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Chapter 0: Problem 108

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

Short Answer

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

Step by step solution

01

- Label Substitution

Let \( a = (x-y) \). So the expression simplifies to \( a^{-1} + a^{-2} \). This could be written as \( \frac{1}{a} + \frac{1}{a^{2}} \).
02

- 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: \( \frac{a}{a^{2}} + \frac{1}{a^{2}} = \frac{a+1}{a^{2}} \).
03

- Substitute Back

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檚 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.

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Most popular questions from this chapter

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