College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 2 - Section 2.5 - Equations of Lines and Linear Models - 2.5 Exercises: 7

Answer

$3x+2y=-7$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the different forms of linear equations to find the equation of the line with the following given characteristcs: \begin{array}{l}\require{cancel} \text{through } (-5,4) \\\text{m}=-\dfrac{3}{2} .\end{array} Use the properties of equality to express the equation in the standard form. $\bf{\text{Solution Details:}}$ Let $x_1=-5,$ $y_1=4,$ and $m=-\dfrac{3}{2}.$ 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 is \begin{array}{l}\require{cancel} y-4=-\dfrac{3}{2}(x-(-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-4=-\dfrac{3}{2}(x+5) \\\\ 2(y-4)=\left[-\dfrac{3}{2}(x+5)\right]2 \\\\ 2(y-4)=-3(x+5) \\\\ 2(y)+2(-4)=-3(x)-3(5) \\\\ 2y-8=-3x-15 \\\\ 3x+2y=-15+8 \\\\ 3x+2y=-7 .\end{array}
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