Problem 60 - Algebra for College Students

Question

Simplify each complex fraction. $$ \frac{\frac{2}{x-y}+\frac{3}{x+y}}{\frac{5}{x+y}-\frac{1}{x^{2}-y^{2}}} $$

Step-by-Step Solution

Answer

The simplified form is \( \frac{5x-y}{5x-5y-1} \).

Step 1: Identify and Rewrite the Common Denominators

First, observe the complex fraction: \( \frac{\frac{2}{x-y}+\frac{3}{x+y}}{\frac{5}{x+y}-\frac{1}{x^{2}-y^{2}}} \). Notice there are fractions in both the numerator and the denominator. Recognize that the term \( x^2-y^2 \) in the denominator can be rewritten as the difference of squares: \((x-y)(x+y)\). This helps in finding common denominators.

Step 2: Find the LCD for the Numerator

In the numerator, we have the fractions \( \frac{2}{x-y} \) and \( \frac{3}{x+y} \). The least common denominator (LCD) for these two fractions is \((x-y)(x+y)\). Rewrite each fraction with this common denominator: \[\frac{2}{x-y} = \frac{2(x+y)}{(x-y)(x+y)}\]\[\frac{3}{x+y} = \frac{3(x-y)}{(x-y)(x+y)}\]Thus, the combined numerator becomes: \[\frac{2(x+y) + 3(x-y)}{(x-y)(x+y)}\]

Step 3: Simplify the Numerator

Simplify the expression in the numerator: \[2(x+y) + 3(x-y) = 2x + 2y + 3x - 3y\]\[= 5x - y\]Thus, the simplified numerator is: \[\frac{5x-y}{(x-y)(x+y)}\]

Step 4: Find the LCD for the Denominator

In the denominator, we have the fractions \( \frac{5}{x+y} \) and \( \frac{1}{x^2-y^2} \). Their LCD is also \((x-y)(x+y)\). Rewrite each fraction with this common denominator:\[\frac{5(x-y)}{(x-y)(x+y)} = \frac{5}{x+y} \text{ as is}\]\[\frac{1}{x^2-y^2} = \frac{1}{(x-y)(x+y)} \text{ remains as is}\]Thus, the combined denominator becomes:\[\frac{5(x-y) - 1}{(x-y)(x+y)}\]

Step 5: Simplify the Denominator

Simplify the expression in the denominator:\[5(x-y) - 1 = 5x - 5y - 1\]Thus, the simplified denominator is: \[\frac{5x-5y-1}{(x-y)(x+y)}\]

Step 6: Simplify the Entire Complex Fraction

Since both the numerator and the denominator now share a common denominator of \((x-y)(x+y)\), you can simplify by multiplying the entire fraction by the reciprocal of the denominator:\[= \frac{5x-y}{(x-y)(x+y)} \cdot \frac{(x-y)(x+y)}{5x-5y-1}\]Cancel out the common denominators:\[= \frac{5x-y}{5x-5y-1}\]This is the fully simplified form of the complex fraction.

Key Concepts

Simplifying Fractions

Simplifying fractions is a fundamental skill in mathematics that involves reducing a fraction to its simplest form. This means that the numerator and the denominator have no common factors other than 1. Let's take a closer look at how we simplify fractions:

• Check if there are common numerical factors between the numerator and the denominator. Divide them by these factors to reduce the fraction.
• For algebraic expressions, factor both the numerator and the denominator completely.
• Cancel out any common terms present in both expressions.

In the context of complex fractions, simplification involves dealing with a fraction in which the numerator, the denominator, or both are also fractions.
This process requires 1. finding a common denominator 2. combining the fractions 3. simplifying the resulting expression. In the exercise, this was applied to simplify both the numerator and the denominator of the complex fraction. Remember, simplifying fractions is all about making expressions as straightforward and compact as possible.

Least Common Denominator (LCD)

The Least Common Denominator (LCD) is crucial when simplifying complex fractions. The LCD is the smallest expression that can serve as a common denominator for two or more fractions. It allows you to combine fractions by giving them a uniform base. Here’s how you determine the LCD for a set of fractions:

• Identify the denominators of each fraction. In our example, we look at \( x-y \), \( x+y \), and \( x^2-y^2 \).
• Factorize the denominators as much as possible. Recognize expressions like the difference of squares, \( x^2-y^2 = (x-y)(x+y) \).
• The LCD is the product of the highest powers of all factors appearing in any of the denominators.

In the original problem, both the numerator and the denominator required the LCD as \((x-y)(x+y)\), which allowed us to rewrite all terms with a common base. This step is critical for making addition or subtraction of fractions feasible.

Difference of Squares

The difference of squares is a specific algebraic technique used to factor expressions like \( x^2-y^2 \). This method exploits the formula \( a^2-b^2 = (a+b)(a-b) \). Recognizing and applying this identity can greatly simplify expressions. Here are the steps to handle difference of squares:

• Identify the expression as a difference of squares if it's in the form \( a^2-b^2 \).
• Apply the formula to factor it into \( (a+b)(a-b) \). In our case, \( (x+y)(x-y) \).
• This factoring helps in finding common denominators and simplifying the complex fraction.

In the exercise, recognizing \( x^2-y^2 \) as a difference of squares was pivotal for rewriting denominators. This simple factorization opens the door to simplifying more complex algebraic fractions by finding common terms and reducing them.