Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 7 - Rational Functions - 7.5 Solving Rational Equations - 7.5 Exercises - Page 598: 3



Work Step by Step

Multiply $2w+3$ to both sides of the equation to obtain: $5(2w+3) = \dfrac{4}{2w+3} \cdot (2w+3) \\5(2w)+5(3)=4 \\10w+15=4$ Subtract $15$ to both sides of the equation to obtain: $10w=4-15 \\10w=-11$ Divide $10$ to both sides to obtain: $w = -\frac{11}{10}$ Cheking: $5=\dfrac{4}{2w+3}? \\5 = \dfrac{4}{2\cdot(-\frac{11}{10})+3}? \\5=\dfrac{4}{-\frac{11}{5}+3}? \\5=\dfrac{4}{-\frac{11}{5} + \frac{15}{5}}? \\5=\dfrac{4}{\frac{4}{5}}? \\5=4 \cdot \frac{5}{4}? \\5=5$ Thus, the solution to the given equation is: $\color{blue}{-\dfrac{11}{10}}$.
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