Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 1 - 1.2 - Linear Equations in One Variable - 1.2 Exercises - Page 87: 41

Answer

x = 5

Work Step by Step

Solve : $ \frac{1}{x-3} + \frac{1}{x+3} = \frac{10}{x^{2}-9}$ $ \frac{x+3}{x^{2}-9} + \frac{x-3}{x^{2}-9} = \frac{10}{x^{2}-9}$ as (x+3)(x-3) = ${x^{2}-9}$ x+3 + x-3 = 10 2x = 10 x = 5
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