Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 3 - Section 3.6 - Graphing Linear Inequalities in Two Variables - Practice - Page 259: 2

Answer

Please see the graph, made using graphing software.

Work Step by Step

We treat $x-y>3$ as $x-y=3$ to find where to draw the boundary line. Let $x=1$ $x-y=3$ $1-y=3$ $1=3+y$ $-2=y$ Let $x=2$ $x-y=3$ $2-y=3$ $-1-y=0$ $-1=y$ Let $x=3$ $x-y=3$ $3-y=3$ $3=3+y$ $0=y$ $(1,-2), (2,-1)$, and $(3,0)$ are on the boundary line. We try one point not on the line to determine what side of the boundary line to shade. $(0,0)$ $x-y>3$ $0-0>3$ $0 > 3$ is a false statement, so we shade the side of the line without the point $(0,0)$.
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