Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 1 - Section 1.5 - Subtracting Real Numbers - Exercise Set - Page 46: 62

Answer

$\frac{10}{3}$

Work Step by Step

We are asked to evaluate the given expression when $x=-5$, $y=4$, and $t=10$. We can plug in the given variables directly to evaluate. $\frac{15-x}{y+2}=\frac{15-(-5)}{(4)+2}=\frac{15+5}{4+2}=\frac{20}{6}=\frac{20\div2}{6\div2}=\frac{10}{3}$
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