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: 46

Answer

x = 5 y = 0

Work Step by Step

simplify to first equation 5x + 2y -4x - 2y = 2 ( 2y + 6 ) -7 x = 4y + 12 -7 use the distributive property x = 4y + 5 use the distributive equation to second equation 3 ( 2x - y ) - 4x = 1 + 9 6x - 3y - 4x = 10 6x - 4x - 3y = 10 2x - 3y = 10 substitude now x = 4y + 5 2 ( 4y + 5 ) - 3y = 10 use the distributive property 8y + 10 - 3y = 10 simplify 5y + 10 = 10 5y = 10 - 10 solve 5y = 0 y = 0 substitude y = 0 to x = 4y + 5 x = $4\times0$ + 5 x = 0 + 5 x = 5
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