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: 30

Answer

x = 3 y = -2

Work Step by Step

6x + 3y = 12 9x + 6y = 15 to the first equation,divide by 3 2x + y = 4 minus 2x to both sides 2x + y - 2x = 4 - 2x y = 4 - 2x substitude y to the second equation 9x + 6 ( 4 - 2x ) = 15 9x + 24 - 12x = 15 -3x + 24 = 15 minus 24 to both sides -3x + 24 - 24 = 15 - 24 - 3x = -9 divide by -3 $\frac{-3}{-3}$ = $\frac{-9}{-3}$ x = 3 substitude x to y = 4 - 2x y = 4 - $2\times3$ y = 4 - 6 y = -2
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