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 9 - Inequalities and Problem Solving - 9.3 Absolute-Value Equations and Inequalities - 9.3 Exercise Set - Page 600: 95



Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Two-Point form of linear equations to find the equation of the line with the given characteristics: \begin{array}{l}\require{cancel} \text{contains } (-4,-3) \text{ and } (-1,3) .\end{array} Then use the properties of equality to convert the equation to Slope-Intercept form. $\bf{\text{Solution Details:}}$ Using $y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)$ or the Two-Point Form of linear equations, where \begin{array}{l}\require{cancel} x_1=-4 ,\\x_2=-1 ,\\y_1=-3 ,\\y_2=3 ,\end{array} the equation of the line is \begin{array}{l}\require{cancel} y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1) \\\\ y-(-3)=\dfrac{-3-3}{-4-(-1)}(x-(-4)) \\\\ y+3=\dfrac{-3-3}{-4+1}(x+4) \\\\ y+3=\dfrac{-6}{-3}(x+4) \\\\ y+3=2(x+4) .\end{array} Using the properties of equality, in $y=mx+b$ or the Slope-Intercept form, the equation above is equivalent to \begin{array}{l}\require{cancel} y+3=2(x+4) \\\\ y+3=2(x)+2(4) \\\\ y+3=2x+8 \\\\ y=2x+8-3 \\\\ y=2x+5 .\end{array}
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