## Intermediate Algebra (12th Edition)

$3i+1$
$\bf{\text{Solution Outline:}}$ To divide the given expression, $\dfrac{3-i}{-i} ,$ multiply both the numerator and the denominator by $i$. Use $i^2=-1.$ $\bf{\text{Solution Details:}}$ Multiplying both the numerator and the denominator by $i,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{3-i}{-i}\cdot\dfrac{i}{i} \\\\= \dfrac{(3-i)i}{-i^2} .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{(3)i-(i)i}{-i^2} \\\\= \dfrac{3i-i^2}{-i^2} .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{3i-(-1)}{-(-1)} \\\\= \dfrac{3i+1}{1} \\\\= 3i+1 .\end{array}