Problem 108 - Algebra and Trigonometry

Question

perform the indicated operations. $$ (x-y)^{-1}+(x-y)^{-2} $$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{x-y+1}{(x-y)^{2}} \) .

Step 1 - Label Substitution

Let \( a = (x-y) \). So the expression simplifies to \( a^{-1} + a^{-2} \). This could be written as \( \frac{1}{a} + \frac{1}{a^{2}} \).

Step 2 - Common Denominator

To add two fractions, a common denominator is needed. The common denominator here is \( a^{2} \). Convert the fractions with this common denominator: \( \frac{a}{a^{2}} + \frac{1}{a^{2}} = \frac{a+1}{a^{2}} \).

Step 3 - Substitute Back

Now substitute back \( a = (x-y) \) into the expression to get the final answer: \( \frac{x-y+1}{(x-y)^{2}} \)

Key Concepts

Exponents

Understanding exponents is essential when working with mathematical expressions. An exponent, such as in the term \((x-y)^{-1}\), shows how many times a number, called the base, is multiplied by itself. Negative exponents, like

• \((x-y)^{-1}\) mean you take the reciprocal
• \((x-y)^{-2}\) means the reciprocal squared

So, \(x-y^{-1} = \frac{1}{x-y}\) and \(x-y^{-2} = (\frac{1}{x-y})^2 = \frac{1}{(x-y)^{2}}\). By realigning exponents as fractions, calculations remain straightforward and consistent.

Fraction Addition

When adding fractions, it's crucial to have a common denominator. This makes it possible for the fractions to be combined seamlessly:

• Identify the denominators in each fraction
• Find the Least Common Denominator (LCD)
• Re-write each fraction using the LCD

In the example, \(\frac{1}{a} + \frac{1}{a^{2}}\) requires a common denominator. Here, it's \(a^{2}\), transforming the fractions to share the same base for easy addition:

• Re-write \(\frac{1}{a}\) as \(\frac{a}{a^{2}}\)
• With both as \(\frac{\text{something}}{a^{2}}\), they combine effortlessly

The result: \(\frac{a+1}{a^{2}}\). This approach simplifies operations and enables accurate calculations.

Substitution Method

Substitution simplifies complex expressions to make them more manageable. By letting \(a = (x-y)\), we substitute and simplify calculations. Here’s how it works:

• Choose a simple variable, say \(a\), to replace the complex part (\(x-y\))
• Reevaluate the expression using \(a\), such as transforming \((x-y)^{-1} + (x-y)^{-2}\) into \(a^{-1} + a^{-2}\)
• Perform operations easily with this simpler form
• Substitute back the original expression to complete

By substituting and reverting back, we manage to simplify what seems intricate into comprehensible segments, ultimately arriving at \(\frac{x-y+1}{(x-y)^{2}}\). Simplifying expressions is a key algebraic strategy to handle complex tasks efficiently.