Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 57

Answer

$k=\dfrac{-2t-3s}{b-1}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 3s+bk=k-2t ,$ for $ k ,$ put all expressions with $ k $ on one side and all other expressions on the other side. Then use the properties of equality to isolate and solve for the variable. $\bf{\text{Solution Details:}}$ Putting all variables with $k$ on the left side, the equation above is equivalent to \begin{array}{l}\require{cancel} bk-k=-2t-3s .\end{array} Factoring $k$ on the left side and using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} k(b-1)=-2t-3s \\\\ \dfrac{k(b-1)}{(b-1)}=\dfrac{-2t-3s}{(b-1)} \\\\ k=\dfrac{-2t-3s}{b-1} .\end{array}
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