Problem 26 - Intermediate Algebra

Question

Simplify each complex fraction. See Example 4. $$ \frac{\frac{1}{a}+\frac{1}{b}}{\frac{a}{b}-\frac{b}{a}} $$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{1}{a-b} \).

Step 1: Identify the complex fraction

We need to simplify the expression \( \frac{\frac{1}{a}+\frac{1}{b}}{\frac{a}{b}-\frac{b}{a}} \) by evaluating both the numerator and the denominator individually.

Step 2: Simplify the numerator

The numerator is \( \frac{1}{a} + \frac{1}{b} \). Find a common denominator which is \( ab \) and rewrite each fraction: \[ \frac{1}{a} = \frac{b}{ab}, \quad \frac{1}{b} = \frac{a}{ab} \]. Adding these gives \( \frac{b + a}{ab} \).

Step 3: Simplify the denominator

The denominator is \( \frac{a}{b} - \frac{b}{a} \). Find a common denominator which is \( ab \) and rewrite each fraction: \[ \frac{a}{b} = \frac{a^2}{ab}, \quad \frac{b}{a} = \frac{b^2}{ab} \]. Subtracting these gives \( \frac{a^2 - b^2}{ab} \).

Step 4: Divide the numerators

The problem now becomes \( \frac{\frac{b+a}{ab}}{\frac{a^2-b^2}{ab}} \). To divide fractions, multiply by the reciprocal of the denominator: \[ \frac{b+a}{ab} \times \frac{ab}{a^2-b^2} \].

Step 5: Simplify further

Since the \( ab \) terms cancel each other out, you're left with \( \frac{b+a}{a^2-b^2} \).

Step 6: Factor the denominator

Notice that \( a^2 - b^2 \) can be rewritten as a difference of squares: \( (a-b)(a+b) \). Thus, the expression becomes \( \frac{b+a}{(a-b)(a+b)} \).

Step 7: Simplify the expression

The \( b+a \) in the numerator is equivalent to \( a+b \), allowing us to cancel this out with the \( a+b \) in the denominator, leaving: \( \frac{1}{a-b} \).

Key Concepts

Simplifying Fractions

Understanding how to simplify fractions, especially complex ones, can seem daunting at first. But, by methodically breaking down the problem into smaller and more manageable parts, the task becomes much simpler.
Complex fractions often involve fractions both in the numerator and the denominator. To simplify them:

• First, simplify the numerator and denominator separately to single fractions.
• Once simplified, divide the top fraction by the bottom fraction. This means flipping (finding the reciprocal of) the denominator fraction and multiplying it by the numerator.
• Clear out common factors to further simplify the expression.

Simplifying requires careful attention to detail, and it often helps to rewrite all parts of the expression with a common denominator to make addition or subtraction straightforward. Patience and practice are key!

Algebraic Expressions

Algebraic expressions form the basis of most algebra problems. They consist of variables, coefficients, and operators like addition and subtraction. Here, the complex fraction can be seen as a large algebraic expression that includes smaller fractions as parts of it.
To handle these effectively:

• Identify each part of the expression: What are the terms, what are the coefficients, and how are these combined?
• Look for opportunities to factor expressions. This means looking for patterns like common factors among terms or recognizable differences of squares.
• Rewrite expressions to have common denominators if they are fractions. This simplifies the process of combining or cancelling terms later on.

Understanding the properties and behaviors of algebraic expressions allows you to manipulate them with greater ease, leading to correct simplifications.

Difference of Squares

The difference of squares is a specific algebraic identity that is very useful when simplifying expressions. This identity states that for any two terms, say \(a\) and \(b\), the difference \(a^2 - b^2\) can be factored as \((a-b)(a+b)\).
This principle was applied in simplifying the complex fraction:

• Recognizing \(a^2 - b^2\) in the denominator as a difference of squares, it was rewritten as \((a-b)(a+b)\).
• Factoring this helps in cancelling terms with the numerator, refining the expression.

Remember: spotting opportunities to apply such identities will significantly simplify the problem at hand. It reduces larger expressions into simpler forms, enabling you to move towards the solution more directly.