Problem 35 - Intermediate Algebra

Question

Multiply, and then simplify, if possible. See Example 4. $$ 15 x\left(\frac{x+1}{15 x}\right) $$

Step-by-Step Solution

Answer

The simplified expression is \(x + 1\).

Step 1: Apply Distributive Property

Start by applying the distributive property to the expression. Distribute the term outside the parentheses to each term inside the parentheses: \[ 15x \left( \frac{x+1}{15x} \right) = 15x \cdot \frac{x}{15x} + 15x \cdot \frac{1}{15x} \]

Step 2: Simplify Each Term

Now simplify each part of the expression by canceling out terms within the fractions.For the first term:\[ 15x \cdot \frac{x}{15x} = \frac{15x^2}{15x} = x \] (since \( 15x \) in the numerator and denominator cancel out)For the second term:\[ 15x \cdot \frac{1}{15x} = \frac{15x}{15x} = 1 \] (since \( 15x \) in the numerator and denominator cancel out)

Step 3: Combine the Simplified Terms

Combine the simplified terms from Step 2.\[ x + 1 \] Thus, the expression simplifies to \( x + 1 \).

Key Concepts

Distributive Property

The distributive property is a fundamental concept in algebra that allows us to simplify expressions. It is used when you have a term being multiplied by a sum or difference inside a set of parentheses. Here, it is applied as follows:

• If you have an expression like \( a(b + c) \), you use the distributive property to expand it: \( a(b + c) = ab + ac \).

In our exercise, we have \( 15x \) being multiplied by \( \frac{x+1}{15x} \). To overcome the fractions and unpack it efficiently, we distribute \( 15x \) across the terms inside the parenthesis, which are \( \frac{x}{15x} \) and \( \frac{1}{15x} \). This step is akin to multiplying \( 15x \) by each component within the parentheses, maximizing the clarity of our expression.
Remember, applying the distributive property is about breaking down a more complicated expression into simpler parts, making it easier to handle.

Simplification

Simplification is the process of making an expression as simple as possible. In algebra, this often means reducing fractions, combining like terms, or eliminating unnecessary parts of an expression. In our problem:

• We simplify the first part by observing that \( 15x \cdot \frac{x}{15x} \) results in \( \frac{15x^2}{15x} \) where \( 15x \) cancels out, leaving us with just \( x \).
• The second part, \( 15x \cdot \frac{1}{15x} \), turns into \( \frac{15x}{15x} \), which simplifies to 1.

The key to simplification is to look for terms that can cancel each other out or combine to form a simplified value.
It involves operations such as factoring, distributing, and simply observing where similar elements in the numerator and denominator can reduce. This is crucial for solving algebraic equations efficiently.

Rational Expressions

Rational expressions are expressions that involve fractions containing polynomials in the numerator, the denominator, or both. They resemble fractions but can include variables. Understanding how to manipulate these correctly is vital in algebra because they appear frequently in equations and functions.
In our example, \( \frac{x+1}{15x} \) is a rational expression.

• The numerator is a polynomial (\( x + 1 \)), and the denominator is a mono-term polynomial (\( 15x \)).

The flexibility in rational expressions allows us to use operations such as addition, subtraction, multiplication, and division while respecting the rules of algebraic expressions.
When multiplying rational expressions like in our step-by-step solution, ensure to streamline or eliminate common factors in the numerator and denominator to simplify. Rational expressions require practice to master, especially in accurately identifying opportunities for simplification.