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: 16

Answer

x = -3 y = 1

Work Step by Step

3y - x = 6 4x +12y = 0 to the first equation minus 3y to both sides 3y - x -3y = 6 - 3y -x = 6 - 3y multiply by -1 to both sides x = -6 + 3y substitude x to the second equation 4 ( -6 + 3y ) + 12y = 0 use the distributive property -24 + 12y + 12y = 0 simplify -24 + 24y = 0 add 24 to both sides -24 + 24y + 24 = 0 + 24 24y = 24 divide both sides by 24 y = 1 substitude y for 1 to the x = -6 + 3y x = -6 + $3\times1$ x = -6 +3 x = -3
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