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: 16

Answer

(3, 5, 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 4 and the second equation by -3 and add to obtain: $-2x +10y = 44 $ 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 2 and the third equation by -3 and add to obtain: $ -7x + 2y = -11 $ Plugging $x = 5y -22$ into this equation, we obtain: $-33y + 154 = -11 \\ y =5$ Now, we plug this value into one of the equations that only has x and y in them to find: $ x = 3$ 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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