Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.7 Inequalities - 1.7 Exercises - Page 159: 22


The solution is $\Big(-\infty,\dfrac{1}{2}\Big]$

Work Step by Step

$\dfrac{2x-5}{-8}\le1-x$ Change the sign of the denominator and the sign of the fraction on the left side: $-\dfrac{2x-5}{8}\le1-x$ Rearrange: $x-1\le\dfrac{2x-5}{8}$ Take $8$ to multiply the left side: $8(x-1)\le2x-5$ $8x-8\le2x-5$ Take $2x$ to the left side and $8$ to the right side: $8x-2x\le-5+8$ $6x\le3$ Take $6$ to divide the right side and simplify: $x\le\dfrac{3}{6}$ $x\le\dfrac{1}{2}$ Expressing the solution in interval notation: $\Big(-\infty,\dfrac{1}{2}\Big]$
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