Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 3 - Solving Inequalities - 3-5 Working with Sets - Mixed Review - Page 199: 63



Work Step by Step

$\frac{2d-3}{5}\longrightarrow$ substitute 9 for d =$\frac{2(9)-3}{5}\longrightarrow$ multiply =$\frac{18-3}{5}\longrightarrow$ subtract =$\frac{15}{5}\longrightarrow$ divide =$3$
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