Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.5 Inequalities - R.5 Exercises - Page R-21: 47

Answer

The solution is $[-8,5)$

Work Step by Step

$\frac{2y+3}{y-5} \leq 1$ First solve the corresponding equation $\frac{2y+3}{y-5}=1$ $2y+3=y-5$ $y=-8$ For $(-\infty,-8)$, choose $-10$: $\frac{2(-10)+3}{(-10)-5}=\frac{17}{15} \gt 1$ For $(-8,5)$, choose $0$: $\frac{2(0)+3}{0-5}=\frac{-3}{5} \lt 1$ For $(5,\infty)$, choose $6:$, $\frac{2(6)+3}{6-5}=15\gt1$ The solution is $[-8,5)$
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