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

Published by Pearson

# Chapter R - Elementary Algebra Review - R.3 Introduction to Graphing - R.3 Exercise Set: 42

#### Answer

$y=-\dfrac{1}{3}x+\dfrac{8}{3}$

#### Work Step by Step

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= 5 ,\\x_2= 8 ,\\y_1= 1 ,\\y_2= 0 ,\end{array} the equation of the line is \begin{array}{l}\require{cancel} y-1=\dfrac{1-0}{5-8}(x-5) \\\\ y-1=\dfrac{1}{-3}(x-5) \\\\ y-1=-\dfrac{1}{3}(x-5) .\end{array} In $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y-1=-\dfrac{1}{3}(x-5) \\\\ y-1=-\dfrac{1}{3}(x)-\dfrac{1}{3}(-5) \\\\ y-1=-\dfrac{1}{3}x+\dfrac{5}{3} \\\\ y=-\dfrac{1}{3}x+\dfrac{5}{3}+1 \\\\ y=-\dfrac{1}{3}x+\dfrac{5}{3}+\dfrac{3}{3} \\\\ y=-\dfrac{1}{3}x+\dfrac{8}{3} .\end{array}

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