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: 2

Answer

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

Work Step by Step

$\bf{\text{Solution Outline:}}$ Substitute the given points in the given inequality, $ x+y\gt0 .$ 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} x+y\gt0 \\\\ 0+0\gt0 \\\\ 0\gt0 \text{ (FALSE)} .\end{array} Hence, $( 0,0 )$ is NOT a solution. b) Substituting the given point, $( -2,1 )$ in the given inequality results to \begin{array}{l}\require{cancel} x+y\gt0 \\\\ -2+1\gt0 \\\\ -1\gt0 \text{ (FALSE)} .\end{array} Hence, $( -2,1 )$ is NOT a solution. c) Substituting the given point, $( 2,-1 )$ in the given inequality results to \begin{array}{l}\require{cancel} x+y\gt0 \\\\ 2+(-1)\gt0 \\\\ 2-1\gt0 \\\\ 1\gt0 \text{ (TRUE)} .\end{array} Hence, $( 2,-1 )$ is a solution. d) Substituting the given point, $( -4,6 )$ in the given inequality results to \begin{array}{l}\require{cancel} x+y\gt0 \\\\ -4+6\gt0 \\\\ 2\gt0 \text{ (TRUE)} .\end{array} Hence, $( -4,6 )$ is a solution. Hence, \begin{array}{l}\require{cancel} \text{a) NOT a solution }\\\text{b) NOT a solution }\\\text{c) solution }\\\text{d) solution } \end{array}
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