Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 4 - Section 4.2 - Solving Systems of Linear Equations by Substitution - Exercise Set: 23

Answer

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

Work Step by Step

6x - 3y = 5 x + 2y = 0 minus 2y to both sides to the second equation x + 2y - 2y = 0 - 2y x = - 2y substitude x = -2y to the first equation 6 ( -2y ) - 3y = 5 -12y -3y = 5 simplify - 15y = 5 $\frac{-15 y}{-15}$ =$\frac{5}{-15}$ divide by -15 y = - $\frac{1}{3}$ divide by 5 x = -2y substitude y to x x = -2 ( $\frac{ - 1}{3}$) y = $\frac{2}{3}$
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