Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 11 - Additional Topics - 11.2 - 3 x 3 Systems of Equations - Problem Set 11.2 - Page 484: 10

Answer

(-4, 3, 1)

Work Step by Step

In order to solve systems of three linear equations, we multiply the first and second equation by values that will cancel out a variable when they are added. Thus, we multiply the first equation by -2 and the second equation by 1 and add to obtain: $-8x -7y = 11$ We now multiply the first and third equation by values that will cancel out a variable when they are added. Thus, we multiply the first equation by 3 and the third equation by one and add to obtain: $ 16x + 5y = -49 $ Plugging $x = -7/8y -11/8$ into this equation, we obtain: $-9y -22 = -49 \\ y = 3 $ Now, we plug this value into one of the equations that only has x and y in them to find: $ x = -4$ Finally, we plug the values of x and y into the first equation listed in the book to find: $ z = 1$
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