## Elementary Algebra

Published by Cengage Learning

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

(-1, 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 1 and the second equation by -4 and add to obtain: $-18x +9y = 45$ 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 one and the third equation by 2 and add to obtain: $8x +7y = 13$ Plugging $x = .5y -2.5$ into this equation, we obtain: $11y -20 = 13 \\ y =3$ Now, we plug this value into one of the equations that only has x and y in them to find: $x = -1$ 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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