# Chapter 11 - Additional Topics - 11.2 - 3 x 3 Systems of Equations - Concept Quiz 11.2 - Page 483: 10

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#### 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: $x -5y = -20$ 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 one and add to obtain: $3x + 3y = 12$ Plugging $x = 5y -20$ into this equation, we obtain: $18y - 60 = 12 \\ y =4$ Now, we plug this value into one of the equations that only has x and y in them to find: $x = 0$ Finally, we plug the values of x and y into the first equation listed in the book to find: $z = 8$

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