Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 2 - Equations, Inequalities, and Problem Solving - 2.2 Using the Principles Together - 2.2 Exercise Set - Page 94: 56



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 0.91-0.2z=1.23-0.6z ,$ remove first the decimal numbers by multiplying both sides by a power of $10.$ Then use the properties of equality to isolate the variable. Do checking of the solution. $\bf{\text{Solution Details:}}$ Since the largest number of decimal places a term has is $2,$ multiply both sides by $100.$ This results to \begin{array}{l}\require{cancel} 0.91-0.2z=1.23-0.6z \\\\ 100(0.91-0.2z)=100(1.23-0.6z) \\\\ 91-20z=123-60z .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} 91-20z=123-60z \\\\ -20z+60z=123-91 \\\\ 40z=32 \\\\ z=\dfrac{32}{40} \\\\ z=\dfrac{\cancel8(4)}{\cancel8(5)} \\\\ z=\dfrac{4}{5} \\\\ z=0.8 .\end{array} Checking: If $z=0.8,$ then \begin{array}{l}\require{cancel} 0.91-0.2z=1.23-0.6z \\\\ 0.91-0.2(0.8)=1.23-0.6(0.8) \\\\ 0.91-0.16=1.23-0.48 \\\\ 0.75=0.75 \text{ (TRUE) } .\end{array} Hence, the solution is $ z=0.8 .$
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