To solve and check , the given equation has no solution of 2x−1−xx+3=6x2+2x−3

Expert Solution & Answer

Answer to Problem 21WE

The answer is {0,3}

Explanation of Solution

Given that: : 2x−1−xx+3=6x2+2x−3

First factor the x2+2x−3

x2+2x−3=x2+(3−1)x−3=x2+3x−x−3=x(x+3)−1(x+1)=(x+3)(x−1)

Rewrite the given equation

2x−1−xx+3=6(x−1)(x+3)

Multiply by (x−1)(x+3) , LCM of (x−1) and (x+3)

(x+3)(x−1)(2x−1−xx+3)=(x+3)(x−1)(6(x−1)(x+3))

Apply distributive property

2(x+3)−x(x−1)=62x+6−x2+x=63x+6−x2=6x2−3x=0

Apply the factor method and Apply zero product property to find the solution of equation

When x equal to 1 and -3, a denominator in the original equation has a value of 0. Therefore,

1 and 5 are not in the domain of x and cannot be solution.

Check:

For x=323−1−33+3=69+6−31−12=1212=12LHS=RHS

For x=020−1−00+3=60+0−3−2−0=−2−2=−2LHS=RHS

Hence, the solution of equation is {0,3}

Chapter 5.9, Problem 20WE

Chapter 5.9, 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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