Intermediate Algebra (12th Edition)

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

Chapter 2 - Mixed Review Exercises: 6

Answer

$x+2y=6$

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 following characteristics: \begin{array}{l}\require{cancel} \text{Through } (0,3) \text{ and } (-2,4) .\end{array} Express the answer in Standard 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=0 ,\\x_2=-2 ,\\y_1=3 ,\\y_2=4 ,\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{3-4}{0-(-2)}(x-0) \\\\ y-3=\dfrac{3-4}{0+2}(x) \\\\ y-3=\dfrac{-1}{2}(x) \\\\ y-3=-\dfrac{1}{2}x .\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-3=-\dfrac{1}{2}x \\\\ 2(y-3)=2\left( -\dfrac{1}{2}x \right) \\\\ 2y-6=-x \\\\ x+2y=6 .\end{array}
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