Chapter 10 - Exponents and Radicals - 10.8 The Complex Numbers - 10.8 Exercise Set - Page 686: 65

$3-i$

Multiplying by the conjugate of the denominator, the given expression, $ \dfrac{10}{3+i} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{10}{3+i}\cdot\dfrac{3-i}{3-i} \\\\= \dfrac{10(3)+10(-i)}{9-i^2} \\\\= \dfrac{30-10i}{9-(-1)} \\\\= \dfrac{30-10i}{10} \\\\= \dfrac{10(3-i)}{10} \\\\= \dfrac{\cancel{10}(3-i)}{\cancel{10}} \\\\= 3-i .\end{array}