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 19WE

To determine

To find the open sentence of y(2y−1)4+310=y(y+2)5

Expert Solution & Answer

Answer to Problem 19WE

The value of open sentence is y={32,23}

Explanation of Solution

Given open sentence is y(2y−1)4+310=y(y+2)5

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

y(2y−1)4+310=y(y+2)5

Multiply both the sides by 20, the LCM of 4,5 and 10

20(y(2y−1)4+310)=20(y(y+2)5)

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

5y(2y−1)+6=4y(y+2)10y2−5y+6=4y2+8y

Arrange both sides

10y2−4y2−5y−8y+6=06y2−13y+6=0

Use factorization and simplify

6y2−13y+6=0Apply quadratic root equation−b±b2−4ac2a where a=6, b=−13 and c=6y1,2=−(−13)±(−13)2−4(6)(6)2(6)y1,2=13±169−14412=13±2512=13±512y1=13+512=1812=32Similarlyy2=13−512=812=23

Hence, the open sentence variable values are y={32,23}

Chapter 5.8, Problem 18WE

Chapter 5.8, Problem 20WE

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 ) 4 + 3 10 = y ( y + 2 ) 5

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To find the open sentence of y(2y−1)4+310=y(y+2)5

Answer to Problem 19WE

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Chapter 5 Solutions

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