Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 3 - Review - Page 197: 114


see graph

Work Step by Step

Changing the given inequality, $ 5x-3y\lt10 ,$ to equality and then isolating $y$ result to \begin{array}{l}\require{cancel} 5x-3y=10 \\\\ -3y=-5x+10 \\\\ y=\dfrac{-5}{-3}x+\dfrac{10}{-3} \\\\ y=\dfrac{5}{3}x-\dfrac{10}{3} .\end{array} Use the table of values below to graph this line. Since the inequality used is "$ \lt $", use a broken line. Using the testpoint $( 0,0 )$, then \begin{array}{l}\require{cancel} 5(0)-3(0)\lt10 \\\\ 0\lt10 \text{ (TRUE)} .\end{array} Since the solution above ended with a $\text{ TRUE }$ statement, then the testpoint is $\text{ part }$ of the solution set.
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