## Elementary Algebra

Published by Cengage Learning

# Chapter 11 - Additional Topics - 11.2 - 3 x 3 Systems of Equations - Problem Set 11.2 - Page 484: 12

(2, 1, -3)

#### 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 2 and add to obtain: $10x + 3y = 23$ 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 one and add to obtain: $-6x +5y =-7$ Plugging $x = -.3y + 2.3$ into this equation, we obtain: $6.8y -13.8 = -7 \\ y =1$ Now, we plug this value into one of the equations that only has x and y in them to find: $x = 2$ Finally, we plug the values of x and y into the first equation listed in the book to find: $z = -3$

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