Problem 28 - Algebra 1: Concepts and Skills

Question

Use the distributive property to rewrite the expression without parentheses. $$\frac{1}{5}(10+15 r)$$

Step-by-Step Solution

Answer

The expression \(\frac{1}{5}(10+15 r)\) without parentheses using the distributive property is \(2+3r\).

Step 1: Identify the Terms Inside the Parentheses

The terms inside the parentheses are \(10\) and \(15r\). The factor outside the parentheses is \(\frac{1}{5}\). Our goal is to apply the distributive property, which states that \(a(b+c) = ab + ac\), to the given expression.

Step 2: Apply Distributive Property to the First Term

Applying the distributive law to the first term, we multiply \(\frac{1}{5}\) by \(10\) which gives us \(2\).

Step 3: Apply Distributive Property to the Second Term

Next, we multiply \(\frac{1}{5}\) by \(15r\), which gives us \(3r\) as a result.

Step 4: Combine the Results

The final step is to combine the results from the distributed terms to form the final expression without parentheses. In this case, this gives us the expression \(2+3r\).

Key Concepts

Simplifying Algebraic Expressions

Understanding how to simplify algebraic expressions is crucial in algebra. Essentially, this process involves reducing an expression to its most basic form without changing its value. To achieve this, we combine like terms and use mathematical operations in a systematic manner.

Consider the expression \(\frac{1}{5}(10 + 15r)\). Simplifying it requires us to remove the parentheses, which enclose the terms we want to unite with the factor of \(\frac{1}{5}\) outside. By distributing the \(\frac{1}{5}\) across each term inside the parentheses, we equivalently apply the same operation to the entire internal sum, ensuring the expression's value remains unaltered.

What Are Like Terms?

Like terms are elements within an algebraic expression that have the same variable raised to the same power. For instance, in \(2+3r\), '2' is a constant term and \(3r\) is a linear term, and they cannot be combined because they are not like terms. However, if we had \(3r+4r\), we could simplify it to \(7r\) because \(3r\) and \(4r\) are like terms. Simplification of expressions is a foundational skill that will be used throughout algebra including more complex operations with polynomials.

Applying Distributive Law

The distributive law, also known as the distributive property, is an essential mathematical tool in algebra that allows us to simplify expressions that involve parentheses. It states that for any numbers or expressions \(a\), \(b\), and \(c\), the equation \(a(b + c) = ab + ac\) holds true.

In our exercise, the distributive law is utilized to express \(\frac{1}{5}(10 + 15r)\) without the parentheses. Applying the property involves multiplying the factor outside the parentheses (\(\frac{1}{5}\)) by each term inside the parentheses individually. The key steps include:

• Multiplying \(\frac{1}{5}\) by '10' to get '2'
• Multiplying \(\frac{1}{5}\) by \(15r\) to get \(3r\)

Once each term has been distributed, we combine the products to finalize the expression as \(2 + 3r\). The application of the distributive law makes handling complex algebraic expressions more manageable and prepares students for algebraic operations that involve polynomial expansion and factoring.

Solving Linear Equations

Solving linear equations is a core component of algebra, involving finding the value of the variable that makes the equation true. A linear equation is one where the highest power of the variable is one. The process usually entails isolating the variable on one side of the equation and simplifying the terms.

For example, if our exercise resulted in an equation such as \(2 + 3r = 17\), we would proceed to solve for \(r\) by first subtracting '2' from both sides to isolate the \(3r\) term, yielding \(3r = 15\). Dividing both sides by '3' would then give us \(r = 5\), which is the solution to this linear equation.

Stages of Solving Linear Equations

Typically, the stages are:

• Rearranging the equation to get 'like terms' on the same side
• Simplifying both sides of the equation if necessary
• Isolating the variable on one side using inverse operations
• Checking the solution by substituting it back into the original equation

While the present exercise does not result in an equation to solve, understanding the principles of solving linear equations is crucial as it provides the tools necessary to tackle a wide range of math problems beyond simple expression simplification.