Problem 31 - Introductory and Intermediate Algebra for College Students 4th

Question

Use a form of the distributive property to rewrite each algebraic expression without parentheses. $$\frac{1}{3}(12+6 r)$$

Step-by-Step Solution

Answer

The expression without parentheses is \(4 + 2r\).

Step 1: Apply the distributive property to the constant

Multiply the constant outside the parentheses, which is \(\frac{1}{3}\), by both of the terms inside the parentheses - \(12\) and \(6r\). This gives us \(\frac{1}{3} \cdot 12 + \frac{1}{3} \cdot 6r\)

Step 2: Simplify both terms

We can now simplify this to \(4 + 2r\) by performing the multiplication.

Key Concepts

Algebraic Expressions

Algebraic expressions are a foundational concept in algebra, denoting a combination of numbers, variables, and arithmetic operations like addition, subtraction, multiplication, and division. For instance, \(\frac{1}{3}(12+6r)\) is an algebraic expression comprising the constant \(\frac{1}{3}\), a variable \(r\), and arithmetic operations (addition and multiplication).

Understanding the components of algebraic expressions is crucial because they represent mathematical ideas and relationships. Variables like \(r\) take on different values, and constants are fixed numbers. When you see parentheses, they typically organize operations and indicate that you should consider the terms inside as a group. Mastering the interpretation of these expressions plays a vital role in solving algebra problems and creating a strong mathematical foundation.

Simplifying Algebraic Expressions

Simplifying algebraic expressions involves reducing them to their most basic form without changing their value. It is similar to cleaning up a messy room so that it's easier to navigate. To simplify an expression like \(\frac{1}{3}(12+6r)\), one must use the distributive property, which is a method to eliminate the parentheses.

Applying the Distributive Property

The distributive property allows us to multiply a single term by each term inside the parentheses. In our example, the constant \(\frac{1}{3}\) distributes over \(12\) and \(6r\), yielding \(\frac{1}{3} \cdot 12 + \frac{1}{3} \cdot 6r\). After distributing, we can simplify by multiplying: \(4 + 2r\).

This process helps in interpreting and solving expressions easily. Moreover, simplifying expressions can assist in solving equations, comparing expressions, and understanding the underlying relationships between variables and constants.

Applying Constants to Variables

In algebra, applying constants to variables is an integral part of working with expressions. Constants are fixed values, whereas variables can change. Multiplying a constant with a variable combines these two elements to alter the expression's value.

The Importance of Constants

Constants play a pivotal role in determining the weight of a variable within an expression. For example, in the expression \(6r\), the constant \(6\) acts as a multiplier that indicates how much \(r\)'s value will contribute to the overall expression.

When you see a problem like \(\frac{1}{3} \cdot 6r\), the constant \(\frac{1}{3}\) is distributed to the variable \(r\), yielding \(2r\) upon simplification. This shows us how \(r\)'s influence in the expression is reduced by a third. Understanding how to apply constants to variables is essential in manipulating algebraic expressions to showcase relationships, solve equations, and create functions that effectively describe real-world scenarios.