Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 3 - Linear Systems - 3-5 Systems With Three Variables - Practice and Problem-Solving Exercises - Page 171: 11

Answer

$(2, 1, -5)$

Work Step by Step

Label the original equations first: 1. $-2x + y - z = 2$ 2. $-x - 3y + z = -10$ 3. $3x + 6z = -24$ Divide equation $3$ by $3$ so that the terms are the simplest possible: 4. $x + 2z = -8$ Add the first and second equations and modify them such that one variable is the same in both equations but differs only in sign. Since equation $3$ is missing a $y$ term, modify the first and second equations so that the $y$ term can be eliminated as well. Multiply equation $1$ by $3$: 5. $3(-2x + y - z) = 3(2)$ Use distributive property: 5. $-6x + 3y - 3z = 6$ Combine equations $2$ and $5$: 2. $-x - 3y + z = -10$ 5. $-6x + 3y - 3z = 6$ Add the equations: 6. $-7x - 2z = -4$ Combine equations $4$ and $6$ to eliminate a variable: 4. $x + 2z = -8$ 6. $-7x - 2z = -4$ Add the two equations together: $-6x = -12$ Divide both sides of the equation by $-6$: $x = 2$ Substitute this $x$ value into equation $4$ to find the value for $z$: 4. $2 + 2z = -8$ Subtract $2$ from both sides of the equation: $2z = -10$ Divide both sides by $2$: $z = -5$ Substitute the values for $x$ and $z$ into one of the original equations to find $y$. Let's use the second equation: 2. $-2 - 3y + (-5) = -10$ Combine like terms on the left side of the equation: $-7 - 3y = -10$ Add $7$ to each side of the equation: $-3y = -3$ Divide both sides of the equation by $-3$ to solve for $y$: $y = 1$ The solution is $(2, 1, -5)$. To check the solution, plug in the three values into one of the original equations. Use equation $1$: $-2(2) + 1 - (-5) = 2$ $-4 + 1 + 5 = 2$ $2 = 2$ The sides are equal to one another; therefore, the solution is correct.
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