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: 90

Answer

$x=\dfrac{31}{14}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 0.9(2x-8)=4-(x+5) ,$ 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 $1,$ multiply both sides by $10.$ This results to \begin{array}{l}\require{cancel} 10[0.9(2x-8)]=10[4-(x+5)] \\\\ 9(2x-8)=40-10(x+5) .\end{array} Using the Distributive Property and the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} 9(2x-8)=40-10(x+5) \\\\ 9(2x)+9(-8)=40-10(x)-10(5) \\\\ 18x-72=40-10x-50 \\\\ 18x+10x=40-50+72 \\\\ 28x=62 \\\\ x=\dfrac{62}{28} \\\\ x=\dfrac{31}{14} .\end{array} Checking: If $x=\dfrac{31}{14},$ then \begin{array}{l}\require{cancel} 0.9(2x-8)=4-(x+5) \\\\ 0.9\left( 2\cdot\dfrac{31}{14}-8\right)=4-\left(\dfrac{31}{14}+5\right) \\\\ \dfrac{9}{10}\left( \dfrac{62}{14}-8\right)=4-\left(\dfrac{31}{14}+5\right) \\\\ \dfrac{9}{10}\left( \dfrac{62}{14}-\dfrac{112}{14}\right)=\dfrac{56}{14}-\left(\dfrac{31}{14}+\dfrac{70}{14}\right) \\\\ \dfrac{9}{10}\left( \dfrac{50}{14}\right)=\dfrac{56}{14}-\dfrac{101}{14} \\\\ \dfrac{9}{\cancel{10}}\left( \dfrac{\cancel{10}(5)}{14}\right)=\dfrac{56}{14}-\dfrac{101}{14} \\\\ \dfrac{45}{14}=\dfrac{45}{14} \text{ (TRUE) } .\end{array} Hence, the solution is $ x=\dfrac{31}{14} .$
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