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Chapter 7: Problem 29

Simplify. If possible, use a second method, evaluation, or a graphing calculator as a check. $$ \frac{y+y^{-1}}{y-y^{-1}} $$

Short Answer

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

Step by step solution

01

Rewrite the Expression

Rewrite the numerator and denominator by expressing the negative exponents as fractions. \[ \frac{y + \frac{1}{y}}{y - \frac{1}{y}} \ \text{Numerator: } y + \frac{1}{y} \ \text{Denominator: } y - \frac{1}{y} \]
02

Find a Common Denominator

To simplify, find a common denominator for the terms in both the numerator and the denominator. The common denominator is \( y \). \[ \text{Numerator: } \frac{y^2+1}{y} \] \[ \text{Denominator: } \frac{y^2-1}{y} \]
03

Simplify the Fraction

Now, divide the numerator by the denominator. Since both parts have a common denominator \( y \), they cancel each other out. \[ \frac{\frac{y^2+1}{y}}{\frac{y^2-1}{y}} = \frac{y^2+1}{y^2-1} \]
04

Factor the Denominator (if possible)

Factor the denominator as a difference of two squares. \[ y^2-1 = (y-1)(y+1) \]
05

Write Final Simplified Expression

Combine all the simplified components to write the final expression. Since \( y^2 + 1 \) cannot be factored easily, leave it as is over the factored denominator. \[ \frac{y^2+1}{(y-1)(y+1)} \]

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Key Concepts

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

Negative Exponents
Let's start with understanding negative exponents. A negative exponent indicates that the base, instead of being multiplied, should be divided.
In simpler terms, if you see an expression like this: \(a^{-n}\), it means the same as \(\frac{1}{a^n}\).
For example, \(y^{-1}\) is equivalent to \(\frac{1}{y}\).

This step is super crucial because it helps transform complicated expressions.
In our exercise, \(\frac{y + y^{-1}}{y - y^{-1}}\), we change \(y^{-1}\) into fractions: \(y + \frac{1}{y} \ and \ y - \frac{1}{y}\).

By converting negative exponents, problems become more manageable and easier to simplify.
Common Denominator
To simplify algebraic fractions, finding a common denominator is key.
Why? Because it makes combining fractions less complex.

In our example, the least common denominator (LCD) for \(y\) and \(\frac{1}{y}\) is simply \(y\).
So, we rewrite the numerator and denominator by putting everything over \(y\):
  • Numerator: \(y + \frac{1}{y} \rightarrow \frac{y^2 + 1}{y} \)
  • Denominator: \(y - \frac{1}{y} \rightarrow \frac{y^2 - 1}{y} \)


This 'preparation' step sets us up for easy-like-Sunday-morning simplification.
Difference of Squares
A very useful thing to remember in algebra is the difference of squares formula: \(a^2 - b^2 = (a - b)(a + b)\).
This formula allows us to factor expressions that fit this pattern.

In our exercise, the denominator \(y^2 - 1\) can be factored using this formula:
\(y^2 - 1 = (y - 1)(y + 1) \).

Factoring makes further simplification straightforward and easier to understand.

Just remember: the difference of squares is your friend whenever you see \(a^2 - b^2\). It breaks complex expressions into simpler, factorable parts.

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Most popular questions from this chapter

The length of a rectangular picture window is 3 yd greater than the width. The area of the rectangle is \(10 \mathrm{yd}^{2} .\) Find the perimeter.

If the LCM of two third-degree polynomials is a sixth-degree polynomial, what can be concluded about the two polynomials?

Find the LCM. $$10 t^{3}, 5 t^{4}$$

Find the LCM. $$10 x^{2} y, 6 y^{2} z, 5 x z^{3}$$

Find the LCM. $$9 n^{2}-9,\left(5 n^{2}-10 n+5\right)^{2}, 15 n-15$$
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