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: 20

Answer

$x>\displaystyle \frac{1}{3}$
1507398495

Work Step by Step

Applying the properties of inequality, we can P1. add any number to both sides, P2. multiply (or divide) both sides with a positive $\quad $ number to arrive at a valid inequality. $\quad $ If we P3. multiply (or divide) both sides with a negative number, we must change the direction of the inequality sign, to arrive at a valid inequality.. Our goal is to, step by step, isolate the unknown on one side and interpret the result (which, if any, will be an interval) ----------------------------- $x+5(x+1)>4(2-x)+x\qquad \qquad $... parentheses $x+5x+5>8-4x+x \qquad $... simplify $6x+5>8-3x \qquad \qquad $P1: ...$/-5$ $6x>3-3x \qquad \qquad $P1: ...$/+3x$ $9x>3 \qquad \qquad $P$2$: ...$/\div 9$ $x>\displaystyle \frac{1}{3}$ In interval notation:$\qquad (\displaystyle \frac{1}{3}$ , $\infty)$.
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