## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 2 - Section 2.4 - Linear Inequalities in Two Variables - 2.4 Exercises - Page 183: 1

#### Answer

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

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ Substitute the given points in the given inequality, $x-2y\le4 .$ 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-2y\le4 \\\\ 0-2(0)\le4 \\\\ 0-0\le4 \\\\ 0\le4 \text{ (TRUE)} .\end{array} Hence, $( 0,0 )$ is a solution. b) Substituting the given point, $( 2,-1 )$ in the given inequality results to \begin{array}{l}\require{cancel} x-2y\le4 \\\\ 2-2(-1)\le4 \\\\ 2+2\le4 \\\\ 4\le4 \text{ (TRUE)} .\end{array} Hence, $( 2,-1 )$ is a solution. c) Substituting the given point, $( 7,1 )$ in the given inequality results to \begin{array}{l}\require{cancel} x-2y\le4 \\\\ 7-2(1)\le4 \\\\ 7-2\le4 \\\\ 5\le4 \text{ (FALSE)} .\end{array} Hence, $( 7,1 )$ is NOT a solution. d) Substituting the given point, $( 0,2 )$ in the given inequality results to \begin{array}{l}\require{cancel} x-2y\le4 \\\\ 0-2(2)\le4 \\\\ 0-4\le4 \\\\ -4\le4 \text{ (TRUE)} .\end{array} Hence, $( 7,1 )$ is a solution. Hence, \begin{array}{l}\require{cancel} \text{a) solution }\\\text{b) solution }\\\text{c) NOT a solution }\\\text{d) solution } \end{array}

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