Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 2 - Section 2.4 - Linear Inequalities in Two Variables - 2.4 Exercises: 4

Answer

$\text{a) solution }\\\text{b) solution }\\\text{c) solution }\\\text{d) NOT a solution }$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Substitute the given points in the given inequality, $ y\le1 .$ If the inequality is satisfied, then the given point is a solution. Otherwise, the given point is not a solution. $\bf{\text{Solution Details:}}$ a) Substituting the given point, $( 0,0 )$ in the given inequality results to \begin{array}{l}\require{cancel} y\le1 \\\\ 0\le1 \text{ (TRUE)} .\end{array} Hence, $( 0,0 )$ is a solution. b) Substituting the given point, $( 3,1 )$ in the given inequality results to \begin{array}{l}\require{cancel} y\le1 \\\\ 1\le1 \text{ (TRUE)} .\end{array} Hence, $( 3,1 )$ is a solution. c) Substituting the given point, $( 2,-1 )$ in the given inequality results to \begin{array}{l}\require{cancel} y\le1 \\\\ -1\le1 \text{ (TRUE)} .\end{array} Hence, $( 2,-1 )$ is a solution. d) Substituting the given point, $( -3,3 )$ in the given inequality results to \begin{array}{l}\require{cancel} y\le1 \\\\ 3\le1 \text{ (FALSE)} .\end{array} Hence, $( -3,3 )$ is NOT a solution. Hence, \begin{array}{l}\require{cancel} \text{a) solution }\\\text{b) solution }\\\text{c) solution }\\\text{d) NOT a solution } \end{array}
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