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: 57



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 0.76+0.21t=0.96t-0.49 ,$ 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} 0.76+0.21t=0.96t-0.49 \\\\ 100(0.76+0.21t)=100(0.96t-0.49) \\\\ 76+21t=96t-49 .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} 76+21t=96t-49 \\\\ 21t-96t=-49-76 \\\\ -75t=-125 \\\\ t=\dfrac{-125}{-75} \\\\ t=\dfrac{\cancel{-25}(5)}{\cancel{-25}(3)} \\\\ t=\dfrac{5}{3} .\end{array} Checking: If $t=\dfrac{5}{3},$ then \begin{array}{l}\require{cancel} 0.76+0.21t=0.96t-0.49 \\\\ 0.76+0.21\left( \dfrac{5}{3} \right) =0.96\left( \dfrac{5}{3} \right) -0.49 \\\\ 0.76+0.35 =1.6-0.49 \\\\ 1.11 =1.11 \text{ (TRUE) } .\end{array} Hence, the solution is $ t=\dfrac{5}{3} .$
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