## Intermediate Algebra (12th Edition)

$k=\dfrac{-2t-3s}{b-1}$
$\bf{\text{Solution Outline:}}$ 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}