you've set up the equation, you're on your way to finding the solution.
Solving Linear Equations
Solving linear equations is the process of finding the value of the variable that makes the equation true.
For the simple equation 5\(x\) = 35, we want to isolate \(x\) to find out what number it represents.

The basic steps to solve linear equations:
  • Isolate the variable on one side of the equation by performing inverse operations.
  • Simplify the equation at each step to keep it manageable.
  • Check your solution by plugging the value back into the original equation.
In our case, dividing both sides by 5 yields \(x = 7\) which is the solution since 5 times 7 equals 35. Learning to solve linear equations is fundamental, as it forms the basis for more complex algebraic concepts.

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Most popular questions from this chapter

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. $$\text { Simplify: } \frac{1}{4}-6(2+8) \div\left(-\frac{1}{3}\right)\left(-\frac{1}{9}\right)$$

Evaluate both expressions for \(x=4 .\) What do you observe? $$3(x+5) ; 3 x+15$$

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