# Chapter 10 - Exponents and Radicals - 10.8 The Complex Numbers - 10.8 Exercise Set - Page 686: 76

$2-i$

#### Work Step by Step

Multiplying the denominator by $i$, the given expression, $\dfrac{6i+3}{3i} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{6i+3}{3i}\cdot\dfrac{i}{i} \\\\= \dfrac{6i(i)+3(i)}{3i(i)} \\\\= \dfrac{6i^2+3i}{3i^2} \\\\= \dfrac{6(-1)+3i}{3(-1)} \\\\= \dfrac{-6+3i}{-3} \\\\= \dfrac{-3(2-i)}{-3} \\\\= \dfrac{\cancel{-3}(2-i)}{\cancel{-3}} \\\\= 2-i .\end{array}

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