Algebra and Trigonometry: Structure and Method, Book 2

2000th Edition

ISBN: 9780395977255

Author: MCDOUGAL LITTEL

Publisher: McDougal Littell

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Equations and Inequations

Equations and inequalities describe the relationship between two mathematical expressions.

Linear Functions

A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.

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Question

Chapter 5.8, Problem 15WE

To determine

To find the open sentence of y(2y−1)2=1−y3

Expert Solution & Answer

Answer to Problem 15WE

The value of open sentence is y={−12,23}

Explanation of Solution

Given open sentence is y(2y−1)2=1−y3

Value of open sentence means a sentence is it true and depend upon the variable values .

y(2y−1)2=1−y3

Multiply both the sides by 6, the LCM of 3 and 2

6(y(2y−1)2)=6(1−y3)

Apply distributive property and simplify [a(b±c)=a∗b±a*c]

3y(2y−1)=2(1−y)6y2−3y=2−2y

Add -2 to both sides

6y2−3y−2=2−2−2y6y2−3y−2=2y

Add 2y to both sides

6y2−3y−2+2y=2y−2y6y2−y−2=0

Use factorization and simplify

6y2−y−2=06y2−y(4−3)−2=06y2−4y+3y−2=02y(3y−2)+1(3y−2)=0(3y−2)(2y+1)=0

Now, check the variable value satisfied the sentence with zero product property.

(3y−2)=03y=2y=23Similarly(2y+1)=02y=−1y=−12

Hence, the open sentence variable values are y={−12,23}

Chapter 5.8, Problem 14WE

Chapter 5.8, Problem 16WE

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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To find the open sentence of y ( 2 y − 1 ) 2 = 1 − y 3

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To find the open sentence of y(2y−1)2=1−y3

Answer to Problem 15WE

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Chapter 5 Solutions

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