Intermediate Algebra (6th Edition)

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

Chapter 2 - Section 2.5 - Compound Inequalities - Exercise Set: 69

Answer

$(\frac{-4}{3},\frac{7}{3})$

Work Step by Step

$\frac{1}{15} \lt \frac{8-3x}{15} \lt \frac{4}{5}$ Using the properties of inequalities, Multiply all parts by the LCD 15. $15(\frac{1}{15}) \lt 15(\frac{8-3x}{15}) \lt 15(\frac{4}{5})$ $1 \lt 8-3x \lt 12$ Add $-8$ to all parts. $1-8 \lt 8-3x-8 \lt 12-8$ $-7 \lt -3x \lt4$ Divide by $-3$ and reverse the inequality symbol $\frac{-7}{-3} \gt \frac{-3x}{-3} \gt \frac{4}{-3}$ $\frac{7}{3} \gt x \gt \frac{-4}{3}$ This is equal to $\frac{-4}{3} \lt x \lt \frac{7}{3}$ Interval Notation: $(\frac{-4}{3},\frac{7}{3})$
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