Algebra 2 Common Core

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

Chapter 3 - Linear Systems - Chapter Review - Page 186: 24

Answer

The solution is $(-4, 1, -5)$.

Work Step by Step

Label the original equations: 1. $-x + y + 2z = -5$ 2. $5x + 4y - 4z = 4$ 3. $x - 3y - 2z = 3$ The first step is to choose two equations to work with where one variable can be eliminated. Let's choose equations $1$ and $3$. Modify the equations by multiplying them by a non-zero factor so that one variable is the same in both equations but differing in sign so that this variable can be eliminated when the two equations are added together. In this case, no modifications are needed: 1. $-x + y + 2z = -5$ 3. $x - 3y - 2z = 3$ Add the equations: $-2y = -2$ In this case, two variables were eliminated. Divide both sides of the equation by $-2$ to solve for $y$: $y = 1$ Now, choose another two equations and modify them. Modify these equations such that the $y$ variable can be eliminated. This modified equation will be equation $4$ and will be added to equation $2$ to try to eliminate another variable. Use equations $1$ and $2$. Multiply equation $1$ by $2$ and leave equation $2$ as-is: 1. $2(-x + y + 2z) = 2(-5)$ 2. $5x + 4y - 4z = 4$ Distribute and multiply to simplify: $-2x + 2y + 4z = -10$ $5x + 4y - 4z = 4$ Add the equations together. This will become equation $5$: 5. $3x + 6y = -6$ Substitute the value for $y$ we just found into equation $5$ to find the value of $x$: $3x + 6(1) = -6$ $3x + 6 = -6$ $3x = -12$ $x = -4$ Substitute the values for $x$ and $y$ into one of the original equations to find $z$. Use equation $1$: $-(-4) + 1 + 2z = -5$ $4 + 1 + 2z = -5$ $5 + 2z = -5$ $2z = -10$ $z = -5$ The solution is $(-4, 1, -5)$.
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