Algebra and Trigonometry: Structure and Method, Book 2
Algebra and Trigonometry: Structure and Method, Book 2
2000th Edition
ISBN: 9780395977255
Author: MCDOUGAL LITTEL
Publisher: McDougal Littell
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Concept explainers

Percentage
A percentage is a number indicated as a fraction of 100. It is a dimensionless number often expressed using the symbol %.
Algebraic Expressions
In mathematics, an algebraic expression consists of constant(s), variable(s), and mathematical operators. It is made up of terms.
Numbers
Numbers are some measures used for counting. They can be compared one with another to know its position in the number line and determine which one is greater or lesser than the other.
Subtraction
Before we begin to understand the subtraction of algebraic expressions, we need to list out a few things that form the basis of algebra.
Addition
Before we begin to understand the addition of algebraic expressions, we need to list out a few things that form the basis of algebra.
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Question
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Chapter 5.6, Problem 23WE
To determine

To simplify the given expression.

Expert Solution & Answer
Check Mark

Answer to Problem 23WE

  2(t+1)(t−1)

Explanation of Solution

Given:

Expression: 1t2+t+1t2−t

Calculation:

Consider the given expression 1t2+t+1t2−t ,

Since the denominators of both the fractions are different, find the least common denominator for both the fractions.

Here, least common denominator is t(t+1)(t−1) , so

  1t2+t+1t2−t

  =1t(t+1)+1t(t−1)

  =1⋅(t−1)t(t+1)⋅(t−1)+1⋅(t+1)t(t−1)⋅(t+1)

Now, as the denominators of the two fractions are same, simply add the numerators and put the result on the same denominator.

  =(t−1)t(t+1)(t−1)+(t+1)t(t−1)(t+1)

  =(t−1)+(t+1)t(t+1)(t−1)=t−1+t+1t(t+1)(t−1)=2tt(t+1)(t−1)

  =2(t+1)(t−1)

Conclusion:

Therefore, after simplification, the answer is 2(t+1)(t−1) .

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Chapter 5.6, Problem 22WE
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Chapter 5.6, Problem 24WE

Chapter 5 Solutions

Algebra and Trigonometry: Structure and Method, Book 2

Ch. 5.1 - Prob. 1OECh. 5.1 - Prob. 2OECh. 5.1 - Prob. 3OECh. 5.1 - Prob. 4OECh. 5.1 - Prob. 5OECh. 5.1 - Prob. 6OECh. 5.1 - Prob. 7OECh. 5.1 - Prob. 8OECh. 5.1 - Prob. 9OECh. 5.1 - Prob. 10OE
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Prob. 1ECh. 5.1 - Prob. 2ECh. 5.2 - Prob. 1OECh. 5.2 - Prob. 2OECh. 5.2 - Prob. 3OECh. 5.2 - Prob. 4OECh. 5.2 - Prob. 5OECh. 5.2 - Prob. 6OECh. 5.2 - Prob. 7OECh. 5.2 - Prob. 8OECh. 5.2 - Prob. 9OECh. 5.2 - Prob. 10OECh. 5.2 - Prob. 11OECh. 5.2 - Prob. 12OECh. 5.2 - Prob. 13OECh. 5.2 - Prob. 14OECh. 5.2 - Prob. 15OECh. 5.2 - Prob. 16OECh. 5.2 - Prob. 1WECh. 5.2 - Prob. 2WECh. 5.2 - Prob. 3WECh. 5.2 - Prob. 4WECh. 5.2 - Prob. 5WECh. 5.2 - Prob. 6WECh. 5.2 - Prob. 7WECh. 5.2 - Prob. 8WECh. 5.2 - Prob. 9WECh. 5.2 - Prob. 10WECh. 5.2 - Prob. 11WECh. 5.2 - Prob. 12WECh. 5.2 - Prob. 13WECh. 5.2 - Prob. 14WECh. 5.2 - Prob. 15WECh. 5.2 - Prob. 16WECh. 5.2 - Prob. 17WECh. 5.2 - Prob. 18WECh. 5.2 - Prob. 19WECh. 5.2 - Prob. 20WECh. 5.2 - Prob. 21WECh. 5.2 - Prob. 22WECh. 5.2 - Prob. 23WECh. 5.2 - Prob. 24WECh. 5.2 - Prob. 25WECh. 5.2 - Prob. 26WECh. 5.2 - Prob. 27WECh. 5.2 - Prob. 28WECh. 5.2 - Prob. 29WECh. 5.2 - Prob. 30WECh. 5.2 - Prob. 31WECh. 5.2 - Prob. 32WECh. 5.2 - Prob. 33WECh. 5.2 - Prob. 34WECh. 5.2 - 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