Algebra and Trigonometry: Structure and Method, Book 2

2000th Edition

ISBN: 9780395977255

Author: MCDOUGAL LITTEL

Publisher: McDougal Littell

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Question

Chapter 5.5, Problem 21WE

To determine

To simplify the given expression.

Expert Solution & Answer

Answer to Problem 21WE

(x+y)(x+3y)

Explanation of Solution

Given:

Expression: 3x2+xy−2y23x2−xy−2y2÷3x2+7xy−6y23x2−2xy−y2÷3x+y3x+2y

Concept used:

To multiply two fractions, simply multiply numerators by each other and denominators by each other and then reduce the fraction if possible.

ab⋅cd→a×cb×d

To divide two fractions, simply flip the second fraction into a reciprocal and change the division sign to a multiplication sign and then multiply the two fractions by multiplying the numerators by each other and denominators by each other. As illustrated below:

ab÷cd→ab×dc→a×db×c

Calculation:

Consider the given expression,

⇒3x2+xy−2y23x2−xy−2y2÷3x2+7xy−6y23x2−2xy−y2÷3x+y3x+2y

⇒3x2+xy−2y23x2−xy−2y2⋅3x2−2xy−y23x2+7xy−6y2⋅3x+2y3x+y [Flipping second and third fractions]

⇒3x2+3xy−2xy−2y23x2−3xy+2xy−2y2×3x2−3xy+xy−y23x2+9xy−2xy−6y2×3x+2y3x+y [Factorizing the fractions]

⇒3x(x+y)−2y(x+y)3x(x−y)+2y(x−y)×3x(x−y)+y(x−y)3x(x+3y)−2y(x+3y)×3x+2y3x+y

⇒(x+y)(3x−2y)(x−y)(3x+2y)×(x−y)(3x+y)(x+3y)(3x−2y)×3x+2y3x+y

⇒(x+y)1×1(x+3y)×11 [Reducing the fractions]

⇒(x+y)×1×11×(x+3y)×1⇒(x+y)(x+3y)

Conclusion:

Therefore, after simplification, the answer is (x+y)(x+3y) .

Chapter 5.5, Problem 20WE

Chapter 5.5, Problem 22WE

Chapter 5 Solutions

Algebra and Trigonometry: Structure and Method, Book 2

Ch. 5.1 - Prob. 1OE Ch. 5.1 - Prob. 2OE Ch. 5.1 - Prob. 3OE Ch. 5.1 - Prob. 4OE Ch. 5.1 - Prob. 5OE Ch. 5.1 - Prob. 6OE Ch. 5.1 - Prob. 7OE Ch. 5.1 - Prob. 8OE Ch. 5.1 - Prob. 9OE Ch. 5.1 - Prob. 10OE

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