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Ch. R - Review of Basic Concepts
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
Chapter 1, Problem 92a

Perform all indicated operations, and write each answer with positive integer exponents. [ (x-2 + y-2)/ (x-2 - y-2) ] * [ (x+y)/(x-y) ]

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1
Start by rewriting the expression clearly: \(\left( \frac{x^{-2} + y^{-2}}{x^{-2} - y^{-2}} \right) \times \left( \frac{x + y}{x - y} \right)\).
Recognize that \(x^{-2} = \frac{1}{x^2}\) and \(y^{-2} = \frac{1}{y^2}\). Rewrite the numerator and denominator of the first fraction using positive exponents: \(\frac{\frac{1}{x^2} + \frac{1}{y^2}}{\frac{1}{x^2} - \frac{1}{y^2}}\).
Find a common denominator for both the numerator and denominator of the large fraction: For the numerator, combine \(\frac{1}{x^2} + \frac{1}{y^2}\) as \(\frac{y^2 + x^2}{x^2 y^2}\); for the denominator, combine \(\frac{1}{x^2} - \frac{1}{y^2}\) as \(\frac{y^2 - x^2}{x^2 y^2}\).
Rewrite the large fraction as \(\frac{\frac{y^2 + x^2}{x^2 y^2}}{\frac{y^2 - x^2}{x^2 y^2}}\). Since both numerator and denominator share the same denominator \(x^2 y^2\), simplify by multiplying the numerator by the reciprocal of the denominator: \(\frac{y^2 + x^2}{x^2 y^2} \times \frac{x^2 y^2}{y^2 - x^2} = \frac{y^2 + x^2}{y^2 - x^2}\).
Now multiply this result by the second fraction \(\frac{x + y}{x - y}\): \(\frac{y^2 + x^2}{y^2 - x^2} \times \frac{x + y}{x - y}\). Notice that \(y^2 - x^2\) can be factored as \((y - x)(y + x)\). Use this factorization to simplify the expression further, and rewrite all terms with positive exponents.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Laws of Exponents

The laws of exponents govern how to simplify expressions involving powers, including negative exponents. A negative exponent indicates the reciprocal of the base raised to the positive exponent, e.g., x^-2 = 1/x^2. Understanding these laws is essential to rewrite expressions with positive exponents.
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