Problem 62 - Algebra A Combined Function

Question

\(\frac{9 z+5}{15} \cdot \frac{5 z}{81 z^{2}-25}\)

Step-by-Step Solution

Answer

\( \frac{z}{3(9z - 5)} \)

Step 1: Simplify the fractions

First, notice that the expression is the product of two fractions: \( \frac{9z + 5}{15} \) and \( \frac{5z}{81z^2 - 25} \). We will simplify each fraction separately if possible. For the first fraction, \( \frac{9z + 5}{15} \), there are no common factors between the numerator and the denominator. For the second fraction, \( \frac{5z}{81z^2 - 25} \), we can simplify the denominator by recognizing it as a difference of squares: \( 81z^2 - 25 = (9z)^2 - 5^2 = (9z - 5)(9z + 5) \). Thus, the second fraction becomes \( \frac{5z}{(9z - 5)(9z + 5)} \).

Step 2: Cancel common factors

Now, we look for common factors in the expression \( \frac{9z + 5}{15} \cdot \frac{5z}{(9z - 5)(9z + 5)} \). We can cancel the \(9z + 5\) factor from the first fraction with the \(9z + 5\) factor in the denominator of the second fraction. The expression now becomes: \( \frac{5z}{15(9z - 5)} \).

Step 3: Simplify the remaining fraction

Now let's simplify the fraction \( \frac{5z}{15(9z - 5)} \). Note that 15 in the denominator can be factored as \(3 \times 5\). Therefore, we can simplify \( \frac{5z}{3 \times 5 \times (9z - 5)} \) by canceling the factor 5 from the numerator and denominator: \( \frac{z}{3(9z - 5)} \).

Step 4: Final expression

After canceling out the common factors and simplifying, the expression simplifies to \( \frac{z}{3(9z - 5)} \).

Key Concepts

Algebraic Fractions

An algebraic fraction is like a regular fraction, but instead of numbers, it involves polynomials or algebraic expressions. Think of it as a way to represent division of one algebraic expression by another. Algebraic fractions are important because they allow us to perform operations such as addition, subtraction, multiplication, and division on polynomials in a systematic way.

• If the numerator or denominator of a fraction is an algebraic expression, it is an algebraic fraction.
• The simplest form of an algebraic fraction is reduced, meaning there are no common factors between the numerator and the denominator other than 1.
• Just like with numerical fractions, always look for factors in both the numerator and denominator to simplify the fraction.

In the given exercise, we have two algebraic fractions: \( \frac{9z + 5}{15} \) and \( \frac{5z}{81z^2 - 25} \). These fractions involve both numbers and algebraic terms (involving \(z\)). Our goal is to simplify these fractions as much as possible.

Simplifying Expressions

Simplifying expressions is the process of reducing an expression to its simplest form by eliminating common factors, combining like terms, and performing operations like addition or multiplication. This practice helps in making expressions easier to work with thus saving time and effort in later calculations.

• Always start by looking for common factors in the terms that can be canceled out.
• Simplifying makes the expression cleaner and can reveal more about the relationships between the variables involved.
• Keep in mind that the final simplified form should have no common factors in the numerator and the denominator other than 1.

In our exercise, we initially identify common factors from the expression \( \frac{9z + 5}{15} \cdot \frac{5z}{(9z - 5)(9z + 5)} \). Upon simplifying, we use multiplication rules and cancel like components in the board expression to arrive at \( \frac{z}{3(9z - 5)} \). This step reduces the complexity of the expression.

Difference of Squares

The difference of squares is a special algebraic expression structured as \( a^2 - b^2 \), which can be factored to \((a + b)(a - b)\). Recognizing this pattern is essential when simplifying polynomial expressions as it often simplifies the manipulation required.

• The pattern relies on perfect squares and the difference between them, making it easy to factorize expressions with this structure.
• When you spot an expression that matches \( a^2 - b^2 \), apply the trick to break it into a product, which often reveals simplifications.
• This is particularly useful when dealing with large polynomial expressions since it reduces the polynomial's degree.

In the given problem, the denominator of one of our fractions, \( 81z^2 - 25 \), is indeed a difference of squares. Recognizing this, we factor it into \( (9z + 5)(9z - 5) \). By factoring, we are allowed to cancel out the common terms in the main expression, aiding in reducing fractions and simplifying further calculations.