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

Published by Pearson

# Chapter 2 - Equations, Inequalities, and Problem Solving - 2.2 Using the Principles Together - 2.2 Exercise Set: 89

#### Answer

$x=-\dfrac{1}{31}$

#### Work Step by Step

Multiplying both sides by $10$, then the solution to the given equation, $0.7(3x+6)=1.1-(x-3)$, is \begin{array}{l} 7(3x+6)=11-10(x-3) \\\\ 21x+42=11-10x+30 \\\\ 21x+10x=11+30-42 \\\\ 31x=-1 \\\\ x=-\dfrac{1}{31} .\end{array} CHECKING: \begin{array}{l} 0.7\left(3\left( -\dfrac{1}{31} \right)+6\right)=1.1-\left(-\dfrac{1}{31}-3\right) \\\\ \dfrac{7}{10}\left(-\dfrac{3}{31}+6\right)=\dfrac{11}{10}-\left(-\dfrac{94}{31}\right) \\\\ \dfrac{7}{10}\left(\dfrac{183}{31}\right)=\dfrac{11}{10}+\dfrac{94}{31} \\\\ \dfrac{1281}{310}=\dfrac{1281}{310} \text{ (TRUE)} .\end{array} Hence, the solution is $x=-\dfrac{1}{31}$.

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