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Chapter 1: Problem 83

Explain why \(6 x\) and \(5 y\) are not like terms.

Short Answer

Expert verified
The terms \(6x\) and \(5y\) are not like terms because they have different variables.

Step by step solution

01

Define like terms

Like terms are terms that have identical variable parts raised to the same power. This means that both the variable and its exponent must be the same for the terms to be considered like terms.
02

Identify the terms

Identify the terms you are comparing. In this exercise, the terms are \(6x\) and \(5y\).
03

Compare the variable parts

Examine the variable parts of the terms. The term \(6x\) has the variable \(x\) while \(5y\) has the variable \(y\).
04

Conclusion

Since the variables \(x\) and \(y\) are different, the terms \(6x\) and \(5y\) are not like terms.

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Key Concepts

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

Identical Variable Parts
In algebra, identifying like terms is essential for combining and simplifying expressions.

Like terms are terms that have identical variable parts, which means both the variables and their exponents need to match.
For example, in the terms \(3x\) and \(5x\), the variable part is the same, and both terms include the same variable raised to the same power.

With identical variable parts, you can perform addition and subtraction easily, consolidating the terms for simpler expressions.
Variable Comparison
To determine whether two terms are like terms, a crucial step is the comparison of their variable parts.

Let's consider the given example of \(6x\) and \(5y\). The variables in these terms are different: one has \(x\) and the other has \(y\). Since the variables are not the same, they cannot be combined as like terms.

Therefore, it is important always to examine the variables closely and ensure they are the same before attempting to add or subtract terms in algebra.
Algebraic Terms
Understanding the composition of algebraic terms is fundamental in algebra.

An algebraic term consists of a coefficient (a numerical part) and a variable part. For instance, in the term \(6x\), \(6\) is the coefficient, and \(x\) is the variable part.

By analyzing the given terms \(6x\) and \(5y\), we see they differ in their variable parts, making them unlike terms.
Recognizing and differentiating the parts of algebraic terms helps in performing accurate algebraic operations and simplifying expressions efficiently.

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Most popular questions from this chapter

Problem: Complete the "Understand the problem," "Make a plan," and "Complete the plan" steps. A 20 -acre field is going to be divided into building lots. The area of each lot is \(\frac{2}{5}\) acre. Find the number of lots in the development. Incorrect Answer: The unknown is the number of lots: \(L\) = number of lots. To find the number of lots, multiply the area by the number of lots. $$ \begin{aligned} &L=\left(\frac{2}{5}\right)(20) \\ &L=8 \text { lots } \end{aligned} $$

Evaluate \(12 \mathrm{~kg}+54 \mathrm{~g}\). Write the answer in kilograms.

$$ 9000\( earning \)5 \%$ annual simple interest for 8 years.

Evaluate \(8 \mathrm{~km}+4 \mathrm{~m}\). Write the answer in meters.

The original price of a phone is \(\$ 199\). If it is marked down \(32 \%\), find the sale price.
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