College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.6 - Page 86: 30

Answer

$ 1 ; x \ne 2,-2,-3,-5$

Work Step by Step

Factor each expression in the denominator and the numerator. We get $=\frac{(x+2)(x-2)}{(x-2)(x+5)} \div \frac{(x+3)(x+2)}{(x+5)(x+3)} ; x \ne 2,-2,-3,-5;$ To divide, invert the divisor and then multiply. $=\frac{(x+2)(x-2)}{(x-2)(x+5)} \times \frac{(x+5)(x+3)}{(x+3)(x+2)} ; x \ne 2,-2,-3,-5;$ Divide common factors. The result is $ = 1 ; x \ne 2,-2,-3,-5$
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