Problem 32 - Introductory and Intermediate Algebra for College Students 4th

Question

Use a form of the distributive property to rewrite each algebraic expression without parentheses. $$\frac{1}{4}(12+8 r)$$

Step-by-Step Solution

Answer

The simplified form of the expression \(\frac{1}{4}(12+8 r)\) without parentheses is \(3 + 2r\)

Step 1: Distribute the \(\frac{1}{4}\) to both terms inside the parentheses

To distribute the \(\frac{1}{4}\) to both terms, it needs to be multiplied with the individual terms. So the new distributed expression would be \(\frac{1}{4} * 12 + \frac{1}{4} * 8r\)

Step 2: Simplify the Multiplication

Now, calculate the product of \(\frac{1}{4}\) and 12 and also the product of \(\frac{1}{4}\) and \(8r\). That gives \(3 + 2r\)

Step 3: Final Answer

Without the parentheses, the final simplified expression is \(3 + 2r\).

Key Concepts

Simplifying Algebraic Expressions

Understanding how to simplify algebraic expressions is a fundamental skill in algebra. The process often involves distributing factors and combining like terms. In our exercise example, we deal with the simplification of the expression \( \frac{1}{4}(12 + 8r) \).

To tackle this, we first apply the distributive property, which allows us to remove the parentheses by distributing \( \frac{1}{4} \) to both the constant and the variable inside the bracket. The operation yields \( \frac{1}{4} \times 12 + \frac{1}{4} \times 8r \). This turns the problem into an arithmetic task involving multiplication.

• \( \frac{1}{4} \times 12 = 3 \)
• \( \frac{1}{4} \times 8r = 2r \)

Once we have our individual multiplications, we combine them to get the simplified expression, \( 3 + 2r \), free of parentheses and in its simplest form. This skill isn't limited to numbers; it can also be applied to more complex variables and coefficients, making it a versatile tool in the arsenal of algebra.

Multiplying Fractions

Multiplying fractions is another essential concept that often appears in algebraic expressions. When faced with a fraction, such as \( \frac{1}{4} \), being multiplied by a whole number or a term with a variable, you simply multiply the numerator (the top number) by the whole number or the term, while keeping the denominator (the bottom number) the same.

In our example, \( \frac{1}{4}(12 + 8r) \), the fraction \( \frac{1}{4} \) is being multiplied by both 12 and \( 8r \). This results in:

• \( \frac{1}{4} \times 12 = \frac{12}{4} = 3 \)
• \( \frac{1}{4} \times 8r = \frac{8r}{4} = 2r \)

Remembering to simplify the fractions after multiplication is crucial. This often means dividing both the numerator and the denominator by their greatest common divisor. In the case of \( \frac{12}{4} \) and \( \frac{8r}{4} \), the number 4 is the common divisor, simplifying the fractions to 3 and \( 2r \) respectively, just as we found in our example.

Linear Expressions

Linear expressions are algebraic expressions that do not have variable exponents higher than one or contain variables being multiplied by themselves. They are usually presented in the form \( ax + b \) or a similar variation where \( a \) and \( b \) are constants, and \( x \) is a variable. Our goal with linear expressions is often to solve for the variable or simplify the expression as much as possible.

In the given exercise, \( \frac{1}{4}(12 + 8r) \), once we distribute and multiply as previously discussed, we end up with a simplified linear expression, \( 3 + 2r \). This is linear because \( r \) is raised to the first power, aligning with the definition. Linear expressions are crucial in algebra since they are the building blocks for creating equations and functions that describe lines on a graph. Recognizing them and understanding how to manipulate them is key to mastering basic algebra and moving on to more complex mathematical concepts.