Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.7 - Complex Numbers - Exercise Set - Page 463: 48

Answer

$\dfrac{11-8i}{5}$

Work Step by Step

Multiplying the numerator and the denominator by the conjugate of the denominator, the given expression, $ \dfrac{6-i}{2+i} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{6-i}{2+i}\cdot\dfrac{2-i}{2-i} \\= \dfrac{(6-i)(2-i)}{(2+i)(2-i)} \\= \dfrac{6(2)+6(-i)-i(2)-i(-i)}{(2)^2-(i)^2} \\= \dfrac{12-6i-2i+i^2}{4-i^2} \\= \dfrac{12-8i+i^2}{4-i^2} .\end{array} Since $i^2=-1,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{12-8i+(-1)}{4-(-1)} \\= \dfrac{12-8i-1}{4+1} \\= \dfrac{11-8i}{5} .\end{array}
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