Problem 11 - Intermediate Algebra

Question

Simplify each expression. a. \(\frac{3 \cdot 5 \cdot x \cdot y \cdot y}{5 \cdot 7 \cdot x \cdot x \cdot x \cdot y}\) b. \(\frac{(x+8)(x-3)}{(x+2)(x+8)}\) c. \(\frac{a^{3}(a-9)}{(9-a)(9+a)}\)

Step-by-Step Solution

Answer

a. \(\frac{3y}{7x^2}\), b. \(\frac{x-3}{x+2}\), c. \(-\frac{a^3}{9+a}\).

Step 1: Simplify by Canceling Common Factors (a)

For part a, identify common factors in both the numerator and the denominator. The expression is \( \frac{3 \cdot 5 \cdot x \cdot y \cdot y}{5 \cdot 7 \cdot x \cdot x \cdot x \cdot y} \). Cancel the common factor 5 between the numerator and the denominator. This yields \( \frac{3 \cdot x \cdot y \cdot y}{7 \cdot x \cdot x \cdot x \cdot y} \). Next, cancel \( x \) from both the numerator and the denominator: \( \frac{3 \cdot y \cdot y}{7 \cdot x \cdot x \cdot y} \). Finally, cancel \( y \) from both, resulting in \( \frac{3 \cdot y}{7 \cdot x \cdot x} \). Simplified further, the expression becomes \( \frac{3y}{7x^2} \).

Step 2: Apply Factoring and Cancel Common Factors (b)

For part b, rewrite the expression \( \frac{(x+8)(x-3)}{(x+2)(x+8)} \). The common factor \( (x+8) \) in the numerator and the denominator can be canceled out, resulting in \( \frac{(x-3)}{(x+2)} \). The expression is now simplified to its smallest form.

Step 3: Recognize Negative Identity and Cancel Common Factors (c)

In part c, handle \( \frac{a^{3}(a-9)}{(9-a)(9+a)} \). Notice that \( a-9 \) can be rewritten as \( -(9-a) \). Therefore, the expression becomes \( -\frac{a^3 (9-a)}{(9-a)(9+a)} \). Cancel the \( 9-a \) from the numerator and denominator, yielding \( -\frac{a^3}{9+a} \). The expression is now in its simplest form.

Key Concepts

Factoring

Factoring is a method used in algebra to simplify expressions or solve equations by breaking numbers or expressions into their "factors," which are pieces or parts that, when multiplied together, give the original number or expression.
In the context of our exercise, consider part b, where we need to simplify the expression \( \frac{(x+8)(x-3)}{(x+2)(x+8)} \). Here, each parenthetical term is a binomial factor of the polynomial expression.
By factoring, we can identify common factors in the numerator and the denominator, which allows us to simplify the expression efficiently.

• Identify the complete factors of each part of the numerator and denominator.
• Look for common factors that appear in both the numerator and denominator.

In step 2, the common factor \((x+8)\) allowed us to cancel out terms, leading directly to the simplified result \( \frac{(x-3)}{(x+2)} \). Remember, after canceling common factors, it’s important to verify that no additional simplification can be made.

Canceling Common Factors

Canceling common factors is a fundamental technique in simplifying rational expressions. It involves identifying and eliminating the same expressions or elements present in both the numerator and denominator.
Let's break it down with the example from part a. Initially, our expression is: \[\frac{3 \cdot 5 \cdot x \cdot y \cdot y}{5 \cdot 7 \cdot x \cdot x \cdot x \cdot y}\]Here is how we apply the method:

• First, reduce any identical numerical coefficients. In this example, the number 5 appears in both the numerator and the denominator, so we cancel it.
• Next, identify and cancel other repeating factors. Variable terms like \(x\) and \(y\) are in both parts, so we also cancel them to the extent there are matching numbers in both lines.

After canceling all possible common factors, the expression simplifies to \( \frac{3y}{7x^2} \). This side-by-side reduction process continues until the expression cannot be simplified further, ensuring that only the irreducible form remains.

Negative Identity in Algebra

The negative identity in algebra is used to represent expressions that can be rewritten or simplified using properties of negative numbers. Specifically, it leverages the relationship between numbers and their opposites.
In part c of our exercise, understanding this identity becomes crucial. To simplify the expression \( \frac{a^{3}(a-9)}{(9-a)(9+a)} \), notice how the terms \(a-9\) and \(9-a\) are related:

• \(a-9\) is actually \(-(9-a)\). This trick allows the expression to be rewritten as \(-\frac{a^3 (9-a)}{(9-a)(9+a)}\).

Applying this understanding, we can cancel \((9-a)\) in both the numerator and denominator. This yields \(-\frac{a^3}{9+a} \), a much simpler form.
Recognizing and utilizing the negative identity efficiently simplifies challenging expressions, as it reveals opportunities to cancel out terms and reduce complexity.