College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 5 - Systems of Equations and Inequalities - Exercise Set 5.1 - Page 528: 54


$y=\displaystyle \frac{1}{3}x-2$ $y=\displaystyle \frac{1}{3}x+2$

Work Step by Step

A graphical solution to a system of linear equations is a point of intersection of two lines. If the solution set is empty, the lines do not intersect. They are parallel. From the graph, the pair of parallel lines have equations $ x-3y=6\qquad$and$\quad x+3y=-6$ Slope-intercept form: solve each equation for $y$: $\left[\begin{array}{lll} x-3y=6 & & x+3y=-6\\ -3y=-x+6 & & -3y=-x-6\\ y=\frac{1}{3}x-2 & & y=\frac{1}{3}x+2 \end{array}\right]$ The system: $y=\displaystyle \frac{1}{3}x-2$ $y=\displaystyle \frac{1}{3}x+2$
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