Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Review Exercises: 58

Answer

$w=\dfrac{7}{z-3}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ z=\dfrac{3w+7}{w} ,$ for $ w ,$ put all expressions with $ w $ on one side and all other expressions on the other side. Then use the properties of equality to isolate and solve for the variable. $\bf{\text{Solution Details:}}$ By cross-multiplication, the given equation is equivalent to \begin{array}{l}\require{cancel} zw=3w+7 .\end{array} Putting all variables with $ w $ on the left side, the equation above is equivalent to \begin{array}{l}\require{cancel} zw-3w=7 .\end{array} Factoring $ w $ on the left side and using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} w(z-3)=7 \\\\ \dfrac{w(z-3)}{(z-3)}=\dfrac{7}{(z-3)} \\\\ w=\dfrac{7}{z-3} .\end{array}
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