Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 6 - Section 6.5 - Solving Equations Containing Rational Expressions - Exercise Set - Page 378: 42



Work Step by Step

Multiplying both sides by the $LCD= x(x+3) ,$ then the solution to the given equation, $ 2+\dfrac{3}{x}=\dfrac{2x}{x+3} ,$ is \begin{array}{l}\require{cancel} x(x+3)(2)+(x+3)(3)=x(2x) \\\\ 2x^2+6x+3x+9=2x^2 \\\\ (2x^2-2x^2)+(6x+3x)=-9 \\\\ 9x=-9 \\\\ x=-\dfrac{9}{9} \\\\ x=-1 .\end{array}
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