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: 48

Answer

The solution is $(-\infty,-\frac{3}{2}) \cup [\frac{-13}{9},\infty)$

Work Step by Step

$\frac{a+2}{3+2a} \leq 5$ First solve the corresponding equation $\frac{a+2}{3+2a} = 5$ $a+2=15+10a$ $9a=-13$ $a=\frac{-13}{9}$ For $(-\infty,-\frac{3}{2})$, choose $-3$: $\frac{-3+2}{3+2(-3)}=\frac{1}{3} \lt 5$ For $(-\frac{3}{2},\frac{-13}{9})$, choose $-1.45$: $\frac{-1.45+2}{3+2(-1.45)}=\frac{11}{2}\gt5$ For $(\frac{-13}{9},\infty)$, choose $4:$, $\frac{4+2}{3+2(4)}=\frac{6}{11}\lt5$ The solution is $(-\infty,-\frac{3}{2}) \cup [\frac{-13}{9},\infty)$
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