Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 2 - Chapter 2 Test - Page 212: 14

Answer

$\text{a) Slope-Intercept Form: } y=-\dfrac{3}{5}x-\dfrac{11}{5} \\\\\text{b) Standard Form: } 3x+5y=-11$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Find the slope of the given equation, $ 3x+5y=6 .$ Then use this slope and the given point, $ (-7,2) ,$ to find the equation of the needed line. Give the equation in the Slope-Intercept Form and in the Standard Form. $\bf{\text{Solution Details:}}$ In the form $y=mx+b$ or the Slope-Intercept Form (where $m$ is the slope), the given equation is equivalent to \begin{array}{l}\require{cancel} 3x+5y=6 \\\\ 5y=-3x+6 \\\\ \dfrac{5y}{5}=\dfrac{-3x}{5}+\dfrac{6}{5} \\\\ y=-\dfrac{3}{5}x+\dfrac{6}{5} .\end{array} Hence, the slope of the given line is $ m=-\dfrac{3}{5} .$ Since parallel lines have the same slope, then the needed line has the following characteristics: \begin{array}{l}\require{cancel} m=-\dfrac{3}{5} \\\text{Through: } (-7,2) .\end{array} Using $y-y_1=m(x-x_1)$ or the Point-Slope Form of linear equations, the equation of the line with the given conditions, \begin{array}{l}\require{cancel} y_1=2 ,\\x_1=-7 ,\\m=-\dfrac{3}{5} ,\end{array} is \begin{array}{l}\require{cancel} y-y_1=m(x-x_1) \\\\ y-2=-\dfrac{3}{5}(x-(-7)) \\\\ y-2=-\dfrac{3}{5}(x+7) .\end{array} Using the properties of equality, in the form $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y-2=-\dfrac{3}{5}(x)-\dfrac{3}{5}(7) \\\\ y-2=-\dfrac{3}{5}x-\dfrac{21}{5} \\\\ y=-\dfrac{3}{5}x-\dfrac{21}{5}+2 \\\\ y=-\dfrac{3}{5}x-\dfrac{21}{5}+\dfrac{10}{5} \\\\ y=-\dfrac{3}{5}x-\dfrac{11}{5} .\end{array} Using the properties of equality, in the form $ax+by=c$ or the Standard Form, the equation above is equivalent to \begin{array}{l}\require{cancel} y=-\dfrac{3}{5}x-\dfrac{11}{5} \\\\ 5(y)=5\left( -\dfrac{3}{5}x-\dfrac{11}{5} \right) \\\\ 5y=-3x-11 \\\\ 3x+5y=-11 .\end{array} The equation of the line is \begin{array}{l}\require{cancel} \text{a) Slope-Intercept Form: } y=-\dfrac{3}{5}x-\dfrac{11}{5} \\\\\text{b) Standard Form: } 3x+5y=-11 .\end{array}
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