Linear Algebra: A Modern Introduction

Published by Cengage Learning
ISBN 10: 1285463242
ISBN 13: 978-1-28546-324-7

Chapter 1 - Vectors - 1.1 The Geometry and Algebra of Vectors - Exercises 1.1 - Page 17: 44

Answer

$x=4$

Work Step by Step

Since $4+3=7$ in $\mathbb{Z}$, and $7$ divided by $5$ leaves a remainder of $2$, $4+3=2$ in $\mathbb{Z}_5$. Thus, by inspection, there is a solution to the equation $x+3=2$.
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