Intermediate Algebra (6th Edition)

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

Chapter 3 - Review - Page 197: 114

Answer

see graph
1500767987

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.
Small 1500767987
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.